diff --git "a/data_all_eng_slimpj/shuffled/split2/finalzzjwka" "b/data_all_eng_slimpj/shuffled/split2/finalzzjwka" new file mode 100644--- /dev/null +++ "b/data_all_eng_slimpj/shuffled/split2/finalzzjwka" @@ -0,0 +1,5 @@ +{"text":"\\section{Introduction}\n\nGravitational lensing is recognized as a unique tool to conduct many important investigations in modern astrophysics \\cite{Liebes:1964,Schneider-Ehlers-Falco:1992,Schneider-etal:2006}. Since the beginning of the 21st century, it is used to study the distribution of matter in stelar structures, to probe the dark matter distribution in the universe, even to search for exoplanets orbiting distant stars \\cite{Refsdal:1964,Blandford-Narayan:1992,Wambsganss:1998,Gaudi:2012}.\n\nDue to the nonlinear nature of the equations involved, most relevant efforts were constrained to monopole lenses, where the gravitational field is taken to be that of a structureless point source \\cite{Herlt-Stephani:1976,Deguchi-Watson:1986,Narayan-Bartelmann:1996}. Nevertheless, there have been attempts to model extended lenses, including quadrupole and general shear distortions of the lensing potential \\cite{Kovner:1987,Schneider-Ehlers-Falco:1992,Erdl-Schneider:1993,Gould:2001} as well as to describe binaries \\cite{Congdon-Keeton-book:2018}. It was recognized that such deviations from spherical symmetry lead to the formation of caustics, which complicates image formation \\cite{Ohanian:1983,Blandford-Kovner:1988,Nambu:2013,Chu:2016}. Most of these investigations were conducted using the geometric optics approximation, which is known to be of limited utility when it comes to describing light amplification, especially in the presence of caustics \\cite{Gaudi-Petters:2001,Gaudi-Petters:2002}, where such results have singularities. As caustics appear naturally in the point-spread function (PSF) characterizing the optical properties of an extended gravitational lens, there is a need to address these shortcomings. In particular, it was recognized that a wave-optical treatment of gravitational lensing is needed \\cite{Nakamura-Deguchi:1999,Nambu:2012}. Until recently, such a description of an extended lens was not available.\n\nMeanwhile, the solar gravitational lens (SGL) gained attention as a possible means to obtain resolved images of exoplanets \\cite{Turyshev:2017,Turyshev-Toth:2017,Turyshev-Toth:2020-im-extend,Toth-Turyshev:2020}. From the beginning, our efforts to describe the SGL were conducted within the Mie theory \\cite{Mie:1908,Born-Wolf:1999}, aiming to describe diffraction of electromagnetic (EM) waves by a gravitational field. Such an approach solves a Schr\\\"odinger-like wave equation for Debye potentials, yielding a wave-optical description for the lens\\footnote{In Ref. \\cite{Turyshev-Toth:2021-multipoles} we show that although similar results may be obtained within a scalar theory by using a general Fresnel--Kirchhoff diffraction formula or the path integral formalism of quantum field theory, the Mie-inspired solution covers a method to treat vector fields. Such an approach is advantageous from a practical standpoint as it allows us to deal with directly observable quantities and evaluate detection sensitivities (i.e., signal-to-noise ratio \\cite{Turyshev-Toth:2020-im-extend,Toth-Turyshev:2020}) while preserving the vectorial nature of the EM field in a weak gravitational field.}. At first, the lens was modeled as a gravitational monopole. This established a good foundation on which increasingly refined models could be constructed. These refinements were needed to capture the fact that the Sun is not a perfect sphere: its rotation and the resulting oblateness result in small axisymmetric perturbations of its otherwise spherically symmetric gravitational field in the form of the quadrupole and, to a lesser extent, higher-order zonal harmonics.\n\nAlthough our initial objective was to capture only the dominant quadrupole perturbation captured by the $J_2$ zonal harmonics, we were able to do much more. We developed a wave-optical treatment that we call the angular eikonal method, which can be used to describe gravitational lensing by any axisymmetric gravitational field that is dominated by the monopole potential, but perturbed by an infinite set of zonal harmonics \\cite{Turyshev-Toth:2021-multipoles}.\nThe resulting wave-optical treatment of gravitational lensing focuses on evaluating the eikonal phase shift that an EM wave acquires as it travels from the source to the image plane. This phase shift now may be evaluated for any gravitational potential that can be modeled as a perturbed monopole gravitational field. The new method may in fact be used to recover the multipole moments that characterize the mass distribution of the lens, and thus recover its basic geometry and structure. This wave-optical approach is especially useful to describe imaging with realistic axisymmetric astrophysical lenses \\cite{Turyshev-Toth:2021-imaging}.\n\nOur prior work on gravitational lensing focused on the strong interference region of the lens that exists in the vicinity of its primary optical axis (see Fig.~\\ref{fig:regions}). As we move further away from that axis, we enter the weak interference region and then the region of geometric optics (see description in \\cite{Turyshev-Toth:2017}.) Clearly, the farther we are from the optical axis, the less is the impact of perturbations to the monopole gravitational field. At some distance from the optical axis, the behavior of an extended gravitational lens becomes indistinguishable from that of a monopole lens. In any case, a complete description of the transition process between various regions is needed to fully understand the behavior of the PSF of the lens. The images seen by a telescope at different distances from the optical axis are also of interest. A similar discussion in the context of a monopole lens was presented in \\cite{Turyshev-Toth:2019-extend,Turyshev-Toth:2020-im-extend}. We can now extend these results to the case of a generic lens that can be described as a perturbed gravitational monopole.\n\nIn this paper, we apply our new approach beyond the strong interference region, describing gravitational lensing in all lensing regimes. This paper is organized as follows:\nIn Section~\\ref{sec:EM-field} we summarize the wave-optical solution we call \\emph{the angular eikonal method} \\cite{Turyshev-Toth:2021-multipoles} that allows us to determine the EM field in all regions behind the extended axisymmetric SGL, including the regions of strong and weak interference as well as the region of geometric optics.\nIn Section~\\ref{sec:v-large-disp} we address imaging with the SGL of the extended Sun, where we describe the signal received at the focal plane of an imaging telescope which moves in the image plane.\nIn Section~\\ref{sec:sims} we demonstrate the power of our formalism by presenting results that show the view of a point source, projected by the SGL and observed at various distances from the optical axis by an imaging telescope. We also present simulations of an extended source modeling light from a distant resolved star.\nIn Section~\\ref{sec:end} we present our conclusions and identify next steps.\nIn Appendix~\\ref{sec:lim-cases} we consider some limiting cases and demonstrate agreement between previously obtained results and the results in the present paper.\n\n\n\\section{The EM field all regions behind the lens}\n\\label{sec:EM-field}\n\nIn Ref.~\\cite{Turyshev-Toth:2021-multipoles}, we studied diffraction of EM waves in the presence of gravity. For this we considered a Mie problem with the electromagnetic field propagating in the vicinity of an extended gravitational lens in the first post-Newtonian approximation of the general theory of relativity. We were able to reduce the problem to a Schr\\\"odinger-like equation describing the Debye potential and then derived a complete solution for the EM field on an image plane positioned behind the lens. The resulting EM field was used to compute the energy flux in various regions behind that lens by calculating the Poynting vector -- the quantity that is needed to study the optical properties of the lens\\footnote{To simplify the material and keep the focus of this paper on its broader objectives, here we only summarize the solution, inviting the reader to consult \\cite{Turyshev-Toth:2021-multipoles} for technical details if needed.}.\n\n\\subsection{Summary of the solution}\n\n\\begin{figure}\n\\includegraphics[scale=0.27]{regions}\n\\caption{\\label{fig:regions}The different optical regions of the SGL\n(adapted from \\cite{Turyshev-Toth:2019-extend}).\n}\n\\end{figure}\n\nWe use a heliocentric spherical coordinate system $(r,\\theta,\\phi)$ and consider a source positioned at a distance $r_0$ from a lens. In \\cite{Turyshev-Toth:2021-multipoles}, we studied propagation of a light ray with impact parameter $b$ with respect to the lens and determined the components of the EM field that would be observed on an image plane at distance $r$ from the lens. For a high-frequency EM wave (i.e., neglecting terms $\\propto(kr)^{-1}$) and for $r\\gg r_g$ (with $r_g=2GM\/c^2$ being the Schwarzschild radius of the lens), we derived the EM field that is needed to estimate the flux through the image plane. Following the logic of solving the Mie problem \\cite{Born-Wolf:1999,Turyshev-Toth:2017}, this field can be given to the required order in the following form \\cite{Turyshev-Toth:2021-multipoles}:\n{}\n\\begin{eqnarray}\n \\left( \\begin{aligned}\n{D}_\\theta& \\\\\n{B}_\\theta& \\\\\n \\end{aligned} \\right) = \\left( \\begin{aligned}\n{B}_\\phi& \\\\\n-{D}_\\phi& \\\\\n \\end{aligned} \\right)&=&\n \\left( \\begin{aligned}\n \\cos\\phi& \\\\\n \\sin\\phi& \\\\\n \\end{aligned} \\right)\\,e^{-i\\omega t}\\gamma(r, \\theta)+{\\cal O}(r_g^2, \\theta^2, b\/r_0),\n \\label{eq:DB-sol-rho_go}\n\\end{eqnarray}\nwith the term $\\gamma(r, \\theta)$ given to ${\\cal O}\\big({r_g}\/{r},r_g^2\\big)$ as\n{}\n\\begin{eqnarray}\n \\gamma(r,\\theta,\\phi) &=& \\frac{E_0}{r_0}\\frac{ue^{ik(r+r_0+r_g\\ln 4k^2rr_0)}}{ikr}\\sum_{\\ell=kR^\\star_\\odot}^\\infty\\frac{\\ell+{\\textstyle\\frac{1}{2}}}{\\ell(\\ell+1)}e^{i\\big(2\\sigma_\\ell+\\frac{\\ell(\\ell+1)}{2k\\tilde r}+2\\xi_b\n\\big)}\n\\Big\\{\\frac{\\partial P^{(1)}_\\ell(\\cos\\theta)}\n{\\partial \\theta} +\\frac{P^{(1)}_\\ell(\\cos\\theta)}{\\sin\\theta}\n \\Big\\},\n \\label{eq:beta*1*}\n\\end{eqnarray}\nwhere $1\/\\tilde r=1\/r+1\/r_0$ (as discussed in \\cite{Turyshev-Toth:2019-extend}) and $\\sigma_\\ell$ is the Coulomb phase shift (see details in \\cite{Turyshev-Toth:2017}). The summation is conducted over the partial momenta $\\ell$ that, in a semiclassical analogy, is related to the impact parameter $b$ as $\\ell=kb$. Also, the sum in (\\ref{eq:beta*1*}) starts at $\\ell=kR^\\star_\\odot$, that corresponds to applying a fully-absolving boundary condition, capturing the fact that light rays with impact parameters $0\\leq b < R^\\star_\\odot=R_\\odot+r_g$ are completely absorbed by the opaque Sun.\nAs usual, $P^{(1)}_\\ell(\\cos\\theta)$ are Legendre polynomials of the first kind \\cite{Abramovitz-Stegun:1965}. The radial components of the EM wave behave as $({D}_r, {B}_r)\\sim {\\cal O}({\\rho}\/{z},b\/r_0)$; thus they are negligibly small compared to the other two components (\\ref{eq:DB-sol-rho_go}).\n\nThe quantity $\\xi_b$ in the phase of (\\ref{eq:beta*1*}) is the eikonal phase shift that is acquired by an EM wave as it travels in the vicinity of an extended axisymmetric gravitational lens (such as our Sun). To establish the form of this quantity, in \\cite{Turyshev-Toth:2021-multipoles} we used a heliocentric coordinate system with its $z$-axis aligned with the wave vector $\\vec k$ of the incident wave, so that $\\vec k=(0,0,1)$, and introduce the vector of the impact parameter, $\\vec b=b \\vec n_\\xi$. Using $z$ to denote the heliocentric distance of the image plane, we define $\\vec x$ to mark a position in the image plane. Lastly, we introduce a unit vector in the direction of the solar rotation axis, $\\vec s$. These quantities are given as:\n{}\n\\begin{eqnarray}\n{\\vec b}&=&b(\\cos\\phi_\\xi,\\sin \\phi_\\xi,0),\n\\label{eq:note-b}\\\\\n{\\vec x}&=&\\rho(\\cos\\phi,\\sin \\phi,0),\n\\label{eq:note-x}\\\\\n{\\vec s}&=&(\\sin\\beta_s\\cos\\phi_s,\\sin\\beta_s\\sin\\phi_s,\\cos\\beta_s).\n\\label{eq:note}\n\\end{eqnarray}\nWith this parametrization, the additional eikonal phase shift $\\xi_b$ induced by an extended, axisymmetric and rotating gravitational lens characterized in terms of zonal harmonics was determined \\cite{Turyshev-Toth:2021-multipoles} to have the form\n{}\n\\begin{eqnarray}\n\\xi_b=-kr_g\\sum_{n=2}^\\infty\\frac{J_n}{n}\\Big(\\frac{R_\\odot}{b}\\Big)^n \\sin^n\\beta_s\\cos[n(\\phi_\\xi-\\phi_s)],\n \\label{eq:xi_b}\n\\end{eqnarray}\nwhere $J_n$ are the zonal harmonic coefficients of the gravitational field of the lens, such as the SGL.\n\nIn \\cite{Turyshev-Toth:2021-multipoles}, we considered solution (\\ref{eq:DB-sol-rho_go})--(\\ref{eq:xi_b}) only in the strong interference region that lies in the proximity of the primary optical axis where $\\theta\\simeq \\sqrt{2r_g\/r}$. Our objective for this paper is to use the solution above and derive results that will be applicable in all the gravitational lensing regions that are formed behind the else that also include the weak interference region and that of the geometric optics.\n\n\\subsection{Eikonal correction for the azimuthal term}\n\nTo evaluate expression (\\ref{eq:beta*1*}), following \\cite{Turyshev-Toth:2021-multipoles}, we use the asymptotic representation for $P_\\ell(\\cos\\theta)$ and $\\ell\\gg1$ from\\footnote{For an improved, explicit, uniformly valid two-term asymptotic form of this expression, see \\cite{Bakaleinikov:2020}} \\cite{Bateman-Erdelyi:1953,Korn-Korn:1968,Kerker-book:1969,Abramovitz-Stegun:1965}:\n{}\n\\begin{eqnarray}\nP_\\ell(\\cos\\theta)&=& \\sqrt{\\frac{\\theta}{\\sin\\theta}} J_0\\big(\\ell \\theta\\big)+{\\cal O}(\\theta^2).\n\\label{eq:Bess0}\n\\end{eqnarray}\n\nNext, we use expression\n{}\n\\begin{eqnarray}\nP^{(1)}_\\ell(\\cos\\theta)=-\\frac{\\partial P_\\ell(\\cos\\theta)}{\\partial\\theta}=\\ell J_1(\\ell\\theta)+\n{\\textstyle\\frac{1}{6}}\\theta J_0(\\ell\\theta) +{\\cal O}(\\theta^2),\n\\label{eq:Bess}\n\\end{eqnarray}\nalongside with the recurrence relations for the Bessel functions \\cite{Abramovitz-Stegun:1965}\n{}\n\\begin{eqnarray}\n\\frac{2n}{x}J_n(x)=J_{n-1}(x)+J_{n+1}(x),\n\\label{eq:Bess-rec}\n\\end{eqnarray}\nand derive the following two well-known \\cite{Born-Wolf:1999} and useful relations\n{}\n\\begin{eqnarray}\n\\frac{P^{(1)}_\\ell(\\cos\\theta)}{\\sin\\theta}&=& {\\textstyle\\frac{1}{2}}\\ell^2\\Big(J_0(\\ell \\theta)+J_2(\\ell \\theta)\\Big),\n\\label{eq:pi-l=}\n\\qquad\n\\frac{dP^{(1)}_\\ell(\\cos\\theta)}{d\\theta}= {\\textstyle\\frac{1}{2}}\\ell^2\\Big(J_0(\\ell \\theta)-J_2(\\ell \\theta)\\Big).\n\\label{eq:tau-l=}\n\\end{eqnarray}\n\nSubstituting (\\ref{eq:pi-l=}) in expression (\\ref{eq:beta*1*}), and, following the approach that we presented in \\cite{Turyshev-Toth:2019-extend}, we consider the case of the large partial momenta, $\\ell\\gg1$, which is certainly valid here, as the integration is done from $\\ell=kR_\\odot\\gg1$ to infinity. In this case, the term $\\gamma(r, \\theta,\\phi)$ is determined from the following integral:\n{}\n\\begin{eqnarray}\n\\gamma(r,\\theta,\\phi) =\n\\frac{E_0}{r_0}\\frac{ue^{ik(r+r_0+r_g\\ln 4k^2rr_0)}}{ikr}\\int_{\\ell=kR^\\star_\\odot}^\\infty\n\\ell d\\ell e^{i\\big(2\\sigma_\\ell+\\frac{\\ell^2}{2k\\tilde r}+\n2\\xi_b\\big)}\\Big(\nJ_0(\\ell\\theta) +{\\cal O}\\big(\\theta^2,\\frac{r_g}{r},r_g^2\\big)\\Big).\n \\label{eq:gamma**1*}\n\\end{eqnarray}\n\nTo evaluate this integral we used the \\emph{angular eikonal method} presented in \\cite{Turyshev-Toth:2021-multipoles}. For that, we first recognize that in the case of a point mass (i.e, when only monopole is present), the resulting gravitational field is spherically symmetric \\cite{Turyshev-Toth:2017,Turyshev-Toth:2019}. However, once we include the field from the gravitational multipoles, that symmetry is broken as the eikonal phase shift acquires an azimuthal term, namely $\\xi_b=\\xi_b(b, \\theta,\\phi)$. However, we found a way to develop the treatment of the problem even in this generic case. First, we recall that the eikonal phase shift $\\xi_b$ was obtained through an iterative process involving the eikonal approximation that originates from the field of high-energy particle physics but is also applicable in the optics domain.\nNext, we recognize that for a spherically symmetric field (which is used as the starting point of our iterative process), the Bessel function $J_0(\\ell \\theta)$ can be used in its integral form\\footnote{Note that we can use the same representation of this function with the positive sign in the phase, but the result is identical as it will be integrated over the entire range of the azimuthal angle $\\phi_\\xi$. }\n{}\n\\begin{eqnarray}\nJ_0(\\ell\\theta)&=& \\frac{1}{2\\pi}\\int_0^{2\\pi} d\\phi_\\xi e^{-i\\ell\\theta\\cos(\\phi_\\xi-\\phi)}.\n\\label{eq:J0}\n\\end{eqnarray}\nThis is the natural step that captures the spherical symmetry of the field of a gravitational monopole. So, the iterative process used to derive the eikonal phase is conducted under this integral over all the azimuthal angles, $\\phi_\\xi$. However, in the case of the multipoles the azimuthal symmetry is broken. The presence of this integral over $d\\phi_\\xi$ allows us to account for this azimuthal dependence within the angular eikonal approximation; hence the name of the method.\n\nWe now substitute (\\ref{eq:J0}) into (\\ref{eq:gamma**1*}) and see that expression (\\ref{eq:gamma**1*}), to the order of ${\\cal O}\\big(\\theta^2,{r_g}\/{r},r_g^2\\big)$, transforms as\n{}\n\\begin{eqnarray}\n\\gamma(r,\\theta,\\phi) &=&\n\\frac{E_0}{r_0} \\frac{ue^{ik(r+r_0+r_g\\ln 4k^2rr_0)}}{ikr} \\frac{1}{2\\pi}\\int_0^{2\\pi} d\\phi_\\xi \\int_{\\ell=kR^\\star_\\odot}^\\infty\n\\ell d\\ell e^{i\\big(2\\sigma_\\ell+\\frac{\\ell^2}{2k\\tilde r}+2\\xi_b-\\ell\\theta\\cos(\\phi_\\xi-\\phi)\\big)}.\n \\label{eq:beta*3}\n\\end{eqnarray}\n\nIn this form, the integral over $d\\phi_\\xi$ properly acts not only on the monopole term represented by the term $2\\sigma_\\ell+\\frac{\\ell^2}{2k\\tilde r}-\\ell\\theta\\cos(\\phi_\\xi-\\phi)$ in the phase of the expression \\ref{eq:beta*3}, but on the entire phase $2\\sigma_\\ell+\\frac{\\ell^2}{2k\\tilde r}+2\\xi_b-\\ell\\theta\\cos(\\phi_\\xi-\\phi)$, which now includes contributions from nonspherical parts of the gravitational potential via the eikonal phase term, $2\\xi_b$. This process constitutes the \\emph{angular eikonal method}, valid for weak gravitational fields, which allows us to study the scattering of light on nonspherical potentials under the eikonal approximation.\n\n\\subsection{Taking the integral over $b$ with the method of stationary phase}\n\nTo develop a solution for (\\ref{eq:beta*3}), and for convenience, we use (\\ref{eq:xi_b}) and introduce quantity $\\psi(\\vec b)$, as\n{}\n\\begin{eqnarray}\n\\xi_b(\\vec b)=-kr_g\\psi(\\vec b),\\, \\qquad\n\\psi(\\vec b)=\\sum_{n=2}^\\infty\\frac{J_n}{n}\\Big(\\frac{R_\\odot}{b}\\Big)^n \\sin^n\\beta_s\\cos[n(\\phi_\\xi-\\phi_s)].\n \\label{eq:psi}\n\\end{eqnarray}\n\n\nFurthermore, for $\\ell\\gg kr_g$, evaluate $\\sigma_\\ell$ as \\cite{Turyshev-Toth:2018-grav-shadow}:\n{}\n\\begin{eqnarray}\n\\sigma_\\ell&=& -kr_g\\ln \\ell.\n\\label{eq:sig-l*}\n\\end{eqnarray}\nThis form agrees with the other known forms of $\\sigma_\\ell$ \\cite{Cody-Hillstrom:1970,Barata:2009ma} that are approximated for large $\\ell$ (see discussion in \\cite{Turyshev-Toth:2017,Turyshev-Toth:2019-image}).\n\nWe rely on the semiclassical approximation (see relevant discussion in \\cite{Turyshev-Toth:2017,Turyshev-Toth:2019}) that connects the partial momenta, $\\ell$, to the impact parameter $b$:\n{}\n\\begin{equation}\n\\ell\\simeq kb,\n\\label{eq:S-l-pri-p-g}\n\\end{equation}\nwhich is applicable for small angles $\\theta$ (or, large distances from the Sun, $R_\\odot\/r=\\frac{c}{4\\pi}\\frac{E_0^2}{(r+r_0)^2}\n\\big<\\overline{\\big({\\rm Re}\\big[{ B}({\\vec x})e^{i\\Omega (t)}\\big]\\big)^2}\\big>=\\frac{c}{8\\pi}\\frac{E_0^2}{(r+r_0)^2}\n|B({\\vec x})|^2,\n \\label{eq:S_z*6z}\n\\end{eqnarray}\nwhere $|B({\\vec x})|^2=B({\\vec x})B^*({\\vec x})$, with $B^*(\\vec x)$ being the complex conjugate of $B({\\vec x})$. Note that ${\\bar S}_\\rho= {\\bar S}_\\phi=0$ for all practical purposes. Defining light amplification as usual \\cite{Turyshev-Toth:2017,Turyshev-Toth:2019,Turyshev-Toth:2019-extend}, $\\mu_z({\\vec x})=S_z({\\vec x})\/|\\vec S_0({\\vec x})|$, where $\\vec S_0({\\vec x})=(c\/8\\pi){E_0^2}\/{(r+r_0)^2}\\, \\vec k$ being the Poynting vector carried by a plane wave in the vacuum in flat spacetime, we have the light amplification factor of the lens that, for short wavelengths (i.e., $kr_g\\gg1$) is given by\n {}\n\\begin{eqnarray}\n\\mu_z({\\vec x})=|B({\\vec x})|^2.\n \\label{eq:S_mu}\n\\end{eqnarray}\n We recognize that the quantity $\\mu_z({\\vec x})$ is the PSF of the SGL that is scaled by the amplification factor and it describes all lensing regimes with this extended lens. In Appendix~\\ref{sec:small-dev} we show that, in some cases, the amplification factor explicitly multiplies the PSF, but, in general, the PSF (\\ref{eq:S_mu}) is implicitly scaled by the amplification factor via (\\ref{eq:DB-sol-in-cc+}).\n\nIn Appendix~\\ref{sec:lim-cases} we consider limiting cases of $B\\big(\\vec x) $ from (\\ref{eq:DB-sol-in-cc+}). Those cases include very small deviations $\\rho$ from the optical axis, namely $\\rho\/r\\equiv \\theta\\ll \\sqrt{2r_g\/r}$; very large deviations $\\rho\/r\\equiv \\theta\\gg \\sqrt{2r_g\/r}$; and those in-between. We show that far from the optical axis, the PSF that is constructed with the help of $B({\\vec x})$ from (\\ref{eq:DB-sol-in-cc+}) exhibits the behavior of the monopole PSF \\cite{Turyshev-Toth:2017}, but as we come closer to the optical axis, the effect of multipoles becomes more pronounced, ultimately bringing us to the caustic region, discussed in \\cite{Turyshev-Toth:2021-multipoles}.\n\n\\section{Imaging with the SGL of the extended Sun}\n\\label{sec:v-large-disp}\n\nThe complex amplitude (\\ref{eq:DB-sol-in-cc+}) developed in the previous section describes the EM field in the image plane. This field, however, is not what is usually observed. Rather, observations are made with an imaging telescope looking back in the direction of the lens. Our formalism also grants us the ability to accurately describe the image that forms in the focal plane of such a telescope: i.e., the actual observable.\n\n \\subsection{Description of the imaging geometry}\n\\label{sec:geom}\n\n With the knowledge of the EM field in the image plane behind an extended gravitational lens (\\ref{eq:DB-sol-in-cc8=})--(\\ref{eq:DB-sol-in-cc+}) and following the approach developed in \\cite{Turyshev-Toth:2019-image,Turyshev-Toth:2020-im-extend,Turyshev-Toth:2021-imaging}, we can now describe what an imaging telescope would detect on its focal plane. Such telescopic capability is important as it characterizes the measured optical signal \\cite{Turyshev-Toth:2021-imaging,Turyshev-Toth:2021-quartic}.\n\nSimilarly to \\cite{Turyshev-Toth:2019-image,Turyshev-Toth:2021-imaging}, we describe the geometry of the observation using ${\\vec x}$ to represent the current position of an optical telescope in the SGL's image plane, ${\\vec x}'$, denoting any point in the same plane, and ${\\vec x}_i$, representing a point on the focal plane of the optical telescope. These positions are given as\n{}\n\\begin{eqnarray}\n\\{{\\vec x}\\}&\\equiv& (x,y,0)=\\rho\\,\\big(\\cos\\phi,\\sin\\phi,0\\big)=\\rho{\\vec n}, \\label{eq:coord'}\\\\\n\\{{\\vec x'}\\}&\\equiv& (x',y',0)=\\rho'\\big(\\cos\\phi',\\sin\\phi',0\\big)=\\rho'{\\vec n'},\n\\label{eq:x-im} \\\\\n \\{{\\vec x}_i\\}&\\equiv& (x_i,y_i,0)=\\rho_i\\big(\\cos\\phi_i,\\sin\\phi_i,0\\big)=\\rho_i{\\vec n}_i.\n \\label{eq:coord}\n\\end{eqnarray}\n\nTo convolve the PSF of the SGL with that of a thin lens that represents an aperture of a telescope, we first need to establish an appropriate form of the PSF for point sources. Examining (\\ref{eq:DB-sol-inA}), we see that it contains the expression $\\rho\\cos(\\phi_\\xi-\\phi)$, which may be transformed as\n{}\n\\begin{eqnarray}\n\\rho\\cos(\\phi_\\xi-\\phi)=({\\vec n}_\\xi\\cdot{\\vec x}).\n\\label{eq:rn}\n\\end{eqnarray}\n\nWe now transition from the current position $\\vec x$ of the telescope to an arbitrary location within the telescope's aperture by the substitution\n{}\n\\begin{eqnarray}\n\\vec x \\qquad \\Rightarrow \\qquad {\\vec x}+\\vec x'.\n\\label{eq:rn0}\n\\end{eqnarray}\nTherefore, we may write\n{}\n\\begin{eqnarray}\n({\\vec n}_\\xi\\cdot{\\vec x}) \\qquad \\rightarrow \\qquad\n({\\vec n}_\\xi\\cdot{\\vec x}) +({\\vec n}_\\xi\\cdot{\\vec x}').\n\\label{eq:rn2}\n\\end{eqnarray}\nWe note that $\\vec x'$ varies only with the aperture, whereas $\\vec x$ can be anywhere in the SGL image plane. In much of the image plane, we have $\\rho' \\ll \\rho$. This allows us to expand (\\ref{eq:DB-sol-inA}) in terms of the small parameter $\\rho'\/\\rho$, keeping only terms of the first order in $\\rho'\/\\rho$. In addition, we recognize that the vector $\\vec \\theta=\\theta(\\cos\\phi,\\sin\\phi,0)=\\vec x\/r$, with $r$ being the distance to the image plane, may be transformed as\n{}\n\\begin{eqnarray}\n({\\vec n}_\\xi\\cdot{\\vec \\theta})=({\\vec n}_\\xi\\cdot{\\vec x})\/r \\qquad \\rightarrow \\qquad\n({\\vec n}_\\xi\\cdot{\\vec x})\/r +({\\vec n}_\\xi\\cdot{\\vec x}')\/r=\\frac{\\rho}{r}\\cos(\\phi_\\xi-\\phi)+\\frac{\\rho'}{r}\\cos(\\phi_\\xi-\\phi').\n\\label{eq:the}\n\\end{eqnarray}\n\nThis approximation yields the following result for the complex amplitude, ${ A}_{\\tt }({\\vec x})$, from (\\ref{eq:DB-sol-inA}), but given with the shifted argument according to (\\ref{eq:rn0}):\n{}\n\\begin{eqnarray}\n{ A}_{\\tt }({\\vec x},{\\vec x}')&=&\na_{\\tt }(\\vec x,\\vec n_\\xi)\n\\exp\\Big[i\\Big(\\delta\\varphi_{\\tt }(\\vec x,{\\vec n}_\\xi)-\\nu{\\tt }({\\vec n}_\\xi\\cdot{\\vec x'})\\Big)\\Big],\n \\label{eq:amp-Ain}\n\\end{eqnarray}\nwith the amplitude factor $a_{\\tt }(\\vec x)$ and phase $\\delta\\varphi_{\\tt }(\\vec x) $ given as\n{}\n \\begin{eqnarray}\na_{\\tt }(\\vec x,\\vec n_\\xi) &=& \\sqrt{\\pi k\\tilde r}\\Bigg[\\frac{\\Big(\\sqrt{\\big({\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r \\big)^2+\\frac{2r_g}{ \\tilde r}}+{\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r \\Big)^3}{\\sqrt{\\big({\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r \\big)^2+\\frac{2r_g}{ \\tilde r}}}\\Bigg]^{1\/2}+{\\cal O}(\\vec x\/r),\n \\label{eq:q_insc}\\\\\n \\delta\\varphi_{\\tt }\n(\\vec x,\\vec n_\\xi) &=& -k\\bigg\\{{\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x}) \\Big( \\sqrt{\\big({\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r \\big)^2+\\frac{2r_g}{\\tilde r}}+{\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r\\Big)+\n\\nonumber\\\\\n&+&\n2r_g\\bigg(\\ln \\Big(\\sqrt{\\big({\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r\\big)^2+\\frac{2r_g}{\\tilde r}}+ {\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r\\Big)+\n\\sum_{n=2}^\\infty \\frac{J_n}{n}\\frac{R^n_\\odot \\sin^n\\beta_s\\cos[n(\\phi_\\xi-\\phi_s)]}{\\Big(\\sqrt{\\big({\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\\big)^2+2r_g \\tilde r}+ {\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\\Big)^n}\\bigg)\\bigg\\}.~~~~~\n \\label{eq:Ain-d_ph}\n\\end{eqnarray}\nWe note that when the angle $\\theta$ is large, $\\theta \\gg \\sqrt{2r_g\/\\tilde r}$ and thus, $\\rho\\gg \\sqrt{2r_g\\tilde r}$, and we get back the PSF of a monopole. Thus, the integral (\\ref{eq:beta*3}) may be taken using the method of stationary phase applied to the double integral. In that case, the factors $a_{\\tt in\/sc}$ in (\\ref{eq:q_insc}) take their known values (see \\cite{Turyshev-Toth:2019-extend} for details), namely $a^2_{\\tt in}(\\rho,\\tilde r)=1+{\\cal O}(r_g\\theta^2,r_g^2)$ and $a^2_{\\tt sc}(\\rho,\\tilde r) =({2r_g\\tilde r}\/{\\rho^2})^2 +{\\cal O}(r_g\\theta^2,r_g^2)$. However, our new expressions (\\ref{eq:q_insc}) allow studying the cases when $\\rho\\simeq \\sqrt{2r_g\\tilde r}$ anywhere in the image plane.\nThe last quantity present in (\\ref{eq:amp-Ain}) is the spatial frequency $\\nu=\\nu_{\\tt }(\\vec x,\\vec n_\\xi)$, defined as\n{}\n \\begin{eqnarray}\n\\nu_{\\tt }(\\vec x,\\vec n_\\xi)&=&k\\Big(\\sqrt{\\big({\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r\\big)^2+\\frac{2r_g}{\\tilde r}}+ {\\textstyle\\frac{1}{2}} ({\\vec n}_\\xi\\cdot{\\vec x})\/r\\Big).\n \\label{eq:betapm}\n \\end{eqnarray}\n\nThe quantities (\\ref{eq:amp-Ain})--(\\ref{eq:betapm}) describe the complex amplitude of the EM field, ${B}({\\vec x},{\\vec x}')$ from (\\ref{eq:DB-sol-in-cc=}), as measured in the focal plane of an imaging telescope.\n\n \\subsection{The EM field in the telescope's focal plane}\n\\label{sec:im-tele}\n\nThe focal plane of the optical telescope is located at the focal distance $f$ of the lens, centered on ${\\vec x}'$. Using the Fresnel--Kirchhoff diffraction formula, the amplitude of the image field in the optical telescope's focal plane at a location ${\\vec x}_i=(x_i,y_i)$ is derived from (\\ref{eq:DB-sol-in-cc=}) and is given by \\cite{Wolf-Gabor:1959,Richards-Wolf:1959,Born-Wolf:1999}:\n{}\n\\begin{eqnarray}\n{B}({\\vec x},{\\vec x}_i)=\\frac{i}{\\lambda}\\iint \\displaylimits_{|{\\vec x}'|^2\\leq (d\/2)^2} \\hskip -7pt B({\\vec x},{\\vec x}')e^{-i\\frac{k}{2f}|{\\vec x}'|^2}\\frac{e^{iks'}}{s'}d^2{\\vec x}'.\n \\label{eq:amp-w-f0}\n\\end{eqnarray}\n\nThe function $\\exp[-i\\frac{k}{2f}|{\\vec x}'|^2]=\\exp[-i\\frac{k}{2f}(x'^2+y'^2)]$ in (\\ref{eq:amp-w-f0}) represents the action of the convex lens that transforms incident plane waves to spherical waves, focusing at the focal point. Assuming that the focal length is sufficiently greater than the radius of the lens, we may approximate the optical path $s'$ as $s'=\\sqrt{(x'-x_i)^2+(y'-y_i)^2+f^2}\\sim f+\\big((x'-x_i)^2+(y'-y_i)^2\\big)\/2f$. This allows us to present (\\ref{eq:amp-w-f0}) as\n{}\n\\begin{eqnarray}\n{B}({\\vec x},{\\vec x}_i)&=&\n- \\frac{e^{ikf(1+{{\\vec x}_i^2}\/{2f^2})}}{i\\lambda f}\\iint\\displaylimits_{|{\\vec x}'|^2\\leq (\\frac{1}{2}d)^2} d^2{\\vec x}'\n B({\\vec x},{\\vec x}') e^{-i\\frac{k}{f}({\\vec x}'\\cdot{\\vec x}_i)}.\n \\label{eq:amp-w-f+}\n\\end{eqnarray}\n\nExpressions (\\ref{eq:amp-Ain})--(\\ref{eq:betapm}) allow us to consider imaging of point sources with the SGL, now treated as that produced by a gravitating body that is axisymmetric and rotating, thus admitting characterization of its external gravitational field by zonal harmonics. To accomplish this, following \\cite{Turyshev-Toth:2019-image,Turyshev-Toth:2020-im-extend}, we use the expression for $A({\\vec x},{\\vec x}')$ from (\\ref{eq:amp-Ain}) and present the Fresnel--Kirchhoff diffraction formula as\n{}\n\\begin{eqnarray}\n{\\cal A}_{\\tt }({\\vec x},{\\vec x}_i)&=&\n- \\frac{e^{ikf(1+{{\\vec x}_i^2}\/{2f^2})}}{i\\lambda f}\\iint\\displaylimits_{|{\\vec x}'|^2\\leq (\\frac{1}{2}d)^2} d^2{\\vec x'}\\,\n{ A}_{\\tt }({\\vec x},{\\vec x}') e^{-i\\eta_i({\\vec n}_i\\cdot{\\vec x}')}=\\nonumber\\\\\n&=&- \\frac{e^{ikf(1+{{\\vec x}_i^2}\/{2f^2})}}{i\\lambda f}\n a_{\\tt}(\\vec x,\\vec n_\\xi)\ne^{i\\delta\\varphi_{\\tt }(\\vec x,\\vec n_\\xi)}\\,\n\\iint\\displaylimits_{|{\\vec x}'|^2\\leq (d\/2)^2}\\hskip -8pt\n d^2{\\vec x}'\\,\ne^{i\\big(-\\nu_{\\tt }({\\vec n}_\\xi\\cdot{\\vec x}')-\\eta_i({\\vec n}_i\\cdot{\\vec x}')\\big)},\n \\label{eq:amp-w-f}\n\\end{eqnarray}\nwhere the spatial frequency $\\nu=\\nu_{\\tt }(\\vec x,\\vec n_\\xi)$ is given by (\\ref{eq:betapm}). Also, for a telescope with focal length of $f$ and for a radial pixel position $\\rho_i$, the factor $ \\eta_i$ has the form \\cite{Turyshev-Toth:2019-image,Turyshev-Toth:2020-im-extend,Turyshev-Toth:2021-imaging}\n{}\n\\begin{eqnarray}\n \\eta_i=k\\frac{\\rho_i}{f}.\n\\label{eq:zerJ}\n\\end{eqnarray}\n\nTherefore, to derive the amplitudes of the EM field in the focal plane of the optical telescope, corresponding to (\\ref{eq:amp-Ain}), we need to evaluate an integral of the type\n{}\n\\begin{eqnarray}\n\\iint\\displaylimits_{|{\\vec x}'|^2\\leq (d\/2)^2}\\hskip -8pt\n d^2{\\vec x}'\\,e^{i\\big(- \\nu_{\\tt }({\\vec n}_\\xi\\cdot{\\vec x}')-\\eta_i({\\vec n}_i\\cdot{\\vec x}')\\big)}.\n \\label{eq:amp-int}\n\\end{eqnarray}\nTo evaluate this integral, we present the phase in (\\ref{eq:amp-int}) as\n{}\n\\begin{eqnarray}\n- \\nu_{\\tt }({\\vec n}_\\xi\\cdot{\\vec x}')-\\eta_i({\\vec n}_i\\cdot{\\vec x}')=-u\\,\\rho' \\cos\\big(\\phi'-\\sigma\\big)+{\\cal O}(\\rho^2),\n \\label{eq:ph4s}\n\\end{eqnarray}\nwhere, for convenience, we defined\n{}\n\\begin{eqnarray}\nu&=&\\sqrt{\\nu^2+2\\nu\\eta_i\\cos\\big(\\phi_\\xi-\\phi_i\\big)+\\eta_i^2},\n\\qquad\n\\cos\\sigma=\n\\frac{\\nu \\cos\\phi_\\xi+\\eta_i\\cos\\phi_i}{u},\n\\qquad\n\\sin\\sigma=\\frac{\\nu \\sin\\phi_\\xi+\\eta_i\\sin\\phi_i}{u}.\n\\label{eq:vpms}\n\\end{eqnarray}\n\nWith these definitions, and using the parameterization given in (\\ref{eq:x-im}), the integral (\\ref{eq:amp-int}) may be evaluated as\n{}\n\\begin{eqnarray}\n \\int_0^{2\\pi} \\hskip -4pt d\\phi' \\int_0^{d\/2} \\hskip -4pt \\rho' d\\rho'\\,\n e^{-iu\\rho'\\cos(\\phi'-\\sigma)}=\\pi\\Big(\\frac{d}{2}\\Big)^2\\, \\frac{\n2J_1(u\\frac{1}{2}d)}{u\\frac{1}{2}d}.\n \\label{eq:amp-int*}\n\\end{eqnarray}\nAs a result, using (\\ref{eq:amp-Ain}) in (\\ref{eq:amp-w-f}) leads to the following amplitude of the EM wave on the optical telescope's image plane:\n{}\n\\begin{eqnarray}\n{\\cal A}_{\\tt }({\\vec x},{\\vec x}_i,\\vec n_\\xi)&=&\n\\Big(\\frac{kd^2}{8f}\\Big)\\, \\Big\\{\na_{\\tt }\n\\Big(\\frac{\n2J_1(u\\frac{1}{2}d)}{u\\frac{1}{2}d}\\Big)e^{i\\big(kf(1+{{\\vec x}_i^2}\/{2f^2})+\\delta\\varphi_{\\tt }\n(\\vec x,\\vec n_\\xi) +\\frac{\\pi}{2}\\big)}+{\\cal O}(r_g^2)\\Big\\}.\n \\label{eq:amp-Aind}\n\\end{eqnarray}\n\nTherefore, the Fourier-transformed complex amplitude (\\ref{eq:amp-w-f0}) takes the from\n{}\n\\begin{eqnarray}\n&& \\frac{1}{2\\pi}\\int_0^{2\\pi} d\\phi_\\xi \\, {\\cal A}_{\\tt }({\\vec x},{\\vec x}_i,\\vec n_\\xi)=\n \\Big(\\frac{kd^2}{8f}\\Big)e^{i\\big(kf(1+{{\\vec x}_i^2}\/{2f^2}) +\\frac{\\pi}{2}\\big)}\n {\\cal B} \\big(\\vec x,\\vec x_i\\big),~~~~~\n \\label{eq:FtB}\n\\end{eqnarray}\nwhere ${\\cal B} \\big(\\vec x,\\vec x_i\\big) $ is given as\n{}\n\\begin{eqnarray}\n {\\cal B} \\big(\\vec x,\\vec x_i\\big) &=&\n \\frac{1}{2\\pi}\\int_0^{2\\pi} d\\phi_\\xi \\, \\Big\\{\n a_{\\tt }(\\vec x,\\vec n_\\xi)\n \\Big(\\frac{\n2J_1(u({\\vec x}_i,\\vec x,\\vec n_\\xi)\\frac{1}{2}d)}{u({\\vec x}_i,\\vec x,\\vec n_\\xi) \\frac{1}{2}d}\\Big)\ne^{i\\delta\\varphi_{\\tt }(\\vec x,\\vec n_\\xi)}\\Big\\},~~~~~\n \\label{eq:Binsc}\n\\end{eqnarray}\nwhere $a_{\\tt }(\\vec x,\\vec n_\\xi)$, $\\delta\\varphi_{\\tt }(\\vec x,\\vec n_\\xi)$, and $u({\\vec x}_i,\\vec x,\\vec n_\\xi)$ are given by (\\ref{eq:q_insc}), (\\ref{eq:Ain-d_ph}), and (\\ref{eq:vpms}), correspondingly.\n\nUsing this result together with (\\ref{eq:DB-sol-in-cc8=}), we obtain the EM field on the detector that is given as below\n{}\n\\begin{eqnarray}\n \\left( \\begin{aligned}\n{E}_\\rho& \\\\\n{H}_\\rho& \\\\\n \\end{aligned} \\right)_{\\tt} = \\left( \\begin{aligned}\n{H}_\\phi& \\\\\n-{E}_\\phi& \\\\\n \\end{aligned} \\right)_{\\tt}&=&\n\\frac{E_0}{r+r_0}\n e^{i\\big(\\Omega(t)+\\frac{\\pi}{2}+kf(1+{{\\vec x}_i^2}\/{2f^2})\\big)}\n \\Big(\\frac{kd^2}{8f}\\Big)\n {\\cal B} \\big(\\vec x,\\vec x_i\\big)\n \\bigg( \\begin{aligned}\n\\cos\\overline \\phi& \\\\\n\\sin\\overline \\phi& \\\\\n \\end{aligned} \\bigg).\n \\label{eq:EM-det}\n\\end{eqnarray}\n\nAfter time averaging, we derive the Poynting vector of the EM wave in the focal plane of the imaging telescope:\n {}\n\\begin{eqnarray}\nS_{\\tt }({\\vec x},{\\vec x}_i)&=&\\frac{c}{8\\pi}\n\\frac{E_0^2}{(r+r_0)^2}\n \\Big(\\frac{kd^2}{8f}\\Big)^2\n {\\cal B}^2 \\big(\\vec x,\\vec x_i\\big) .\n \\label{eq:PV}\n\\end{eqnarray}\n\nAs a result, the intensity on the focal plane, ${\\cal I}({\\vec x},{\\vec x}_i)$, of the system that includes the SGL and a thin lens is given in the form as below:\n {}\n\\begin{eqnarray}\n{\\cal I}({\\vec x},{\\vec x}_i)&=&\n {\\cal B}^2 \\big(\\vec x,\\vec x_i\\big),\n \\label{eq:PFT}\n\\end{eqnarray}\nwhere the Fourier-transformed complex amplitude $ {\\cal B}\\big(\\vec x,\\vec x_i\\big)$ from (\\ref{eq:Binsc}).\nWe emphasize that $\\mu_z({\\vec x})$ from (\\ref{eq:S_mu}) is the PSF of the extended SGL. It describes the image of a point source projected on the image plane at the SGL's focal region. At the same time, the quantity ${\\cal I}({\\vec x},{\\vec x}_i)$ from (\\ref{eq:PFT}) is the intensity of light received on the focal plane of an imaging telescope. This is a directly observable quantity that is accessible to an optical telescope. As such, it is of most importance for any practical applications of the SGL.\nThe resulted expression for the intensity on the focal plane allows considering imaging of various sources with the SGL of an extended Sun. We will do that next.\n\n\\section{Application of results}\n\\label{sec:sims}\n\n\\begin{figure}\n\\includegraphics{psf4m}\n\\caption{\\label{fig:astroid}An illustrative example of the SGL PSF, appearing as the astroid caustic projected into the image plane by the SGL, with its recognizable cusps (vertices) and folds. As an imaging telescope enters this region in the image plane, its view of a distant source transitions from a pair of images (the primary and secondary image) into some variation of an Einstein cross or Einstein ring, depending on the size of the astroid, the imaging wavelength, and the size of the light source. Adapted from \\cite{Turyshev-Toth:2021-multipoles}.}\n\\end{figure}\n\nThe formalism developed in the preceding section opens the route to simulate the effects of the SGL beyond the immediate vicinity of the optical axis in its strong interference region (see Fig.~\\ref{fig:regions}). There is, however, first our obstacle: evaluation of the remaining integral in our final expression (\\ref{eq:Binsc}).\n\n\\subsection{Evaluation method}\n\nEquation~(\\ref{eq:Binsc}) describes the view seen by an imaging telescope of a distant source, both near and far from the optical axis of the gravitational lens. To use this equation, it is necessary to evaluate the remaining integral in the regions of interest. Examining it more closely, we note that the integral has finite integration limits, which makes numerical evaluation easier. However, it is still an oscillatory integral. Moreover, at large distances from the optical axis, the oscillations become very rapid. This makes direct numerical evaluation challenging.\n\nOn the other hand, a rapidly oscillating integral implies the possible use of the method of stationary phase once again. This is precisely what we have accomplished in \\cite{Turyshev-Toth:2021-quartic}, for the case when $J_4$ and higher order zonal harmonics can be safely neglected, thus leaving only the astroid caustic due to $J_2$. The result, expressed through the roots of a quartic equation, works reliably everywhere in the region of strong interference, only showing occasional rounding errors in the immediate vicinity of the caustic boundary of the projected astroid pattern of a quadrupole lens (Fig.~\\ref{fig:astroid} \\cite{Turyshev-Toth:2021-multipoles}).\n\nBeyond the region of strong interference, the contribution of the zonal harmonics is negligible and we can use previously developed monopole solutions for efficient evaluation.\n\nUsing this combination of methods, we are now in the position to evaluate (\\ref{eq:Binsc}) everywhere in the image plane, constructing simulated views of point sources as seen by an imaging telescope through the SGL.\n\n\\subsection{Simulated approach to the optical axis}\n\n\\begin{figure}\n\\includegraphics{ptearly-new}\n\\caption{\\label{fig:appr-pre}View of a distant compact source by a telescope approaching the SGL optical axis associated with that source. The telescope is positioned at $3\\times 10^5$~km, $1\\times 10^5$~km and $4\\times 10^3$~km from the optical axis. Note that at $3\\times 10^5$~km, the secondary image of the source is still obscured by the solar disk (shown as a yellow circle). By the time we reach $4\\times 10^3$~km, the images become indistinguishable, even as light amplification increases nearly hundredfold. For the full animation, see \\protect\\url{https:\/\/www.vttoth.com\/CMS\/physics-notes\/361}.}\n\\end{figure}\n\n\n\\begin{figure}\n\\includegraphics{ptfinal-new}\n\\caption{\\label{fig:approach1}View of a distant point source by a telescope near the optical axis, at 2~m, 1~m and positioned on the axis. The optical axis is at 5.74$^\\circ$ from the solar axis of rotation, a direction chosen because it representatively shows the development of an Einstein cross during this final approach. The view is that of a telescope with a 1~m aperture; light amplification is of ${\\cal O}(10^8)$.\nFor the full animation, see \\protect\\url{https:\/\/www.vttoth.com\/CMS\/physics-notes\/361}.}\n\\end{figure}\n\n\nTo demonstrate the power of the approach captured by the expression (\\ref{eq:Binsc}), we chose to simulate the view of a distant point source, as seen by an imaging telescope that is approaches the optical axis of that star with respect to the SGL.\n\nWe were able to assemble a series of still images, ultimately in the form of animations\\footnote{See \\protect\\url{https:\/\/www.vttoth.com\/CMS\/physics-notes\/361} for a full set of animations.}, which show how an imaging telescope would see the distant source as it was approaches the optical axis that corresponds to that source. Select frames from this animation are presented in this section.\n\nWe began the simulation with the imaging telescope located at $10^6$ km from the optical axis, looking in the direction of the Sun (see Fig.~\\ref{fig:appr-pre}). This distance was chosen because it is comparable in magnitude to the solar radius, thus placing the imaging telescope firmly in the region of geometric optics.\n\nAt the beginning, the source's ``primary image'' is outside the telescope's field of view, and no noticeable ``secondary image'' forms yet on the opposite side of the Sun. At $6\\times 10^5$~km from the optical axis, a faint secondary image emerges, or rather, would emerge if the Sun were transparent. In reality, light from that secondary image is yet blocked by the opaque disk of the Sun. When the telescope is at $3\\times 10^5$~km from the optical axis (less than half the solar radius) the primary image becomes clearly visible within the imaging area. This is the unobstructed view of the distant source, already amplified by the SGL, so its peak central brightness is $\\sim$1.8 times the brightness of the unamplified image. The secondary image, now less faint, is still obscured by the solar disk.\n\nWhen the telescope is only $\\sim1\\times 10^5$~km from the optical axis, the secondary image emerges from behind the Sun. Light amplification is becoming significant: the primary image's peak brightness is now more than four times as bright as the unamplified star. When the telescope approaches within $\\sim 2\\times 10^4$~km of the optical axis, the primary and secondary images are already nearly identical in appearance, at symmetric positions, settling at a distance from the solar limb that corresponds to the radius of a yet-to-form Einstein ring. Light amplification is substantial: the peak brightness that the imaging telescope sees is nearly 20 times the intensity of light from the unamplified star. Even so, the images remain point-like in appearance: This is dictated by the diffraction-limited resolution of the imaging telescope itself.\n\nAt this stage, the position of the two images of the point source is final. As the telescope continues to approach the optical axis, however, light amplification increases across several orders of magnitude.\n\nFor the purposes of this simulation, we chose to place the optical axis very near the solar axis of rotation, in order to keep the contribution of the $J_2$ zonal harmonic small. Figure \\ref{fig:approach1} shows the telescope's final approach to an optical axis that is at $5.74^\\circ$ from the solar axis of rotation, which corresponds to $\\sin\\beta_s=0.1$. This yields an astroid PSF that is relatively small, convenient for visual presentation.\n\nOnce the telescope is within a distance comparable to the size of the astroid caustic (in this case, within 10 meters), the secondary image begins to widen into an arc. Even closer to the optical axis, the arc splits into three distinct spots of light. As the telescope settles on the optical axis, these spots migrate to their final positions on the circumference of the Einstein ring, resulting in a fully formed Einstein cross. (This simulation assumed that the telescope approaches from one of the principal directions of the astroid caustic, i.e., one of the cusps. To see what happens when the telescope approaches from a different angle, see, e.g., \\cite{Turyshev-Toth:2021-quartic}.)\n\nIt is remarkable that all these animation frames are simply surface density plots of the integral expression given by Eq.~(\\ref{eq:Binsc}), which accurately describes an axisymmetric gravitational lens dominated by a spherically symmetric gravitational potential in all regions, both near and far the optical axis. We can generate with equal ease images seen through a telescope that is positioned as far as a million kilometers or more from the optical axis or a telescope that is at the optical axis or its immediate vicinity.\n\n\\subsection{Viewing an extended object}\n\n\\begin{figure}\n\\includegraphics{hoststar-new}\n\\caption{\\label{fig:hoststar-new}View of a distant star by a telescope approaching the SGL optical axis, at distances of 1,000~km, $\\sim$200~km and $\\sim$150~km. The geometric projection of the start to the image plane would yield a disk with a radius of 200~km. As the imaging telescope approaches this distance, a full Einstein-ring forms; subsequently, the ring brightens and becomes uniform as the telescope settles on the optical axis.}\n\\end{figure}\n\nThe PSF of a lens represents its impulse response: the image that forms when the light source is a point source. An extended object can, of course, be considered as a collection of point sources. The most straightforward method (though computationally inefficient) of convolving an extended source with the PSF of the lens is by dividing the source into point sources and iterating through them.\n\nTo demonstrate this, we considered an extended source in the form of a uniformly illuminated disk, which could represent a host star. We chose a disk that would be geometrically projected to an image with a 200~km radius in the image plane. With the image plane at 650~AU, this would correspond to a Sun-sized star at $\\sim$36 light-years.\n\nFor computational efficiency, we modeled the extended source using a simple adaptive mesh implementation, refining the resolution for regions that are projected close to the telescope's location in the image plane. This approach was sufficient to create a series of animation frames\\footnote{For the full animation, see \\url{https:\/\/www.vttoth.com\/CMS\/physics-notes\/360}.}, several of which are shown in Fig.~\\ref{fig:hoststar-new}.\n\nWhen the telescope looking at such an extended object is far from the optical axis, the telescopic image appears similar to that produced by a point source (see Fig.~\\ref{fig:appr-pre}). However, when the telescope begins to approach the projected image area corresponding to the extended source, a very different picture emerges. Instead of developing into an Einstein cross, the view of the telescope shows a fully formed Einstein ring. We may think of this Einstein ring as a collection of a large number of overlapping Einstein crosses at various orientations, corresponding to the point sources constituting the extended source. Thus, instead of being dominated by light from a single point-like region in the source, the Einstein ring now contains a mix of light from many different regions of the extended source.\n\n\\section{Discussion and Conclusions}\n\\label{sec:end}\n\nWe studied the optical properties of an extended axisymmetric gravitational lens. The gravitational potential for such a lens can be described using an infinite series of zonal harmonics. We extended the description of the SGL optical properties from the strong interference region to all lensing regimes. The new results can now also describe lensing in the weak interference region and that in the geometric optics region.\n\nThe expressions that we obtained can be used to describe the light field that is created by the SGL in its focal region. It can also be convolved with a representation of an optical telescope (modeled as a thin lens telescope) to show the view seen by such a telescope. The results are ``actionable'' in the sense that they are reduced to a single integral expression that can be evaluated in many cases using direct numerical methods.\n\nMoreover, when used in conjunction with our earlier work \\cite{Turyshev-Toth:2021-caustics} in which we obtained a closed form expression of the SGL PSF monopole and quadrupole contribution (ignoring higher-order zonal harmonics that contribute little) the new formalism allows us to compute the light field of the SGL or the view seen by a model telescope without resorting to numerical integration, and thus not hindered by the properties of rapidly oscillating integrals.\n\nWe put these results to use, in particular, by creating a series of multiframe animations that show the view of a pont source through a telescope that is approaching the SGL optical axis from afar. The strength of our formalism is powerfully demonstrated when we consider that the same expression can model the (essentially unamplified) view of a distant object when the telescope is still far from the SGL optical axis; the emergence of a secondary image from behind the solar limb; and the eventual widening of these images into arcs and their transition to form an Einstein cross around the Sun. We can also simulate light from extended objects, showing how, even in the presence of multipole moments, such objects still form an Einstein ring around the Sun.\n\nFinally, we note that although our focus remains the SGL that can be represented elegantly using zonal harmonics, our approach can be readily extended to other gravitational lenses that can be represented using symmetric trace-free (STF) tensors \\cite{Turyshev-Toth:2021-multipoles}. The resulting formalism covers every gravitational lens that can be described by small deviations from the spherically symmetric gravitational field of a mass monopole. Our approach, therefore, is the most comprehensive wave-theoretical treatment of gravitational lensing in a weak gravitational field to date.\n\nConcluding, we emphasize that the analytical expressions derived in this paper are presented in terms of physically observable quantities and, as such, they are directly suitable for realistic data analysis. To that extent, we can use them to process, e.g., time series brightness data available from the OGLE\\footnote{\\url{https:\/\/en.wikipedia.org\/wiki\/Optical_Gravitational_Lensing_Experiment}} and MACHO\\footnote{\\url{https:\/\/en.wikipedia.org\/wiki\/MACHO_Project}}\nprojects, the upcoming Roman Space Telescope\\footnote{\\url{https:\/\/roman.gsfc.nasa.gov\/}}, or other microlensing projects that may benefit from the improved modeling. In addition, the results presented in this paper offer a solution for establishing a local reference frame that can be used to achieve the required navigational precision for future missions to the SGL's focal region for high-resolution exoplanet imaging \\cite{Turyshev-etal:2020-PhaseII}. The corresponding efforts are under way; results, when available, will be published elsewhere.\n\n\\begin{acknowledgments}\nThis work in part was performed at the Jet Propulsion Laboratory, California Institute of Technology, under a contract with the National Aeronautics and Space Administration.\nVTT acknowledges the generous support of Plamen Vasilev and other Patreon patrons.\n\\end{acknowledgments}\n\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section*{Introduction}\n\nAlgebraic classification (up to isomorphism) of algebras of small dimension from a certain variety defined by a family of polynomial identities is a classic problem in the theory of non-associative algebras. There are many results related to algebraic classification of small dimensional algebras in varieties of Jordan, Lie, Leibniz, Zinbiel and other algebras.\nAnother interesting approach of studying algebras of a fixed dimension is to study them from a geometric point of view (that is, to study degenerations and deformations of these algebras). The results in which the complete information about degenerations of a certain variety is obtained are generally referred to as the geometric classification of the algebras of these variety. There are many results related to geometric classification of Jordan, Lie, Leibniz, Zinbiel and other algebras \\cite{ack, kv17,ikv17, kv16}.\n\nIn 1972, Kantor introduced the notion a conservative algebra as a generalization of Jordan algebras \\cite{Kantor72}. Unlike other classes of non-associative algebras, this class is not defined by a set of identities. To introduce the notion of a conservative algebra, we need some notation. \n Let $\\mathbb V$ be a vector space, let $A$ be a linear operator on $\\mathbb V,$ and let $B$ and $C$ be bilinear operators on $\\mathbb V.$ For all $x,y,z \\in \\mathbb V,$ put\n\\begin{gather*}\n[B,x](y)=B(x,y),\\\\\n[A,B](x,y)= A(B(x,y))- B(A(x),y)-B(x,A(y)),\\\\\n[B,C](x,y,z)=B(C(x,y),z) + B(x,C(y,z))+ B(y,C(x,z))\\\\\n-C(B(x,y),z)- C(x,B(y,z))-C(y,B(x,z)).\n\\end{gather*}\n \nConsider an algebra as a vector space $\\mathbb V$ over a field $\\mathbb{C}$, together with an\n element $\\mu$ of $\\operatorname{Hom}(\\mathbb V \\otimes \\mathbb V, \\mathbb V),$ so that $a \\cdot b =\\mu(a \\otimes b).$ \nFor an algebra $(\\mathbb V, \\mathcal P)$ with a multiplication $\\mathcal P$ and $x\\in \\mathbb V$ we denote by $L_x^{\\mathcal P}$ the operator of left multiplication by $x.$\nThus, Kantor defines conservative algebras as follows:\n\n\\begin{definition}\nAn algebra $(\\mathbb V, \\mathcal P),$ where $\\mathbb V$ is the vector space and $\\mathcal P$ is the multiplication, is called a conservative algebra if there is a new multiplication $\\mathcal P^*: \\mathbb V\\times \\mathbb V\\rightarrow \\mathbb V$ such that \n\\begin{equation}\\label{uno}\n[L_b^{\\mathcal P},[L_a^{\\mathcal P}, {\\mathcal P}]]=-[L_{{\\mathcal P^*}(a,b)}^{\\mathcal P},{\\mathcal P}] \n\\textrm{, for all $a, b \\in \\mathbb V.$}\n\\end{equation}\nSimple calculations take us to the following identity with an additional multiplication $\\mathcal P^*,$ which must hold for all $a, b, x, y\\in \\mathbb V$:\n\\begin{equation}\\label{dos}\n\\begin{split}\nb(a(xy)-(ax)y-x(ay)) - a((bx)y) + (a(bx))y+(bx)(ay) -a(x(by))+(ax)(by)+x(a(by))= \\\\\n= - \\mathcal P^*(a,b)(xy)+(\\mathcal P^*(a,b)x)y + x(\\mathcal P^*(a,b)y).\n\\end{split}\n\\end{equation}\n\\end{definition}\n\nThe class of conservative algebras is very vast. It includes\n all associative algebras, \n all quasi-associative algebras,\n all Jordan algebras, \n all Lie algebras, \n all (left) Leibniz algebras,\n all (left) Zinbiel algebras, \n and many other classes of algebras.\nOn the other side, all conservative algebras are \"rigid\" (by Cantarini and Kac) algebras \\cite{kacan}.\n\n\nHowever, this class is very hard to study and for now even the basic general questions about it remain unanswered, so it is a good idea to study its subclasses which are sufficiently wide but easier to deal with. In \\cite{Kantor89} Kantor, studying the generalized TKK functor, introduced the class of terminal algebras which is a subclass of the class of conservative algebras. \n\n\\begin{definition}\nAn algebra $(\\mathbb V, \\mathcal P),$ where $\\mathbb V$ is a vector space and $\\mathcal P$ is a multiplication, is called a terminal algebra if for all $a\\in \\mathbb V$ we have\n\\begin{equation}\\label{tress}\n[[[{\\mathcal P},a],{\\mathcal P}],{\\mathcal P}]=0.\n\\end{equation}\n\n\\end{definition}\n\nNote that we can expand the relation (\\ref{tress}), obtaining an identity of degree 4. Therefore, the class of terminal algebras is a variety.\nAlso, about terminal and conservative algebras see, for example \\cite{Kantor90, KPP18, cfk19}.\n \nThe following remark is obtained by straightforward calculations.\n\n\\begin{remark}\nA commutative algebra satisfying (\\ref{tress}) is a Jordan algebra. \n\\end{remark}\n\nAside from Jordan algebras, the class of terminal algebras includes all Lie algebras, all (left) Leibniz algebras and some other types of algebras.\n\nThe following characterization of terminal algebras, proved by Kantor \\cite[Theorem 2]{Kantor89}, provides a description of this class as a subclass of the class of conservative algebras.\n\n\\begin{remark}\nAn algebra $(\\mathbb{V}, {\\bf P})$ is terminal if and only if it is conservative\n and the multiplication in the associated superalgebra ${\\bf P}^*$ can be defined by\n \\begin{equation}\n\\label{terminal_associated} \n {\\bf P}^*(x,y)=\\frac{2}{3}{\\bf P}(x,y)+\\frac{1}{3}{\\bf P}(y,x).\n \\end{equation}\n\\end{remark}\n\n\nOur method of classification of nilpotent terminal algebras is based on the calculation of central extensions of smaller nilpotent algebras from the same variety. The algebraic study of central extensions of Lie and non-Lie algebras has a very big story \\cite{omirov,hac16,ss78}.\nSkjelbred and Sund used central extensions of Lie algebras for a classification of nilpotent Lie algebras \\cite{ss78}. After that, using the method described by Skjelbred and Sund were described all non-Lie central extensions of all $4$-dimensional Malcev algebras \\cite{hac16}, all non-associative central extensions of $3$-dimensional Jordan algebras, all anticommutative central extensions of $3$-dimensional anticommutative algebras,\nall central extensions of $2$-dimensional algebras \\cite{cfk18}.\n\nThe algebraic classification of nilpotent algebras will be achieved by the calculation of central extensions of algebras from the same variety which have a smaller dimension.\nCentral extensions of algebras from various varieties were studied, for example, in \\cite{ss78,omirov}.\nSkjelbred and Sund \\cite{ss78} used central extensions of Lie algebras to classify nilpotent Lie algebras.\nUsing the same method, \nall non-Lie central extensions of all $4$-dimensional Malcev algebras \\cite{hac16},\nall non-associative central extensions of all $3$-dimensional Jordan algebras,\nall anticommutative central extensions of $3$-dimensional anticommutative algebras,\nall central extensions of $2$-dimensional algebras \\cite{cfk18}\nand some others were described.\nOne can also look at the classification of\n$3$-dimensional nilpotent algebras \\cite{fkkv},\n$4$-dimensional nilpotent associative algebras \\cite{degr1},\n$4$-dimensional nilpotent Novikov algebras,\n$4$-dimensional nilpotent bicommutative algebras,\n$4$-dimensional nilpotent commutative algebras in \\cite{fkkv},\n$5$-dimensional nilpotent restricted Lie agebras \\cite{usefi1},\n$5$-dimensional nilpotent Jordan algebras,\n$5$-dimensional nilpotent anticommutative algebras \\cite{fkkv},\n$6$-dimensional nilpotent Lie algebras \\cite{degr3, degr2},\n$6$-dimensional nilpotent Malcev algebras \\cite{hac18},\n$6$-dimensional nilpotent Tortkara algebras,\n$6$-dimensional nilpotent binary Lie algebras \\cite{ack}.\n\n\nIn the present paper, we classify nilpotent terminal algebras of dimension less than or equal to 4, and obtain the complete description of degenerations of these algebras.\n\n\n\\section{The algebraic classification of nilpotent terminal algebras}\n\\subsection{Method of classification of nilpotent algebras}\nThroughout this paper, we use the notation and methods described in \\cite{hac16,cfk18}\nand adapted for the terminal case with some modifications (see also \\cite{Jac} for a discussion of extensions of algebras in an arbitrary nonassociative variety). Therefore, all statements in this subsection are given without proofs, which can be found in the papers cited above.\n\nLet ${\\bf A}$ be a terminal algebra over $\\mathbb C$ and $\\mathbb V$ a vector space of dimension $s$ over the same base field. Then the space ${\\rm Z_T^2}(\\bf A,\\mathbb V )$ is defined as the set of all maps $\\theta :{\\bf A} \\times {\\bf A} \\to {\\mathbb V}$ such that \n\\[\\theta(b,a(xy) - (ax)y - x(ay)) - \\theta(a,(bx)y) + \\theta(a(bx),y) + \\theta(bx,ay)\\]\n\\[- \\theta(a,x(by)) + \\theta(ax,(by)) + \\theta(x,(ab)y) = -\\theta({\\bf P}^*(a,b),xy) + \\theta({\\bf P}^*(a,b)x,y) + \\theta(x,{\\bf P}^*(a,b)y),\\]\nwhere ${\\bf P}^*$ is given by (\\ref{terminal_associated}). Its elements will be called \\textit{cocycles}. For a linear map $f: \\bf A \\to \\mathbb V$ define $\\delta f: {\\bf A} \\times\n{\\bf A} \\to {\\mathbb V}$ by $\\delta f (x,y ) =f(xy).$ One can check that $\\delta f\\in {\\rm Z_T^2}({\\bf A},{\\mathbb V} ).$ Therefore, ${\\rm B^2}({\\bf A},{\\mathbb V}) =\\left\\{ \\theta =\\delta f\\ :f\\in \\operatorname{Hom}({\\bf A},{\\mathbb V}) \\right\\}$ is a subspace of ${\\rm Z_T^2}({\\bf A},{\\mathbb V})$ whose elements are called \\textit{coboundaries}. We define the \\textit{second cohomology space} ${\\rm H_T^2}({\\bf A},{\\mathbb V})$ as the quotient space ${\\rm Z_T^2}({\\bf A},{\\mathbb V}) \\big\/{\\rm B^2}({\\bf A},{\\mathbb V}).$\nThe equivalence class of $\\theta \\in {\\rm Z^2}({\\bf A},{\\mathbb V})$ in ${\\rm H^2}({\\bf A},{\\mathbb V})$ will be denoted by $[\\theta].$\n\n\\bigskip\n\nFor a bilinear map $\\theta :{\\bf A} \\times {\\bf A} \\to {\\mathbb V}$ define on the linear space ${\\bf A}_{\\theta }:={\\bf A}\\oplus {\\mathbb V}$ a bilinear product $[-,-]_{{\\theta}}$ by \n\\[[x+x^{\\prime },y+y^{\\prime }]_{{\\theta}}= xy +\\theta (x,y) \\mbox{ for all } x,y\\in {\\bf A},x^{\\prime },y^{\\prime }\\in {\\mathbb V}.\\]\nThen ${\\bf A}_{\\theta }$ is an algebra called an $s${\\it{-dimensional central extension}} of ${\\bf A}$ by ${\\mathbb V}.$ The following statement can be verified directly:\n\n\\begin{lemma}\nThe algebra ${\\bf A_{\\theta}}$ is terminal \\textit{if and only if} $\\theta \\in {\\rm Z_T^2}({\\bf A}, {\\mathbb V}).$\n\\end{lemma}\n\nRecall that the {\\it{annihilator}} of ${\\bf A}$ is defined as the ideal $\\operatorname{Ann}({\\bf A} ) =\\{ x\\in {\\bf A}: x{\\bf A}+{\\bf A}x=0\\}.$ Given $\\theta \\in {\\rm Z_T^2}({\\bf A}, {\\mathbb V}),$ we call the set $\\operatorname{Ann}(\\theta)=\\left\\{ x\\in {\\bf A}:\\theta (x, {\\bf A} )+\\theta ({\\bf A},x ) = 0 \\right\\}$ the {\\it{annihilator}} of $\\theta.$ \n\n\\begin{lemma}\n\\label{ann_of_ext}\n$\\operatorname{Ann}({\\bf A}_{\\theta }) = (\\operatorname{Ann}(\\theta) \\cap \\operatorname{Ann}({\\bf A}))\n \\oplus {\\mathbb V}.$\n\\end{lemma}\n\n\\bigskip\n\nTherefore, $0 \\neq \\mathbb{V} \\subseteq \\operatorname{Ann}({\\bf A}_{\\theta}).$ The following lemma shows that every algebra with a nonzero annihilator can be obtained in the way described above:\n\n\\begin{lemma}\nLet ${\\bf A}$ be an $n$-dimensional terminal algebra such that $\\dim(\\operatorname{Ann}({\\bf A}))=m\\neq0.$ Then there exists, up to isomorphism, a unique $(n-m)$-dimensional terminal algebra ${\\bf A}^{\\prime }$ and a bilinear map $\\theta \\in {\\rm Z_T^2}({\\bf A}, {\\mathbb V})$ with $\\operatorname{Ann}({\\bf A}')\\cap \\operatorname{Ann}(\\theta)=0,$ where $\\mathbb V$ is a vector space of dimension $m,$ such that ${\\bf A}\\cong {\\bf A}^{\\prime }_{\\theta}$ and ${\\bf A}\/\\operatorname{Ann}({\\bf A})\\cong {\\bf A}^{\\prime }.$\n\\end{lemma}\n\nIn particular, any finite-dimensional nilpotent algebra is a central extension of another nilpotent algebra of strictly smaller dimension. Thus, to classify all nilpotent terminal algebras of a fixed dimension, we need to classify cocycles of nilpotent terminal algebras ${\\bf A}'$ of smaller dimension (with an additional condition $\\operatorname{Ann}({\\bf A}')\\cap \\operatorname{Ann}(\\theta)=0$) and central extensions that arise from them.\n\n\\bigskip\n\nWe can reduce the class of extensions that we need to consider.\n\n\n\\begin{definition}\nLet ${\\bf A}$ be an algebra and $I$ be a subspace of ${\\rm Ann}({\\bf A})$. If ${\\bf A}={\\bf A}_0 \\oplus I$\nthen $I$ is called an {\\it annihilator component} of ${\\bf A}$.\n A central extension of an algebra $\\bf A$ without an annihilator component is called a non-split central extension.\n\\end{definition} \n\nClearly, we are only interested in non-split extensions (in the contrary case we can cut off annihilator components lying in $\\mathbb{V}$ until we obtain a non-split extension). \n\nLet us fix a basis $e_{1},\\ldots,e_{s}$ of ${\\mathbb V},$ and $\\theta \\in {\\rm Z_T^2}({\\bf A},{\\mathbb V}).$ Then $\\theta$ can be uniquely written as $\\theta (x,y) =\\theta (x,y) =\\sum_{i=1}^s\\theta_{i}(x,y) e_{i},$ where $\\theta_{i}\\in {\\rm Z_T^2}({\\bf A},\\mathbb C).$ Moreover, $\\operatorname{Ann}(\\theta)=\\operatorname{Ann}(\\theta_{1})\\cap \\operatorname{Ann}(\\theta_{2})\\cap\\cdots \\cap \\operatorname{Ann}(\\theta_{s}).$ Further, $\\theta \\in {\\rm B^2}({\\bf A},{\\mathbb V}) $\\ if and only if all $\\theta_{i}\\in {\\rm B^2}({\\bf A},\\mathbb C).$ Using this presentation, one can determine whether the extension corresponding to a cocycle $\\theta$ is split:\n\n\\begin{lemma} \\cite[Lemma 13]{hac16}\n\\label{split_lindep}\nLet $\\theta (x,y) =\\sum_{i=1}^s\\theta_{i}(x,y) e_{i}\\in {\\rm Z_T^2}({\\bf A},{\\mathbb V})$ be such that $\\operatorname{Ann}(\\theta)\\cap \\operatorname{Ann}({\\bf A}) = 0.$ Then ${\\bf A}_{\\theta }$ has an annihilator component if and only if $[\\theta_{1}],[\\theta_{2}],\\ldots,[\\theta_{s}]$ are linearly dependent in ${\\rm H_T^2}({\\bf A},\\mathbb C).$\n\\end{lemma} \n\n\\bigskip\n\nSome cocycles give rise to isomorphic extensions:\n\n\\begin{lemma}\nLet $\\theta, \\vartheta \\in {\\rm Z_T^2}({\\bf A}, {\\mathbb V})$ be such that $[\\theta] = [\\vartheta].$ Then ${\\bf A}_\\theta \\cong {\\bf A}_\\vartheta.$ \n\\end{lemma}\n\n\n\nBy above, the isomorphism classes of extensions correspond to certain equivalence classes on ${\\rm H_T^2}({\\bf A},{\\mathbb V}).$ These classes can be given in terms of actions of certain groups on this space. In particular, let $\\operatorname{Aut}({\\bf A})$ be the automorphism group of ${\\bf A},$ let $\\phi \\in \\operatorname{Aut}({\\bf A}),$ and let $\\psi \\in \\operatorname{GL}(\\mathbb V).$ For $\\theta \\in {\\rm Z_T^2}({\\bf A},{\\mathbb V})$ define \n\\[\\phi \\theta(x,y) =\\theta (\\phi(x), \\phi(y)), ~ \\psi\\theta (x,y) = \\psi(\\theta(x,y)).\\] \nThen $\\phi\\theta, \\psi\\theta \\in {\\rm Z_T^2}({\\bf A},{\\mathbb V}).$ Hence, $\\operatorname{Aut}({\\bf A})$ and $\\operatorname{GL}(\\mathbb V)$ act on ${\\rm Z_T^2}({\\bf A},{\\mathbb V}).$ It is easy to verify that ${\\rm B^2}({\\bf A},{\\mathbb V})$ is invariant under both actions. Therefore, we have induced actions on ${\\rm H_T^2}({\\bf A},{\\mathbb V}).$\n\n\\begin{lemma}\nLet $\\theta, \\vartheta \\in {\\rm Z_T^2}({\\bf A},{\\mathbb V})$ be such that $\\operatorname{Ann}({\\bf A}_\\theta) = \\operatorname{Ann}({\\bf A}_\\vartheta) = \\mathbb{V}.$ Then ${\\bf A}_\\theta \\cong {\\bf A}_\\vartheta$ if and only if there exist a $\\phi \\in \\operatorname{Aut}({\\bf A}), \\psi \\in \\operatorname{GL}(\\mathbb V)$ such that $[\\phi\\theta] = [\\psi\\vartheta].$\n\\end{lemma}\n\n\\bigskip\n\nNow we rewrite the above lemma in a form more suitable for computations.\n\nLet ${\\mathbb U}$ be a finite-dimensional vector space over $\\mathbb C.$ The {\\it{Grassmannian}} $G_{k}({\\mathbb U})$ is the set of all $k$-dimensional linear subspaces of ${\\mathbb V}.$ Let $G_{s}({\\rm H_T^2}({\\bf A},\\mathbb C) )$ be the Grassmannian of subspaces of dimension $s$ in ${\\rm H_T^2}({\\bf A},\\mathbb C).$ There is a natural action of $\\operatorname{Aut} ({\\bf A})$ on $G_{s}({\\rm H_T^2}({\\bf A},\\mathbb C)):$ for $\\phi \\in \\operatorname{Aut} ({\\bf A}), W=\\langle [\\theta_{1}],[\\theta_{2}],\\dots,[\\theta_{s}] \\rangle \\in G_{s}({\\rm H_T^2}({\\bf A},\\mathbb C)),$ define \n\\[\\phi W=\\langle [\\phi \\theta_{1}], [\\phi \\theta_{2}],\\dots,[\\phi \\theta_{s}]\\rangle.\\]\nNote that this action is compatible with the action of $\\operatorname{Aut} ({\\bf A})$ on ${\\rm H_T^2}({\\bf A},{\\mathbb V})$ and the above presentation of a cocycle as a collection of $s$ elements of ${\\rm H_T^2}({\\bf A},\\mathbb C).$ Denote the orbit of $W$ under the action of $\\operatorname{Aut} ({\\bf A})$ by $\\mathrm{Orb}(W).$ It is easy to check that given two bases of a subspace \n\\begin{equation*}\nW=\\langle [\\theta_{1}],[\\theta_{2}],\\dots,[\\theta_{s}] \\rangle =\\langle [\\vartheta_{1}],[\\vartheta_{2}],\\dots,[\\vartheta_{s}]\n\\rangle \\in G_{s}({\\rm H^2}({\\bf A},\\mathbb C)),\n\\end{equation*}\nwe have $\\cap_{i=1}^s\\operatorname{Ann}(\\theta_{i})\\cap \\operatorname{Ann}({\\bf A}) =\\cap_{i=1}^s\\operatorname{Ann}(\\vartheta_{i})\\cap \\operatorname{Ann}({\\bf A}).$ Therefore, we can introduce the set\n\n\\begin{equation*}\nT_{s}({\\bf A}) =\\left\\{ W=\\left\\langle [\\theta_{1}],%\n[\\theta_{2}],\\dots,[\\theta_{s}] \\right\\rangle \\in\nG_{s}({\\rm H^2}({\\bf A},\\mathbb C) ) : \\cap_{i=1}^s \\operatorname{Ann}(\\theta_{i})\\cap \\operatorname{Ann}({\\bf A}) =0\\right\\},\n\\end{equation*}\nwhich is stable under the action of $\\operatorname{Aut}({\\bf A}).$\n\n\\medskip\n\nLet us denote by $E({\\bf A},{\\mathbb V})$ the set of all {\\it non-split} central extensions of ${\\bf A}$ by ${\\mathbb V}.$ By Lemmas \\ref{ann_of_ext} and \\ref{split_lindep}, we can write\n\\begin{equation*}\nE({\\bf A},{\\mathbb V}) =\\left\\{ {\\bf A}_{\\theta }:\\theta (x,y) =\\sum_{i=1}^s\\theta_{i}(x,y) e_{i}\\mbox{ and }\\langle [\\theta_{1}],[\\theta_{2}],\\dots,\n[\\theta_{s}] \\rangle \\in T_{s}({\\bf A})\\right\\}.\n\\end{equation*}\nAlso, we have the next result, which can be proved as \\cite[Lemma 17]{hac16}.\n\n\\begin{lemma}\n Let $\\theta(x,y) =\\sum_{i=1}^s \\theta_{i}(x,y) e_{i}$ and $\\vartheta(x,y) = \\sum_{i=1}^s\\vartheta_{i}(x,y) e_{i}$ be such that ${\\bf A}_{\\theta},{\\bf A}_{\\vartheta }\\in E({\\bf A},{\\mathbb V}).$ Then the algebras ${\\bf A}_{\\theta }$ and ${\\bf A}_{\\vartheta }$ are isomorphic if and only if \n\\[\\mathrm{Orb}\\langle [\\theta_{1}], [\\theta_{2}],\\dots,[\\theta_{s}] \\rangle = \\mathrm{Orb}\\langle [\\vartheta_{1}],[\\vartheta\n_{2}],\\dots,[\\vartheta_{s}] \\rangle. \\]\n\\end{lemma}\n\nThus, there exists a one-to-one correspondence between the set of $\\operatorname{Aut}({\\bf A}) $-orbits on $T_{s}({\\bf A})$ and the set of isomorphism classes of $E({\\bf A},{\\mathbb V}).$ Consequently, we have a procedure that allows us, given a terminal algebra ${\\bf A}^{\\prime }$ of dimension $n,$ to construct all non-split central extensions of ${\\bf A}^{\\prime }.$ This procedure is as follows:\n\n\\medskip\n\n{\\centerline{\\it Procedure}}\n\n\\begin{enumerate}\n\\item For a given (nilpotent) terminal algebra $\\bf{A}^{\\prime }$\nof dimension $n-s,$ determine ${T}_{s}(\\bf{A}^{\\prime })$\nand $\\operatorname{Aut}(\\bf{A}^{\\prime }).$\n\n\\item Determine the set of $\\operatorname{Aut}(\\bf{A}^{\\prime })$-orbits on $%\n{T}_{s}(\\bf{A}^{\\prime }).$\n\n\\item For each orbit, construct the terminal algebra corresponding to one of its\nrepresentatives.\n\\end{enumerate}\n\nThe above described method gives all (Leibniz and non-Leibniz) terminal algebras. But we are interested in developing this method in such a way that it only gives non-Leibniz terminal algebras, because the classification of all Leibniz algebras is given in \\cite{demir}. Clearly, any central extension of a non-Leibniz terminal algebra is non-Leibniz. But a Leibniz algebra may have extensions which are not Leibniz algebras. More precisely, let ${\\bf L}$ be a Leibniz algebra and $\\theta \\in {\\rm Z_T^2}({\\bf L}, {\\mathbb C}).$ Then ${\\bf L}_{\\theta }$ is a Leibniz algebra if and only if \n\\begin{equation*}\n\\theta ( x, yz ) = \\theta (xy, z )+ \\theta (y, xz ) \n\\end{equation*}\nfor all $x,y,z\\in {\\bf L}.$ Define the subspace ${\\rm Z_L^2}({\\bf L},{\\mathbb C})$ of ${\\rm Z_T^2}({\\bf L},{\\mathbb C})$ by\n\\begin{equation*}\n{\\rm Z_L^2}({\\bf L},{\\mathbb C}) =\\left\\{\\begin{array}{c} \\theta \\in {\\rm Z_T^2}({\\bf L},{\\mathbb C}) : \\theta ( x, yz ) = \\theta (xy, z ) + \\theta (y, xz ) \\text{ for all } x, y,z\\in {\\bf L}\\end{array}\\right\\}.\n\\end{equation*}\nObserve that ${\\rm B^2}({\\bf L},{\\mathbb C})\\subseteq{\\rm Z_L^2}({\\bf L},{\\mathbb C}).$\nLet ${\\rm H_L^2}({\\bf L},{\\mathbb C}) =%\n{\\rm Z_L^2}({\\bf L},{\\mathbb C}) \\big\/{\\rm B^2}({\\bf L},{\\mathbb C}).$ Then ${\\rm H_L^2}({\\bf L},{\\mathbb C})$ is a subspace of $%\n{\\rm H_T^2}({\\bf L},{\\mathbb C}).$ Define \n\\begin{eqnarray*}\n{\\bf R}_{s}({\\bf L}) &=&\\left\\{ {\\bf W}\\in {T}_{s}({\\bf L}) :{\\bf W}\\in G_{s}({\\rm H_L^2}({\\bf L},{\\mathbb C}) ) \\right\\}, \\\\\n{\\bf U}_{s}({\\bf L}) &=&\\left\\{ {\\bf W}\\in {T}_{s}({\\bf L}) :{\\bf W}\\notin G_{s}({\\rm H_L^2}({\\bf L},{\\mathbb C}) ) \\right\\}.\n\\end{eqnarray*}\nThen ${T}_{s}({\\bf L}) ={\\bf R}_{s}(\n{\\bf L})$ $\\mathbin{\\mathaccent\\cdot\\cup}$ ${\\bf U}_{s}(\n{\\bf L}).$ The sets ${\\bf R}_{s}({\\bf L}) $\nand ${\\bf U}_{s}({\\bf L})$ are stable under the action\nof $\\operatorname{Aut}({\\bf L}).$ Thus, the terminal algebras\ncorresponding to the representatives of $\\operatorname{Aut}({\\bf L}) $%\n-orbits on ${\\bf R}_{s}({\\bf L})$ are Leibniz algebras,\nwhile those corresponding to the representatives of $\\operatorname{Aut}({\\bf L}%\n) $-orbits on ${\\bf U}_{s}({\\bf L})$ are not\nLeibniz algebras. Hence, we may construct all non-split non-Leibniz terminal algebras $%\n\\bf{A}$ of dimension $n$ with $s$-dimensional annihilator \nfrom a given terminal algebra $\\bf{A}%\n^{\\prime }$ of dimension $n-s$ in the following way:\n\n\\begin{enumerate}\n\\item If $\\bf{A}^{\\prime }$ is non-Leibniz, then apply the Procedure.\n\n\\item Otherwise, do the following:\n\n\\begin{enumerate}\n\\item Determine ${\\bf U}_{s}(\\bf{A}^{\\prime })$ and $%\n\\operatorname{Aut}(\\bf{A}^{\\prime }).$\n\n\\item Determine the set of $\\operatorname{Aut}(\\bf{A}^{\\prime })$-orbits on ${\\bf U%\n}_{s}(\\bf{A}^{\\prime }).$\n\n\\item For each orbit, construct the terminal algebra corresponding to one of its\nrepresentatives.\n\\end{enumerate}\n\\end{enumerate}\n\n\n\n\\medskip\n\n\\subsection{Notations}\nLet us introduce the following notations. Let ${\\bf A}$ be a terminal algebra with\na basis $e_{1},e_{2}, \\ldots, e_{n}.$ Then by $\\Delta_{ij}$\\ we will denote the\nbilinear form\n$\\Delta_{ij}:{\\bf A}\\times {\\bf A}\\longrightarrow \\mathbb C$\nwith $\\Delta_{ij}(e_{l},e_{m}) = \\delta_{il}\\delta_{jm}.$\nThe set $\\left\\{ \\Delta_{ij}:1\\leq i, j\\leq n\\right\\}$ is a basis for the linear space of \nbilinear forms on ${\\bf A},$ so every $\\theta \\in\n{\\rm Z^2}({\\bf A},\\bf \\mathbb V )$ can be uniquely written as $%\n\\theta ={\\sum }_{i,j=1}^nc_{ij}\\Delta_{ij},$ where $%\nc_{ij}\\in \\mathbb C.$\nWe also denote by\n\n\n$$\\begin{array}{lll}\n\\T {i*}{j}& \\mbox{the }j\\mbox{th }i\\mbox{-dimensional nilpotent Leibniz algebra}, \\\\\n\\T i{j}& \\mbox{the }j\\mbox{th }i\\mbox{-dimensional nilpotent non-Leibniz terminal algebra}, \\\\\n{\\mathfrak{N}}_i& \\mbox{the }i\\mbox{-dimensional algebra with zero product}, \\\\\n({\\bf A})_{i,j} & j\\mbox{th }i\\mbox{-dimensional central extension of }\\bf A. \\\\\n\\end{array}$$\n\n\nAlso, it is easy to see that every central extension of $\\mathfrak{N}_i$ is a Leibniz algebra.\n\n\n\n\\subsection{The algebraic classification of 3-dimensional nilpotent terminal algebras}\n\nObserve that a 2-dimensional nilpotent algebra is either $\\mathfrak{N}_2,$ or is isomorphic to \n\\begin{align*\n \\begin{array}{llllll}\n \\T {2*}{01} &:& e_1 e_1 = e_2. \n \\end{array}\n\\end{align*}\nBoth of these algebras are nilpotent of index less than 4, and hence are terminal. All central extensions of 2-dimensional nilpotent algebras were described in~\\cite{cfk18}. By a direct verification, we have the following list of all 3-dimensional nilpotent terminal algebras with annihilator of codimension 1 or 2\n\\begin{align}\\label{3-dim-term}\n \\begin{array}{lllllllll}\n \\T {3*}{01}&:&\\T {2*}{01}\\oplus\\mathfrak{N}_1 &:& e_1 e_1 = e_2; \\\\\n \\T {3*}{02}&:&(\\mathfrak{N}_2)_{3,1} &:& e_1 e_1 = e_3, & e_2 e_2=e_3; \\\\\n \\T {3*}{03}&:& (\\mathfrak{N}_2)_{3,2} &:& e_1 e_2=e_3, & e_2 e_1=-e_3; \\\\\n \\T {3*}{04}(\\lambda)&:&(\\mathfrak{N}_2)_{3,3} &:& e_1 e_1 = \\lambda e_3, & e_2 e_1=e_3, & e_2 e_2=e_3; \\\\\n \\T {3*}{05}&:&(\\T {2*}{01})_{3,1} &:& e_1 e_1 = e_2, & e_1 e_2=e_3; \\\\\n \\T 3{01}(\\lambda)&:&(\\T {2*}{01})_{3,2} &:& e_1 e_1 = e_2, & e_1 e_2=\\lambda e_3, & e_2 e_1=e_3. \n \\end{array}\n\\end{align}\nNotice that only $\\T 3{01}(\\lambda)$ from list \\cref{3-dim-term} is non-Leibniz. \n\n\\subsection{The algebraic classification of $4$-dimensional nilpotent terminal algebras}\n\n\nAnalyzing the list of all the 4-dimensional nilpotent algebras with annihilator of codimension 2, we have only one terminal non-Leibniz algebra:\n$$\n\\begin{array}{lllllllll}\n \n \n \n \n \n \n \\T 4{02}&:&(\\T {2*}{01} )_{4,1} &:& e_1 e_1 = e_2, & e_1 e_2=e_4, & e_2 e_1=e_3.\n\\end{array}\n$$\nTo complete the algebraic classification of 4-dimensional nilpotent terminal algebras, we need to describe all the 1-dimensional terminal non-Leibniz extensions of the algebras from the table \\cref{3-dim-term}. This is done in Subsections~\\ref{aut-and-H^2}--\\ref{ext-T_05^3*} and summarized in the main theorem of the first part of the paper.\n\n\\begin{theorem}\\label{main-alg}\nLet $\\mathbf A$ be a 4-dimensional nilpotent non-Leibniz terminal algebra over $\\mathbb C$. Then $\\mathbf A$ is isomorphic to one of the algebras $\\T 4{01}-\\T 4{44}$ or $\\D 4{01}-\\D 4{40}$ found below.\n\\end{theorem}\n\\begin{Proof}\n The proof is split into several steps presented in Subsections~\\ref{aut-and-H^2}--\\ref{ext-T_05^3*} below.\n\\end{Proof}\n\n\\subsubsection{Automorphism and cohomology groups of $3$-dimensional nilpotent terminal algebras}\\label{aut-and-H^2}\n\n\\[\n\\tiny\n\\begin{tabular}{|c|c|c|c|}\n\\hline \n$\\mathbf{A}$ & $\\aut\\mathbf{A}$ & ${\\rm Z_T^2}(\\mathbf{A})$ & ${\\rm H_T^2}(\\mathbf{A})$\\\\\n\\hline\n\n$\\T {3*}{01}$ \n&\n$\\begin{pmatrix}\nx & 0 & 0\\\\\ny & x^2 & u\\\\\nz & 0 & v\n\\end{pmatrix}$ \n& \n$\\Big\\langle\n\\begin{array}{l}\n \\Dt 11, \\Dt 12, \\Dt 13, \\Dt 21,\\\\\n \\Dt 23, \\Dt 31, \\Dt 32, \\Dt 33\n\\end{array}\n\\Big\\rangle$ \n&\n$\\Big\\langle\n\\begin{array}{l}\n \\Dl 12, \\Dl 13, \\Dl 21, \\Dl 23\\\\\n \\Dl 31, \\Dl 32, \\Dl 33\n\\end{array}\n\\Big\\rangle$\n\\\\\n\\hline\n\n$\\T {3*}{02}$ \n&\n$\\begin{pmatrix}\nx & y & 0\\\\\n(-1)^{n+1} y & (-1)^n x & 0\\\\\nz & u & x^2+y^2\n\\end{pmatrix}$ \n&\n$\\Big\\langle\n\\begin{array}{l}\n \\Dt 11, \\Dt 12, \\Dt 13, \\Dt 21\\\\\n \\Dt 22, \\Dt 23, \\Dt 31, \\Dt 32\n\\end{array}\n\\Big\\rangle$ \n& \n$\\Big\\langle\n\\begin{array}{l}\n \\Dl 11, \\Dl 12, \\Dl 13, \\Dl 21\\\\\n \\Dl 23, \\Dl 31, \\Dl 32\n\\end{array}\n\\Big\\rangle$ \n\\\\\n\\hline\n\n$\\T {3*}{03}$ \n&\n$\\begin{pmatrix}\nx & y & 0\\\\\nz & u & 0\\\\\nv & w & xu-yz\n\\end{pmatrix}$ \n&\n$\\Big\\langle\n\\begin{array}{l}\n \\Dt 11, \\Dt 12, \\Dt 13, \\Dt 21\\\\\n \\Dt 22, \\Dt 23, \\Dt 31, \\Dt 32\n\\end{array}\n\\Big\\rangle$ \n&\n$\\Big\\langle\n\\begin{array}{l}\n \\Dl 11, \\Dl 12, \\Dl 13, \\Dl 22\\\\\n \\Dl 23, \\Dl 31, \\Dl 32\n\\end{array}\n\\Big\\rangle$\n\\\\\n\\hline\n\n$\\T {3*}{04}$ \n&\n $\\begin{pmatrix}\n x & y & 0\\\\\n -\\lambda y & x-y & 0\\\\\n z & u & x^2-xy+\\lambda y^2\n \\end{pmatrix}$\n&\n$\\Big\\langle\n\\begin{array}{l}\n \\Dt 11, \\Dt 12, \\Dt 13, \\Dt 21\\\\\n \\Dt 22, \\Dt 23, \\Dt 31, \\Dt 32\n\\end{array}\n\\Big\\rangle$ \n&\n$\\Big\\langle\n\\begin{array}{l}\n \\Dl 11, \\Dl 12, \\Dl 13, \\Dl 21\\\\\n \\Dl 23, \\Dl 31, \\Dl 32\n\\end{array}\n\\Big\\rangle$\n\\\\\n\\hline\n\n$\\T {3*}{05}$ \n&\n$\\begin{pmatrix}\nx & 0 & 0\\\\\ny & x^2 & 0\\\\\nz & xy & x^3\n\\end{pmatrix}$ \n& \n$\\Big\\langle\n\\begin{array}{l}\n \\Dt 11, \\Dt 12, \\Dt 13,\\\\\n \\Dt 21, \\Dt 22 - 3\\Dt 31\n\\end{array}\n\\Big\\rangle$ \n&\n$\\Big\\langle\n\\begin{array}{l}\n \\Dl 13, \\Dl 21,\\\\\n \\Dl 22 - 3\\Dl 31\n\\end{array}\n\\Big\\rangle$\n\\\\\n\\hline\n\n$\\T 3{01}$ \n&\n$\\begin{pmatrix}\nx & 0 & 0\\\\\ny & x^2 & 0\\\\\nz & (\\lambda+1)xy & x^3\n\\end{pmatrix}$ \n& \n$\\begin{cases}\n\\Big\\langle\n\\begin{array}{l}\n \\Dt 11, \\Dt 12, -\\Dt 13 + 3\\Dt 31,\\\\\n \\Dt 21, {\\color{black}\\Dt 13 + \\Dt 22}, \\Dt 23 \n\\end{array}\n\\Big\\rangle & \\lambda=0\\\\\n\\Big\\langle\n\\begin{array}{l}\n \\Dt 11, \\Dt 12, (\\lambda-1)\\Dt 13 + 3\\Dt 31,\\\\\n \\Dt 21, \\color{black}{\\Dt 13 + \\Dt 22}\n\\end{array}\n\\Big\\rangle & \\lambda\\ne 0\n\\end{cases}$\n&\n\n$\\begin{cases}\n\\Big\\langle\n\\begin{array}{l}\n \\Dl 12, -\\Dl 13 + 3\\Dl 31,\\\\\n {\\color{black}\\Dl 13 + \\Dl 22}, \\Dl 23 \n\\end{array}\n\\Big\\rangle & \\lambda=0\\\\\n\\Big\\langle\n\\begin{array}{l}\n \\Dl 12, (\\lambda-1)\\Dl 13 + 3\\Dl 31,\\\\\n \\color{black}{\\Dl 13 + \\Dl 22}\n\\end{array}\n\\Big\\rangle & \\lambda\\ne 0\n\\end{cases}$\n\\\\\n\\hline\n\n\\end{tabular}\n\\] \n\n\\vskip0.5cm\n\nSince the algebras $\\T {3*}{01}, \\T {3*}{02},\\T {3*}{03},\\T {3*}{04}$ and $\\T {3*}{05}$ are Leibniz, it is natural to find the relation between the Leibniz and terminal cohomologies of these algebras in order to exclude those cocycles which give Leibniz algebras. \n\n\\vskip0.5cm\n\n$$\\begin{tabular}{|c|c|c|}\n\t\\hline \n\t$\\mathbf{A}$ & ${\\rm H_L^2}(\\mathbf{A})$ & ${\\rm H_T^2}(\\mathbf{A})$\\\\\n\t\\hline\n\t\n\t$\\T {3*}{01}$ \n\t&\n\t$\\langle \\Dl 12, \\Dl 13, \\Dl 31, \\Dl 33 \\rangle$\n\t&\n\t${\\rm H_L^2}(\\mathbf{A})\\oplus \\langle \\Dl 21, \\Dl 23, \\Dl 32 \\rangle$\n\t\\\\\n\t\\hline\n\t\n\t$\\T {3*}{02}$ \n\t&\n\t$\\langle \\Dl 11, \\Dl 12, \\Dl 21 \\rangle$ \n\t& \n\t${\\rm H_L^2}(\\mathbf{A})\\oplus\\langle \\Dl 13, \\Dl 23, \\Dl 31, \\Dl 32\\rangle$ \n\t\\\\\n\t\\hline\n\t\n\t$\\T {3*}{03}$ \n\t&\n\t$\\Big\\langle\n\t\\begin{array}{l}\n\t\\Dl 11, \\Dl 12, \\Dl 13 - \\Dl 31,\\\\\n\t\\Dl 22, \\Dl 23 - \\Dl 32\n\t\\end{array}\n\t\\Big\\rangle$ \n\t&\n\t${\\rm H_L^2}(\\mathbf{A})\\oplus\\langle \\Dl 31, \\Dl 32 \\rangle$\n\t\\\\\n\t\\hline\n\t\n\t$\\T {3*}{04}$ \n\t&\n\t$\\begin{cases}\n\t\\langle \\Dl 11, \\Dl 12, \\Dl 21 \\rangle, & \\lambda\\ne 0\\\\\n\t\\langle\\Dl 11, \\Dl 12, \\Dl 21, \\Dl 23\\rangle, & \\lambda=0\n\t\\end{cases}$\n\t&\n\t$\n\t\\begin{cases}\n\t{\\rm H_L^2}(\\mathbf{A})\\oplus \\langle \\Dl 13, \\Dl 23, \\Dl 31, \\Dl 32 \\rangle, & \\lambda\\ne 0\\\\\n\t{\\rm H_L^2}(\\mathbf{A})\\oplus \\langle \\Dl 13, \\Dl 31, \\Dl 32 \\rangle, & \\lambda=0\n\t\\end{cases}$\n\t\\\\\n\t\\hline\n\t\n\t$\\T {3*}{05}$ \n\t&\n\t$\\langle \\Dl 13\\rangle$\n\t&\n\t${\\rm H_L^2}(\\mathbf{A})\\oplus \\langle \\Dl 21,\\Dl 22 - 3\\Dl 31 \\rangle$\n\t\\\\\n\t\\hline\n\t\n\t\\end{tabular}$$\n\n\n\\subsubsection{$1$-dimensional central extensions of $\\T {3*}{01}$}\n\nLet us use the following notations: \n\\begin{align*}\n \\nb 1 = \\Dl 12, \\nb 2 = \\Dl 13, \\nb 3 = \\Dl 31, \\nb 4 = \\Dl 33, \\nb 5 = \\Dl 21, \\nb 6 = \\Dl 23, \\nb 7 = \\Dl 32. \n\\end{align*}\nTake $\\theta=\\sum_{i=1}^7\\alpha_i\\nb i\\in {\\rm H_T^2}(\\T {3*}{01}).$\nIf \n$$\n\\phi=\n\\begin{pmatrix}\nx & 0 & 0\\\\\ny & x^2 & u\\\\\nz & 0 & v\n\\end{pmatrix}\\in\\aut{\\T {3*}{01}},\n$$\nthen\n$$\n\\phi^T\\begin{pmatrix}\n0& \\alpha_1& \\alpha_2\\\\\n\\alpha_5& 0& \\alpha_6\\\\\n\\alpha_3& \\alpha_7& \\alpha_4\n\\end{pmatrix} \\phi=\n\\begin{pmatrix}\n\\alpha^*& \\alpha_1^*& \\alpha_2^*\\\\\n\\alpha_5^*& \\alpha^{**}& \\alpha_6^*\\\\\n\\alpha_3^*& \\alpha_7^*& \\alpha_4^*\n\\end{pmatrix},\n$$\nwhere\n\\begin{align*}\n \\alpha_1^* &= x^2(\\alpha_1x + \\alpha_7z),\\\\\n \\alpha_2^* &= u(\\alpha_1x + \\alpha_7z) + v(\\alpha_2x + \\alpha_6y + \\alpha_4z),\\\\\n \\alpha_3^* &= u(\\alpha_5x + \\alpha_6z) + v(\\alpha_3x + \\alpha_7y + \\alpha_4z),\\\\\n \\alpha_4^* &= v((\\alpha_6 + \\alpha_7)u + \\alpha_4v),\\\\\n \\alpha_5^* &= x^2(\\alpha_5x + \\alpha_6z),\\\\\n \\alpha_6^* &= \\alpha_6x^2v,\\\\\n \\alpha_7^* &= \\alpha_7x^2v.\n\\end{align*}\nHence, $\\phi\\langle\\theta\\rangle=\\langle\\theta^*\\rangle,$ where $\\theta^*=\\sum\\limits_{i=1}^7 \\alpha_i^* \\nb i.$\n \n We are interested in $\\theta$ with $(\\alpha_5,\\alpha_6,\\alpha_7) \\neq (0,0,0).$ Moreover, the condition $\\theta \\in \\mathbf{T}_1 (\\T{3*}{01})$ gives us $(\\alpha_2, \\alpha_3, \\alpha_4, \\alpha_6, \\alpha_7) \\neq (0,0,0,0,0)$ and \n $(\\alpha_1 + \\gamma\\alpha_2, \\alpha_5 + \\gamma\\alpha_3, \\alpha_6, \\alpha_7, \\alpha_6 + \\gamma\\alpha_4, \\alpha_7 + \\gamma\\alpha_4) \\neq (0,0,0,0,0,0)$ for all $\\gamma \\in \\mathbb{C}.$\n \n\\begin{enumerate}\n\n \n\\item Let $\\alpha_6 \\neq 0, \\alpha_7 \\neq 0$ and $\\alpha_1 \\alpha_6 - \\alpha_5 \\alpha_7=0.$\n Taking $v =\\frac{1}{x^2 \\alpha_7},$ \n $z=-\\frac{x \\alpha_1}{\\alpha_7}$ and $y=\\frac{x (\\alpha_1 \\alpha_4 - \\alpha_3 \\alpha_7)}{\\alpha_7^2},$\n we get the family of representatives \n $$ \n \\langle \\alpha_2^\\star \\nabla_2 + \\alpha_4^\\star \\nabla_4 + \\alpha_6^\\star \\nabla_6+ \\nabla_7\\rangle,\n $$\n where \n\n\\begin{align*}\n \\as 2 &= \\frac 1{\\alpha_7^3x}(\\alpha_1\\alpha_4\\alpha_6 - \\alpha_1\\alpha_4\\alpha_7 - \\alpha_3\\alpha_6\\alpha_7 + \\alpha_2\\alpha_7^2),\\\\\n \\as 4 &= \\frac 1{\\alpha_7^2x^4}(ux^2\\alpha_7(\\alpha_6 + \\alpha_7) + \\alpha_4),\\\\\n \\alpha_6^\\star &= \\frac{\\alpha_6}{\\alpha_7}.\n\\end{align*}\n\n \\begin{enumerate}\n \\item $\\alpha_6\\neq -\\alpha_7$\n and $\\alpha_1\\alpha_4\\alpha_6 - \\alpha_1\\alpha_4\\alpha_7 - \\alpha_3\\alpha_6\\alpha_7 + \\alpha_2\\alpha_7^2\\neq 0$.\nThen choosing $u =-\\frac{\\alpha_4}{x^2\\alpha_7(\\alpha_6 + \\alpha_7)}$, where $x$ is such that $\\as 2=1$, we have the family of representatives \n $ \\langle \\nabla_2 + \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha \\neq -1,0}.$ \n \n \\item $\\alpha_6\\neq -\\alpha_7$\n and $\\alpha_1\\alpha_4\\alpha_6 - \\alpha_1\\alpha_4\\alpha_7 - \\alpha_3\\alpha_6\\alpha_7 + \\alpha_2\\alpha_7^2 = 0$. Then\n choosing $u = -\\frac{\\alpha_4}{x^2 \\alpha_7 (\\alpha_6 + \\alpha_7)},$ we have the family of representatives \n $ \\langle \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha \\neq -1,0}.$ \n\n\n\n \\item $\\alpha_6=- \\alpha_7,$ $\\alpha_4=0$\n and $\\alpha_2 \\neq - \\alpha_3$. Then choosing $x = \\frac{\\alpha_2 + \\alpha_3}{\\alpha_7},$ we have the representative \n $ \\langle \\nabla_2 - \\nabla_6+ \\nabla_7\\rangle,$ which will be joined with the family $ \\langle \\nabla_2 + \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha \\neq -1,0}.$ \n\n\n\n \\item $\\alpha_6=- \\alpha_7,$ $\\alpha_4=0$\n and $\\alpha_2 = - \\alpha_3$. Then\n we have the representative \n $ \\langle - \\nabla_6+ \\nabla_7\\rangle,$\n which will be joined with the family $ \\langle \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha \\neq -1,0}.$ \n\n \\item $\\alpha_6=- \\alpha_7$ and $\\alpha_4\\neq 0$.\n Then choosing $x = \\sqrt[4]{\\frac{\\alpha_4}{\\alpha_7^2}},$ we have the family of representatives \n $ \\langle \\alpha \\nabla_2 +\\nabla_4 - \\nabla_6+ \\nabla_7\\rangle.$\n It gives only two distinct orbits with representatives \n $ \\langle \\nabla_2 +\\nabla_4 - \\nabla_6+ \\nabla_7\\rangle$ and \n $ \\langle \\nabla_4 - \\nabla_6+ \\nabla_7\\rangle.$\n\n\\end{enumerate}\n\n\\item Let $\\alpha_6 \\neq 0, \\alpha_7 \\neq 0$ and $\\alpha_1 \\alpha_6 - \\alpha_5 \\alpha_7\\ne 0.$\n Taking \n $z=-\\frac{\\alpha_5x}{\\alpha_6},$ \n $v = \\frac x{\\alpha_6\\alpha_7}(\\alpha_1\\alpha_6 - \\alpha_5\\alpha_7),$\n $y = \\frac x{\\alpha_6\\alpha_7}(\\alpha_4\\alpha_5 - \\alpha_3\\alpha_6),$ \n and \n $u = \\frac x{\\alpha_6\\alpha_7^2}(\\alpha_3\\alpha_6^2 - \\alpha_2\\alpha_6\\alpha_7 + \\alpha_4\\alpha_5(\\alpha_7 -\\alpha_6)),$\n we get the family of representatives \n $$ \\langle \\nabla_1 + \\alpha_4^\\star \\nabla_4 + \\alpha_6^\\star \\nabla_6+ \\nabla_7\\rangle,$$\n where \n\n\\begin{align*}\n \\alpha_4^\\star &= \\frac 1{\\alpha_7^3x}(\\alpha_3\\alpha_6^2 + \\alpha_1\\alpha_4\\alpha_7 + \\alpha_3\\alpha_6\\alpha_7 - \\alpha_2\\alpha_6\\alpha_7 - \\alpha_2\\alpha_7^2 - \\alpha_4\\alpha_5\\alpha_6), \\\\\n \\alpha_6^\\star &= \\frac{\\alpha_6}{\\alpha_7}.\n\\end{align*}\nIt gives two families of representatives of distinct orbits \n $ \\langle \\nabla_1 + \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha\\neq 0} $\n and \n $ \\langle \\nabla_1 + \\nabla_4 + \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha\\neq 0}$ \n depending on whether $\\alpha_3\\alpha_6^2 + \\alpha_1\\alpha_4\\alpha_7 + \\alpha_3\\alpha_6\\alpha_7 - \\alpha_2\\alpha_6\\alpha_7 - \\alpha_2\\alpha_7^2 - \\alpha_4\\alpha_5\\alpha_6=0$ or not.\n \n \n \n \n \n \n \n \n \n \n \\item Let $\\alpha_6 = 0, \\alpha_7 \\neq 0.$\n Taking \n $y = \\frac x{\\alpha_7^2} (\\alpha_1 \\alpha_4 + \\alpha_4 \\alpha_5 - \\alpha_3 \\alpha_7),$ $z = -\\frac{x \\alpha_1}{\\alpha_7},$ \n and $u = -\\frac{\\alpha_4v}{\\alpha_7} $\n we get a family of representatives \n $$ \\langle \\alpha_2^\\star \\nabla_2 + \\alpha_5^\\star \\nabla_5 + \\nabla_7\\rangle,$$\nwhere \n\\begin{align*}\n \\alpha_2^\\star &= \\frac 1{x \\alpha_7^2}( \\alpha_2 \\alpha_7-\\alpha_1 \\alpha_4), \\\\ \n\\alpha_5^\\star &= \\frac{\\alpha_5x}{\\alpha_7v}. \n\\end{align*}\n\\begin{enumerate}\n \\item $\\alpha_5 = 0$ and $\\alpha_1\\alpha_4 - \\alpha_2\\alpha_7=0$. Then \n we have the representative $\\langle\\nabla_7\\rangle,$\n which will be joined with the family $ \\langle \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha \\neq 0}.$ \n\n\n \\item $\\alpha_5 = 0$ and $\\alpha_1\\alpha_4 - \\alpha_2\\alpha_7\\ne 0$. Then \n choosing $x= \\frac 1{\\alpha_7^2}(\\alpha_2 \\alpha_7-\\alpha_1 \\alpha_4),$ \n we have the representative $\\langle\\nabla_2 + \\nabla_7\\rangle.$\n which will be joined with the family $ \\langle \\nabla_2 + \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha \\neq 0}$, \n\n \n \\item $\\alpha_5 \\neq 0$ and $\\alpha_1\\alpha_4 - \\alpha_2\\alpha_7\\ne 0$. Then \n choosing $x= \\frac 1{\\alpha_7^2}(\\alpha_2 \\alpha_7-\\alpha_1 \\alpha_4)$ and\n $v=\\frac{\\alpha_5\\alpha_7x^2}{\\alpha_2\\alpha_7-\\alpha_1\\alpha_4}$\n we get the representative $\\langle \\nabla_2 + \\nabla_5 + \\nabla_7 \\rangle.$\n \n \n \\item $\\alpha_5 \\neq 0$ and $\\alpha_1\\alpha_4 - \\alpha_2\\alpha_7=0$. Then \n choosing \n $v = \\frac {\\alpha_5x}{\\alpha_7}$\n we get the representative $\\langle \\nabla_5 + \\nabla_7 \\rangle.$\n \\end{enumerate}\n \n \n \n \n \\item $\\alpha_6 \\neq 0$ and $\\alpha_7 = 0.$\n Taking \n $y = \\frac x{\\alpha_6^2} (\\alpha_1 \\alpha_4 + \\alpha_4 \\alpha_5 - \\alpha_2 \\alpha_6),$ \n $z = -\\frac{x \\alpha_5}{\\alpha_6},$ and $u = -\\frac{\\alpha_4v}{\\alpha_6}$\n we get the family of representatives \n $$ \\langle \\alpha_1^\\star \\nabla_1 + \\alpha_3^\\star \\nabla_3 + \\nabla_6\\rangle,$$\nwhere \n\\begin{align*}\n \\alpha_1^\\star &= \\frac{\\alpha_1x}{\\alpha_6v}, \\\\ \n\\alpha_3^\\star &= \\frac 1{x \\alpha_6^2}(\\alpha_3 \\alpha_6-\\alpha_4 \\alpha_5). \n\\end{align*}\n \n \\begin{enumerate}\n \\item $\\alpha_1\\neq 0$ and $\\alpha_3 \\alpha_6- \\alpha_4 \\alpha_5\\ne 0$. Then\n we have the representative \n $ \\langle \\nabla_1 + \\nabla_3 + \\nabla_6\\rangle.$\n\n \\item $\\alpha_1 = 0$ and $\\alpha_3 \\alpha_6- \\alpha_4 \\alpha_5\\ne 0$. Then\n we have the representative $ \\langle \\nabla_3 + \\nabla_6\\rangle.$\n\n \\item $\\alpha_1 \\neq 0$ and $\\alpha_3 \\alpha_6 - \\alpha_4 \\alpha_5=0$. Then\n we have the representative $ \\langle \\nabla_1 + \\nabla_6\\rangle.$\n \n \n \\item $\\alpha_1 = 0$ and $\\alpha_3 \\alpha_6 - \\alpha_4 \\alpha_5=0$. Then\n we have the representative $ \\langle \\nabla_6\\rangle.$\n\n \\end{enumerate}\n \\item $\\alpha_5 \\neq 0$, $\\alpha_6=0$ and $\\alpha_7=0.$ \n Taking \n $u =\\frac{-v x \\alpha_3 - v z \\alpha_4}{x \\alpha_5}$ and $x = \\frac{1}{\\sqrt[3]{\\alpha_5}},$\n we get a family of representatives \n $$ \\langle \\alpha_1^\\star \\nabla_1 + \\alpha_2^\\star \\nabla_2 + \\alpha_4^\\star \\nabla_4 + \\nabla_5\\rangle,$$\nwhere \n\\begin{align*}\n \\alpha_1^\\star &= \\frac{\\alpha_1}{\\alpha_5}, \\\\ \n \\alpha_2^\\star &= \\frac v{\\alpha_5^2x^3}((\\alpha_2\\alpha_5 -\\alpha_1\\alpha_3)x + \\alpha_4(\\alpha_5 - \\alpha_1)z), \\\\\n \\alpha_4^\\star &= \\frac{\\alpha_4v^2}{\\alpha_5x^3}. \n\\end{align*}\n\n \\begin{enumerate}\n \\item $\\alpha_4 \\neq 0$ and $\\alpha_1 - \\alpha_5\\ne 0$. Then \n choosing $z= -\\frac{(\\alpha_2\\alpha_5 -\\alpha_1\\alpha_3)x}{\\alpha_4 (\\alpha_1 -\\alpha_5)}$ and $v = \\sqrt{\\frac{\\alpha_5x^3}{\\alpha_4}}$\n we have the family of representatives of distinct orbits\n $ \\langle \\alpha \\nabla_1 + \\nabla_4 + \\nabla_5\\rangle_{\\alpha \\neq 1}.$ \n\n \\item $\\alpha_4 = 0$, $\\alpha_1 - \\alpha_5\\ne 0$ and $\\alpha_2\\alpha_5-\\alpha_1\\alpha_3=0$. Then \n we have the family of representatives of distinct orbits\n $ \\langle \\alpha \\nabla_1 + \\nabla_5\\rangle_{\\alpha \\neq 1}$. The corresponding extensions are split. \n \n \\item $\\alpha_4 = 0$, $\\alpha_1 - \\alpha_5\\ne 0$ and $\\alpha_2\\alpha_5-\\alpha_1\\alpha_3=0$. Then \n we have the family of representatives of distinct orbits\n $ \\langle \\alpha \\nabla_1 +\\nabla_2 +\\nabla_5\\rangle_{\\alpha \\neq 1}.$ \n\n \\item $\\alpha_4 \\neq 0$, $\\alpha_1 - \\alpha_5=0$ and $\\alpha_2 - \\alpha_3\\ne 0$. Then \n choosing $x = \\frac 1{\\alpha_4\\alpha_5}(\\alpha_2-\\alpha_3)^2$,\n $v=\\frac 1{\\alpha_4^2\\alpha_5}(\\alpha_2- \\alpha_3)^3$,\n $u = -\\frac {\\alpha_2 - \\alpha_3}{\\alpha_4^2\\alpha_5^2}(\\alpha_4^2\\alpha_5z + (\\alpha_2 - \\alpha_3)^2\\alpha_3)$\n we have the representative \n $ \\langle \\nabla_1 + \\nabla_2 +\\nabla_4+ \\nabla_5\\rangle.$ \n \n\n \\item $\\alpha_4 \\neq 0$, $\\alpha_1 - \\alpha_5=0$ and $\\alpha_2 - \\alpha_3= 0$. Then \n choosing $v=\\sqrt {\\frac{\\alpha_5x^3}{\\alpha_4}}$ and $u=-\\frac v{\\alpha_5x}(\\alpha_3x + \\alpha_4z)$\n we have the representative \n $ \\langle \\nabla_1 + \\nabla_4+ \\nabla_5\\rangle,$\n which will be joined with the family $\\langle \\alpha \\nabla_1 + \\nabla_4+ \\nabla_5 \\rangle_{\\alpha \\neq 1}$.\n\n \\item $\\alpha_4 = 0$, $\\alpha_1 - \\alpha_5=0$ and $\\alpha_2 - \\alpha_3\\ne 0$. Then \n choosing $u=-\\frac{\\alpha_3v}{\\alpha_5}$ and $v=\\frac{\\alpha_5x^2}{\\alpha_2 - \\alpha_3}$\n we have the representative \n $ \\langle \\nabla_1 + \\nabla_2+ \\nabla_5\\rangle,$\n which will be joined with the family $\\langle \\alpha \\nabla_1 + \\nabla_2+ \\nabla_5 \\rangle_{\\alpha \\neq 1}$.\n\n \\item $\\alpha_4 = 0$, $\\alpha_1 - \\alpha_5=0$ and $\\alpha_2 - \\alpha_3= 0$. Then choosing $u=-\\frac{\\alpha_3v}{\\alpha_5}$\n we have the representative \n $ \\langle \\nabla_1 + \\nabla_5\\rangle,$\n which gives a split extension. \n\n\n\n \\end{enumerate}\n \\end{enumerate}\n\n \\ \n \n \n \n \n \n \nSummarizing, we have the following representatives of distinct orbits: \n\n$ \n\\begin{array}{l}\n \\langle \\nabla_1 + \\nabla_4 + \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha\\neq 0},\\\\\n \\langle \\nabla_1 + \\alpha \\nabla_6+ \\nabla_7\\rangle_{\\alpha\\neq 0},\\\\\n \\langle \\nabla_2 +\\nabla_4 - \\nabla_6+ \\nabla_7\\rangle,\\\\\n \\langle \\nabla_2 + \\nabla_5 + \\nabla_7 \\rangle,\n\\end{array}\n\\begin{array}{l}\n \\langle \\nabla_2 + \\alpha \\nabla_6+ \\nabla_7\\rangle,\\\\\n \\langle \\nabla_4 - \\nabla_6+ \\nabla_7\\rangle,\\\\\n \\langle \\nabla_5 + \\nabla_7 \\rangle,\\\\\n \\langle \\alpha \\nabla_6+ \\nabla_7\\rangle, \n\\end{array}\n\\begin{array}{l}\n \\langle \\nabla_1 + \\nabla_6\\rangle,\\\\\n \\langle \\nabla_1 + \\nabla_3 + \\nabla_6\\rangle,\\\\ \n \\langle \\nabla_3 + \\nabla_6\\rangle,\\\\\n \\langle \\nabla_6\\rangle,\n\\end{array}\n\\begin{array}{l}\n \\langle \\alpha \\nabla_1 +\\nabla_2 +\\nabla_5\\rangle,\\\\ \n \\langle \\nabla_1 + \\nabla_2 +\\nabla_4+ \\nabla_5\\rangle,\\\\\n \\langle \\alpha \\nabla_1 + \\nabla_4 + \\nabla_5\\rangle.\n\\end{array}\n$\n\n\nThe corresponding algebras are:\n\\[\\begin{array}{lllllllllll}\n\\T {4}{03}(\\alpha)_{\\alpha\\neq 0}&:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3=\\alpha e_4,& e_3e_2=e_4,& e_3e_3=e_4; \\\\\n\\T {4}{04}(\\alpha)_{\\alpha\\neq 0}&:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3=\\alpha e_4,& e_3e_2=e_4; \\\\\n\\T {4}{05} &:& e_1e_1 = e_2,& e_1e_3=e_4,& e_2e_3=- e_4,& e_3e_2=e_4,& e_3e_3=e_4; \\\\\n\\T {4}{06} &:& e_1e_1 = e_2,& e_1e_3=e_4,& e_2e_1= e_4,& e_3e_2=e_4; \\\\\n\\T {4}{07}(\\alpha) &:& e_1e_1 = e_2,& e_1e_3=e_4,& e_2e_3=\\alpha e_4,& e_3e_2=e_4; \\\\\n\\T {4}{08} &:& e_1e_1 = e_2,& e_2e_3=- e_4,& e_3e_2=e_4,& e_3e_3=e_4; \\\\\n\\T {4}{09} &:& e_1e_1 = e_2,& e_2e_1= e_4,& e_3e_2=e_4; \\\\\n\\T {4}{10}(\\alpha) &:& e_1e_1 = e_2,& e_2e_3=\\alpha e_4,& e_3e_2=e_4; \\\\\n\\T {4}{11} &:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3= e_4; \\\\\n\\T {4}{12} &:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3= e_4,& e_3e_1=e_4; \\\\\n\\T {4}{13} &:& e_1e_1 = e_2,& e_2e_3= e_4,& e_3e_1=e_4; \\\\\n\\T {4}{14} &:& e_1e_1 = e_2,& e_2e_3= e_4; \\\\\n\\T {4}{15}(\\alpha) &:& e_1e_1 = e_2,& e_1e_2= \\alpha e_4,& e_1e_3= e_4,& e_2e_1=e_4; \\\\\n\\T {4}{16} &:& e_1e_1 = e_2,& e_1e_2= e_4,& e_1e_3= e_4,& e_2e_1=e_4,& e_3e_3=e_4; \\\\\n\\T {4}{17}(\\alpha) &:& e_1e_1 = e_2,& e_1e_2= \\alpha e_4,& e_2e_1=e_4,& e_3e_3= e_4.\n\n\\end{array}\\]\n\n\n\\subsubsection{$1$-dimensional central extensions of $\\T {3*}{02}$}\nLet us use the following notations: \n\\begin{align*}\n \\nb 1 = \\Dl 11, \\nb 2 = \\Dl 12, \\nb 3 = \\Dl 21, \\nb 4 = \\Dl 13, \\nb 5 = \\Dl 23, \\nb 6 = \\Dl 31, \\nb 7 = \\Dl 32. \n\\end{align*}\nTake $\\theta=\\sum_{i=1}^7\\alpha_i\\nb i\\in {\\rm H_T^2}(\\T {3*}{02}).$\nIf \n$$\n\\phi=\n\\begin{pmatrix}\nx & y& 0 \\\\\n(-1)^{n+1}y& (-1)^nx& 0\\\\\nz & u & x^2+y^2\n\\end{pmatrix}\\in\\aut{\\T {3*}{02}},\n$$\nthen\n$$\n\\phi^T\\begin{pmatrix}\n\\alpha_1& \\alpha_2& \\alpha_4\\\\\n\\alpha_3& 0& \\alpha_5\\\\\n\\alpha_6& \\alpha_7& 0\n\\end{pmatrix} \\phi=\n\\begin{pmatrix}\n\\alpha_1^*+\\alpha^*& \\alpha_2^*& \\alpha_4^*\\\\\n\\alpha_3^*& \\alpha^*& \\alpha_5^*\\\\\n\\alpha_6^*& \\alpha_7^*& \\alpha^{**}\n\\end{pmatrix},\n$$\nwhere\n\\begin{align*}\n \\alpha_1^* &= \\alpha_1(x^2 - y^2) + 2(-1)^{n+1}(\\alpha_2 + \\alpha_3)xy+(\\alpha_4x+(-1)^{n+1}\\alpha_5y+\\alpha_6x+ (-1)^{n+1}\\alpha_7y)z\\\\\n\t&\\quad - (\\alpha_4y +(-1)^n\\alpha_5x+\\alpha_6y+(-1)^n\\alpha_7x)u,\\\\\n\t\\alpha_2^* &= (-1)^n\\alpha_2x^2 + \\alpha_1xy + (-1)^{n+1}\\alpha_3y^2 + (\\alpha_4x + (-1)^{n+1}\\alpha_5y)u + (\\alpha_6y + (-1)^n\\alpha_7x)z,\\\\\n\t\\alpha_3^* &= (-1)^n\\alpha_3x^2 + \\alpha_1xy + (-1)^{n+1}\\alpha_2y^2 + (\\alpha_4y + (-1)^n\\alpha_5x)z + (\\alpha_6x + (-1)^{n+1}\\alpha_7y)u,\\\\\n\t\\alpha_4^* &= (\\alpha_4x+(-1)^{n+1}\\alpha_5y)(x^2+y^2),\\\\\n\t\\alpha_5^* &= (\\alpha_4y+(-1)^n\\alpha_5x)(x^2 + y^2),\\\\\n\t\\alpha_6^* &= (\\alpha_6x+(-1)^{n+1}\\alpha_7y)(x^2+y^2),\\\\\n\t\\alpha_7^* &= (\\alpha_6y + (-1)^n\\alpha_7x)(x^2+y^2).\n\\end{align*}\nHence, $\\phi\\langle\\theta\\rangle=\\langle\\theta^*\\rangle,$ where $\\theta^*=\\sum\\limits_{i=1}^7 \\alpha_i^* \\nb i.$\n \n We are interested in $\\theta$ with $(\\alpha_4,\\alpha_5,\\alpha_6,\\alpha_7) \\neq (0,0,0,0).$ The condition $\\theta \\in \\mathbf{T}_1 (\\T{3*}{02})$ does not give us any new restrictions on parameters.\n\n\\begin{enumerate}\n \\item $\\alpha_6=0$ and $\\alpha_7=0.$ Then $\\alpha_4\\ne 0$ or $\\alpha_5\\ne 0$. If $\\alpha_4\\ne 0$ and $\\alpha_5=0$, then choosing $x=y=1$, we have $\\alpha^*_4\\ne 0$ and $\\alpha^*_5\\ne 0$. The same holds for $\\alpha_4=0$ and $\\alpha_5\\ne 0$. Thus, we shall assume from the very beginning that $\\alpha_4\\ne 0$ and $\\alpha_5 \\neq 0$. \n \\begin{enumerate}\n \\item $\\alpha^2_4+\\alpha^2_5\\ne 0$. Then choosing $x=(-1)^{n+1}\\frac{\\alpha_4y}{\\alpha_5},\n z = \\frac{(-1)^ny}{\\alpha_5(\\alpha_4^2 + \\alpha_5^2)}(\\alpha_1\\alpha_4^2 + 2(\\alpha_2 + \\alpha_3)\\alpha_4\\alpha_5 - \\alpha_1\\alpha_5^2)$ and $u = \\frac y{\\alpha_5(\\alpha_4^2 + \\alpha_5^2)}(\\alpha_2\\alpha_4^2 -\\alpha_3\\alpha_5^2 - \\alpha_1\\alpha_4\\alpha_5)$ we obtain the representative $\\langle\\as 3\\nb 3+\\as 4\\nb 4\\rangle$, where\n \\begin{align*}\n \\as 3 &= \\frac{(-1)^ny^2}{\\alpha_5^2}(\\alpha_3\\alpha_4^2 - \\alpha_1\\alpha_4\\alpha_5 - \\alpha_2\\alpha_5^2),\\\\\n \\as 4 &= \\frac{(-1)^{n+1}y^3}{\\alpha_5^3}(\\alpha_4^2 + \\alpha_5^2)^2.\n \\end{align*}\n Thus, we have two representatives $\\langle \\nabla_4 \\rangle $ and $\\langle \\nabla_3 + \\nabla_4 \\rangle $ depending on whether $\\alpha_3\\alpha_4^2 - \\alpha_1\\alpha_4\\alpha_5 - \\alpha_2\\alpha_5^2=0$ or not.\n \\item $\\alpha^2_4+\\alpha^2_5=0$. If $\\alpha_4=-i\\alpha_5$, then choosing $x=0$ and $y=n=1$, we obtain $\\alpha^*_4=i\\alpha^*_5$, so we shall assume from the very beginning that $\\alpha_4= i \\alpha_5.$ \n Taking $n=0$,\n $u=\\frac{i(\\alpha_2x^2 + \\alpha_1xy - \\alpha_3y^2)}{\\alpha_5(x + iy)}$ and \n $z=-\\frac{\\alpha_3x^2 + \\alpha_1xy - \\alpha_2y^2}{\\alpha_5(x + iy)}$\n we have the representative $\\langle \\alpha_1^{\\star} \\nabla_1 + i \\nabla_4 + \\nabla_5 \\rangle$, where\n \\begin{align*}\n \\as 1 &= \\frac{x - iy}{\\alpha_5(x + iy)^2}(\\alpha_1 - i\\alpha_2 - i\\alpha_3).\n \\end{align*}\nThus, we have two representatives $\\langle i \\nabla_4 + \\nabla_5 \\rangle$ and $\\langle \\nabla_1 + i \\nabla_4 + \\nabla_5 \\rangle$ depending on whether $\\alpha_1 - i\\alpha_2 - i\\alpha_3=0$ or not.\n\n \n \\end{enumerate}\n \n\\item $\\alpha_6\\neq 0$ or $\\alpha_7\\neq0$, and $\\alpha_6^2+\\alpha_7^2\\ne 0$. \n Then we may make $\\alpha_6\\ne 0$ and $\\alpha_7\\neq 0$. After that, if we choose $x=(-1)^{n+1}\\frac{\\alpha_6y}{\\alpha_7}\\ne 0,$ then $x^2+y^2=\\frac {y^2}{\\alpha^2_7}(\\alpha_6^2+\\alpha_7^2)\\ne 0$, $\\alpha^*_6=\\frac{(-1)^{n+1}y^3}{\\alpha_7^3}(\\alpha_6^2 + \\alpha_7^2)^2\\ne 0$ and $\\alpha_7^* = 0$. Thus, we shall assume that $\\alpha_6\\ne 0$ and $\\alpha_7=0$ from the very beginning. Taking $y = 0,$ we get $\\alpha^*_7=0$ and\n \\begin{align*}\n \\alpha_1^* &= x (x \\alpha_1 + (-1)^{n + 1} u \\alpha_5 + z (\\alpha_4 + \\alpha_6)),\\\\\n\t\\alpha_2^* &= x ((-1)^n x \\alpha_2 + u \\alpha_4),\\\\\n\t\\alpha_3^* &= x ((-1)^n x \\alpha_3 + (-1)^n z \\alpha_5 + u \\alpha_6),\\\\\n\t\\alpha_4^* &= x^3 \\alpha_4,\\\\\n\t\\alpha_5^* &= (-1)^n x^3 \\alpha_5,\\\\\n\t\\alpha_6^* &= x^3 \\alpha_6.\n\\end{align*}\nConsider $\\alpha_2^* = 0, \\alpha_3^* = 0$ as a system of linear equations on $z$ and $u.$ Its determinant is $(-1)^{n+1}x^2\\alpha_4\\alpha_5,$ so we have the following cases (each of which defines an $\\operatorname{Aut}{\\T {3*}{02}}$-invariant subset):\n\\begin{enumerate}\n \\item $\\alpha_4 \\neq 0$ and $\\alpha_5 \\neq 0$. Taking $u = \\frac{(-1)^{n+1} x \\alpha_2}{\\alpha_4}, z = \\frac{x (\\alpha_2 \\alpha_6-\\alpha_3 \\alpha_4)}{\\alpha_4 \\alpha_5},$ we get $\\alpha_2^* = 0, \\alpha_3^* = 0$ and \n \\begin{align*}\n \\alpha_1^* &= \\frac{x^2}{\\alpha_4\\alpha_5}(\\alpha_2\\alpha_5^2 - \\alpha_3\\alpha_4^2 + \\alpha_1\\alpha_4\\alpha_5 + \\alpha_2\\alpha_4\\alpha_6 - \\alpha_3\\alpha_4\\alpha_6 +\\alpha_2\\alpha_6^2),\\\\\n\t\\alpha_4^* &= x^3 \\alpha_4,\\\\\n\t\\alpha_5^* &= (-1)^n x^3 \\alpha_5,\\\\\n\t\\alpha_6^* &= x^3 \\alpha_6.\n\\end{align*}\nThus, we obtain two families of representatives of distinct orbits $\\langle \\nabla_1 +\\alpha \\nabla_4+\\beta \\nabla_5+\\nabla_6 \\rangle_{\\alpha\\beta\\neq 0, \\beta \\in \\mathbb C_{\\geq 0}}$ and $\\langle \\alpha \\nabla_4+\\beta \\nabla_5+\\nabla_6 \\rangle_{\\alpha\\beta\\neq 0, \\beta \\in \\mathbb C_{\\geq 0}}. $\n \\item $\\alpha_4 \\neq 0$ and $\\alpha_5 = 0$. Taking $u = \\frac{(-1)^{n+1} x \\alpha_2}{\\alpha_4}$, we get $\\alpha_2^* = 0$ and \n \\begin{align*}\n \\alpha_1^* &= x (x \\alpha_1 + z (\\alpha_4 + \\alpha_6)),\\\\\n \\alpha_3^* &= \\frac{(-1)^nx^2}{\\alpha_4}(\\alpha_3\\alpha_4 - \\alpha_2\\alpha_6),\\\\\n\t\\alpha_4^* &= x^3 \\alpha_4,\\\\\n\t\\alpha_6^* &= x^3 \\alpha_6.\n\\end{align*}\n \\begin{enumerate}\n \\item $\\alpha_4 + \\alpha_6 \\neq 0$. Taking $z = -\\frac{x\\alpha_1}{\\alpha_4 + \\alpha_6},$ we get two series of representatives of distinct orbits $\\langle \\nabla_3 +\\alpha \\nabla_4 +\\nabla_6\\rangle_{\\alpha\\neq -1,0}$ and $\\langle \\alpha \\nabla_4 +\\nabla_6\\rangle_{\\alpha\\neq -1,0}.$\n \\item $\\alpha_4 + \\alpha_6 = 0.$ In this case we get the series of representatives of distinct orbits $\\langle \\nabla_1 +\\alpha \\nabla_3 - \\nabla_4+\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}}$ and two distinct representatives $\\langle \\nabla_3 - \\nabla_4+\\nabla_6 \\rangle$ and $\\langle - \\nabla_4+\\nabla_6 \\rangle$, which will be joined with the families $\\langle \\nabla_3 +\\alpha \\nabla_4 +\\nabla_6\\rangle_{\\alpha\\neq -1,0}$ and $\\langle \\alpha \\nabla_4 +\\nabla_6\\rangle_{\\alpha\\neq -1,0}$ found above.\n \\end{enumerate}\n \\item $\\alpha_4 = 0$ and $\\alpha_5 \\neq 0.$ Consider $\\alpha_1^* = 0, \\alpha_3^* = 0$ as a system of linear equations on $z,u.$ Its determinant is $x^2(\\alpha_5^2 + \\alpha_6^2).$ \n \\begin{enumerate}\n \\item $\\alpha_5^2 + \\alpha_6^2 \\neq 0.$ Taking $u = \\frac{(-1)^n x (\\alpha_1 \\alpha_5 - \\alpha_3 \\alpha_6)}{ \\alpha_5^2 + \\alpha_6^2}, z = -\\frac{x ( \\alpha_3 \\alpha_5 + \\alpha_1 \\alpha_6)}{ \\alpha_5^2 + \\alpha_6^2},$ we get $\\alpha^*_1 = \\alpha^*_3 = 0$ and\n \\begin{align*}\n \\alpha_2^* &= (-1)^n x^2 \\alpha_2,\\\\\n\t\\alpha_5^* &= (-1)^n x^3 \\alpha_5,\\\\\n\t\\alpha_6^* &= x^3 \\alpha_6.\n\\end{align*}\nTherefore, we have two series of representatives of distinct orbits $\\langle \\nabla_2 + \\alpha \\nabla_5+\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}, \\alpha \\neq 0, i}$ and $\\langle \\alpha \\nabla_5+\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}, \\alpha \\neq 0, i}.$\n \\item $\\alpha_5^2 + \\alpha_6^2 = 0.$ We may assume that $\\alpha_5 = i \\alpha_6$ (if $\\alpha_5=-i\\alpha_6$, then choose $n=-1$). Taking $n=0$, $u = -\\frac{x\\alpha_3}{\\alpha_6}$and $z = 0,$ we get $\\alpha_3 = 0$ and \n \\begin{align*}\n \\alpha_1^* &= x^2 (\\alpha_1+i\\alpha_3),\\\\\n \\alpha_2^* &= x^2 \\alpha_2,\\\\\n\t\\alpha_5^* &= ix^3 \\alpha_6,\\\\\n\t\\alpha_6^* &= x^3 \\alpha_6.\n\\end{align*}\nTherefore, we get the family of representatives of distinct orbits $\\langle \\nabla_1 + \\alpha \\nabla_2 + i \\nabla_5 + \\nabla_6 \\rangle$ and two representatives $\\langle \\nabla_2 + i \\nabla_5 + \\nabla_6 \\rangle, \\langle i \\nabla_5 + \\nabla_6 \\rangle$, which will be joined with the families $\\langle \\nabla_2 + \\alpha \\nabla_5+\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}, \\alpha \\neq 0, i}$ and $\\langle \\alpha \\nabla_5+\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}, \\alpha \\neq 0, i}$ found above. \n \\end{enumerate}\n \\item $\\alpha_4 = 0$ and $\\alpha_5 = 0.$ Taking $z = -\\frac{x\\alpha_1}{\\alpha_6}$ and $u = \\frac{(-1)^{n+1}x\\alpha_3}{\\alpha_6},$ we get $\\alpha_1^* = \\alpha_3^* = 0$ and \n \\begin{align*}\n \\alpha_2^* &= (-1)^nx^2 \\alpha_2,\\\\\n\t\\alpha_6^* &= x^3 \\alpha_6.\n\\end{align*}\nTherefore, we get two representatives $\\langle \\nabla_2 + \\nabla_6 \\rangle$ and $\\langle \\nabla_6 \\rangle$, which will be joined with the families $\\langle \\nabla_2 + \\alpha \\nabla_5+\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}, \\alpha \\neq 0, i}$ and $\\langle \\alpha \\nabla_5+\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}, \\alpha \\neq 0, i}$ found above. Note that representatives $\\langle \\nabla_2 + \\nabla_6 \\rangle$ and $\\langle \\nabla_6 \\rangle$ define the same orbit.\n\\end{enumerate}\n\n\n\n\n\n\n\n \n\n\n\n\n\n\n\n\n\n\n\n \n \n\n \n \n\n\\item $\\alpha_6\\neq 0$ or $\\alpha_7\\neq0$, and $\\alpha_6^2+\\alpha_7^2=0$. Then we may assume that $\\alpha_6= i \\alpha_7\\ne 0.$\n\n\n\\begin{enumerate}\n \\item $\\alpha_4 \\neq \\pm i \\alpha_5.$ We may assume that $\\alpha_4 \\neq 0$ (otherwise $\\alpha_5\\ne 0$, so we may take $x=y=1$ to make $\\alpha^*_4\\ne 0$). Then choosing $n=0$ and $y=-\\frac{x\\alpha_5}{\\alpha_4}x,$ we get $\\alpha^*_4=\\frac{x^3}{\\alpha_4^3}(\\alpha_4^2 + \\alpha_5^2)^2\\ne 0$ and $\\alpha_5^* = 0$, so we shall assume that $\\alpha_4\\ne 0$ and $\\alpha_5=0$ from the very beginning. Now, choosing $n=y=0$, $u = \\frac{i x \\alpha_3}{\\alpha_7}, z = -\\frac{x (i \\alpha_3 \\alpha_4 + \\alpha_2 \\alpha_7)}{\\alpha_7^2},$ we get $\\alpha_2^* = \\alpha_3^* = \\alpha^*_5=0$ and \n \\begin{align*}\n \\alpha_1^* &= \\frac{x^2}{\\alpha_7^2}(\\alpha_3\\alpha_4\\alpha_7 + \\alpha_1\\alpha_7^2 - \\alpha_2\\alpha_4\\alpha_7 -i\\alpha_3\\alpha_4^2 - i\\alpha_2\\alpha_7^2 -i\\alpha_3\\alpha_7^2),\\\\\n \\alpha_4^* &= x^3 \\alpha_4,\\\\\n\t\\alpha_6^* &= ix^3 \\alpha_7,\\\\\n\t\\alpha_7^* &= x^3 \\alpha_7.\n\\end{align*}\nThus, we get two series of representatives of distinct orbits $\\langle \\nabla_1 + \\alpha \\nabla_4 + i \\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq 0}$ and $\\langle \\alpha \\nabla_4 + i \\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq 0}.$\n \n \n \n \n \n \n\n \n \n \\item $\\alpha_4= i \\alpha_5$. Then choosing $n=0$ we have \n \\begin{align*}\n \\alpha_1^* &= x^2 \\alpha_1 - x (2 y (\\alpha_2 + \\alpha_3) + (u - i z) (\\alpha_5 + \\alpha_7)) - y (y \\alpha_1 + (i u + z) (\\alpha_5 + \\alpha_7)),\\\\\n \\alpha_2^* &= x^2 \\alpha_2 - y (y \\alpha_3 + u \\alpha_5 -i z \\alpha_7) + x (y \\alpha_1 + i u \\alpha_5 + z \\alpha_7),\\\\\n \\alpha_3^* &= (x + i y) z \\alpha_5 + x (y \\alpha_1 + x \\alpha_3 + i u \\alpha_7) - y (y \\alpha_2 + u \\alpha_7),\\\\\n \\alpha_4^* &= i (x + i y) (x^2 + y^2) \\alpha_5,\\\\\n \\alpha_5^* &= (x + i y) (x^2 + y^2) \\alpha_5,\\\\\n\t\\alpha_6^* &= i (x + i y) (x^2 + y^2) \\alpha_7,\\\\\n\t\\alpha_7^* &= (x + i y) (x^2 + y^2) \\alpha_7.\n\\end{align*}\n Consider $\\alpha_1^* = 0, \\alpha_3^* = 0$ as a system of linear equations in $u,z.$ Its determinant is $i(x+iy)^2(\\alpha_7^2 - \\alpha_5^2).$\n \\begin{enumerate}\n \\item $\\alpha_5^2 - \\alpha_7^2 \\neq 0$. Then by choosing the appropriate values of $u$ and $z$ we get $\\alpha_1^* = 0, \\alpha_3^* = 0$, $\\alpha_2^* = i(i\\alpha_1 + \\alpha_2 + \\alpha_3)(x - iy)^2$ and $\\alpha^*_4,\\alpha^*_5,\\alpha^*_6,\\alpha^*_7$ as above. Therefore, we have two series of representatives of distinct orbits $\\langle \\nabla_2 + i \\alpha \\nabla_4 + \\alpha \\nabla_5 + i \\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq \\pm 1}$ and $\\langle i \\alpha \\nabla_4 + \\alpha \\nabla_5 + i \\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq \\pm 1}$.\n \\item $\\alpha_5 = \\alpha_7$. Then taking $n=0$, $z = \\frac{-x y \\alpha_1 + y^2 \\alpha_2 - x^2 \\alpha_3 - i u x \\alpha_7 + u y \\alpha_7}{(x + i y)\\alpha_7}$ we get $\\alpha_3^* = 0$ and\n \\begin{align*}\n \\alpha_1^* &= (x - i y) (x (\\alpha_1- 2i\\alpha_3) - y (i \\alpha_1 + 2 \\alpha_2)),\\\\\n \\alpha_2^* &= (x^2 + y^2) (\\alpha_2-\\alpha_3),\\\\\n \\alpha_4^* &= i (x + i y) (x^2 + y^2) \\alpha_7,\\\\\n \\alpha_5^* &= (x + i y) (x^2 + y^2) \\alpha_7,\\\\\n\t\\alpha_6^* &= i (x + i y) (x^2 + y^2) \\alpha_7,\\\\\n\t\\alpha_7^* &= (x + i y) (x^2 + y^2) \\alpha_7.\n\\end{align*}\n\\begin{enumerate}\n \\item $i\\alpha_1 + \\alpha_2 + \\alpha_3\\ne 0$, $\\alpha_2 - \\alpha_3\\ne 0$ and $i\\alpha_1 + 2\\alpha_2\\ne 0$. Then taking $y = \\frac{(\\alpha_1 - 2i\\alpha_3)x}{i\\alpha_1 + 2\\alpha_2}$ we get the representative $\\langle\\as 2\\nb 2+i \\nabla_4 + \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle$, where\n \\begin{align*}\n \\as 2 &= \\frac{(i\\alpha_1 + 2\\alpha_2)(\\alpha_2 - \\alpha_3)}{2\\alpha_7 x(i\\alpha_1 + \\alpha_2 + \\alpha_3)}. \n \\end{align*}\nSo, choosing the appropriate value of $x$, we get the representative $\\langle \\nabla_2 + i \\nabla_4 + \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle.$\n \\item $i\\alpha_1 + \\alpha_2 + \\alpha_3 = 0$. Then we have \n \\begin{align*}\n \\alpha_1^* &= i(x^2+y^2)(\\alpha_2-\\alpha_3),\\\\\n \\alpha_2^* &= (x^2+y^2)(\\alpha_2-\\alpha_3),\n\\end{align*}\nTherefore, we get two representatives $\\langle i\\nabla_1 + \\nabla_2 + i \\nabla_4 + \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle$ and $\\langle i \\nabla_4 + \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle.$\n \\item $\\alpha_2-\\alpha_3 = 0$. Then $\\alpha_1^* = (x-iy)^2(\\alpha_1 - 2i\\alpha_3)$, and we have only one new representative $\\langle \\nabla_1 + i \\nabla_4 + \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle.$\n \\item $i\\alpha_1 + 2\\alpha_2=0$. Then\n $\n \\alpha_1^* = x(x - i y) (\\alpha_1- 2i\\alpha_3),\n $\n so choosing $x=0$ we have two representatives $\\langle \\nabla_2 + i \\nabla_4 + \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle$ and $\\langle i \\nabla_4 + \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle$ found above.\n\\end{enumerate}\n\\item $\\alpha_5 = -\\alpha_7$. Then taking $n=0$ and $z = \\frac{x y \\alpha_1 - y^2 \\alpha_2 + x^2 \\alpha_3 + i u x \\alpha_7 - u y \\alpha_7}{(x + i y) \\alpha_7}$ we get $\\alpha_3^*=0$ and\n \\begin{align*}\n \\alpha_1^* &= (x^2 - y^2)\\alpha_1 - 2 x y (\\alpha_2+\\alpha_3),\\\\\n \\alpha_2^* &= (x^2 - y^2)(\\alpha_2+\\alpha_3) +2 x y \\alpha_1,\\\\\n \\alpha_4^* &= -i(x + i y) (x^2 + y^2) \\alpha_7,\\\\\n \\alpha_5^* &= -(x + i y) (x^2 + y^2) \\alpha_7,\\\\\n\t\\alpha_6^* &= i (x + i y) (x^2 + y^2) \\alpha_7,\\\\\n\t\\alpha_7^* &= (x + i y) (x^2 + y^2) \\alpha_7.\n\\end{align*}\n \\begin{enumerate}\n \\item $(\\alpha_1,\\alpha_2+\\alpha_3) \\neq (0,0)$ and $\\alpha_1^2 + (\\alpha_2+\\alpha_3)^2 \\neq 0$. Then we may assume that $\\alpha_1\\ne 0$ and $\\alpha_2+\\alpha_3\\ne 0$. In this case the equality $\\alpha^*_1=0$ has two distinct roots $y_1=\\mu_1x$ and $y_2=\\mu_2x$, where at least one of $\\mu_1,\\mu_2$ is different from $\\pm i$ (otherwise $\\alpha_2+\\alpha_3=0$). Let $\\mu_1\\ne\\pm i$. Then choosing $y = \\mu_1x$ we get $\\alpha_1^* = 0$. Observe that in this case $x^2 + y^2\\ne 0$ and $\\alpha_1 = \\frac{2(\\alpha_2 + \\alpha_3)\\mu_1}{1 - \\mu_1^2}$. Substituting this into $\\alpha^*_2$, we obtain $\\alpha_2^* = \\frac{x^2(1 + \\mu_1^2)^2(\\alpha_2 + \\alpha_3)}{1-\\mu_1^2}\\ne 0.$ Therefore, we have the representative $\\langle \\nabla_2 - i \\nabla_4 - \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle.$\n \\item $\\alpha_2+\\alpha_3 \\neq 0$ and $\\alpha_1 = \\pm i (\\alpha_2+\\alpha_3)\\ne 0$. Then we get \n \\begin{align*}\n \\alpha_1^* &= \\pm i (x \\pm i y)^2 (\\alpha_2+\\alpha_3),\\\\\n \\alpha_2^* &= (x \\pm i y)^2 (\\alpha_2+\\alpha_3),\n \\end{align*}\n so we obtain two representatives $\\langle i\\nabla_1 + \\nabla_2 - i \\nabla_4 - \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle$ and $\\langle -i\\nabla_1 + \\nabla_2 - i \\nabla_4 - \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle.$\n \\item $(\\alpha_1,\\alpha_2+\\alpha_3) = (0,0)$. Then we get the representative $\\langle - i \\nabla_4 - \\nabla_5 + i \\nabla_6 + \\nabla_7\\rangle.$\n \\end{enumerate}\n \n\n\n\n \n\n \\end{enumerate}\n \\item $\\alpha_4 = -i\\alpha_5$. Choosing $n=0$, we have \n \\begin{align*}\n \\alpha_1^* &= x^2 \\alpha_1 - x (2 y (\\alpha_2 + \\alpha_3) + i z (\\alpha_5 - \\alpha_7) + u (\\alpha_5 + \\alpha_7)) - y (y \\alpha_1 - i u (\\alpha_5 - \\alpha_7) + z (\\alpha_5 + \\alpha_7)),\\\\\n \\alpha_2^* &= x^2 \\alpha_2 - y (y \\alpha_3 + u \\alpha_5 - i z \\alpha_7) + x (y \\alpha_1 - i u \\alpha_5 + z \\alpha_7),\\\\\n \\alpha_3^* &= (x - i y) z \\alpha_5 + x (y \\alpha_1 + x \\alpha_3 + i u \\alpha_7) - y (y \\alpha_2 + u \\alpha_7),\\\\\n \\alpha_4^* &= -i(x - iy) (x^2 + y^2) \\alpha_5,\\\\\n \\alpha_5^* &= (x - i y) (x^2 + y^2) \\alpha_5,\\\\\n\t\\alpha_6^* &= i (x + i y) (x^2 + y^2) \\alpha_7,\\\\\n\t\\alpha_7^* &= (x + i y) (x^2 + y^2) \\alpha_7.\n\\end{align*}\nConsider $\\alpha_2^* = 0, \\alpha_3^* = 0$ as a linear system in $u, z.$ Its determinant is $-i((x - i y)^2 \\alpha_5^2 + (x + i y)^2 \\alpha_7^2).$ So, we may choose $x,y$ such that $x^2 + y^2 \\neq 0, -i((x - i y)^2 \\alpha_5^2 + (x + i y)^2 \\alpha_7^2) \\neq 0$ and $u$ and $z$ to make $\\alpha_2^* = \\alpha_3^*=0$. Observe that this does not change the conditions on $\\alpha_4,\\alpha_5,\\alpha_6,\\alpha_7$. So, we may assume that $\\alpha_2=\\alpha_3=0$ from the very beginning. \nWe may also suppose that $\\alpha_5 + \\alpha_7 \\neq 0.$ \n\\begin{enumerate}\n \\item $\\alpha_5 \\neq 0$. Then taking $y = \\frac{\\alpha_5 - \\alpha_7}{i(\\alpha_5+\\alpha_7)}x$ we may make $\\alpha_5^* = \\alpha_7^*$, so we shall assume $\\alpha_5=\\alpha_7$. Now, choosing $n=y = u=z=0,$ we get \n \\begin{align*}\n \\alpha_1^* &= x^2 \\alpha_1,\\\\\n \\alpha_4^* &= -ix^3 \\alpha_7,\\\\\n \\alpha_5^* &= x^3 \\alpha_7,\\\\\n\t\\alpha_6^* &= i x^3 \\alpha_7,\\\\\n\t\\alpha_7^* &= x^3 \\alpha_7.\n\\end{align*}\nTherefore, we get two representatives $\\langle \\nabla_1 - i \\nabla_4 + \\nabla_5 + i\\nabla_6 + \\nabla_7 \\rangle$ and $\\langle - i \\nabla_4 + \\nabla_5 + i\\nabla_6 + \\nabla_7 \\rangle.$\n \\item $\\alpha_5 = 0$. Then $\\alpha_4 = 0$, so that $\\alpha^*_4=\\alpha^*_5=0$. Choosing $n=0$ and the appropriate values of $u$ and $z$ we have $\\alpha^*_2=\\alpha^*_3=0$ and\n \\begin{align*}\n \\alpha_1^* &= (x - i y)^2 \\alpha_1,\\\\\n\t\\alpha_6^* &= i (x + i y) (x^2 + y^2) \\alpha_7,\\\\\n\t\\alpha_7^* &= (x + i y) (x^2 + y^2) \\alpha_7.\n\\end{align*}\nTherefore, we get two representatives $\\langle \\nabla_1 + i\\nabla_6 + \\nabla_7 \\rangle$ and $\\langle i\\nabla_6 + \\nabla_7 \\rangle,$ which will be joined with the series $\\langle \\nabla_1 + \\alpha \\nabla_4 + i \\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq 0}$ and $\\langle \\alpha \\nabla_4 + i \\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq 0}.$ Note that by above representatives $\\langle \\nabla_1 + i\\nabla_6 + \\nabla_7 \\rangle$ and $\\langle \\nabla_2 + i\\nabla_6 + \\nabla_7 \\rangle$ define the same orbit.\n\\end{enumerate}\n \n\\end{enumerate}\n\n\\end{enumerate}\n\nSummarizing, \nwe have the following distinct orbits\n\n\\[ \\langle \\nabla_4\\rangle,\n\\langle \\nabla_3+ \\nabla_4\\rangle,\n\\langle \\nabla_1+i\\nabla_4+\\nabla_5\\rangle,\n\\langle i\\nabla_4 +\\nabla_5\\rangle,\\]\n\n\\[ \\langle \\alpha \\nabla_4+\\beta\\nabla_5+\\nabla_6 \\rangle_{ \\beta \\in \\mathbb C_{\\geq 0}}, \\langle \\nabla_1+ \\alpha \\nabla_4+\\beta\\nabla_5+\\nabla_6 \\rangle_{\\alpha\\beta \\neq 0, \\beta \\in \\mathbb C_{\\geq 0}},\\]\n\\[ \\langle \\nabla_3+\\alpha\\nabla_4+\\nabla_6 \\rangle_{\\alpha \\neq 0}, \\langle \\nabla_1 +\\alpha\\nabla_3 - \\nabla_4 +\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}}, \\langle \\nabla_1 + \\alpha \\nabla_2 + i \\nabla_5 + \\nabla_6 \\rangle, \\langle \\nabla_2 +\\alpha\\nabla_5 +\\nabla_6 \\rangle_{\\alpha \\in \\mathbb C_{\\geq 0}}\\]\n\n\\[\\langle \\nabla_1+\\alpha\\nabla_4+i\\nabla_6 + \\nabla_7 \\rangle, \\langle \\alpha\\nabla_4+i\\nabla_6 + \\nabla_7 \\rangle,\\]\n\\[\\langle \\nabla_2+i\\alpha\\nabla_4 + \\alpha\\nabla_5+i\\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq 0}, \\langle i\\alpha\\nabla_4 + \\alpha\\nabla_5+i\\nabla_6 + \\nabla_7 \\rangle_{\\alpha \\neq 0},\\]\n\\[\\langle i \\nabla_1 + \\nabla_2 + i \\nabla_4 + \\nabla_5 + i\\nabla_6 + \\nabla_7\\rangle, \\langle \\pm i \\nabla_1 + \\nabla_2 - i \\nabla_4 - \\nabla_5 + i\\nabla_6 + \\nabla_7\\rangle\\]\n\nThe corresponding algebras are\n\n$$ \\begin{array}{lllllllllll}\n \\T {4}{18}&:& e_1 e_1 = e_3, & e_1e_3=e_4, & e_2 e_2=e_3; \\\\\n \\T {4}{19}&:& e_1 e_1 = e_3, & e_1e_3=e_4,& e_2e_1=e_4, & e_2 e_2=e_3; \\\\\n \\T {4}{20}&:& e_1 e_1 = e_3+e_4, & e_1e_3=ie_4, & e_2 e_2=e_3, & e_2e_3=e_4; \\\\\n \\T {4}{21}&:& e_1 e_1 = e_3, & e_1e_3=ie_4, & e_2 e_2=e_3,& e_2e_3=e_4; \\\\\n \\T {4}{22}(\\alpha,\\beta)_{\\beta\\in \\mathbb C_{\\geq0}}&:& e_1 e_1 = e_3, & e_1e_3=\\alpha e_4,& e_2 e_2=e_3, &e_2e_3=\\beta e_4,& e_3e_1=e_4; \\\\\n \\T {4}{23}(\\alpha,\\beta)_\n \\beta \\in \\mathbb C_{\\geq0}}&:& e_1 e_1 = e_3+e_4, & e_1e_3=\\alpha e_4,& e_2 e_2=e_3, &e_2e_3=\\beta e_4,& e_3e_1=e_4; \\\\\n \\T {4}{24}(\\alpha\n &:& e_1 e_1 = e_3,& e_1e_3 = \\alpha e_4, & e_2e_1 = e_4, & e_2e_2 = e_3, & e_3e_1 = e_4; \\\\\n \\T {4}{25}(\\alpha)_{\\alpha \\in \\mathbb C_{\\geq0}}&:& e_1 e_1 = e_3+e_4, & e_1e_3=- e_4,& e_2e_1 = \\alpha e_4, & e_2e_2 = e_3, & e_3e_1 = e_4; \\\\\n \\T {4}{26}(\\alpha)&:& e_1 e_1 = e_3+e_4, & e_1e_2=\\alpha e_4, & e_2 e_2=e_3, &e_2e_3=i e_4,& e_3e_1=e_4; \\\\\n \\T {4}{27}(\\alpha)_{\\alpha \\in \\mathbb C_{\\geq0}}&:& e_1 e_1 = e_3, & e_1e_2= e_4, & e_2 e_2=e_3,& e_2e_3=\\alpha e_4,& e_3e_1=e_4; \\\\\n \\T {4}{28}(\\alpha)&:& e_1 e_1 = e_3 + e_4, & e_1e_3=\\alpha e_4, & e_2 e_2=e_3,& e_3e_1=ie_4,& e_3e_2=e_4; \\\\\n \\T {4}{29}(\\alpha)&:& e_1 e_1 = e_3, & e_1e_3=\\alpha e_4, & e_2 e_2=e_3,& e_3e_1=ie_4,& e_3e_2=e_4; \\\\\n \\T {4}{30}(\\alpha\n &:& e_1 e_1 = e_3, & e_1 e_2 = e_4, & e_1e_3=i\\alpha e_4, & e_2 e_2=e_3,& e_2e_3=\\alpha e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\\\\n \\T {4}{31}(\\alpha\n &:& e_1 e_1 = e_3, & e_1e_3=i\\alpha e_4, & e_2 e_2=e_3,& e_2e_3=\\alpha e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\\\\n \\T {4}{32}&:& e_1 e_1 = e_3 + ie_4, & e_1 e_2 = e_4, & e_1e_3=i e_4, & e_2 e_2=e_3,& e_2e_3= e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\\\\n \\T {4}{33}&:& e_1 e_1 = e_3 + ie_4, & e_1 e_2 = e_4, & e_1e_3= -ie_4, & e_2 e_2=e_3,& e_2e_3= -e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\\\\n \\T {4}{34}&:& e_1 e_1 = e_3 -ie_4, & e_1 e_2 = e_4, & e_1e_3= -ie_4, & e_2 e_2=e_3,& e_2e_3= -e_4, & e_3e_1=ie_4,& e_3e_2=e_4.\n \\end{array}$$\n\n \nThe algebras above are pairwise non-isomorphic, except $\\T {4}{23}(\\alpha,0) \\cong \\T {4}{22}(\\alpha,0)$ for $\\alpha \\neq -1,$ $\\T {4}{23}(-1,0) \\cong \\T {4}{25}(0),$ $\\T {4}{23}(0,\\beta) \\cong \\T {4}{22}(0,\\beta)$ for $\\beta \\neq i,$ $\\T {4}{23}(0,i) \\cong \\T {4}{26}(0), \\T {4}{24}(0) \\cong \\T {4}{22}(0,0), \\T {4}{30}(0) \\cong \\T {4}{28}(0), \\T {4}{31}(0) \\cong \\T {4}{29}(0).$ \n\n\\newpage \n\\subsubsection{$1$-dimensional central extensions of $\\T {3*}{03}$}\nLet us use the following notations \n\\begin{align*}\n \\nb 1 = \\Dl 11, \\nb 2 = \\Dl 12, \\nb 3 = \\Dl 13 - \\Dl 31, \\nb 4 = \\Dl 22, \\nb 5 = \\Dl 23 - \\Dl 32, \\nb 6 = \\Dl 31, \\nb 7 = \\Dl 32. \n\\end{align*}\nTake $\\theta=\\sum_{i=1}^7\\alpha_i\\nb i\\in {\\rm H_T^2}(\\T {3*}{03}).$\nIf \n$$\n\\phi=\n\\begin{pmatrix}\nx & y & 0\\\\\nz & u & 0\\\\\nv & w & xu-yz\n\\end{pmatrix}\\in\\aut{\\T {3*}{03}},\n$$\nthen\n$$\n\\phi^T\\begin{pmatrix}\n\\alpha_1& \\alpha_2& \\alpha_3\\\\\n 0& \\alpha_4& \\alpha_5\\\\\n\\alpha_6-\\alpha_3& \\alpha_7-\\alpha_5& 0\n\\end{pmatrix} \\phi=\n\\begin{pmatrix}\n\\alpha_1^*& \\alpha_2^*-\\alpha^*& \\alpha_3^*\\\\\n\\alpha^*& \\alpha_4^*& \\alpha_5^*\\\\\n\\alpha_6^*-\\alpha_3^*& \\alpha_7^*-\\alpha_5^*& \\alpha^{**}\n\\end{pmatrix},\n$$\nwhere\n\\begin{align*}\n\\alpha^*_1 &= \\alpha_1x^2 + \\alpha_2xz + \\alpha_4z^2 + v(\\alpha_6x + \\alpha_7z),\\\\\n\\alpha^*_2 &= x(2\\alpha_1y + \\alpha_2u) + z(\\alpha_2y + 2\\alpha_4u) + w(\\alpha_6x +\\alpha_7z) + v(\\alpha_6y + \\alpha_7u),\\\\\n\\alpha^*_3 &= (\\alpha_3x+\\alpha_5z)(xu - yz),\\\\\n\\alpha^*_4 &= y(\\alpha_1y + \\alpha_2u) + \\alpha_4u^2 + w(\\alpha_6y + \\alpha_7u),\\\\\n\\alpha^*_5 &= (\\alpha_3y+\\alpha_5u)(xu - yz),\\\\\n\\alpha^*_6 &= (\\alpha_6x+\\alpha_7z)(xu - yz),\\\\\n\\alpha^*_7 &= (\\alpha_6y+\\alpha_7u)(xu - yz).\n\\end{align*}\nHence, $\\phi\\langle\\theta\\rangle=\\langle\\theta^*\\rangle,$ where $\\theta^*=\\sum\\limits_{i=1}^7 \\alpha_i^* \\nb i.$\n\n We are interested in $\\theta$ with $(\\alpha_6,\\alpha_7) \\neq (0,0).$ Moreover, the condition $\\theta \\in \\mathbf{T}_1 (\\T{3*}{03})$ gives us $(\\alpha_3,\\alpha_5, \\alpha_6 -\\alpha_3, \\alpha_7 - \\alpha_5) \\neq (0,0,0,0).$\n\n If $\\alpha_7\\neq 0$, then taking $u=-\\frac{\\alpha_6y}{\\alpha_7}$ we have $\\alpha^*_7=0$, so we shall assume that $\\alpha^*_6\\ne 0$ and $\\alpha^*_7=0$ from the very beginning. Choosing $y=0$, $z=-\\frac{\\alpha_3x}{\\alpha_5}$, $u= \\frac{\\alpha_6x}{\\alpha_5}$, $v = \\frac x{\\alpha_5^2\\alpha_6}(\\alpha_2\\alpha_3\\alpha_5-\\alpha_3^2\\alpha_4 - \\alpha_1\\alpha_5^2)$, $w = \\frac x{\\alpha_5^2}(2\\alpha_3\\alpha_4 - \\alpha_2\\alpha_5)$, we get the family of representatives \n $ \\langle \\frac{\\alpha_4}{\\alpha_5 x}\\nabla_4 + \\nabla_5+ \\nabla_6\\rangle$. It gives two distinct representatives $ \\langle \\nabla_5+ \\nabla_6\\rangle$ and $ \\langle \\nabla_4 + \\nabla_5+ \\nabla_6\\rangle$ depending on whether $\\alpha_4=0$ or not.\n \n The algebras corresponding to $ \\langle \\nabla_4 + \\nabla_5+ \\nabla_6\\rangle$ and $ \\langle \\nabla_5+ \\nabla_6\\rangle$ are:\n \n \n \n \n \n \n \n \n \n\n\n\n\n\n\n \n \n\n\n\n \n\n\n\n\n\n\n$$ \n\\begin{array}{lllllllll}\n\\T {4}{35} &:& e_1 e_2=e_3, & e_2 e_1=-e_3,& e_2e_2=e_4,& e_2e_3=e_4,& e_3e_1=e_4,& e_3e_2=-e_4;\\\\\n\\T {4}{36} &:& e_1 e_2=e_3, & e_2 e_1=-e_3,& e_2e_3=e_4,& e_3e_1=e_4,& e_3e_2=-e_4.\n \\end{array} \n $$\n\n\n\n\n\n\\subsubsection{$1$-dimensional central extensions of $\\T {3*}{04}$}\nLet us use the following notations \n\\begin{align*}\n \\nb 1 = \\Dl 11, \\nb 2 = \\Dl 12, \\nb 3 = \\Dl 13, \\nb 4 = \\Dl 21,\n\t\\nb 5 = \\Dl 23, \\nb 6 = \\Dl 31, \\nb 7 = \\Dl 32. \n\\end{align*}\nTake $\\theta=\\sum_{i=1}^7\\alpha_i\\nb i\\in {\\rm H_T^2}(\\T {3*}{04}).$\nIf \n$$\n\\phi=\n \\begin{pmatrix}\n x & y & 0\\\\\n -\\lambda y & x-y & 0\\\\\n z & u & x^2-xy+\\lambda y^2\n \\end{pmatrix}\\in\\aut{\\T 3{04}},\n$$\nthen\n$$\n\\phi^T\\begin{pmatrix}\n\t\\alpha_1 & \\alpha_2 & \\alpha_3\\\\\n\t\\alpha_4 & 0 & \\alpha_5\\\\\n\t\\alpha_6 & \\alpha_7 & 0\n\t\\end{pmatrix} \\phi=\n\t\\begin{pmatrix}\n\t\\alpha_1^*+\\lambda\\alpha^* & \\alpha_2^* & \\alpha_3^*\\\\\n\t\\alpha_4^*+\\alpha^* & \\alpha^* & \\alpha_5^*\\\\\n\t\\alpha_6^* & \\alpha_7^* & 0\n\t\\end{pmatrix},\n$$\nwhere\n\\begin{align*}\n\t\\alpha^*_1 &= \\alpha_1x^2 + \\lambda(-\\alpha_1 + \\alpha_2 + \\alpha_4) y^2 -2\\lambda(\\alpha_2 + \\alpha_4) xy + (\\alpha_3 + \\alpha_6)xz - (\\alpha_5 + \\alpha_7) yz\\\\\n\t&\\quad- \\lambda(\\alpha_5 + \\alpha_7) ux + \\lambda(- \\alpha_3 + \\alpha_5 - \\alpha_6 + \\alpha_7) uy,\\\\\n \\alpha^*_2 &= \\alpha_2x^2 - \\lambda\\alpha_4 y^2 + (\\alpha_1 - \\alpha_2)xy + \\alpha_7xz + (\\alpha_6 - \\alpha_7)yz -\\alpha_5\\lambda uy + \\alpha_3ux,\\\\\n \\alpha^*_3 &= (x^2 - xy + \\lambda y^2)(\\alpha_3x -\\lambda\\alpha_5 y),\\\\\n \\alpha^*_4 &= \\alpha_4x^2 + (- \\alpha_1 + (1-\\lambda)\\alpha_2 + \\alpha_4)y^2 + (\\alpha_1 - \\alpha_2 - 2\\alpha_4)xy + (\\alpha_3 - \\alpha_5)yz + \\alpha_5xz\\\\\n &\\quad - \\alpha_7\\lambda uy + (- \\alpha_5 + \\alpha_6 - \\alpha_7)ux + (- \\alpha_3 + \\alpha_5 - \\alpha_6 + \\alpha_7)uy,\\\\\n \\alpha^*_5 &= (x^2 - xy + \\lambda y^2)(\\alpha_5x + (\\alpha_3 - \\alpha_5)y),\\\\\n \\alpha^*_6 &= (x^2 - xy + \\lambda y^2)(\\alpha_6x -\\lambda\\alpha_7 y),\\\\\n \\alpha^*_7 &= (x^2 - xy + \\lambda y^2)(\\alpha_7x + (\\alpha_6-\\alpha_7)y).\n\t\\end{align*}\nHence, $\\phi\\langle\\theta\\rangle=\\langle\\theta^*\\rangle,$ where $\\theta^*=\\sum\\limits_{i=1}^7 \\alpha_i^* \\nb i.$\n\n We are interested in $\\theta$ with $(\\alpha_3,\\alpha_5,\\alpha_6,\\alpha_7) \\neq (0,0,0,0)$ (if $\\lambda \\neq 0$) and $(\\alpha_3,\\alpha_6,\\alpha_7) \\neq (0,0,0)$ (if $\\lambda = 0$). Moreover, the condition $\\theta \\in \\mathbf{T}_1 (\\T{3*}{04})$ gives us $(\\alpha_4,\\alpha_5,\\alpha_6,\\alpha_7) \\neq (0,0,0,0).$\n \n \\begin{enumerate}\n \n \\item $\\alpha_7\\ne 0$. We have the following subcases.\n\t\\begin{enumerate}\n\t \\item $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2\\ne 0$. Choosing $x=-\\frac{(\\alpha_6-\\alpha_7)y}{\\alpha_7}$, we have $\\alpha^*_7=0$. Since $(\\alpha^*_6)^2 - \\alpha^*_6\\alpha^*_7 + \\lambda(\\alpha^*_7)^2=(\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2)(x^2 - xy + \\lambda y^2)^3$, the condition $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2\\ne 0$ is invariant under the action of the automorphism group. Thus, we may assume that $\\alpha_7=0$ and $\\alpha_6\\ne 0$ from the very beginning. Then choosing $y=0$ we obtain $\\alpha^*_7=0$ and\n \\begin{align*}\n\t \\alpha^*_1 &= x(\\alpha_1x + (\\alpha_3 + \\alpha_6)z - \\lambda\\alpha_5u),\\\\\n \\alpha^*_2 &= x(\\alpha_2x + \\alpha_3u),\\\\\n \\alpha^*_3 &= \\alpha_3 x^3,\\\\\n \\alpha^*_4 &= x(\\alpha_4x + \\alpha_5z + (- \\alpha_5 + \\alpha_6)u),\\\\\n \\alpha^*_5 &= \\alpha_5 x^3,\\\\\n \\alpha^*_6 &= \\alpha_6 x^3.\n\t \\end{align*}\n\t \\begin{enumerate}\n\t \\item $\\alpha_5\\ne 0$ and $\\alpha_3\\ne 0$. Then we choose $u=-\\frac{\\alpha_2x}{\\alpha_3}$ and $z=-\\frac{\\alpha_4x+(- \\alpha_5 + \\alpha_6)u}{\\alpha_5}$ and obtain the family of representatives $\\langle\\alpha\\nb 1+\\beta\\nb 3+\\gamma\\nb 5+\\nb 6\\rangle_{\\beta,\\gamma\\ne 0}$. It gives two families of representatives of distinct orbits: $\\langle\\nb 1+\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0}$ and $\\langle\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0}$.\n\t \n\t \\item $\\alpha_5\\ne 0$ and $\\alpha_3=0$. Then $\\alpha^*_3=0$ and $\\alpha^*_4$, $\\alpha^*_5$, $\\alpha^*_6$ are as above and\n\t \\begin{align*}\n\t \\alpha^*_1 &= x(\\alpha_1x + \\alpha_6z - \\lambda\\alpha_5u),\\\\\n \\alpha^*_2 &= \\alpha_2x^2.\n\t \\end{align*}\n\t Choosing $z=\\frac{\\lambda\\alpha_5u-\\alpha_1x}{\\alpha_6}$, we have $\\alpha^*_1=0$. Now we have the following subcases:\n\t \\begin{enumerate}\n\t \\item $\\alpha_6^2 - \\alpha_5\\alpha_6 + \\lambda\\alpha_5^2\\ne 0$. Then choosing $u=\\frac{(\\alpha_1\\alpha_5 - \\alpha_4\\alpha_6)x}{\\alpha_6^2 - \\alpha_5\\alpha_6 + \\lambda\\alpha_5^2}$ we have $\\alpha^*_4=0$, so we obtain the family of representatives $\\langle\\alpha\\nb 2+\\beta\\nb 5+\\nb 6\\rangle_{1-\\beta+\\lambda\\beta^2\\ne 0}$. It determines two families of representatives of distinct orbits: $\\langle\\nb 2+\\alpha\\nb 5+\\nb 6\\rangle_{\\alpha \\neq 0, 1-\\alpha+\\lambda\\alpha^2\\ne 0}$ and $\\langle\\alpha\\nb 5+\\nb 6\\rangle_{\\alpha \\neq 0, 1-\\alpha+\\lambda\\alpha^2\\ne 0}$.\n\t \\item $\\alpha_6^2 - \\alpha_5\\alpha_6 + \\lambda\\alpha_5^2=0$. \n\t \n\t If $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we have two families of representatives of distinct orbits\n\t $\\Big\\langle \\lambda\\nb 2+\\lambda\\alpha\\nb 4+\\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 5+\\lambda\\nb 6\\Big\\rangle, \\Big\\langle\\nb 2+\\alpha\\nb 4+\\frac{2}{1+\\sqrt{1-4\\lambda}}\\nb 5+\\nb 6\\Big\\rangle$ and 4 separate representatives\n\t $\\Big\\langle\\lambda \\nb 4+\\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 5+\\lambda \\nb 6\\Big\\rangle, \\Big\\langle\\nb 4+\\frac{2}{1+\\sqrt{1-4\\lambda}}\\nb 5+\\nb 6\\Big\\rangle,$ $\\Big\\langle\\frac{1\\pm\\sqrt{1-4\\lambda}}{2\\lambda}\\nb 5+\\nb 6\\Big\\rangle,$ the last two belonging to the family above.\n\t \n\t If $\\lambda=\\frac 14$, then we have the family of representatives of distinct orbits $\\langle\\nb 2+\\alpha\\nb 4+2\\nb 5+\\nb 6\\rangle$ and two separate representatives $\\langle\\nb 4+2\\nb 5+\\nb 6\\rangle$ and $\\langle 2\\nb 5+\\nb 6\\rangle$. \n\t \n\t If $\\lambda=0$, then we have $\\alpha_5=\\alpha_6$, so we obtain the family of representatives of distinct orbits $\\langle\\nb 2+\\alpha\\nb 4+\\nb 5+\\nb 6\\rangle$ and two separate representatives $\\langle\\nb 4+\\nb 5+\\nb 6\\rangle$ and $\\langle\\nb 5+\\nb 6\\rangle$.\n\t \n\t \\end{enumerate}\n\t \n\t \\item $\\alpha_3\\ne 0$ and $\\alpha_5=0$. Then $\\alpha^*_5=0$, $\\alpha^*_2$, $\\alpha^*_3$ and $\\alpha^*_6$ are as above and\n\t \\begin{align*}\n\t \\alpha^*_1 &= x(\\alpha_1x + (\\alpha_3 + \\alpha_6)z),\\\\\n \\alpha^*_4 &= x(\\alpha_4x + \\alpha_6u).\n\t \\end{align*}\n\t Choosing $u=-\\frac{\\alpha_4x}{\\alpha_6}$, we get $\\alpha^*_4=0$. Now we have two subcases:\n\t \\begin{enumerate}\n\t \\item $\\alpha_3 + \\alpha_6\\ne 0$. Then choosing $z=-\\frac{\\alpha_1x}{\\alpha_3 + \\alpha_6}$, we obtain two families of representatives of distinct orbits $\\langle \\nb 2+\\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\not\\in \\{0,-1\\}}$ and $\\langle \\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\not\\in \\{0,-1\\}}$.\n\t \\item $\\alpha_3 + \\alpha_6=0$. Then choosing $z=0$, we have the family of representatives of distinct orbits $\\langle\\nb 1+\\alpha\\nb 2-\\nb 3+\\nb 6\\rangle$ and two separate representatives $\\langle\\nb 2-\\nb 3+\\nb 6\\rangle$ and $\\langle-\\nb 3+\\nb 6\\rangle$ which will be joined with the families $\\langle \\nb 2+\\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\not\\in \\{0,-1\\}}$ and $\\langle \\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\not\\in \\{0,-1\\}}$ found above.\n\t \\end{enumerate}\n\t \n\t \\item $\\alpha_3=\\alpha_5=0$. Then $\\alpha^*_3=\\alpha^*_5=0$, $\\alpha^*_6$ is as above and \n\t \\begin{align*}\n\t \\alpha^*_1 &= x(\\alpha_1x + \\alpha_6z),\\\\\n \\alpha^*_2 &= \\alpha_2x^2,\\\\\n \\alpha^*_4 &= x(\\alpha_4x + \\alpha_6u).\n\t \\end{align*}\n\t Thus, choosing $z=-\\frac{\\alpha_1x}{\\alpha_6}$ and $u=-\\frac{\\alpha_4x}{\\alpha_6}$, we have two representatives depending on whether $\\alpha_2=0$ or not: $\\langle\\nb 2+\\nb 6\\rangle$ and $\\langle\\nb 6\\rangle$. They will be joined with the families $\\langle\\nb 2+\\alpha\\nb 5+\\nb 6\\rangle_{\\alpha \\neq 0, 1-\\alpha+\\lambda\\alpha^2\\ne 0}$ and $\\langle\\alpha\\nb 5+\\nb 6\\rangle_{\\alpha \\neq 0, 1-\\alpha+\\lambda\\alpha^2\\ne 0}$ found above.\n\t \\end{enumerate}\n\t \n\t \\item $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2=0$ and $\\alpha_5\\ne 0$ and $\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2\\ne 0$. Then choosing $x=-\\frac{(\\alpha_3 - \\alpha_5)y}{\\alpha_5}$, we get $\\alpha^*_5=0$. Since $(\\alpha^*_3)^2 - \\alpha^*_3\\alpha^*_5 + \\lambda(\\alpha^*_5)^2=(\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2)(x^2-xy+\\lambda y^2)^3\\ne 0$, we may assume that $\\alpha_5=0$ and $\\alpha_3\\ne 0$ from the very beginning. Then choosing $y=0$ and $z=-\\frac{\\alpha_2x+ \\alpha_3u}{\\alpha_7}$ we have $\\alpha^*_2=\\alpha^*_5=0$ and \n\t \\begin{align*}\n\t \\alpha^*_1 &= \\frac x{\\alpha_7}((\\alpha_1\\alpha_7- \\alpha_2\\alpha_3 - \\alpha_2\\alpha_6)x - (\\alpha_3^2 + \\alpha_3\\alpha_6 - \\alpha_6^2 + \\alpha_6\\alpha_7)u),\\\\\n \\alpha^*_3 &= \\alpha_3x^3,\\\\\n \\alpha^*_4 &= x(\\alpha_4x - (\\alpha_6 - \\alpha_7)u),\\\\\n \\alpha^*_6 &= \\alpha_6x^3,\\\\\n \\alpha^*_7 &= \\alpha_7x^3.\n\t \\end{align*}\n\t \\begin{enumerate}\n\t \\item $\\alpha_6 - \\alpha_7\\ne 0$. Then choosing $u=\\frac{\\alpha_4x}{\\alpha_6 - \\alpha_7}$ we have $\\alpha^*_4=0$.\n\t \n\t If $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we have four families of representatives of distinct orbits $\\Big\\langle \\nb 1+\\alpha\\nb 3+ \\frac{1\\pm\\sqrt{1-4\\lambda}}{2}\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0}$, $\\Big\\langle \\alpha\\nb 3+ \\frac{1\\pm\\sqrt{1-4\\lambda}}{2}\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0}$.\n\t \n\t \tIf $\\lambda=\\frac 14$, then $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2=0$ implies $\\alpha_6=\\frac 12\\alpha_7$, so $\\alpha_6 - \\alpha_7\\ne 0$ is satisfied. Thus, we have two families of representatives of distinct orbits $\\langle\\nb 1+\\alpha\\nb 3+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0}$ and $\\langle\\alpha\\nb 3+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0}$.\n\t \t \n\t If $\\lambda = 0$, then $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2=0$ and $\\alpha_6 - \\alpha_7\\ne 0$ imply that $\\alpha_6=0$. Hence, we have two families of representatives of distinct orbits $\\langle\\nb 1+\\alpha\\nb 3+\\nb 7\\rangle_{\\alpha\\ne 0}$ and $\\langle\\alpha\\nb 3+\\nb 7\\rangle_{\\alpha\\ne 0}$.\n\t \n\t \\item $\\alpha_6 - \\alpha_7=0$. Then $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2=0$ and $\\alpha_7\\ne 0$ imply that $\\lambda = 0$. Now,\n\t \\begin{align*}\n\t \\alpha^*_1 &= \\frac x{\\alpha_7}((\\alpha_1\\alpha_7- \\alpha_2\\alpha_3 - \\alpha_2\\alpha_7)x - \\alpha_3(\\alpha_3 + \\alpha_7)u),\\\\\n \\alpha^*_3 &= \\alpha_3x^3,\\\\\n \\alpha^*_4 &= \\alpha_4x^2,\\\\\n \\alpha^*_6 &= \\alpha_7x^3,\\\\\n \\alpha^*_7 &= \\alpha_7x^3.\n\t \\end{align*}\n\t \\begin{enumerate}\n\t \\item $\\alpha_3 + \\alpha_7\\ne 0$. Then choosing $u=\\frac{(\\alpha_1\\alpha_7- \\alpha_2\\alpha_3 - \\alpha_2\\alpha_7)x}{\\alpha_3(\\alpha_3 + \\alpha_7)}$, we have $\\alpha^*_1=0$. Thus, we obtain two families of representatives of distinct orbits $\\langle\\alpha\\nb 3+\\nb 4+\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0,-1}$ and $\\langle\\alpha\\nb 3+\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0,-1}$.\n\t \\item $\\alpha_3 + \\alpha_7=0$. Then we obtain the family of representatives of distinct orbits $\\langle\\nb 1-\\nb 3+\\alpha\\nb 4+\\nb 6+\\nb 7\\rangle$ and two separate representatives $\\langle-\\nb 3+\\nb 4+\\nb 6+\\nb 7\\rangle$ and $\\langle-\\nb 3+\\nb 6+\\nb 7\\rangle,$ which will be joined with the families $\\langle\\alpha\\nb 3+\\nb 4+\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0,-1}$ and $\\langle\\alpha\\nb 3+\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0,-1}$ found above.\n\t \\end{enumerate}\n\t \\end{enumerate}\n\t \\item $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2=0$ and $\\alpha_5\\ne 0$ and $\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2=0$. Choosing $x,y$ such that $(x^2 - xy + \\lambda y^2)(\\alpha_7x + (\\alpha_6-\\alpha_7)y)=1$, we have $(\\alpha_3 - \\alpha_5)y + \\alpha_5x\\ne 0$, since otherwise $x^2 - xy + \\lambda y^2=(\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2)\\frac{y^2}{\\alpha_5^2}=0$. Now, the suitable value of $z$ gives $\\alpha^*_4=0$, so we shall assume $\\alpha_7=1$ and $\\alpha_4=0$ from the very beginning. The equality $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2=0$ takes the form $\\alpha_6^2 - \\alpha_6 + \\lambda=0$, whence $\\lambda=\\alpha_6-\\alpha_6^2$. On the other hand, $\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2=0$ implies $\\lambda=\\frac{\\alpha_3\\alpha_5-\\alpha_3^2}{\\alpha_5^2}$. Therefore, $\\alpha_6-\\alpha_6^2=\\frac{\\alpha_3\\alpha_5-\\alpha_3^2}{\\alpha_5^2}$, i.e. $\\alpha_3^2-\\alpha_3\\alpha_5+\\alpha_5^2(\\alpha_6-\\alpha_6^2)=0$. This equation has two solutions in $\\alpha_3$, namely, $\\alpha_3=\\alpha_5\\alpha_6$ and $\\alpha_3=\\alpha_5(1-\\alpha_6)$.\n\t \n\t \\begin{enumerate}\n\t \t\\item $\\alpha_3=\\alpha_5\\alpha_6$. Then\n\t \t\\begin{align*}\n\t \t\t\\alpha^*_2 &= (x + (\\alpha_6-1)y)z + \\alpha_5\\alpha_6(x + (\\alpha_6-1)y)u + \\alpha_2x^2 + \\alpha_1xy - \\alpha_2xy,\\\\\n\t \t\t\\alpha^*_4 &= \\alpha_5(x + (\\alpha_6-1)y)z + ((\\alpha_6 - \\alpha_5 - 1)x + (\\alpha_6^2 - \\alpha_5\\alpha_6 + \\alpha_5 - 2\\alpha_6 + 1)y)u\\\\\n\t \t\t&\\quad + (\\alpha_1-\\alpha_2)xy + (\\alpha_2 + \\alpha_2\\alpha_6^2 - \\alpha_1 - \\alpha_2\\alpha_6)y^2. \n\t \t\\end{align*}\n\t \tWe consider $\\alpha^*_2=\\alpha^*_4=0$ as a system of linear equations in $z$ and $u$. Its determinant is $(\\alpha_5 + 1)(\\alpha_6 - \\alpha_5\\alpha_6 - 1)( x + (\\alpha_6-1)y)^2$. So, we have the following cases:\n\t \t\\begin{enumerate}\n\t \t\t\\item $\\alpha_5 + 1\\ne 0$ and $\\alpha_6 - \\alpha_5\\alpha_6 - 1\\ne 0$. Then choosing $x,y$ such that $x + (\\alpha_6-1)y\\ne 0$, $x - \\alpha_6y\\ne 0$ and $z,u$ such that $\\alpha^*_2=\\alpha^*_4=0$ we obtain \n\t \t\t\\begin{align*}\n\t \t\t\t\\alpha^*_1 &= (\\alpha_1 - \\alpha_2\\alpha_6)(x - \\alpha_6y)^2,\\\\\n\t \t\t\t\\alpha^*_3 &= \\alpha_5\\alpha_6(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t \t\t\t\\alpha^*_5 &= \\alpha_5(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t \t\t\t\\alpha^*_6 &= \\alpha_6(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t \t\t\t\\alpha^*_7 &= (x - \\alpha_6y)(x + (\\alpha_6-1)y)^2.\n\t \t\t\\end{align*}\n\t \t\t\\begin{itemize}\n\t \t\t \\item Let $\\alpha_1 - \\alpha_2\\alpha_6\\ne 0$. Taking $y=0$ and $x=\\alpha_1 - \\alpha_2\\alpha_6$ we obtain the family of representatives $\\langle\\nb 1+\\alpha\\beta\\nb 3+\\alpha\\nb 5+\\beta\\nb 6+\\nb 7\\rangle$, where $\\alpha\\not\\in \\{0,-1\\}$, $\\beta-\\alpha\\beta-1\\ne 0$ and $\\beta^2 - \\beta + \\lambda=0$.\n\t \t\t\n\t \t\tIf $\\lambda\\not\\in\\{0,\\frac 14\\}$, then $\\beta=\\frac{1\\pm\\sqrt{1-4\\lambda}}2$. Observe that $\\beta\\ne 0$, so the condition $\\beta-\\alpha\\beta-1\\ne 0$ is equivalent to $\\alpha\\ne 1-\\frac 1\\beta=\\frac{\\sqrt{1-4\\lambda}\\mp 1}{\\sqrt{1-4\\lambda}\\pm 1}$. Thus, we obtain two families $\\Big\\langle\\nb 1+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\not\\in\\left\\{0,-1,\\frac{\\sqrt{1-4\\lambda}\\mp 1}{\\sqrt{1-4\\lambda}\\pm 1}\\right\\}}$ of representatives of distinct orbits.\n\t \t\t\n\t \t\tIf $\\lambda=\\frac 14$, then $\\beta=\\frac 12$, and $\\alpha\\ne 1-\\frac 1\\beta$ becomes $\\alpha\\ne -1$. Thus, we obtain the family $\\langle\\nb 1+\\frac 12\\alpha\\nb 3+\\alpha\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\not\\in\\{0,-1\\}}$ of representatives of distinct orbits.\n\t \t\t\n\t \t\tIf $\\lambda=0$, then $\\beta=0$ or $\\beta=1$. If $\\beta=0$, then the condition $\\beta-\\alpha\\beta-1\\ne 0$ becomes $-1\\ne 0$; and if $\\beta=1$, then it is $\\alpha\\ne 0$. Thus, we obtain two families $\\langle\\nb 1+\\alpha\\nb 5+\\nb 7\\rangle_{\\alpha\\not\\in\\{0,-1\\}}$ and $\\langle\\nb 1+\\alpha\\nb 3+\\alpha\\nb 5+\\nb 6+\\nb 7\\rangle_{\\alpha\\not\\in\\{0,-1\\}}$ of representatives of distinct orbits.\n\t \t\t \\item Let $\\alpha_1 - \\alpha_2\\alpha_6=0$. Then we obtain the same families as in the previous case, but without $\\nb 1$.\n\t \t\t\\end{itemize}\n\t \n\t \t\t\\item $\\alpha_6 - \\alpha_5\\alpha_6 - 1=0$. Then clearly $\\alpha_6\\ne 0$ and $\\alpha_5=1-\\frac 1{\\alpha_6}$, so $\\alpha_3=\\alpha_6-1$. Moreover, $\\alpha_6\\ne 1$ since $\\alpha_5\\ne 0$. In particular, $\\lambda\\ne 0$. Choosing $z$ such that $\\alpha^*_4=0$, we obtain\n\t \t\t\\begin{align*}\n\t \t\t\t\\alpha^*_1 &= \\frac{x - \\alpha_6y}{\\alpha_6 - 1}(\\alpha_1(\\alpha_6-1)x + \\alpha_6(\\alpha_2(2\\alpha_6^2 + 1) - \\alpha_6(\\alpha_1 + 2\\alpha_2))y),\\\\\n\t \t\t\t\\alpha^*_2 &= \\frac{x - \\alpha_6y}{\\alpha_6 - 1}(\\alpha_2(\\alpha_6-1)x + (\\alpha_2\\alpha_6(\\alpha_6-1) - \\alpha_1 + \\alpha_2)y),\\\\\n\t \t\t\t\\alpha^*_3 &= (\\alpha_6 - 1)(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t \t\t\t\\alpha^*_5 &= \\left(1 - \\frac 1{\\alpha_6}\\right)(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t \t\t\t\\alpha^*_6 &= \\alpha_6(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t \t\t\t\\alpha^*_7 &= (x - \\alpha_6y)(x + (\\alpha_6-1)y)^2.\n\t \t\t\\end{align*}\n\t \t\tLet $(\\alpha_1,\\alpha_2) =(0,0)$. Then we get the representative $\\Big\\langle(\\alpha-1)\\nb 3+\\left(1-\\frac 1{\\alpha}\\right)\\nb 5+\\alpha\\nb 6+\\nb 7\\Big\\rangle$, where $\\alpha^2 - \\alpha + \\lambda=0$.\n\t\t\t\t\n\t\t\t\tIf $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we obtain two representatives $\\Big\\langle\\frac{-1\\pm\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}\\mp 1}{\\sqrt{1-4\\lambda}\\pm 1}\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle$ which will be joined with the families $\\Big\\langle\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\not\\in\\left\\{0,-1,\\frac{\\sqrt{1-4\\lambda}\\mp 1}{\\sqrt{1-4\\lambda}\\pm 1}\\right\\}}$.\n\t\t\t\t\n\t\t\t\tIf $\\lambda=\\frac 14$, then we obtain the representative $\\langle -\\frac 12\\nb 3-\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle$ which will be joined with the family $\\langle\\frac 12\\alpha\\nb 3+\\alpha\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\not\\in\\{0,-1\\}}.$\n\t\t\t\n\t\t\t\t\n\t\t\t\tLet $(\\alpha_1,\\alpha_2) \\neq (0,0).$ If $\\alpha_2 =0,$ then \n\t\t\t\t\\begin{align*}\n\t\t\t\t\\alpha^*_1 &= \\frac{x - \\alpha_6y}{\\alpha_6 - 1}((\\alpha_6-1)x - \\alpha_6^2y)\\alpha_1,\\\\\n\t\t\t\t\\alpha^*_2 &= -\\frac{x - \\alpha_6y}{\\alpha_6 - 1}y\\alpha_1,\n\t\t\t\t\\end{align*}\n\t\t\t\tso choosing $y \\neq 0$ we may suppose for the rest of the case that $\\alpha_2 \\neq 0.$\n\\begin{itemize}\n \\item Let $\\alpha_1-\\alpha_2\\alpha_6\\ne 0$ and $\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6\\ne 0$. Then choosing $z=\\frac{\\alpha_6(\\alpha_2\\alpha_6(\\alpha_6-1) - \\alpha_1 + \\alpha_2)y}{\\alpha_6 - 1}-(\\alpha_6-1)u$ and $x=\\frac{(\\alpha_1 - \\alpha_2 -\\alpha_2\\alpha_6(\\alpha_6-1))y}{\\alpha_2(\\alpha_6 - 1)}$ we obtain $\\alpha^*_2=\\alpha^*_4=0$ and \n\t \t\t\\begin{align*}\n\t \t\t\t\\alpha^*_1 &= (\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6)^2(\\alpha_1 - \\alpha_2\\alpha_6)\\frac{y^2}{\\alpha_2^2(\\alpha_6 - 1)^2},\\\\\n\t \t\t\t\\alpha^*_3 &= (\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6)(\\alpha_1 - \\alpha_2\\alpha_6)^2\\frac{y^3}{\\alpha_2^3(\\alpha_6 - 1)^2},\\\\\n\t \t\t\t\\alpha^*_5 &= (\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6)(\\alpha_1 - \\alpha_2\\alpha_6)^2\\frac{y^3}{\\alpha_2^3\\alpha_6(\\alpha_6 - 1)^2},\\\\\n\t \t\t\t\\alpha^*_6 &= (\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6)(\\alpha_1 - \\alpha_2\\alpha_6)^2\\frac{\\alpha_6y^3}{\\alpha_2^3(\\alpha_6 - 1)^3},\\\\\n\t \t\t\t\\alpha^*_7 &= (\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6)(\\alpha_1 - \\alpha_2\\alpha_6)^2\\frac{y^3}{\\alpha_2^3(\\alpha_6 - 1)^3}.\n\t \t\t\\end{align*}\n\t \t\tChoosing $y=\\frac{\\alpha_2(\\alpha_6 - 1)(\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6)}{\\alpha_1 - \\alpha_2\\alpha_6}$ we obtain the representative $\\Big\\langle\\nb 1+(\\alpha-1)\\nb 3+\\left(1-\\frac 1{\\alpha}\\right)\\nb 5+\\alpha\\nb 6+\\nb 7\\Big\\rangle$, where $\\alpha^2 - \\alpha + \\lambda=0$.\n\t \t\t\n\t \t\tIf $\\lambda\\not\\in\\{0,\\frac 14\\}$, then $\\alpha=\\frac{1\\pm\\sqrt{1-4\\lambda}}2$, so $\\alpha-1=\\frac{-1\\pm\\sqrt{1-4\\lambda}}2$ and $1-\\frac 1{\\alpha}=\\frac{\\sqrt{1-4\\lambda}\\mp 1}{\\sqrt{1-4\\lambda}\\pm 1}$, and we obtain the following two representatives $\\Big\\langle \\nb 1+\\frac{-1\\pm\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}\\mp 1}{\\sqrt{1-4\\lambda}\\pm 1}\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle$ which will be joined with the families $\\Big\\langle\\nb 1+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\not\\in\\left\\{0,-1,\\frac{\\sqrt{1-4\\lambda}\\mp 1}{\\sqrt{1-4\\lambda}\\pm 1}\\right\\}}$ found above.\n\t \t\t\n\t \t\tIf $\\lambda=\\frac 14$, then $\\alpha=\\frac 12$, so $\\alpha-1=-\\frac 12$ and $1-\\frac 1{\\alpha}=-1$, and we obtain the representative $\\langle\\nb 1-\\frac 12\\nb 3-\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle$ which will be joined with the family $\\langle\\nb 1+\\frac 12\\alpha\\nb 3+\\alpha\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\not\\in\\{0,-1\\}}$ found above.\n \\item Let $\\alpha_1 - \\alpha_2 - 2\\alpha_2\\alpha_6^2 + 2\\alpha_2\\alpha_6=0$. Then $\\alpha_1=(2\\alpha_6^2 - 2\\alpha_6 + 1)\\alpha_2$.\n\t \n\t \n\t \n\t\n\t\n\n\n\t\n\t\n\t \t\t\n\t\n\t \t\t\n\t\n\t \n\t \n\tIn this case $\\alpha_3^*, \\alpha_5^*, \\alpha_6^*, \\alpha_7^*$ are as above and\n\t\t\\begin{align*}\n\t \t\t\\alpha^*_1 &= \\alpha_2 (x - y \\alpha_6)^2 (2\\alpha_6^2 - 2\\alpha_6 + 1)\\\\\n\t \t\t\\alpha^*_2 &=\\alpha_2 (x - y \\alpha_6)^2.\n\t \t\t\\end{align*}\n\tTherefore, we obtain the representative $\\Big\\langle (2 \\alpha^2-2\\alpha + 1)\\nabla_1+\\nabla_2+ (\\alpha-1)\\nabla_3 + (1-\\frac{1}{\\alpha})\\nabla_5 +\\alpha\\nabla_6+\\nabla_7\\Big\\rangle$, where $\\alpha^2 - \\alpha + \\lambda=0$. Note that $2 \\alpha^2-2\\alpha + 1 = 1- 2\\lambda.$\n\t\n\tIf $\\lambda\\not\\in\\{0,\\frac 14\\},$ then we obtain two representatives $\\Big\\langle (1-2\\lambda)\\nabla_1 + \\nabla_2 +\\frac{-1+\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}- 1}{\\sqrt{1-4\\lambda}+ 1}\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle, \\Big\\langle (1-2\\lambda)\\nabla_1 + \\nabla_2 +\\frac{-1-\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}+ 1}{\\sqrt{1-4\\lambda}- 1}\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle = \\Big\\langle (1-2\\lambda)\\lambda\\nabla_1 + \\lambda\\nabla_2 -\\lambda\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\Big(\\frac{\\sqrt{1-4\\lambda}+ 1}{2}\\Big)^2\\nb 5+\\lambda\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\lambda\\nb 7 \\Big\\rangle$.\n\t\n\t\tIf $\\lambda=\\frac 14$, then we obtain the representative $\\langle \\frac 12\\nb 1 + \\nabla_2 -\\frac 12\\nb 3-\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle$.\n\t\n \\item \tLet $\\alpha_1-\\alpha_2\\alpha_6=0$. We may assume that $\\alpha_6\\ne\\frac 12$ (and hence $\\lambda\\ne\\frac 14$), since otherwise $\\alpha_1=(2\\alpha_6^2 - 2\\alpha_6 + 1)\\alpha_2$, which has already been considered.\n \n \n\n\n\n\n\n\n\n\t\t\t\t\n\n\t\t\t\t\n\n\t\t\t\t\n\n\t\t\t\t\n\tIn this case $\\alpha_3^*, \\alpha_5^*, \\alpha_6^*, \\alpha_7^*$ are as above and\n\t\t\\begin{align*}\n\t \t\t\\alpha^*_1 &= \\alpha_2\\alpha_6 (x - y \\alpha_6)(x+(\\alpha_6 - 1)y)\\\\\n\t \t\t\\alpha^*_2 &=\\alpha_2 (x - y \\alpha_6)(x+(\\alpha_6 - 1)y).\n\t \t\t\\end{align*}\n\tTherefore, we obtain the representative $\\Big\\langle \\alpha\\nabla_1+\\nabla_2+ (\\alpha-1)\\nabla_3 + (1-\\frac{1}{\\alpha})\\nabla_5 +\\alpha\\nabla_6+\\nabla_7\\Big\\rangle$, where $\\alpha^2 - \\alpha + \\lambda=0$. \n\t\n\t\tIf $\\lambda\\not\\in\\{0,\\frac 14\\},$ then we obtain two representatives $\\Big\\langle \\frac{1+\\sqrt{1-4\\lambda}}2\\nabla_1 + \\nabla_2 +\\frac{-1+\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}- 1}{\\sqrt{1-4\\lambda}+1}\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle, \\Big\\langle \\frac{1-\\sqrt{1-4\\lambda}}2\\nabla_1 + \\nabla_2 -\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}+ 1}{\\sqrt{1-4\\lambda}- 1}\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle = \\Big\\langle \\lambda\\frac{1-\\sqrt{1-4\\lambda}}2\\nabla_1 + \\lambda\\nabla_2 -\\lambda\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\Big(\\frac{\\sqrt{1-4\\lambda}+ 1}{2}\\Big)^2\\nb 5+\\lambda\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\lambda\\nb 7 \\Big\\rangle$\n\t\n\t \n\t \n\t\n\t\n\n\\end{itemize}\n\t\t\t\t\\item $\\alpha_5 + 1=0$. Then $\\alpha_3=-\\alpha_6$. Observe that we may assume that $\\alpha_6\\ne\\frac 12$ (and hence $\\lambda\\ne\\frac 14$), since otherwise $\\alpha_6 - \\alpha_5\\alpha_6 - 1=0,$ which was considered above. Choosing $z$ such that $\\alpha^*_4=0$, we have\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\alpha^*_1 &= \\alpha_1x^2 + 2\\alpha_2\\alpha_6(\\alpha_6 - 1)xy + \\alpha_6(\\alpha_6-1)(\\alpha_1 - \\alpha_2)y^2,\\\\\n\t\t\t\t\t\\alpha^*_2 &= \\alpha_2x^2 + 2(\\alpha_1 - \\alpha_2)xy + (\\alpha_2\\alpha_6(\\alpha_6-1) - \\alpha_1 + \\alpha_2)y^2,\\\\\n\t\t\t\t\t\\alpha^*_3 &= -\\alpha_6(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t\t\t\t\t\\alpha^*_5 &= -(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t\t\t\t\t\\alpha^*_6 &= \\alpha_6(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t\t\t\t\t\\alpha^*_7 &= (x - \\alpha_6y)(x + (\\alpha_6-1)y)^2.\n\t\t\t\t\\end{align*}\n\t\t\t\t\n\t\t\t\tIf $(\\alpha_1,\\alpha_2)=(0,0)$, then we obtain the representative $\\langle-\\beta\\nb 3-\\nb 5+\\beta\\nb 6+\\nb 7\\rangle$, where $\\beta^2 - \\beta + \\lambda=0$.\n\t\t\t\t\n\t\t\t\tIf $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we have two representatives $\\Big\\langle\\frac{-1\\mp\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle$ which will be joined with the families $\\Big\\langle\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0,-1}$ found above.\n\t\t\t\t\n\t\t\t\t\n\t\t\t\tIf $\\lambda=0$, then we have two representatives $\\langle-\\nb 3-\\nb 5+\\nb 6+\\nb 7\\rangle$ and $\\langle-\\nb 5+\\nb 7\\rangle$ which will be joined the families $\\langle\\alpha\\nb 3+\\alpha\\nb 5+\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0,-1}$ and $\\langle\\alpha\\nb 5+\\nb 7\\rangle_{\\alpha\\ne 0,-1}$ found above.\n\t\t\t\t\n\t\t\t\tLet $(\\alpha_1,\\alpha_2) \\neq (0,0).$ Suppose that $\\alpha_2=0$. Then we have $\\alpha^*_2 = \\alpha_1 y (2x - y).$ So, whenever $(\\alpha_1,\\alpha_2) \\neq (0,0)$, we may find $x,y$ such that $\\alpha^*_2\\neq 0$ and $\\alpha^*_7=1$. In the rest of the case we will suppose that $\\alpha_2 \\neq 0.$\n\t\t\t\t\n\t\t\t\tThe determinant of the equation $\\alpha^*_2=0$ is $4(\\alpha_2\\alpha_6 + \\alpha_1 - \\alpha_2)(\\alpha_1-\\alpha_2\\alpha_6)$.\n\t\t\t\t\n\t\t\t\t\\begin{itemize}\n\t\t\t\t \\item Let $(\\alpha_2\\alpha_6 + \\alpha_1 - \\alpha_2)(\\alpha_1-\\alpha_2\\alpha_6)\\ne 0.$ \n\t\t\t\t \n\t\t\t\t Suppose that $\\alpha_2 = 2\\alpha_1.$ Observe that in this case $(\\alpha_2\\alpha_6 + \\alpha_1 - \\alpha_2)(\\alpha_2\\alpha_6 - \\alpha_1)\\ne 0$ is equivalent to $\\alpha_1^2(2\\alpha_6 - 1)^2\\ne 0$. Then $\\alpha_2^* - 2\\alpha_1^* = -2\\alpha_1(2x-y)y(2\\alpha_6-1)^2,$ so choosing appropriate $x,y$ we may suppose that $\\alpha_2\\ne 2\\alpha_1.$\n\t\t\t\t \n\t\t\t\t The equation $\\alpha^*_2=0$ has two solutions $x_1=\\mu_1 y$ and $x_2=\\mu_2 y$, where $\\mu_1,\\mu_2\\in{\\mathbb C}$, $\\mu_1\\ne\\mu_2$. As in the case 1(a) if $\\mu_1^2 - \\mu_1 + \\lambda=\\mu_2^2 - \\mu_2 + \\lambda=0$, then $\\mu_1+\\mu_2=1$. But $\\mu_1+\\mu_2=-\\frac{2(\\alpha_1 - \\alpha_2)}{\\alpha_2}$, whence $2(\\alpha_1 - \\alpha_2)=-\\alpha_2$, which contradicts the assumption that $\\alpha_2\\ne 2\\alpha_1$. Thus, we may choose $x=\\mu_iy$ to make $\\alpha^*_2=0$. Since in this case the condition $(\\alpha_1,\\alpha_2,\\alpha_4)\\ne (0,0,0)$ is invariant under automorphisms, we have $\\alpha^*_1\\ne 0$ for such a choice of $x$. Thus, we obtain the family of representatives $\\langle\\alpha\\nb 1-\\beta\\nb 3-\\nb 5+\\beta\\nb 6+\\nb 7\\rangle$, where $\\beta^2 - \\beta + \\lambda=0$ and $\\alpha\\ne 0$. Then taking $y=0$ and $x=\\alpha$, we obtain the representative $\\langle\\nb 1-\\beta\\nb 3-\\nb 5+\\beta\\nb 6+\\nb 7\\rangle.$ It will be joined with the family $\\langle\\nb 1+\\alpha\\beta\\nb 3+\\alpha\\nb 5+\\beta\\nb 6+\\nb 7\\rangle_{\\alpha\\not\\in \\{0,-1\\}}.$\n\t\t\t\t\\item \tLet $\\alpha_1-\\alpha_2\\alpha_6=0$. Then\n\t\t\t\t\\begin{align*}\n\t\t\t\t\\alpha^*_1 =\\alpha_2\\alpha_6(x + (\\alpha_6-1)y)^2,\n\t\t\t\t\\alpha^*_2 =\\alpha_2(x + (\\alpha_6-1)y)^2.\n\t\t\t\t\\end{align*}\n\t\t\t \tSo, taking $x=\\alpha_6y+\\alpha_2$, we obtain the representative $\\langle\\beta\\nb 1+\\nb 2-\\beta\\nb 3-\\nb 5+\\beta\\nb 6+\\nb 7\\rangle$, where $\\beta^2 - \\beta + \\lambda=0$. If $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we have two representatives $\\Big\\langle\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 1+\\nb 2-\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle$.\n\t\t\t \tIf $\\lambda=0$, then we have two representatives $\\langle\\nb 1+\\nb 2-\\nb 3-\\nb 5+\\nb 6+\\nb 7\\rangle$ and $\\langle\\nb 2-\\nb 5+\\nb 7\\rangle$.\n\t\t\t \t\\item \t\tLet $\\alpha_2\\alpha_6 + \\alpha_1 - \\alpha_2=0$. Then\n\t\t\t \t\\begin{align*}\n\t\t\t \t\\alpha^*_1 =\\alpha_2(1-\\alpha_6)(x - \\alpha_6y)^2,\n\t\t\t \t\\alpha^*_2 =\\alpha_2(x - \\alpha_6y)^2.\n\t\t\t \t\\end{align*}\n\t\t\t \tTaking $x,y$ such that $(x + (\\alpha_6-1)y)^2=\\alpha_2(x - \\alpha_6y)\\ne 0$, we obtain the representative $\\langle(1-\\beta)\\nb 1+\\nb 2-\\beta\\nb 3-\\nb 5+\\beta\\nb 6+\\nb 7\\rangle$, where $\\beta^2 - \\beta + \\lambda=0$. If $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we have two representatives $\\Big\\langle\\frac{1\\mp\\sqrt{1-4\\lambda}}2\\nb 1+\\nb 2-\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle$. \n\t\t\t \n\t\t\t \tIf $\\lambda=0$, then we have two representatives $\\langle\\nb 2-\\nb 3-\\nb 5+\\nb 6+\\nb 7\\rangle$ and $\\langle\\nb 1+\\nb 2-\\nb 5+\\nb 7\\rangle$.\n\t\t\t\t\\end{itemize}\n\t\t\t\n\t\t\t\n\t\t\t\n\t\t\t\n\t\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\t\t\n\t\t\n\t\t\t \t\n\t \t\\end{enumerate}\n \t\n \t\t\\item $\\alpha_3=\\alpha_5(1-\\alpha_6)$. Observe that we may assume that $\\alpha_6\\ne\\frac 12$ (and hence $\\lambda\\ne\\frac 14$), since otherwise $\\alpha_3=\\alpha_5\\alpha_6$, which has already been considered.\n \t\t\n \t\tConsider $\\alpha^*_2=\\alpha^*_4=0$ as a system of linear equations in $z$ and $u$. Then its determinant is the following polynomial in $x$ and $y$:\n \t\t\\begin{align*}\n \t\t\tD(x,y)&=(\\alpha_5^2\\alpha_6 - \\alpha_5^2 - \\alpha_5 + \\alpha_6 - 1)x^2\\\\\n \t\t\t&\\quad-(2\\alpha_5^2\\alpha_6^2 - 2\\alpha_5^2\\alpha_6 - 2\\alpha_6^2 - \\alpha_5 + 4\\alpha_6 - 2)xy\\\\\n \t\t\t&\\quad+(\\alpha_5^2\\alpha_6^2 + \\alpha_5\\alpha_6 + \\alpha_6^2 - 2\\alpha_6 + 1)(\\alpha_6 - 1)y^2.\n \t\t\\end{align*}\n \t\t\n \t\tSuppose that $\\alpha_5^2\\alpha_6 - \\alpha_5^2 - \\alpha_5 + \\alpha_6 - 1 = 0.$ In this case $\\alpha_5^2+1\\ne 0$, since otherwise $(\\alpha_5^2+1)\\alpha_6 = \\alpha_5^2 + \\alpha_5 + 1$ would imply $\\alpha_5=0$, whence $\\alpha_5^2+1=1$, a contradiction. Thus, $\\alpha_6=\\frac{\\alpha_5^2 + \\alpha_5 + 1}{\\alpha_5^2+1}$ and hence $\\alpha_3=-\\frac{\\alpha_5^2}{\\alpha_5^2+1}$. Then $D(x,y) = \\frac{y \\alpha_5 (1 + \\alpha_5)^3 ((1-\\alpha_5)x + \\alpha_5y )}{(1 + \\alpha_5^2)^2}.$ If $\\alpha_5+1=0,$ then $\\alpha_5=-1$, $\\alpha_6=\\frac 12$, $\\alpha_3=-\\frac 12=\\alpha_5\\alpha_6$. This case has been considered above. Therefore, we may suppose that $D(x,y)$ is not identically zero, and hence we may solve $\\alpha^*_2=\\alpha^*_4=0$ in $z$ and $u$. Choosing these values of $z$ and $u$, we may suppose that $\\alpha_2 = \\alpha_4 = 0$ from the very beginning.\n \t\t\nThen taking the appropriate values of $z$ and $u$ we have \t\n\\begin{align*} \t \n \\alpha^*_1&=\\frac{\\alpha_1(\\alpha_5^2\\alpha_6 - \\alpha_5^2 - \\alpha_5 + \\alpha_6 - 1)(x + y(\\alpha_6-1))^2(x-y\\alpha_6)^2}{D(x,y)},\\\\\t\n \\alpha^*_3&=\\alpha_5 (1 - \\alpha_6) (x - y \\alpha_6)^2 (x + y(\\alpha_6-1)),\\\\\n \t\t\t\t\\alpha^*_5&=\\alpha_5 (x - y \\alpha_6)^2 (x + y(\\alpha_6-1)),\\\\\n \t\t\t\t\\alpha^*_6&=\\alpha_6 (x - y \\alpha_6) (x + y(\\alpha_6-1))^2,\\\\\n \t\t\t\t\\alpha^*_7&=(x - y \\alpha_6) (x + y(\\alpha_6-1))^2.\n \t\t\t\\end{align*}\n \t\t\t\n \t\tObserve that for $x,y$ such that $\\alpha_7^* = 1$ and $D(x,y)\\ne 0$ we have $(\\alpha_5^*)^2\\alpha_6^* - (\\alpha_5^*)^2 - \\alpha_5^* + \\alpha_6^* - 1 = \\frac{D(x,y)}{(x + y(\\alpha_6-1))^2}\\ne 0,$ so we may assume that $\\alpha_5^2\\alpha_6 - \\alpha_5^2 - \\alpha_5 + \\alpha_6 - 1 \\neq 0$ from the very beginning.\n \t\t\n \t\tLet $\\alpha_5 \\neq -1$. Taking $x = \\frac{y (\\alpha_6(\\alpha_5-1)+1)}{\\alpha_5+1},$ we have $ D(x,y) = \\frac{y^2 \\alpha_5^2 (2 \\alpha_6-1)^2}{(\\alpha_5+1)^2}\\ne 0$, $x^2 - xy + \\lambda y^2 = -\\frac{y^2 \\alpha_5 (2 \\alpha_6-1)^2}{(\\alpha_5+1)^2}\\ne 0$, $\\alpha_5^* = -\\alpha_7^*$, so we get the representative $\\langle \\alpha \\nabla_1 + (\\beta-1)\\nabla_3 - \\nabla_5 + \\beta \\nabla_6 + \\nabla_7 \\rangle_{\\beta^2 - \\beta + \\lambda=0},$ where\n \\[\n \\alpha = -\\frac{\\alpha_1 (\\alpha_5+1) (\\alpha_5^2\\alpha_6 - \\alpha_5^2 - \\alpha_5 + \\alpha_6 - 1)}{y \\alpha_5^2 (2 \\alpha_6-1)^2}\n \\]\n \n \\begin{itemize}\n \\item If $\\alpha_1 \\neq 0,$ then we take $y = -\\frac{\\alpha_1 (\\alpha_5+1) (\\alpha_5^2\\alpha_6 - \\alpha_5^2 - \\alpha_5 + \\alpha_6 - 1)}{\\alpha_5^2 (2 \\alpha_6-1)^2} \\neq 0,$ so that $x^2 - xy + \\lambda y^2 \\neq 0$, $D(x,y) \\neq 0$, and we get the representative $\\langle \\nabla_1 + (\\alpha-1)\\nabla_3 - \\nabla_5 + \\alpha \\nabla_6 + \\nabla_7 \\rangle_{\\alpha^2 - \\alpha + \\lambda=0}.$\n \n If $\\lambda \\neq 0$, then we get two distinct representatives $\\Big\\langle \\nb 1 - \\frac{1\\mp\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle.$\n \n If $\\lambda = 0$, then we get two distinct representatives $\\Big\\langle \\nb 1 - \\nb 3-\\nb 5 +\\nb 7\\Big\\rangle$ and $\\Big\\langle \\nb 1 -\\nb 5+ \\nb 6+\\nb 7\\Big\\rangle.$\n \n \\item If $\\alpha_1 = 0,$ then we get the same representatives but without $\\nabla_1.$\n \\end{itemize} \n \t\t\n\n\n \n \n \t\t\t\n \n \t\t\t\n \n \n \n\n \n \n \n \n \n \n \n \n \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n \n \n\n\n\n\n \t\t\t\n \n \t\t\t\n \t\t\t\n \n \t\t\t\n \n \t\t\t\n \t\n \t\n \n \t\n\n \t\t\n\n \t\t\n \n \t\t\n \n \n \n \n \t \n \n \n \t\n \n \t\t\n \n \t\t\n \n\n\n \n \n \n \t\t\n \n \n \n \n \n \t\t\n \t\t\t\n \t\t\n \n \t\n \t\t\t\n \n \t\t\t\n \n \t\t\t\n \t\n \t\t\t\n \t\t\n \n \t\t\t\n \n \t\t\t\n \t\n \t\t\t\n \t\n \t\t\t\n \n \n \t \\end{enumerate}\n\t \n\t \\item $\\alpha_6^2 - \\alpha_6\\alpha_7 + \\lambda\\alpha_7^2=0$ and $\\alpha_5=0$. We may assume that $\\alpha_3=0$, since the case $\\alpha_5=0$ and $\\alpha_3\\ne 0$ was considered in 1(b). Then $\\alpha^*_3=\\alpha^*_5=0$. A suitable choice of $x,y$ and $z$ gives $\\alpha^*_7=1$ and $\\alpha^*_2=0$, so we shall assume that $\\alpha_7=1$ and $\\alpha_2=0$. As above, $\\lambda=\\alpha_6-\\alpha_6^2$. Now, choosing $z=-\\frac{\\alpha_1xy + (\\alpha_4\\alpha_6^2 - \\alpha_4\\alpha_6)y^2}{x+(\\alpha_6 - 1)y}$ we obtain $\\alpha^*_2=0$.\n\t \n\t \\begin{enumerate}\n\t \t\\item $\\alpha_6\\ne 1$. Then choosing $u=-\\frac{\\alpha_4x^2 + (\\alpha_1 - 2\\alpha_4)xy - (\\alpha_1 - \\alpha_4)y^2}{(\\alpha_6 - 1)(x + (\\alpha_6 - 1)y)}$ we have $\\alpha^*_4=0$ and\n\t \t\\begin{align*}\n\t \t\t\\alpha^*_1&=(\\alpha_1-\\alpha_4\\alpha_6)(x-\\alpha_6y)^2,\\\\\n\t \t\t\\alpha^*_6 &= \\alpha_6(x - \\alpha_6y)(x + (\\alpha_6-1)y)^2,\\\\\n\t \t\t\\alpha^*_7 &= (x - \\alpha_6y)(x + (\\alpha_6-1)y)^2.\n\t \t\\end{align*}\n\t \t\\begin{enumerate}\n\t \t\t\\item $\\alpha_1-\\alpha_4\\alpha_6\\ne 0$. Then choosing $y=0$ and $x=\\alpha_1-\\alpha_4\\alpha_6$ we obtain the representative $\\langle\\nb 1+\\alpha\\nb 6+\\nb 7\\rangle$, where $\\alpha^2 - \\alpha + \\lambda=0$. If $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we have two representatives $\\Big\\langle\\nb 1+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle$. We will join them with the families $\\Big\\langle\\nb 1+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1\\pm\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0}$found above. If $\\lambda=\\frac 14$, then we have the representative $\\langle\\nb 1+\\frac 12\\nb 6+\\nb 7\\rangle$, which will be joined with the family $\\langle\\nb 1+\\frac 12\\alpha\\nb 3+\\alpha\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0}$ found above. If $\\lambda=0$, then $\\alpha_6=0$, since $\\alpha_6\\ne 1$ by assumption. So, we have the representative $\\langle\\nb 1+\\nb 7\\rangle$. It will be joined with $\\langle\\nb 1+\\alpha\\nb 5+\\nb 7\\rangle_{\\alpha\\ne 0}$.\n\t \t\t\n\t \t\t\\item $\\alpha_1-\\alpha_4\\alpha_6=0$. Then we have the same representatives as above, but without $\\nb 1$. We join them with the families found above. \n\t \t\\end{enumerate}\n \t\n \t\t\\item $\\alpha_6=1$, so that $\\lambda =0$. Then \n \t\t\\begin{align*}\n \t\t\t\\alpha^*_1 &=\\alpha_1x(x - y),\\\\\n \t\t\t\\alpha^*_4 &=(\\alpha_4(x-y) + \\alpha_1y)(x - y),\\\\\n \t\t\t\\alpha^*_6 &= x^2(x - y),\\\\\n \t\t\t\\alpha^*_7 &= x^2(x - y).\n \t\t\\end{align*}\n \t\t\\begin{enumerate}\n \t\t\t\\item $\\alpha_1\\ne 0$ and $\\alpha_1-\\alpha_4\\ne 0$. Then choosing $x=\\alpha_1$ and $y=-\\frac{\\alpha_1\\alpha_4}{\\alpha_1-\\alpha_4}$ we obtain the representative $\\langle\\nb 1+\\nb 6+\\nb 7\\rangle$ which will be joined with the family $\\langle\\nb 1+\\alpha\\nb 3+\\alpha\\nb 5+\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0}$ found above.\n \t\t\t\n \t\t\t\\item $\\alpha_1=0$. Then we have two representatives $\\langle\\nb 4+\\nb 6+\\nb 7\\rangle$ and $\\langle\\nb 6+\\nb 7\\rangle$ depending on whether $\\alpha_4=0$ or not. Both of them belong to the families found above.\n \t\t\t\n \t\t\t\\item $\\alpha_1-\\alpha_4=0$. Then\n \t\t\t\\begin{align*}\n \t\t\t\\alpha^*_1 &=\\alpha_1x(x - y),\\\\\n \t\t\t\\alpha^*_4 &=\\alpha_1x(x - y),\\\\\n \t\t\t\\alpha^*_6 &= x^2(x - y),\\\\\n \t\t\t\\alpha^*_7 &= x^2(x - y).\n \t\t\t\\end{align*}\n \t\t\tThus, we have two representatives $\\langle\\nb 1+\\nb 4+\\nb 6+\\nb 7\\rangle$ and $\\langle\\nb 6+\\nb 7\\rangle$. The second one was found above.\n \t\t\\end{enumerate}\n\t \\end{enumerate}\n\t\\end{enumerate}\n\t\n \\item $\\alpha_7=0$. We may assume that $\\alpha_6=0$, since the case $\\alpha_7=0$ and $\\alpha_6\\ne 0$ was considered in 1(a).\n \\begin{enumerate}\n \\item Let $\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2 \\neq 0.$ If $\\alpha_5 \\neq 0,$ we may take $x = \\frac{(\\alpha_5 - \\alpha_3)y}{\\alpha_5}$ and get $\\alpha^*_5=0$. Observe that $(\\alpha^*_3)^2 - \\alpha^*_3\\alpha^*_5 + \\lambda(\\alpha^*_5)^2=(\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2)(x^2 - xy + \\lambda y^2)^3$, so the condition $\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2\\ne 0$ is invariant under the automorphisms. Thus, we shall assume that $\\alpha_5=0$ and $\\alpha_3\\ne 0$ from the very beginning. Choosing $y=0$, we have\n \\begin{align*}\n\\alpha^*_1 &= x(\\alpha_1x + \\alpha_3z),\\\\\n\\alpha^*_2 &= x(\\alpha_2x + \\alpha_3u),\\\\\n\\alpha^*_3 &= \\alpha_3x^3,\\\\\n\\alpha^*_4 &= \\alpha_4x^2.\n\\end{align*}\nTaking $z = -\\frac{\\alpha_1x}{\\alpha_3}, u = -\\frac{\\alpha_2x}{\\alpha_3},$ we get two representatives $\\langle \\nabla_3 + \\nabla_4 \\rangle$ and $\\langle \\nabla_3 \\rangle$ depending on whether $\\alpha_4=0$ or not. \n \\item Let $\\alpha_3^2 - \\alpha_3\\alpha_5 + \\lambda\\alpha_5^2 = 0, \\alpha_5 \\neq 0.$ Taking $x,y$ such that $(x^2-xy+\\lambda y^2)(\\alpha_5x + (\\alpha_3-\\alpha_5)y) = 1,$ we may suppose that $\\alpha_5^* = 1$ and $\\lambda = \\alpha_3 - \\alpha_3^2.$ Consider $\\alpha_2^* = 0, \\alpha_4^* = 0$ as a linear system in $u, z.$ Its determinant is $\\alpha_3(x + (\\alpha_3-1)y)^2.$ If $\\alpha_3 = 0,$ then $\\lambda = 0$ and we get a Leibniz cocycle. Therefore, we may suppose that this determinant is nonzero and the system $\\alpha_2^* = 0, \\alpha_4^* = 0$ has a unique solution. We get the family of representatives $\\langle \\alpha_1\\nabla_1 + \\alpha_3\\nabla_3 + \\nabla_5 \\rangle,$ where $\\alpha_3^2 - \\alpha_3 + \\lambda = 0,$ except for $(\\lambda,\\alpha_3) = (0,0)$ which gives a Leibniz cocycle. Therefore, we may suppose that $\\alpha_2 = \\alpha_4 = 0$ from the very beginning and\n \\begin{align*}\n \\alpha^*_1 &= \\alpha_1(x-\\alpha_3y)^2,\\\\\n \\alpha^*_3 &= \\alpha_3(x-\\alpha_3y)(x+(\\alpha_3-1)y)^2,\\\\\n \\alpha^*_5 &= (x-\\alpha_3y)(x+(\\alpha_3-1)y)^2.\n \\end{align*}\n If $\\lambda\\not\\in\\{0,\\frac 14\\}$, then we have the following representatives: $\\Big\\langle \\nabla_1 + \\frac{1\\pm\\sqrt{1-4\\lambda}}{2}\\nabla_3 + \\nabla_5 \\Big\\rangle$ and $\\Big\\langle \\frac{1\\pm\\sqrt{1-4\\lambda}}{2}\\nabla_3 + \\nabla_5 \\Big\\rangle$. If $\\lambda=\\frac 14$, then we have the representatives $\\langle\\nb 1+\\frac 12\\nb 3+\\nb 5\\rangle$ and $\\langle\\frac 12\\nb 3+\\nb 5\\rangle$. If $\\lambda=0$, then we have the representatives $\\langle \\nabla_1 + \\nabla_3 + \\nabla_5 \\rangle$ and $\\langle \\nabla_3 + \\nabla_5 \\rangle$. \n \\end{enumerate}\n \\end{enumerate}\n\n\t\\begin{landscape}\n\t\\tiny{\n\t\\begin{tabular}{|c|c|c|}\n\t \\hline\n\t $\\lambda=0$ & $\\lambda=\\frac 14$ & $\\lambda\\not\\in\\{0,\\frac 14\\}$\\\\\n\t \\hline\n\t $\\begin{array}{l}\n\t \\langle\\nb 1+\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0},\\\\\n\t \\langle\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0},\\\\\n\t \\langle\\nb 2+\\alpha\\nb 5+\\nb 6\\rangle_{\\alpha\\ne 1},\\\\\n\t \\langle\\alpha\\nb 5+\\nb 6\\rangle,\\\\\n\t \\langle\\nb 2+\\alpha\\nb 4+\\nb 5+\\nb 6\\rangle,\\\\\n\t \\langle\\nb 4+\\nb 5+\\nb 6\\rangle,\\\\\n\t \\langle \\nb 2+\\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle \\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\nb 1+\\alpha\\nb 2-\\nb 3+\\nb 6\\rangle,\\\\\n\t \\langle\\nb 1+\\alpha\\nb 3+\\nb 7\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\alpha\\nb 3+\\nb 7\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\alpha\\nb 3+\\nb 4+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\alpha\\nb 3+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 1-\\nb 3+\\alpha\\nb 4+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 1+\\alpha\\nb 5+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 1+\\alpha\\nb 3+\\alpha\\nb 5+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\alpha\\nb 5+\\nb 7\\rangle,\\\\\n\t \\langle\\alpha\\nb 3+\\alpha\\nb 5+\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\nb 1+\\nb 2-\\nb 3-\\nb 5+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 2-\\nb 5+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 2-\\nb 3-\\nb 5+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 1+\\nb 2-\\nb 5+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 1-\\nb 3-\\nb 5+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 1-\\nb 5+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle-\\nb 3-\\nb 5+\\nb 7\\rangle,\\\\\n\t \\langle-\\nb 5+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\nb 1+\\nb 4+\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle \\nb 3 + \\nb 4 \\rangle,\\\\\n\t \\langle \\nb 3 \\rangle,\\\\\n\t \\langle \\nb 1 + \\nb 3 + \\nb 5 \\rangle,\\\\\n\t \\langle \\nb 3 + \\nb 5 \\rangle\n\t \\end{array}$\n\t &\n\t $\\begin{array}{l}\n\t \\langle\\nb 1+\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0},\\\\\n\t \\langle\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0},\\\\\n\t \\langle\\nb 2+\\alpha\\nb 5+\\nb 6\\rangle_{\\alpha\\ne 2},\\\\\n\t \\langle\\alpha\\nb 5+\\nb 6\\rangle,\\\\\n\t \\langle\\nb 2+\\alpha\\nb 4+2\\nb 5+\\nb 6\\rangle,\\\\\n\t \\langle\\nb 4+2\\nb 5+\\nb 6\\rangle,\\\\\n\t \\langle \\nb 2+\\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle \\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\nb 1+\\alpha\\nb 2-\\nb 3+\\nb 6\\rangle,\\\\\n\t \\langle\\nb 1+\\alpha\\nb 3+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\alpha\\nb 3+\\frac 12\\nb 6+\\nb 7\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\nb 1+\\frac 12\\alpha\\nb 3+\\alpha\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle\\frac 12\\alpha\\nb 3+\\alpha\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle,\\\\\n\t \\langle \\frac 12\\nb 1+ \\nb 2 -\\frac 12\\nb 3-\\nb 5+\\frac 12\\nb 6+\\nb 7\\rangle,\\\\\n\t \n\t \\langle \\nb 3 + \\nb 4 \\rangle,\\\\\n\t \\langle \\nb 3 \\rangle,\\\\\n\t \\langle\\nb 1+\\frac 12\\nb 3+\\nb 5\\rangle,\\\\\n\t \\langle\\frac 12\\nb 3+\\nb 5\\rangle\n\t \\end{array}$\n\t &\n\t $\\begin{array}{l}\n\t \\langle\\nb 1+\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0},\\\\\n\t \\langle\\alpha\\nb 3+\\beta\\nb 5+\\nb 6\\rangle_{\\alpha,\\beta\\ne 0},\\\\\n\t \\langle\\nb 2+\\alpha\\nb 5+\\nb 6\\rangle_{\\alpha\\ne\\frac{1\\pm\\sqrt{1-4\\lambda}}{2\\lambda}},\\\\\n\t \\langle\\alpha\\nb 5+\\nb 6\\rangle,\\\\\n\t \\Big\\langle \\lambda\\nb 2+\\lambda\\alpha\\nb 4+\\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 5+\\lambda\\nb 6\\Big\\rangle,\\\\\n\t \n\t \\Big\\langle\\nb 2+\\alpha\\nb 4+\\frac{2}{1+\\sqrt{1-4\\lambda}}\\nb 5+\\nb 6\\Big\\rangle,\\\\\n\t \n\t \\Big\\langle\\lambda \\nb 4+\\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 5+\\lambda \\nb 6\\Big\\rangle,\\\\\n\t \\Big\\langle\\nb 4+\\frac{2}{1+\\sqrt{1-4\\lambda}}\\nb 5+\\nb 6\\Big\\rangle,\\\\\n\t \\langle \\nb 2+\\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle \\alpha\\nb 3+\\nb 6\\rangle_{\\alpha\\ne 0},\\\\\n\t \\langle\\nb 1+\\alpha\\nb 2-\\nb 3+\\nb 6\\rangle,\\\\\n\t \\Big\\langle \\nb 1+\\alpha\\nb 3+ \\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0},\\\\\n\t \\Big\\langle \\nb 1+\\alpha\\nb 3+ \\frac{1-\\sqrt{1-4\\lambda}}{2}\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0},\\\\\n\t \\Big\\langle \\alpha\\nb 3+ \\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0},\\\\\n\t \\Big\\langle \\alpha\\nb 3+ \\frac{1-\\sqrt{1-4\\lambda}}{2}\\nb 6+\\nb 7\\Big\\rangle_{\\alpha\\ne 0},\\\\\t \n\t \\Big\\langle\\nb 1+\\frac{1+\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\nb 1+\\frac{1-\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\frac{1+\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\frac{1-\\sqrt{1-4\\lambda}}2\\alpha\\nb 3+\\alpha\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle (1-2\\lambda)\\nb 1+ \\nb 2 + \\frac{-1+\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}- 1}{\\sqrt{1-4\\lambda}+ 1}\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle,\\\\\n\t \\Big\\langle (1-2\\lambda)\\lambda\\nabla_1 + \\lambda\\nabla_2 -\\lambda\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\Big(\\frac{\\sqrt{1-4\\lambda}+ 1}{2}\\Big)^2\\nb 5+\\lambda\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\lambda\\nb 7 \\Big\\rangle\\\\\n\t \n\t \\Big\\langle \\frac{1+\\sqrt{1-4\\lambda}}2\\nb 1+ \\nb 2 + \\frac{-1+\\sqrt{1-4\\lambda}}2\\nb 3+\\frac{\\sqrt{1-4\\lambda}- 1}{\\sqrt{1-4\\lambda}+ 1}\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7 \\Big\\rangle,\\\\\n\t \\Big\\langle \\lambda\\frac{1-\\sqrt{1-4\\lambda}}2\\nabla_1 + \\lambda\\nabla_2 -\\lambda\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\Big(\\frac{\\sqrt{1-4\\lambda}+ 1}{2}\\Big)^2\\nb 5+\\lambda\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\lambda\\nb 7 \\Big\\rangle\\\\\n\t \n\t \\Big\\langle\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 1+\\nb 2-\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 1+\\nb 2-\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 1+\\nb 2-\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 1+\\nb 2-\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\nb 1-\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle\\nb 1-\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle-\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\Big\\langle-\\frac{1+\\sqrt{1-4\\lambda}}2\\nb 3-\\nb 5+\\frac{1-\\sqrt{1-4\\lambda}}2\\nb 6+\\nb 7\\Big\\rangle,\\\\\n\t \\langle \\nb 3 + \\nb 4 \\rangle,\\\\\n\t \\langle \\nb 3 \\rangle,\\\\\n\t \\Big\\langle \\nb 1 + \\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 3 + \\nb 5 \\Big\\rangle,\\\\\n\t \\Big\\langle \\nb 1 + \\frac{1-\\sqrt{1-4\\lambda}}{2}\\nb 3 + \\nb 5 \\Big\\rangle,\\\\\n\t \\Big\\langle \\frac{1+\\sqrt{1-4\\lambda}}{2}\\nb 3 + \\nb 5 \\Big\\rangle,\\\\\n\t \\Big\\langle \\frac{1-\\sqrt{1-4\\lambda}}{2}\\nb 3 + \\nb 5 \\Big\\rangle.\n\t \\end{array}$\n\t \\\\\n\t \\hline\n\t\\end{tabular}\n\t}\n \n\n\t\\end{landscape}\n\n\n\\newpage \n\n\n\nDenote $\\Theta=\\frac{1+\\sqrt{1-4\\lambda}}{2}.$ The orbits above correspond to the following algebras:\t\n\n{\\tiny\n\\[\\begin{array}{lllllllllll}\n\n\n\n \\D{4}{01}(\\lambda,\\alpha,\\beta)&:& e_1 e_1 = \\lambda e_3 + e_4, & e_1 e_3 = \\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \\beta e_4, & e_3e_1 = e_4; \\\\\n\t \\D{4}{02}(\\lambda,\\alpha,\\beta)&:& e_1 e_1 = \\lambda e_3, & e_1 e_3 = \\alpha e_4, & e_2 e_1=e_3 & e_2 e_2 = e_3, & e_2 e_3 = \\beta e_4, & e_3e_1 = e_4; \\\\\n\t \\D{4}{03}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3, & e_1 e_2 = e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \\alpha e_4, & e_3e_1 = e_4; \\\\\n\t \\D{4}{04}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \\alpha e_4, & e_3e_1 = e_4; \\\\\n\t \\D{4}{05}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3, & e_1 e_2 = \\lambda e_4, & e_2 e_1=e_3 + \\lambda\\alpha e_4, & e_2 e_2 = e_3, & e_2 e_3 = \t\\Theta e_4, & e_3e_1 = \\lambda e_4; \\\\\n\t \n\t \\D{4}{06}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3, & e_1 e_2 = e_4, & e_2 e_1=e_3 + \\alpha e_4, & e_2 e_2 = e_3, & e_2 e_3 = \t\\Theta^{-1} e_4, & e_3e_1 = e_4; \\\\\n\t \n\t \\D{4}{07}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_2 e_1=e_3 + \\lambda e_4, & e_2 e_2 = e_3, & e_2 e_3 = \t\\Theta e_4, & e_3e_1 = \\lambda e_4; \\\\\n\t \n\t \\D{4}{08}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3, & e_2 e_3 = \t\\Theta^{-1} e_4, & e_3e_1 = e_4; \\\\\n\t \n\t \\D{4}{09}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3, & e_1 e_2 = e_4,& e_1 e_3 = \\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = e_4; \\\\\n\t \\D{4}{10}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3,& e_1 e_3 = \\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = e_4; \\\\\n \\D{4}{11}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3 + e_4,& e_1e_2 = \\alpha e_4, & e_1 e_3 = -e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = e_4; \\\\\n \\D{4}{12}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3 + e_4,& e_1e_3 = \\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = \\Theta e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{13}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3 + e_4,& e_1e_3 = \\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = (1-\\Theta)e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{14}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3,& e_1e_3 = \\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = \\Theta e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{15}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3,& e_1e_3 = \\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = (1-\\Theta)e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{16}(\\alpha)&:& e_1e_3 = \\alpha e_4, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3, & e_3e_1 = e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{17}(\\alpha)&:& e_1 e_1 = e_4, & e_1e_3 = -e_4, & e_2 e_1=e_3 + \\alpha e_4, & e_2 e_2 = e_3, & e_3e_1 = e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{18}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3 + e_4,& e_1e_3 = \\Theta\\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \\alpha e_4, & e_3e_1 = \\Theta e_4,\\\\\n && e_3e_2 = e_4; \\\\\n \\D{4}{19}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3 + e_4,& e_1e_3 = (1-\\Theta)\\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \\alpha e_4, & e_3e_1 = (1-\\Theta)e_4,\n \\\\&& e_3e_2 = e_4; \\\\\n \\D{4}{20}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3,& e_1e_3 = \\Theta\\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \\alpha e_4, & e_3e_1 = \\Theta e_4,\n \\\\&& e_3e_2 = e_4; \\\\\n \\D{4}{21}(\\lambda,\\alpha)&:& e_1 e_1 = \\lambda e_3,& e_1e_3 = (1-\\Theta)\\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \\alpha e_4, & e_3e_1 = (1-\\Theta)e_4, \\\\&& e_3e_2 = e_4; \\\\\n \\D{4}{22}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + (1-2\\lambda)e_4,& e_1 e_2 = e_4,& e_1e_3 = (\\Theta - 1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = (1-\\Theta^{-1}) e_4,\n \\\\&& e_3e_1 = \\Theta e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{23}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + \\lambda(1-2\\lambda) e_4& e_1 e_2 = \\lambda e_4& e_1e_3 = -\\lambda\\Theta e_4 & e_2 e_1=e_3 & e_2 e_2 = e_3 & e_2 e_3 = -\\Theta^2 e_4, \n \\\\&& e_3e_1 = \\lambda(1-\\Theta)e_4 & e_3e_2 = \\lambda e_4; \\\\\n \\D{4}{24}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + \\Theta e_4,& e_1 e_2 = e_4,& e_1e_3 = (\\Theta - 1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = (1-\\Theta^{-1}) e_4, \n \\\\&& e_3e_1 = \\Theta e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{25}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + \\lambda(1-\\Theta)e_4,& e_1 e_2 = \\lambda e_4,& e_1e_3 = -\\lambda\\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -\\Theta^2 e_4, \n \\\\&& e_3e_1 = \\lambda (1-\\Theta)e_4, & e_3e_2 = \\lambda e_4; \\\\\n \\D{4}{26}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + \\Theta e_4,& e_1 e_2 = e_4,& e_1e_3 = -\\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4,\n \\\\&& e_3e_1 = \\Theta e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{27}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + (1-\\Theta)e_4,& e_1 e_2 = e_4,& e_1e_3 = (\\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4,\n \\\\&& e_3e_1 = (1-\\Theta)e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{28}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + (1-\\Theta)e_4,& e_1 e_2 = e_4,& e_1e_3 = -\\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, \n \\\\&& e_3e_1 = \\Theta e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{29}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + \\Theta e_4,& e_1 e_2 = e_4,& e_1e_3 = (\\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, \n \\\\&& e_3e_1 = (1-\\Theta)e_4, & e_3e_2 = e_4; \\\\\n \\D{4}{30}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + e_4, & e_1e_3 = (\\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = \\Theta e_4,\n \\\\&& e_3e_2 = e_4; \\\\\n \\D{4}{31}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + e_4, & e_1e_3 = -\\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = (1-\\Theta)e_4,\n \\\\&& e_3e_2 = e_4; \\\\\n \\D{4}{32}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_1e_3 = (\\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = \\Theta e_4, \n \\\\&& e_3e_2 = e_4; \\\\\n \\D{4}{33}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_1e_3 = -\\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = (1-\\Theta)e_4, \n \\\\&& e_3e_2 = e_4; \\\\\n \\D{4}{34}&:& e_1 e_1 = e_4, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3, & e_3 e_1 = e_4,& e_3 e_2 = e_4; \\\\\n \\D{4}{35}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_1e_3 = e_4, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3; \\\\\n \\D{4}{36}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_1e_3 = e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3; \\\\\n \\D{4}{37}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + e_4, & e_1e_3 = \\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4; \\\\\n \\D{4}{38}(\\lambda)&:& e_1 e_1 = \\lambda e_3 + e_4, & e_1e_3 = (1-\\Theta)e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4; \\\\\n \\D{4}{39}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_1e_3 = \\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4; \\\\\n \\D{4}{40}(\\lambda)&:& e_1 e_1 = \\lambda e_3, & e_1e_3 = (1-\\Theta)e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4.\n\\end{array}\\]}\n\n\nThe algebras above are pairwise non-isomorphic, except for {\\tiny \\begin{gather*}\n\\D{4}{01}(\\lambda,0,\\beta) \\cong \\D{4}{02}(\\lambda,0,\\beta) \\cong \\D{4}{04}(\\lambda,\\beta),\\quad \\D{4}{01}(\\lambda,\\alpha,0)_{\\alpha \\neq -1} \\cong \\D{4}{02}(\\lambda,\\alpha,0) \\cong \\D{4}{10}(\\lambda,\\alpha),\\quad \\D{4}{01}(\\lambda,-1,0) \\cong \\D{4}{11}(\\lambda,0),\\quad\\\\\n\\D{4}{03}(\\lambda,0) \\cong \\D{4}{09}(\\lambda,0),\\quad \\D{4}{03}\\left(\\lambda,(1-\\Theta)^{-1}\\right)_{\\lambda \\neq 0} \\cong \\D{4}{05}(\\lambda,0)_{\\lambda \\neq 0}, \\D{4}{03}\\left(\\lambda,\\Theta^{-1}\\right)\\cong \\D{4}{06}(\\lambda,0),\\quad \\D{4}{04}(\\lambda,0) \\cong \\D{4}{10}(\\lambda,0),\\quad\\\\\n\\D{4}{05}(1\/4,\\alpha) \\cong \\D{4}{06}(1\/4,\\alpha),\\quad \\D{4}{07}(1\/4) \\cong \\D{4}{08}(1\/4),\\quad\\\\\n\\D{4}{05}(0,\\alpha) \\cong \\D{4}{07}(0) \\cong \\D{4}{23}(0) \\cong \\D{4}{25}(0) \\cong \\D{4}{40}(0),\\quad\\\\\n\\D{4}{12}(\\lambda,0) \\cong \\D{4}{18}(\\lambda,0),\\quad \\D{4}{12}(1\/4,\\alpha) \\cong \\D{4}{13}(1\/4,\\alpha),\\quad \\D{4}{12}(0,\\alpha)_{\\alpha \\neq -1} \\cong \\D{4}{14}(0,\\alpha),\\quad \\D{4}{12}(0,-1) \\cong \\D{4}{17}(0),\\quad\\\\\n\\D{4}{13}(\\lambda,0) \\cong \\D{4}{19}(\\lambda,0),\\quad \\D{4}{14}(\\lambda,0) \\cong \\D{4}{20}(\\lambda,0),\\quad \\D{4}{14}(1\/4,\\alpha) \\cong \\D{4}{15}(1\/4,\\alpha),\\quad \\D{4}{15}(\\lambda,0) \\cong \\D{4}{21}(\\lambda,0),\\quad\\\\\n\\D{4}{18}(1\/4,\\alpha) \\cong \\D{4}{19}(1\/4,\\alpha),\\quad \\D{4}{18}(0,0) \\cong \\D{4}{22}(0) \\cong \\D{4}{24}(0),\\quad \\D{4}{18}(1\/4,-1) \\cong \\D{4}{19}(1\/4,-1) \\cong \\D{4}{30}(1\/4) \\cong \\D{4}{31}(1\/4),\\quad\\\\\n\\D{4}{20}(1\/4,\\alpha) \\cong \\D{4}{21}(1\/4,\\alpha),\\quad \\D{4}{20}(1\/4,-1) \\cong \\D{4}{21}(1\/4,-1) \\cong \\D{4}{32}(1\/4) \\cong \\D{4}{33}(1\/4),\\quad\\\\\n\\D{4}{22}(1\/4) \\cong \\D{4}{23}(1\/4) \\cong \\D{4}{24}(1\/4) \\cong \\D{4}{25}(1\/4) \\cong \\D{4}{26}(1\/4) \\cong \\D{4}{27}(1\/4) \\cong \\D{4}{28}(1\/4) \\cong \\D{4}{29}(1\/4),\\quad\\\\\n \\D{4}{37}(1\/4) \\cong \\D{4}{38}(1\/4),\\quad \\D{4}{39}(1\/4) \\cong \\D{4}{40}(1\/4).\n\\end{gather*}}\n\nMoreover, the algebras $\\D{4}{05}(0,\\alpha) \\cong \\D{4}{07}(0) \\cong \\D{4}{23}(0) \\cong \\D{4}{25}(0) \\cong \\D{4}{40}(0), \\D{4}{38}(0)$ are Leibniz.\n\t\n\n\n\n\n\\newpage \n\n\n\\subsubsection{$1$-dimensional central extensions of $\\T {3*}{05}$}\\label{ext-T_05^3*}\n\tLet us use the following notations \n\t\\begin{align*}\n\t\\nb 1 = \\Dl 13, \\nb 2 = \\Dl 21, \\nb 3 = \\Dl 22 - 3\\Dl 31. \n\t\\end{align*}\n\tTake $\\theta=\\sum_{i=1}^3\\alpha_i\\nb i\\in {\\rm H_T^2}(\\T {3*}{05}).$\n\tIf \n\t$$\n\t\\phi=\n\t\\begin{pmatrix}\n\tx & 0 & 0\\\\\n\ty & x^2 & 0\\\\\n\tz & xy & x^3\n\t\\end{pmatrix}\\in\\aut{\\T {3*}{05}},\n\t$$\n\tthen\n\t$$\n\t\\phi^T\\begin{pmatrix}\n\t0 & 0 & \\alpha_1\\\\\n\t\\alpha_2 & \\alpha_3 & 0\\\\\n\t-3\\alpha_3& 0 & 0\n\t\\end{pmatrix} \\phi=\n\t\\begin{pmatrix}\n\t\\alpha^* & \\alpha^{**} & \\alpha^*_1\\\\\n\t\\alpha^*_2 & \\alpha^*_3 & 0\\\\\n\t-3\\alpha^*_3 & 0 & 0\n\t\\end{pmatrix},\n\t$$\n\twhere\n\t\\begin{align*}\n\t\\alpha^*_1 &= \\alpha_1x^4,\\\\\n\t\\alpha^*_2 &= x^2(\\alpha_2x - 2\\alpha_3y),\\\\\n\t\\alpha^*_3 &= \\alpha_3x^4.\n\t\\end{align*}\n\tHence, $\\phi\\langle\\theta\\rangle=\\langle\\theta^*\\rangle,$ where $\\theta^*=\\sum\\limits_{i=1}^3 \\alpha_i^* \\nb i.$\n\t\n\tWe are interested in $\\theta$ with $(\\alpha_2,\\alpha_3) \\neq (0,0).$ Moreover, the condition $\\theta \\in \\mathbf{T}_1 (\\T{3}{05})$ gives us $(\\alpha_1, \\alpha_3) \\neq (0,0).$\n\t\n\tLet $\\theta$ be as above. Consider two mutually exclusive cases:\n\t\n\t\\begin{enumerate}\n\t \\item $\\alpha_3 = 0.$ The conditions above imply that $\\alpha_1 \\neq 0, \\alpha_2 \\neq 0.$\n\t Then choosing $x=\\frac{\\alpha_2}{\\alpha_1},$ we have the representative $\\langle \\nabla_1+\\nabla_2 \\rangle.$ \n\n \\item $\\alpha_3 \\neq 0.$ Choosing\n $x = \\frac{1}{\\sqrt[4]{\\alpha_3}}$ and $y=\\frac{\\alpha_2}{ 2 \\sqrt[4]{\\alpha_3^5}}$ we have the family of representatives $\\langle\\alpha\\nabla_1 + \\nabla_3 \\rangle.$ \n\\end{enumerate}\n\nSummarizing, we have the following distinct orbits: \n \n\\[ \\langle \\nabla_1+\\nabla_2 \\rangle, \\ \\langle\\alpha\\nabla_1 + \\nabla_3 \\rangle.\\]\n\nThey correspond to the following algebras:\t \n \n\\[\\begin{array}{lllllllllll}\n\n\n\n \\T{4}{37}&:& e_1e_1 = e_2,& e_1e_2 = e_3,& e_1e_3 = e_4,& e_2e_1 = e_4; \\\\\n\t \\T{4}{38}(\\alpha)&:& e_1e_1 = e_2,& e_1e_2 = e_3,& e_1e_3 = \\alpha e_4,& e_2e_2 = e_4,& e_3e_1 = -3e_4.\n\t\n\\end{array}\\]\n\n\\subsubsection{$1$-dimensional central extensions of $\\T {3}{01}(\\lambda)$}\n\n\n\n\\begin{enumerate}\n \\item \t$\\lambda \\neq 0.$\n\n \n Let us use the following notations \n\t\\begin{align*}\n\t\\nb 1 = \\Dl 12, \\nb 2 = (\\lambda-1)\\Dl 13 + 3 \\Dl 31, \\nb 3 = \\color{black}{\\Dl 13 + \\Dl 22}. \n\t\\end{align*}\n\tTake $\\theta=\\sum_{i=1}^3\\alpha_i\\nb i\\in {\\rm H_T^2}(\\T {3}{01}).$\n\tIf \n\t$$\n\t\\phi=\n\t\\begin{pmatrix}\nx & 0 & 0\\\\\ny & x^2 & 0\\\\\nz & (\\lambda+1)xy & x^3\n\\end{pmatrix}\n\\in\\aut{\\T {3}{01}},\n\t$$\n\tthen\n\t$$\n\t\\phi^T\\begin{pmatrix}\n\t0 & \\alpha_1 & (\\lambda-1)\\alpha_2\\color{black}{+\\alpha_3}\\\\\n\t0 & \\color{black}{\\alpha_3} & 0\\\\\n\t\\color{black}{3\\alpha_2}& 0 & 0\n\t\\end{pmatrix} \\phi=\n\t\\begin{pmatrix}\n\t\\alpha^* & \\alpha_1^* + \\lambda\\alpha^{**} & (\\lambda-1)\\alpha_2^*\\color{black}{+\\alpha_3^*}\\\\\n\t\\alpha^{**} & \\color{black}{\\alpha_3^*} & 0\\\\\n\t\\color{black}{3\\alpha_2^*}& 0 & 0\n\t\\end{pmatrix},\n\t$$\n\twhere\n\t\\begin{align*}\n\n\n\t\\alpha^*_1 &= x^2(\\alpha_1x+({\\color{black} 2}\\alpha_3 - (\\lambda + 1)(2\\lambda + 1)\\alpha_2)y),\\\\\n\t\\alpha^*_2 &= \\alpha_2x^4,\\\\\n\t\\alpha^*_3 &= \\alpha_3x^4.\n\t\\end{align*}\n\tHence, $\\phi\\langle\\theta\\rangle=\\langle\\theta^*\\rangle,$ where $\\theta^*=\\sum\\limits_{i=1}^3 \\alpha_i^* \\nb i.$\n\t\n\tThe condition $\\theta \\in \\mathbf{T}_1 (\\T{3}{01}(\\lambda))$ gives us ${\\color{black}(\\alpha_2, \\alpha_3)} \\neq (0,0).$\n\t\n\t\t\\begin{enumerate}\n\t\t\\item $(\\lambda + 1)(2\\lambda + 1)\\ne 0$.\n\t\t\\begin{enumerate}\n\t\t\t\\item $\\alpha_3\\ne 0$.\n\t\t\t\\begin{enumerate}\n\t\t\t\t\\item $\\alpha_2 \\ne \\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\alpha_3$. Then choosing $x=\\frac 1{\\sqrt[4]{\\alpha_3}}$ and $y=-\\frac{\\alpha_1x}{{\\color{black} 2}\\alpha_3 - (\\lambda + 1)(2\\lambda + 1)\\alpha_2}$, we obtain the family of representatives of distinct orbits\n\t\t\t\t$\\langle\\alpha\\nb 2+\\nb 3\\rangle$, where $\\alpha\\ne \\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}$.\n\t\t\t\t\\item $\\alpha_2=\\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\alpha_3$. Then we have two representatives \n\t\t\t\t$\\Big\\langle\\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\nb 2+\\nb 3\\Big\\rangle$ and $\\Big\\langle\\nb 1+\\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\nb 2+\\nb 3\\Big\\rangle$ depending on whether $\\alpha_1=0$ or not. For convenience, we shall join the representative $\\Big\\langle\\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\nb 2+\\nb 3\\Big\\rangle$ with the family $\\langle\\alpha\\nb 2+\\nb 3\\rangle_{\\alpha\\ne \\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}}$ found above. \n\t\t\t\\end{enumerate}\n\t\t\t\\item $\\alpha_3=0$. Then $\\alpha_2\\ne 0$. Choosing $y=\\frac{\\alpha_1x}{(\\lambda + 1)(2\\lambda + 1)\\alpha_2}$, we obtain the representative $\\langle\\nb 2\\rangle$.\n\t\t\\end{enumerate}\n\n\t\\item $(\\lambda + 1)(2\\lambda + 1)=0$. \n\t\t\\begin{enumerate}\n\t\t\t\\item $\\alpha_3\\ne 0$. Then choosing $x=\\frac 1{\\sqrt[4]{\\alpha_3}}$ and $y=-\\frac{\\alpha_1x}{{\\color{black} 2}\\alpha_3}$, we obtain the family of representatives of distinct orbits $\\langle\\alpha\\nb 2+\\nb 3\\rangle$.\n\t\t\t\\item $\\alpha_3=0$. Then $\\alpha_2\\ne 0$. So, we obtain two representatives $\\langle\\nb 2\\rangle$ and $\\langle\\nb 1+\\nb 2\\rangle$ depending on whether $\\alpha_1=0$ or not.\n\t\t\\end{enumerate}\n\n\t\\end{enumerate}\n\n\n\nSummarizing, in the case (1) we have the following distinct orbits: \n$$\n \\langle\\nb 1+\\nb 2\\rangle_{\\lambda\\in\\{-1,-\\frac 12\\}},\n \\left\\langle\\nb 1+\\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\nb 2+\\nb 3\\right\\rangle_{\\lambda\\not\\in\\{-1,-\\frac 12,0\\}},\n \\langle\\nb 2\\rangle_{\\lambda\\ne 0},\n \\langle\\alpha\\nb 2+\\nb 3\\rangle_{\\lambda\\ne 0}.\n$$\n\n\n\n\n\n\n \n\n\n\t\\item $\\lambda=0.$ \n\tLet us use the following notations \n\t\\begin{align*}\n\t\\nb 1 = \\Dl 12, \\nb 2 = -\\Dl 13 + 3 \\Dl 31, \\nb 3 = {\\color{black}\\Dl 13 + \\Dl 22}, \\nb 4 = \\Dl 23. \n\t\\end{align*}\n\tTake $\\theta=\\sum_{i=1}^4\\alpha_i\\nb i\\in {\\rm H_T^2}(\\T {3}{01}(0)).$\n\tIf \n\t$$\n\t\\phi=\n\t\\begin{pmatrix}\n x & 0 & 0\\\\\n y & x^2 & 0\\\\\n z & xy & x^3\n \\end{pmatrix}\n\\in\\aut{\\T {3}{01}(0)},\n\t$$\n\tthen\n\t$$\n\t\\phi^T\\begin{pmatrix}\n\t0 & \\alpha_1 & -\\alpha_2{\\color{black}+\\alpha_3}\\\\\n\t0 & \\color{black}{\\alpha_3} & \\alpha_4\\\\\n\t{\\color{black} 3\\alpha_2}& 0 & 0\n\t\\end{pmatrix} \\phi=\n\t\\begin{pmatrix}\n\t\\alpha^* & \\alpha^*_1 & -\\alpha^*_2{\\color{black}+\\alpha_3^*}\\\\\n\t\\alpha^{**} & {\\color{black}\\alpha^*_3} & \\alpha^*_4\\\\\n\t{\\color{black} 3\\alpha^*_2}& 0 & 0\n\t\\end{pmatrix},\n\t$$\n\twhere\n\t\\begin{align*}\n\t\\alpha^*_1 &= x (x^2 \\alpha_1 - x y (\\alpha_2 {\\color{black}-2}\\alpha_3) + y^2 \\alpha_4),\\\\\n \\alpha^*_2 &= {\\color{black} \\alpha_2 x^4},\\\\\n \\alpha^*_3 &= x^3 (x \\alpha_3 {\\color{black}+} y \\alpha_4),\\\\\n\t\\alpha^*_4 &= \\alpha_4x^5.\n\t\\end{align*}\n\tHence, $\\phi\\langle\\theta\\rangle=\\langle\\theta^*\\rangle,$ where $\\theta^*=\\sum\\limits_{i=1}^4 \\alpha_i^* \\nb i.$\n\t\n\tThe condition $\\theta \\in \\mathbf{T}_1 (\\T{3}{01}(0))$ gives us $(\\alpha_2, {\\color{black}\\alpha_3},\\alpha_4) \\neq (0,0,0).$\n\t\n\t\n\t\\begin{enumerate}\n\t \\item $\\alpha_4\\neq0$. Then choosing $y={\\color{black}-}\\frac{\\alpha_3x}{\\alpha_4}$ we have the family of representatives \n\t $\\langle \\alpha_1^{\\star} \\nabla_1 + {\\color{black}\\alpha_2^*} \\nabla_2+ \\alpha^*_4\\nabla_4 \\rangle$,\n\t where\n\t \\begin{align*}\n\t \\alpha^\\star_1=& \\frac{x^3}{\\alpha_4}(\\alpha_1\\alpha_4{\\color{black}+}\\alpha_2\\alpha_3{\\color{black}-\\alpha_3^2}).\n\t \\end{align*}\n\t \n\t \\begin{enumerate}\n\t \\item ${\\color{black}\\alpha_2}\\ne 0$. Then we have the family of representatives of distinct orbits $\\langle \\alpha \\nabla_1+ \\nabla_2+ \\nabla_4 \\rangle$.\n\t \n\t \\item ${\\color{black}\\alpha_2}=0$. Then we have two representatives $\\langle \\nabla_4 \\rangle$ and $\\langle \\nabla_1+ \\nabla_4 \\rangle$ depending on whether $\\alpha_1\\alpha_4{\\color{black}+}\\alpha_2\\alpha_3{\\color{black}-\\alpha_3^2}=0$ or not.\n \n\t \\end{enumerate}\n\n\t \\item $\\alpha_4=0.$ \n\t \\begin{enumerate}\n\t \\item $\\alpha_2 {\\color{black}-2}\\alpha_3\\ne 0$. Then choosing $y = \\frac{\\alpha_1x}{\\alpha_2 {\\color{black}-2}\\alpha_3}$ we have the representative $\\langle\\nb 2\\rangle$ and the family of representatives of distinct orbits $\\langle\\alpha\\nb 2+\\nb 3\\rangle_{\\alpha\\ne{\\color{black} 2}}$ depending on whether $\\alpha_3=0$ or not.\n\t \\item $\\alpha_2 {\\color{black}-2}\\alpha_3=0$. Then we have two representatives $\\langle{\\color{black} 2}\\nb 2+\\nb 3\\rangle$ and $\\langle\\nb 1{\\color{black} +2}\\nb 2+\\nb 3\\rangle$ depending on whether $\\alpha_1=0$ or not. The representative $\\langle{\\color{black} 2}\\nb 2+\\nb 3\\rangle$ will be joined with the family $\\langle\\alpha\\nb 2+\\nb 3\\rangle_{\\alpha\\ne{\\color{black} 2}}$.\n\t \\end{enumerate}\n\t \\end{enumerate}\n\t \nSummarizing, in the case (2) we have the following distinct orbits: \n$$ \n\\langle\\nb 1{\\color{black} +2}\\nb 2+\\nb 3\\rangle_{\\lambda=0}, \\langle \\alpha \\nabla_1+ \\nabla_2+ \\nabla_4 \\rangle_{\\lambda=0}, \\langle \\nabla_1+ \\nabla_4 \\rangle_{\\lambda=0}, \\langle\\nb 2\\rangle_{\\lambda=0}, \\langle\\alpha\\nb 2+\\nb 3\\rangle_{\\lambda=0}, \\langle \\nabla_4 \\rangle_{\\lambda=0}.\n$$\n\\end{enumerate}\nNow, taking into account the both cases (1) and (2), we have the following distinct orbits:\n\n$\n\\begin{array}{l}\n \\langle\\nb 1+\\nb 2\\rangle_{\\lambda\\in\\{-1,-\\frac 12\\}},\\\\ \n \\left\\langle\\nb 1+\\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\nb 2+\\nb 3\\right\\rangle_{\\lambda\\not\\in\\{-1,-\\frac 12\\}},\n\\end{array}\n\\begin{array}{l}\n \\langle \\alpha \\nabla_1+ \\nabla_2+ \\nabla_4 \\rangle_{\\lambda=0},\\\\\n \\langle \\nabla_1+ \\nabla_4 \\rangle_{\\lambda=0},\\\\\n\\end{array}\n\\begin{array}{l}\n \\langle\\nb 2\\rangle,\\\\\n \\langle\\alpha\\nb 2+\\nb 3\\rangle,\n\\end{array}\n\\begin{array}{l}\n \\langle \\nabla_4 \\rangle_{\\lambda=0}.\n\\end{array}\n$\n\nObserve that $\\left\\langle\\nb 1+\\frac{\\color{black} 2}{(\\lambda + 1)(2\\lambda + 1)}\\nb 2+\\nb 3\\right\\rangle=\\langle(\\lambda + 1)(2\\lambda + 1)\\nb 1+{\\color{black} 2}\\nb 2+(\\lambda + 1)(2\\lambda + 1)\\nb 3\\rangle$, the latter being $\\langle\\nb 2\\rangle$, when $\\lambda\\in\\{-1,-\\frac 12\\}$. Now, if $\\lambda\\not\\in\\{-1,-\\frac 12\\}$, then $\\mathrm{Orb}\\langle\\nb 2\\rangle=\\mathrm{Orb}\\langle\\nb 1+\\nb 2\\rangle$. Thus, we may reorganize our orbits as follows:\n\n$\n\\begin{array}{l}\n \\langle\\nb 1+\\nb 2\\rangle,\\\\ \n \\langle(\\lambda + 1)(2\\lambda + 1)\\nb 1+{\\color{black} 2}\\nb 2+(\\lambda + 1)(2\\lambda + 1)\\nb 3\\rangle,\n\\end{array}\n\\begin{array}{l}\n \\langle \\alpha \\nabla_1+ \\nabla_2+ \\nabla_4 \\rangle_{\\lambda=0},\\\\\n \\langle \\nabla_1+ \\nabla_4 \\rangle_{\\lambda=0},\n\\end{array}\n\\begin{array}{l}\n \\langle\\alpha\\nb 2+\\nb 3\\rangle,\\\\\n \\langle\\nb 4 \\rangle_{\\lambda=0}.\n\\end{array}\n$\n\nThe corresponding algebras are:\t \n\n\n$$\n\\begin{array}{lllllllll}\n \\T 4{39}(\\lambda) &:& e_1 e_1 = e_2, & e_1 e_2=\\lambda e_3+e_4, & e_1e_3 = (\\lambda-1)e_4,\\\\\n && e_2 e_1=e_3, & e_3e_1 = 3e_4;\\\\ \n \n \\T 4{40}(\\lambda) &:& e_1 e_1 = e_2, & e_1 e_2=\\lambda e_3+(\\lambda + 1)(2\\lambda + 1)e_4, & e_1e_3 = {\\color{black}(2\\lambda^2 + 5\\lambda - 1)}e_4,\\\\\n && e_2 e_1=e_3, & e_2e_2={\\color{black}(\\lambda + 1)(2\\lambda + 1)}e_4, & e_3e_1 = {\\color{black} 6}e_4;\\\\ \n \n \\T 4{41}(\\alpha) &:& e_1 e_1 = e_2, & e_1 e_2=\\alpha e_4, & e_1e_3=-e_4,\\\\\n && e_2 e_1=e_3, & e_2e_3 = e_4, & e_3e_1=3e_4;\\\\\n \n \\T 4{42} &:& e_1 e_1 = e_2, & e_1 e_2=e_4, & e_2 e_1=e_3, & e_2e_3 = e_4;\\\\\n \n \\T 4{43}(\\lambda,\\alpha) &:& e_1 e_1 = e_2, & e_1 e_2=\\lambda e_3, & e_1e_3 = (\\alpha(\\lambda-1){\\color{black}+1})e_4,\\\\\n && e_2 e_1=e_3, & e_2e_2={\\color{black} e_4}, & e_3e_1 = 3{\\color{black}\\alpha} e_4;\\\\ \n \n \\T 4{44} &:& e_1 e_1 = e_2, & e_2 e_1=e_3, & e_2e_3 = e_4.\n\\end{array}\n$$\n\n\n \\section{The geometric classification of nilpotent terminal algebras}\n\n\\subsection{Degenerations of algebras}\nGiven an $n$-dimensional vector space ${\\bf V}$, the set ${\\rm Hom}({\\bf V} \\otimes {\\bf V},{\\bf V}) \\cong {\\bf V}^* \\otimes {\\bf V}^* \\otimes {\\bf V}$ \nis a vector space of dimension $n^3$. This space inherits the structure of the affine variety $\\mathbb{C}^{n^3}.$ \nIndeed, let us fix a basis $e_1,\\dots,e_n$ of ${\\bf V}$. Then any $\\mu\\in {\\rm Hom}({\\bf V} \\otimes {\\bf V},{\\bf V})$ is determined by $n^3$ structure constants $c_{i,j}^k\\in\\mathbb{C}$ such that\n$\\mu(e_i\\otimes e_j)=\\sum_{k=1}^nc_{i,j}^ke_k$. A subset of ${\\rm Hom}({\\bf V} \\otimes {\\bf V},{\\bf V})$ is {\\it Zariski-closed} if it can be defined by a set of polynomial equations in the variables $c_{i,j}^k$ ($1\\le i,j,k\\le n$).\n\n\nThe general linear group ${\\rm GL}({\\bf V})$ acts by conjugation on the variety ${\\rm Hom}({\\bf V} \\otimes {\\bf V},{\\bf V})$ of all algebra structures on ${\\bf V}$:\n$$ (g * \\mu )(x\\otimes y) = g\\mu(g^{-1}x\\otimes g^{-1}y),$$ \nfor $x,y\\in {\\bf V}$, $\\mu\\in {\\rm Hom}({\\bf V} \\otimes {\\bf V},{\\bf V})$ and $g\\in {\\rm GL}({\\bf V})$. Clearly, the ${\\rm GL}({\\bf V})$-orbits correspond to the isomorphism classes of algebras structures on ${\\bf V}$. Let $T$ be a set of polynomial identities which is invariant under isomorphism. Then the subset $\\mathbb{L}(T)\\subset {\\rm Hom}({\\bf V} \\otimes {\\bf V},{\\bf V})$ of the algebra structures on ${\\bf V}$ which satisfy the identities in $T$ is ${\\rm GL}({\\bf V})$-invariant and Zariski-closed. It follows that $\\mathbb{L}(T)$ decomposes into ${\\rm GL}({\\bf V})$-orbits. The ${\\rm GL}({\\bf V})$-orbit of $\\mu\\in\\mathbb{L}(T)$ is denoted by $O(\\mu)$ and its Zariski closure by $\\overline{O(\\mu)}$.\n\n\n\nLet ${\\bf A}$ and ${\\bf B}$ be two $n$-dimensional algebras satisfying the identities from $T$ and $\\mu,\\lambda \\in \\mathbb{L}(T)$ represent ${\\bf A}$ and ${\\bf B}$ respectively.\nWe say that ${\\bf A}$ {\\it degenerates} to ${\\bf B}$ and write ${\\bf A}\\to {\\bf B}$ if $\\lambda\\in\\overline{O(\\mu)}$.\nNote that in this case we have $\\overline{O(\\lambda)}\\subset\\overline{O(\\mu)}$. Hence, the definition of a degeneration does not depend on the choice of $\\mu$ and $\\lambda$. It is easy to see that any algebra degenerates to the algebra with zero multiplication. If ${\\bf A}\\to {\\bf B}$ and ${\\bf A}\\not\\cong {\\bf B}$, then ${\\bf A}\\to {\\bf B}$ is called a {\\it proper degeneration}. We write ${\\bf A}\\not\\to {\\bf B}$ if $\\lambda\\not\\in\\overline{O(\\mu)}$ and call this a {\\it non-degeneration}. Observe that the dimension of the subvariety $\\overline{O(\\mu)}$ equals $n^2-\\dim\\mathfrak{Der}({\\bf A})$. Thus if ${\\bf A}\\to {\\bf B}$ is a proper degeneration, then we must have $\\dim\\mathfrak{Der}({\\bf A})>\\dim\\mathfrak{Der}({\\bf B})$.\n\nLet ${\\bf A}$ be represented by $\\mu\\in\\mathbb{L}(T)$. Then ${\\bf A}$ is {\\it rigid} in $\\mathbb{L}(T)$ if $O(\\mu)$ is an open subset of $\\mathbb{L}(T)$.\nRecall that a subset of a variety is called {\\it irreducible} if it cannot be represented as a union of two non-trivial closed subsets. A maximal irreducible closed subset of a variety is called an {\\it irreducible component}.\nIt is well known that any affine variety can be represented as a finite union of its irreducible components in a unique way.\nThe algebra ${\\bf A}$ is rigid in $\\mathbb{L}(T)$ if and only if $\\overline{O(\\mu)}$ is an irreducible component of $\\mathbb{L}(T)$. \n\n\n\n\n\n\n\nIn the present work we use the methods applied to Lie algebras in \\cite{GRH,GRH2}.\nTo prove\ndegenerations, we will construct families of matrices parametrized by $t$. Namely, let ${\\bf A}$ and ${\\bf B}$ be two algebras represented by the structures $\\mu$ and $\\lambda$ from $\\mathbb{L}(T)$, respectively. Let $e_1,\\dots, e_n$ be a basis of ${\\bf V}$ and $c_{i,j}^k$ ($1\\le i,j,k\\le n$) be the structure constants of $\\lambda$ in this basis. If there exist $a_i^j(t)\\in\\mathbb{C}$ ($1\\le i,j\\le n$, $t\\in\\mathbb{C}^*$) such that the elements $E_i^t=\\sum_{j=1}^na_i^j(t)e_j$ ($1\\le i\\le n$) form a basis of ${\\bf V}$ for any $t\\in\\mathbb{C}^*$, and the structure constants $c_{i,j}^k(t)$ of $\\mu$ in the basis $E_1^t,\\dots, E_n^t$ satisfy $\\lim\\limits_{t\\to 0}c_{i,j}^k(t)=c_{i,j}^k$, then ${\\bf A}\\to {\\bf B}$. In this case $E_1^t,\\dots, E_n^t$ is called a {\\it parametric basis} for ${\\bf A}\\to {\\bf B}$.\n\n\nTo prove a non-degeneration ${\\bf A}\\not\\to {\\bf B}$ we will use the following lemma (see \\cite{GRH}).\n\n\\begin{lemma}\\label{main}\nLet $\\mathcal{B}$ be a Borel subgroup of ${\\rm GL}({\\bf V})$ and $\\mathcal{R}\\subset \\mathbb{L}(T)$ be a $\\mathcal{B}$-stable closed subset.\nIf ${\\bf A} \\to {\\bf B}$ and ${\\bf A}$ can be represented by $\\mu\\in\\mathcal{R}$ then there is $\\lambda\\in \\mathcal{R}$ that represents ${\\bf B}$.\n\\end{lemma}\n\n\nIn particular, it follows from Lemma \\ref{main} that ${\\bf A}\\not\\to {\\bf B}$, whenever $\\dim({\\bf A}^2)<\\dim({\\bf B}^2)$.\n\n\nWhen the number of orbits under the action of ${\\rm GL}({\\bf V})$ on $\\mathbb{L}(T)$ is finite, the graph of primary degenerations gives the whole picture. In particular, the description of rigid algebras and irreducible components can be easily obtained. Since the variety of $4$-dimensional nilpotent terminal algebras contains infinitely many non-isomorphic algebras, we have to fulfill some additional work. Let ${\\bf A}(*):=\\{{\\bf A}(\\alpha)\\}_{\\alpha\\in I}$ be a family of algebras and ${\\bf B}$ be another algebra. Suppose that, for $\\alpha\\in I$, ${\\bf A}(\\alpha)$ is represented by a structure $\\mu(\\alpha)\\in\\mathbb{L}(T)$ and ${\\bf B}$ is represented by a structure $\\lambda\\in\\mathbb{L}(T)$. Then by ${\\bf A}(*)\\to {\\bf B}$ we mean $\\lambda\\in\\overline{\\cup\\{O(\\mu(\\alpha))\\}_{\\alpha\\in I}}$, and by ${\\bf A}(*)\\not\\to {\\bf B}$ we mean $\\lambda\\not\\in\\overline{\\cup\\{O(\\mu(\\alpha))\\}_{\\alpha\\in I}}$.\n\nLet ${\\bf A}(*)$, ${\\bf B}$, $\\mu(\\alpha)$ ($\\alpha\\in I$) and $\\lambda$ be as above. To prove ${\\bf A}(*)\\to {\\bf B}$ it is enough to construct a family of pairs $(f(t), g(t))$ parametrized by $t\\in\\mathbb{C}^*$, where $f(t)\\in I$ and $g(t)=\\left(a_i^j(t)\\right)_{i,j}\\in {\\rm GL}({\\bf V})$. Namely, let $e_1,\\dots, e_n$ be a basis of ${\\bf V}$ and $c_{i,j}^k$ ($1\\le i,j,k\\le n$) be the structure constants of $\\lambda$ in this basis. If we construct $a_i^j:\\mathbb{C}^*\\to \\mathbb{C}$ ($1\\le i,j\\le n$) and $f: \\mathbb{C}^* \\to I$ such that $E_i^t=\\sum_{j=1}^na_i^j(t)e_j$ ($1\\le i\\le n$) form a basis of ${\\bf V}$ for any $t\\in\\mathbb{C}^*$, and the structure constants $c_{i,j}^k(t)$ of $\\mu\\big(f(t)\\big)$ in the basis $E_1^t,\\dots, E_n^t$ satisfy $\\lim\\limits_{t\\to 0}c_{i,j}^k(t)=c_{i,j}^k$, then ${\\bf A}(*)\\to {\\bf B}$. In this case, $E_1^t,\\dots, E_n^t$ and $f(t)$ are called a {\\it parametric basis} and a {\\it parametric index} for ${\\bf A}(*)\\to {\\bf B}$, respectively. In the construction of degenerations of this sort, we will write $\\mu\\big(f(t)\\big)\\to \\lambda$, emphasizing that we are proving the assertion $\\mu(*)\\to\\lambda$ using the parametric index $f(t)$.\n\n\nThrough a series of degenerations summarized in the table below by the corresponding parametric bases and indices, we obtain the main result of the second part of the paper.\n\n\n\\begin{theorem}\\label{main-geo}\nThe variety of $4$-dimensional complex nilpotent terminal algebras has 3 irreducible components: one of dimension $17$ determined by the family of algebras $\\D 4{01}(\\lambda,\\alpha,\\beta)$ and two of dimension 15 determined by the families of algebras $\\T 4{41}(\\alpha)$ and $\\T 4{43}(\\lambda,\\alpha)$. \n\\end{theorem}\n\n\n\n\\begin{proof}[{\\bf Proof}]\nThanks to \\cite{kppv} the algebras \n$\\mathfrak{L}_2,$ $\\mathfrak{L}_5,$ $\\mathfrak{L}_{11}$ and $\\mathfrak{N}_3(\\alpha)$ \n define irreducible components in the variety of $4$-dimensional nilpotent Leibniz algebras. \n Note that in \\cite{kppv} right Leibniz algebras were considered, and here we use their opposite versions which are left Leibniz algebras (and hence are terminal):\n$$ \n \\begin{array}{lllllllll}\n \n\\mathfrak{N}_3(\\alpha) & :& e_1e_1 = e_4 & e_1e_2 = -\\alpha e_4 & e_2e_1 = \\alpha e_4 & e_2e_2 = e_4& e_3e_3 = e_4 \\\\\n\n\\mathfrak{L}_2 &: & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\\\ \n\n\\mathfrak{L}_5 &:& e_1e_1 = e_3& e_1e_2 = e_3& e_2e_2 = e_4& e_1e_3=e_4 \\\\ \n\n\\mathfrak{L}_{11} &:& e_1e_1 = e_4& e_1e_2 = e_3 & e_1e_3=e_4 & e_2e_1=-e_3 & e_2e_2=e_4 & e_3e_1=-e_4. \\\\ \n\\end{array}\n$$\n The list of all $4$-dimensional nilpotent non-Leibniz terminal algebras was found in Theorem~\\ref{main-alg}. \n All these algebras degenerate from one of the families: $\\D 4{01}(\\lambda,\\alpha,\\beta)$, $\\T 4{41}(\\alpha)$ or $\\T 4{43}(\\lambda,\\alpha)$, as it is shown in the table below. By a straightforward computation we have $\\dim\\mathfrak{Der}(\\D 4{01}(\\lambda,\\alpha,\\beta))=2$ for $\\alpha,\\beta\\ne 0$, and $\\dim\\mathfrak{Der}(\\D 4{01}(\\lambda,\\alpha,\\beta))>2$ otherwise. Hence, the closure of the orbit of $\\D 4{01}(\\lambda,\\alpha,\\beta)$ has dimension $4^2-2+3=17$. A similar argument yields that the dimensions of the orbit closures of $\\T 4{41}(\\alpha)$ and $\\T 4{43}(\\lambda,\\alpha)$ are both equal to $15<17$. In particular, this shows that $\\D 4{01}(\\lambda,\\alpha,\\beta)$ cannot degenerate from $\\T 4{41}(\\alpha)$ or $\\T 4{43}(\\lambda,\\alpha)$. Moreover, $\\T 4{41}(\\alpha)$ and $\\T 4{43}(\\lambda,\\alpha)$\\ do not degenerate from $\\D 4{01}(\\lambda,\\alpha,\\beta)$, since the squares of $\\T 4{41}(\\alpha)$ and $\\T 4{43}(\\lambda,\\alpha)$ have dimension 3, while the square of $\\D 4{01}(\\lambda,\\alpha,\\beta)$ has dimension $2<3$. This completes the proof of the theorem.\n\n\\end{proof}\n\n\n\n\n\n\n{\\tiny \n\n\n\\begin{longtable}{|lcl|ll|}\n\\multicolumn{5}{c}{ {\\bf Table.} {\\it Degenerations of $4$-dimensional nilpotent terminal algebras.}}\\\\ \n\n\\hline\n$\\T {4}{03}\\left(\\frac{1}{1 - t}\\right)$ &$\\to$& $\\mathfrak{N}_3(\\alpha)$&\n$E^t_1= \\frac{2 \\alpha t}{\\sqrt{t-1}} e_1 - \\alpha t \\sqrt{t-1} e_2 + \\alpha t \\sqrt{t-1} e_3$ &\n$E^t_3= t e_3 + \\frac{t}{4} \\left(1 + \\frac{1 - t}{\\alpha^2} - t\\right) e_4$\\\\\n\n&&&\n$E^t_2= -t \\sqrt{t-1} e_2 + t \\sqrt{t-1} e_3$ &\n$E^t_4= t^2 e_4$ \\\\\n\n\n\\hline\n$\\T {4}{38}\\left(t^{-1}\\right)$ &$\\to$& $\\mathfrak{L}_2$&\n\n$E^t_1= e_1$ &\n$E^t_3= e_3$ \\\\\n&&& \n$E^t_2= e_2$&\n$E^t_4= t^{-1} e_4$\\\\\n\n\\hline\n$\\D {4}{01}\\left( 2t^{-1}, t^{-1}, t \\right)$ &$\\to$& $\\mathfrak{L}_5$&\n\n$E^t_1= 2 e_1 - 2 e_2 - 2 e_3$ &\n$E^t_3= 4 t e_3 - 4t^{-1} e_4$\\\\\n&&& \n$E^t_2= -2 t e_2 - 2 e_3$ &\n$E^t_4= 8 e_4$\\\\\n\n\\hline\n$\\D {4}{01}\\left( -\\frac 12, 1, \\frac{1 + t^2}{4}\\right)$ &$\\to$& $\\mathfrak{L}_{11}$&\n$E^t_1= 4t^{-1} e_1 - 2t^{-1} e_2$ &\n$E^t_3= -8t^{-1} e_3 + 32t^{-3} e_4$\\\\\n&&&\n$E^t_2= 4 e_2 - 8t^{-2} e_3$ &\n$E^t_4= 32t^{-2} e_4$\\\\\n\n\\hline\n\n\n\\color{black} $\\T {4}{43} \\left(0,0\\right)$ &\\color{black} $\\to$& \\color{black} $\\T {4}{01}$ & \n\n\\color{black} $E^t_1= te_1$ &\n\\color{black} $E^t_3= e_3$ \\\\\n&&& \n\\color{black} $E^t_2= t^2e_2$&\n\\color{black} $E^t_4= e_4$\\\\\n\\hline\n\n\n\\color{black}$\\T {4}{43} \\left(0, 0\\right)$ &\\color{black}$\\to$& \\color{black}$\\T {4}{02}$ & \n\n\\color{black} $E^t_1= te_1 + \\frac 12 t^{-2}e_2$ &\n\\color{black} $E^t_3= t^3e_3 + \\frac 12 e_4$ \\\\ \n&&& \n\\color{black} $E^t_2= t^2e_2 + \\frac 12 t^{-1}e_3 + \\frac 14 t^{-4}e_4$ &\n\\color{black} $E^t_4= e_4$\\\\\n\\hline\n\n$\\D {4}{01} \\left(0, t + \\alpha^{-1},t\\right)$ &$\\to$& $\\T {4}{03} (\\alpha)$ & \n$E^t_1= \\alpha^2 t e_2$ &\n$E^t_3= \\alpha^3 t^2 (e_1 - e_2)$\\\\\n&&&\n$E^t_2= \\alpha^4 t^2 e_3$ &\n$E^t_4= \\alpha^6 t^4 e_4$\n \\\\\n\\hline\n\n$\\T {4}{03} \\left(\\alpha \\right)$ &$\\to$& $\\T {4}{04} (\\alpha)$ & \n$E^t_1= t^{-1} e_1$ &\n$E^t_3= t^{-1} e_3$\\\\\n&&&\n$E^t_2= t^{-2} e_2$ &\n$E^t_4= t^{-3} e_4$\n \\\\\n\\hline\n\n$\\T {4}{07} \\left(\\frac{1}{t-1} \\right)$ &$\\to$& $\\T {4}{05}$ & \n\n$E^t_1= \\frac{(t-1)^2}{1 - 3 t + t^2} e_1 + \\frac{(t-1)^3}{(1 - 3 t + t^2)^2} e_2+ t e_3$ &\n\n$E^t_2= \\frac{(t-1)^4}{ (1 - 3 t + t^2)^2} e_2 + \\frac{(t-1)^4 t}{ (1 - 3 t + t^2)^2} e_4$ \\\\\n&&&\n$E^t_3= - \\frac{(t-1)^4}{ t (1 - 3 t + t^2)^2} e_2+e_3$ &\n$E^t_4= -\\frac{(t-1)^3}{ (1 - 3 t + t^2)^2} e_4$\n\n \\\\\n\\hline\n\n$\\T {4}{03} \\left(t \\right)$ &$\\to$& $\\T {4}{06}$ & \n\n$E^t_1= \\frac{t - 1}{1 + t^2} e_1 + \\frac{t^2 - 1}{(1 + t^2)^2} e_2 + \\frac{1 - t}{1 + t^2} e_3$ &\n$E^t_3= \\frac{(1 - t) t}{1 + t^2} e_3$ \\\\\n&&& \n$E^t_2= \\frac{(t - 1)^2}{(1 + t^2)^2} e_2 + \\frac{(1 - t)^3}{(1 + t^2)^3} e_4$ &\n$E^t_4= \\frac{(1 - t)^3 t}{(1 + t^2)^3} e_4$\\\\\n\\hline\n\n$\\T {4}{03} \\left( \\alpha \\right)$ &$\\to$& $\\T {4}{07} (\\alpha)$ & \n$E^t_1= -\\frac{1}{ 1 +\\alpha} e_1 - \\frac{1}{ (1 +\\alpha)^2} e_2 + \\frac{1}{ 1 +\\alpha} e_3$ &\n$E^t_2= \\frac{1}{(1 +\\alpha)^2} e_2 + \\frac{1}{ (1 +\\alpha)^3} e_4$\\\\\n&&&\n$E^t_3= t^{-1} (e_2-(1 +\\alpha) e_3)$ &\n$E^t_4= -\\frac{1}{ t(1 +\\alpha)} e_4$ \\\\\n\n\\hline\n\n\n$\\T {4}{03} \\left(-1 \\right)$ &$\\to$& $\\T {4}{08}$ & \n\n$E^t_1= t^{-1} e_1$ & $E^t_3= t^{-2} e_3$ \\\\\n&&&\n$E^t_2= t^{-2} e_2$ &\n$E^t_4= t^{-4} e_4$\\\\\n\n\n\\hline\n\n$\\T {4}{03} \\left(t+t^4 \\right)$ &$\\to$& $\\T {4}{09}$ & \n$E^t_1= \\frac{t-1}{t^2 + t^5} e_1 + \\frac{t-1}{t^3 (1 + t^3)^2}e_2 + \\frac{1 - t}{t^2 + t^5} e_3$ &\n$E^t_3= -\\frac{(t-1 ) (-1 + t - t^2 - t^3 + 2 t^4 + t^7)}{ t (1 + t^3)^2 (-1 + t + t^4)} e_3$ \\\\\n&&& \n$E^t_2= \\frac{(t-1)^2}{t^4 (1 + t^3)^2} e_2 + \\frac{t-1}{(1 + t^3)^2} e_3$& \n$E^t_4= -\\frac{(t-1)^2 (-1 + t - t^2 - t^3 + 2 t^4 + t^7)}{ t^5 (1 + t^3)^4} e_4$\\\\ \n\\hline\n \n$\\T {4}{03} (\\alpha )$ &$\\to$& $\\T {4}{10}(\\alpha )$ & \n\n$E^t_1= t^{-2} e_1$ & $E^t_3= t^{-3} e_3$\\\\\n&&&\n$E^t_2= t^{-4} e_2$ & $E^t_4= t^{-7} e_4$\\\\\n\\hline\n \n$\\T {4}{03} (1+t^{-2})$ &$\\to$& $\\T {4}{11}$ & \n$E^t_1= t^3 e_1$ & $E^t_3= t^5 e_3$ \\\\\n\n&&&\n$E^t_2= t^6 e_2 - t^6 e_3$ &\n$E^t_4= t^9 e_4$\\\\\n\\hline\n\n$\\T {4}{41} (t^{-1})$ &$\\to$& $\\T {4}{12}$ & \n\n$E^t_1= 3 e_1 + 3 e_2$ &\n$E^t_3= 3t^{-1} e_3 + 9t^{-1} e_4$ \\\\\n&&& \n$E^t_2= 9 e_2 + 9 e_3 + 9t^{-1} e_4$ &\n$E^t_4= 27t^{-1} e_4$\\\\\n\\hline\n\n\n\n$\\T {4}{12}$ &$\\to$& $\\T {4}{13}$ & \n\n$E^t_1= e_1$ & $E^t_3= t^{-1} e_3$ \\\\\n&&& $E^t_2= e_2$ & $E^t_4= t^{-1} e_4$\\\\\n\\hline\n\n\n$\\T {4}{03} (1\/t )$ &$\\to$& $\\T {4}{14}$ & \n\n$E^t_1= t^{-1} e_1$ & $E^t_3= t^{-2} e_3$ \\\\\n&&& $E^t_2= t^{-2} e_2$ & $E^t_4= t^{-5} e_4$\\\\\n\\hline\n\n\n$\\T {4}{03} (\\frac{1 - t}{\\alpha} )$ &$\\to$& $\\T {4}{15}(\\alpha)$ & \n\n$E^t_1= e_1 + \\frac{\\alpha}{ t ( \\alpha -1+ t)} e_2 - \\frac{\\alpha}{t ( \\alpha-1 + t)} e_3$ &\n$E^t_3= \\frac{1 - t}{\\alpha-1 + t} e_3 + \\frac{\\alpha (\\alpha t -1+ t^2)}{ t^3 ( \\alpha-1 + t)^2} e_4$ \\\\\n&&& \n$E^t_2= e_2 + \\frac{(t-1) t}{\\alpha-1 + t} e_3$ &\n$E^t_4= \\frac{-1 + t}{t ( \\alpha-1 + t)} e_4$\\\\\n\\hline\n\n\n$\\T {4}{03} (1 + t )$ &$\\to$& $\\T {4}{16}$ & \n\n\n$E^t_1= \\frac{t}{(1 + t) X^2} e_1 + \\frac{1 - t}{(1 + t) X^3} e_2 + \\frac{1}{(1 + t) X^2}e_3$ &\n$E^t_3= \\frac{t}{(1 + t) X^3} e_3$ \\\\\n\\multicolumn{3}{|l|}{$X=-1 + t + t^2$}&\n$E^t_2= \\frac{t^2}{(1 + t)^2 X^4} e_2 + \\frac{ 1 + t - t^2}{(1 + t)^2 X^5} e_4$ &\n$E^t_4= \\frac{t^2}{(1 + t)^2 X^6} e_4$\\\\\n\\hline\n\n\n\n$\\T {4}{03} (\\frac{1}{1 - t^2} )$ &$\\to$& $\\T {4}{17}(\\alpha)$ & \n\n$E^t_1= t^2 X e_1 + t^2(-1 + t^2) X^2 e_2 + (t^2 - t^4) X^2 e_3$ &\n$E^t_3= t^3 X^2 e_3$ \\\\\n\n\\multicolumn{3}{|l|}{$X=(-1 + \\alpha + t^2)^{-1}$}&\n$E^t_2= t^4 X^2 e_2 + t^4 ( t^2-1)(\\alpha + t^2) X^4 e_4$ &\n$E^t_4= t^6 X^4 e_4$\\\\\n\\hline\n\n\n\n$\\T 4{19}$ &$\\to$& $\\T 4{18}$ & \n\n$E_1^t = \\sqrt{1-t^2}e_1 + te_2 - te_3$ & $E_3^t = e_3$\\\\\n&&& $E_2^t = e_2$ & $E_4^t = \\sqrt{1-t^2}e_4$\\\\\n\\hline\n\n$\\T 4{23}(t^{-1},-t-1)$ &$\\to$& $\\T 4{19}$ & \n\n$E_1^t = t^2e_1 - \\frac{t^3}{t + 1}e_3$ & $E_3^t = t^4e_3$ \\\\\n&&& $E_2^t = t^2e_2$ & $E_4^t = t^5e_4$\\\\\n\\hline\n\n$\\T 4{23}(it^{-1},t^{-1})$ &$\\to$& $\\T 4{20}$ & \n\n$E_1^t = te_1$ & $E_3^t = t^2e_3$ \\\\\n&&& $E_2^t = te_2$ & $E_4^t = t^2e_4$\\\\\n\\hline\n\n\n$\\T 4{20}$ &$\\to$& $\\T 4{21}$ & \n\n$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\T 4{23}(\\alpha,\\beta)$ &$\\to$& $\\T 4{22}(\\alpha,\\beta)$ & \n\n$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$ \\\\\n&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\beta^2t^{-2},t+\\alpha,t\\right)$ &$\\to$& $\\T 4{23}(\\alpha,\\beta)$ & \n\n$E_1^t = t^2\\beta^{-2}(e_1 - e_2)$ & $E_3^t = t^2\\beta^{-2}e_3$ \\\\\n&&& $E_2^t = t\\beta^{-1}e_2$ & $E_4^t = t^4\\beta^{-4}e_4$\\\\\n\\hline\n\n$\\T 4{23}(\\alpha,-(\\alpha + 1)t)$ &$\\to$& $\\T 4{24}(\\alpha)$ & \n\n$E_1^t = te_1 - t(\\alpha + 1)^{-1}e_3$ & $E_3^t = t^2e_3$ \\\\\n&&& $E_2^t = te_2$ & $E_4^t = t^3e_4$\\\\\n\\hline\n\n$\\T 4{23}(-1,t)$ &$\\to$& $\\T 4{25}(\\alpha)$ & \n\n$E_1^t = e_1 + \\alpha t^{-1}e_3$ & $E_3^t = e_3$ \\\\\n&&& $E_2^t = e_2$ & $E_4^t = e_4$\\\\\n\\hline\n\n$\\T 4{23}(t,i)$ &$\\to$& $\\T 4{26}(\\alpha)$ & \n\n$E_1^t = \\frac i{\\alpha + i}e_1 -\\frac{\\alpha i}{(\\alpha + i)^2t}e_3$ & $E_3^t = -\\frac 1{(\\alpha + i)^2}e_3 + \\frac{\\alpha}{(\\alpha + i)^3t}e_4$ \\\\\n&&& $E_2^t = \\frac i{\\alpha + i}e_2 - \\frac{\\alpha}{(\\alpha + i)^2t}e_3$ & $E_4^t = -\\frac i{(\\alpha + i)^3}e_4$\\\\\n\\hline\n\n$\\T 4{23}\\left(\\frac{(\\alpha^2 + 1)t}{\\alpha - t},\\alpha\\right)$ &$\\to$& $\\T 4{27}(\\alpha)$ & \n\n$E_1^t = te_1 +\\frac{(t-\\alpha)t}{\\alpha(\\alpha^2 + 1)}e_3$ & $E_3^t = t^2e_3 - \\frac{\\alpha(t-\\alpha)t^2}{\\alpha^2 + 1}e_4$ \\\\\n&&& $E_2^t = te_2 - \\frac{(t-\\alpha)t}{\\alpha^2 + 1}e_3$ & $E_4^t = t^3e_4$\\\\\n\\hline\n\n$\\T 4{23}\\left(\\frac{i\\alpha(1-t)}t,\\frac{\\alpha\\sqrt{1-t}}t\\right)$ &$\\to$& $\\T 4{28}(\\alpha)$ & \n\n$E_1^t = \\frac{\\alpha^2(t - 1)}t\\left(-i e_1 + \\frac 1{\\sqrt{1-t}} e_2 - (\\alpha - i) e_3\\right)$ & $E_3^t = -\\frac{\\alpha^4(t - 1)}t e_3$ \\\\\n&&& $E_2^t = \\frac{\\alpha^2(t - 1)}t(-e_1 + i\\sqrt{1-t} e_2 + e_3)$ & $E_4^t = \\frac{\\alpha^6(t - 1)^2}{t^2}e_4$\\\\\n\\hline\n\n$\\T 4{28}(\\alpha)$ &$\\to$& $\\T 4{29}(\\alpha)$ & \n\n$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$ \\\\\n&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\T 4{23}\\left(1,(1 - \\alpha)\\sqrt{t - 1}\\right)$ &$\\to$& $\\T 4{30}(\\alpha)$ & \n \\multicolumn{2}{l|}{$E_1^t = \\frac{(\\alpha - 1)t(t - 1)}D\\left(-e_1+\\frac 1{\\sqrt{t - 1}} e_2-\\frac{(\\alpha - 2)t - \\alpha + 1}D e_3\\right)$}\\\\\n \\multicolumn{3}{|l|}{$D=\\alpha^2t^2 - 2\\alpha^2t - 2\\alpha t^2 + \\alpha^2 + 2\\alpha t + t^2 + 2t - 1$}\n& \\multicolumn{2}{l|}{$E_2^t = \\frac{i(\\alpha - 1)t(t - 1)}D\\left(e_1+\\sqrt{t - 1}e_2-\\frac{(\\alpha-1)t^2 - (2\\alpha-3)t + \\alpha - 1}D e_3\\right)$}\\\\\n&&& \\multicolumn{2}{l|}{$E_3^t = \\frac{(\\alpha - 1)^2(t - 1)t^3}{D^2}\\left(e_3+\\frac{((\\alpha-1) t - \\alpha)(\\alpha - 1)(t - 1)}De_4\\right)$} \\\\\n&&& \\multicolumn{2}{l|}{$E_4^t = \\frac{i(\\alpha - 1)^3(t - 1)^2t^4}{D^3}e_4$}\\\\\n\\hline\n\n$\\T 4{30}(\\alpha)$ &$\\to$& $\\T 4{31}(\\alpha)$ & \n\n$E_1^t = -\\frac{i(\\alpha^2 - 1)}{\\alpha t}e_1 - \\frac i{\\alpha t}e_3$ & $E_3^t = -\\frac{(\\alpha^2 - 1)^2}{\\alpha^2t^2}e_3 - \\frac{i(\\alpha + 1)^2(\\alpha - 1)}{\\alpha t^2}e_4$\\\\\n&&& $E_2^t = -\\frac{i(\\alpha^2 - 1)}{\\alpha t}e_2 + \\frac 1t e_3$ & $E_4^t = i\\frac{(\\alpha^2 - 1)^3}{\\alpha^3 t^3}e_4$\\\\\n\\hline\n\n$\\T 4{23}(\\sqrt{t-1} + 1,1)$ &$\\to$& $\\T 4{32}$ & \n\n$E_1^t = \\frac 1{(\\sqrt{t-1} + 1)^2}\\left(e_1 - \\sqrt{t-1}e_2 -\\frac{\\sqrt{t-1}}{(\\sqrt{t-1} + 1)^2}e_3\\right)$ & $E_3^t = \\frac t{(\\sqrt{t-1} + 1)^4}e_3$ \\\\\n&&& $E_2^t = \\frac i{(\\sqrt{t-1} + 1)^2}\\left(-e_1 - \\frac 1{\\sqrt{t-1}}e_2 + \\frac {\\sqrt{t - 1}}{(\\sqrt{t - 1} + 1)^2}e_3\\right)$ & $E_4^t = -\\frac {it}{(\\sqrt{t - 1} + 1)^6}e_4$\\\\\n\n\\hline\n$\\T 4{23}(\\sqrt{t-1} - 1,1)$ &$\\to$& $\\T 4{33}$ & \n\n$E_1^t = \\frac 1t\\left(e_1 - \\sqrt{t - 1}e_2 -\\frac{\\sqrt{t - 1}}t e_3\\right)$ & $E_3^t = \\frac 1t e_3$ \\\\\n&&& $E_2^t = \\frac it\\left(-e_1 -\\frac 1{\\sqrt{t - 1}}e_2 + \\frac{\\sqrt{t - 1}}t e_3\\right)$ & $E_4^t = -\\frac i{t^2}e_4$\\\\\n\\hline\n\n$\\T 4{23}(\\sqrt{t-1} - 1,1)$ &$\\to$& $\\T 4{34}$ & \n\n\\multicolumn{2}{l|}{$E_1^t = \\frac{\\sqrt{t-1} - 1}{t\\sqrt{t-1} - t + 2}\\left(e_1 -\\sqrt{t-1}e_2 -\\frac{\\sqrt{t-1}(t\\sqrt{t-1} - t + 4)}{t(t\\sqrt{t-1} - t + 2)}e_3\\right)$}\\\\\n&&& \n\\multicolumn{2}{l|}{$E_2^t = \\frac{i(\\sqrt{t-1}-1)}{t\\sqrt{t-1} - t + 2}\\left(-e_1 -\\frac 1{\\sqrt{t-1}}e_2 +\\frac{t^2 - t\\sqrt{t-1} - 3t + 4\\sqrt{t-1}}{t(t\\sqrt{t-1} - t + 2)}e_3\\right)$}\\\\\n&&& \n\\multicolumn{2}{l|}{$E_3^t = \\frac{(\\sqrt{t-1} - 1)^2}{(t\\sqrt{t-1} - t + 2)^2}\\left(te_3 +\\frac{2\\sqrt{t-1}(t - \\sqrt{t-1})}{t\\sqrt{t-1} - t + 2}e_4\\right)$}\\\\\n&&& \n\\multicolumn{2}{l|}{$E_4^t = -\\frac{it(\\sqrt{t-1} - 1)^3}{(t\\sqrt{t-1} - t + 2)^3}e_4$}\\\\\n\\hline\n\n$\\D 4{01}\\left(\\frac 14,\\frac 1{2t},\\frac 1t\\right)$ &$\\to$& $\\T 4{35}$ & \n\n$E_1^t = -t^{-1}(2e_1 - e_2 - 2e_3)$ & $E_3^t = -2t^{-1}e_3$ \\\\\n&&& $E_2^t = t^{-1}(2e_1 - (2t + 1)e_2 - 4e_3)$ & $E_4^t = 4t^{-2}e_4$\\\\\n\\hline\n\n$\\T 4{35}$ &$\\to$& $\\T 4{36}$ & \n\n$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$ \\\\\n&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\T 4{38}\\left(t^{-1}\\right)$ &$\\to$& $\\T 4{37}$ & \n\n$E_1^t = e_1 - \\frac 1{2t}e_2$ & $E_3^t = e_3 - \\frac 1{2t}(1 + \\frac 1t)e_4$ \\\\\n&&& $E_2^t = e_2 - \\frac 1{2t}e_3 + \\frac 1{4t^2}e_4$ & $E_4^t = \\frac 1te_4$\\\\\n\\hline\n\n\\color{black} $\\T 4{43}\\left((1-\\alpha)(t + 1)t^{-1}, t(\\alpha - 1)^{-1}\\right)$ &\\color{black} $\\to$& \\color{black} $\\T 4{38}(\\alpha)$ & \n\n\\color{black} $E^t_1= e_1$ & \n\\color{black} $E^t_3= (1-\\alpha)t^{-1}e_3$ \\\\\n&&& \n\\color{black} $E^t_2= e_2$ &\n\\color{black} $E^t_4= e_4$\\\\\n\\hline\n\n\n\\color{black} $\\T 4{43}\\left(\\lambda+t, -t^{-1}\\right)$ &\\color{black} $\\to$& \\color{black} $\\T 4{39}(\\lambda)$ & \n\n\\multicolumn{2}{l|}{\\color{black} $E_1^t = e_1 - De_2$}\\\\\n \\multicolumn{3}{|l|}{\\color{black} $D=(2t^2 + (4\\lambda +5)t + 2\\lambda^2 + 3\\lambda + 1)^{-1}$}\n& \\multicolumn{2}{l|}{\\color{black} $E_2^t = e_2 - (t + \\lambda + 1)De_3 + D^2 e_4$}\\\\\n&&& \\multicolumn{2}{l|}{\\color{black} $E_3^t = e_3 + t^{-1}(2t + 3\\lambda + 3)De_4$} \\\\\n&&& \\multicolumn{2}{l|}{\\color{black} $E_4^t = -t^{-1}e_4$}\\\\\n\\hline\n\\color{black}\n$\\T 4{43}\\left(\\lambda + t, \\frac 2{(\\lambda + 1)(2\\lambda + 1)}\\right)$ &\\color{black}$\\to$& \\color{black}$\\T 4{40}(\\lambda)$ & \n\n \\multicolumn{2}{l|}{\\color{black} $E_1^t = e_1 - \\frac 12(\\lambda + 1)(2\\lambda + 1)D e_2$}\\\\\n \\multicolumn{3}{|l|}{\\color{black} $D=(t(2t + 4\\lambda + 3))^{-1}$}\n& \\multicolumn{2}{l|}{\\color{black} $E_2^t = e_2 - \\frac 12(\\lambda + 1)(2\\lambda + 1)(t + \\lambda + 1)De_3 + \\frac 14(\\lambda + 1)^2(2\\lambda + 1)^2D^2e_4$}\\\\\n&&& \\multicolumn{2}{l|}{\\color{black} $E_3^t = e_3 - \\frac 12(6t + 2\\lambda^2 + 9\\lambda + 7)De_4$} \\\\\n&&& \\multicolumn{2}{l|}{\\color{black} $E_4^t = \\frac 1{(\\lambda + 1)(2\\lambda + 1)}e_4$}\\\\\n\\hline\n\n$\\T 4{41}(t^{-2})$ &$\\to$& $\\T 4{42}$ & \n\n$E_1^t = t^{-1}e_1 + t^{-1}e_2$ & $E_3^t = t^{-3}e_3 + 3t^{-3}e_4$\\\\\n&&& $E_2^t = t^{-2}e_2 + t^{-2}e_3 + t^{-4}e_4$ & $E_4^t = t^{-5}e_4$\\\\\n\\hline\n\n$\\T 4{41}(0)$ &$\\to$& $\\T 4{44}$ & \n\n$E_1^t = t^{-1}e_1 + t^{-1}e_2$ & $E_3^t = t^{-3}e_3 + 3t^{-3}e_4$ \\\\\n&&& $E_2^t = t^{-2}e_2 + t^{-2}e_3$ & $E_4^t = t^{-5}e_4$\\\\\n\\hline\n\n\n$\\D 4{01}\\left(\\lambda, \\alpha,\\beta\\right)$ &$\\to$& $\\D 4{02}\\left(\\lambda, \\alpha,\\beta\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left( \\frac{\\lambda - t}{(t-1)^2}, t - \\alpha t ,\\alpha - \\alpha t \\right)$ &$\\to$& $\\D 4{03}\\left(\\lambda, \\alpha \\right)$ \n& \n\n\\multicolumn{2}{l|}{$E^t_1= \\frac{(1 - \\alpha ) \\alpha (-1 + t)^2 t}{1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t)} e_1 + \n\\frac{(-1 + \\alpha ) \\alpha (-1 + t) t^2}{1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t)} e_2 - \n\\frac{(-1 + \\alpha )^2 \\alpha (-1 + t)^2 t}{ (1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t))^2} e_3$} \\\\\n\n&&& \\multicolumn{2}{l|}{$E^t_2= \\frac{(-1 + \\alpha ) \\alpha (-1 + t) t}{ 1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t)}e_2 - \n\\frac{(-1 + \\alpha ) \\alpha ^2 (-1 + t)^2 t}{ (1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t))^2} e_3$ }\\\\\n\n&&&\n\\multicolumn{2}{l|}{$E^t_3= \\frac{(-1 + \\alpha )^2 \\alpha ^2 (-1 + t)^2 t^2}{(1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t))^2} e_3 + \n\\frac{ (-1 + \\alpha )^2 \\alpha ^4 (-1 + t)^4 t^2}{ (1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t))^3} e_4$}\\\\\n\n\n&&&\\multicolumn{2}{l|}{$E^t_4= -\\frac{ (-1 + \\alpha )^3 \\alpha ^3 (-1 + t)^4 t^3}{ (1 + \\alpha ^2 \\lambda + t - \\alpha (1 + t))^3} e_4$ }\\\\\n \n \n\\hline\n\n\n$\\D 4{01}\\left(\\lambda, 0,\\alpha \\right)$ &$\\to$& $\\D 4{04}\\left(\\lambda, \\alpha\\right)$ \n\n\n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n \\hline\n \n \n$\\D 4{01}\\left(\\lambda, t , \\frac{1 + \\sqrt{1 - 4 \\lambda}}{2 \\lambda }\\right)$ &$\\to$& $\\D 4{05}\\left(\\lambda, \\alpha\\right)$ \n&\n\n$E^t_1= -\\frac{1 + \\sqrt{1 - 4 \\lambda}}{D} e_1 + \n\\frac{(1 + \\sqrt{1 - 4 \\lambda}) (D-2 \\lambda \\alpha )}{ t D^2} e_3$ &\n \n$E^t_3= \\frac{(1 + \\sqrt{1 - 4 \\lambda})^2}{ D^2} e_3 - \\frac{(1 + \\sqrt{1 - 4 \\lambda})^4}{ 2 \\lambda t D^3} e_4$ \\\\\n\n\\multicolumn{3}{|l|}{$D=1 + \\sqrt{1 - 4 \\lambda} + 2 \\lambda (-1 + \\alpha + \\alpha t)$}&\n$E^t_2= -\\frac{1 + \\sqrt{1 - 4 \\lambda}}{ D} e_2 + \\frac{(1 + \\sqrt{1 - 4 \\lambda})^2}{ t D^2} e_3$ &\n\n$E^t_4= -\\frac{(1 + \\sqrt{1 - 4 \\lambda})^3}{ D^3} e_4$\n \\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda, t , \\frac{1 -\\sqrt{1 - 4 \\lambda}}{2 \\lambda }\\right)$ &$\\to$& $\\D 4{06}\\left(\\lambda, \\alpha\\right)$ \n&\n\n$E^t_1= -\\frac{1 - \\sqrt{1 - 4 \\lambda}}{D} e_1 + \n\\frac{(1 - \\sqrt{1 - 4 \\lambda}) (D-2 \\lambda \\alpha )}{ t D^2} e_3$ &\n \n\n$E^t_3= \\frac{(1 - \\sqrt{1 - 4 \\lambda})^2}{ D^2} e_3 - \\frac{(1 - \\sqrt{1 - 4 \\lambda})^4}{ 2 \\lambda t D^3} e_4$ \\\\\n\n\\multicolumn{3}{|l|}{$D=1 - \\sqrt{1 - 4 \\lambda} + 2 \\lambda (-1 + \\alpha + \\alpha t)$}&\n$E^t_2= -\\frac{1 - \\sqrt{1 - 4 \\lambda}}{ D} e_2 + \\frac{(1 - \\sqrt{1 - 4 \\lambda})^2}{ t D^2} e_3$ &\n\n\n$E^t_4= -\\frac{(1 - \\sqrt{1 - 4 \\lambda})^3}{ D^3} e_4$ \n \\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda, t , \\frac{1 + \\sqrt{1 - 4 \\lambda}}{2 \\lambda }\\right)$ &$\\to$& $\\D 4{07}\\left(\\lambda\\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1=-\\frac{D - (1 + \\sqrt{1 - 4 \\lambda}) \\lambda}{D \\lambda (1 + t)} e_1 + \n\\frac{2 (D -(1 + \\sqrt{1 - 4 \\lambda}) \\lambda) (D -2 (1 + \\sqrt{1 - 4 \\lambda})\\lambda + 2 \\lambda^2)}{D^3 \\lambda (1 + t)^2} e_3$} \\\\\n \n\\multicolumn{3}{|l|}{$D=1 + \\sqrt{1 - 4 \\lambda} - 2 \\lambda$}&\n\\multicolumn{2}{l|}{$E^t_2= -\\frac{D - (1 + \\sqrt{1 - 4 \\lambda}) \\lambda}{D \\lambda (1 +t)}e_2$} \\\\\n\n&&&\n$E^t_3= \\frac{(D - (1 + \\sqrt{1 - 4\\lambda}) \\lambda)^2}{D^2 \\lambda^2 (1 + t)^2} e_3$ &\n$E^t_4= \\frac{(D - (1 + \\sqrt{1 - 4 \\lambda}) \\lambda)^3}{ D^3 \\lambda^3 (1 + t)^3} e_4$\\\\\n\\hline\n \n$\\D 4{01}\\left(\\lambda, t , \\frac{1 - \\sqrt{1 - 4 \\lambda}}{2 \\lambda }\\right)$ &$\\to$& $\\D 4{08}\\left(\\lambda\\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1=-\\frac{D - (1 - \\sqrt{1 - 4\\lambda}) \\lambda}{D\\lambda (1 + t)} e_1 + \n\\frac{2 (D - (1 - \\sqrt{1 - 4\\lambda}) \\lambda) (D - 2 (1 - \\sqrt{1 - 4\\lambda}) \\lambda + 2\\lambda^2)}{D^3\\lambda (1 + t)^2} e_3$} \\\\\n \n\\multicolumn{3}{|l|}{$D=1 - \\sqrt{1 - 4\\lambda} - 2\\lambda$}&\n\\multicolumn{2}{l|}{$E^t_2= -\\frac{D - (1- \\sqrt{1 - 4\\lambda})\\lambda}{D\\lambda (1 +t)}e_2$} \\\\\n\n&&&\n$E^t_3= \\frac{(D - (1 - \\sqrt{1 - 4\\lambda})\\lambda)^2}{D^2\\lambda^2 (1 + t)^2} e_3$ &\n$E^t_4=- \\frac{(D - (1 - \\sqrt{1 - 4\\lambda})\\lambda)^3}{ D^3\\lambda^3 (1 + t)^3} e_4$\\\\\n\\hline\n \n$\\D 4{01}\\left( \\frac{\\lambda - t}{(t-1)^2}, \\alpha - t^2, t - t^2 \\right)$ &$\\to$& $\\D 4{09}\\left(\\lambda, \\alpha \\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1=(1-t)^2 t ( t^2-\\alpha) D e_1 + (1 -t) t^2 ( t^2-\\alpha) D e_2 + ( t-1 )^3\n t (\\alpha - t^2) D^2 e_3$ } \\\\\n \n\\multicolumn{3}{|l|}{$D=(-1 + \\alpha (-1 + t) + t - \\lambda t^2)^{-1}$}&\n\\multicolumn{2}{l|}{$E^t_2= (1 - t) t (-\\alpha + t^2) D e_2 - (-1 + t)^2 t^2 (-\\alpha + t^2) D^2 e_3$ } \\\\\n\n&&&\n\\multicolumn{2}{l|}{$E^t_3=(1 - t)^2 t^2 (\\alpha - t^2)^2 D^2 e_3 + (1 - t)^4 t^4 (\\alpha - t^2)^2 D^3 e_4$ } \\\\\\\n\n\n&&&\n\\multicolumn{2}{l|}{$E^t_4=( 1-t )^4 ( t^3-\\alpha t)^3 D^3 e_4$}\\\\\n\\hline\n \n\n$\\D 4{09}\\left(\\lambda, \\alpha \\right)$ &$\\to$& $\\D 4{10}\\left(\\lambda, \\alpha \\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n\n\n$\\D 4{01}\\left( \\frac{\\lambda - t}{(t-1)^2}, t^2-1, t^2 - t\\right)$ &$\\to$& $\\D 4{11}\\left(\\lambda, \\alpha \\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1=(1 - t)^3 (1 + t) D e_1 + (1 - t)^2 t (1 + t) D e_2 - \\alpha (1 - t)^3 (1 + t)^2 D^2t^{-1} e_3$ } \\\\\n \n\\multicolumn{3}{|l|}{$D=(1 + \\alpha \\lambda t - t^2)^{-1}$}&\n\\multicolumn{2}{l|}{$E^t_2= (1 - t)^2 (1 + t) D e_2 - \\alpha (1 - t)^3 (1 + t) D^2 e_3$ } \\\\\n\n&&&\n\\multicolumn{2}{l|}{$E^t_3= (1 - t)^4 (1 + t)^2 D^2 e_3 + \\alpha (1 - t)^6 t (1 + t)^2 D^3 e_4 $} \\\\\\\n\n\n&&&\n\\multicolumn{2}{l|}{$E^t_4=(1 - t)^7 (1 + t)^3 D^3 e_4$}\\\\\n\\hline\n\n\n$\\D 4{01}\\left(\\lambda\\left(1-\\frac {\\alpha t}{\\Theta^2}\\right), \\frac 1\\Theta\\left(\\alpha+\\frac{\\lambda}{t}\\right), \\frac 1t\\right)$ &$\\to$& $\\D 4{12}\\left(\\lambda,\\alpha\\right)$\n& $E_1^t = -\\alpha\\Theta t^{-1}e_1 + \\alpha\\lambda t^{-1}e_2 + \\alpha(\\Theta - \\alpha)t^{-1}e_3$ & $E_3^t = -\\alpha^3 t^{-1}e_3$\\\\\n &&& $E_2^t = -\\alpha t^{-1}e_1 + \\alpha(\\alpha t + \\lambda)(\\Theta t)^{-1}e_2 + \\alpha t^{-1}e_3$ & $E_4^t = \\alpha^4 t^{-2}e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1-\\frac {\\alpha t}{\\Psi^2}\\right), \\frac 1{\\Psi}\\left(\\alpha+\\frac{\\lambda}{t}\\right), \\frac 1t\\right)$ &$\\to$& $\\D 4{13}\\left(\\lambda,\\alpha\\right)$ \n& $E_1^t = -\\alpha\\Psi t^{-1}e_1 + \\alpha\\lambda t^{-1}e_2 + \\alpha(\\Psi - \\alpha)t^{-1}e_3$ & $E_3^t = -\\alpha^3 t^{-1}e_3$\\\\\n\\multicolumn{3}{|l|}{$\\Psi=1-\\Theta$}&\n $E_2^t = -\\alpha t^{-1}e_1 + \\alpha(\\alpha t + \\lambda)(\\Psi t)^{-1}e_2 + \\alpha t^{-1}e_3$ & $E_4^t = \\alpha^4 t^{-2}e_4$\\\\\n\\hline\n\n$\\D 4{12}\\left(\\lambda,\\alpha\\right)$ &$\\to$& $\\D 4{14}\\left(\\lambda,\\alpha\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{13}\\left(\\lambda,\\alpha\\right)$ &$\\to$& $\\D 4{15}\\left(\\lambda,\\alpha\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(t^2, \\alpha , t^{-1}\\right)$ &$\\to$& $\\D 4{16}\\left(\\alpha\\right)$ \n&\n\n$E^t_1= -\\alpha D e_1 + \\alpha(\\alpha t - 1) D^2 e_3$ &\n \n$E^t_3= -\\alpha^3 t D^2 e_3 -\\alpha^3 D^3 e_4$ \\\\\n\n\\multicolumn{3}{|l|}{$D=((\\alpha^2 + \\alpha + 1)t - \\alpha - 1)^{-1}$}&\n$E^t_2= -\\alpha D e_1 + \\alpha^2t D e_2 + \\alpha(\\alpha t - 1) D^2 e_3$ &\n\n$E^t_4= \\alpha^4 t D^3 e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(t^2, -1 , t^{-1}\\right)$ &$\\to$& $\\D 4{17}\\left(\\alpha\\right)$ \n&\n\n$E^t_1= D e_1 + ((\\alpha - 2) t + \\alpha)D^2 e_3$ &\n \n$E^t_3= tD^2 e_3 + \\alpha D^3 e_4$ \\\\\n\n\\multicolumn{3}{|l|}{$D=((\\alpha + 1)t)^{-1}$}&\n$E^t_2= D e_1 + tD e_2 + ((\\alpha - 1)t + \\alpha )D^2 e_3$ &\n\n$E^t_4= tD^3 e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Theta^2}\\right), \\alpha + \\Psi, 1\\right)$ &$\\to$& $\\D 4{18}\\left(\\lambda,\\alpha\\right)$ \n&\n\\multicolumn{2}{l|}{$E^t_1= t\\Theta^3D(\\alpha + \\Psi)\\left( e_1 - \\Psi e_2 - \\Theta(\\Theta^2\\alpha + \\alpha t + \\lambda) D e_3\\right)$}\\\\\n\n\\multicolumn{3}{|l|}{$D=(\\alpha t^2+\\Theta\\alpha(\\alpha + \\Theta) t+\\Theta^2(\\alpha + 1)(\\Theta\\alpha + \\Psi))^{-1}$}&\n\n\\multicolumn{2}{l|}{$E^t_2= t\\Theta(\\alpha + \\Psi)D\\left(\\Theta e_1+(t-\\lambda)e_2 - (t^2+(\\alpha - \\Psi)\\Theta t+\\Theta(\\Theta\\alpha + \\Psi)\\Theta^2) D e_3\\right)$}\\\\\n\n\\multicolumn{3}{|l|}{$\\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_3= t^3\\Theta^4(\\alpha + \\Psi)^2D^2\\left(e_3 -\\Theta(\\Theta\\alpha + t) D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t^4\\Theta^6(\\alpha + \\Psi)^3D^3e_4$}\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Psi^2}\\right), \\alpha + \\Theta, 1\\right)$ &$\\to$& $\\D 4{19}\\left(\\lambda,\\alpha\\right)$ \n&\n\\multicolumn{2}{l|}{$E^t_1= t\\Psi^3D(\\alpha + \\Theta)\\left( e_1 - \\Theta e_2 - \\Psi(\\Psi^2\\alpha + \\alpha t + \\lambda) D e_3\\right)$}\\\\\n\n\\multicolumn{3}{|l|}{$D=(\\alpha t^2+\\Psi\\alpha(\\alpha + \\Psi) t+\\Psi^2(\\alpha + 1)(\\Psi\\alpha + \\Theta))^{-1}$}&\n\n\\multicolumn{2}{l|}{$E^t_2= t\\Psi(\\alpha + \\Theta)D\\left(\\Psi e_1+(t-\\lambda)e_2 - (t^2+(\\alpha - \\Theta)\\Psi t+\\Psi(\\Psi\\alpha + \\Theta)\\Psi^2) D e_3\\right)$}\\\\\n\n\\multicolumn{3}{|l|}{$\\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_3= t^3\\Psi^4(\\alpha + \\Theta)^2D^2\\left(e_3 -\\Psi(\\Psi\\alpha + t) D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t^4\\Psi^6(\\alpha + \\Theta)^3D^3e_4$}\\\\\n\\hline\n\n$\\D 4{18}\\left(\\lambda,\\alpha\\right)$ &$\\to$& $\\D 4{20}\\left(\\lambda,\\alpha\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{19}\\left(\\lambda,\\alpha\\right)$ &$\\to$& $\\D 4{21}\\left(\\lambda,\\alpha\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Theta^2}\\right), -\\frac{\\lambda^2}{\\Theta^3}, 1\\right)$ &$\\to$& $\\D 4{22}\\left(\\lambda\\right)$ \n&\n\\multicolumn{2}{l|}{$E^t_1= \\Theta^2D(\\Theta e_1 - \\lambda e_2 - \\Theta\\lambda D e_3)$}\\\\\n\n\\multicolumn{3}{|l|}{$D=(\\Psi(t-2\\lambda+3\\Theta-1))^{-1}$}&\n\n\\multicolumn{2}{l|}{$E^t_2= \\Theta D\\left(\\Theta e_1 + (t-\\lambda)e_2 - \\Theta^3\\lambda^{-1}((2\\Theta - 1)t - \\Theta (2\\lambda + 1) + 1) D e_3\\right)$}\\\\\n\n\\multicolumn{3}{|l|}{$\\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_3= \\Theta^4D^2\\left(t e_3 - \\Theta^3\\lambda^{-1}(2\\Theta - 1)(t - \\Psi) D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t\\Theta^6 D^3 e_4$}\n \n \\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Psi^2}\\right), -\\frac{\\lambda^2}{\\Psi^3}, 1\\right)$ &$\\to$& $\\D 4{23}\\left(\\lambda\\right)$ \n&\n\\multicolumn{2}{l|}{$E^t_1= \\Psi^2D(\\Psi e_1 - \\lambda e_2 - \\Psi\\lambda D e_3)$}\\\\\n\n\\multicolumn{3}{|l|}{$D=(\\Theta(t-2\\lambda+3\\Psi-1))^{-1}$}&\n\n\\multicolumn{2}{l|}{$E^t_2= \\Psi D\\left(\\Psi e_1 + (t-\\lambda)e_2 - \\Psi^3\\lambda^{-1}((2\\Psi - 1)t - \\Psi (2\\lambda + 1) + 1) D e_3\\right)$}\\\\\n\n\\multicolumn{3}{|l|}{$\\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_3= \\Psi^4D^2\\left(t e_3 - \\Psi^3\\lambda^{-1}(2\\Psi - 1)(t - \\Theta) D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t\\Psi^6 \\lambda^{-1} D^3 e_4$}\n \n \\\\\n \n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1-\\frac t{\\Theta^2}\\right), -\\frac{\\lambda^2}{\\Theta^3}, 1\\right)$ &$\\to$& $\\D 4{24}\\left(\\lambda\\right)$ \n&\n\n$E^t_1= -\\Theta D\\left(\\Theta e_1 - \\lambda e_2 + \\Theta^2 D e_3\\right)$ &\n \n$E^t_3= -t\\Theta^2 D^2 e_3$ \\\\\n\n\\multicolumn{3}{|l|}{$D=(t - 2\\Theta + 1)^{-1}$}&\n$E^t_2= -D\\left(\\Theta e_1 - (t+\\lambda) e_2 + \\Theta^2 D e_3\\right)$ &\n\n$E^t_4= t\\Theta^3 D^3 e_4$ \\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1-\\frac t{\\Psi^2}\\right), -\\frac{\\lambda^2}{\\Psi^3}, 1\\right)$ &$\\to$& $\\D 4{25}\\left(\\lambda\\right)$ \n&\n\n$E^t_1= -\\Psi D\\left(\\Psi e_1 - \\lambda e_2 + \\Psi^2 D e_3\\right)$ &\n \n$E^t_3= -t\\Psi^2 D^2 e_3$ \\\\\n\n\\multicolumn{3}{|l|}{$D=(t - 2\\Psi + 1)^{-1},\\ \\Psi=1-\\Theta$}&\n$E^t_2= -D\\left(\\Psi e_1 - (t+\\lambda) e_2 + \\Psi^2 D e_3\\right)$ &\n\n$E^t_4= t\\Psi^3\\lambda^{-1} D^3 e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Theta^2}\\right), -\\Theta, 1\\right)$ &$\\to$& $\\D 4{26}\\left(\\lambda\\right)$ \n& $E_1^t = \\Theta t^{-1}\\left(\\Theta e_1 - \\lambda e_2 - \\Theta t^{-1} e_3\\right)$ & $E_3^t = \\Theta^2 t^{-1} e_3$\\\\\n &&& $E_2^t = t^{-1}\\left(\\Theta e_1 + (t-\\lambda) e_2 - \\Theta^2 t^{-1} e_3\\right)$ & $E_4^t = \\Theta^3 t^{-2} e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Psi^2}\\right), -\\Psi, 1\\right)$ &$\\to$& $\\D 4{27}\\left(\\lambda\\right)$\n& $E_1^t = \\Psi t^{-1}\\left(\\Psi e_1 - \\lambda e_2 - \\Psi t^{-1} e_3\\right)$ & $E_3^t = \\Psi^2 t^{-1} e_3$\\\\\n &&& $E_2^t = t^{-1}\\left(\\Psi e_1 + (t-\\lambda) e_2 - \\Psi^2 t^{-1} e_3\\right)$ & $E_4^t = \\Psi^3 t^{-2} e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Theta^2}\\right), -\\Theta, 1\\right)$ &$\\to$& $\\D 4{28}\\left(\\lambda\\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1= \\Theta^3D\\left(\\Theta e_1 - \\lambda e_2 - t^{-1}\\Theta^2(\\Psi^2t + 4\\Theta\\lambda - \\Theta) D e_3\\right)$}\\\\\n \n\\multicolumn{3}{|l|}{$D=(\\Psi t-2\\lambda+\\Theta)^{-1},\\ \\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_2= \\Theta^2D\\left(\\Theta e_1 + (t-\\lambda) e_2 + t^{-1}\\Theta((1 - 2\\Theta)t^2 + \\Theta(3\\lambda-2\\Psi)t -\\Theta^2(4\\lambda-1)) D e_3\\right)$}\\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_3= \\Theta^6 D^2\\left(t e_3 + (2\\lambda-\\Theta)(t - \\Theta) D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t\\Theta^9 D^3 e_4$}\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1+\\frac t{\\Psi^2}\\right), -\\Psi, 1\\right)$ &$\\to$& $\\D 4{29}\\left(\\lambda\\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1= \\Psi^3D\\left(\\Psi e_1 - \\lambda e_2 - t^{-1}\\Psi^2(\\Theta^2t + 4\\Psi\\lambda - \\Psi) D e_3\\right)$}\\\\\n \n\\multicolumn{3}{|l|}{$D=(\\Theta t-2\\lambda+\\Psi)^{-1},\\ \\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_2= \\Psi^2D\\left(\\Psi e_1 + (t-\\lambda) e_2 + t^{-1}\\Psi((1 - 2\\Psi)t^2 + \\Psi(3\\lambda-2\\Theta)t -\\Psi^2(4\\lambda-1)) D e_3\\right)$}\\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_3= \\Psi^6 D^2\\left(t e_3 + (2\\lambda-\\Psi)(t - \\Psi) D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t\\Psi^9 D^3 e_4$}\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1-\\left(2+\\frac{2\\lambda-1}{\\Theta^2}\\right)t\\right), \\frac\\lambda{\\Theta^2}\\left(1-\\frac\\Theta t\\right), \\frac 1t\\right)$ &$\\to$& $\\D 4{30}\\left(\\lambda\\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1= -D\\left(\\Theta e_1 - \\lambda e_2 + t\\Psi^{-1}D e_3\\right)$}\\\\\n \n\\multicolumn{3}{|l|}{$D=\\frac{\\lambda(t-\\Theta)}{t\\left((1-2\\Theta)t+\\Theta^2\\right)},\\ \\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_2= -D\\left(e_1 - \\Theta^{-1}((2\\Theta-1)t + \\lambda) e_2 + t\\Psi^{-1}(2t-\\Theta)(t-\\Theta)^{-1} D e_3\\right)$}\\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_3= tD^2\\left((1-2\\Theta)e_3 + \\Theta^2\\lambda^{-1}(t-\\Theta)^{-1}D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t D^3(2\\Theta - 1) e_4$}\n\n \\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda\\left(1-\\left(2+\\frac{2\\lambda-1}{\\Psi^2}\\right)t\\right), \\frac\\lambda{\\Psi^2}\\left(1-\\frac\\Psi t\\right), \\frac 1t\\right)$ &$\\to$& $\\D 4{31}\\left(\\lambda\\right)$ \n&\n\n\\multicolumn{2}{l|}{$E^t_1= -D\\left(\\Psi e_1 - \\lambda e_2 + t\\Theta^{-1}D e_3\\right)$}\\\\\n \n\\multicolumn{3}{|l|}{$D=\\frac{\\lambda(t-\\Psi)}{t\\left((1-2\\Psi)t+\\Psi^2\\right)},\\ \\Psi=1-\\Theta$}&\n\n\\multicolumn{2}{l|}{$E^t_2= -D\\left(e_1 - \\Psi^{-1}((2\\Psi-1)t + \\lambda) e_2 + t\\Theta^{-1}(2t-\\Psi)(t-\\Psi)^{-1} D e_3\\right)$}\\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_3= tD^2\\left((1-2\\Psi)e_3 + \\Psi^2\\lambda^{-1}(t-\\Psi)^{-1}D e_4\\right)$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= t D^3(2\\Psi - 1) e_4$}\n\n \\\\\n\\hline\n\n$\\D 4{30}\\left(\\lambda\\right)$ &$\\to$& $\\D 4{32}\\left(\\lambda\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{31}\\left(\\lambda\\right)$ &$\\to$& $\\D 4{33}\\left(\\lambda\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\frac t{(t + 1)2}, 1 , 1 + \\frac 1t\\right)$ &$\\to$& $\\D 4{34}$\n&\n\n\\multicolumn{2}{l|}{$E^t_1= -(t + 1)^2 D e_1 + (t + 1)t D e_2 + (t + 1)^2(t - 1)D^2 e_3$}\\\\\n \n\\multicolumn{3}{|l|}{$D=(t^2 + t - 1)^{-1}$}&\n\n\\multicolumn{2}{l|}{$E^t_2= -(t + 1)^2 D e_1 + (t + 1)^2t D e_2 + (t + 1)^2 D e_3$}\\\\\n\n&&&\n\\multicolumn{2}{l|}{$E^t_3= -(t + 1)^2t^2 D^2 e_3 - (t + 1)^4t D^3 e_4$} \\\\\n\n&&&\n\n\\multicolumn{2}{l|}{$E^t_4= (t + 1)^4t^2 D^3 e_4$}\n \\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda, t^{-1}, 1\\right)$ &$\\to$& $\\D 4{35}\\left(\\lambda\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3 + (\\lambda t^2)^{-1}e_4$\\\\\n &&& $E_2^t = t^{-1}e_2 + (\\lambda t)^{-1}e_3$ & $E_4^t = t^{-4}e_4$\\\\\n\\hline\n\n$\\D 4{35}\\left(\\lambda\\right)$ &$\\to$& $\\D 4{36}\\left(\\lambda\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda,\\Theta t^{-1},t^{-1}\\right)$ &$\\to$& $\\D 4{37}\\left(\\lambda\\right)$ \n& $E_1^t = te_1$ & $E_3^t = t^2e_3$\\\\\n &&& $E_2^t = te_2$ & $E_4^t = t^2e_4$\\\\\n\\hline\n\n$\\D 4{01}\\left(\\lambda,(1-\\Theta) t^{-1},t^{-1}\\right)$ &$\\to$& $\\D 4{38}\\left(\\lambda\\right)$ \n& $E_1^t = te_1$ & $E_3^t = t^2e_3$\\\\\n &&& $E_2^t = te_2$ & $E_4^t = t^2e_4$\\\\\n\\hline\n\n$\\D 4{37}\\left(\\lambda\\right)$ &$\\to$& $\\D 4{39}\\left(\\lambda\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n$\\D 4{38}\\left(\\lambda\\right)$ &$\\to$& $\\D 4{40}\\left(\\lambda\\right)$ \n& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\\\\n &&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\\\\n\\hline\n\n\\end{longtable}\n}\n\n\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section{Perfect matchings in the operator formalism}\n\\label{sec:perf-match-oper}\n\n\nIn Section~\\ref{sec:loop-expans-ferm}, we found the loop states to be\ncombinations of two perfect matchings. It would be interesting to\nrecover the same result in terms of the operator formalism of\nSection~\\ref{sec:loops-as-fermionic}. To do so, we start by\ndistinguishing between even and odd plaquettes. We will say that a\nplaquette with coordinates $\\mathbf{z} = (z, w) $ is \\emph{even}\n(resp. \\emph{odd}) if $z + w $ is even (resp. odd). We use the\ndefinition of the height function from the superposition of two\nperfect matchings $\\mathrm{PM}_1-\\mathrm{PM}_2$ given in Section\n\\ref{sec:preliminaries}. If an odd plaquette contains either an upward\narrow on the left side, a downward arrow on the right side, a right\npointing arrow on the bottom side or a left pointing arrow on the\nupper side, it belongs to the first perfect matching $\\mathrm{PM}_1$,\nin the converse case, it belongs to $\\mathrm{PM}_2$. For an even\nplaquette, the arrows are reversed. The strategy to construct an\noperator which only preserves the operators corresponding to one of\nthe perfect matchings, say $\\mathrm{PM}_1$, when acting on a state\n$\\ket{\\Psi}$ is simple. Consider \\emph{e.g.} all the black nodes. For each of\nthem add all the annihilators corresponding to the arrows belonging to\n$\\mathrm{PM}_2$. If $\\ket{\\Psi}$ is an allowed state, only one\noperator is paired for each node which gives a $1$ as a result of the\nanticommutation, while all the others will just annihilate the\nstate. This results in only the operators belonging to $\\mathrm{PM}_1$\nsurviving and so yielding the result we are looking for. We can define\nthe operators ${PM}_1$ and ${PM}_2$ explicitly as follows:\n\\begin{subequations}\n \\label{eq:loop-splitting}\n \\begin{align}\n {PM}_1 &= \\prod_{z = 1}^{N} \\prod_{w = 1}^{M} \\frac{1 + \\left( -1 \\right)^{z + w}}{2} \\left[ a(z,w) + d(z, w) + b(z, w - 1) + c(z-1,w) \\right] \\, ,\\\\\n {PM}_2 &= \\prod_{z = 1}^{z} \\prod_{w = 1}^{w} \\frac{1 +\n \\left( -1 \\right)^{z + w}}{2} \\left[ b(z,w) + c(z,\n w) + a(z, w - 1) + d(z-1,w) \\right] \\, .\n \\end{align}\n\\end{subequations}\nA state $\\ket{\\Psi} = \\kket{m_1} \\otimes \\kket{m_2}$ is decomposed\ninto its two constituent perfect matchings by their action:\n\\begin{equation}\n PM_i \\ket{\\Psi} = \\kket{m_i} \\, , \\hspace{1em} i = 1,2 \\, .\n\\end{equation}\nNow we are able to give a consistency condition for a state to\ncorrespond to a loop appearing in the expansion of the fermionic\naction. The idea is that the state should be decomposable into two\nmatchings, which both touch each node once. Let us therefore define\n\\begin{subequations}\n \\begin{align}\n P_1 (z,w) &= \\textstyle{\\frac{1 + \\left( -1 \\right)^{z + w}}{2}}\n \\left[ N_a(z,w) + N_d(z, w) + N_b(z, w - 1) +\n N_c(z-1,w)\n - 1\\right] \\, , \\\\\n P_2 (z,w) &= \\textstyle{\\frac{1 + \\left( -1 \\right)^{z + w}}{2}}\n \\left[ N_b(z,w) + N_c(z, w) + N_a(z, w - 1) +\n N_d(z-1,w) - 1\\right] \\, .\n \\end{align}\n\\end{subequations}\nAn allowed state $\\ket{\\Psi}$ must satisfy the local conditions\n\\begin{equation}\n \\label{eq:state-consistency}\n P_1 (\\mathbf{z}) \\ket{\\Psi} = P_2 (\\mathbf{z}) \\ket{\\Psi} = 0 \\, , \\hspace{1em} \\forall \\mathbf{z} \\, . \n\\end{equation}\n\n\n\nThe action of the height operators in Eq.~\\eqref{eq:global-height} is\nstill defined on the perfect matching states $\\kket{\\Psi}$. To be\nconsistent with the constructions in Section \\ref{sec:preliminaries},\nwe introduce a sign to distinguish between $\\kket{m_1}$ and\n$\\kket{m_2}$:\n\\begin{subequations}\n\\label{eq:matching-charges}\n \\begin{align}\n H_z \\kket{m_i} &= \\left(-1 \\right)^{i+1} \\left[ \\sum_{\\zeta =\n 1}^{N} N_a ( \\zeta, \\bar w) - N_b ( \\zeta, \\bar w)\n \\right] \\kket{m_i} \\, , \\\\\n H_w \\kket{m_i} &= \\left(-1 \\right)^{i+1} \\left[ \\sum_{\\omega =\n 1}^{M} N_c ( \\bar z, \\omega) - N_d ( \\bar z, \\omega)\n \\right] \\kket{m_i} \\, .\n \\end{align}\n\\end{subequations}\nWe find that each loop can be identified by four charges,\ncorresponding to the eigenvalues of the winding operators acting on\nthe two perfect matchings. The relation between the winding of a loop\nand the weights of the constituent perfect matchings is\n\\begin{subequations}\n \\begin{align}\n H_z \\ket{\\Psi} &= H_z \\kket{m_1} - H_z \\kket{m_2} \\, ,\\\\\n H_w \\ket{\\Psi} &= H_w \\kket{m_1} - H_w \\kket{m_2} \\, .\n \\end{align}\n\\end{subequations}\nExtending the definition of the height function to perfect matchings\nrequires some care, since the equality in\nEq.~\\eqref{eq:height-equality} does not hold for perfect matchings. A\npossible way out consists in symmetrizing the sum of the two\nexpressions and define\n\\begin{equation}\n h (z, w ) \\kket{m_i} = \\frac{\\left(-1 \\right)^{i+1}}{2} \\left[ \\sum_{\\zeta = 0}^{z} N_a ( \\zeta, w) - N_b ( \\zeta, w) + \\sum_{\\omega = 0}^{w} N_c ( z, \\omega) - N_d ( z, \\omega) \\right] \\kket{m_i} \\, . \n\\end{equation}\nIn this way, one recovers once more\n\\begin{equation}\n h (\\mathbf{z}) \\ket{\\Psi} = h (\\mathbf{z}) \\kket{m_1} - h (\\mathbf{z}) \\kket{m_2} \\, .\n\\end{equation}\n\n\n\n\n\n\n\\section{Generating functions}\n\\label{sec:generating-functions}\n\nIn this section, we give the generating function which contains the full\ninformation of the fermion loop gas, \\emph{i.e.} all the\nloop states. The dimer model partition function and the Witten index\ncan be recovered from it.\n\n\n\\subsection{Generating function for the fermion loops on the cylinder}\n\\label{sec:generating-function}\n\nWe compute the generating functions\nfor the loops defined as\n\\begin{equation}\n G (q, z, y) = \\Tr \\left[ q^H z^K y^F \\right] \\, , \n\\end{equation}\nwhich keeps track of all the information. Expanding the trace one finds\nan explicit expression depending on the matching degeneracies $N_k$:\n\\begin{equation}\n \\label{eq:generating-cylinder}\n G( q, z, y ) = \\sum_{k=0}^N N_k\\, y^{2k} z^k + \\left( 1 + y \\right) \\left\\{ \\frac{1}{2}\\sum_{k=0}^N N_k \\left( N_k - 1 \\right) y^{2k} z^k + \\sum_{l=1}^N q^l \\sum_{k=l}^N N_k\\, N_{k-l}\\, y^{2k} z^k \\right\\} \\, . \n\\end{equation}\n\nAll the physical information about the system is contained in the\ngenerating function $G$. For example:\n\\begin{itemize}\n\\item the Euler number for the complex in Eq.~\\eqref{eq:SQM-complex}\n is $\\,\\chi = \\Tr \\left[ \\left( -1 \\right)^F \\right] = G(1,1,-1)$ ;\n\\item the Poincar\\'e polynomial for the complex is $\\,\\chi (z) = \\Tr \\left[ \\left( -1\n \\right)^F z^k \\right] = G(1,z,-1)$\\,;\n\\item the total number of perfect matchings is $N = \\Tr \\left[ \\left(\n -1 \\right)^{K+F} \\right] = G(1,-1,-1)$~;\n\\item the multiplicity of weight-$k$ matchings is $N_k = \\frac{1}{k!}\n \\left. \\frac{\\del^k}{\\del z^k } G(q,\\frac{z}{y},y)\\right|_{q=1,z=0,y=-1}$~;\n\\item the dimension of the $k$-th chain group is $\\,\\dim C_k =\n \\frac{1}{k!} \\left. \\frac{\\del^k}{\\del y^k}\n G(q,z,y)\\right|_{q=1,z=1,y=0}$~;\n\\item the number of loops with winding number $h$ is $L_h =\n \\frac{1}{h!} \\left. \\frac{\\del^h}{\\del q^h}\n G(q,z,y)\\right|_{q=0,z=1,y=1}$~.\n\\end{itemize}\n\n\n\\subsection{Generating function for the fermion loops on the torus}\n\\label{sec:generating-function-1}\n\nLet us now extend the definition of generating function for the loops\nto graphs embedded on the torus. Following the formula for the\ncylinder in Eq.~\\eqref{eq:generating-cylinder} we define:\n\\begin{equation}\n G (\\mathbf{q}, \\mathbf{z}, y) = \\Tr \\left[ \\mathbf{q}^{\\mathbf{H}} \\mathbf{z}^{\\mathbf{K}} y^F \\right] \\, ,\n\\end{equation}\nwhere the notation $\\mathbf{q}^{\\mathbf{H}} $ is a shortcut for\n$\\mathbf{q}^{\\mathbf{H}} \\doteq q_z^{H_z} q_w^{H_w}$ and\n$\\mathbf{z}^{\\mathbf{k}} \\doteq z^{k_z} w^{k_w}$. Expanding in the loop basis we find:\n\\begin{multline}\n G ( \\mathbf{q}, \\mathbf{z}, y ) %\n\n = \\sum_{\\mathbf{k}=(0,0)}^{\\mathbf{N}} N_{\\mathbf{k}}\n y^{2 k_z} \\mathbf{z}^\\mathbf{k} + \\frac{ 1 + y}{2} \\left\\{\n \\sum_{\\mathbf{k} = (0,0)}^{\\mathbf{N}}\n N_{\\mathbf{k}} \\left( N_{\\mathbf{k}} - 1 \\right) y^{2k_z} \\mathbf{z}^{\\mathbf{k}} + \\right. \\\\\n \\left. + \\sideset{}{'}\\sum_{\\mathbf{l} = (0,0)}^{\\mathbf{N}} \\mathbf{q}^{\\mathbf{l}} \\sum_{\\mathbf{k}\n = \\mathbf{l}}^{\\mathbf{N}} \\left( N_{\\mathbf{k}}\n N_{\\mathbf{k}-\\mathbf{l}} + N_{k_z (k_w-l_w)} N_{(k_z-l_z)k_w}\n \\right) y^{2 k_z} \\mathbf{z}^{\\mathbf{k}} \\right\\} \\, ,\n\\end{multline}\nwhere the primed sum $\\displaystyle{\\sideset{}{'}\\sum_{\\mathbf{l}}}$\nmeans that the value $\\mathbf{l}=(0,0)$ is to be omitted. All the\nphysics of the system is encoded into $G ( \\mathbf{q}, \\mathbf{z}, y\n)$. In particular:\n\\begin{itemize}\n\\item the Euler number for the complex in Eq.~\\eqref{eq:torus-complex}\n is $\\,\\chi(w)=\\Tr \\left[ (-1)^F w^{K_w} \\right]$ \\\\ $= G((1,1),(1,w),-1)$~;\n\\item the Poincar\\'e characteristic for the complex is $ \\chi (z,w) = \\Tr \\left[ (-1)^F\n \\mathbf{z}^{\\mathbf{k}} \\right]$\\\\ $= G((1,1),\\mathbf{z},-1)$~;\n\\item the number of perfect matchings with height $\\mathbf{k}$ is\n $N_{\\mathbf{k}} = \\frac{1}{\\mathbf{k}!}\n \\left. \\frac{\\del^{\\abs{\\mathbf{k}}}}{\\del \\mathbf{z}^{\\mathbf{k}}}\n G \\right|_{\\mathbf{q}= (1,1), \\mathbf{z} = (0,0), y = -1}$~, where\n we introduced the notation $\\mathbf{k}! \\doteq k_z! k_w !$~,\n $\\abs{\\mathbf{k}} \\doteq k_z + k_w $~, and\n $\\frac{\\del^{\\abs{\\mathbf{k}}}}{\\del \\mathbf{z}^{\\mathbf{k}}} \\doteq\n \\frac{\\del^{k_z + k_w}}{\\del z^{k_w} \\del w^{k_w}}$~;\n\\item the dimension of the $C_{\\mathbf{k}}$ chain group is $\\dim\n C_{\\mathbf{k}} = \\frac{1}{\\mathbf{k!}}\n \\left. \\frac{\\del^{\\abs{\\mathbf{k}}}}{\\del y^{k_z} \\del w^{k_w}} G\n \\right|_{\\mathbf{q}= (1,1), \\mathbf{z} = (0,1), y = 1}$~;\n\\item the number of loops with winding $\\mathbf{h}$ is $L_{\\mathbf{h}}\n = \\frac{1}{\\mathbf{h}!} \\left. \\frac{\\del^{\\abs{\\mathbf{h}}}}{\\del\n \\mathbf{q}^{\\mathbf{h}}} G \\right|_{\\mathbf{q}= (0,0),\n \\mathbf{z} = (1,1), y = 1}$~.\n\\end{itemize}\n\n\n\n\n\\section{Example: One square on the torus}\n\\label{sec:ex_torus}\n\n\nTo illustrate the construction given in Section~\\ref{sec:sqm_torus}, we now present the smallest possible example\non the torus, consisting only of one square. The eight possible\nmatchings and their quantum numbers are shown in Figure\n\\ref{fig:ex_matchings}.\n\\begin{figure}[h!]\n \\begin{center}\n \\includegraphics[width=110mm]{ex_matchings-T}\n \\caption{Matchings for the square on the torus}\n \\label{fig:ex_matchings-T}\n \\end{center}\n\\end{figure}\nThey combine into 64 loop configurations, which are, sorted into their\nco--chain groups according to total winding number, depicted in Table~\\ref{chaingroupsT}.\n\nThe partition function of the dimer model is the one given in (\\ref{eq:partition_square}),\nthe generating function for this example reads\n\\begin{eqnarray}\n G( \\mathbf{q}, \\mathbf{z}, y ) &=& w + y^2 z + 4 w y^2 z + w^2 y^2 z + w y^4 z^2 +\\nonumber \\\\\n && +\\left( 1 + y \\right) \\left[ \\left( 6 w y^2 z + 4 q_z + 4 q_w + q_zq_w \\right) w y^2 z + \\left( 4 q_w + q_w^2 + q_zq_w \\right) w^2 y^2 z + \\right. \\nonumber\\\\\n && \\left. + \\left( 4 q_z + q_z^2 + q_zq_w \\right) w y^4 z^2 + q_zq_w w^2 y^4 z^2 \\right].\n\\end{eqnarray}\n\n\\afterpage{%\n \\begin{center}\n \\begin{longtable}{ccccccc}\n \\caption[Co--chain groups for the square on the torus]{Co--chain groups\n for the square on the torus}\n \\label{chaingroupsT}\\\\\n \n \\toprule {$\\,K_w\\,$}& $C^0$ &{$C^1$} \n &{$C^2$}& {$C^3$} & {$C^4$} & {$C^5$} \\cr \\midrule\n \\endfirsthead\n\n \\multicolumn{7}{l}\n {\\tablename\\ \\thetable{} -- continued from previous page} \\\\\n \\toprule\n $\\,K_w\\,$& $C^0$ &{$C^1$} & {$C^2$}& {$C^3$} & {$C^4$} & {$C^5$} \\cr\n \\midrule\n \\endhead\n \n \\midrule \\multicolumn{7}{r}{{Continued on next page}} \\\\ \\bottomrule\n \\endfoot\n \n \\bottomrule\n \\endlastfoot\n \n \\rowcolor[gray]{.95} 2& -& - & \\pb{$\\ket{1,2,1;0,1,1}_+$ \\includegraphics[height=10mm]{121211}} & \\pb{$\\ket{1,2,1;0,1,1}_{-}$ \\includegraphics[height=10mm]{211121}} & - & - \\cr\n 2 & -& - & \\pb{$\\ket{1,2,1;1,0,1}_+$ \\includegraphics[height=10mm]{121101}}&\\pb{$\\ket{1,2,1;1,0,1}_{-}$ \\includegraphics[height=10mm]{101121}} & - & - \\cr\n \\rowcolor[gray]{.95} 2 & -& - & \\pb{$\\ket{1,2,1;1,1,4}_+$ \\includegraphics[height=10mm]{121114}} &\\pb{$\\ket{1,2,1;1,1,4}_{-}$ \\includegraphics[height=10mm]{114121}} & - & - \\cr\n 2 & -& - & \\pb{$\\ket{1,2,1;1,1,3}_+$ \\includegraphics[height=10mm]{121113}}&\\pb{$\\ket{1,2,1;1,1,3}_{-}$ \\includegraphics[height=10mm]{113121}} & - & - \\cr\n \\rowcolor[gray]{.95} 2 & -& - & \\pb{$\\ket{1,2,1;1,1,2}_+$ \\includegraphics[height=10mm]{121112}}&\\pb{$\\ket{1,2,1;1,1,2}_{-}$ \\includegraphics[height=10mm]{112121}} & - & - \\cr\n 2 & -& - & \\pb{$\\ket{1,2,1;1,1,1}_+$ \\includegraphics[height=10mm]{121111}} &\\pb{$\\ket{1,2,1;1,1,1}_{-}$ \\includegraphics[height=10mm]{111121}} & - & - \\cr\n \\rowcolor[gray]{.95} 2 & -& - & \\pb{$\\ket{1,2,1;1,2,1}$ \\includegraphics[height=9mm]{121121}}&-&\\pb{$\\ket{2,1,1;1,2,1}_+$ \\includegraphics[height=10mm]{011121}} &\\pb{$\\ket{2,1,1;1,2,1}_-$ \\includegraphics[height=10mm]{121011}} \\cr\n \\midrule\n 1 & -& - & \\pb{$\\ket{1,1,4;0,1,1}_+$ \\includegraphics[height=10mm]{114211}} & \\pb{$\\ket{1,1,4;0,1,1}_-$ \\includegraphics[height=10mm]{211114}} & - &-\\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,3;0,1,1}_+$ \\includegraphics[height=10mm]{113211}}& \\pb{$\\ket{1,1,3;0,1,1}_-$ \\includegraphics[height=10mm]{211113}}& -&-\\cr\n 1 & -& - & \\pb{$\\ket{1,1,2;0,1,1}_+$ \\includegraphics[height=10mm]{112211}} & \\pb{$\\ket{1,1,2;0,1,1}_-$ \\includegraphics[height=10mm]{211112}} & -&-\\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,1;0,1,1}_+$ \\includegraphics[height=10mm]{111211}}& \\pb{$\\ket{1,1,1;0,1,1}_-$ \\includegraphics[height=10mm]{211111}}& -&-\\cr\n 1 & -& - & \\pb{$\\ket{1,0,1;0,1,1}_+$ \\includegraphics[height=10mm]{101211}}& \\pb{$\\ket{1,0,1;0,1,1}_-$ \\includegraphics[height=10mm]{211101}}& -&-\\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,4;1,0,1}_+$ \\includegraphics[height=10mm]{101114}}& \\pb{$\\ket{1,1,4;1,0,1}_-$ \\includegraphics[height=10mm]{114101}}& -&- \\cr\n 1 & -& - & \\pb{$\\ket{1,1,3;1,0,1}_+$ \\includegraphics[height=10mm]{113101}}& \\pb{$\\ket{1,1,3;1,0,1}_-$ \\includegraphics[height=10mm]{101113}}& -&-\\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,2;1,0,1}_+$ \\includegraphics[height=10mm]{112101}} & \\pb{$\\ket{1,1,2;1,0,1}_-$ \\includegraphics[height=10mm]{101112}}& -&-\\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,1;1,0,1}_+$ \\includegraphics[height=10mm]{101111}} & \\pb{$\\ket{1,1,1;1,0,1}_-$ \\includegraphics[height=10mm]{111101}}& -&-\\cr\n 1 & -& - & \\pb{$\\ket{1,1,4;1,1,3}_+$ \\includegraphics[height=10mm]{113114}}& \\pb{$\\ket{1,1,4;1,1,3}_-$ \\includegraphics[height=10mm]{114113}}& -&-\\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,4;1,1,2}_+$ \\includegraphics[height=10mm]{114112}}& \\pb{$\\ket{1,1,4;1,1,2}_-$ \\includegraphics[height=10mm]{112114}}& -&-\\cr\n 1 & -& - & \\pb{$\\ket{1,1,4;1,1,1}_+$ \\includegraphics[height=10mm]{114111}} & \\pb{$\\ket{1,1,4;1,1,1}_-$ \\includegraphics[height=10mm]{111114}}& -&-\\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,3;1,1,2}_+$ \\includegraphics[height=10mm]{113112}}& \\pb{$\\ket{1,1,3;1,1,2}_-$ \\includegraphics[height=10mm]{112113}} & \\pb{$\\ket{2,1,1;0,1,1}_+$ \\includegraphics[height=10mm]{011211}} &\\pb{$\\ket{2,1,1;0,1,1}_-$ \\includegraphics[height=10mm]{211011}} \\cr\n 1 & -& - & \\pb{$\\ket{1,1,3;1,1,1}_+$ \\includegraphics[height=10mm]{113111}} & \\pb{$\\ket{1,1,3;1,1,1}_-$ \\includegraphics[height=10mm]{111113}} & \\pb{$\\ket{2,1,1;1,0,1}_+$ \\includegraphics[height=10mm]{011101}} & \\pb{$\\ket{2,1,1;1,0,1}_-$ \\includegraphics[height=10mm]{101011}} \\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,2;1,1,1}_+$ \\includegraphics[height=10mm]{112111}} & \\pb{$\\ket{1,1,2;1,1,1}_-$ \\includegraphics[height=10mm]{111112}} &\\pb{$\\ket{2,1,1;1,1,4}_+$ \\includegraphics[height=10mm]{011114}} &\\pb{$\\ket{2,1,1;1,1,4}_-$ \\includegraphics[height=10mm]{114011}} \\cr\n 1 & -& - & \\pb{$\\ket{1,1,4;1,1,4}$ \\includegraphics[height=10mm]{114114}} &- & \\pb{$\\ket{2,1,1;1,1,3}_+$ \\includegraphics[height=10mm]{011113}} &\\pb{$\\ket{2,1,1;1,1,3}_-$ \\includegraphics[height=10mm]{113011}} \\cr\n \\rowcolor[gray]{.95} 1 & -& - & \\pb{$\\ket{1,1,3;1,1,3}$ \\includegraphics[height=10mm]{113113}} &- & \\pb{$\\ket{2,1,1;1,1,2}_+$ \\includegraphics[height=10mm]{011112}} &\\pb{$\\ket{2,1,1;1,1,2}_-$ \\includegraphics[height=10mm]{112011}} \\cr\n 1 & -& - & \\pb{$\\ket{1,1,2;1,1,2}$ \\includegraphics[height=10mm]{112112}} &-& \\pb{$\\ket{2,1,1;1,1,1}_+$ \\includegraphics[height=10mm]{011111}} &\\pb{$\\ket{2,1,1;1,1,1}_-$ \\includegraphics[height=10mm]{111011}} \\cr\n \\rowcolor[gray]{.95} 1 & \\pb{ $\\ket{0,1,1;0,1,1}$ \\includegraphics[height=10mm]{211211}} & -& \\pb{$\\ket{1,1,1;1,1,1}$ \\includegraphics[height=10mm]{111111}}&-& {\\pb{$\\ket{2,1,1;2,1,1}$ \\includegraphics[height=10mm]{001001}}}&-\\cr\n \\midrule\n 0 & -& - & \\pb{$\\ket{1,0,1;1,0,1}$ \\includegraphics[height=10mm]{101101}} & - &- &- \\cr\n \\end{longtable}\n \\end{center}\n}\n\n\n\\section{Categorification of the Newton polynomial}\n\\label{sec:categorification}\n\nWhat follows is inspired by the categorification programme for knot\npolynomials of Khovanov~\\cite{Khovanov1, Khovanov2, Khovanov3}. The\nbasic idea of his work is that for a knot $K$, a doubly graded\nhomology theory $H_{i, j}(K)$ can be constructed, whose graded Euler\ncharacteristic with respect to one of the gradings yields the knot\npolynomial in question, \\emph{e.g.} the Jones polynomial, or the\nAlexander polynomial. The homology groups are constructed through a\ncategorification process which starts with a state sum representation\nof the invariant polynomial. A group is constructed for each term in\nthe summation, and differential maps between these groups are\nappropriately defined. Similar categorifications were subsequently\nperformed for other invariants with state sums, \\emph{e.g.} various\ngraph polynomials, such as the chromatic polynomial~\\cite{Helme1,\n Helme2}, and the more general Tutte polynomial~\\cite{Rong}.\n\nBut exactly what is meant by the term categorification? When we\ncategorify, we replace set theoretic concepts by category theoretic\nones. Sets become categories, functions become functors, and equations\nbetween functions become natural isomorphisms between functors\n(fulfilling certain relations). The opposite process,\ndecategorification, is much more familiar to us. When we decategorify,\nwe are basically throwing away extra information by taking all isomorphic\nobjects to be equal. When we say that all vector spaces of the same\ndimension are the same, we decategorify and are left with a set of\nisomorphism classes. Obviously, the reverse process of inventing the\nmissing information is much more difficult. The category\n$\\mathrm{FinSet}$ of all finite sets for example is a categorification\nof the set of natural numbers. For an easy to read introduction to the\nconcept of categorification, we refer the reader to~\\cite{Baez}.\n\n\\bigskip\n\nAfter this general digression, we return to our own system of\nfermion loops. Its partition function, the Newton polynomial, is, once\nmore, a state sum. From the point of view of the dimer model, its\ninteger coefficients and polynomial nature are obvious. From the point\nof view of the free fermion, this is much less so. Giving this\npartition function the interpretation of a graded Euler characteristic\nexplains its integer nature and gives it a geometric interpretation,\nwhile at the same time suggesting its relation to an index.\n\n\nThe task at hand is obvious: construct a bigraded co--chain complex from the\nfermion loop states with a differential map preserving the grading, such\nthat the Euler characteristic of this co--chain complex yields the Newton\npolynomial.\n\nThe fact that we can associate four different charges to each loop\nstate living on the torus (see Eq.~\\eqref{eq:matching-charges}) gives\nus a certain freedom in the choice of the gradings in the double\ncomplex. Our emphasis is placed on the matchings, so we choose to sort\nthe loop states into co--chain groups according to the maximum winding\nnumber of the two constituent matchings in one direction ($\\max_i H_p\n\\kket{m_i}$) and to define an internal grading based on the maximum\nweight in the other direction ($\\max_i H_q \\kket{m_i}$). This\nseemingly arbitrary choice just amounts to picking a horizontal and a\nvertical direction in the double complex. The Euler characteristic,\nfor example, is not sensitive to this choice.\n\n\nWhen studying the loop states, there is an obvious observation to\nmake: because of the two possible orientations of the loops, all\nstates are paired with the exception of the double line perfect\nmatchings. This very much smells of supersymmetric Quantum mechanics\n(\\textsc{sqm}), where the supersymmetric ground states are the only ones which\nneed not be paired. Indeed, if one takes the total weight of a\nconfiguration to be its energy, the perfect matchings, having zero\nwinding, become ground states.\n\nThe Witten index,\n\\begin{equation}\n \\label{eq:wittenindex}\n \\mathrm{dim}\\mathcal{H}^B_{(0)}-\\dim\\mathcal{H}^F_{(0)}=\\Tr(-1)^F,\n\\end{equation}\ncounts the number of bosonic minus fermionic ground states and is an\ninvariant of the system. This is, indeed, also what the Newton\npolynomial is doing! Once we re--interpret our fermion loop gas as a\n\\textsc{sqm} system and construct a co--chain complex such that its boundary\noperator can be identified with the $Q$ operator and states with\nopposite orientations become superpartners, the partition function\nbecomes the generalized Witten index of the\nsystem.\n\nThis will be done explicitly in the following sections.\n\n\n\n\n\n\\section{Conclusions}\n\\label{sec:conclusions} \n\nIn this note, we relate a statistical mechanical system, the dimer\nmodel on a torus, to a system of lattice quantum field theory,\n\\emph{i.e.} the massless free fermion. The loop configurations\nobtained after a diagrammatic expansion of the fermion determinant\n(Sec.~\\ref{sec:loop-expans-ferm}) can be interpreted as states in\n\\textsc{sqm} after employing categorification techniques inspired by\nKhovanov's work (Sec.~\\ref{sec:categorification}). The Newton\npolynomial of the dimer model becomes the generalized Euler\ncharacteristic of a co--chain complex (Sec.~\\ref{sec:sqm_torus}). The\nstates with vanishing overall winding become the supersymmetric ground\nstates of the system. Since all loop states except the double line\nperfect matchings are paired, taking the Witten index of this system\nagain counts the number of perfect matchings\n(Eq.~\\eqref{eq:Witten-torus}).\\\\\nIn the following, we make use of the dimer model--quiver gauge theory\ncorrespondence to give the loop configurations an interpretation in\nyet another picture. Interpreting the loops as maps from one perfect\nmatching to another (Sec.~\\ref{sec:loops-as-operators}), maps between\ninternal perfect matchings associated to the same point in the toric\ndiagram become changes of foundation of the associated helix and\nsequences of double Seiberg dualities in the acyclic quiver\n(Sec.~\\ref{sec:doubleseiberg}). When specializing to the honeycomb\ngraph, the plaquette moves which generate all $(0,0)$--maps correspond\ndirectly to the \\textsc{bps} black hole configurations which are parametrized\nby crystal melting configurations (Sec.~\\ref{sec:melting}).\n\nMapping between matchings associated to different internal points is a\nmore severe change and cannot be achieved as a sequence of Seiberg\ndualities. In the language of exceptional collections, we map from one\nhelix to another.\n\n\nIn Table~\\ref{table:dictionary}, we summarize the results of this\npaper by giving a dictionary between the loop states, the fermion\nstates, the gauge theory and the geometry.\n\n\n\\bigskip\n\nFrom this point, one can go to several different directions for\nfurther research.\n\\begin{itemize}\n\\item The dimer model admits a description as a free fermion model\n based on a transfer matrix approach\n (see~\\cite{lieb:2339,Sutherland:1968,alet:041124} and\n also~\\cite{Okounkov:2003sp}) which, at first sight, is different\n from the one we introduced in this note. It would be very\n interesting to clarify the link between these two descriptions.\n\\item Furthermore, it would be interesting to use a more general\n definition for the Dirac operator on a graph than the one in\n Eq.~\\eqref{eq:Lattice-Dirac}, to prove the equivalence to the dimer\n model completely independently of the type of graph and its\n representation. Indeed, this could provide an elegant solution to\n the problem of choosing a Dirac operator on a random graph that\n depends only on the the adjacency.\n\\item The algebraic structure we introduce in\n Sec.~\\ref{sec:categorification} also deserves further\n investigation. A better understanding of the complex and the mapping\n to \\textsc{sqm} might indeed cast a new light on the dual quiver\n gauge theory.\n\\item A final word must be spent on the geometric interpretation in\n Sec.~\\ref{sec:geometry}. To our knowledge, the meaning of the\n boundary perfect matchings (and, correspondingly, of the\n infinite-dimensional representations of the path algebras) in terms\n of derived category is not yet clear. This is an obvious\n prerequisite to a complete understanding of the geometrical meaning\n of all the loop states that appear in the theory.\n\\end{itemize}\n\nThe study of the dimer model, which due to its combinatorial nature provides a relatively easy playground, has proven to be very fruitful thanks to its manifold connections to other physical systems.\nWe believe that these inter--relations have not yet been fully exploited and deserve further attention.\n\n\\begin{landscape}\n \\begin{table}[]\n \\begin{center}\n \\begin{tabular}{cccp{4.5cm}p{4cm}}\n \\toprule\n & \\textbf{Loop state} & \\textbf{Gauge theory on $S$}&\n \\begin{minipage}{4.4cm}\n \\center \\textbf{Geometry}\n \\end{minipage} &\n \\begin{minipage}{3.9cm}\n \\center \\textbf{Free fermion}\n \\end{minipage}\n \\\\ \\midrule\n \\rowcolor[gray]{.95}\n \\cellcolor[gray]{1} & double line $\\mathrm{PM}$ & Identity &\n \\begin{minipage}{4.4cm}\n \\center Identity\n \\end{minipage}\n & time evolution with no local charge creation\\\\[.3cm]\n \\cellcolor[gray]{1} $\\{\\mathrm{PM}^{int}\\} \\to \\{\\mathrm{PM}^{int}\\}$ &$(0,0)$&Seiberg duality & change of foundation in helix & time evolution in which no net charge is created\\\\[0.3cm]\n \\rowcolor[gray]{.95}\n \\cellcolor[gray]{1} & $(p,q)\\neq (0,0)$ & general gauge duality &\n general sequence of mutation, change of helix, birational\n transformation\n &\n initial conditions with gradient $q$ and creation of net charge $p$\n \\\\ \\midrule\n $\\{\\mathrm{PM}^{int}\\} \\leftrightarrow \\{\\mathrm{PM}^{bd}\\}$ & no corner $ \\mathrm{PM}$ & ? & birational transformation & \\hspace{1.8cm}\" \\\\ \\midrule\n \\rowcolor[gray]{.95}\n \\cellcolor[gray]{1}$\\set{\\mathrm{PM}^{bd}} \\to \\set{\\mathrm{PM}^{bd}}$ &\n \\begin{minipage}{3.3cm}\n \\tabtop adjacent boundary matchings. Zig-zag path\n \\end{minipage}\n & \\begin{minipage}{3.5cm}\n gauge-invariant multi--trace operator\n \\end{minipage}\n & & \\hspace{1.8cm}\" \\\\[.7cm]\n & no corner PM & \n & birational transformation & \\hspace{1.8cm}\" \\\\ \\bottomrule\n \\end{tabular}\n \\caption{Dictionary}\n \\label{table:dictionary}\n \\end{center}\n \\end{table}\n\\end{landscape}\n\n\n\n\n\\section{Dimer Model Preliminaries}\n\\label{sec:preliminaries}\n\n\n\n\nTake a plane \\emph{bipartite} graph $\\gra$, \\emph{i.e.} one in which all\nvertices can be colored black and white, such that each black vertex\nis only connected by links to white vertices and vice versa. Let $M$\nbe a subset of the set $E$ of edges of $\\gra$. $M$ is called a\nmatching, if its elements are links and no two of them are\nadjacent. If every vertex of $\\gra$ is saturated under $M$, the\nmatching is called \\emph{perfect}. Such a link which joins a black and\na white vertex is also called a \\emph{dimer}. The dimer model describes the\nstatistical mechanics of a system of random perfect matchings. In the\nsimplest case, we ask for the number of close packed dimer\nconfigurations, \\emph{i.e.} the number of perfect matchings.\n\nLet us label the vertices of the underlying graph $\\gra$ on which the\ndimer model lives consecutively. The topology of $\\gra$ can be encoded\nin its adjacency matrix $A$, where $A_{x,y}=1$ if the vertices $x$ and\n$y$ are joined by an edge, and $A_{x,y}=0$ otherwise. The number of\nperfect matchings is given by the number of ways in which the vertices\ncan be partitioned into adjacent pairs.\n\nThe \\emph{Hafnian} of a symmetric matrix $A$, introduced by Caianiello\n\\cite{Caianiello}, is defined as\n\\begin{equation}\n \\label{eq:hafnian}\n \\Hf(A)=\\sum_{\\pi \\in \\mathcal{S}}A_{\\pi(1),\\pi(2)}A_{\\pi(3),\\pi(4)} \\dots A_{\\pi(n-1),\\pi(n)},\n\\end{equation}\nwhere $\\mathcal{S}$ is the set of all permutations $\\pi$ of $\\set{1, \\dots \n n}$ satisfying $\\pi(1)<\\pi(3)< \\dots <\\pi(n-1)$ and $\\pi(2i-1)<\\pi(2i)$\n for $1\\leq i \\leq n$, $n$ even. If we take $A$ to be the adjacency\n matrix of ${\\gra}$, $\\Hf(A)$ counts the perfect matchings, since\n terms in (\\ref{eq:hafnian}) which contain non--adjacent pairs are\n zero. The \\emph{permanent} of an $n\\times n$ matrix $A$ is, like the\n determinant, given by the sum over all permutations, except that the\n sign of the permutations is not taken into account:\n\\begin{equation}\n \\label{eq:permanent}\n \\perm(A)=\\sum_{\\pi\\in S_n}\\prod_{i=1}^n A_{i,\\pi(i)}.\n\\end{equation}\nWhile the determinant can be evaluated in polynomial time via Gaussian\nelimination, computing the permanent is $\\#P$--complete\n\\cite{Valiant}. The Hafnian can be regarded as a generalization of\nthe permanent. In fact~\\cite{Kuperberg1},\n\\begin{equation}\n \\Hf(A)=\\perm(B)\\ \\ \\text{for $A$ of the form}\\ \\ A= \\left( \\begin{array}{c|c}\n 0&B\\\\ \\hline\n B^t&0\n \\end{array} \\right).\n\\end{equation}\nThis identity accounts for the fact that sometimes, the number of\nperfect matchings is given as the permanent of the adjacency matrix in\nthe literature. Like the permanent, also the Hafnian is difficult to\nevaluate and does not satisfy any useful identities.\n\nThe counterparts of the Hafnian and permanent are the \\emph{Pfaffian} and determinant, which take into account the sign of the permutations:\n\\begin{align}\n \\label{eq:pfaffian}\n \\Pf(A)&= \\sum_{\\pi \\in \\mathcal{S}}\\sign(\\pi)\\,A_{\\pi(1),\\pi(2)}A_{\\pi(3),\\pi(4)} \\dots A_{\\pi(n-1),\\pi(n)},\\\\\n \\label{eq:det}\n \\det(A)&=\\sum_{\\pi\\in S_n}\\sign(\\pi)\\prod_{i=1}^n A_{i,\\pi(i)},\n\\end{align}\nwhere the permutations contributing to the Pfaffian are subject to the same conditions as for the Hafnian.\nHere, we have\n\\begin{equation}\n \\Pf(A')=(-1)^{\\binom{n}{2}}\\det(B)\\ \\ \\text{for $A'$ of the form}\\ \\ A'=\\left(\\begin{array}{c|c}\n 0 & B \\\\ \\hline\n \\! \\! -B^t & 0 \\end{array} \\right).\n\\end{equation}\nMoreover, for every anti--symmetric matrix of even order, the relation\n\\begin{equation}\n \\label{eq:rel}\n \\det A= \\left( \\Pf A \\right)^2\n\\end{equation}\nholds. Already in 1913, Polya suggested to change some signs in a\nmatrix $A$, such that the determinant of the new matrix $\\tilde A$\nwould be the permanent of $A$~\\cite{polya}. Kasteleyn\n\\cite{Kasteleyn1, Kasteleyn2} actually does exactly this: he\nintroduces an orientation on \\gra, which leads to a signed adjacency\nmatrix $K$, now called the \\emph{Kasteleyn matrix}. The Pfaffian of $K$\ngives the number of perfect matchings.\n\nA \\emph{Kasteleyn orientation} fulfills the following\ncondition: the product of all edge weights around a face must equal $-1$\nif the number of edges around the face is $0 \\mod 4$. If the number\nof edges equals $2 \\mod 4$, the product must equal $1$~\\cite{Kenyon2}. One\ncan choose an orientation by consistently assigning arrows to\nthe edges of the graph, as originally suggested by Kasteleyn\n\\cite{Kasteleyn1, Kasteleyn2}. For our purposes, it turns out to be\nmore convenient to allow the roots of 1 as (complex) edge weights.\n\nAlready Caianiello~\\cite{Caianiello} remarked that the expectation\nvalue of any product of free Fermi fields is a Pfaffian, while the\nexpectation value of any product of free Bose fields is a Hafnian. In\nthis sense Kasteleyn's method of attaching signs to the edges of a graph \\gra\\\ncould be seen as a fermionization of a bosonic system. Given the\ncorrespondence of the dimer model to the free fermion on \\gra, one\ncould look for its bosonic equivalent, but this goes beyond the scope\nof the present work.\n\n\nIn the following, we will consider graphs embedded on a cylinder or a\ntorus. The treatment can be straightforwardly generalized to any genus\n$g$ Riemann surface. In essence, all of the above remains true. The\nonly change is that on the torus (cylinder), we have two (one)\nnon--trivial cycles, which we will denote by $z$ and $w$. In the case\nof the plane graph, the edge weights originated solely from the\nKasteleyn orientation. More generally, the Kuperberg flatness condition must be respected~\\cite{Kuperberg2}. We choose a positive direction on the dimers,\nsay $\\bullet\\to\\circ$. Now we assign the weight $z$ ($w$) to each edge\nwhich crosses the cycle $z$ ($w$) in positive direction and the weight\n$1\/z$ ($1\/w$) to each edge which crosses it in negative\ndirection. While the Pfaffian of the Kasteleyn matrix yielded a number\nin the case of the plane graph, it now becomes a polynomial in $z$ and\n$w$, the so--called characteristic polynomial or \\emph{Newton\npolynomial} of the graph. The coefficient of each monomial $z^pw^q$\ngives the number of matchings with \\emph{weight} $(z,w)=(p,q)$.\nThese are matchings with the number of dimers crossing $z$ in positive\ndirection minus the number of dimers crossing $z$ in negative\ndirection equal to $p$ (analogous for $q$). In the literature,\nwhat we call the weight is usually referred to as the slope of a height\nfunction defined on the composition of two matchings. The height\nfunction is defined as follows. Choose a reference matching $\\mathrm{PM}_0$. To\nfind the slope of a matching $\\mathrm{PM}$, compose it with the reference\nmatching, $\\mathrm{PM}-\\mathrm{PM}_0$, where the minus serves to change the orientation of\n$\\mathrm{PM}_0$ to $\\circ\\to\\bullet$. This results in closed loops (composition\ncycles) and double line dimers. The rule is that when an edge in $\\mathrm{PM}$\nbelonging to a closed loop is crossed such that the black node is to\nits left (right), the height changes by $+1$ $(-1)$. If an edge\nbelonging to $\\mathrm{PM}_0$ is crossed, the signs are reversed. This height\nfunction is defined up to the choice of the reference matching\n$\\mathrm{PM}_0$. Crossing the boundary of the fundamental region of the torus, this function\ncan jump. If the height function jumps by $p$ units crossing $z$, it\nis associated to the power $z^p$ in the Newton polynomial of the graph\n(and equivalently for $w$). Choosing a different reference matching\nresults in a common prefactor of $z^{p_0}w^{q_0}$ for all\nmonomials. Our method of assigning weights to a matching corresponds\nto choosing a reference matching of weight $(0,0)$ that does not\nintersect the $z$ or $w$ cycle. The matching shown in Figure\n\\ref{fig:example} has weight $(1,0)$, where $1=2-1$.\n\\begin{figure}[h!]\n \\begin{center}\n \\includegraphics[width=50mm]{example}\n \\caption{Example of a square graph on the torus}\n \\label{fig:example}\n \\end{center}\n\\end{figure}\n\\\\\nThe characteristic, or Newton polynomial, of the dimer model on\nthe torus takes the form\n\\begin{equation}\n\\label{eq:torus-partition}\n P(z,w)=\\sum_{n_z, n_w} N_{n_z,n_w}\\,(-1)^{n_z+n_w+n_zn_w}z^{n_z}w^{n_w},\n\\end{equation}\nwhere the $N_{n_z,n_w}$ count the number of matchings that have weight\n(height change) $(n_z,n_w)$.\nFurthermore, the partition function which counts the total number of matchings for the square graph on the torus is given by\n\\begin{equation}\\label{eq:totalnumber}\nZ=\\frac{1}{2}\\left(-P(1,1)+P(1,-1)+P(-1,1)+P(-1,-1)\\right),\n\\end{equation}\nwhere the first term is always zero. $P(z,w)$ evaluated in $z,w=\\pm1$ corresponds to the contributions of the four different spin structures.\n\n\n\n\n\n\\subsection*{Acknowledgements}\n\nWe would like to thank Luis \\'Alvarez-Gaum\\'e, David Cimasoni, Davide Forcella,\nEmanuel Scheidegger, and Alberto Zaffaroni for enlightening\ndiscussions. Furthermore, we would like to thank David Cimasonifor detailed comments on the manuscript.\n\nD.O. and S.R. would like to thank \\textsc{CERN} for hospitality, where\npart of this work was carried out.\n\nThe research of R.D. was supported by a NWO Spinoza grant and the FOM program \"String Theory and Quantum Gravity.\"\nD.O. is supported in part by \\textsc{INFN} and \\textsc{MIUR} under contract 2005-024045 and by \nthe European Community's Human Potential Program \\textsc{MRTN}-\\textsc{CT}-2004-005104.\nS.R. is supported by the EC's Marie Curie Research Training Network under the contract \\textsc{MRTN}-\\textsc{CT}-2004-512194 \"Superstrings\".\n\n\n\\section{The dimer model as a free fermion}\n\\label{sec:fermions-dimers}\n\nAs already mentioned in the introduction, many clues point towards the\nequivalence of the dimer model and the free fermion living on the\nsame lattice. On one hand, we know that the dimer model corresponds to\nthe two--dimensional Ising model. In the continuum limit, the critical\nIsing model corresponds in turn to the free Majorana fermion. Another clue is\nthat the partition function of the dimer model on a genus $g$ Riemann\nsurface is a linear combination of $2^{2g}$ Pfaffians, one for each of\nthe different boundary conditions. The one--to--one correspondence\nbetween the $2^{2g}$ equivalence classes of Kasteleyn orientations and\nthe spin structures was proved in~\\cite{cimasoni-2006}. \n\nNote that our construction given in the following is different from the transfer matrix approach used\nin~\\cite{lieb:2339,Sutherland:1968,alet:041124}.\n\n\n\\subsection{The Dirac operator and the Kasteleyn matrix}\n\\label{sec:dirac}\n\nIn this section, we show that the dimer model on a square lattice can be\nnaturally mapped to the dynamics of a free massless fermion on the\nsame lattice. \nMore precisely, we will see how introducing a Kasteleyn orientation on\nthe graph can be thought of as the projection of the $\\del$ operator\non $\\gamma $ matrices, see also~\\cite{KenyonDirac, Kenyon2}.\n\nLet us consider the theory for a free massless fermion on a bipartite\nsquare graph\\footnote{ A different definition for the Dirac operator,\n such as the one given in~\\cite{Rabin:1981qj,Becher:1982ud} would be\n needed to deal with a general random graph. This goes beyond the\n scope of the present note.}. At this stage, the construction is\nindependent of the boundary conditions, \\emph{i.e.} we can think of\nthe model as living on the infinite plane. The graph being bipartite,\nit is natural to use staggered fermions~\\cite{PhysRevD.16.3031} and\nassociate a two--component real spinor to each pair of nodes:\n\\begin{equation}\\label{eq:twocomponent}\n \\Psi =\n \\begin{pmatrix}\n \\chi_{\\bullet} \\\\ \\chi_{\\circ}\n \\end{pmatrix} .\n\\end{equation}\nThe naive discretization of the free massless fermion action becomes then\n\\begin{equation}\n S = \\int \\di x \\: \\bar \\Psi (x) \\slashed{\\del} \\Psi (x) = \\sum_{ \\braket{x, y}} \\chi (x)\\, \\slashed{\\del}(x,y)\\, \\chi(y) ,\n\\end{equation}\nwhere the sum runs over all neighbouring vertices $ x $ and $y$\non the lattice. The discretized Dirac operator\ncan be cast into the form\n\\begin{equation}\n\\label{eq:Lattice-Dirac}\n \\slashed{\\del}(x,y) = \\gamma^\\mu \\frac{ x_\\mu - y_\\mu}{\\abs{x-y}} \\, , \n\\end{equation}\nwhere $\\gamma^\\mu$ are two matrices satisfying the Clifford algebra\n$\\set{ \\gamma^\\mu, \\gamma^\\nu} = 2\\, \\delta^{\\mu \\nu}$ in two\nEuclidean dimensions. On the square lattice, each node has\nfour neighbours, separated by unit vectors, so the action reads\n\\begin{equation}\n S = \\sum_{x \\in E (\\gra)} \\sum_{k=1}^4 \\chi(x)\\, \\gamma^\\mu e\\ud{k}{\\mu}\\, \\chi (x + e^k) \\, ,\n\\end{equation}\nwhere $e^k$ is the vector $e^k= ( \\cos \\frac{2k \\pi}{4},\\, \\sin \\frac{2k\n \\pi}{4})$. If $x$ is a $\\bullet $ site, $x \\pm e^k$ is a\n$\\circ$ site and vice versa. This means that we can split the action\ninto two pieces:\n\\begin{equation}\n \\label{eq:Dirac-action}\n S = \\left(\n \\begin{array}{c|c}\n \\vec{\\chi}_\\bullet & \\vec{\\chi}_\\circ \n \\end{array} \\right) \\left(\n \\begin{array}{c|c}\n 0 & \\slashed{\\del}_{\\bullet \\circ} \\\\ \\hline \n \\slashed{\\del}_{\\circ \\bullet} & 0 \n \\end{array} \\right) \\left(\n \\begin{array}{c}\n \\vec{\\chi}_\\bullet \\\\ \\hline\n \\vec{\\chi}_\\circ \n \\end{array} \\right) = \\vec{\\chi}_\\bullet \\cdot \\slashed{\\del}_{\\bullet \\circ} \\cdot \\vec{\\chi}_\\circ + \\vec{\\chi}_\\circ \\cdot \\slashed{\\del}_{\\circ \\bullet} \\cdot \\vec{\\chi}_\\bullet \\, ,\n\\end{equation}\nwhere $\\vec{\\chi}_\\bullet$ is the vector of Grassmanian variables\nliving on the $\\bullet $ nodes. Let us consider the first term. The\noperator $\\slashed{\\del}_{\\bullet \\circ}$ is non--vanishing if and only\nif $\\bullet $ and $\\circ$ are neighbours, and takes values\nproportional to the components of the $\\gamma $ matrices. More\nprecisely, if we think of $\\slashed{\\del}_{\\bullet \\circ}$ as a\nmatrix, each line will contain four non--vanishing entries $\\set{\n \\gamma_{\\bullet \\circ}^1, \\gamma_{\\bullet \\circ}^2, -\n \\gamma_{\\bullet \\circ}^1, - \\gamma_{\\bullet \\circ}^2}$. Choosing the\nrepresentation\n\\begin{align}\n \\gamma^1 =\n \\begin{pmatrix}\n 0 & 1 \\\\ 1 & 0\n \\end{pmatrix} \\, ,&&\n \\gamma^2 =\n \\begin{pmatrix}\n 0 & \\imath \\\\ - \\imath & 0\n \\end{pmatrix} \\, ,\n\\end{align}\nthese components read $\\set{1, \\imath, -1, - \\imath}$. These are the\nweights of the four links around each $\\bullet$ node. Since their\nproduct equals to $\\left( - 1 \\right)$ and on this graph the product of weights around a node is the same as the product around a plaquette, this is precisely a Kasteleyn\norientation (see Fig.~\\ref{fig:Dirac-Kasteleyn})\\footnote{This argument\n can be easily generalized to any regular tessellation of the\n hyperbolic plane (leading to higher--genus graphs). One can show that\n for an $n$-gon, the product of the weights around a node is $\\left(\n -1 \\right)$ if $n = 0 \\mod 4$ and $\\left(+1 \\right)$ if $n = 2\n \\mod 4$. For the hexagonal lattice (genus one), the\n weights around a $\\bullet$ site read $\\set{ 1, e^{2 \\pi \\imath \/3},\n e^{-2\\pi \\imath \/3} }$ and their product is $\\left( + 1\n \\right)$. It is worth to remark that for higher genus surfaces, there is no unique global choice of complex structure.}\n\n\\begin{figure}\n \\centering\n \\includegraphics{Dirac-Kasteleyn}\n \\caption{Dirac operator}\n \\label{fig:Dirac-Kasteleyn}\n\\end{figure}\n\nThe same construction can of course be repeated for the second term in\nEq.~\\eqref{eq:Dirac-action}. In this case, we get for each line of the\noperator $\\slashed{\\del}_{\\circ \\bullet}$ the four non--vanishing elements\n$\\{ \\gamma_{\\circ \\bullet}^1, \\gamma_{\\circ \\bullet}^2,$ $ -\n \\gamma_{\\circ \\bullet}^1, - \\gamma_{\\circ \\bullet}^2\\}$ which, in the\nsame representation for the $\\gamma$ matrices that we used above, read\n$\\set{ 1, - \\imath, - 1, \\imath} $, yielding again a Kasteleyn\norientation.\n\nAdjacency matrices with equivalent Kasteleyn orientations have the same determinant (up to an overall sign), therefore we have\n\\begin{equation}\n \\det ( \\slashed{\\del}_{\\bullet \\circ} ) = \\det ( \\slashed{\\del}_{\\circ \\bullet}) = \\Pf ( K )\\, ,\n\\end{equation}\nwhere $K$ is a Kasteleyn matrix for the graph.\nWriting the partition function for the free massless fermion gives\n\\begin{equation}\n \\label{eq:fermion-path-integral}\n Z_{\\text{fermion}} = \\int \\di \\psi \\: e^{-S[\\psi]} = \\sqrt{ \\det (\\slashed{\\del})} = \\sqrt{ \\det (\\slashed{\\del}_{\\bullet \\circ} ) \\det ( \\slashed{\\del}_{\\circ \\bullet}) } = \\Pf (K) \\, ,\n\\end{equation}\nwhich is precisely the number of perfect matchings on the lattice for a given boundary condition.\n\nExplicit expressions for both the dimer model on a square graph and\nthe free fermion theory are known. \nTake a triangular lattice describing a torus with\nperiodicity\n\\begin{align}\n z = z + i N + j M e^{\\imath \\theta} \\, ,&& i,j \\in \\setZ,\\ z\\in\\setC, \n\\end{align}\n\\emph{i.e.} with modular parameter $\\tau = \\tfrac{M}{N} e^{\\imath\n \\theta}$. In~\\cite{Nash:1996kn}, it is shown that the partition\nfunction for a vortex with winding $(a,b)$ on this lattice,\nwhich can be mapped to a free fermion, is given by the infinite product\n\\begin{multline}\n Z_{\\text{vortex}} \\oao{a}{b} (M,N, \\theta) = \\prod_{k=-M\/2}^{M\/2} \\prod_{l=-N\/2}^{N\/2} 2 \\abs{ \\alpha \\cos \\left( 2 \\pi \\frac{k}{M} + \\pi \\frac{a}{M} \\right) + \\beta \\cos \\left( 2 \\pi \\frac{l}{N} + \\pi \\frac{b}{N} \\right) + \\right. \\\\\n \\left. + \\gamma \\cos \\left( 2 \\pi \\left( \\frac{k}{M} - \\frac{l}{N} \\right) + \\pi \\left( \\frac{a}{M} - \\frac{b}{N} \\right) \\right) - \\sigma}, \n\\end{multline}\nwhere\n\\begin{align}\n \\alpha = \\beta = \\frac{1 - \\cos \\theta}{\\sin^2 \\theta}, && \\gamma = \\frac{\\cos \\theta}{\\sin^2 \\theta}, && \\sigma = \\frac{2 - \\cos \\theta}{\\sin^2 \\theta} . \n\\end{align}\nSpecializing this formula for a square torus with $\\theta = \\pi \/2 $ such that\n$\\alpha = \\beta = 1, \\gamma = 0, \\ \\sigma =2 $, we find:\n\\begin{equation}\n Z_{\\text{vortex}} \\oao{a}{b} (M,N, \\tfrac{\\pi}{2}) = \\prod_{k=-M\/2}^{M\/2} \\prod_{l=-N\/2}^{N\/2} 2 \\abs{ \\cos \\left( \\frac{2k \\pi}{M} + a \\frac{\\pi}{M} \\right) + \\cos \\left( \\frac{2 l \\pi}{N} + b \\frac{\\pi}{N} \\right) - 2},\n\\end{equation}\nwhich is precisely the partition function of the dimer model.\n\n\n\n\n\\subsection{The scaling limit}\n\\label{sec:scaling-limit-theta}\n\nHaving shown that the dimer model on a graph is equivalent to a free\nMajorana fermion theory, we expect the torus partition function to be\nmodular invariant in the thermodynamical limit, and to consist of\ntheta functions.\n\nSince this is a universal property, we specialize to a\nparticular lattice, for example the $M \\times N$ square graph. The\npartition function for this model was discussed in\n\\cite{Kasteleyn2,Kenyon1} and takes the form\n\\begin{equation}\n Z_{MN} = \\tfrac{1}{2} \\sum_{a,b=0}^1 \\left( -1 \\right)^{a+b+ab} Z_{MN} \\oao{a}{b} \\, ,\n\\end{equation}\nwhere\n\\begin{equation}\n\\label{eq:partition-square-mn}\n Z_{MN} \\oao{a}{b} = \\prod_{k=-M\/2}^{M\/2} \\prod_{l=-N\/2}^{N\/2} 2 \\abs{ \\cos \\left( \\frac{2k \\pi}{M} + a \\frac{\\pi}{M} \\right) + \\cos \\left( \\frac{2 l \\pi}{N} + b \\frac{\\pi}{N} \\right) - 2 } \\, .\n\\end{equation}\nIf we regard the $M \\times N$ lattice as a discretization of a square\ntorus with modular parameter $\\tau = \\imath t$ into $M \\times N$\nsquares, the scaling limit is obtained by taking\n\\begin{align}\n M, N \\to \\infty && \\frac{M}{N} = t , \\quad \\text{$t\\,$ fixed}. \n\\end{align}\nThis corresponds to sending the lattice size $\\frac{1}{M}$ to zero.\n\nAs we have previously pointed out, this is precisely the same\nexpression found for a soliton on a triangular lattice~\\cite{Nash:1995ba}. After a\ncareful analysis, the authors prove that in the continuum limit, this\nreproduces the Ray--Singer result for the $\\bar \\del$--torsion of the\ntorus~\\cite{Ray:1973sb} (see also~\\cite{Ferdinand1,Ferdinand2}):\n\\begin{equation}\n Z_{MN} \\oao{a}{b} \\xrightarrow[\\genfrac{}{}{0pt}{}{M,N \\to\n \\infty}{M\/N = - \\imath \\tau }]{} \\left[\n \\frac{\\vartheta \\oao{a}{b} (\\tau)}{\\eta(\\tau)} \\right]^2 \\, .\n\\end{equation}\n\n\nInstead of reproducing the proof,\nwe will limit ourselves here to a heuristic argument. In the infrared\nlimit, the cosine can be approximated by a parabola and the functions\nabove behave as follows:\n\\begin{multline}\n Z_{MN} \\oao{a}{b} = \\prod_{k=-M\/2}^{M\/2} \\prod_{l=-N\/2}^{N\/2} 2\n \\abs{\\, \\cos \\left( \\frac{2k \\pi}{M} + a \\frac{\\pi}{M} \\right) +\n \\cos \\left( \\frac{2 l \\pi}{N} + b \\frac{\\pi}{N} \\right) - 2\\, }\n \\sim \\\\\n \\sim \\prod_{k=-M\/2}^{M\/2} \\prod_{l=-N\/2}^{N\/2} \\frac{8 \\pi^2}{M^2}\n \\abs{ \\left(k+ \\frac{a}{2} \\right)^2 + \\left(l+ \\frac{b}{2}\n \\right)^2 t^2} \\, .\n\\end{multline}\nIn the $\\zeta$--function regularization, one recognizes the usual product\nexpansion for the theta functions. More precisely, introducing $\\tau =\n\\imath t$ we find %\n\\begin{comment}\n\n\n\nIt is a known fact that $\\vartheta \\oao{a}{b} (\\tau)\/ \\eta(\\tau)$ can be expanded as a double product as follows:\n\\begin{equation}\n \\frac{\\vartheta \\oao{a}{b} (\\tau)}{\\eta(\\tau)} = \\prod_{k \\in \\setZ} \\prod_{l \\in \\setZ} \\abs{ k + \\frac{a}{2} + \\left( l + \\frac{b}{2} \\right) \\tau} .\n\\end{equation}\nIf we decompose\n\\begin{equation}\n \\abs{ \\left(k+ \\frac{a}{2} \\right)^2 + \\left(l+ \\frac{b}{2} \\right)^2 t^2} = \\abs{ \\left(k+ \\frac{a}{2} \\right) + \\imath \\left(l+ \\frac{b}{2} \\right) t } \\abs{ \\left(k+ \\frac{a}{2} \\right) - \\imath \\left(l+ \\frac{b}{2} \\right) t } ,\n\\end{equation}\nit is immediate to see that in $\\zeta$-function regularization, given\nthat\n\\begin{align}\n \\prod_{k=1}^\\infty a = \\sqrt{a}, && \\prod_{k = -\\infty}^\\infty a =\n a^{2 \\zeta(0) + 1} = 1,\n\\end{align}\nwe find:\n\\end{comment}\n\\begin{gather}\n Z_{MN} \\oao{a}{b} \\xrightarrow[\\genfrac{}{}{0pt}{}{M,N \\to\n \\infty}{M\/N = - \\imath \\tau }]{} \\left[ \\prod_{k \\in \\setZ}\n \\prod_{l \\in \\setZ} \\abs{ k + \\frac{a}{2} + \\left( l + \\frac{b}{2}\n \\right) \\tau} \\right]^2= \\left[\n \\frac{\\vartheta \\oao{a}{b} (\\tau)}{\\eta(\\tau)} \\right]^2 \\, , \\\\\n Z_{MN} = \\tfrac{1}{2} \\sum_{a,b=0}^1 \\left( -1 \\right)^{a+b+ab}\n Z_{MN} \\oao{a}{b} \\xrightarrow[\\genfrac{}{}{0pt}{}{m,n \\to\n \\infty}{m\/n = - \\imath \\tau }]{} \\tfrac{1}{2} \\sum_{a,b=0}^1\n \\left( -1 \\right)^{a+b+ab} \\left[ \\frac{\\vartheta \\oao{a}{b}\n (\\tau)}{\\eta(\\tau)} \\right]^2 = Z_{\\text{Dirac}} (\\tau).\n\\end{gather}\nThe final result is that in the scaling limit, the partition function\nis the one of a Dirac fermion, or equivalently, for two Majorana\nfermions. It is not surprising that we encounter a fermion doubling effect~\\cite{Nielsen:1981hk, Nielsen:1980rz} since we have started with a periodic\ndispersion relation.\n\n\n\n\\section{Gauge theory interpretation: Preliminaries}\n\\label{sec:geometry}\n\n\nAs mentioned in the introduction, there exists a correspondence\nbetween quiver gauge theories describing $D$--branes probing singular\ntoric surfaces and the dimer\nmodel~\\cite{Franco:2005rj,Franco:2005sm}. We will make use of it to\nre--interpret the loops in terms of the quiver gauge theory and the\ntoric geometry. We will quickly summarize the necessary knowledge of\nquiver gauge theories and this correspondence in the following.\n\n\\subsection{The quiver gauge theory}\n\nConsider $D3$--branes in type IIB string theory probing a Calabi--Yau\n$X$. They correspond to BPS $B$--type branes given by points on\n$X$. If the $D$--branes are placed at a singularity, they are expected\nto decay into a collection of so--called \\emph{fractional} branes.\nThe resulting world--volume gauge theory can be summarized in a\n\\emph{quiver} graph $Q_X$ as follows. Each constituent brane appearing\nwith multiplicity $N_i$ corresponds to a $U(N_i)$ gauge group which is\nrepresented as a node in the quiver graph. The massless open strings\nstretching between the fractional branes correspond to chiral fields\nin the $(\\overline{N}_i, N_j)$ representation and are depicted as\narrows pointing from the $U(N_i)$--node to the $U(N_j)$--node in the\ndiagram. For the theory to be anomaly free, the number of arrows\n$a_{ik}$ going from node $N_i$ to node $N_k$ (in our convention\n$a_{ki} = - a_{ik}$) must fulfill\n\\begin{equation}\n \\label{eq:anomaly}\n \\sum_i N_i a_{ik} = 0 \\, , \\hspace{2em} \\forall k \\, .\n\\end{equation}\nIf the ranks of all gauge groups are equal, this reduces to the number\nof incoming and outgoing arrows being equal at each node.\n\nWe will be concentrating on the case of the Calabi--Yau $X$ being the cone over a singular toric surface $S$.\n\nTo fully capture the physics of the $D$--branes, they must be described in terms of the (bounded) derived category of coherent sheaves $D^\\flat(X)$. A full explanation of the machinery of the derived category is beyond the scope of this paper, the reader is therefore referred to~\\cite{Aspinwall:2004jr}. Luckily, we can avoid working directly in $D^\\flat(X)$ by using so--called \\emph{exceptional collections} of sheaves supported on a partial resolution of the singular surface $S$. In our case, these sheaves can be mostly thought of as line bundles. The exceptional collections form a convenient basis for the fractional branes and it is possible to construct the quiver from a given exceptional collection and vice versa~\\cite{Aspinwall:2004vm}. The whole treatment is based on the fact that the derived category of coherent sheaves is equivalent to the derived category of quiver representations.\n\n\nThe open strings in the topological sector between two $B$--branes $A$\nand $B$ are parametrized by $\\oplus_p \\mathrm{Ext}^p(A,B)$. The mass\nof such an open string between two branes with the same grade is given\nin string units by\n\\begin{equation}\\label{eq:mass}\nm^2=\\tfrac{1}{2}(p-1).\n\\end{equation}\nFrom this, we learn that the massless open strings (\\emph{i.e.} the arrows in\nthe quiver) are counted by $\\mathrm{Ext}^1$, while $\\mathrm{Ext}^0$\ncounts the tachyons. For $p>1$, the strings are very massive and are\nnot seen by the quiver gauge theory.\n\n\nIf oriented loops appear in a quiver (which is always the case for our\nquiver gauge theories because of the anomaly--freedom condition in\nEq.~\\eqref{eq:anomaly}), the quiver representations become infinite\ndimensional, causing the methods employed in~\\cite{Aspinwall:2004vm}\nto fail. Fortunately, this problem can be evaded by considering\nacyclic subquivers. Some of the arrows of the gauge theory quiver\n$Q_X$ are linked to the properties of the surface $S$ itself, while\nthe others parametrize the embedding of $S$ in the Calabi--Yau $X$.\nDeleting these latter arrows removes all oriented loops. The resulting\nquiver, $Q_S$, allows an ordering relation and is also called the\nBeilinson or Bondal quiver in the literature.\n\nThe derived category of coherent sheaves on $S$, $D^\\flat(S)$, is\nequivalent to the derived category of quiver representations for\n$Q_S$, denoted by $D^\\flat(A-{\\mathrm{mod}})$ (where $A$ is the path\nalgebra of $Q_S$). The basic representations $L_i$ of the path algebra\n$A$ are associated to the fractional branes at the nodes of the\nquiver. The arrows in the quiver are thus associated with the\n$\\mathrm{Ext}^1(L_i, L_j)$. The $\\mathrm{Ext}^2(L_i, L_j)$ correspond\nto relations in the quiver which the paths must obey. There is another\nquiver representation given by the projective objects $P_i$, which are\nthe subspaces of $A$ generated by all paths starting at node $i$. The\nsheaves in the exceptional collections we consider are the projective\nobjects $P_i$, which are in some sense dual to the fractional branes\n$L_i$.\n\nLet us illuminate a bit the connection between the quivers $Q_S$ and\n$Q_X$ and its geometrical meaning. Consider the embedding of the\nsurface $S$ into $X$ via $i:\\,S\\hookrightarrow X$. A $D$--brane in $S$\ncan obviously also be regarded as a $D$--brane in $X$, \\emph{i.e.} we can map\nobjects in $D^\\flat(S)$ injectively to objects in $D^\\flat(X)$ using\n$i_*$. \nSince there might be more morphisms in $D^\\flat(X)$ than $D^\\flat(S)$,\nsome of the open strings between two branes in $S$ may live in $X$ but\noutside of $S$. This fact is captured by the relation\n\\begin{equation}\\label{eq:Ext_decomposition}\n\\mathrm{Ext}^1_X(i_*L_i,i_*L_j)=\\mathrm{Ext}^1_S(L_i,L_j)\\oplus\\mathrm{Ext}^2_S(L_j,L_i).\n\\end{equation}\nThus, the decomposition of the arrows of the full quiver $Q_X$ into $\\mathrm{Ext}^1$ and $\\mathrm{Ext}^2$ on $S$ accounts for the embedding of $S$ in $X$. \n\n\n\n\\subsection{Exceptional collections and helices}\n\nIn this section, we collect the definitions on exceptional collections\nnecessary for later chapters. An (ordered) collection of sheaves\n$\\{\\mathcal{F}_0, \\dots , \\mathcal{F}_{n-1}\\}$ on $S$ is called\n\\emph{exceptional} if \n\\begin{subequations}\n \\label{eq:exceptional}\n \\begin{align}\n \\mathrm{Ext}^0_S(\\mathcal{F}_i, \\mathcal{F}_i)&=\\setC \\\\\n \\mathrm{Ext}^p_S(\\mathcal{F}_i, \\mathcal{F}_i)&=0, \\quad p\\geq1 \\\\\n \\mathrm{Ext}^p_S(\\mathcal{F}_i, \\mathcal{F}_j)&=0, \\quad i>j.\n \\end{align}\n\\end{subequations}\nThe collection is called \\emph{strongly} exceptional, if\n$\\mathrm{Ext}^p_S(\\mathcal{F}_i, \\mathcal{F}_j)=0$ for $p\\neq0$ and\n\\emph{complete} if it generates $D^\\flat(S)$. Since the $D$--branes on\n$S$ can wrap $0-$, $2-$ and $4-$cycles, a collection must contain $n$\nsheaves to be complete, where $n=\\,$sum over all Betti numbers\n($=\\chi(S)$). For physics, the collection being strong means that the\nbasis of fractional branes it generates does not contain ghost matter.\n\nExceptional collections can be transformed into new exceptional\ncollections by left and right \\emph{mutations}, which represent an\naction of the braid group on the set of possible collections. On a\nneighboring pair of sheaves in an exceptional collection, the\nmutations act as\n\\begin{align}\n \\label{eq:mutation}\n L_i &: (E_i, E_{i+1})\\mapsto(L_{E_i}E_{i+1}, E_i),\\\\\n R_i &: (E_i, E_{i+1})\\mapsto(E_{i+1}, R_{E_{i+1}}E_i).\n\\end{align}\nThe sheaves $L_{E_i}E_{i+1}$ and $R_{E_{i+1}}E_i$ are defined via\nshort exact sequences, see~\\cite{Rudakov,Zaslow:1994nk}. Left and\nright mutations can be seen as braiding and unbraiding. A\n\\emph{helix} $\\mathcal{H}$ is an infinite collection of coherent\nsheaves $\\{\\mathcal{F}_i\\}_{i\\in\\setZ}$ such that\n\\begin{itemize}\n\\item[(a)] for any $i\\in\\setZ$,\n $\\{\\mathcal{F}_{i+1},\\dots,\\mathcal{F}_{i+n}\\}$ is an exceptional\n collection.\n\\item[(b)] $R^{n-1}\\mathcal{F}_i=\\mathcal{F}_{i+n}$.\n\\end{itemize}\nThe name stems from the fact that after moving $n$ steps to the right,\none is back at the original place up to a translation. $n$ is called\nthe \\emph{period} $n$ of $\\mathcal{H}$. Each collection of the form\n$\\{\\mathcal{F}_{i+1}, \\dots,\\mathcal{F}_{i+n}\\}$ is called a\n\\emph{foundation} of the helix and $\\mathcal{H}$ is uniquely determined by\neach of its foundations.\n\n\n\n \n\n\n\n\n\\subsection{The correspondence}\n\nHere we quickly summarize the correspondence worked out in\n\\cite{Franco:2005rj, Franco:2005sm}.\n\n\\begin{enumerate}\n\\item Start with a toric quiver gauge theory, given by a quiver graph\n and its tree--level superpotential.\n\\item Draw the quiver diagram on a torus, such that each of the terms\n of the superpotential corresponds to a plaquette. The resulting\n graph is called the \\emph{periodic quiver}.\n\\item Take the graph dual of this graph (i.e. vertices become faces,\n faces become vertices and edges remain edges). Colour the vertices\n coming from the faces associated to a superpotential term with\n negative sign black; the ones coming from a positive term colour\n white. The resulting graph is bipartite and also lives on a torus.\n\\item Solve the dimer model on this graph by taking the Pfaffian of\n the Kasteleyn matrix. This yields the Newton polynomial. The\n associated \\emph{Newton polygon} is given by taking the exponents of\n the monomials in $zw$ to be coordinates in the $(z,w)$--plane. The\n resulting polygon is exactly the toric diagram of the surface the\n quiver gauge theory comes from!\n\\end{enumerate}\n\nThe singular surface $S$ is specified by the corners of the toric diagram. These corner vertices each correspond to Weil divisors and therefore to line bundles. The fully resolved surface is given by adding all the points lying in the intersection of the convex hull of the polygon and the integer $\\setZ^2$--lattice to the toric diagram. Each such vertex corresponds to an exceptional divisor which is blown up to resolve the singularity. The surface $S$ can also be resolved partially by blowing up only part of the exceptional divisors. Passing from one partial resolution $S'$ corresponding to the exceptional divisor $E'$ being blown up to another partial resolution $S''$ with $E''$ being blown up corresponds to a sequence of birational transformations (blow down $E'$ and blow up $E''$ instead).\n \nThe vertices of the toric diagram can be labeled with the multiplicities of the perfect matchings with the corresponding weight. There are certain patterns in the multiplicities which are worth pointing out:\n\\begin{itemize}\n\\item The vertices at the corners of the toric diagram always have multiplicity one, since they correspond to the highest or lowest weight states in either $z$ or $w$, which are unique.\n\\item The multiplicities on the edges of the graph follow the rule of Pascal's triangle (i.e. $1,\\,2,\\,1$ or $1,\\,3,\\,3,\\,1$, etc.). Indeed, the polynomial consisting only of the monomials associated to the vertices of an edge has the form $(z+w)^n$, or can be brought to this form by multiplication with a prefactor $z^{n_0}w^{m_0}$ or by redefinition of the variables.\n\\end{itemize}\nFor the multiplicities of the vertices in the interior of the toric diagram, a pattern is much less obvious.\n\nWe illustrate the above by the example of one square on the torus, which corresponds to the quiver gauge theory on $\\mathbb{F}_0$ in Figure \\ref{table:corr}.\n\\begin{comment}\n \\begin{table}[h!]\n \\begin{center}\n \\begin{tabular}{ccc}\n \\includegraphics[width=28mm]{torusquiver} &\n \\begin{minipage}{6em}\n \\vspace{-5em} ${\\text{graph dual}}$ \\\\ \\vspace{5em}\n \\end{minipage}\n & \\includegraphics[width=25mm]{one_square} \\\\\n $\\updownarrow$ & & $\\downarrow$ \\\\\n \\includegraphics[width=24mm]{quiver} & \\begin{minipage}{6em}\n \\vspace{-3em} $\\longleftrightarrow$ \\\\ \\vspace{3em}\n \\end{minipage} & \\includegraphics[width=32mm]{F0}\n \\end{tabular}\n \\caption{Web of correspondences for one square on the torus\/\n quiver gauge theory on $\\mathbb{F}_0$}\n \\label{table:corr}\n \\end{center}\n \\end{table}\n\\end{comment}\n\n\n\\newarrow{Corresponds}<--->\n\\begin{figure}\n \\centering\n \\begin{diagram}\n \\begin{minipage}{29mm}\n \\includegraphics[width=28mm]{torusquiver}\n \\end{minipage} & \\rCorresponds^{\\text{\\hspace{1em}graph dual}} &\n \\begin{minipage}{26mm}\n \\includegraphics[width=25mm]{one_square}\n \\end{minipage} \\\\\n \\\\\n \\uTo^{\\text{on the torus}} & & \\dTo^{\\det(K)} \\\\\n \\\\\n \\begin{minipage}{23mm}\n \\includegraphics[width=22mm]{quiver}\n \\end{minipage} & \\rCorresponds &\n \\begin{minipage}{33mm}\n \\includegraphics[width=32mm]{F0}\n \\end{minipage}\n \\end{diagram}\n\\caption{Web of correspondences for one square on the torus\/ quiver gauge theory on $\\mathbb{F}_0$}\n \\label{table:corr}\n\\end{figure}\nThe tree-level superpotential is given by\n\\begin{equation}\\label{eq:supo}\nW=-2\\,X_{DC}X_{CA}X_{AB}X_{BD}+2\\,X_{CA}X_{AB}X_{BD}X_{DC},\n\\end{equation}\nthe Newton polynomial is \n\\begin{equation}\n \\label{eq:partition_square}\n P (z,w) = zw \\left(- \\frac{1}{z}-\\frac{1}{w} + 4 - z -w \\right) \\, .\n\\end{equation}\n\n\n\n\n\n\n\\section{Introduction}\n\\label{sec:intro}\n\nThe dimer model is concerned with the statistical mechanics of close\npacked dimer arrangements on a bipartite graph. The physical system\nwhich can be thought of as a real world representation of the dimer\nmodel is the adsorption of diatomic molecules on a crystal surface. \n\nIn the 1960s, the question of how many perfect matchings exist on a plane\ngraph was solved independently by Kasteleyn~\\cite{Kasteleyn1,\nKasteleyn2}, and Temperley and Fisher~\\cite{Temperley, Fisher}: the total number is given by the Pfaffian of a signed, weighted adjacency matrix of the graph (the Kasteleyn matrix).\nMuch of\nthe original interest in the dimer model arose because it provides a simple\nand elegant solution for the 2--dimensional Ising model~\\cite{Hurst}.\n\nThe problem of enumerating perfect matchings is of course a classical problem in graph theory and combinatorics, see \\emph{e.g.}~\\cite{Kenyon2}. It can also be phrased in terms of domino tilings~\\cite{Kenyon1}.\nDuring the last years, the interest in the dimer model was revived\nthanks its manifold connections to other branches of\nmathematics and physics. The dimer model is related to\n\\begin{itemize}\n\\item configurations of a melting crystal corner and the\ntopological string A--model \\cite{Okounkov:2003sp, Iqbal:2003ds},\n\\item real algebraic geometry~\\cite{okounkov1,okounkov2}, \n\\item BPS black holes from $D$--branes wrapping collapsed cycles~\\cite{Heckman:2006sk}.\n\\end{itemize}\nFurthermore, a correspondence between the dimer model and quiver\ngauge theories arising from $D3$--branes probing a singular toric\nsurface was discovered and worked out in great detail\n\\cite{Hanany:2005ve, Franco:2005rj, Franco:2005sm, Hanany:2005ss,\nHanany:2006nm}. An explanation of this\ncorrespondence via mirror symmetry was given in~\\cite{Feng:2005gw}.\n\n\\bigskip\nThat the dimer model must be related to the free massless fermion on\nthe same lattice has been known for a long time already. Its relation\nto the Ising model, which in turn corresponds to a free fermion model\nclearly indicates this. Moreover, Kasteleyn also showed that the\npartition function of the dimer model living on a graph embeddable on\na genus $g$ Riemann surface is given by a linear combination of\n$2^{2g}$ Pfaffians. This is, of course, reminiscent of the number of\nspin structures on a Riemann surface. In fact, a one-to-one\ncorrespondence between equivalence classes of Kasteleyn orientations\nand spin structures was proved in~\\cite{cimasoni-2006}. \n\nIn the following, we will make the connection between the dimer model and the\nfree fermion precise by identifying the Dirac operator with the Kasteleyn matrix and relating the expansion of the determinant\nappearing in the free fermion path integral to loop configurations\ncomposed of two perfect matchings. This picture of free fermions as a\nloop gas already appeared in a different context, for example in\n\\cite{Fermionloops}. \n\nThe perfect matchings form a\nbasis for the space of fermion loop configurations. If the loop gas lives on a surface of non--trivial homotopy, non--contractible loops of non--zero winding\nnumber can occur, and the Pfaffian of the Kasteleyn matrix becomes a\npolynomial. Drawing inspiration from the categorification programme of\nKhovanov~\\cite{Khovanov1, Khovanov2, Khovanov3}, we \ninterpret this polynomial as the Euler characteristic of a co--chain\ncomplex. \nThe loop configurations of the free fermion model can\nbe naturally classified by the\nweight of the constituent matchings of the loop state and used to generate Abelian groups. %\nFor the Euler\ncharacteristic to coincide with the Newton polynomial, a differential\noperator must be constructed accordingly. In our fermion loop gas,\nall loop states are paired since they come in two opposite\norientations, except for states consisting only of double line perfect\nmatchings. This is reminiscent of the supersymmetric ground states in\nsupersymmetric quantum mechanics. Indeed, it is possible to\nre--interpret the fermion loop gas as \\textsc{sqm} and to construct a differential\noperator which can be identified with the $Q$ operator. Like this, the\nNewton polynomial becomes the generalized Witten index of the system.\nSince the fermion loop states are bilinear in the perfect matchings, they can also be interpreted as maps from one matching to another. \n\nThanks to the correspondence between the dimer model and quiver gauge theory, \nwe know that the perfect matchings, \\emph{i.e.} our zero modes, parametrize the toric surface being probed by the $D$--branes. \nIn the following, we make use of this correspondence to give the loops states an interpretation in terms of the quiver gauge theory. Loops of vanishing net winding can be related to sequences of double Seiberg dualities on the associated acyclic quivers. In the special case of the hexagon graph, they correspond to certain BPS black hole configurations which are parametrized by melting crystal configurations~\\cite{Heckman:2006sk}. \n\n\\bigskip The plan of this paper is as follows. In Section\n\\ref{sec:preliminaries}, we review the necessary background and\ndefinitions of the dimer model. In Section \\ref{sec:dirac},\nwe relate the Kasteleyn matrix to the discretized Dirac operator\nof the free fermion. In Section~\\ref{sec:scaling-limit-theta}, we go to the limit of vanishing lattice\nspacing in which we recover the partition function of the Dirac\nfermion, which in the light of the interpretation of the dimer model\nas a free fermion model is no longer surprising. In Section~\\ref{sec:loop-expans-ferm}, we\nexpand the determinant arising in the free fermion path integral into\ncyclic permutations, which correspond to closed loops on the graph. Furthermore, we give an interpretation of the loops as fermionic states in Subsection~\\ref{sec:loops-as-fermionic}.\n\nIn Section \\ref{sec:categorification}, the basic idea of\ncategorification is explained, and in Section \\ref{sec:sqm}, the\nfermion loop system is re--interpreted as supersymmetric quantum\nmechanics. The simpler case of a square graph on the cylinder\nis worked out first, together with a simple explicit example in Section~\\ref{sec:example:-one-square}.\nIn Section~\\ref{sec:loops-as-operators}, we interpret the loops as operators mapping from one perfect matching to another. Finally, Section~\\ref{sec:sqm_torus} gives the full construction on the torus.\n\nIn Section \\ref{sec:geometry}, the basics of quiver gauge theories and the\ncorrespondence to the dimer model is reviewed. The interpretation of\nthe loop states in terms of the physics of $D$--branes probing a singular toric CY is attempted in Section \\ref{sec:results}. In Section~\\ref{sec:acyclic}, the relation between perfect matchings and corresponding acyclic quiver graphs is clarified. In Section~\\ref{sec:doubleseiberg}, we show that sequences of double Seiberg dualities correspond to cyclic permutations of the nodes of the acyclic quiver and to isomorphisms of the full gauge quiver. With these results, we can interpret the loops with vanishing overall winding in Section~\\ref{sec:0-0-loops}. The loops with non--trivial winding are discussed in Section~\\ref{sec:winding_loops}.\n\nThe results of this paper are summarized in a dictionary given in\nSection~\\ref{sec:conclusions}, where we close with some concluding\nremarks.\n\n\n\n\n\n\n\\subsection{The loop expansion of the fermion determinant}\n\\label{sec:loop-expans-ferm}\n\nAfter having shown the equivalence of the discretized Dirac operator\nand the Kasteleyn orientations and the appearance of the Dirac fermion\npartition function in the scaling limit, we will now directly relate\nthe dimer model to the free fermion via a diagrammatic expansion of\nthe fermion determinant.\n\nConsider a Grassmanian field $\\chi$ living on the nodes $V(\\gra)$ of a\ngraph $\\gra $ described by a quadratic action:\n\\begin{equation}\n S[\\Psi] = \\sum_{x,y \\in V(\\gra)} \\chi_x A_{x,y} \\chi_y . \n\\end{equation}\nAs we have seen in Eq.~(\\ref{eq:twocomponent}), we can associate a\ntwo--component Majorana fermion $\\Psi ={\\chi_\\bullet \\choose\n \\chi_\\circ}$ to each black--white pair in a bipartite graph. The\npath integral for such an action is easily computed and is given by\nthe Pfaffian of the matrix $A$:\n\\begin{equation}\n Z_{\\text{fermion}} = \\int \\left[ \\prod_{x \\in V(\\gra )} \\di \\chi_x \\right] e^{\\frac{1}{2} \\vec{\\chi} A \\vec{\\chi}} = \\Pf (A),\n\\end{equation}\nwhere $\\vec{\\chi}$ is the vector $\\vec{\\chi} = \\set{ \\chi_1, \\chi_2,\n \\dots, \\chi_{\\abs{V(\\gra)}} }$. In fact, expanding the exponential\n(and using the fact that $\\chi_x$ are Grassmanian variables), we find:\n\\begin{equation}\n Z_{\\text{fermion}} = \\int \\left[ \\prod_{x \\in V(\\gra )} \\di \\chi_x \\right]\n \\prod_{x0$, because in addition to the one--cycles generated by\nthe plaquettes, there are $2g$ equivalence classes of cycles of\nnon--trivial holonomy. In this case, the winding cycles in\n$\\gra^\\prime$ are generated by the zig--zag paths. It follows that\nbeing a perfect matching in $\\gra$ is only a necessary condition for a\nset of edges to be a \\textsc{fas} in $\\gra^\\prime$. It was shown\nin~\\cite{Hanany:2006nm} that a perfect matching corresponding to an\ninternal point in the toric diagram is always a \\textsc{fas}. Removing\na boundary matching, on the other hand, always preserves at least one\nzig--zag path.\n\n\\bigskip\n\nAs mentioned before, the gauge quiver $Q_X$ is the graph dual of the\ngraph $\\gra$ on which the dimer model lives. The acyclic quiver\n$Q_{S_n}$ is obtained from $Q_X$ by deleting all the edges contained\nin the $n$th perfect matching\n$\\mathrm{PM}_n\\in\\{\\mathrm{PM}^{int}\\}$. The deleted arrows are\nrepresented by dashed lines and correspond to relations in the quiver,\n\\emph{i.e.} $\\mathrm{Ext}^2$s. Such an acyclic quiver allows an\nordering of its nodes such that there are no arrows between the nodes\n$i$ and $j$ if $il$,} \\\\ \n \\ket{k,a; l,b}_{-} &=& \\kket{l,b} \\otimes \\kket{k,a} & \\text{if $k>l$,} \\\\ \n \\ket{k,a; k,b}_{+} &=& \\kket{k,a} \\otimes \\kket{k,b} & \\text{if $a>b$,} \\\\ \n \\ket{k,a; k,b}_{-} &=& \\kket{k,b} \\otimes \\kket{k,a} & \\text{if $a>b$,} \\\\ \n \\ket{k,a; k,a}_{+} &=& \\kket{k,a} \\otimes \\kket{k,a} \\, ,\\\\ \n \\ket{k,a; k,a}_{-} &=& 0 \\, .\n \\end{array}\n \\end{equation}\n Opposite polarizations $\\epsilon$ correspond to opposite orientations\n of the same loop configuration. States of the type $\\ket{k,a;\n k,a}_+$ (which are in one--to--one correspondence with the perfect\n matchings) are singled out because they are not paired.\n\\end{itemize}\n\n\n\n\\begin{comment}\nOur system of loops on a graph can be thought of as an example of\nsupersymmetric quantum mechanics. Then the partition function for the\nperfect matchings becomes a generalization of the Witten index.\n\nA generic loop state is characterized by five quantum numbers:\n\\begin{align}\n \\ket{\\text{loop}} = \\ket{k,a; l,b}_\\epsilon , && l \\le k, \\, a = 0,\n \\dots, N_k - 1, \\, b = 0, \\dots, N_l - 1, \\, \\epsilon = \\pm 1,\n\\end{align}\nwhere $k$ and $l$ are the weights of the matchings that make up the\nloop, $N_k$ is the degeneracy of matchings with weight $k$, and\n$\\epsilon $ is the polarization. Opposite polarizations correspond to\ntwo loops with reversed orientations. We define a loop with\ncounterclockwise orientation to have positive polarization,\ni.e. $\\epsilon=1$.\n\\end{comment}\n\n\nSince each loop configuration with a given orientation can be uniquely\ndecomposed into matchings, these quantum numbers can be assigned to a\ngiven configuration by inspection. The weight of a matching is given\nby the number of dimers crossing the boundary of the fundamental\nregion in positive direction minus the number of dimers crossing the\nboundary of the fundamental region in negative direction (where we\nchoose $\\bullet\\to\\circ$ as orientation). $\\epsilon \\left( k-l\n\\right)$ is the total winding number of the loop configuration (which\ncoincides with the slope of the height function as defined in\nSec.~\\ref{sec:preliminaries}). States of the type $\\ket{k,a ; k,a}$\nare in one--to--one correspondence with the perfect matchings. All other\nstates are paired (i.e. there are two polarizations). From now on, we\nwill omit to specify the degeneracy quantum numbers and write a loop\nas $\\ket{k ; l}_\\epsilon$ (note the semicolon).\n\n\nThe quantum numbers can be seen as eigenvalues for the following operators:\n\\begin{align}\n H \\ket{k;l}_\\epsilon &= \\left( k - l \\right) \\ket{k; l}_\\epsilon \\, ,\\\\\n K \\ket{k;l}_\\epsilon &= k \\ket{k;l}_\\epsilon \\, ,\\\\\n \\Pi \\ket{k; l}_\\epsilon &= \\epsilon \\ket{k;l}_\\epsilon \\, .\n\\end{align}\nThey commute on the above loop configurations by definition. It\nfollows that $H$ (which we want to identify with the energy) commutes\nalso with the operator\n\\begin{equation}\n F = 2 K - \\frac{1}{2} \\left( \\Pi - 1 \\right) \\, ,\n\\end{equation}\nwhich we identify with the fermion number operator.\n\nIn terms of the loops, the above operators have the following\ninterpretation. $H$ counts the (unsigned) total winding number of the\nstate. $K$ counts the maximum attainable total monodromy of a loop\nconfiguration. The state $\\ket{k; l}_\\epsilon$ allows for maximally\n$k$ overall (i.e. positively minus negatively oriented) homologically\nnon--trivial windings, of which $k-l$ are actually realized. If $H$ is\nidentified with the energy, $k$ gives the maximum possible\nenergy. Each loop with non--trivial winding which is added to the\nconfiguration raises the energy of the state by one unit. Ground\nstates do not contain non--contractible loops. The matchings, in\nparticular, contain no loops at all. %\n\\begin{comment}\n In the simple case of a product of spin chains on the cylinder, the\n states $\\ket{k; l}_\\epsilon,\\ \\, l\\leq k$ can be viewed as a system\n with $k$ boxes, into which $k-l$ particles (non--trivial loops) are\n put.\n\\end{comment}\nThe eigenvalue of the polarization operator $\\Pi$ is the sign of the\nwinding number. The fermion number operator $F$ is constructed such,\nthat positive polarization states $\\ket{k; l}_+$ have even fermion\nnumber $2k$, while the fermion number of the corresponding negative\npolarization states $\\ket{k; l}_-$ is raised by one, \\emph{i.e.} is\n$2k+1$. The operator $\\left(-1\\right)^F$ gives a $\\IZ_2$ grading\naccording to the fermion parity. States with positive polarization\nhave even parity, while states negative polarization have odd parity.\n\n\\bigskip\n\nTo obtain the structure of a \\textsc{sqm}, we need to construct two\nconserved supercharges $Q$ and $Q^\\dag$ which satisfy the following\nalgebra (see \\emph{e.g.} Chapter 10 of~\\cite{Hori}):\n\\begin{align}\n \\acomm{Q, Q} &= 0 & \\acomm{Q^\\dag, Q^\\dag} &= 0 \\\\\n \\acomm{Q, Q^\\dag} &= 2 H \\\\\n \\comm{F, Q} &= Q & \\comm{F, Q^\\dag} &= - Q^\\dag .\n\\end{align}\nA possible choice is the following:\n\\begin{align}\n\\label{eq:Q-cylinder}\n \\begin{cases}\n Q \\ket{k; l}_+ = \\sqrt{2 \\left( k - l \\right)} \\ket{k; l}_- ,\\\\\n Q \\ket{k; l}_- = 0 .\n \\end{cases} &&\n \\begin{cases}\n Q^\\dag \\ket{k; l}_+ = 0 ,\\\\\n Q^\\dag \\ket{k; l}_- = \\sqrt{2 \\left( k - l \\right)} \\ket{k; l}_+ .\n \\end{cases}\n\\end{align}\nIt is also useful to define the operator $Q_1$ as in\nEq.~\\eqref{eq:Q1-creator}:\n\\begin{equation}\n Q_1 = Q + Q^\\dag.\n\\end{equation}\nIt reverses the polarization of a state,\n\\begin{equation}\n Q_1 \\ket{k ; l}_\\epsilon = \\sqrt{2 \\left(k -l \\right)} \\ket{k; l}_{-\\epsilon} , \n\\end{equation}\nand squares to the energy:\n\\begin{equation}\n Q_1^2 = 2 H . \n\\end{equation}\n\n\nThe space of states can be decomposed into subspaces according to the fermion number. In the following, we will concentrate our attention on $Q$.\n\\begin{subequations}\n \\begin{align}\n C^p &= \\Sp \\set{\\ket{k,a; l,b}_\\epsilon | F \\ket{k,a;\n l,b}_\\epsilon = p\n \\ket{k,a; l,b}_\\epsilon } \\, , \\\\\n C^{2k} &= \\Sp \\set{ \\ket{k,a; l, b}_+ } \\, ,\\\\\n C^{2k+1} &= \\Sp \\set{\\ket{k,a ; l, b}_- } \\, .\n \\end{align}\n\\end{subequations}\nWe also define\n\\begin{subequations}\n \\begin{align}\n C^+ &= \\bigoplus_{k} C^{2k} \\, , \\\\\n C^- &= \\bigoplus_{k} C^{2k + 1} \\, , \\\\\n C^* &= C^+ \\oplus C^- = \\bigoplus_{p} C^{p} \\, .\n \\end{align}\n\\end{subequations}\nOne verifies easily that\n\\begin{equation}\n Q : C^p \\to C^{p+1} \n\\end{equation}\n(and conversely $Q^\\dag : C^p \\to C^{p-1}$). We can now construct the co--chain complex\n\\begin{equation}\n\\label{eq:SQM-complex}\n 0 \\hookrightarrow C^0 \\xrightarrow{Q} C^1 \\xrightarrow{Q} \\dots \\xrightarrow{Q} C^N \\xrightarrow{Q} 0 \\, . \n\\end{equation}\n\nLet us now study the cohomology of such a complex. The $0$--energy states\n$\\ket{k,a; k,b}_\\epsilon$ and the negative polarization states\n$\\ket{k,a;l,b}_-$ are $Q$--closed. On the other hand,\nnegative polarization states (with $k\\neq l$) are also $Q$--exact, such that the\ncohomology coincides with the set of supersymmetric ground states:\n\\begin{equation}\n \\bigoplus_p H^p(Q) = \\Sp \\set{ \\ket{k, a; l, b}_\\epsilon| H \\ket{k, a; l, b}_\\epsilon = 0} = \\Sp \\set{\\ket{k,a;k,b}_\\epsilon} .\n\\end{equation}\nMoreover, all the states with $a \\neq b$ are paired (opposite parities). This means that when taking the Euler number, only the $\\ket{k,a;k,a}$ states (perfect matchings) contribute:\n\\begin{equation}\n \\chi = \\Tr \\left( - 1 \\right)^F = \\sum_p \\left( - 1\\right)^p \\dim \\left[ H^p (Q) \\right] = \\sum_p N_p . \n\\end{equation}\nAs we remarked above, diagrammatically, ground states are\nconfigurations with vanishing total winding number. The ones\ncontributing to the Euler number are only those that do not contain\nany loops.\n\nA different invariant can be constructed by taking the trace of the $\\left(\n - 1\\right)^F z^K$ operator (which is still a conserved quantity since\nit is a function of conserved operators). It is easy to see that it is\ngiven by\n\\begin{equation}\n \\chi (z)= \\Tr \\left[ \\left( - 1\\right)^F z^K \\right] = \\sum_{k,\\epsilon} \\left( - 1\\right)^p z^k \\dim \\left[ H^p (Q) \\right] = \\sum_k z^k N_k,\n\\end{equation}\nwhere $p = 2k - \\tfrac{1}{2}(\\epsilon -1)$. This is the Poincar\\'e\npolynomial for the sequence in Eq.~(\\ref{eq:SQM-complex}) and in\nparticular, $\\chi(1) = \\chi$. Moreover, $\\chi(-z)$ is by construction\nthe partition function for the free Majorana fermion on the lattice.\n\nIt is worth stressing that even if, as it is often the case, the\nzero--modes alone are enough to describe important features of the\nphysics of the system such as the partition function, we have complete\ncontrol over all the states in the theory. In Appendix~\\ref{sec:generating-functions}, the generating functions capturing all loop states are given.\n\n\n\\subsection{Example: One square on the cylinder}\n\\label{sec:example:-one-square}\n\nTo illustrate the above, we now present the smallest possible example\non the cylinder, consisting only of one square. The five possible\nmatchings and their quantum numbers are shown in Figure\n\\ref{fig:ex_matchings}.\n\\begin{figure}[h!]\n \\begin{center}\n \\includegraphics[width=110mm]{ex_matchings}\n \\caption{Matchings for the square on the cylinder}\n \\label{fig:ex_matchings}\n \\end{center}\n\\end{figure}\nThey combine into 25 loop configurations, which are, sorted into their\nco--chain groups according to total winding number, depicted in Table\n\\ref{table:chaingroups}.\n\nThe partition function can easily be found by inspection and reads:\n\\begin{equation}\n P (z) = z \\left(- \\frac{1}{z} + 3 - z \\right) \\, .\n\\end{equation}\nThe generating function for this example reads\n\\begin{equation}\n G( \\mathbf{q}, \\mathbf{z}, y ) = 1 + 3 y^2 z + y^4 z^2 + \\left( 1 + y \\right) \\left[ 3 y^2 z + 3 q \\left( y^2 z + y^4 z^2 \\right) + q^2 y^4 z^2 \\right] \\, .\n\\end{equation}\n\n\n\n\\begin{table}[h!]\n \\begin{center}\n \\begin{tabular}{ccccccc}\n \\toprule\n \\boldmath $k - l$& $C^0$ & $C^1$ & $C^2$ & $C^3$ & $C^4$ & $C^5$ \\unboldmath \\cr \\midrule\n \\rowcolor[gray]{.95} 2 & - & -& - & -& \\pb{$\\ket{2,1;0,1}_+$ \\includegraphics[width=13mm]{2101plus}} &\\pb{$\\ket{2,1;0,1}_{-}$ \\includegraphics[width=13mm]{2101minus}} \\cr \n \\midrule\n 1 & - & - & \\pb{$\\ket{1,3;0,1}_+$ \\includegraphics[width=13mm]{1301plus}} & \\pb{$\\ket{1,3;0,1}_-$ \\includegraphics[width=13mm]{1301minus}}& \\pb{$\\ket{2,1;1,3}_+$ \\includegraphics[width=13mm]{2113plus}}&\\pb{$\\ket{2,1;1,3}_{-}$ \\includegraphics[width=13mm]{2113minus}} \\cr\n \\rowcolor[gray]{.95} 1 & - & -& \\pb{$\\ket{1,2;0,1}_+$ \\includegraphics[width=13mm]{1201plus} }& \\pb{$\\ket{1,2;0,1}_-$ \\includegraphics[width=13mm]{1201minus}}& \\pb{$\\ket{2,1;1,2}_+$ \\includegraphics[width=13mm]{2112plus}}&\\pb{$\\ket{2,1;1,2}_{-}$ \\includegraphics[width=13mm]{2112minus}}\\cr\n 1 & - & - & \\pb{$\\ket{1,1;0,1}_+$ \\includegraphics[width=13mm]{1101plus}} & \\pb{$\\ket{1,1;0,1}_-$ \\includegraphics[width=13mm]{1101minus}}& \\pb{$\\ket{2,1;1,1}_+$ \\includegraphics[width=13mm]{2111plus}}&\\pb{$\\ket{2,1;1,1}_{-}$ \\includegraphics[width=13mm]{2111minus}} \\cr\n \\midrule\n \\rowcolor[gray]{.95} 0 & - & - & \\pb{$\\ket{1,3;1,2}_+$ \\includegraphics[width=13mm]{1312plus}} & \\pb{$\\ket{1,3;1,2}_-$ \\includegraphics[width=13mm]{1312minus}} &- &- \\cr\n 0 & - & - & \\pb{$\\ket{1,3;1,1}_+$ \\includegraphics[width=13mm]{1311plus}} & \\pb{$\\ket{1,3;1,1}_-$ \\includegraphics[width=13mm]{1311minus}} &- &- \\cr\n \\rowcolor[gray]{.95} 0 & - & -& \\pb{$\\ket{1,2;1,1}_+$ \\includegraphics[width=13mm]{1211plus}} & \\pb{$\\ket{1,2;1,1}_-$ \\includegraphics[width=13mm]{1211minus}} &- &- \\cr\n 0 & - & - & \\pb{$\\ket{1,3;1,3}$ \\includegraphics[width=13mm]{1313}} & - &- &- \\cr\n \\rowcolor[gray]{.95} 0 & - & - & \\pb{$\\ket{1,2;1,2}$ \\includegraphics[width=13mm]{1212}} & - &- &- \\cr\n 0 & \\pb{$\\ket{0,1;0,1}$ \\includegraphics[width=13mm]{00}} & - & \\pb{$\\ket{1,1;1,1}$ \\includegraphics[width=13mm]{1111}} & -& {\\pb{$\\ket{2,1;2,1}$ \\includegraphics[width=13mm]{2121}} } &- \\cr \\bottomrule\n \\end{tabular}\n \\caption{Co--chain groups for the square on the cylinder}\n \\label{table:chaingroups}\n \\end{center}\n\\end{table}\n\n\n\n\n\n\\subsection{Loops as operators}\n\\label{sec:loops-as-operators}\n\nIn this section, we introduce a different notation for the loops that\nwill allow us to make contact with the geometric and gauge theory descriptions in\nSec.~\\ref{sec:geometry} and \\ref{sec:results}. Each perfect matching can be seen as a pure\nstate in a quantum system obeying a normalization condition\n$\\braket{k,a|l,b} = \\delta_{kl} \\delta_{ab}$. The general (quantum)\nconfiguration for the system is obtained as a linear combination\n\\begin{equation}\n \\ket{\\Psi} = \\sum_k \\sum_a \\lambda_{k,a} \\ket{k,a} \\, ,\n\\end{equation}\nwhere $\\lambda_{k,a}$ are complex parameters normalized as usual\n$\\sum_{k,a} \\abs{\\lambda_{k,a}}^2 =1 $.\n\nUsing this point of view, a loop can be identified with a map going\nfrom a state $\\ket{k,a}$ to a state $\\ket{l,b}$ and as such is\nrepresented in bracket notation by $\\ketbra{l,b}{k,a}$ acting as\n\\begin{equation}\n \\left(\\ketbra{l,b}{k,a} \\right) \\ket{k',a'} = \\delta_{k k'} \\delta_{a a'} \\ket{l,b} \\, .\n\\end{equation}\nThe linear combination of two loops is still well defined, but a new\noperation is naturally defined by the product of two loops:\n\\begin{equation}\n \\left( \\ketbra{l,b}{k,a} \\right) \\times \\left( \\ketbra{l',b'}{k',a'} \\right) \\mapsto \\ketbra{l,b}{k',a'} \\delta_{k l'} \\delta_{ab'} \\, .\n\\end{equation}\nNote that this is the usual path algebra on a complete graph whose\nnodes are labelled by $\\ket{k,a}$. It is useful to introduce a new\nrepresentation for our states. Consider a graph with $N$ nodes\narranged on the lattice $\\setN^2$ at coordinates $\\left( k, a \\right)$,\nand represent the loop $\\ketbra{l,b}{k,a}$ as the arrow from\n$\\left(k,a \\right)$ to $\\left(l,b \\right)$. Our system of loops\ncorresponds to a complete graph with double lines pointing in\nopposite directions plus a loop on top of each vertex (this is a state graph,\nsee Fig.~\\ref{fig:state-graph-cylinder}(a)). The co--chain groups\n(Abelian, freely generated by the loops) receive the following\ninterpretation:\n\\begin{itemize}\n\\item $C^{2p}$ is the group generated by the loops mapping from a\n state with winding $p$ to a state with winding smaller than $p$\n (arrows pointing left), from a state with winding $p$ to another one\n with a smaller degeneracy number (arrows pointing down), or from a\n state to itself (loops). Fig.~\\ref{fig:state-graph-cylinder}(b);\n\\item $C^{2p+1}$ is the group generated by the loops mapping from a\n state with winding smaller than $p$ to one with winding $p$ (arrows\n pointing right), or from a state with winding $p$ to another with\n the same winding and higher degeneracy number (arrows pointing\n up). Fig.~\\ref{fig:state-graph-cylinder}(c).\n\\end{itemize}\nIn terms of the state graph, this means that the differential map\n$Q_{2p} : C^{2p} \\to C^{2p+1}$ maps every arrow with a non--vanishing\nleft component to its opposite and annihilates all the others:\n\\begin{gather}\n Q_{2p}: \\ketbra{l,b}{p,a} \\mapsto \\ketbra{p,a}{l,b} \\, , \\\\\n Q_{2p}: \\ketbra{p,b}{p,a} \\mapsto 0 \\, .\n\\end{gather}\nOn the other hand, $Q_{2p+1} : C^{2p+1} \\to C^{2p+2}$ annihilates every map\n\\begin{equation}\n Q_{2p + 1}: \\ketbra{p,a}{l,b} \\mapsto 0 \\, . \n\\end{equation}\nIt is immediate to see that when considering the cohomology one finds:\n\\begin{itemize}\n\\item $H^{2p} \\subset C^{2p}$ is generated by downwards pointing arrows and\n loops. Fig.~\\ref{fig:state-graph-cylinder}(d);\n\\item $H^{2p+1} \\subset C^{2p+1}$ is generated by upwards pointing\n arrows. Fig.~\\ref{fig:state-graph-cylinder}(e).\n\\end{itemize}\nThis implies in turn that taking the Euler character, only the loops\nsurvive since upward and downward pointing arrows are paired and\ncounted with opposite signs \n(Fig.~\\ref{fig:state-graph-cylinder}(f)). They are in one--to--one\ncorrespondence with the nodes and all counted with $\\left(+ \\right)$\nsign, such that\n\\begin{equation}\n \\chi = \\sum_p \\left( - 1\\right)^p \\dim (H^p) = \\sum_p N_p = N \\, . \n\\end{equation}\n\n\\begin{figure}\n \\centering\n \\subfigure[Complete state graph $C^\\ast$]{\\includegraphics[width=.4\\textwidth]{StateGraph-131-crop}} \\hfill\n \\subfigure[$C^+ = \\bigoplus_{p} C^{2p}$]{\\includegraphics[width=.4\\textwidth]{StateGraph-131-Cp-crop}} \\vfill\n \\subfigure[$C^- = \\bigoplus_{p} C^{2p+1}$]{\\includegraphics[width=.35\\textwidth]{StateGraph-131-Cm-crop}} \\hfill\n \\subfigure[$H^+ = \\bigoplus_{p} H^{2p}$]{\\includegraphics[width=.4\\textwidth]{StateGraph-131-Hp-crop}} \\vfill\n \\subfigure[$H^- = \\bigoplus_{p} H^{2p+1}$]{\\includegraphics[width=.35\\textwidth]{StateGraph-131-Hm-crop}} \\hfill\n \\subfigure[Loops contributing to the Euler number]{\\includegraphics[width=.4\\textwidth]{StateGraph-131-Eu-crop}}\n \\caption{State graph for the dimer model in\n Sec.~\\ref{sec:example:-one-square}}\n \\label{fig:state-graph-cylinder}\n\\end{figure}\n\n\n \n\n\n\n\n\\subsection{The dimer model on the torus and supersymmetric QM}\n\\label{sec:sqm_torus}\n\nWe are now ready to pass to the actual case of interest, \\emph{i.e.}\nto extend the categorification construction of Sec.~\\ref{sec:sqm} to\ngraphs embedded on a torus. Now we have two non--trivial cycles, so\neach matching has two weights, $k_z$ and $k_w$, as explained in\nSec.~\\ref{sec:preliminaries}. These weights will be treated on a\ndifferent footing, but the final result will restore the expected\nsymmetry.\n\nThe notation is modified as follows:\n\\begin{itemize}\n\\item a perfect matching is labelled as $\\kket{\\mathbf{k},a}$, where\n $\\mathbf{k}$ is the vector $\\mathbf{k} = \\left( k_z, k_w \\right) $\n and $a = 1, \\dots, N_{\\mathbf{k}}$;\n\\item a loop is the combination of two matchings and is labelled by\n $\\ket{\\mathbf{k},a;\\mathbf{l},b}_\\epsilon$, where $k_z \\geq l_z$,\n $\\epsilon = \\pm 1$.\n \\begin{equation}\n \\begin{array}{ccll}\n \\ket{\\mathbf{k},a; \\mathbf{l},b}_{+} &=& \\kket{\\mathbf{k},a} \\otimes \\kket{\\mathbf{l},b} & \\text{if $\\mathbf{k} \\succ \\mathbf{l}$,} \\\\ \n \\ket{\\mathbf{k},a; \\mathbf{l},b}_{-} &=& \\kket{\\mathbf{l},b} \\otimes \\kket{\\mathbf{k},a} & \\text{if $\\mathbf{k} \\succ \\mathbf{l}$,} \\\\ \n \\ket{\\mathbf{k},a; \\mathbf{k},b}_{+} &=& \\kket{\\mathbf{k},a} \\otimes \\kket{\\mathbf{k},b} & \\text{if $a>b$,} \\\\ \n \\ket{\\mathbf{k},a; \\mathbf{k},b}_{-} &=& \\kket{\\mathbf{k},b} \\otimes \\kket{\\mathbf{k},a} & \\text{if $a>b$,} \\\\ \n \\ket{\\mathbf{k},a; \\mathbf{k},a}_{+} &=& \\kket{\\mathbf{k},a} \\otimes \\kket{\\mathbf{k},a} \\, ,\\\\ \n \\ket{\\mathbf{k},a; \\mathbf{k},a}_{-} &=& 0 \\, ,\n \\end{array}\n \\end{equation}\n where $\\mathbf{k} \\succ \\mathbf{l} $ if $k_z > l_z $ or if $\\set{ k_z = l_z, \\,k_w>l_w}$. This corresponds to lexicographic order.\n\\end{itemize}\nAgain, the loops can be seen as eigenstates for the following operators:\n\\begin{align}\n H \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon &= \\left( k_z - l_z \\right)\\ket{\\mathbf{k}; \\mathbf{l}}_\\epsilon \\, ,\\\\\n K_z \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon &= k_z \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon \\, ,\\\\\n K_w \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon &= \\max (k_w, l_w) \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon \\, ,\\\\\n \\Pi \\ket{\\mathbf{k}; \\mathbf{l}}_\\epsilon &= \\epsilon \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon \\, .\n\\end{align}\nAs opposed to the example on the cylinder, a loop state now comes with\ntwo natural signs, $\\epsilon_z=\\sign{(k_z - l_z)}$, and\n$\\epsilon_w=\\sign{(k_w- l_w)}$. For the grading by the fermion number,\nwe assign only one sign $\\epsilon$ to each loop state according to\nTable \\ref{table:signs}.\n\\begin{table}[h!]\n \\begin{center}\n \\begin{tabular}{ccc}\n \\toprule\n \\boldmath $\\epsilon_z$&$\\epsilon_w$&$\\epsilon$ \\unboldmath \\cr \\midrule\n \\rowcolor[gray]{.95} $+$&$+$&$+$\\\\\n $-$&$-$&$-$\\\\\n \\rowcolor[gray]{.95} $+$&$-$&$+$\\\\\n $-$&$+$&$-$\\\\\n \\bottomrule\n \\end{tabular}\n \\caption{Sign table for torus loop states}\n \\label{table:signs}\n \\end{center}\n\\end{table}\nThis is achieved by \n\\begin{equation}\n \\label{eq:signdef}\n \\epsilon=\\epsilon_z+(1-\\epsilon_z^2)\\,\\epsilon_w.\n\\end{equation}\nBy choosing the right order on the perfect matchings of the same\nweight and choosing a sign for some loop configurations of winding\nnumber 0, the signs of the overall winding numbers around the $z$ and\n$w$ cycles can be brought to correspondence with the\nsigns $\\epsilon_z,\\,\\epsilon_w$ defined above. We assign a sign to an oriented loop\nby choosing a right handed (counterclockwise) system to have positive\nsign and a left handed (clockwise) to have negative sign. The signs of\na loop configuration are determined by the overall winding number\naround the cycles. Only in cases with an equal number of loops winding\na cycle in opposite directions, a sign has to be chosen arbitrarily.\n\nThe fermion number operator only depends on $K_z$ and is defined by\n\\begin{equation}\n F = 2\\, K_z - \\frac{1}{2} \\left( \\Pi - 1 \\right) \\, .\n\\end{equation}\n\nStates can be again collected into subspaces according to the\neigenvalues of $F$, but in this case, we have an extra grading given by $k_w$. Since $\\comm{F, K_w} = 0$, we obtain a direct sum structure, and define:\n\\begin{gather}\n C^p = \\bigoplus_{q=0}^M C^{p,\\,q} \\, , \\\\\n C^{p,\\,q} = \\Sp \\set{\\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon | F \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon = p, \\, K_w \\ket{\\mathbf{k};\\mathbf{l}}_\\epsilon = q} \\, .\n\\end{gather}\nThe differential operators (or supercharges) are defined precisely as\nin Eq.~(\\ref{eq:Q-cylinder})\n\\begin{align}\n \\begin{cases}\n Q \\ket{\\mathbf{k}; \\mathbf{l}}_+ = \\sqrt{2 \\left( k_z - l_z \\right)} \\ket{\\mathbf{k}; \\mathbf{l}}_- ,\\\\\n Q \\ket{\\mathbf{k}; \\mathbf{l}}_- = 0 .\n \\end{cases} &&\n \\begin{cases}\n Q^\\dag \\ket{\\mathbf{k}; \\mathbf{l}}_+ = 0 ,\\\\\n Q^\\dag \\ket{\\mathbf{k}; \\mathbf{l}}_- = \\sqrt{2 \\left( k_z - l_z \\right)} \\ket{\\mathbf{k}; \\mathbf{l}}_+ ,\n \\end{cases}\n\\end{align}\nand it is immediate to verify that they respect the grading since\n$\\comm{K_w, Q} = \\comm{K_w, Q^\\dag} = 0$. Again we concentrate on $Q$. It can be decomposed into the sum\n\\begin{align}\n Q_p &= \\sum_{q=0}^M Q_{p,\\,q} \\, , \\\\\n \\intertext{where}\n Q_p &: C^{p,\\,q} \\to C^{p+1,\\,q} \\, ,\\hspace{2em} q = 0, \\dots, M \\, . \n\\end{align}\nAs a result, the $C^\\ast$ complex is decomposed into $C_{\\ast,q}$\ncomplexes:\n\\begin{equation}\n \\label{eq:torus-complex}\n \\begin{array}{rccccc}\n 0 &\\hookrightarrow C^{0,M} &\\xrightarrow{Q_{0,M}} C^{1,M} &\\xrightarrow{Q_{1,M}} \\dots &\\xrightarrow{Q_{N-1,M}} C^{N,M} &\\xrightarrow{Q_{N,M}} 0 \\, , \\\\\n & & & \\dots \\\\\n 0 &\\hookrightarrow C^{0,q} &\\xrightarrow{Q_{0,q}} C^{1,q} &\\xrightarrow{Q_{1,q}} \\dots &\\xrightarrow{Q_{N-1,q}} C^{N,q} &\\xrightarrow{Q_{N,q}} 0 \\, , \\\\\n & & &\\dots \\\\\n 0 &\\hookrightarrow C^{0,0} &\\xrightarrow{Q_{0,0}} C^{1,0} &\\xrightarrow{Q_{1,0}} \\dots &\\xrightarrow{Q_{N-1,0}} C^{N,0} &\\xrightarrow{Q_{N,0}} 0 \\, .\n \\end{array}\n\\end{equation}\n\nThe analysis of the cohomology is the same as in the case of the cylinder and one can easily convince oneself that it consists of states of zero energy:\n\\begin{equation}\n \\bigoplus_{p,q} H^{p,q} = \\Sp \\set{ \\ket{\\mathbf{k},a; \\mathbf{l},b}_\\epsilon | k_z = l_z} . \n\\end{equation}\nOnce more, only the states that are in one--to--one correspondence with the perfect\nmatchings (\\emph{i.e.} the ones that do not contain non--trivial closed loops) are not paired\nand thus survive the projection by the $(-1)^F$ operator:\n\\begin{equation}\n \\chi(w)=\\Tr \\left[ (-1)^F w^{K_w} \\right] = \\sum_{p,q} w^q N_{p,q} \\,,\n\\end{equation}\nwhere we introduced the variable $w$ to keep track of the internal\ngrading. The partition function described in\nEq.~(\\ref{eq:torus-partition}) is obtained as the trace of the\noperator $(-1)^{F + K_z K_w} \\left( - z \\right)^{K_z} \\left( - w \\right)^{K_w}$:\n\\begin{equation}\n\\label{eq:Witten-torus}\n P (z,w) = \\Tr \\left[ (-1)^{F + K_z K_w} \\left( - z \\right)^{K_z} \\left( - w \\right)^{K_w} \\right] = \\sum_{p,q} (-1)^{p + q + pq}\\, z^p w^q\\, N_{p,q} \\, .\n\\end{equation}\nNote the presence of the factor $(-1)^{p+q+p\\,q}$, which coincides with\nthe signature of the spin structure. Because of this, the expression\ndiffers from the Poincar\\'e polynomial for the graded complex\n\\begin{equation}\n \\chi (z,w) = \\sum_{p,q}\\, z^p\\, w^q\\, N_{p,q} \\, .\n\\end{equation}\nThe two polynomials, on the other hand, contain the same information\nand one can pass from one to the other using the identity\n\\begin{equation}\n 2 \\,\\chi (z,w) = - P(z,w) + P(-z,w) + P(z,-w) + P(-z,-w) \\, . \n\\end{equation}\nIn particular, $\\chi(w) = \\chi(-1,w) $ and $\\chi (1,1)$ is the overall\nnumber of perfect matchings on the graph. As already remarked at the\nbeginning of this section, it is worth emphasizing that the choice of\nthe cycles is arbitrary. This means in particular that one could have\nused $K_w$ to construct the Fermion number operator and $K_z$ as an\ninternal grading without affecting the final result.\n\nThe generalization to graphs embedded on higher--genus Riemann surfaces\nis straightforward and leads to a $\\left( 2g -1 \\right)$--graded complex.\n\n\nA simple example to illustrate the above is given in\nAppendix~\\ref{sec:ex_torus}.\n\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section{Introduction}\n\nThe description of physical systems is standardly given in terms of coordinates as defined by reference frames. Thanks to the principle of covariance, stating the equivalence of all descriptions regardless of the choice of the reference frame, it is possible to choose the reference frame where the relevant dynamical quantities can be most conveniently described. For example, it is typically easier to describe the dynamics of a system from the point of view of its rest frame, because only internal degrees of freedom contribute to the dynamics in the rest frame. \n\nWhen the external degrees of freedom (momentum) of the system are in a quantum superposition from the perspective of the laboratory, no classical reference frame transformation can map the description of physics from the laboratory to the rest frame. However, this can be achieved via a quantum reference frame (QRF) transformation between two frames moving in a superposition of velocites relative to one another. In order to achieve such change of quantum reference frame in the nonrelativistic regime, a formalism was introduced in Ref.~\\cite{QRF} to change the description to a refence frame which is in a quantum relationship with the initial one. This QRF transformation only depends on relational quantities, and it has also been derived starting from a gravity inspired symmetry principle in a perspective neutral model \\cite{perspective1, perspective2} . An immediate consequence of the formalism is that entanglement and superposition are QRF-dependent features. This formalism naturally leads to the possibility of identifying the rest frame of a quantum system in an operational way. \n\nHere, we further develop this approach in the case of a relativistic quantum particle with spin, with the goal of finding an operational description of the spin in a special-relativistic setting. Spin is operationally defined in the rest frame of a particle (or, to a good approximation, for slow velocities) via the Stern-Gerlach experiment. When the particle has relativistic velocities, the spin degree of freedom transforms in a momentum-dependent way. If a standard Stern-Gerlach measurement is performed on a particle in a pure quantum state moving in a superposition of relativistic velocities, the operational identification of the spin fails, because no orientation of the Stern-Gerlach apparatus returns an outcome with unit probability. This happens because, as shown in Ref.~\\cite{peres_first}, the reduced density matrix of the spin degree of freedom is mixed when a Lorentz boost is performed and the momentum is traced out. The question arises whether it is possible to find `covariant measurements' of the spin and possibly momentum, which predict invariant probabilities in different Lorentzian reference frames also for the case of a quantum relativistic particle moving in a superposition of velocities. In this case, it would be possible to map the unambiguous description of spin in the rest frame of the particle to the frame of the laboratory, and therefore derive the corresponding observables to be measured in the laboratory frame to verify spin with probability one.\n\nThe question of finding such covariant measurements is motivated by the ubiquitous applications where the spin degree of freedom is used as a qubit, to encode and transmit quantum information. Such protocols are no longer valid in a relativistic context, thus limiting the range of applicability of techniques involving spin as a quantum information carrier. It is then important to explore possible alternative methods which could overcome this limitation. In the context of relativistic quantum information, this question has been extensively discussed \\cite{peres_first, peresterno_review, alsingmilburn, gingrichadami, leeyoung_entanglementboost, bartlettterno_encoding, vedral_paradox, esteban, friis_relativisticentanglement, caban_reducedspin, vedral_operational, palmer_sterngerlach, markusphilipp_lorentz, terno_localisation} in relation to Wigner rotations \\cite{wigner1939, weinberg, sexlurbantke} and has been related to the problem of identifying a covariant spin operator. The problem of identifying such covariant spin operator has arisen long before the birth of relativistic quantum information, and dates back to the early times of quantum mechanics \\cite{thomas_precession, frenkel_spin, bargmannOnFrenkel}. Since then, a multitude of relativistic spin operators have been proposed \\cite{bauke_relativisticspin}, such as the Frenkel \\cite{frenkel_spin}, the Pauli-Luba\\'{n}ski \\cite{lubanski, ryder_paulilubanski, terno_tworoles}, the Pryce \\cite{pryce_spin}, the Foldy-Wouthuysen \\cite{foldywouthuysen, caban_spinobservable}, the Czachor \\cite{czachor_spin} the Fleming \\cite{fleming_spin} the Chakrabarti \\cite{Chakrabarti_spin}, and the Fradkin-Good \\cite{fradkingood_spin} spin operators. A comparative description of spin observables can be found in Ref.~\\cite{terno_localisation}.\n\nHere, we introduce `superposition of Lorentz boosts' which allow us to ``jump'' into the rest frame of a relativistic quantum particle even if the particle is {\\it not} in a momentum eigenstate. In the rest frame, the spin observables fulfill the spin $\\mathfrak{su}(2)$ algebra (the algebra of a qubit) and are operationally defined through the Stern-Gerlach experiment. We transform the set of spin observables in the rest frame back to an isomorphic set of observables in the laboratory frame. The transformed observables are in general entangled in the spin and momentum degrees of freedom. The set fulfills the $\\mathfrak{su}(2)$ algebra and is operationally defined through a `relativistic Stern-Gerlach experiment': we construct the interaction and the measurement between the spin-momentum degrees of freedom and the electromagnetic field in the laboratory frame which gives the same probabilities as the Stern-Gerlach experiment in the rest frame. This set of observables in the laboratory frame allows us to partition the total Hilbert space into two (highly degenerate) subspaces corresponding to the two outcomes ``spin up'' and ``spin down''. Hence, with QRFs techniques the relativistic spin can effectively be described as a qubit in an operationally well-defined way.\n\n\\section{A relativistic Stern-Gerlach experiment}\n\n\\begin{figure}\n\t\\begin{center}\n\t\t\\includegraphics[scale=0.3]{spinQRF}\n\t\t\\caption{(a) The state of a Dirac particle A with spin $\\tilde{A}$ as seen from the laboratory perspective (C). When the state is in a superposition of relativistic velocities $-v_1$ and $-v_2$, the spin degree of freedom and the momentum degree of freedom are no longer separable. (b) The state of the spin $\\tilde{A}$ and of the laboratory C as seen in the rest frame of the quantum particle A. In this quantum reference frame, the spin is operationally defined by means of the Stern-Gerlach experiment.}\n\t\t\\label{fig:spinQRF}\n\t\\end{center}\n\\end{figure}\n\nIn the following, we build a QRF transformation between the reference frame of a laboratory C of mass $m_C$ and the rest frame of the external degrees of freedom A of a relativistic quantum particle of mass $m_A>0$ with spin degrees of freedom $\\tilde{A}$, as illustrated in Fig.~\\ref{fig:spinQRF}. We allow the particle to have any quantum state, and in particular to move in a superposition of momenta. This implies that there is a non-classical relationship between the initial and the final reference frame, i.e., that the rest frame A and the laboratory frame C are not related by a standard boost transformation. We show in this section how to generalise the boost transformation to this case. Formally, the situation we consider can be described by taking the one-particle sector of the positive-energy solutions of the Dirac equation\\footnote{For simplicity, we only consider spin-$1\/2$ particles, but the method can be straightfowardly applied to arbitrary spin.} in the Foldy-Wouthuysen representation \\cite{foldywouthuysen}. \n\nFollowing Ref.~\\cite{QRF} (see Supplemental Information for a review of the original formalism), when we ``stand'' in the rest frame of a particle, we describe all the systems external to the particle, but not the external degrees of freedom (i.e., the momentum degrees of freedom) of the particle itself. Hence, the quantum state describes the relational information in a given reference frame. In the reference frame in which A is at rest, the quantum state is assigned to the internal degrees of freedom $\\tilde{A}$ and the laboratory C. For simplicity, we consider that the particle and the laboratory are moving with constant, yet not necessarily well-defined, relative velocity and define the $x$ axis along the direction of the relative motion. The total state of the spin and the laboratory is assumed to be\n\\begin{equation} \\label{eq:stateA}\n\t\\left| \\Psi \\right>_{\\tilde{A}C}^{(A)} = \\left|\\vec{\\sigma} \\right>_{\\tilde{A}}\\left| \\psi \\right>_{C},\n\\end{equation}\nwhere $\\left|\\vec{\\sigma} \\right>_{\\tilde{A}}$ is any vector representing the state of the spin in the rest frame A. In the rest frame, the spin state can in principle be tomographically verified by performing a series of standard Stern-Gerlach measurements. The state of the laboratory has a momentum-basis representation along the $x$ direction (at this stage, we neglect the quantum state in the $y$ and $z$ direction) $\\left| \\psi \\right>_{C} = \\int d\\mu_C(\\pi_C) \\psi (\\pi_C) \\left| \\pi_C \\right>_{C}$, where $d\\mu_C(\\pi_C) = \\frac{d\\pi_C}{(2\\pi)^{1\/2}\\sqrt{2(m_C^2 c^2 + \\pi_C^2)}}$ is the Lorentz-covariant integration measure. \n\nWe now construct the transformation corresponding to the ``superposition of Lorentz boosts'' to the QRF of the laboratory. The unitary operator to boost to the QRF C is\n\\begin{equation} \\label{eq:QRFTransf}\n\t\\hat{S}_L = \\mathcal{P}_{CA}^{(v)} U_{\\tilde{A}}(\\hat{\\pi}_C),\n\\end{equation}\n where $U_{\\tilde{A}}(\\hat{\\pi}_C)$ is a unitary transformation acting on the total Hilbert space $\\mathcal{H}_{\\tilde{A}}\\otimes \\mathcal{H}_C$ (notice that $\\hat{\\pi}_C$ is an operator), and $\\mathcal{P}_{AC}^{(v)}$ is the `generalised parity operator' introduced in Ref.~ \\cite{QRF}, whose explicit expression is $\\mathcal{P}_{CA}^{(v)}= P_{AC} \\exp \\left(\\frac{i}{\\hbar}\\log \\sqrt{\\frac{m_A}{m_C}}(\\hat{q}_C \\hat{\\pi}_C + \\hat{\\pi}_C \\hat{q}_C)\\right)$, where $P_{AC}$ is the parity-swap operator mapping $\\hat{x}_A \\rightarrow -\\hat{q}_C$ and $\\hat{p}_A \\rightarrow -\\hat{\\pi}_C$ (and viceversa), where $\\hat{q}_C, \\,\\hat{\\pi}_C$ are canonically-conjugated one-particle operators of C in the reference frame of A and $\\hat{x}_A, \\,\\hat{p}_A$ are canonically-conjugated one-particle operators of A in the reference frame of C. Additionally to the action of $P_{AC}$, the operator $\\mathcal{P}_{AC}^{(v)}$ rescales the momentum of A by the ratio of the masses of A and C, i.e., $\\mathcal{P}_{AC}^{(v)} \\hat{p}_A \\mathcal{P}_{AC}^{(v)\\dagger} = - \\frac{m_A}{m_C} \\hat{\\pi}_C$. This enforces the physical condition that the velocity of A is mapped to the opposite of the velocity of C via the transformation\\footnote{For a relativistic particle the relation between the $i$-th velocity component and the momentum is $v_i = \\frac{p_i}{m_i} \\left( 1+ \\frac{|\\vec{p}|^2}{m_i^2 c^2} \\right)^{-1\/2}$, where $|\\vec{p}|^2$ is the norm of the spatial momentum. Therefore, only the ratio between momentum and mass determines the velocity.}. The operator $\\hat{S}_L$ can be defined via its action on a basis of the total Hilbert space of the spin and the laboratory $\\hat{S}_L |\\vec{\\sigma} \\rangle_{\\tilde{A}}|\\pi\\rangle_C = | -\\frac{m_A}{m_C}\\pi; \\Sigma_{\\pi}\\rangle_{A\\tilde{A}}$, where the state $|p; \\Sigma_{p}\\rangle_{A\\tilde{A}}$ is defined via a standard Lorentz boost $\\hat{U}(L_p)$ from the rest frame as $|p; \\Sigma_{p}\\rangle_{A\\tilde{A}}=\\hat{U}(L_p)| k; \\vec{\\sigma}\\rangle_{A\\tilde{A}}$ and $k= (mc, \\vec{0})$ is the momentum in the rest frame. \n\n In Supplemental Information we derive the transformation $\\hat{S}_L$ in terms of standard Lorentz boosts connecting two relativistic reference frames where the parameter of the boost transformation is promoted to an operator. \n \n The state of $A$ and $\\tilde{A}$ expressed in the laboratory frame is $\\left| \\Psi \\right>_{A\\tilde{A}}^{(C)} = \\hat{S}_L \\left| \\Psi \\right>_{\\tilde{A}C}^{(A)}$, and is explicitly written as\n\\begin{equation} \\label{eq:StateTrans}\n\t\\left| \\Psi \\right>_{A\\tilde{A}}^{(C)} = \\int d\\mu_A(p_A) \\psi\\left(-\\frac{m_C}{m_A}p_A\\right) \\left| p_A; \\Sigma_{p_A} \\right>_{A\\tilde{A}},\n\\end{equation}\nwhere $d\\mu_A(p_A) = \\frac{d p_A}{(2\\pi)^{1\/2}\\sqrt{2(m_A^2 c^2 + p_A^2)}}$ and the spin degree of freedom cannot be separated anymore from the momentum degree of freedom, which means that the state is not a product state in the laboratory frame. Notice that the effect of the $\\hat{S}_L$ transformation is to apply the usual boost transformation conditional on C's momentum degree of freedom. In the laboratory frame C, unless particle A is in a sharp momentum state, no spin measurement in a standard Stern-Gerlach experiment would give a result with probability one, because of two reasons: the spin and momentum are no longer separable, and the relation between the laboratory and the rest frame is not a standard (classical) reference frame transformation. Our goal is to devise a different measurement in the laboratory reference frame, possibly involving both the spin and momentum degrees of freedom, which gives the same probability distribution as a standard Stern-Gerlach would give, if performed in the rest frame.\n\nIn order to devise such measurement we note that, in the laboratory frame, it is possible to define the observables corresponding to the spin operators in the rest frame by transforming the spin, as defined in the rest frame, with a QRF transformation\n\\begin{equation}\n\t\\hat{\\Xi}_i = \\hat{S}_L ( \\hat{\\sigma}_i \\otimes \\mathbb{1}_C) \\hat{S}_L^\\dagger, \\qquad i=x,y,z.\n\\end{equation}\nIn terms of the momenta and of the manifestly covariant Pauli-Luba\\'{n}ski operator $\\hat{\\Sigma}_{\\hat{p}_A} = (\\hat{\\Sigma}^0_{\\hat{p}_A}, \\vec{\\hat{\\Sigma}}_{\\hat{p}_A})$, the operators $\\hat{\\Xi}_i$ are expressed as (see Supplemental Information) $\\vec{\\hat{\\Xi}} = \\vec{\\hat{\\Sigma}}_{\\hat{p}_A} - \\frac{\\hat{\\gamma}_A}{\\hat{\\gamma}_A +1}\\left(\\vec{\\hat{\\Sigma}}_{\\hat{p}_A} \\cdot \\vec{\\hat{\\beta}}_A \\right) \\vec{\\hat{\\beta}}_A$, where $\\hat{\\gamma}_A = \\sqrt{1 + \\frac{\\hat{p}_A^2}{m_A^2 c^2}}$ and $\\vec{\\hat{\\beta}}_A = \\left( \\hat{\\beta}_A^x, \\hat{\\beta}_A^y, \\hat{\\beta}_A^z \\right)$, where each component is $\\hat{\\beta}_A^i = \\frac{\\hat{p}_A^i}{\\sqrt{m_A^2 c^2 + \\vec{\\hat{p}}_A^2}}$ with $i= x, y, z$. The operators $\\hat{\\Xi}_i$ are equivalent to the Foldy-Wouthuysen \\cite{foldywouthuysen} or Pryce spin operator \\cite{pryce_spin}. By definition, these operators satisfy the $\\mathfrak{su}(2)$ algebra $\\left[ \\hat{\\Xi}_i, \\hat{\\Xi}_j \\right] = i \\epsilon_{ijk} \\hat{\\Xi}_k$, and have the same eigenvalues as the Pauli operators $\\hat{\\sigma}_i$, $i=x,y,z$. This last property can be easily checked by choosing an eigenvector $| \\lambda_i \\rangle$ of the operator $\\hat{\\sigma}_i$ in the rest frame A, such that $\\hat{\\sigma}_i | \\lambda_i \\rangle = \\lambda_i | \\lambda_i \\rangle$ and by noting that $\\hat{\\Xi}_i \\hat{S}_L | \\lambda_i \\rangle_{\\tilde{A}}| \\psi \\rangle_{C} = \\lambda_i \\hat{S}_L | \\lambda_i \\rangle_{\\tilde{A}}| \\psi \\rangle_{C}$. Hence, it is possible to partition the total Hilbert space $\\mathcal{H}_A \\otimes \\mathcal{H}_{\\tilde{A}}$ into two equivalence classes, defined as\n\\begin{subequations}\n\t\\begin{equation}\n\t\t\\mathcal{H}_0 = \\left\\lbrace | \\Psi \\rangle_{A\\tilde{A}} \\in \\mathcal{H}_A \\otimes \\mathcal{H}_{\\tilde{A}}\\,\\, \\text{s.t.}\\,\\, | \\Psi \\rangle_{A\\tilde{A}} \\sim \\hat{S}_L | 0 \\rangle_{\\tilde{A}}| \\psi \\rangle_{C}, \\forall \\, | \\psi \\rangle_{C} \\in \\mathcal{H}_C \\right\\rbrace,\n\t\\end{equation}\n\t\\begin{equation}\n\t\t\\mathcal{H}_1 = \\left\\lbrace | \\Phi \\rangle_{A\\tilde{A}} \\in \\mathcal{H}_A \\otimes \\mathcal{H}_{\\tilde{A}}\\,\\, \\text{s.t.}\\,\\, | \\Phi \\rangle_{A\\tilde{A}} \\sim \\hat{S}_L | 1 \\rangle_{\\tilde{A}}| \\phi \\rangle_{C}, \\forall \\, | \\phi \\rangle_{C} \\in \\mathcal{H}_C \\right\\rbrace,\n\t\\end{equation}\n\\end{subequations}\nwhere $| 0 \\rangle_{\\tilde{A}}$ and $| 1 \\rangle_{\\tilde{A}}$ are the eigenvectors of $\\hat{\\sigma}_z$\\footnote{Notice that we could have chosen any other Pauli operator to define this partition.} and two states are said to be equivalent, i.e., $|\\Psi \\rangle_{A\\tilde{A}} \\sim \\hat{S}_L | i \\rangle_{\\tilde{A}}| \\psi \\rangle_{C}$, with $i=0,1$, if they are both eigenvectors of the $\\hat{\\Xi}_z$ operator with the same eigenvalue. We can then build a partition of the Hilbert space into two highly degenerate subspaces, one corresponding to the ``spin up'' and the other to the ``spin down'' eigenvalue, and on which it is possible to define a set of operators satisfying the $\\mathfrak{su}(2)$ algebra, which can be used to encode or decode information of a single qubit.\n\n\\begin{figure}\n\t\\begin{center}\n\t\t\\includegraphics[scale=0.3]{SternGerlachnew}\n\t\t\\caption{The relativistic Stern-Gerlach experiment as seen from the QRF A (above) and from the QRF C (below). In the rest frame of particle A, the spin is operationally defined via the Stern-Gerlach experiment. To measure spin along direction $\\vec{n}$ the spin (Pauli operator) $\\vec{\\sigma}$ is coupled to an inhomogeneous magnetic field oriented along $\\vec{n}$. The particle is then deflected towards the direction $\\vec{n}$ and $-\\vec{n}$ corresponding to outcome ``spin up'' and ``spin down'' respectively. When transforming to the laboratory frame C, the magnetic field and the spin transform with a superposition of Lorentz boosts for $v_1$ and $v_2$. The interaction Hamiltonian is also transformed, giving rise to a coupling between the transformed vector $\\vec{\\mathcal{S}}_\\Lambda(\\vec{B}^{(A)}) = \\hat{\\gamma}_A \\left[ \\vec{B}^{(C)} - \\frac{\\hat{\\gamma}_A}{\\hat{\\gamma}_A+1}\\left(\\vec{\\hat{\\beta}}_A \\cdot \\vec{B}^{(C)}\\right)\\vec{\\hat{\\beta}}_A + \\left(\\vec{\\hat{\\beta}}_A \\times \\vec{E}^{(C)}\\right) \\right]$ aligned in the same direction $\\vec{n}$ as the magnetic field in the rest frame, and the transformed spin operator $\\vec{\\Xi}$. The particle is again deflected either to $\\vec{n}$ or $-\\vec{n}$ corresponding to the outcome ``spin up'' and ``spin down'' respectively. The probability of detecting the outcomes ``spin up'' and ``spin down'' is preserved under change of QRF.}\n\t\t\\label{fig:SternGerlach}\n\t\\end{center}\n\\end{figure}\n\nThe operators $\\vec{\\hat{\\Xi}}$ in general act on both the external and the internal degrees of freedom of the particle. Operationally, they can be defined via a ``relativistic Stern-Gerlach experiment,'' illustrated in Fig.~\\ref{fig:SternGerlach}. Traditionally, in a Stern-Gerlach experiment, the spin measurement is performed by applying a magnetic field, which interacts with the spin as $\\vec{B} \\cdot \\vec{\\sigma}$ and is inhomogeneous along the direction of its orientation, i.e., $\\vec{B} = B(\\vec{r}\\cdot \\vec{n})\\vec{n}$, where $\\vec{n}$ gives the direction and $\\vec{r} = (x, y, z)$. If the magnetic field is aligned precisely in the direction in which the spin state is prepared, the outcome is obtained with certainty. However, if the particle carrying the spin is moving in a superposition of relativistic velocities, no measurement of the spin alone in the laboratory frame will return the result with probability one in general. To treat such a case we set up a hypothetical Stern-Gerlach experiment in the rest frame of the particle, where the interaction Hamiltonian is $H_{int}^{(A)} = \\mu \\vec{B}^{(A)} \\cdot \\vec{\\sigma}$ and $\\mu$ is a coupling constant. We assume that the direction in which the magnetic field is aligned $\\vec{n}$ is orthogonal to the direction of the boost $x$. Formally, this geometric configuration requires to enlarge the Hilbert space of the laboratory to the $z$ direction, which we identify with the direction $\\vec{n}$ of deflection, and modify our previous definition of the state in Eq.~\\eqref{eq:stateA} as $|\\psi \\rangle_C = |\\psi_x \\rangle_C |\\psi_z \\rangle_C$, where $|\\psi_x \\rangle_C$ transforms with $\\hat{S}_L$ and $|\\psi_z \\rangle_C$ is left invariant by the transformation $\\hat{S}_L$, except for the fact that the label is changed from C to A, i.e., $\\hat{S}_L|\\psi_z \\rangle_C = |\\psi_z \\rangle_A$. Additionally, we assume that the motion in the $z$ direction is non relativistic. We then transform the Hamiltonian to the laboratory frame via the QRF transformation $\\hat{S}_L$. Knowing that the magnetic field transforms under superposition of Lorentz boosts as $\\vec{\\hat{\\mathcal{S}}}_\\Lambda(\\vec{B}^{(A)}) = \\hat{\\gamma}_A \\left[ \\vec{B}^{(C)} - \\frac{\\hat{\\gamma}_A}{\\hat{\\gamma}_A+1}\\left(\\vec{\\hat{\\beta}}_A \\cdot \\vec{B}^{(C)}\\right)\\vec{\\hat{\\beta}}_A + \\left(\\vec{\\hat{\\beta}}_A \\times \\vec{E}^{(C)}\\right) \\right]$, we find that the interaction Hamiltonian $H_{int}^{(A)}$ is transformed to\n\\begin{equation} \\label{eq:SternGerlachHamiltonian}\n\n\tH_{int}^{(C)} = \\mu \\hat{\\gamma}_A^{-1} \\vec{\\hat{\\mathcal{S}}}_\\Lambda(\\vec{B}^{(A)}) \\cdot \\vec{\\hat{\\Xi}}.\n\\end{equation}\n\nIt is straightforward to check that the direction of $\\vec{\\hat{\\mathcal{S}}}_\\Lambda (\\vec{B}^{(A)})$ is also $\\vec{n}$, therefore the deflection of the particle in the laboratory frame happens in the same direction as in the rest frame. Notice that, since both the quantum state and the observables transform unitarily, probabilities are automatically conserved after the change of QRF. In particular, if in the rest frame of the particle A the Stern Gerlach measurement detects that the spin is ``up'' with probability one, the ``relativistic Stern-Gerlach'' experiment in the laboratory frame with the interaction Hamiltonian of Eq.~\\eqref{eq:SternGerlachHamiltonian} will also detect ``spin up'' with probability one. Note that the specific form of the electromagnetic field in Eq.~\\eqref{eq:SternGerlachHamiltonian} is not crucial to our result, but we can design the coupling between the particle and the electromagnetic field according to our experimental capabilities in each reference frame. However, it is crucial that the electromagnetic field couples to the operator $\\vec{\\hat{\\Xi}}$, unlike in the standard Stern-Gerlach experiment. In Supplemental Information, we set up a different experiment, where we couple an ihnomogenous magnetic field in the laboratory frame to give an explicit analysis of a relativistic Stern-Gerlach experiment.\n\nIt is worth noting that the interaction Hamiltonian of Eq.~\\eqref{eq:SternGerlachHamiltonian} is covariant, because the quantity $H^{0} : =\\hat{\\gamma}_A H_{int}^{(C)}$ transforms like the zero-component of a $4$-vector. Therefore, the Schr\\\"{o}dinger equation in the reference frame of A, $i \\hbar \\frac{d}{dt_A} \\left| \\psi \\right>_{\\tilde{A}C}^{(A)} = H_{int}^{(A)}\\left| \\psi \\right>_{\\tilde{A}C}^{(A)}$, where $t_A$ is the proper time in the rest frame of A, is mapped to $i \\hbar \\frac{d}{dt_C} \\left| \\psi \\right>_{\\tilde{A}A}^{(C)} = H_{int}^{(C)}\\left| \\psi \\right>_{\\tilde{A}A}^{(C)}$, where $t_C$ is the proper time in the rest frame of C and the relation $t_C = \\hat{\\gamma}_A t_A$ holds. The general, manifestly covariant expression of $H^0$ is\n\\begin{equation}\n\tH^{0} = \\frac{1}{2}\\eta^{0 \\rho}\\epsilon_{\\rho \\mu \\nu \\lambda} \\hat{\\Sigma}_{p_A}^{\\mu}F^{\\nu \\lambda},\n\\end{equation} \nwhere $\\eta^{\\mu \\nu} = \\text{diag}(1, -1, -1, -1)$ is the Minkowski metric, $F^{\\nu \\lambda}$ is the electromagnetic tensor and $\\epsilon_{\\rho \\mu \\nu \\lambda}$ is the totally antisymmetric tensor such that $\\epsilon_{0123}= 1$. \n\nIn order to complete the measurement, we now have to project the position of the particle along the $z$ direction. Formally, this is achieved by defining the two operators $\\hat{\\Pi}_{+}^{(A)} = \\int^{+\\infty}_{0} dz_c | z_C \\rangle_C \\langle z_C |$ and $\\hat{\\Pi}_{-}^{(A)} = \\int_{-\\infty}^{0} dz_c | z_C \\rangle_C \\langle z_C |$, distinguishing whether the particle is respectively deflected upwards or downwards. For a thorough analysis of a concrete detection of spin via the ``relativistic Stern-Gerlach'' proposed here and more details on the measurement, see Supplemental Information.\n\nThe QRF transformation provides the description of the same experiment from the point of view of two different QRFs, which move in a superposition of velocities relative to each other. This treatment of the relativistic Stern-Gerlach experiment makes it possible to associate an operational meaning to the spin of a relativistic quantum particle, thus solving the problem of encoding quantum information in a particle with spin degrees of freedom as in a qubit.\n\n\\section{Conclusions}\n\nIn this paper, we have provided an operational description of the spin of a special-relativistic quantum particle. Such operational description is hard to obtain with standard methods due to the combined effect of special relativity, which makes the spin and momentum not separable, and quantum mechanics, which makes it impossible to jump to the rest frame with a standard reference frame transformation. We have introduced the `superposition of Lorentz boosts' transformation to the rest frame of a quantum particle, moving in a superposition of relativistic velocities from the point of view of the laboratory. We have found how the state transforms under such quantum reference frame transformation and identified a set of observables in the laboratory frame which satisfies the $\\mathfrak{su}(2)$ algebra and has the same eigenvalues as the spin in its rest frame. In addition, this set complies with the desiderata for a relativistic spin operator in Ref. \\cite{bauke_relativisticspin}: it commutes with the free Dirac Hamiltonian, it satisfies the $\\mathfrak{su}(2)$ algebra, and it has the same eigenvalues as the spin in its rest frame. In addition, it has the correct nonrelativistic limit. It can be easily shown, in fact, that our operator $\\vec{\\hat{\\Xi}}$ coincides with the Foldy-Wouthuysen spin operator \\cite{foldywouthuysen, caban_spinobservable}. Thanks to the unitarity of the transformation, probabilities are the same in the rest frame and in the laboratory frame. Finally, we have generalised the Stern-Gerlach to the special-relativistic regime by means of a transformation of the interaction Hamiltonian from the rest frame to the laboratory frame. Such generalisation opens up the possibility of performing quantum information protocols with spin in the special-relativistic regime. \n\n\\acknowledgements{We would like to thank Carlos Pineda for helpful discussions. We acknowledge support from the research platform \"Testing Quantum and Gravity Interface with Single Photons\" (TURIS), the Austrian Science Fund (FWF) through the project I-2526-N27 and I-2906, the \\\"{O}AW Innovationsfonds-Projekt ``Quantum Regime of Gravitational Source Masses'', and the doctoral program \"Complex Quantum Systems\" (CoQuS) under Project W1210-N25. We also acknowledge financial support from the EU Collaborative Project TEQ (Grant Agreement 766900). This work was funded by a grant from the Foundational Questions Institute (FQXi) Fund. This publication was made possible through the support of a grant from the John Templeton Foundation (Project 60609). The opinions expressed in this publication are those of the authors and do not necessarily reflect the views of the John Templeton Foundation.}\n\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section{Introduction}\nThe computational analysis of mIF histopathological images, by enabling simultaneous and quantitative analysis of multiple markers, is becoming key for the development of novel therapeutic drugs. Because most deep-learning based segmentation and detection methods require large datasets of pixel-precise annotation to yield accurate and robust results, methods to transfer annotations or models from one domain to another are of increasing interest \\cite{brieu2019}. However, the standard CycleGAN approach \\cite{zhu2017} for the underlying unsupervised and unpaired domain translation have limitations in preservation of pathology structures, which led to the introduction of segmentation-based guidance \\cite{mahapatra2020}. We propose in this study a novel and simple guidance scheme tailored to the translation from and to the IF domain, based on the conversion from the RGB to Haematoxylin-DAB (HD) colorspace \\cite{ruifrok2001}. As shown qualitatively and quantitatively, our approach yields improved translation results.\n\n\\vspace{-3mm}\n\\section{Methods}\nWe incorporate two losses $\\mathcal{L}^{IF}$ and $\\mathcal{L}^{IHC}$ to the CycleGAN training (cf. Fig.~1). These respectively guide the translation from the IHC domain to the IF domain by generator $G_{AB}$ and the translation from the IF domain to the IHC domain by generator $G_{BA}$. \nWith $x_{IHC}$ an IHC patch input to $\\mathcal{G}_{AB}$ and $x_{IF}$ an IF patch input to $\\mathcal{G}_{BA}$, we define the stain-isolation guidance loss $\\mathcal{L}_g^{IF}$ as the $\\mathcal{L}_1$ norm between the DAB channel of the generated IF image $\\mathcal{G}_{AB}(x_{IHC})$ and the DAB channel of the pseudo-IF image $\\mathcal{F}(x_{IHC})$. The later is derived from the conversion of $x_{IHC}$ to the Haematoxylin-DAB (HD) colorspace followed by its rescaling\\footnote{https:\/\/scikit-image.org\/docs\/stable\/auto\\_examples\/color\\_exposure\/plot\\_ihc\\_color\\_separation.html} using a set of representative IHC images as reference. The inaccuracy of the estimated Haematoxylin channel leads us to consider only the DAB channel into the loss:\n\\vspace{-1mm}\n\\begin{equation}\n\\mathcal{L}_g^{IF} = \\left\\|\\mathcal{G}_{AB}(x_{IHC}).{DAB} - \\mathcal{F}(x_{IHC}).{DAB}\\right\\|_1\n\\end{equation}\n\\vspace{-1mm}\nWe similarly define the pseudo-IHC loss $\\mathcal{L}_g^{IHC}$ as the $\\mathcal{L}_1$ norm between the generated IHC image $\\mathcal{G}_{BA}(x_{IF})$ and the image $\\mathcal{F}^{-1}(x_{IF})$ obtained by color conversion from HD to RGB:\n\\vspace{-1mm}\n\\begin{equation}\n\\mathcal{L}_g^{IHC} = \\left\\|\\mathcal{G}_{BA}(x_{IF}) - \\mathcal{F}^{-1}(x_{IF})\\right\\|_1\n\\end{equation}\n\\vspace{-1mm}\nThe stain isolation and pseudo-IHC losses are added to the following CycleGAN losses: a cycle consistency loss $\\mathcal{L}_{cycle}$, two adversarial losses on the output of the generators $\\mathcal{G}_{AB}$ and $\\mathcal{G}_{BA}$, an identity loss $\\mathcal{L}_{id}$ as well as an embedding loss $\\mathcal{L}_{emb}$. The overall loss reads as:\n\\vspace{-1mm}\n\\begin{equation}\n\\mathcal{L} = \\mathcal{L}_{GAN}^{AB} + \\mathcal{L}_{GAN}^{BA} + \\lambda_1 \\mathcal{L}_{cycle} + \\lambda_2 \\mathcal{L}_{id} + \\lambda_3 \\mathcal{L}_{emb} + \\lambda_4 \\mathcal{L}_g^{IHC} + \\lambda_5 \\mathcal{L}_g^{IF}\n\\label{eq:loss}\n\\end{equation}\n\\vspace{-1mm}\nwith the following fixed weighting parameters: $\\lambda_1=10$, $\\lambda_2=2$ and $\\lambda_3=10$. In the remaining of this study, the two parameters $\\lambda_4$ and $\\lambda_5$ are either switched off or set to 10.\n\n\\begin{figure}[t!]\n\t\\centering\n\t\\includegraphics[width=0.95\\linewidth]{methods.eps}\n\t\\caption{$\\mathcal{L}_g^{IF}$ and $\\mathcal{L}_g^{IHC}$ guide the training of $\\mathcal{G}_{AB}$ and $\\mathcal{G}_{BA}$. Note the improved DAPI channel of $\\mathcal{G}_{AB}(x_{IHC})$ vs. $\\mathcal{F}(x_{IHC})$ and the more realistic color of $\\mathcal{G}_{BA}(x_{IF})$ vs. ${F}^{-1}(x_{IF})$.\\vspace{-2mm}}\n\t\\label{fig:methods}\n\\end{figure}\n\n\\vspace{-2mm}\n\\section{Results}\nThe unpaired IF and IHC datasets respectively contains 11K and 160K patches of $256\\times256$px with $0.5\\mu$m\/px resolution. On the IF images, the PDL1 and DAPI channels are selected out of the seven initially available channels. The batch size is 1, the learning rates 0.0001 for the generators and 0.0005 for the discriminators. Adam optimizer ($\\beta_1=0.5$, $\\beta_2=0.999$) is used to minimize the loss (cf. Eq.~\\ref{eq:loss}) for 100K iterations. We qualitatively study the following configurations: (a) no guidance ($\\lambda_4=0$, $\\lambda_5=0$), (b) a virtual-IHC guidance ($\\lambda_4=10$, $\\lambda_5=0$), (c) a stain-isolation guidance ($\\lambda_4=0$, $\\lambda_5=10$) and (d) a combine guidance ($\\lambda_4=10$, $\\lambda_5=10$).\nFig.~2A shows the difficulty of baseline (a) to translate highly saturated regions: dark-brown pixels in IHC are wrongly translated into high DAPI signal in IF while some high DAPI signal in IF is wrongly translated into brown pixels in IHC. While these are partially corrected by the virtual-IHC loss (b), the stain-isolation loss (c) yields better translation. Combining both losses (d) does not improve over (c). This is quantitatively confirmed in Fig.~2B: proposed guidance schemes (b)-(d) prevent the generation of high DAPI signal in membrane regions, otherwise observed in (a).\n\n\\vspace{-2mm}\n\\section{Discussion}\nWe propose two novel loss functions based on color conversion in order to provide guidance at training time for the CycleGAN-based translation between the IF and IHC stain domains. Doing so, we are able to prevent cycle-consistent but non structure preserving translation errors. Future work include the inclusion of the proposed guidance losses to the second translation steps of each cycle as well as the downstream task of transferring segmentation and detection models from the IHC domain to the IF domain.\n\n\\begin{figure}[t!]\n\t\\centering\n\t\\includegraphics[width=0.99\\linewidth]{results2.eps}\n\t\\caption{A - Examples of $\\mathcal{G}_{AB}(x_{IHC})$ and $\\mathcal{G}_{BA}(x_{IF})$ generated using: (a) the baseline CycleGAN, (b) the virtual-IHC guidance, (c) the stain-isolation guidance, and (d) combined guidance; B - Respective histograms of the DAPI values of $\\mathcal{G}_{AB}(x_{IHC})$ measured on saturated ($S>0.90$) pixels of annotated membrane regions of 222 unseen IHC images $x_{IHC}$. \\vspace{-2mm}}\n\\end{figure}\n\n\\vspace{-3mm}\n\\midlacknowledgments{\\vspace{-1mm}We thank the mIF and IHC-HER2 teams of AstraZeneca Translation Medicine - Oncology R\\&D, for their support in generating the image datasets used in this study.}\n\n\\vspace{-2mm}\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}}