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2025-03-21T14:48:31.744529
| 2020-08-11T16:13:05 |
368885
|
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|
Stack Exchange
|
For which number fields we know the nonexistence of Stark zeros?
Let $L$ be a number field and let $\zeta_L(s)$ be its associated Dedekind zeta function. It is known that $\zeta_L(s)$ has at most one zero in the region
$$1 - \frac1{4 \log d_L} \leq \sigma \leq 1, \qquad |t| \leq \frac1{4\log d_L},$$
(as usual, $s=\sigma + it$) where $d_L$ is the discriminant of $L / \mathbb{Q}$. If such a zero exists, then it is called Stark zero of $L$, and it is known that it must be real and simple.
The Stark zero of $L$ appears in the error term of the unconditional form of Chebotarev theorem given by Lagarias and Odlyzko [1]. In particular, if $L$ has no Stark zero, then the error term is smaller.
My question is: For which number field $L$ we know that the Stark zero does not exists? I mean both general theorems and concrete examples (e.g., is there a table of quadratic fields with no stark zero?)
[1] J. Lagarias and A. Odlyzko. Effective versions of the Chebotarev density theorem. Algebraic Number Fields (A. Frohlich, editor), Academic Press, New York, 1977, pp. 409-464
I think what you are calling a "Stark zero" is more commonly known as a Siegel zero, in which case it is known (and you essentially indicated as much in your post) that they can only occur for the $L$-function of a real quadratic character. Any number field whose Galois closure does not contain a real quadratic field will not have to worry about a Siegel zero
@StanleyYaoXiao Aren't Siegel zeros defined for Dirichlet L-functions and not Dedekind zeta functions? I think that's the distinction. Anyway, I read that if L is an abelian extension of the rationals then its Dedekind zeta function is the product of L-functions, so you comment might be useful, although I guess most of such L do contain a real quadratic field
In Real zeros of real odd Dirichlet $L$-functions, Mark Watkins showed in 2003 that $L(s,\chi_d)$ (as in the title) has no positive real zeros for $d<300,000,000$. (I think this is still state of the art.). This answers the questions for the corresponding complex quadratic fields $\mathbb Q(\sqrt{-d})$.
|
2025-03-21T14:48:31.744741
| 2020-08-11T16:26:58 |
368886
|
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|
Stack Exchange
|
Weyl algebra as an Azumaya algebra over its centre
Assume that $k$ is an algebraically closed field of positive characteristic $p$. On page 3 (page 6 of the PDF file) of Bezrukavnikov, Mirković, and Rumynin - Localisation of Modules for a semisimple Lie Algebra in prime characteristic, we have the following sentence:
The sheaf $D_X$ of crystalline differential operators on a smooth variety $X$ over $k$ has a non-trivial center, canonically identified with the sheaf of functions on the Frobenius twist $T^∗X^{(1)}$ of the cotangent bundle. Moreover $D_X$ is an Azumaya algebra over $T^∗X^{(1)}$.
Instead of going through the general proof given, I only want to understand, in as simple a manner as possible, the situation when $X$ is the affine $n$-space over $k$. In this case, $D_X$ is simply the Weyl algebra and the Azumaya property, if I understand correctly, means that the quotient of the Weyl algebra by its centre is isomorphic to some matrix algebra over $k$. Is there a way to construct such a matrix algebra and a corresponding isomorphism to the quotient explicitly? Any help, even in the case of the affine line, would be highly appreciated.
P.S. If my understanding is incorrect, could you please point out the flaw(s) and how the question could be turned into something reasonable?
Major Edit It has been pointed out that my understanding of the Azumaya property is incorrect. But my question remains the same: is there a direct way to prove the claim of the paper in the case when $X$ is the affine $n$-space (or even the affine line) over $k$.
Your definition of Azumaya algebra is incorrect — see Azumaya algebra.
Thank you very much for the quick reply. Even if that is the case, will the quotient of the Weyl algebra by its centre be isomorphic to a matrix algebra?
The trivial Azumaya algebra (say, over a field) is a matrix algebra, so the quotient by its center is not a matrix algebra.
@abx If I consider $n=1$ and consider $k[x]/x^p$ and let $x$ and $\partial$ define the standard actions on $k[x]/x^p$. Then since the centre of the Weyl algebra is just $k[x^p, \partial^p]$, each element of the quotient will define an endomorphism of $k[x]/x^p$. Will/can this be an isomorphism?
No. Even briefer version of my previous argument: $k[x, \partial]/k[x^p, \partial^p]$, or even just $k[x, \partial]/k$, is Abelian, so can't be $\operatorname{End}_k(k[x]/(x^p))$.
@LSpice Thank you! Is there any simple way to prove the statement from the paper for the Weyl algebra?
@abx Do you have any ideas for an elementary proof of the statement in the case of the Weyl algebra?
@LSpice What if I quotient it by the ideal generated by $x^p$ and $\partial^p$? My hunch is that this will become a matrix algebra. Do you think this makes sense? If so, do you see any direct way of proving it?
Ah, sorry. I do not know an elementary proof off the top of my head. My initial comment, which I have now deleted, did not make much sense; it was basically about taking the quotient by the ideal, but I confused myself partway. Anyway, if the quotient were a matrix algebra, it would have to be $\mathfrak{gl}_p$, with $X$ and $\partial$ as regular nilpotents with commutator $1$. I think that this can be realised by $X$ the matrix with $1$s on the superdiagonal, $\partial$ the matrix with subdiagonal $1, 2, 3, \dotsc, p - 1$.
(More conceptually—and maybe that's what you wanted from your comment?—this is the realisation of $k[x, \partial]/(x^p, \partial^p)$ as $\operatorname{End}_k(k[x]/(x^p))$ via the natural action.)
@LSpice I am very sorry, but could you please elaborate a bit more on that? I am having difficulty understanding exactly what you mean.
Hmm, I thought links to chat were added automatically, but it doesn't seem to be happening here. We moved the discussion to a chat room: https://chat.stackexchange.com/rooms/info/111685/discussion-between-user11235813-and-lspice .
That was my fault! Very, very sorry! I initially opted to move it to the chat and then deleted the (automatic) comment because I thought that having a discussion here was better, only to change my mind again!
But my question remains the same: is there a direct way to prove the
claim of the paper in the case when is the affine -space (or even
the affine line) over .
Yes, read the proof of Proposition 1 in "The Jacobian Conjecture is stably equivalent to the Dixmier Conjecture" by Alexei Belov-Kanel and Maxim Kontsevich.
If you are only interested in the $n=1$ case $A=k \langle X,Y \mathrel| XY-YX=1 \rangle$, then give $X$ degree $1$ and $Y$ degree $-1$, so that $A$ becomes a $\mathbb{Z}$-graded ring. It is even strongly graded, meaning $A_1\cdot A_{-1} = A_0$, so all graded info is determined by the zero-part $A_0=k[XY]$. In particular, if you divide out a graded maximal ideal you get a strongly graded ring with part of degree $0$ a field and finite over its center, and these must be central simple algebras (in this case, just $p \times p$ matrices) proving that $A$ is what is called a graded Azumaya algebra. Now, for the fun part, as there exist homogeneous central identities, a graded Azumaya algebra is a genuine Azumaya algebra. Done.
Apologies for the late reply. Thank you so much for the reference and the explanation!
|
2025-03-21T14:48:31.745109
| 2020-08-11T17:23:17 |
368889
|
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|
Stack Exchange
|
Why is density and separability needed for uniqueness of weak (time) derivatives?
Let $X,Y$ be Banach spaces with $X \subset Y$. Recall that $u \in L^1(0,T;X)$ has weak derivative $g \in L^1(0,T;Y)$ if
$$\int_0^T u(t)\phi'(t) = -\int_0^T g(t)\phi(t) \qquad\forall \phi \in C_c^\infty(0,T).$$
Suppose that $u$ also has a weak derivative $h \in L^1(0,T;Z)$ where $Y \subset Z$.
In Boyer and Fabrie's book on Navier-Stokes, page 95, he states that if $Y \subset Z$ is dense and $Z'$ is separable, then $g=h$. My question, why is the density and separability needed? Isn't the argument this simple:
Since $g$ and $h$ are weak derivatives of $u$, we have
$$\int_0^T (g(t)-h(t))\phi(t) = 0\qquad\forall \phi \in C_c^\infty(0,T)$$
and by the fundamental lemma of the calculus of variations, it follows that $g(t) = h(t)$ in $Z$ for almost every $t$.
Isn't this enough? What do I miss? Does anyone know another source for this uniqueness claim where the derivatives lie in different spaces?
The point for separability of $Z'$ might be the reduction to the scalar case by applying continuous linear functionals to the equation $\int_0^T(g(t)-h(t))\phi(t)dt$. Then the exceptional null sets would depend on the functional and you only can take the union of countably many.
@JochenWengenroth Do you mean that he assumes separability for simplicity, because the separable case of the fundamental lemma of calculus of variations is well known? Even though separability is not needed.
$\def\bbR{\mathbb R}\def\inc{\subseteq}$The requirements on density or separability are superfluous because of the following
Lemma. Let $J$ be a real open interval, and let $E$ be any real or complex Banach space. Let the function $f$ in $L^1(J,E)$ be such that $\int_J(\varphi\,f)=0_E$ holds for all compactly supported smooth $\varphi:J\to\bbR$. Then $f(t)=0_E$ holds for almost all $t\in J$.
Proof. Let $x\mapsto\|x\|$ be a norm for $E$. Having $f$ a.e. the limit of a sequence of simple functions, there are a Lebesgue null set
$N_1\inc J$ and a separable closed linear subspace $S$ in $E$ such that $f(t)\in S$ holds for all $t\in J\setminus N_1$. By Lemma 8.15.1 (p. 573) in R. E. Edward's Functional Analysis there is a countable set $D$ in the unit ball of the dual of $E$ such that
$\|x\|=\sup\{|u(x)|:u\in D\}$ holds for all $x\in S$. By classical results, it follows existence of a Lebesgue null set $N_0\inc J$ such that $u\circ f(t)=0$ holds for all $t\in J\setminus N_0$ and $u\in D$. It follows that $\|f(t)\|=0$ holds for all $t\in J\setminus N_0$.
I do not know whether there is a published reference where the above Lemma would be explicitly stated or proved.
It is also proved in Corollary 6.33, page 202 of these notes by Hunter.
Thanks for the reference. Hunter seems to first generalize Lebesgue's differentiation theorem to Banach space valued functions, and then get the result as a corollary to it.
|
2025-03-21T14:48:31.745333
| 2020-08-11T18:06:50 |
368892
|
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|
Stack Exchange
|
A question on moduli space of Hitchin's equations
I am reading Hitchin's Self-Duality paper. In section 5 (page 85), he is trying to prove that $Dim H^1=12(g-1)$. In doing so, he defines an operator $d^*_2+d_1$, where $d^*_2$ and $d_1$ are given by
$d_1\dot{\psi}=(d_{A}\dot{\psi},[\Phi, \dot{\psi}])$
$d_2(\dot{A},\dot{\Phi})=(d_A\dot{A}+[\dot{\Phi},\Phi^*]+[\Phi,\dot{\Phi}^*], d^{\prime\prime}_{A}\dot{\Phi}+[\dot{A}^{0,1},\Phi])$
Then, he claims that $(d^*_2+d_1)(\psi_1,\psi_2)=0$ if and only if
$d^{\prime\prime}_{A}\psi_1
+[\Phi^*,\psi_2]=0$
$d^{\prime}_{A}\psi_2
+[\Phi,\psi_1]=0$
I am not able to derive this fact, and I spent quiet a lot of time on this, but unfortunately was not able to prove it. He says that he obtains this by calculating the explicit form of adjoint of $d_2$. I was not able to perform this calculation. I am new to the subject, and I would really appreciate any help, or ideas on how to prove this. Thanks!
P.S. I know that a co-differential is defined by $ d^{*}=(−1)^{n(k-1)+1}*d*:\Omega^{k}\to \Omega^{k-1}$ where $*$ in the definition is Hodge star, and this is the adjoint of exterior derivative with respect to $L^2$ norm. But how would one apply hodge operator in this setting, or even use $L^2$ to get $d^*$.
The operator
$$ d_2^*+d_1\colon \Omega^0(M,ad P\otimes\mathbb C)\oplus\Omega^0(M,ad P\otimes \mathbb C)\to\Omega^{0,1}(M,ad P\otimes \mathbb C)\oplus\Omega^{1,0}(M,ad P\otimes \mathbb C)$$
is just the sum of $d_1$ and $d_2^*,$ and it suffices to describe these two operators using the identification $$\Omega^{0,1}(M,ad P\otimes\mathbb C)=\Omega^1(M,ad P).$$
We decompose $\psi_1\in\Omega^0(M,ad P\otimes\mathbb C)$ into real and imaginary parts $$\psi_1=\omega+i \eta$$ for $\omega,\eta\in \Omega^0(M,ad P),$ and let $d_1$ act on the real part ($\omega$) and $d_2^*$ on the imaginary part ($\eta$).
Then, we have $$d_1(\psi_1,\psi_2)=((d_A\omega)^{0,1},[\Phi,\omega])=(d_A''\omega,[\Phi,\omega])$$
by definition. The dual of the operator $$d_A''\colon\Omega^{1,0}(M,ad P\otimes \mathbb C)\to \Omega^2(M,ad P\otimes \mathbb C); \; \psi_2\mapsto d_A''\psi_2$$ is by Serre duality the operator $$d_A''\colon \Omega^0(M,ad P\otimes \mathbb C)\to \Omega^{0,1}(M,ad P\otimes \mathbb C);\; \psi_2\mapsto d_A''\psi_2,$$ and using the hermitian metric and the fact that $d_A$ is unitary the adjoint operator gets identified with
$$d_A'\colon \Omega^0(M,ad P\otimes \mathbb C)\to \Omega^{1,0}(M,ad P\otimes \mathbb C);\; \psi_2\mapsto d_A'\psi_2.$$
The adjoint of the operator
$$\phi\in\Omega^2(M,ad P)\mapsto ([\phi,\Phi^*]+[\Phi,\phi^*])^{0,1}\in\Omega^{0,1}(M,ad P)$$ becomes
$$\psi_2\mapsto -[\psi_2,\Phi^*]$$ ( taking the $i-$ part into account).
It remains to describe the operator $d_2^*$ acting on the imaginary part $i\eta$ of $\psi_1.$ As above, this becomes $$d_2^*((i\eta))=(id_A''\eta,i[\Phi,\eta]).$$ Putting together the pieces proves the claim.
Hi Sebastian, thanks a lot for your answer! I just have a few questions regarding your answer. From what you wrote I think the adjoint of $d^{''}_A$ is $d^{'}_A$ and adjoint of $d^{'}A$ is $d^{''}A$ ? If so, the explicit form of action of $d_2$ on $\psi_2$ is $d{2}\psi_2=(-[\psi_2,\Phi^*],d^{'}{A}\psi_2)$.
Using this and your equations for actions of $d_1$ on real part of $\psi_1$ and action of $d^*_2$ on imaginary part of $\psi_1$ I got equations 5.1. However, I'd like to know why $d_1$ does not act on $\psi_2$, also is there a reason why we decompose $\psi_1$, but not $\psi_2$.
Sorry above I meant $d^_2\psi_2=(-[\psi_2,\Phi^],d^{'}_{A}\psi_2)$.
$d_1$ does only act on sections of $ad P$ by its definition. In order to obtain a well-behaved complex linear operator Hitchin added $d_1$ and $d_2^*$ in a very specific way so that one can easily conclude that there is no kernel.
Could you also elaborate a bit more on how you obtain the adjoint operator, $d^{'}A$, using unitarity of $d{A}$ and hermitian product. Thanks.
|
2025-03-21T14:48:31.745611
| 2020-08-11T18:18:40 |
368893
|
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|
Stack Exchange
|
natural metrics for proof length
I am trying to make my way into Homotopy Type Theory(HoTT) where a mathematician may view proofs as paths. Intuitively, this leads me to the idea of a metric on the space of mathematical propositions. Has this been developed?
Specifically, is there a way to analyse short proofs as geodesics within the space of
mathematical propositions from the perspective of HoTT? If so, might
this metric be formulated using Kolmogorov Complexity?
Inspired by the informal notion of Cognitive distance, in 2010 Charles Bennett, Peter Gács, Ming Li, Paul Vitanyí and Wojcech Zurech introduced the notion of Information Distance which was used in the seminal paper Clustering by Compression:
\begin{equation}
ID(x,y)=\min\{|p|:p(x)=y \land p(y)=x\} \tag{1}
\end{equation}
with $p$ a finite binary program for the fixed universal computer $U$ which takes the finite binary strings $x,y$ as inputs. They prove that:
\begin{equation}
ID(x,y)=E(x,y) + \mathcal{O}(\log E(x,y)) \tag{2}
\end{equation}
where the key notion:
\begin{equation}
E(x,y) = \max \{K_U(x|y),K_U(y|x) \} \tag{3}
\end{equation}
satisfies the criteria of a metric up to the additive term $\mathcal{O}(\log E(x,y))$.
References:
Bennett, Charles H.; Gács, Péter; Li, Ming; Vitányi, Paul M. B.; Zurek, Wojciech H., Information distance, IEEE Trans. Inf. Theory 44, No. 4, 1407-1423 (1998). ZBL0964.94010.
Cilibrasi, Rudi; Vitányi, Paul M. B., Clustering by compression, IEEE Trans. Inf. Theory 51, No. 4, 1523-1545 (2005). ZBL1297.68097. ArXiv:cs:0312044
You might be interested in a notion of semantic space of logic, which is similar in meaning to cognitive distance: https://archive.org/details/semantic_space_of_logic Disclaimer: If you find any errors or anything unclear, please let me know, so that I can correct it.
The short answer is no.
The notion of length is related to distance and metric, and these are all concepts that topology obliterates, let alone homotopy theory. The popular explanations of homotopy type theory in terms of topological paths (continuous maps $[0,1] \to X$) are meant to help the intuition, but should not be and cannot be understood literally. Topological spaces do not form a model of homotopy type theory.
Since you are trying to make your way into homotopy type theory, you could draw your intuition from several places, depending on your background, in decreasing order of abstraction and correctness:
Higher category theory: an $(\infty,1)$-topos is more or less a model of homotopy type theory, while an $\infty$-groupoid is like a single type.
Homotopy theory: Simplicial sets, and more precisely Kan complexes form a model of homotopy type theory, so you can think of a type as a Kan complex. In general, certain kinds of model categories can be used to interpret (parts of) homotopy type theory.
Groupoids: a groupoid is like a 1-type in homotopy type theory. You can think of $\mathsf{Id}_A(a,b)$ as the collection of all isomorphisms from $a$ to $b$ in a groupoid $A$.
In all cases, the "paths" are of an abstract nature. For instance, the isomorphisms in a groupoid need not be paths, just like morphisms in a category need not be functions.
There are adaptations of homotopy type theory that do relate the abstract homotopy-theoretic models with topological, and even smooth models. These go under the name cohesive homotopy type theory, but I am not sure that's the best starting point for learning homotopy type theory.
Let me attempt a longer answer. (@Andrej Bauer 's answer is mostly about learning homotopy type theory, as opposed to delving deeper into your question.)
Unsurprisingly, the answer is still no. One could try to define the length of a path by the size of the smallest of its witnesses. But that definition relies on syntax, i.e. you need to have a 'language' in which you express your witnesses. The thing is, while HoTT does give rise to programming languages rather naturally (see my work with Amr Sabry), they still are not really 'canonical'. While $\left(\infty,1\right)$-topos certainly point in the right direction, it's not settled yet if we don't in fact want a bit more structure (see the work of Shulman, Riehl, etc if you want to dive in the really deep end of that). So we don't even know what outer structure to work in, never mind the internal language that we'll end up having inside that structure.
Even if all of that settled down, why would you expect the internal language to be Turing-complete? That is one of the cornerstones of the validity of Kolmogorov Complexity. Without it, length becomes a rather whimsical notion. Even with Kolmogorov Complexity, length remains a 'fuzzy' notion, because it's only defined up to a constant. So even if it existed, it would not tell you much about 'short' proofs, it only really tells you something interesting when you have one proof which is significantly shorter than the others. Certainly there is no hope that such a length notion would be a metric, never mind carve out geodesics.
Nevertheless, I do hold out some hope that there will be some notion of 'size' that will turn out to be informative and meaningful. It's just not going to be simple, or as informative as one would like.
“Without it, length becomes a rather whimsical notion.” I don’t think that’s true. Take the problems of minimizing DFAs, NFAs, and regular expressions, which are certainly valid problems that are studied in the literature.
@user76284 That would be because when analysing DFAs, etc, there is a fixed and adequate meta-language in which length is measured. There is a kind of shift that happens as the computing power gets larger, and some things become undecidable. Size becomes an unstable measure.
I think the binary lambda calculus could play that role. Indeed it was invented for that reason (concrete AIT).
@user One problem with using a coding of the lambda calculus is that 'coding' arithmetic is usually very inefficient, even from the point of view program length. Adding binary-coded naturals makes some programs exponentially shorter. So Kolmogorov-short programs will inevitably end up re-inventing such things for brevity.
|
2025-03-21T14:48:31.746050
| 2020-08-11T18:36:37 |
368895
|
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|
Stack Exchange
|
Next-generation matrix of infectious disease
If the population is classified into $\mathbf{S}$, $\mathbf{E}$, $\mathbf{I}$ and $\mathbf{R}$ compartments such that
\begin{equation}
\label{eq4}
\begin{aligned}
\mathbf{S} &=\dfrac{S_{1}N_{1}+S_{2}N_{2}+ \dotsb + S_{k}N_{k}}{N_{1}+N_{2}+\dots +N_{k}} = \dfrac{1}{N} \sum_{n=1}^{k} S_{n} N_{n} \nonumber \\
\mathbf{E} &=\dfrac{E_{1}N_{1}+E_{2}N_{2}+ \dotsb + E_{k}N_{k}}{N_{1}+N_{2}+\dots +N_{k}} = \dfrac{1}{N} \sum_{n=1}^{k} E_{n} N_{n} \nonumber \\
\mathbf{I} &=\dfrac{I_{1}N_{1}+I_{2}N_{2}+ \dotsb + I_{k}N_{k}}{N_{1}+N_{2}+\dotsb +N_{k}} = \dfrac{1}{N} \sum_{n=1}^{k} I_{n} N_{n} \nonumber \\
\mathbf{R} &=\dfrac{R_{1}N_{1}+R_{2}N_{2}+ \dotsb + R_{k}N_{k}}{N_{1}+N_{2}+\dotsb +N_{k}} = \dfrac{1}{N} \sum_{n=1}^{k} R_{n} N_{n} .
\end{aligned}
\end{equation}
The total normalized population of a region (state) or country is given by
\begin{equation} \label{eq5} \mathbf{S}+\mathbf{E}+\mathbf{I}+\mathbf{R}=1
\end{equation}
The dynamics of SEIR epidemic spreading model that describe the transmission of disease in a population without demography in one region is given by the following system of differential equations :
\begin{align} \label{eq6}
\dfrac{d\mathbf{S}}{dt}&= - \dfrac{1}{N} \sum_{n=1}^{k} \left[ \mathbb{S} \mathbb{K}_{\kappa \lambda} \right]_{n} N_{n} \mathbf{I} \nonumber \\
\dfrac{d\mathbf{E}}{dt}&=\dfrac{1}{N} \sum_{n=1}^{k} \left[ \mathbb{S} \mathbb{K}_{\kappa \lambda} \right]_{n} N_{n} \mathbf{I} - \frac{1}{T_{inc.}} \mathbf{E} \nonumber \\
\dfrac{d \mathbf{I}}{dt}&= \frac{1}{T_{inc.}} \mathbf{E} -\frac{1}{T_{inf.}} \mathbf{I} \nonumber \\
\dfrac{d \mathbf{R}}{dt}&= \frac{1}{T_{inf.}} \mathbf{I}
\end{align}
Where $\mathbb{S}=diag(S_{1},\dots,S_{k})$ and $\mathbb{K}_{\kappa \lambda}$ is $(k\times k)$ mobility infectiousness matrix, with $\mathbf{N}=\mathbf{S}+\mathbf{E}+\mathbf{I}+\mathbf{R}=1$ such that $\mathbf{N}=(N_{1},\dots,N_{k})^T$ and initial condition of dynamics
\begin{align}\label{eq7}
\mathbf{S_{0}}&=\dfrac{S^0_{1}N_{1} +S^0_{2}N_{2} + \dots S^0_{k}N_{K} }{N} , ~~~` \mathbf{E_{0}}=\dfrac{E^0_{1}N_{1} +E^0_{2}N_{2} + \dots E^0_{k}N_{K} }{N} \nonumber \\
\mathbf{I_{0}}&=\dfrac{I^0_{1}N_{1} +I^0_{2}N_{2} + \dots I^0_{k}N_{K} }{N}, ~~~~ \mathbf{R_{0}}=\dfrac{R^0_{1}N_{1} +R^0_{2}N_{2} + \dots R^0_{k}N_{K} }{N}
\end{align}
Since the $\mathbf{S}$, $\mathbf{E}$, $\mathbf{I}$ and $\mathbf{R}$ compartments are scalar by above formula,
and the family mobility infectiousness $ \left[ \mathbb{S} \mathbb{K}_{\kappa \lambda} \right]$ is $k\times k$ matrix, I don't know how to determine the next-generation matrix of this dynamics. Please, I need help!
If you're using the next-generation matrix to find $R_0$, it may be possible to find a formula $R_0$ using induction.
What do you mean? Your differential equations make sense only if you view $S$, etc. as $k$-component vectors $(S_1,\dots,S_k)$, etc. (in which case everything starts to fit together but the problem stops being one about $\mathbb R^4$ dynamics and becomes one about $\mathbb R^{4k}$ dynamics). In other words, your averaged values alone at the moment $0$ are totally insufficient to determine the full dynamics in a unique way.
I have edited the question. All variables (S, E, I, R) are positive.
As I said: your initial conditions involving the averages only are insufficient for determining the dynamics, so you have to honestly pose and solve the IVP for a system with $4k$ variables.
|
2025-03-21T14:48:31.746361
| 2020-08-11T19:04:12 |
368896
|
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|
Stack Exchange
|
On a sum of squares representation
We know $p a^2+q b^2+r ab$ can be represented as square (trivially) when $$p,q\geq0$$
$$r^2=|4pq|$$
holds and as a sum of squares (again trivially) of form $(m a+n b)^2$ under readily explainable conditions on $p,q,r$.
Are there other non trivial sum of squares form with higher powers being cancelled off and leaving only $p a^2+q b^2+r ab$ in final sum for other regimes of $p,q,r$?
Notice that for a form to be expressible as a sum of squares it must be nonnegative everywhere. In particular setting $a=0$ or $b=0$ tells you $p,q\geq 0$ and setting $a=\pm \sqrt{q}, b=\sqrt{p}$ tells you $|r|\le 2\sqrt{pq}$. Therefore the only regimes of $p,q,r$ where your form is a sum of squares are the ones you already knew by looking at sums of terms of the form $(am+bn)^2$.
An alternative route to the same conclusion is to realize that the coefficient of the deg-lex largest term in the square of a multivariate polynomial is positive, therefore when you take a sum of squares of higher degree there will always be one term of degree higher than 2 that does not cancel out.
|
2025-03-21T14:48:31.746465
| 2020-08-11T19:08:43 |
368897
|
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|
Stack Exchange
|
Creating a Multi-Variable equation to create a "Most Favorable Supplier Index"
I wanted to create a 'most favorable supplier index', where I want to assign a value to each supplier, ranging from 1-5 or 1-10, whichever, from the most and to the least preferred suppliers. There are couple of different metrics, that need to be taken into consideration to come at this value, which includes, current supplier business per day, since when has the business been serving its customers, supplier customer reviews, supplier cost per product etc.
Wanted to know how I can create an equation to incorporate all these and other factors to generate the "score" for each supplier, also as given that all these will have separate and unique units, how can I incorporate that?
Would be great if I could just get a direction on how to go about it.
Thanks.. :)
This sounds like a business question, not a math question.
|
2025-03-21T14:48:31.746568
| 2020-08-11T19:16:37 |
368898
|
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"Sylvain JULIEN",
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|
Stack Exchange
|
Gödel's ontological proof & Benzmüller's work
For a decade or so, Christoph Benzmüller from Berlin has explored Gödel's ontological proof (and variants) of existence of God. He uses the proof assistant Isabelle/HOL. He recently posted a preprint, which was highlighted by the cover of the French magazine Science et Vie.
Well, I am not familiar with AI, yet even less with applications to metaphysics. But many practitioners of MO must be. I should like to know how serious is Benzmüller's work considered in this community. Is it controversial or is this considered a respectable research activity ?
My recollection is that Goedel's proof is fine, given his axioms, and that there is a single axiom that almost all philosophers do not find convincing.
You say "I recently posted a preprint", but the link is to a paper "A (Simplified) Supreme Being Necessarily Exists, says the Computer: Computationally Explored Variants of Gödel's Ontological Argument" by Benzmüller (not, I guess, you). Is that intentional?
The Ontological argument, in its modern modal versions, hinges on the premise of whether God is possible or not, the rest being in my view mostly distractions including discussions of S5 and what not. As a fairly old-fashioned Thomist, I think Aquinas' criticism is decisive which can be construed as saying that the proof works but only God knows it.
@G.Rodrigues: Dipping our toes into philosophy, one can say that according to Descartes, if God thinks about the proof, then he exists; but following a more Douglas Adams route, if God knows that he exists, then he doesn't have to think about it, and therefore disappears from existence!
@LSpice. Oups! Of course, I meant He recently posted.
The title of Science and Vie is "Existence de Dieu : les mathématiques ont enfin la réponse", that is clearly a big deformation of the (mathematical) onthological argument. My personal point of view, with all due respect for Goedel, is that "Mathematics" and "existence of God" should not stay in the same sentence.
If you look at the article, he says one could rephrase the statement to be about the following property: "An entity $x$ is maximally-rational ($\mathcal{G}$) iff it has all rational/consistent properties." What looks particularly mathematically interesting is that an automated theorem prover was used to explore ways to trim down the assumptions and the needed logical strength to go from them to the conclusion. In this way Benzmüller also reduced unintended side effects of the conclusion that followed from the stronger logical axioms.
A very naive question. Why do we need modal logic to formalize the ontological argument? Already Gödel uses modal logic, but why is the argument nor formalizable usual logic?
@Joël : One of the traditional formulations of the ontological argument relies heavily on the word "necessary." So it is natural, when formalizing the argument, to use a logic where the word "necessary" has a direct formalization.
I'd be personally interested in a proof of the equivalence between the existence of God and the truth of RH :-)
@AsafKaragila God is calculating the question for the answer 42. He or She's busy. Therefore He or She (or It ?) exists.
@PhilippeGaucher: God would not be busy calculating the answer to the wrong question. People always forget that part.
Gödel’s proof has the nice feature that one can cleanly separate the logical core of the argument (which is uncontroversial—but see the next paragraph) from its alleged application to theology (which of course is going to be controversial). My opinion is that the main significance of Gödel’s proof is that it shows that there is something to the ontological argument; many people (including my teenage self), when first encountering the ontological argument, perceive it to be total nonsense. Immanuel Kant’s “existence is not a predicate” objection was taken by many to be a decisive refutation. So showing that the ontological argument isn’t completely free of content is already a significant intellectual achievement.
The work of Benzmüller (and Paleo) is interesting because it showed that a (fixable) mistake in Gödel’s proof was more serious than people had previously realized. See The Inconsistency in Gödel’s Ontological Argument: A Success Story for AI in Metaphysics for more details.
Aside from Kant's objection on ontological arguments, I think the strongest criticism to Gödel's ontological proof is that his axioms imply modal collapse, that is, $\varphi \rightarrow \square \varphi$, which somehow undermines the point of modal logic. (That Gödel's axioms imply modal collapse was also verified in another work of Benzmüller and Paleo.)
I know Benzmüller's work from a slightly different context, formalizing other interesting systems of modal logic. Hadn't heard of this specific project, but it seems very admirable.
Formalization of proofs is a highly respectable activity
Any "proof of existence of God" written by Gödel is bound to inspire curiosity and interest in logic among the public, thus contributing to education
|
2025-03-21T14:48:31.746960
| 2020-08-11T19:38:53 |
368900
|
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|
Stack Exchange
|
If $n = 18k+5$ is composite, there are at least 9 divisors of $\phi(n)$ which do not divide $n-1$
If $n$ is a composite of the form $18k+5$, there at least 9 divisors of $\phi(n)$ which do not divide $n-1$. Is this true in general or if not, what is the smallest counter example? The conjecture has been verified for $n \le 1 \times 10^7$.
Related question.
Note: The question was posted in MSE a year ago but got no answer. Hence posting in MO
@GeoffRobinson "If two of the $p_i$ divide $n$ to the third power or higher, you are done" - why? I count only 8 divisors of $\varphi(n)$ not dividing $n-1$.
@FedorPetrov: You are right, I was careless. I have rewritten the comment.
Any divisor of both $n-1$ and $\phi(n)$ is a divisor of $\prod_{i=1}^{k}(p_{i}-1)$, where $p_{1},p_{2}, \ldots ,p_{k}$ are the distinct primes which divide $n$.
Positive answer to this question would imply that there are no solutions to Lehmer's totient problem of the form $18k+5$. I'm not aware of known restrictions of this kind.
|
2025-03-21T14:48:31.747072
| 2020-08-11T20:05:56 |
368901
|
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|
Stack Exchange
|
Close integer solutions to $ab-cd=1$?
I am looking for infinite set of Diophantine solutions.
Suppose we require
$$0<\min(a,d)<\max(a,d)<\min(b,c)<\max(b,c)\leq\sqrt 2\min(a,d)$$
$$a,b,c,d\in\mathbb Z$$
then can we still find solutions to
$$ab-cd=1?$$
Is it possible to do this if only
$$a,b,c,d\in\mathbb Z$$
$$0<\min(a,d)<\max(a,d)<\min(b,c)<\max(b,c)$$
$$ad<ab+cd\leq2ad\leq2bc$$ holds?
If not what is the closest permissible?
What is a good bound?
Note 2. is not necessarily weaker and is a different statement if $ab+cd<bc$ does not hold.
It is possible with any constant $\lambda>1$ on the place of $\sqrt{2}$. Take $a=t^2$, $d=t^2+t-1$, $b=t^2+2t+1$, $c=t^2+t+1$ for large $t$.
nicely done!...
@FedorPetrov Is it also possible to do 2. or at least do it with a different example? In your example $ab+cd<bc$ does not hold.
@VS. I am confused, what exact conditions do you require now? I thought that 2. is $ad<ab+cd\leqslant 2ad$ which clearly holds for this example.
is impossible. Assume $b=d+\delta_1$ and $c=a+\delta_2$. Then $ab+cd=2ad+a\delta_1+d\delta_2$ and since $\max(b,c)\leq\min(a,d)$ we have $\delta_1,\delta_2>0$.
|
2025-03-21T14:48:31.747423
| 2020-08-11T20:59:56 |
368904
|
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|
Stack Exchange
|
Manifolds with boundary admitting no closed embedded minimal hypersurface
The following Theorem is proved in the paper entitled "Compactness of the space of embedded minimal surfaces with free boundary in three-manifolds with nonnegative Ricci curvature and convex boundary", by A. Fraser and M. Li:
Let $M^n$ be a compact $n$-dimensional Riemannian manifold with nonempty boundary $\partial M$. Suppose $M$ has nonnegative Ricci curvature and the boundary $\partial M$ is strictly mean convex with respect to the inward unit normal. Then, $M$ contains no smooth, closed, embedded minimal hypersurface.
My question is: what are examples of compact Riemannian $3$-manifolds with nonnegative scalar curvature (but not nonnegative Ricci curvature) and mean convex boundary that don't admit closed embedded minimal surfaces?
A solid torus should work. Choose cylindrical coordinates $(r,\theta, \lambda)$, $0\leq r \leq r_0 < \pi/2, 0\leq \theta \leq 2\pi, 0\leq \lambda \leq l$, where we equate $(r,\theta,0)\sim (r,\theta, l)$ and $(0,\theta, \lambda)\sim (0,0,\lambda)$. Put a Riemannian metric on this solid torus of the form $dr^2+ f(r)^2 d\theta^2 + g(r)^2 d\lambda^2$, and $f(r)=\sin(r), g(r)=\cosh(\epsilon r)$, where $0 < \epsilon$ is small.
The sectional curvatures of such a metric are computed in Lemma 2.3 of this paper as
$$K_{\theta\lambda}=-\frac{f'g'}{fg},\ K_{r\theta}=-\frac{f''}{f},\
K_{r\lambda}=-\frac{g''}{g}$$
and the mean curvature of the level $r$ torus is
$$\frac12(\frac{f'}{f}+\frac{g'}{g}).$$
From the sectional curvatures, we get the scalar
curvature as
$$R= -2 (\epsilon \cos(r) \sinh(r) -\sin(r) \cosh(\epsilon r) +\epsilon^2 \sin(r) \cosh(\epsilon r))/fg,$$
and mean curvature of the level surface at height $r$ as
$$ \frac12(\cos(r)\cosh(\epsilon r) +\epsilon\sin(r) \sinh(\epsilon r))/fg.$$
We see that the level surfaces $r=c$ are mean convex tori, and the scalar curvature is positive for $\epsilon$ and $r_0$ small. Hence this metric contains no closed minimal surface: the maximal $r$ value for such a surface would be tangent to a level surface which is mean convex, contradicting the maximum principle.
Here's how I found this metric: given your criteria, the double of the manifold admits a metric with positive scalar curvature (this is a trick of Hubert Bray; cf. also Pengzi Miao). Such manifolds are connect sums of space forms and $S^2\times S^1$. The reflection symmetry quotient then gives a handlebody (a space form with non-trivial fundamental group cannot admit a reflection symmetry with fixed set a surface). Hence the simplest non-trivial case is a solid torus. The above metrics are invariant under the $S^1\times S^1$ action on a solid torus (a ``double-warped product"). The $(r,\theta)$ slice is a spherical cap, and the $(r,\lambda)$ slices are scaled hyperbolic metrics on a cyinder. Then we adjust $\epsilon$ and $r_0$ to make the scalar curvature positive, the positive sectional curvature of the spherical cap dominating the negative curvature of the hyperbolic annulus. I suspect one can use the techniques in Codá-Marques' paper to realize any handlebody with such properties.
|
2025-03-21T14:48:31.747623
| 2020-08-11T21:40:33 |
368908
|
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"Joseph O'Rourke",
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|
Stack Exchange
|
Construct pairs of $n$-dimensional convex bodies with given ratios ($p$) of volumes
Given a dimension $n$ and a number $p \in (0,1)$, to what extent is it possible (in what cases) to construct a convex set $A$--not a hypersphere--and a "snugly" inscribed (InscribedFigure) convex set $B$, such that
(Euclidean) volume(B)=$p$ volume(A)? What if $p$ is rational? (pRational)
Why not just use for $B$ a scaled down version of $A$?
Or why not slice $A$ with a hyperplane $H$ to reach $p$ times the volume of $A$ in the remainder? One could start with $H$ outside and move it parallel inward continuously until $p , \textrm{vol}(A)$ is reached.
Joseph O'Rourke--that would surely do the trick (though not what I was envisioning). Perhaps I should exempt this possibility --to add a further challenge.
Can you define what you mean by "inscribed"?
Thanks for the comment--M. Winter. I did provide that link to the Wikipedia article "Inscribed Figure". Maybe, it's a little tricky. The term "snugly" seems to convey the concept. Perhaps, the relevant idea here is that any similarity transformation of B expanding it, without any translation of B, would yield points outside A.
To continue, perhaps the test should be that any vertex of B (though it may not have any) has to lie on the boundary of A.
|
2025-03-21T14:48:31.747751
| 2020-08-11T21:43:21 |
368909
|
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|
Stack Exchange
|
'Monodromy' for relative homology group
Let $A$ and $X$ be topological manifolds. Denote by $\mathbb {Emb}(A,X)$ the space of all topological embeddings $A\to X$.
A loop $f_s:A\to X$ ($s\in[0,1]$) in $\mathbb{Emb}(A,X)$ should give rise to an isomorphism
$$\Phi: H_k(X,f_0(A))\cong H_k(X,f_0(A))$$
by identifying $H_k(X,f_s(A))$ with $H_k(X,f_{s+\epsilon}(A))$ for all $s$ and sufficiently small $\epsilon>0$.
Suppose the space $\mathbb{Emb}(A,X)$ is not simply-connected and the loop $f_s$ is not contractible in this space.
Then, we may think of $\Phi$ as a 'monodromy' map. For simplicity, we may further assume $A$ is compact and $k\ge 2$, etc.
Question: By what conditions can we ensure the 'monodromy' is trivial, i.e. $\Phi=id$?
The example in my mind is as follows: we have a elliptic K3 surface $g:X\to\mathbb P^1$. Most fibers of $g$ are topologically torus $T^2$ and there are finitely many singular fibers which are 'pinched sphere'. Let $z_0\in \mathbb P^1$ be a singular value and take a small loop $\gamma$ centered at $z_0$; then the torus fibers over $\gamma$ gives rise to a loop in $\mathbb{Emb}(T^2,X)$ and I would like to study $H_2(X, g^{-1}(z))$ for $z\in \gamma$.
Nitpick: Is the monodromy in the traditional sense (the map $A \to A$ you get by moving around that loop in the fiber) trivial for the K3 surface? If not, what you mean is not $\text{Emb}(A,X)$ but rather $\text{Sub}(A,X)$, the space of subspaces homeomorphic to $A$. Equivalently, $\text{Emb}(A,X)/\text{Homeo}(A)$. Loops in this space still give rise to monodromy on relative homology.
|
2025-03-21T14:48:31.747885
| 2020-08-11T21:56:57 |
368910
|
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|
Stack Exchange
|
Zero of the exponential p-adic
Consider the $p$-adic exponential defined over $\mathbb C_p$. One knows $\exp$ is analytic in the domain $\mathcal D=\{z\in\mathbb C_P\mid v_p(z)>\frac1{p-1}\}$. Does it exist an element $z_0\in\mathcal D$ such that $\exp(z_0)=0$?
Thanks in advance.
The exponential function satisfies $\exp \left({x + y}\right) = \exp\left(x \right) \exp\left(y \right)$ for $x, y$ in the convergence domain. It also satisfies $\exp \left( 0 \right) = 1$. So if $\exp \left( z_0 \right) = 0$, then $0 = \exp \left( z_0 \right) \cdot \exp\left(-z_0 \right) = \exp \left( z_0 + (-z_0) \right) = \exp\left( 0 \right) = 1$, which is a contradiction.
But what if $\exp\left(-z_0 \right) = \infty$?
$\exp(-z_0)=\infty$ is impossible in the domain on which $\exp$ is analytic, by definition of analytic (and $\mathcal D$ is invariant under $z_0\mapsto-z_0$).
It’s hard to see in what sense it could be true that the exponential function is “defined over $\Bbb C_p$”, since the logarithm is defined on the whole open unit disk there, and has so very many zeros.
If you look closely, you can see that for all $z\in\mathcal D$, we have $v_p(e^z - 1)=v_p(z)$. This obtains quite independently of any multiplicativity of the exponential.
The OP surely just means by "defined over $\mathbf C_p$" that its domain is being considered on $\mathbf C_p$, in the same way one might refer to the exponential function on $1 + 5\mathbf Z_5$ as the exponential function "defined over/on $\mathbf Q_5$". I agree this use of "defined over" is awkward, and the idea that an exponential function in any setting might take the value $0$ is peculiar.
You don't mean $v_p(e^z) = v_p(z)$ (try $z = 0$!), but rather $v_p(e^z - 1) = v_p(z)$.
Of course, @KConrad. You may imagine that I was thinking of the formal-group formulation. I’ll correct.
|
2025-03-21T14:48:31.748035
| 2020-08-11T22:52:46 |
368913
|
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|
Stack Exchange
|
Homotopy descent and cohomology
I've been reading "Structured Brown representability via concordance" by D.Pavlov (https://dmitripavlov.org/concordance.pdf) and I'm strugeling with a point and was wondering if someone could help me with my confusion. In the text there is a criterion which says that if a simplicial presheaf
$$F:Man^{op}\rightarrow \text{sSet}$$
satisfies homotopy descent, where $Man$ is the category of smooth manifolds, then there exists a K such that
$$F\cong [-,K].$$
As a sanity check, or more to see if I could actually use this criterion, I wanted to show that singular cohomology satisfies this condition. Let $(U_i\rightarrow M)$ be an open cover and denote by $$U_{\underline{i}}=U_{i_0}\cap \ldots \cap U_{i_m}.$$By section 5.8 in https://pages.uoregon.edu/ddugger/hocolim.pdf, we know that
$$[\text{hocolim } U_{\underline{i}},K(\mathbb{Z},n)]\cong \text{holim }[U_{\underline{i}},K(\mathbb{Z},n)]$$
so singular cohomology should satisfy homotopy descent. I would like to prove this without using the existence of Eilenberg-Maclane spaces. As an example, choose some manifold $M$ and an open cover $U,V$. Then we need to show that
$$H^n(\text{hocolim }\left( U\leftarrow U\cap V\rightarrow V\right),\mathbb{Z})\cong \text{holim }\left[H^n(U,\mathbb{Z})\rightarrow H^n(U\cap V,\mathbb{Z})\leftarrow H^n(V,\mathbb{Z})\right].$$
In section 18 of https://pages.uoregon.edu/ddugger/hocolim.pdf, we are given spectal sequences computing the cohomology of a homotopy colimit. In the case of a homotopy pushout, this is just the information we get from the Mayer-Vietoris sequence, i.e.
$$H^n(\text{hocolim }\left( U\leftarrow U\cap V\rightarrow V\right),\mathbb{Z})\cong \text{coker}\left( H^{n-1}(U)\oplus H^{n-1}(V)\rightarrow H^{n-1}(U\cap V) \right)\oplus \text{ker}\left(H^n(U)\oplus H^n(V)\rightarrow H^n(U\cap V)\right). $$
So now I'm left with showing that this Mayer-Vietoris data is weakly equivalent to $\text{holim }\left[H^n(U,\mathbb{Z})\rightarrow H^n(U\cap V,\mathbb{Z})\leftarrow H^n(V,\mathbb{Z})\right]$, but that is where I get stuck. I would very much appreciate any guidence, thanks!
I think you need to use singular cochains in your holim, not singular cohomology.
if a simplicial presheaf F:Man^op→sSet satisfies homotopy descent, where Man is the category of smooth manifolds, then there exists a K such that F≅[−,K].
Here one must also mention that F is required to be concordance-invariant (alias R-local), i.e., the map F(X)→F(R⨯X) must be a weak equivalence.
By section 5.8 in https://pages.uoregon.edu/ddugger/hocolim.pdf, we know that [hocolim U_i,K(Z,n)]≅holim [U_i,K(Z,n)]
Here [-,-] must be the whole mapping space, not just the set of homotopy classes
of maps, just like in the cited text by Dugger.
So in particular, one must use the whole mapping space into K(Z,n) (equivalently,
the whole singular cochain complex) instead of H^n (the homology group).
Concerning the homotopy descent property for singular cochains,
note that the cited paper provides an independent proof in Proposition 2.8,
by applying the main theorem in reverse.
First, it is a classical result (proved using subdivisions)
that the cochain complex C of sheaves of local singular cochains is locally weakly equivalent to the cochain complex of presheaves of singular cochains.
The remainder of the proof uses a simple observation that the simplicial
object k↦C^n_closed(Δ^k⨯X), once converted to a chain complex
using the Dold–Kan correspondence, becomes quasi-isomorphic
to the n-truncated singular cochain complex via an explicit
map that is constructed using subdivisions.
Theorem 0.2 then provides the desired homotopy descent property
by showing that the n-truncated singular cochain complex
is homotopy representable, hence satisfies homotopy descent.
|
2025-03-21T14:48:31.748254
| 2020-08-11T22:59:05 |
368914
|
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|
Stack Exchange
|
Integral of Legendre's function
Is there any formula for computing the following integral $$\int_a^1(P^m_l)^2(x)\,dx \, ,\text{with} -1<a<1$$
where $P^m_l$ is the associated Legendre's function (of the first kind) of order $m\in\mathbb{N}$ and of degree $l=-0.5+it$ with $t\in\mathbb{R}$.
Similar question: https://mathoverflow.net/q/366081
with that $a$ in there, you are asking for the indefinite integral.
the variable $a$ belongs to $(-1;1)$.
|
2025-03-21T14:48:31.748322
| 2020-08-12T00:06:28 |
368918
|
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|
Stack Exchange
|
Closed convex hull in infinite dimensions vs. continuous convex combinations
tl;dr: When is the closed convex hull of a set $K$ equal to the set of "continuous" convex combinations of $K$?
I am essentially asking for the most general, infinite-dimensional analogue of this related question.
Update: I forgot to specicy that $K$ is compact. As @GeraldEdgar points out below, for noncompact $K$, the answer is trivially "no".
Suppose $K\subset E$ where $E$ is a topological vector space (as far as I can tell, this is the most general kind of space for which this question makes sense). Obviously we can define the closed convex hull $\overline{\text{conv} K}$ of $K$ as usual. Now consider the set
$$
K^* =
\{ \int_K x\,d\mu(x) : \mu \in\mathcal{P}(K)\},
$$
where $\mathcal{P}(K)$ is the set of (say, Borel) probability measures over $K$ and integral here is to be understood in the weak (Pettis) sense.
I would like to know when $\overline{\text{conv} K} = K^*$. If $E$ is finite-dimensional, there is equality. What are the most general assumptions on $E$ and $K$ for which this equality continues to hold?
(For the curious, the inspiration for this question came from trying to understand when $K^*$ is compact.)
A preliminary important question is what you mean by the integral, here. Bochner? Pettis? Or is it somehow part of the question (e.g., you want some kind of integral for which there will be equality in the most general setting possible)?
I was thinking of Lebesgue integration. If there is any reason to consider more exotic integrals, I would be interested to learn more.
I don't know what the “Lebesgue integral” means in an infinite-dimensional vector space — is what I was trying to point out. The Bochner and Pettis integrals are way to extend it to this context.
Now I see the confusion. I am used to the convention where such integrals are generally understood as weak integrals, which I suppose is the same as the "Pettis" integral.
No. Even in one dimension. Say $K$ is the open interval $(0,1)$. Show $0 \notin K^*$. Let $\mu$ be a probability measure with support contained in $(0,1)$. Indeed,
$$
r(\mu) := \int_K x\,d\mu(x)
$$
is the integral of a positive function. That is, $x > 0$ a.e. So $\int_K x\,d\mu(x) > 0$. Similarly $1 \notin K^*$.
In a locally convex topological vector space $E$, if there is any extreme point of $M = \overline{\text{conv} K}$ that does not already belong to $K$, then it also does not belong to $K^*$. So what if $K$ is the set $\text{ex}\; M$ of extreme points of a closed convex bounded set $M$? Can we recover $M$ as $K^*$?
A very nice little book that discusses this situation is
Phelps, Robert R., Lectures on Choquet’s theorem, Lecture Notes in Mathematics. 1757. Berlin: Springer. 124 p. (2001). ZBL0997.46005.
Choquet's theorem tells us roughly that every point of a compact convex set $M$ is of the form $r(\mu)$ for some probability measure concentrated on the set $\text{ex}\; M$ of extreme points of $M$.
Plug
My first publication to attract any notice was this one, where there is a generalization of Choquet's theorem to certain closed bounded noncompact sets $M$.
Edgar, G. A., A noncompact Choquet theorem, Proc. Am. Math. Soc. 49, 354-358 (1975). ZBL0273.46012.
Oops. $K$ should be compact, otherwise this trivially fails even in finite dimensions, as you point out.
|
2025-03-21T14:48:31.748680
| 2020-08-12T01:18:17 |
368922
|
{
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"Gerry Myerson",
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"Q_p",
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"Sam",
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|
Stack Exchange
|
Rational points on the elliptic curve $y^2 = x^{3} - t^{2}z^3$
What are the rational points on the elliptic curve $y^2 = x^3 - t^{2}z^3$ ? I seem not to find any besides the trivial ones whereby $txyz=0$ or $x= \pm z$.
ADDENDUM 1. I have just noticed that if $z^3 = x$ then there do exist some non-trivial rational point(s) provided that $t$ is a congruent number. Therefore, to avoid this and related scenarios, we impose the condition that the numerators of the reduced forms of $x \neq \pm 1$ and $z \neq \pm 1$ are relatively prime.
ADDENDUM 2. It is usually interesting to generalise questions. So i would also ask for rational points on the general surface $y^2 = x^{3} -t^{2n}z^3$, $n$ being some positive integer.
Over $\mathbb Q(t)$?
But then '$t$ is a congruent number' does not make sense to me ― it is never evaluated as a number, but always functions as a formal variable... Or are you asking for $\mathbf Q$-rational points on this elliptic surface?
First of all, you probably mean $y^2z$ rather than $y^2$. Secondly, from the question it appears that $t$ is allowed to be algebraic over the rationals. In that case, there are many choices of $t$ which will yield rational points.
@Kapil, i have edited the question now.
@R.vanDobbendeBruyn, i'm taking $t$ to be some rational number, not a formal variable, so i think what i'm looking for are $\mathbb{Q}$-rational points on the surface.
(1) Please edit your question to indicate that you are looking for 4-tuples of rational numbers satisfying your equation. (2) Which "elliptic curves" are you talking about. If you plug in a value for $t$, you get a surface, not an elliptic curve. It's a rational surface, as one of the answers indicates. But probably you meant to write $y^2z=x^3-t^2z^3$, and then for $t\ne0$ you get an elliptic curve sitting in $\mathbb P^2$. These curves are related to the congruent number problem, so you'll find lots of information if you search on that term.
12 versions within five hours of first posting.
If we rewrite the equation as $y^2+t^{2}z^3 = x^3$, and let $v=tz$, then $y^2+zv^2=x^3$.
If you factorize over $Q[\sqrt{-z}]$, then $(y+v\sqrt{-z})(y-v\sqrt{-z})=x^3 $.
Let $x=(a+b\sqrt{-z})(a-b\sqrt{-z})$, then $y+v\sqrt{-z}=(a+b\sqrt{-z})^3=(a^3-3ab^2z)+(3a^2b-b^3z)\sqrt{-z}$
Thus $y=a^3-3ab^2z, v=3a^2b-b^3z, x=a^2+b^2z$, and $t=\frac{v}{z}=\frac{3a^2b-b^3z}{z}$
We then have the general parametric solution: $(a^3-3ab^2z)^2=(a^2+b^2z)^3-(\frac{3a^2b-b^3z}{z})^2z^3$
Edit: Of course, if you take $t=\frac{1}{x^3-y^2}, z=x^3-y^2$, you get trivial solutions, but I'm assuming you don't want those.
Really, your equation has too many variables, so writing boring solutions is easy. Are you wanting solutions that are functions of t? If so, you can rearrange my original solution above to make z a function of t, and go from there. From there, your other equations (in Addendum 2) follow by changing $t$ to $t^n$.
It's not (just) about the number of variables, but rather that the equation defines a rational variety (which is in part caused by the fact that it is a degree $3$ equation in $4$ variables). After the change of variables in the first line, can't you finish by writing $z = \frac{x^3-y^2}{v^2}$ as a function of the other variables? (so $t = \frac{v^3}{ x^3-y^2}$ completes the parametric solution.)
True. I left that up to the OP which way (s)he wanted to organize the solutions
Above equation shown below:
$y^2=x^3-t^2z^3$ ----(1)
Equation $(1)$ has parametric solution given below:
$x=m^4-3m^2+3$
$y=m(m^4-3m^2+3)^2$
$z=(m^2-1)(3m^2-m^4-3)$
$t=1$
For, $m=2$, we get:
$(x,y,z,t)=(98,7,-21,1)$
I think you have mixed up $x$ and $y$ in the last line; I also suspect that the method can be generalized rather easily to a 3-parameter family: choose $t, x_0, z_0$; then let $(x, y, z, t) = (x_0 (x_0^3 - t z_0^3), (x_0^3 - t z_0^3)^2, z_0 (x_0^3 - t z_0^3), t)$.
@user44191. Thanks for catching my typo. You have a typo too. In your value's for (x,y,z) the variable 't ' need's to have the power two. There is a difference in your method & my method. I took the sum of two cubes [ (m^2-1)^3+1^3] . Hence we get it= m^2(m^4-3m^2+3). Thus we just need to multiply the LHS by (m^4-3m^2+3) to make the LHS a square.
|
2025-03-21T14:48:31.748976
| 2020-08-12T01:58:29 |
368923
|
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|
Stack Exchange
|
Maximum length of numerator/denominator in calculating RREF
Condition : Given a $4 \times 5$ matrix, where each element is denoted by $p/q$, we have $|p|<10$ and $1\leq |q|<10$.
Example: $A$ given by
-9/7 -5/8 -1 4/5 4
-1 9/7 -6/5 -7 2/9
-1/3 -2 5/7 -2/9 -7
3 0 -5/3 8/9 -2/5
I have found that when I calculate the row reduced echelon form (RREF) of a $4\times 5$ matrix satisfying the condition above, it produces larger numbers than I expected.
For example, the RREF of $A$ is
1 0 0 0 -13325864657/7239264390
0 1 0 0<PHONE_NUMBER>6/3619632195
0 0 1 0 -3549396949/1447852878
0 0 0 1 562768523/482617626
How long is the maximum length of numerator/denominator?
It looks like $a_{1,5}$ has longest length and it's under $10^{10}$, but I can't prove it.
I know it's not what you asked, but this could be rephrased in terms of a "height" of a matrix with rational entries (basically, the maximum of the absolute values of all the denominators and numerators), and how transforming the matrix into various standard forms (eg Smith normal form, RREF, etc) affects the height. You are starting with an arbitrary matrix (of a certain fixed size) of height less than $10$ and are looking at what is the maximum possible height after passing to RREF, or if not that, upper bounds.
In other words you have a 4 by 4 matrix of the above kind and apply its inverse to a vector of the above kind (assuming full rank, otherwise you have a smaller size matrix and life gets easier) and ask how many digits the numerator and the denominator may have after that. Plug {8/9,9/5,9/8,9/7}*inverse({{9,9,9,9},{9/5,-9/8,-9/7,8/9},{-9/8,-9/7,8/9,9/5},{-9/7,8/9,-8/5,9/8}}) into Wolfram Alpha to see that 17 digits for the denominator are possible and that is, actually, the limit. Not voting to close because I cannot find the answer for the numerator in 5 minutes but this is surely MSE, not MO
I'm an idiot: the largest possible numerator is exactly the same as the largest possible denominator, so 17 digits is the answer for both. However that length is impossible to achieve simultaneously, so if this problem arose as an exercise in evil teaching (give your students an online assignment to find the RREF and provide one less position in the input field than what is needed to type the answer), then you are safe with 32 digit places plus one division sign holder and you know the matrix now. If it doesn't get closed, I'll post the argument later.'
Surprisingly the question hasn't been closed yet, so I'm posting the solution, as promised. As I said, bringing the matrix to the row echelon form is equivalent to the left multiplication by the inverse to the $4\times 4$ sub-matrix (if the rank is full) or smaller size sub-matrix if the rank is smaller than $4$. I'll consider the full rank case only. Let's look at the first entry of $A^{-1}x$. By Cramer's rule, it is just the ratio of the determinants of matrices with the same three columns (from where it is obvious that the longest possible numerator is the same as the longest possible denominator).
Now let $C_1,C_2,C_3$ be those common columns. In each column we can write each entry as a fraction with the denominator from $1$ to $9$. We will not care if the fraction is reducible or not. Now for each of these columns we will compute two quantities: the least common multiple of the denominators $D_j$ and the square root $M_j$ of the sum of squares of the entries. Then all $3\times 3$ minors have the common denominator $D=D_1D_2D_3$ and so each of the two full determinants is just an integer divided by $2520D_1D_2D_3$ ($2520$ is the least common multiple of the numbers from $1$ to $9$). This integer can become the numerator/denominator of the result if no further cancellations take place. By the Hadamard inequality (the volume of a parallelepiped does not exceed the product of its sides), that integer is at most $9\times 2\times 2520\times M_1M_2M_3D_1D_2D_3$.
Now let us look at the product $M_1D_1$, say. Notice that there are $4$ primes up to $9$: $2,3,5,7$. Having the same prime in two different denominators in $C_1$ is not optimal because we can leave only the highest power and increase $M_1$ without changing $D_1$, so each prime occurs in at most one denominator in the (formal) worst case scenario. If $2$ and $3$ occur together (the denominator $6$), then there is a denominator-free entry and we can increase $6$ to $8$ and put $9$ on that entry, increasing $D_1$ $12$ times and decreasing $M_1$ at most $9$ times. Finally, if we can increase the power of some prime $p$ in the denominator by $1$, we'd better do it because $M_1$ will decrease at most $p$ times but $D_1$ will increase $p$ times. Thus, the largest product $M_1D_1$ is achieved (formally) at the column $(9/5,9/7,9/8,9/9)$, which gives $M_1D_1\le Q=9\times\sqrt{5^{-2}+7^{-2}+8^{-2}+9^{-2}}\times 2520$.
The final estimate for the numerator/denominator is then
$9\times 2520\times 2\times Q^3\le 1.4\times 10^{16}$, so the number of digits cannot be greater than $17$.
Plugging {8/9,9/5,9/8,9/7}*inverse({{9,9,9,9},{9/5,-9/8,-9/7,8/9},{-9/8,-9/7,8/9,9/5},{-9/7,8/9,-9/5,9/8}}) into Wolfram Alpha (or any other computational tool), we see that 17 digits are attainable (the matrix has the same sign pattern as the Hadamard $4\times 4$ matrix of $\pm 1$'s, so our estimate is not too crude and the denominator $12,887,816,516,918,280$ is actually rather close to our upper bound).
|
2025-03-21T14:48:31.749340
| 2020-08-12T01:58:53 |
368924
|
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|
Stack Exchange
|
Riemann hypothesis for exponential sum
Recently I've heard about the Riemann hypothesis for one-variable exponential sums, which states as
For a polynomial $f\in\mathbb{F}_{p^k}[x]$ of degree $d$ and a character $\chi$ of $(\mathbb{F}_{p^k},+)$, provided $(d,p)=1$, we have$$\left|\sum_{x\in\mathbb{F}_{p^k}}\chi(f(x))\right|\le(d-1)\sqrt{p^k}$$
I also know the $L$-function associated to $f$ is$$L(f,T)=\exp\left(\sum_{n\ge1}S_n(f,\chi)\frac{T^n}{n}\right)$$
where $S_n(f,\chi)=\sum_{x\in\mathbb{F}_{p^{kn}}}\chi(\operatorname{tr}_{\mathbb{F}_{p^{kn}}/\mathbb{F}_{p^{k}}}(f(x)))$.
My question is, what is the relation of the Riemann hypothesis for one-variable exponential sums, and Riemann hypothesis for the associated $L$-function? I guess they are equivalent forms, but I can't prove it.
My thought: This is similar to the relation in the elliptic curve version. The Hasse theorem
For an elliptic curve $E$ over $\mathbb{F}_p$, $|\#E(\mathbb{F}_p)-p-1|\le 2\sqrt p$.
is equivalent to the Riemann hypothesis for elliptic curve:
The zeros of $\zeta(E,s)$ has real part $\frac1{2}$.
This equivalence is due to computing the zeta function $\zeta(E,s)$, involving Riemann-Roch theorem. But as for the exponential sum case, I have no idea how to compute the $L$-function associated to $f\in\mathbb{F}_{p^k}[x]$.
Any help will be appreciated.
You need to change $T$ to $q^{-s}$ to relate to the real part of the zeros. $L(f,T)$ is a factor of the zeta function of the Artin-Schreier curve $y^q-y=f(x)$. See, e.g., Weil "Basic Number Theory".
@FelipeVoloch Thanks, I think I get the idea.
One can show by a nontrivial but elementary argument that $L(f,T)$ is a polynomial in $T$ of degree $d-1$.
The Riemann hypothesis in this case says all zeroes of $L(f,T)$ have absolute value $p^{-k/2}$.
It follows from the Riemann hypothesis that we can write $L(f,T) = \prod_{i=1}^{d-1} (1 - \alpha_i T)$ where $|\alpha_i|= p^{k/2}$. The bound $S_n(f,x) \leq (d-1) \sqrt{p^{kn}}$ follows from this by taking logs.
Conversely, if we have the bound $S_n(f,x) \leq (d-1) \sqrt{p^{kn}}$ for all $n$, we can check that the roots have absolute value $\geq p^{-k/2}$ using the radius of convergence for the power series, and then check that they have absolute value $p^{-k/2}$ by using the functional equation. Unlike in the elliptic curve case, we need all $n$ here instead of just one.
|
2025-03-21T14:48:31.749669
| 2020-08-12T05:39:04 |
368931
|
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|
Stack Exchange
|
Counting the number of simple labelled bipartite graphs , with edges such that 1 vertices have degree 1
I have tried to count the number of simple labelled bipartite graphs $G_{n,m}$ with $k$ edges such that $d_1$ vertices have degree 1.
Has this problem been studied?
So far the only related paper I found shows a generating function to count all simple labelled graphs where monovalent vertices are represented by u:
$M(u,x) = \sum_{k \geq 0} \frac{x^k}{k!}2^{{k\choose 2}}(e^{ux})^{k}e^{u^2x^2/2}$
This paper also has some results on bipartite graphs, but as I'm not familiar with generating functions, I don't know how to adapt these results to my original problem. Any help would be appreciated.
do you want any bipartite graphs or connected bipartite graphs?
any bipartite graphs
|
2025-03-21T14:48:31.749744
| 2020-08-12T05:47:26 |
368932
|
{
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"url": "https://mathoverflow.net/questions/368932"
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|
Stack Exchange
|
Reference for multivariate generalised CLT
I know that one can generalise the classical CLT in terms of heavy tail distributions, namely, for any i.i.d. random variables $X_i$,
$$\frac{X_1+\cdots+X_n}{n^{1/\alpha}}\rightarrow S(\alpha,\beta,\gamma,\delta)$$
in distribution sense, whenever $X_i$ belongs to the domain of attraction of its corresponding limit. When $\alpha$ takes $2$, this law becomes the classical CLT.
To push this to multivariate versions, I also see https://en.wikipedia.org/wiki/Multivariate_stable_distribution, in which is multivariate stable distribution is introduced.
Where can I find a reference for such multivariate generalised central limit theorem? A textbook reference would be great.
P.S. This is also an existing question in MSE: https://math.stackexchange.com/questions/3786947/reference-for-multivariate-generalised-clt?noredirect=1#comment7800922_3786947
$\newcommand\al\alpha\newcommand\R{\mathbb R}$The domains of attraction to multidimensional stable distributions are characterized by Rvačeva's Theorems 4.1 (p. 194, for $\al=2$) and 4.2 (p. 196, for $\al<2$).
The domains of strict attraction to multidimensional strictly stable distributions are characterized by Shimura's Theorems 3.1 (p. 354, for $\al=1$), 3.3 (p. 356, for $\al\in(0,1)$), 3.4 (p. 356, for $\al\in(1,2)$), and 3.5 (p. 356, for $\al=2$).
|
2025-03-21T14:48:31.749858
| 2020-08-12T06:29:17 |
368935
|
{
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"Gordon Royle",
"Patrick Schnider",
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|
Stack Exchange
|
Graph which is maximal triangle-free 3-colorable, but not maximal triangle-free
I am looking for a graph which is maximal triangle-free 3-colorable, but not maximal triangle-free. Here a graph which is maximal triangle-free 3-colorable is a graph where the addition of any edge violates either the triangle-free condition or the 3-colorable condition.
The possible counterexample necessarily has a pair of vertices with distances $\geq 3$ (in fact no two vertices can have distance greater than 3) where every 3-coloring of the graph has those two vertices being the same color. Some stronger things can be proven about the counterexample, such as every neighbor of one of those aforementioned vertices must be connected to a neighbor of the other vertex, and vice versa (otherwise the distance between a vertex and a neighbor of the other is $\geq 3$, so they must be the same color, but this contradicts that the original two vertices are always the same color). If this explanation is unclear I can elaborate more.
It seems if there exists a triangle-free graph which is 3-colorable and two vertices have to be different colors, one could use this to define the desired graph. Another note is that the graph must be at least 11 vertices - adding in the edge between the two vertices results in a triangle-free graph with chromatic number 4, the smallest example of which is the Groetzsch graph.
You have repeated maximal triangle free twice, could you clarify?
By maximal triangle-free, I mean a graph which is triangle-free, and the addition of any edge creates a triangle, not maximal in terms of edge count (which are the complete bipartite graphs)
From your title you seem to be looking for a graph which is maximal triangle-free, but not maximal triangle-free. Do you maybe mean something else for one of these?
The graph should be maximal with respect to both properties - triangle-free and 3-colorable
dosent a $5$-cycle do the job?
So you want a triangle-free graph of diameter at least 3 (this makes it not maximal-triangle-free) and chromatic number 3 such that joining any pair of vertices at distance at least 3 increases the chromatic number to 4.
@GordonRoyle I think the property of being both maximally triangle free and maximally $3$- colorable is satisfied by the odd cycle of length greater than $7$, isnt it?
@GordonRoyle yes that's the desired example
@vidyarthi the 5-cycle doesn't work, because adding any edge will create a triangle. An odd cycle of length greater than 7 doesn't work because it's not maximal triangle-free, 3-colorable - there are edges that can be added that don't create a triangle or force the graph to have chromatic number 4
Here's an example of such a graph. If you remove the top vertex, this is just the Grötzsch graph with an edge removed. Since the the Grötzsch graph is minimal with respect to having chromatic number 4, the vertices of the removed edge must be the same color in any 3-coloring.
We need this extra vertex at the top to make the graph maximal with respect to being triangle-free and 3-colorable . Since this is not maximal with respect to just being triangle-free, this graph has the property you want. (It's pretty easy to check the above claims, since this graph has a unique 3-coloring up to swapping the colors)
|
2025-03-21T14:48:31.750094
| 2020-08-12T07:30:26 |
368938
|
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|
Stack Exchange
|
The size of monochromatic submatrix
We say a matrix $(a_{ij})$ is 0-1 matrix if $a_{ij}\in \{0,1\}$ for all $i,j$. We say a matrix $(a_{ij})$ is monochromatic if for some $a$, $a_{ij} = a$ for all $i,j$.
Question: Let $c\geq 1/2$ be a constant and $n$ be very large. Given a $n\times n$ 0-1 matrix $M$, must there be a $c\log_2 n\times c\log_2 n$ submatrix $M'$ of $M$ that is monochromatic?
When $c<1/2$ such submatrix $M'$ clearly exists (see below) but we wonder the case of $c\geq 1/2$.
The case $c<1/2$: It suffices to construct a $2c \log_2 n\times 2c \log_2 n$ matrix $\hat{M}$ so that each row of $\hat{M}$ is monochromatic (but different row may have different "color"). The row of $\hat{M}$ is simply the first $2c \log_2 n$ rows of $M$. To select columns of $\hat{M}$, inductively screen out columns of $M$. The first row screen out a set $K$ of columns of $M$ with $|K|\leq n/2$ where $a_{0j} = a$ for all $j\in K$ and $a_{0j} = 1-a$ for all $j\notin K$. The second row screen out a set $K'\subseteq K$ of columns with $|K'|\leq |K|/2$ where $a_{1j} = a'$ for all $j\in K'$ and $a_{1j} = 1-a'$ for all $j\in K\setminus K'$. Keep doing this for each rows of $\hat{M}$ (which is $2c\log_2 n$ many times) and there will be at least $n^{1-2c}$ many columns remained.
can you provide the argument for $c<1/2$ as part of the question?
Even $c=1-\varepsilon$ works when $n$ is large enough. This question is about bounding the diagonal bipartite Ramsey, see the recent achievement of David Conlon here
http://people.maths.ox.ac.uk/~conlond/Bipartite.pdf
|
2025-03-21T14:48:31.750218
| 2020-08-12T08:09:18 |
368939
|
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|
Stack Exchange
|
Question about basis of $\text{Der}_{k}(k[X])$
Let $k[X] = k[x_1,\ldots, x_n]$ be the polynomial ring over a field of characteristic zero.
Assume that $(D_1,\ldots, D_n)$ is a $k[X]$-basis of $\text{Der}_k(k[X])$. Suppose that the vector space $\left<1, x_1,\ldots, x_n\right>$ is invariant by each $D_i$. Define $M_i = D_i|_{\left<1, x_1,\ldots, x_n\right>}$ and let $\left<M_1,\ldots, M_n\right>$ consists of nilpotent linear operators on $\left<1, x_1,\ldots, x_n\right>$.
Is it true that $\left<M_1,\ldots, M_n\right>^k = 0$ for some $k$?
Does $\langle\rangle$ mean $k$-linear span, ideal generated over $k[X]$, or something else?
It means $k$-linear span
Do you know an example where this happens and the change-of-basis matrix from the standard basis $(\partial_1, \dotsc, \partial_n)$ is not scalar?
$\partial_x$, $x\partial_y + \partial_z$ and $\partial_z$ in $k[x, y, z]$.
I edited the question. The previous one was not true beacuse we can consider $(\partial_x,\partial_y,y\partial_x+\partial_z,x\partial_y+\partial_t)$-basis of $\text{Der}_k(k[x, y, z, t])$
I don't understand the question as it stands. Is there now a new assumption that the $k$-linear span $\langle M_1, \dotsc, M_n\rangle$ consists entirely of nilpotent operators? Then you've got a collection of nilpotent operators on an $(n + 1)$-dimensional space, so $\langle M_1, \dotsc, M_n\rangle^{n + 1}$ equals $0$. Or do you mean that $\langle M_1, \dotsc, M_n\rangle$ is defined to be the collection of nilpotent operators in the span?
I'm sorry! I seem to have been thinking of the statement that a Lie algebra of nilpotent endomorphisms of a f.d. space behaves this way (Engel's theorem). I don't know the statement for mere vector spaces of endomorphisms, and agree it's probably not true in that generality.
|
2025-03-21T14:48:31.750364
| 2020-08-12T08:30:58 |
368940
|
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"authors": [
"Pietro Majer",
"https://mathoverflow.net/users/6101"
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|
Stack Exchange
|
Analytic approximations of smooth vector fields
Let $M$ be the set of smooth divergence-free vector fields $u$ on $\mathbb{R}^3$ with
$$|\partial_x^{\alpha} u(x)| \leq C_{\alpha K}(1+|x|)^{-K}$$
on $\mathbb{R}^3$ for any $\alpha,K$.
Further, we consider the subset of analytic functions $M_0 \subset M$. The question is now, whether for each $u \in M$ and $\varepsilon >0$, there is an $v \in M_0$, such that
$$|u(x)-v(x)| < \varepsilon$$ for each $x \in \mathbb{R}^3$.
As a second question consider $M$ to be the set of smooth divergence-free vector fields $u$ on $\mathbb{R}^3$ with
$$u(x+e_j) = u(x)$$
for all $x \in \mathbb{R}^3$ and $1 \leq j \leq 3$.
Again, consider the subset of analytic functions $M_0 \subset M$. The question here is now again, whether for each $u \in M$ and $\varepsilon >0$, there is an $v \in M_0$, such that
$$|u(x)-v(x)| < \varepsilon$$ for each $x \in \mathbb{R}^3$.
These questions are fairly similar to the Stone-Weierstrass theorem, but it seems it is not possible to derive it directly from it.
I believe the most natural approach to this particular question is via Fourier analysis. In the periodic case we have the series
$$u(x)=\sum_{k\in\mathbb{Z}^3}u_k e^{2\pi i (k,x)},$$
and the condition $\nabla\cdot u=0$ simply means $(u_k,k)=0$.
Taking
$$v(x)=\sum_{|k|<K}u_k e^{2\pi i (k,x)}$$
for sufficiently large $K>0$ we can approximate $u$ as close as we want because the Fourier series of a smooth function converges uniformly and very quickly. Obviously, a polynomial $v(x)$ is analytic.
The same trick works in the first case too except the Fourier series has to be replaced by the Fourier integral, and the fact that $v$ is analytic may be a little less obvious (but still true).
In fact in problem1, one can also approximate by convolution with the Weierstrass kernel $C\exp(-||x||^2/\epsilon)$
Here is how to derive both results from the Stone-Weierstras theorem. As you say, it's not direct, but not a long way either. Recall these simple applications of the S-W theorem, to be used in PB1 resp. in PB2
The algebra $A$ of rapidly decreasing real analytic functions on $\mathbb{R}^3$ is uniformly dense in the space of continuous functions vanishing at infinity on $\mathbb{R}^3$. This follows from the S-W theorem applied to the one-point compactification $\mathbb{R}^3\cup\{\infty\}$, the $3$-sphere. To check that the above algebra separates points it is sufficient to consider the function $\exp(-\|x\|^2)$ and its translates.
The algebra $A$ of $\mathbb{Z}^3$-periodic real analytic functions on $\mathbb{R}^3$ is uniformly dense in the space of $\mathbb{Z}^3$-periodic continuous functions on $\mathbb{R}^3$. This follows from the S-W theorem applied to the quotient $\mathbb{R}^3/\mathbb{Z}^3$, the $3$-torus. To check that the above algebra separates points it is sufficient to consider the functions $\sin(2\pi x_1)$, $\sin(2\pi x_2)$, $\sin(2\pi x_3)$ and their translates.
Now, given $u\in M$ and $\epsilon>0$, we find $w_i\in A$, such that $\|u_i-w_i\|_{\infty}\le\epsilon$ (for $i=1,..,3$). To define a divergence-free approximation $v=(v_1,v_2,v_3)\in M_0$, we may then take $v_1:=w_1$, $v_2=w_2$ and for all $(x,y,z)\in\mathbb{R}^3$
$$v_3(x,y,z):=w_3(x,y,0)-\int_0^z\big\{\partial_1w_1(x,y,s)+\partial_2w_2(x,y,s)\big\}ds.$$
It is easy to check that, both in problem 1 and 2, $v_3\in A$, that ${\rm div\, } v=0$, and that $v$ is still uniformly close to $u$.
|
2025-03-21T14:48:31.750680
| 2020-08-12T10:06:58 |
368947
|
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|
Stack Exchange
|
Technics for computing weights of kernel, image or cokernel in category $\mathcal{O}$
As in the title I looking for technics to compute weights of kernels, images or cokernels in category $\mathcal{O}$ besides checking everything directly by hand.
To be more concrete, consider the following example coming from the strong BGG resolution of $L((0,0,0))$ (based on Morphism of Verma modules)
\begin{equation}
M((-2,1,1)) \oplus M((-1,-1,2)) \stackrel{\delta_{2_1}}{\rightarrow} M((0,-1,1))
\end{equation}
where we have for respectively maximal vectors $v_{(-2,1,1)}$, $v_{(-1,-1,2)}$ and $v_{(0,-1,1)}$ that
\begin{equation*}
\delta_{2_1}(v_{(-2,1,1)},v_{(-1,-1,2)})= (b_1y_{\alpha_1}^2+b_2(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2}))v_{(0,-1,1)}
\end{equation*}
with $b_i \in \{-1,1\}$.
As we have relations in $U(\mathfrak{n}^-)$, I already struggle to compute for example the kernel directly and in general it will get even harder. So I was wondering if there are other ways to get the weights of the kernel etc.?
|
2025-03-21T14:48:31.750781
| 2020-08-12T10:21:53 |
368948
|
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"authors": [
"Francesco Polizzi",
"Jorge E. Cardona",
"Pierre PC",
"erz",
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|
Stack Exchange
|
Are there minimal topological conditions on a space for it to have a countable separating set?
Are there minimal topological conditions on a space $X$ for it to have a countable separating set?
A separating set here is a set $D \subset C(X)$ (where $C(X)$ is the space of continuous functions from $X$ to $\mathbb{R}$) such that for every pair of points $x \neq y$ there is a function $f \in D$ satisfying $f(x) \neq f(y)$. I know that second-countable and normal Hausdorff are sufficient to have a countable separating set, but if one takes $X$ to be a reflexive and separable Banach space with the weak topology, there is a countable separating set despite not being even second-countable. So second-countability is not necessary.
What is $C(X)$ for you? The set of continuous functions $X \to \mathbb{R}$?
@FrancescoPolizzi Yes, exactly that.
how about existence of weaker second countable topology?
@erz I guess that if there is a weaker, second-countable topology, there is a countable separating set and the functions there are also continuous with the original topology, hence the original topology is second-countable. Is that what you mean? Is it any easier to prove the existence of a weaker second-countable topology? Thanks.
I don't think seound-countability is sufficient: take any space with the trivial topology, or a countable space with the cofinite topology. Otherwise, a weaker second countable topology is always given by $\lbrace\varnothing, X\rbrace$. The topology obviously has to be Hausdorff, and usually to construct functions one uses that $X$ is normal, but I'm not sure it is necessary.
After thinking about it more, I think $X$ has a countable separating set if and only if $X$ admits a weaker topology that is Hausdorff, regular and second countable.
@PierrePC Ok, you're right we need the space to be Hausdorff as well, from "On convergence determining and separating classes of functions" by Blout and Kouritzin Normality would be needed as well to construct an initial separating set from which a countable one can be extracted. Thanks.
Having a countable separating family means that you have a continuous injection into $R^N$. Equivalently, you have a weaker topology that makes your space homeomorphic to a subset of $R^N$. Such subset has to be metrizable and separable. Perhaps Perre PC is talking about something similar
@erz Absolutely. By Urysohn's metrisation theorem, a Hausdorff space is second countable regular if and only if it is separable metrisable.
With the help of the comments by erz, I will prove the following fact:
$(X,\tau)$ admits a countable separating function set if and only if there exists a weaker topology $\tau^*\subset\tau$ such that $(X,\tau^*)$ is Hausdorff regular (i.e. $T_3$) and second countable.
Comments
Let me first make a few comments.
Regular second countable spaces are completely normal, so is it equivalent that $\tau^*$ is Hausdorff second countable completely normal (i.e. $T_5$).
In terms of open sets of $\tau$, the condition can be rephrased as such: there exists a collection of open sets $U_i$, $i\in I$ such that
(base) for every $x\in U_i\cap U_j$, there exists $k$ such that $x\in U_k$ and $U_k\subset U_i\cap U_j$
(Hausdorff) it separates points, i.e. for each pair $x\neq y$ there are disjoints sets $U_i$, $U_j$ such that $x\in U_i$ and $y\in U_j$;
(regular) for each $x\in U_i$, there exists $j$ such that $x\in U
_j$ and for all $y\in U_i^\complement$, $y\in U_k\subset U_j^\complement$ for some $k=k(y)$ (think $\overline{U_j}\subset U_i$, but take the closure with respect to $\tau^*$).
(second countable) $I$ is countable.
Indeed, if such a family exists, then the topology it generates gives a suitable $\tau^*$, and if a Hausdorff regular second countable $\tau^*$ exists, any of its countable bases gives a suitable $U_i$.
Urysohn's metrisation theorem asserts that a Hausdorff regular second countable space is metrisable. In particular, it means that a Hausdorff space is regular second countable if and only if it is metrisable separable. In other words, a space $(X,\tau)$ admits a countable separating function set if and only if there exists a weaker $\tau^*$ that is metrisable separable, i.e. it admits a distance $d$ such that the associated open balls are open in $\tau$ and there exists a countable subset of $X$ that intersects every open ball.
Proof (open sets)
$(\Rightarrow)$ For the direct implication, suppose that we are given a countable $D\subset C(X)$ that separates points. Then we can define the family $\mathcal V$ of open sets of the form $f^{-1}(a,b)$, for $f\in D$ and $a,b\in\mathbb Q$, and the family $\mathcal U$ of finite intersections of elements of $\mathcal V$. Let us show that the topology $\tau^*\subset\tau$ generated by $\mathcal U$ is Hausdorff regular second countable. As discussed above, we can reduce the proof to statements about $\mathcal U$.
(base) $\mathcal V$ is stable by finite intersection.
(Hausdorff) For a given pair $x\neq y$, because $D$ separates points, we have $f(x)\neq f(y)$ for some $f\in D$; without loss of generality, $a<f(x)<b<f(y)<c$ for some $a,b,c\in\mathbb Q$, and $f^{-1}(a,b)$, $f^{-1}(b,c)$ are disjoint sets in $\mathcal U$ containing respectively $x$ and $y$.
(regular) Let $U_1,\ldots,U_n$ be elements of $\mathcal V$, i.e. $U_i=f_i^{-1}(a_i,b_i)$, $f_i\in D$, $a_i,b_i\in\mathbb Q$. If $x$ belongs to the intersection $U$ of the $U_i$, then $a_i<f_i(x)<b_i$ and we can find $\alpha_i,\beta_i\in\mathbb Q$ such that $a_i<\alpha_i<f_i(x)<\beta_i<b_i$. Then the intersection $U'$ of the sets $U'_i:=f_i^{-1}(\alpha_i,\beta_i)$ contains $x$. Suppose $y$ is not in $U$, for instance $f_1(y)\geq b_1$. Then $y\in f_1^{-1}(\beta_1,M)\subset (U')^\complement$ for some $M\in\mathbb Q$ large enough. Other possibilities for $y$ are treated similarly.
(second countable) Elements of $\mathcal V$ are described by finite sequences of elements of $\mathcal U$, which in turn are described by elements of $D\times\mathbb Q\times\mathbb Q$.
$(\Leftarrow)$ In the other direction, let $\tau^*\subset\tau$ be a Hausdorff regular second countable topology on $X$, and $(U_n)_{n\geq0}$ a countable basis of $\tau^*$. For each $(n,m)$, choose if possible a continuous $f_{nm}:(X,\tau)\to\mathbb R$ such that $(f_{nm})_{|U_n}\equiv 0$, $(f_{nm})_{|U_m}\equiv 1$. If there is no such function, have $f_{nm}\equiv 1/2$. The set $D:=\lbrace f_{nm},n,m\in\mathbb N\rbrace$ is obviously countable; let us show that it separates points.
We work in $\tau^*$ in this paragraph. Choose any $x\neq y$ in $X$. Because $X$ is Hausdorff, there exist $U,V$ disjoint open sets such that $x\in U$ and $y\in V$. Because it is regular, we have $x\in U'\subset\overline{U'}\subset U$ for some open set $U'$, and similarly for $y$. Since $(U_n)_{n\geq0}$ is a basis, we find $n,m$ such that $x\in U_n\subset U'$ and $y\in U_m\subset V'$. It follows that the closures $\overline {U_n}$ and $\overline {U_m}$ are disjoint (they belong to $\overline{U'}\subset U$ and $\overline{V'}\subset V$ respectively). Since $X$ is normal (regular second countable spaces are completely normal hence normal), Urysohn's lemma shows that there exists some continuous function $f:(X,\tau^*)\to\mathbb R$ such that $f_{|\overline{U_n}}\equiv 0$ and $f_{|\overline{U_m}}\equiv 1$. But then $f:(X,\tau)\to\mathbb R$ is continuous, so $f_{nm}$ is not 1/2 but a function that is 0 (resp. 1) when restricted to $U_n$ (resp. $U_m$). In particular, $f_{nm}(x)=0\neq1=f_{nm}(y)$ for some $f_{nm}\in D$.
Proof (metric spaces)
As discussed, the condition on $(X,\tau)$ is equivalent to the existence of some separable metrisable $\tau^*\subset\tau$.
$(\Rightarrow)$ This elegant proof is due to erz. Let $D$ be a countable separating function set. There is an obvious continuous function $(X,\tau)\to\mathbb R^D$ that sends $x$ to the collection of $f(x)$ for $f\in D$. Let $\tau^*$ be the pulls back of the topology of $\mathbb R^D$. Because $D$ separates points, this map is injective, so $(X,\tau^*)$ has the topology of a subset of $\mathbb R^D$ (its image). Since second countability and metrisability are hereditary properties (a subset of a metric/second countable space is metric/second countable) and a separable metric space is second countable, it suffices to show that $\mathbb R^D$ is metrisable separable. This is well known: $d(x,y):=\sum_{k\geq0}\min(|y(f_k)-x(f_k)|,2^{-k})$, for $D=\lbrace f_k\rbrace_{k\geq0}$, is a metric generating the topology, and the set $\mathbb Q^{(D)}$ of rational sequences with finite support is countable dense.
$(\Leftarrow)$ Take $D=\lbrace y\mapsto d(x_n,y) \rbrace$, for $d$ a metic generating $\tau^*$ and $x_n$ a dense sequence with respect to $\tau^*$.
For fun
There is no explicit use of Urysohn's metrisation theorem in the proof above, but one can suspect it is lurking in the shadows. Indeed, the proof I know of this result goes as follows. Suppose $(X,\tau^*)$ is Hausdorff regular second countable. Construct a countable family $(f_n)_{n\geq0}$ of functions that separates points, by following the proof given above. Then $d(x,y):=\sum_{n\geq0}\min(|f(y)-f(x)|,2^{-k})$ is a distance inducing $\tau^*$.
|
2025-03-21T14:48:31.751287
| 2020-08-12T10:37:08 |
368950
|
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|
Stack Exchange
|
Weird claims and conclusions in "Introduction to Shape Optimization"
I'm trying to understand the notions of Euler and Hadamard derivatives of shape functionals. All the lecture notes and papers on this topic that I've found seem to build up on the books Shapes and Geometries and Introduction to Shape Optimization, which are both co-authored by Zolésio.
I've got a hard time trying to understand what they are doing. And if I'm not totally wrong (which is not unlikely) many things they are claiming don't make sense.
The basic idea should be to consider what happens to shape functions under an infinitesimal perturbation of the shape. So, it make sense to consider families $(T_t)_t$ of transformations $T_t$. But here starts the pain. The following excerpt is taken from section 2.9 of Introduction to Shape Optimization:
I don't even know where to start:
I heavily doubt that the conclusion in the line immediately below (2.74) is correct (and it's weird that they use $C([0,\epsilon))$ on the rhs of (2.74), since this is usually a space of real-valued functions)
How can (2.75) be well-defined if $t\mapsto T_t(x)$ is not even assumed to be differentiable for fixed $x$?
And even if we assume that both $t\mapsto T_t(x)$ and $t\mapsto T_t^{-1}(x)$ are $C^1$-differentiable (which they do in a few sections before), I don't think that we can conclude (2.76); neither as stated with $C(0,\epsilon,C^k(\overline D,\mathbb R^N)$ nor with $C^1(0,\epsilon,C^k(\overline D,\mathbb R^N)$ as they seem to assume later on.
Now let's take a look at the definition of the Euler derivative:
I don't know what the space $\mathcal D(\mathbb R^N,\mathbb R^N)$ is, since they haven't defined this space at any point. From the notation it seems to be a space of distributions, but from its usage this doesn't seem to be the case. It's not clear to me how their notion of "shape differentiable in direction $V$" depends on $k$ and I actually don't even understand why we need $V\in C(0,\epsilon;V^k(D))$. In fact, it should be sufficient to assume that $T_t$ is any family of $C^1$-diffeomorphisms on $\mathbb R^N$ for $t\in[0,\epsilon)$ with $T_0=\operatorname{id}_{\mathbb R^N}$, $[0,\tau)\ni t\mapsto T_t(x)$ is differentiable for $x\in\mathbb R^N$ and $V_t:=\frac\partial{\partial t}T_t\circ T_t^{-1}$ for $t\in[0,\tau)$.
I guess, in analogy to the Fréchet derivative on Banach spaces, one wants to obtain a bounded linear operator $V\mapsto{\rm d}J(\Omega;V)$ and that's why we need to take $V$ from a suitable function spaces. I've seen other sources taking $V$ from some kind of Lipschitz functions or to be independent of time and from some Sobolev space. I'm really lost at this point by these apparently conflicting definitions.
Is there any better reference on this topic? I don't want to dive to deep into this stuff. It's sufficient to me to have rigorous treatment of basic shape functionals given by basic domain and boundary integrals which may or may not depend on the shape itself.
I don't know about your main issues, but what I can tell you is that $\mathcal D$ usually stands for the space of smooth, compactly supported test-functions, while the (dual) space of distributions is usually denoted as $\mathcal D'$. So I guess in this context $\mathcal D^k(\mathbb R^N;\mathbb R^N)$ should stand for the space of compactly supported $C^k$ functions from $\mathbb R^N$ into $\mathbb R^N$. (This seems consistent with the use that the authors seem to make in your displayed excerpts of the paper)
@leomonsaingeon Thank you for your comment. Yeah, that's basically what I thought, but at other places they consider the velocities to belong to $C^k(\mathbb R^N,\mathbb R^N)$. Maybe they think on shape functionals given by integral functionals whose integrands are locally integrable and that's why they assume compact support. Oh, and they explicitly define the spaces $C_0^\infty(\Omega,\mathbb R^d)$ as the space of smooth functions which are compactly supported. Shouldn't they denote this space by $\mathcal D^\infty(\Omega,\mathbb R^d)$ or $\mathcal D(\Omega,\mathbb R^d)$? Confusing ...
I think this is the original paper: https://core.ac.uk/download/pdf/82336011.pdf
It does things the other way round (i.e. start with the vector field and define the 1-parameter family via solutions to the associated ODE with different initial conditions).
@DCM The content of this is paper is also in the book. But it does not explain, for example, why $V(t,x)$ is differentiable with resepect $x$ as claimed in (2.76).
@DCM better to also give a human-readable reference, in case that url breaks: M. C. Delfour, Structure of Shape Derivatives for Nonsmooth Domains, J. Funct. Analysis 104 (1992) 1-33 and here's a stable doi link: https://doi.org/10.1016/0022-1236(92)90087-Y
Not an answer, but too long for a comment. The general idea with this stuff seems to be to pair your family $\mathscr{D}$ of admissible domains with a(ny) suitable normed-space $\mathscr{V}$ of vector fields and then insist that the 'shape derivative' be the element of $\mathscr{V}^*$ such that
$$
J(\Omega+V) = J(\Omega) + J'(\Omega)V + o(\Vert V\Vert)
$$
as $\Vert V \Vert\to 0$ in $\mathscr{V}$ (where $\Omega+V$ is either $\{x+V(x):x\in \Omega\}$ or something similar). This seems like the minimal property which a 'derivative' should satisfy in an affine setting (domains are 'points', vector fields are 'vectors').
As for what 'suitable' means in this context will - I think - generally depends on what sort of regularity you want for the associated flow. It's common to choose $\mathscr{V}$ so that its elements are Lipschitz continuous because then you can apply the Picard–Lindelöf theorem to associate a unique $C^1$ path germ with every point of $\Omega$.
The one-parameter flow seems like a bit of a distraction in all this - choose $\mathscr{V}$ right and you'll get the properties you want from $(T_t)_{t>0}$ from an appropriate ODE existence theorem.
Note: I've gone for the 'full' (Frechet-like) shape derivative above, you could also work just in terms of directional derivatives $\nabla_VJ(.):\mathscr{D}\to \mathbb{R}$ defined by requiring that
$$
J(\Omega+tV) =J(\Omega)+t(\nabla_VJ)(\Omega)+o(t) \;\;\mbox{as $t\to 0$},
$$
Either way, I think it's best to choose $\mathscr{D}$ and $\mathscr{V}$ based on where you want to go, and define 'derivatives' in terms of their essential property of being the linear bit of a first-order Taylor expansion (and not get too hung up on the setup used in any particular book).
This is all just my opinion of course :)
One other constraint to think about if you prefer to obtain the shape derivative via quotients of the form $(J(T_t(\Omega))-J(\Omega))/t$ is that you need $T_t(\Omega)\in \mathscr{D}$ for all $\Omega\in \mathscr{D}$; I expect this is probably the main reason for insisting on spatial $C^{k,\alpha}$ regularity when $\mathscr{D}$ is (as I expect it usually will be) some family of $C^{k,\alpha}$ domains.
Thank you for your answer. The picture is getting clearer now. However, I'm still having trouble to understand, for example, how the authors conclude that the transformations are $C^\alpha$-differentiable, when the velocity is $C^\alpha$-differentiable. I'm only able to prove that the transformations are $C^{\alpha-1}$-differentiable, when the velocity is $C^\alpha$-differentiable, but I guess this can be fixed. Maybe you can take a look at the question I've asked for that: https://mathoverflow.net/q/369299/91890.
|
2025-03-21T14:48:31.751783
| 2020-08-12T11:18:53 |
368951
|
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|
Stack Exchange
|
On sums of powers
I was considering the Fermat Catalan conjecture, where the equation $a^m+b^n=c^k$ has only finitely many nontrivial solutions (with coprime $a, b, c$) with $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}=1$ (and it is generally believed that the ten known solutions are the only ones). If the sum of the reciprocals if above one, there are families of parametric solutions, and if the sum equals precisely 1, there is only the Catalan solution $2^3+1^6=3^2$
This leads to the obvious (to me) question of what happens when there are four powers? More specifically, my questions are:
Are there always parametric solutions to $a^m\pm b^n\pm c^k=d^l$ if $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}>1$?
If $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}>1$, are solutions generally rare/hard to find?
In the thirteen cases where $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}=1$, are there known solutions?
Certain modular restrictions might force two of the powers to share a prime factor, so I will relax the coprimality constraint to say that the greatest common factor of all four powers is 1.
I am aware of Noam Elkies' solution to $w^4+x^4+y^4=z^4$, which showed that there are infinitely many solutions, and in fact, the solutions are "dense". However, it also seemed to also show that finding solutions to the cases I'm considering is hard, given that the smallest solutions are quite large, and that Euler himself considered the problem and found no solutions. Also, the analogous problem for fifth (and higher) powers is unsolved, leading me to conclude that the sum of reciprocals heuristic is still potentially valid for four powers.
One equation in particular I would like to look into is $w^6+x^6+y^6=z^2$
Edit: It seems like for various values of $(m,n,k,l)$, there are easily derivable solutions which contradict the idea of using the heuristic in full generality. Also, I consider a solution trivial if either one of the powers or a sum of two powers is zero. So I'm going to write out each configuration where the reciprocals sum to one and what is known so far:
$(2,3,7,42):$ Reduces to $x^2+y^3+z^7=1$. Unknown if a parametric solution exists
$(2,3,8,24):$ Reduces to $x^2+y^3+z^8=1$. Unknown if a parametric solution exists
$(2,3,9,18):$ Reduces to $x^2+y^3+z^9=1$. Unknown if a parametric solution exists
$(2,3,10,15):$ Unknown if a parametric solution exists
$(2,3,12,12):$ Reduces to $x^2+y^3+z^{12}=1$. Unknown if a parametric solution exists
$(2,4,5,20):$ Reduces to $x^2+y^4+z^5=1$. Unknown if a parametric solution exists
$(2,4,6,12):$ Reduces to $x^2+y^4+z^6=1$. Unknown if a parametric solution exists
$(2,4,8,8):$ Reduces to $x^2+y^4+z^8=1$. I believe a parametric solution may be derived here from Elkies' work
$(2,5,5,10):$ Reduces to $x^2+y^5+z^5=1$. Unknown if a parametric solution exists
$(2,6,6,6):$ Unknown if any nontrivial solutions exist
$(3,3,6,6):$ Unknown if any nontrivial solutions exist
$(3,4,4,6):$ Parametric solution derived in Max's answer
$(4,4,4,4):$ Elkies has derived infinitely many solutions
It seems like there are two general categories these fall into. The first is equations of the form $x^2+y^3+z^n=1$ or $x^2+y^4+z^n=1$. I feel like these are relatively easy to solve using elliptic curves (although coprimality could be an issue).
The second broad class is the remaining cases of $(2,5,5,10),(2,6,6,6),(3,3,6,6), (3,4,4,6)$, and $(4,4,4,4)$.
I believe that the reason the $(3,4,4,6)$ case was so easy to solve (at least, in the form $(4,3,6,4)$) is that the exponents are coprime. Also, the difference $(a+b)^4-(a-b)^4$ splits very nicely into the two terms $8a^3b$ and $8ab^3$, which aren't difficult to set equal to very high powers of numbers, which Max and Will exploited.
The case I am most interested in now is the $(2,6,6,6)$ case, because it seems to be the most immune to quick tricks involving high common factors.
Update: By considering the elliptic curve $x^3+z^6+1=y^2$, and the trivial solutions $(x,y)=(-z^2,1), (-1,z^3)$, and finding the third solution along the line formed by those two, I have discovered the equation:
$(2z^4+4z^3+5z^2+4z+2)^3+(z^2+z)^6+(z+1)^6=(3z^6+9z^5+15z^4+17z^3+15z^2+9z+3)^2$.
This can provide a solution to the equation $a^6+b^6+c^6=d^2$ if $2z^4+4z^3+5z^2+4z+2$ is a square
Therefore this boils down to the elliptic curve $y^2=2x^4+4x^3+5x^2+4x+2$. This has a rational point (-1,1), however this doesn't correspond to a non-trivial solution of the original equation. Are there other rational points on this curve?
What about $a^m+b^n=c^k+d^l$? Is that for some reason less interesting, or perhaps harder?
Rational solutions of $z^2 = w^6 + x^6 +y^6$ gives rational points on the double cover of $\mathbb P^2$ branched at the Fermat sextic $w^6+x^6+y^6=0$. This is a K3 surface, like the case Noam Elkies solved, and one could start looking for rational and elliptic curves on it. I think the Picard rank is reasonably high, which is a good start.
@YaakovBaruch it is also interesting, and worth considering. However, it seems that, at least in the (4,4,4,4) case, your configuration is much easier to solve.
@Thomas: A. Bremner and M. Ulas proved that equation (2,6,6,6) has infinitely many integer solutions.
@Tomita can you provide a link to the relevant paper?
@Thomas: A. Bremner, M. Ulas, ON $x^{a} ± y^{b} ± z^{c} ± w^{d}= 0, 1/a + 1/b + 1/c + 1/d = 1$
International Journal of Number Theory Vol. 7, No. 8 (2011) 2081-2090
@Tomita any reference not behind a pay wall?
I'm afraid I don't know.
They also gave parametric solutions for equation (2,3,7,42),(2,3,8,24),(2,3,9,18),(2,3,12,12),(3,3,6,6).
You started your question with the equation $a^m+b^n+c^k=d^l$, but then you seem to not care about the signs. What was $(4,6,3,4)$ in my answer turned into $(3,4,4,6)$ in your update. How did you swap the 6th and 4th powers between the left- and right-hand sides?
Good point. I don't care too much about the signs. Although, it would be interesting to try to find solutions for all permutations
Andrew Bremner, Maciej Ulas, On certain diophantine equations of diagonal type, is freely available at https://arxiv.org/abs/1311.0717
Note also discussion at https://math.stackexchange.com/questions/2767537/the-diophantine-equation-x-16x-26y6-z2-where-both-x-i-equiv-0-pmod-7
@Thomas: Then it's worth to update the question accordingly to avoid any confusion.
The heurisitc is shady. For example, there are infinitely many coprime solutions for $(m,n,k,l)=(4,6,30,4)$ (where the sum of reciprocals < 1) and for $(m,n,k,l)=(4,6,3,4)$ (where the sum of reciprocals = 1) since
$$(3^{90t+30} - 2^{30t+9})^4 + ( 2^{5t+2}\cdot 3^{45t+15} )^6 + (6^{3t+1})^{30} = (3^{90t+30} + 2^{30t+9})^4$$
for any integer $t$.
Perhaps, focusing on pairwise coprime solutions can save from this kind of examples.
One can simplify and generalize your identity to $ ( a^{30} - 2^9 b^{30})^4 + ( 2^2 a^{15} b^5)^6 + ( 2 a b^3)^{30} = ( a^{30} + 2^9 b^{30})^4$ for $a$ odd and coprime to $b$.
Even more generally $(a^{ (n+3) (3n+3) } - 2^n b^{ (n+3) (3n+3) })^4 + ( 2 a^{ 3 (3n+3) } b^{ 3n+3} )^{n+3} + (2 a^{ n+3} b^{ 3(n+3) } )^{3n+3} = (a^{ (n+3) (3n+3) } + 2^n b^{ (n+3) (3n+3) })^4 $ gives quadruples $(4, n+3, 3n+3, 4)$, i.e. with sums of reciprocals arbitrarily close to $\frac{1}{2}$. I guess $a^n + (-a)^n + b^m = b^m$ for $n$ odd gets sum of reciprocals arbitrarily close to $0$.
@Will: Indeed, there are many variations of this theme. And allowing negative solutions is another Pandora's box.
@MaxAlekseyev your solution is very interesting. I had not considered that there might be simple equations such as that. In regards to pairwise coprimality, that might work in some cases, but modular restrictions come into effect in some cases, like (4,4,4,4), forcing at least two of the powers to be even for example. I don't see how to eliminate the solutions with high common factors without removing all solutions.
@WillSawin I quite like your general solution, taking advantage of the difference of two fourth powers. I shall have to think about how those fit in with what I had originally thought. As for $a^n+(-a)^n+b^m=b^m$, I consider that to be trivial, as two of the powers sum to zero.
@Thomas One can ask for Zariski dense solutions, or pairwise coprime as Max suggested.
|
2025-03-21T14:48:31.752293
| 2020-08-12T11:26:37 |
368954
|
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|
Stack Exchange
|
A class of simplicial complexes defined by arithmetic properties
The purpose of my question is to ask about properties in a certain class of 3-dimensional (and other odd dimensional) simplicial complexes. I will first describe the construction in 3 dimension and then in general odd dimensions.
Let $n,g$ and $h$ be integers. Consider the pure simplicial complex $K(n;g,h)$ on the vertex set $[n]=\{1,2,\dots,n\}$ whose facets are described by quadruples $\{a,b,c,d\}$ such that $$b-a=g ({\rm mod}~ n)$$ and $$d-c=h ({\rm mod}~ n)$$.
Let us also consider a variant $L(n;g,h)$ where we further assume that the ordering of $\{a,b,c,d\}$ is cyclic. (Namely, $a<b<c<d$ or $b<c<d<a$ or $c<d<a<b$ or $d<a<b<c$.)
Question 1:
What can be said about the combinatorics and topology of simplicial complexes of the form $K(n;g,h)$ and $L(n;g,h)$.
Question 2:
The construction can be extended to give similar $2k-1$-dimensional simplicial complexes $K(n;g_1,g_2,\dots , g_k)$ and $L(n;g_1,g_2,\dots , g_k)$. What can be said about the combinatorics and topology of these simplicial complexes
Question 3:
Are there nice extensions to odd dimensions?
Motivation
The constructions are motivated by two classes of constructions.
A) When all the $g_1$s are one this is precisely the construction of the cyclic even-dimensional polytopes.
Like them it looks that the complexes considered above (especially the $K's$ have some chance to be pseudomanifolds. Are they ever manifolds? Is there a way to give an arithmetic definition of this kind to notable odd dimensional triangulations like the 6-vertex RP^2 or Kuhnel's CP_2.
B) There is a remarkable simple arithmetic constructions of acyclic complexes by Linial, Meshulam and Rosenthal (see this paper and this post). In dimension 2 you consider all triples $a,b,c$ modulo $n$ ($n$ a prime) such that $a+b+c$ equals one of three numbers $x,y,z$.
I've tried writing some Sage code to construct these, at least in dimension 3, so I can ask for the homology, or the fundamental group, or which ones are pseudo-manifolds, among other things. Are there any experiments you would like me to run?
@JohnPalmieri Dear John, this is great! I guess I wonder if you get pseudomanifolds this way. I think there ate too many 3-faces (at least for K) that it cannot be a manifold. and of course homology (and f.g.) is very interesting.
This looks like it might resemble the complexes considered in
MR2844711 Catanzaro, Michael J. Generalized Tonnetze.
J. Math. Music 5 (2011), no. 2, 117--139.
He considers 2-dimensional simplicial complexes $L(n_1,n_2,n_3)$ with vertex set $\{1,\ldots,N\}$ and $n_1+n_2+n_3 = N$, whose 2-simplices are $\{a, a+n_1, a+n_1+n_2\}$. These parameterize triads of `shape' $(n_1,n_2,n_3)$ in an $N$ note scale, generalizing the diatonic $N=12$ major and minor triads, $(n_1,n_2,n_3) = (4,3,5)$.
|
2025-03-21T14:48:31.752980
| 2020-08-12T12:47:16 |
368956
|
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"Bastiaan Cnossen",
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|
Stack Exchange
|
Defining (infinity,1)-categories in HoTT using only an interval type
In this article, Emily Riehl and Michael Shulman describe a type theory in which one can do $\infty$-category theory synthetically. Their framework allows them to define simplices $\Delta^n$, and a morphism in a type $A$ is simply a map $\Delta^1 \to A$. Any map $H: \Delta^2 \to A$ witnesses its 'bottom edge' $d_1(H): \Delta^1 \to A$ as a composite of 'top edges' $d_2(H)$ and $d_0(H)$. We can then think of the type $A$ as a (higher) category if it is a Segal type, which is a type in which every two composable morphisms have a contractible choice of composites.
After reading the article, I had the following two questions:
The type theory introduced in the article is a lot more involved than the type theory from the HoTT book. It uses several layers of type theory, using so-called cubes, topes and shapes. To what extent are these extra layers necessary? It would seem to me that one could develop this whole theory in the setting of the HoTT book, with only an additional directed interval type $\mathbb{I}$ (some thoughs on this below.) Does this approach make sense? Has it been worked out by someone already? What are the pros/cons for either approach?
(Perhaps this is already contained in the recent work on the cubical approach to Homotopy Type Theory, with which, I should say, I am not yet really familiar...)
For some types, like the type of groups, we already have a natural notion of morphism around. How can we relate this notion of morphism to the abstract notion of a morphism defined via maps out of $\Delta^1$? Does it make sense to add an axiom about the universe $\mathcal{U}$ saying that for types $A,B:\mathcal{U}$, we have an equivalence
$$
A \to B \simeq \text{hom}_{\mathcal{U}}(A,B)
$$
between the function type $A \to B$ and the morphisms in $\mathcal{U}$ from $A$ to $B$? In this case, how do we make sure that maps $\Delta^2 \to \mathcal{U}$ actually correspond to (homotopy) commutative diagrams? Once you have both of these things, I think it should follow that for example $\text{hom}_{Grp}(G,H)$ is precisely the type of group homomorphisms from $G$ to $H$ as given in the HoTT book.
Some thoughts on synthetic category theory with just an interval
Let me spell out what I had in mind in point 1. Say that instead of these cubes/topes/shapes we only include an interval type $\mathbb{I}$ with constructors $0,1:\mathbb{I}$ and $\lor,\land: \mathbb{I} \to (\mathbb{I} \to \mathbb{I})$, satisfying the axioms of a distributive lattice. (We don't want an inverse $\neg: \mathbb{I} \to \mathbb{I}$, since not all morphisms should be invertible.) It seems that with some modification, one can repeat most of the constructions of Riehl and Shulman in this simple setting. My suggestions:
a morphism $f: \text{hom}_A(a,b)$ in a type $A$ from $a:A$ to $b:A$ is a map $f: \mathbb{I} \to A$ with identifications $p_0: f(0) = a$ and $p_1: f(1) = b$.
the identity $\text{id}_a: \text{hom}_A(a,a)$ on $a:A$ is the constant map $\text{const}_a: \mathbb{I} \to A$ (with twice $\text{refl}_a$);
for $a:A$, the 'under-category' type $a/A$ is the type of morphisms $f: \mathbb{I} \to A$ with $p_0: f(0) = a$.
a map $F: B \to A$ is a covariant fibration if the induced map
$$
F_*: b/B \to F(b)/A
$$
is an equivalence for any $b:B$, i.e. any morphism $f: \text{hom}_A(F(b),a)$ lifts to a 'unique' morphism $\tilde{f}: \text{hom}_B(b,b')$ starting in $b:B$ and projecting to $f$ under $F$. This implies that any morphism $f: \text{hom}_A(a,b)$ induces a map on fibers
$$
f_*: \text{fib}_F(a) \to \text{fib}_F(b),
$$
so the fibers vary 'covariantly' in $A$.
I call a type $A$ is covariant if the 'target map' $t: a/A \to A$ is a covariant fibration for all $a:A$. Since the fiber of $t$ over $b:A$ is $\text{hom}_A(a,b)$, every $g: \text{hom}_A(b,c)$ induces a 'post-composition' map
$$
g_*: \text{hom}_A(a,b) \to \text{hom}_A(a,c).
$$
One now checks that $t: a/A \to A$ is a fibration if and only if every square
$$
\require{AMScd}
\begin{CD}
b @>{g}>> c\\
@AfAA @AAA \\
a @= a
\end{CD}
$$
has a contractible type of 'solutions' $H: \mathbb{I} \times \mathbb{I} \to A$ with $H(0,-) = f$, $H(-,0) = \text{id}_a$ and $H(-,1) = g$. Restriction to the right vertical edge corresponds to $g_*(f): \text{hom}_A(a,c)$.
It follows that for any covariant type, defining $g \circ f :\equiv g_*(f): \text{hom}_A(a,c)$ gives a composition of morphisms that is unital and associative: we have $(\text{id}_b)_*(f) = f$ and $f_*(\text{id}_a) = f$ by using the squares $\lambda (s,t). f(t): \mathbb{I} \times \mathbb{I} \to A$ and $\lambda (s,t). f(s \land t): \mathbb{I} \times \mathbb{I} \to A$:
$$
\require{AMScd}
\begin{CD}
b @= b\\
@AfAA @AAfA \\
a @= a
\end{CD}
\hspace{50pt}
\begin{CD}
a @>{f}>> b\\
@| @AAfA \\
a @= a,
\end{CD}
$$
and for associativity one proves (as in Corollary 5.6 of R+S) that the function type $\mathbb{I} \to A$ is again a covariant type for which we can apply the above square-filling (or 'cube-filling') criterion to the two morphisms (i.e. squares)
$$
\require{AMScd}
\begin{CD}
b @>{g}>> c\\
@AfAA @AA{g\circ f}A \\
a @= a
\end{CD}
\hspace{30pt} \text{ and } \hspace{30pt}
\begin{CD}
c @>{h}>> d\\
@AgAA @AA{h \circ g}A \\
b @= b,
\end{CD}
$$
and the resulting cube gives a proof of $(h \circ g) \circ f = h \circ (g \circ f)$.
One can dualize everything above, defining contravariant fibrations $F: B \to A$ using 'over-category' types $A/a$ and call $A$ a contravariant type if $s: A/a \to A$ is a contravariant fibration for all $a:A$. One can now define a second composition by considering for $f: \text{hom}_A(a,b)$ the induced map
$$
f^*: \text{hom}_A(b,c) \to \text{hom}_A(a,c).
$$
The composition $g \circ_2 f :\equiv f^*(g)$ is again unital and associative.
I would like to define $A$ to be a category if $A$ is both a covariant and a contravariant type, i.e. the type family $\text{hom}_A(a,b)$ is both contravariant in $a$ and covariant in $b$. In this case the two compositions actually agree. To see this, one uses that there is a correspondence between squares of the form
$$
\require{AMScd}
\begin{CD}
b @>{g}>> c\\
@AfAA @AAA \\
a @= a
\end{CD}
\hspace{30pt} \text{ and } \hspace{30pt}
\begin{CD}
b @>{g}>> c\\
@AfAA @| \\
a @>>> c
\end{CD}.
$$
For example the map from right to left is given by sending the map $H: \mathbb{I} \times \mathbb{I} \to A$ to the map $\lambda (s,t). H(s \wedge t,t)$ that basically before applying $H$ first smashes the square onto the its upper left triangle with the right-bottom corner going to the left-bottom corner. These maps are fiberwise maps over the projection to $(\mathbb{I} \to A) \times_A (\mathbb{I} \to A)$ that only remembers $f$ and $g$, so if $A$ is both covariant and contravariant, they are automatically inverse equivalences (as then for each $f$ and $g$ there is an essentially unique such square). Furthermore both maps preserve the diagonal, and since going back and forth gives you another square that restricts to $f$ and $g$, this diagonal must be equal to either of the two composites.
I haven't checked this in detail, but it seems to me that the Yoneda lemma
$$
C(a) \simeq \prod_{x:A} (\text{hom}_A(a,x) \to C(x))
$$
should go through fine by just literally copying what Riehl and Shulman do.
I think you are over estimating the complexcity of this (shape,tope) layer and so one. At the end of the day "all they do" is to add an object $I$ endowed with an order relation making $I$ an internal (i.e. it has top and bottom elements $0$ and $1$ and $x \leqslant y $ or $y \leqslant x$ for all $x,y$) exactly in the same way that you propose to add an object $I$. The reason why they single-out this type $I$ in a different layer of type theory is in order do define the notion of "extention type", which allows to actually define $hom_A(x,y)$ without using identity type.
Yeah I think to my unexperienced eye it seems like a lot of complexity for turning some propositional equalities into judgmental equalities. What would happen if one introduces $\text{hom}_A(x,y)$ as a formal type family, satisfying the rule that every $f: \mathbb{I} \to A$ gives a term $f': \text{hom}_A(f(0),f(1))$ and conversely every $f': \text{hom}_A(x,y)$ gives a map $f: \mathbb{I} \to A$ with judgmental equalities $f(0) \equiv a$ and $f(1)\equiv b$? Where does one run into problems?
So what you are describing is basically the definition of an extention type. Now, the thing is we don't want to be able to consider just the extention type corresponding to $hom_A(x,y)$, i.e. to the boundary inclusion $\Delta[0] \coprod \Delta[0] \to \Delta[1]$, but also these corresponding to more general simplicial maps. For e.g. the extention type along $\Lambda^1[2] \to \Delta[2]$ give you the type of arrows which are a composite of $f,g$ (a type which is contractible in a Segal type, but maybe not in general).
And there are an infinite number of maps you might want to consider extention type along. The role of the first two layer of their type theory is to determine which maps you are allowed to consider extention type along (basically, all inclusion between finite simplicial sets).... Of course I'm not saying that these extention type are neccessary, maybe what you are proposing can also works and extention type are just things that are there to make our lives easier.
The shape/tope type theory is indeed just a "convenience". When I first suggested this approach to synthetic $(\infty,1)$-categories, I took the approach you describe with a simple axiomatic interval. But the coherence paths very quickly became much too complicated to deal with in practice; the shape/tope theory is just to make the endpoint equalities judgmental so that the coherence problems are manageable. We discussed this in the introduction of the paper:
In principle, all of the above theory could be developed within ordinary homotopy type theory,
simply by axiomatically assuming the type 2 and its strict interval structure...These equalities are then data, which have to be carried around everywhere. This is quite tedious,
and the technicalities become nearly insurmountable when we come to define commutative
triangles, let alone commutative tetrahedra.
For your second question, yes it makes sense to add some such axiom, but not about the universe of all types; you need to use a smaller "covariant" universe. The simplest one is a classifying type for covariant fibrations, which semantically represents the $(\infty,1)$-category of $\infty$-groupoids; you can also consider a classifying type for coCartesian fibrations, which would semantically represent the $(\infty,1)$-category of $(\infty,1)$-categories. The universe of all types is neither of these; even the internally defined "universe of Segal types" or "universe of discrete types" doesn't end up semantically with functions as morphisms, rather some kind of span. Various people are working on stating and modeling such axioms; the keyword to search for is directed univalence.
|
2025-03-21T14:48:31.753708
| 2020-08-12T12:53:49 |
368957
|
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|
Stack Exchange
|
Is there a known asymptotic for $A(X):= \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)}$?
My guess is that there exists a constant $C$ such that $A(X) \sim C (\log X)^2$.
See also Theorem 7.1 in https://cs.uwaterloo.ca/journals/JIS/VOL10/Bordelles2/bordelles61.pdf
For refinements and generalizations, see Section 2 of https://arxiv.org/pdf/1805.10877.pdf. I think that "the plausible conjecture (2.7)" in this paper is not too hard, e.g. a detailed analysis of the sum twisted by $\mu(n_1)\cdots\mu(n_k)$ appears in https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/132/3/83091/etude-d-une-somme-arithmetique-multiple-liee-a-la-fonction-de-mobius
Similar questions https://mathoverflow.net/q/356545/5712 and https://mathoverflow.net/q/33600/5712
$$ \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)} = \sum_{1 \leq i,j \leq X} \frac{\mathrm{gcd}(i,j)}{ij} $$
$$ = \sum_{1 \leq i,j \leq X} \frac{\sum_{d|i,j} \phi(d)}{ij}$$
$$ = \sum_{d \leq X} \phi(d) \sum_{1 \leq i,j \leq X: d|i,j} \frac{1}{ij}$$
$$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} \sum_{1 \leq i',j' \leq X/d} \frac{1}{i'j'}$$
$$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} ( \sum_{1 \leq i \leq X/d} \frac{1}{i})^2$$
$$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} ( \log(X/d) + O(1))^2$$
$$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} ( \log^2(X) - 2 \log(d) \log(X) + \log^2(d) + O( \log X ) )$$
$$ = A_0 \log^2 X - 2 A_1 \log X + A_2 + O( \log^2 X )$$
where
$$ A_j := \sum_{d \leq X} \frac{\phi(d) \log^j d}{d^2}.$$
One can compute asymptotics for the $A_j$ by Perron's formula, but we proceed instead by elementary means. Since $\phi(d) = \sum_{d=ab} \mu(a) b$ we have
$$ A_j = \sum_{ab \leq X} \frac{\mu(a) b \log^j(ab)}{a^2b^2}$$
$$ = \sum_{a \leq X} \frac{\mu(a)}{a^2} (\frac{\log^{j+1}(X)}{j+1} + O( (1 + \log^j(a)) \log^j X ) )$$
$$ = \frac{1}{j+1} \log^{j+1} X \sum_{a \leq X} \frac{\mu(a)}{a^2} + O(\log^j X)$$
$$ = \frac{1}{(j+1)\zeta(2)} \log^{j+1} X + O(\log^j X).$$
Hence
$$ \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)} = \frac{1}{3 \zeta(2)} \log^3 X + O(\log^2 X).$$
Your estimate for $A_0$ leads to division by $0$. I think you mean to have $A_j = \frac{1}{(j+1)\zeta(2)} \log^{j+1} X + O(\log^j X)$.
Corrected, thanks - T.
This result also appears (with a secondary main term) in Theorem 7.1 of https://cs.uwaterloo.ca/journals/JIS/VOL10/Bordelles2/bordelles61.pdf
If we are talking about "elementary", then just multiply the original sum by the sum of inverse squares and note that if we have three numbers $a=a'd, b=b'd, n^2$ where $(a',b')=1$, $d,n$ are arbitrary, then $A=a'n, B=b'n, d$ are arbitrary $3$ integers with the product $LCM(a,b)n^2$. The original triple sum has the bounds $a'd, b'd\le X$, $n$ formally unrestricted, but restricting it to $[1,N]$ for fixed $N$ changes the triple sum $1+O(1/N)$ times. But then we can squeeze the bounds on $A,B,d$ between $Ad\le X, Bd\le X$ (treating the implicit $n=(A,B)$ as unrestricted) to get the lower bound and $Ad\le NX, Bd\le NX$ (assuming $n\le N$ now) to get the upper bound, so we get an answer for the triple sum of $\frac 1{ABd}$ between roughly speaking $\frac13\log^3 X$ and $\frac13\log^3(XN)$ with additive error $O(\log^2(XN))$ (no number theory in this sum!), which have the same asymptotic up to $1+O(\frac{\log N}{\log X})$. Taking $N$ about $\sqrt{\log X}$, we get the total multiplicative error $1+O(\frac{1}{\sqrt{\log X}})$, which is, of course, suboptimal but who cares. :-)
|
2025-03-21T14:48:31.753947
| 2020-08-12T12:54:30 |
368958
|
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|
Stack Exchange
|
discontinuous functions on the Sobolev borderline
The Sobolev embedding theorem implies that every function of class $W^{k,p}$ on a reasonable $n$-dimensional domain is continuous if $kp > n$. Cases with $kp=n$ are known as "borderline" cases. In my question I'm going to focus on the case $p=2$ for functions on either ${\mathbb R}^n$ or ${\mathbb T}^n$, so that the Sobolev spaces $H^k = W^{k,2}$ have a nice description in terms of Fourier transforms or series, but answers concerning more general Sobolev spaces are also welcome.
It seems to be tricky to find concrete examples of discontinuous functions that are Sobolev borderline cases. Some searching turned up an example in $H^1({\mathbb R}^2)$, but I have been unable to find a "simple" example in what I intuitively expect to be the easiest case, namely $H^{1/2}(S^1)$, and I was surprised that none of the textbooks I could think to search in give one. My first impulse was to try classic discontinuous functions like the square wave and sawtooth whose Fourier series are easy to compute: these just miss the mark, as they turn out to be in $H^s(S^1)$ for all $s < 1/2$ but not for $s=1/2$. The one thing I have tried that worked was writing down an explicit Fourier series like
$$
f(x) := \sum_{k=2}^\infty \frac{e^{2\pi i k x}}{k \ln k},
\qquad \text{ (here $x \in S^1 := {\mathbb R} / {\mathbb Z}$)}
$$
which one can easily check is in $H^{1/2}(S^1)$, and one can then use summation by parts to estimate $\sum_{k=N}^\infty \frac{e^{2\pi i kx}}{k \ln k}$ for large $N$ and small $|x|$ and thus prove $\lim_{x \to 0} f(x) = \infty$. One can do something similar with a Fourier transform and integration by parts to find a function in $H^{1/2}({\mathbb R})$ that is continuous everywhere except at $x=0$, where it blows up. But this type of construction is a lot trickier than what I was hoping for; expressing a function as a conditionally convergent series or improper integral does not give me the feeling that I can get my hands on it.
So, first question: does anyone know a simpler example of something that is discontinuous and belongs to $H^{1/2}(S^1)$ or $H^{1/2}({\mathbb R})$? Or other interesting examples of Sobolev borderline functions that can be understood without having to search the exercises in Baby Rudin for hints?
Followup question, admittedly a little vague: if you don't know more concrete examples, is there any deep reason why they don't exist, i.e. why every function I can think to write down in a reasonable way turns out to fall short of the borderline case?
There are plenty of examples of discontinuous Sobolev function in $W^{1,n}(\mathbb{R}^n)$. For example $f(x)=\log|\log|x||$ defined in a neighborhood of zero.
Now take $n=2$ and restrict the function to the $x$-axis. You will get a discontinuous function in the trace space which is $H^{1/2}(\mathbb{R})$.
You can use this function to construct quite strange examples.
Taking $x\mapsto f(x-a)$ you can place singularity at any point $a$.
Modifying this example you can assume that $\Vert f\Vert_{1,n}<\epsilon$
and that the function has support in a small neighborhood of $a$.
If $\{a_i\}_i$ is a countable and dense subset of $\mathbb{R}^n$,
and $f_i$ is a function with the singularity as above at the point $a_i$ and $\Vert f_i\Vert_{1,n}<2^{-i}$, then the series
$$
f=\sum_{i=1}^\infty f_i
$$
converges to a function in $W^{1,n}$, because it is a Cauchy series in the norm and $W^{1,n}$ is a Banach space. The function $f$ will have singularities located on a dense subset of $\mathbb{R}^n$ and in particular the essential supremum of $f$ over any open set will be equal $+\infty$.
You can also take $\{a_i\}_i$ to be a dense subset in a subspace $\mathbb{R}^{n-1}$ of $\mathbb{R}^n$ and a similar construction will give you a function that is bad when restricted to that subspace. The trace belongs to $W^{1-1/n,n}(\mathbb{R}^{n-1})$. In particular if $n=1$ you get such a function in $W^{1/2,2}(\mathbb{R})=H^{1/2}(\mathbb{R})$.
|
2025-03-21T14:48:31.754234
| 2020-08-12T14:05:24 |
368963
|
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|
Stack Exchange
|
Intuitively, what does a graph Laplacian represent?
Recently I saw an MO post Algebraic graph invariant $\mu(G)$ which links Four-Color-Theorem with Schrödinger operators: further topological characterizations of graphs? that got me interested. It is about a graph parameter that is derived from the Laplacian of a graph. Its origins are in spectral operator theory, but it is quite strong in characterizing important properties of graphs. So I was quite fascinated by the link it creates between different branches of mathematics.
I went through other posts on MO that discuss this topic as well, and in the meantime I read a few linked articles that work with the graph Laplacian. I understand that they view an (undirected) graph as a metric graph embedded in a surface, and the metric on the graph is approximated by Riemannian metrics which give the edge distance along the edges, and which is close to zero everywhere else on the surface. The eigenvalues of the surface Laplacian approximate the eigenvalues of the graph Laplacian, and a lot of surprisingly useful conclusions follow, about connectivity and embeddability of the graph, and even about minor-monotonicity.
I have gained a technical understanding of what is happening and how these eigenvalues (and their multiplicity) are determined, using the graph Laplacian. I also have a basic understanding of the role of a Laplacian in differential geometry, like the Laplacian of a function $f$ at a point $x$ measures by how much the average value of $f$ over small spheres around $x$ deviates from $f(x)$, or I think of it to represent the flux density of the gradient flow of $f$.
But I am failing to gain or develop such an intuition for the graph Laplacian. Conceptually or intuitively, what does a graph Laplacian represent? I am trying to understand, how can it be so powerful when applied to graphs? (I am aware that the graph Laplacian can be defined using the graph adjacency matrix, but I was unable to link this with my differential geometry intuition)
The graph Laplacian also factors heavily in a more algebraic or algebro-geometric story where graphs are discrete analogs of algebraic curves/Riemann surfaces (see e.g. https://arxiv.org/abs/math/0608360).
The graph Laplacian is the flux density of the gradient flow of a graph (the flow on each edge being the difference between the values on the vertices).
@SamHopkins Thank you very much for this link! I am very interested and will read this article
@WillSawin Thank you for your comment! What I am struggling with, in the articles I was reading, no value was assigned to the vertices (if I understood correctly). So what does the flow actually represent? It seems to come "out of nowhere". Or maybe the vertices are just viewed as abstract sinks / or sources?
What do you mean no value was assigned to the vertices? The graph Laplacian is an operator from $\mathbb R^n$ to $\mathbb R^n$, $n$ the number of vertices, and $\mathbb R^n$ is exactly the space of functions assigning one value to each vertex.
@WillSawin this is a helpful comment! Thanks a lot
I posted an answer to a similar question there ... just in case it might be of interest
The coefficient of $x^k$ of the characteristic polynomial $det(xI + L)$ is the number of ways to take a subset of the edges of the directed graph such that the connected components form k directed trees.
How to understand the Graph Laplacian (3-steps recipe for the impatients)
read the answer here by Muni Pydi. This is essentially a concentrate of a comprehensive article, which is very nice and well-written (see here).
work through the example of Muni. In particular, forget temporarily about the adjacency matrix and use instead the incidence matrix.
Why? Because the incidence matrix shows the relation nodes-edges, and that in turn can be reinterpreted as coupling between vectors (the value at the nodes) and dual vectors (the values at the edges). See point 3 below.
now, after 1 and 2, think of this:
you know the Laplacian in $R^n$ or more generally in differential geometry.
The first step is to discretize: think of laying a regular grid on your manifold and discretize all operations ( derivatives become differences between adjacent points). Now you are already in the realm of graph laplacians. But not quite: the grid is a very special type of graph, for instance the degree of a node is always the same.
So you need to generalize a notch further: forget the underlying manifold, and DEFINE THE DERIVATIVES and the LAPLACIAN directly on the Graph.
If you do the above, you will see that the Laplacian on the Graph is just what you imagine it to be, the Divergence of the Gradient. Except that here the Gradient maps functions on the nodes to functions on the edges (via the discrete derivative , where every edge is a direction..) and the divergence maps the gradient back into a nodes function: the one which measures the value at a node with respect to its neighbors. So, nodes-edges-nodes, that is the way (that is why I said focus on the incidence matrix)
Hope it helps
Thank you for a fantastic answer! I wish I could give more than just +1 votes.
Not so fantastic, a lazy answer (I should have written more). Anyway, there is a FOURTH point, which will complete the journey. This one: above, the Laplacian is defined from the incidence matrix, whereas the standard definition employs the adjacency one. Fact: they are the same. Try to understand why and you become a :Laplacian Jedi . Good luck!
Your answer is great and really helps me with my intuition. I find this topic of graph Laplacians really amazing. Maybe the answer to your Yedi question is: the Laplacian is normally defined $I-A$ but $I-A=DD^T$ where $D$ is the incidence matrix and $DD^T$ corresponds exactly to your nodes-edges-nodes statement. But let me check the signs here I think I am getting something wrong with the signs.
This is just a long comment, adding to the excellent answers above.
There is a great article from László Lovász "Discrete and Continuous:
Two sides of the same?", written around 2000 (https://web.cs.elte.hu/~lovasz/telaviv.pdf) which might be of interest to you. In chapter 5 of this article, Lovász covers the graph Laplacian. He explains the relation to random walks on graphs and also the link to the Colin de Vérdière graph invariant which sparked your interest (your link in the OP).
In your OP, you are asking how can the graph Laplacian be so powerful when applied to graphs? I think two quotes from this article could be of special interest to you, because quote (1) relates to the "power" and quote (2) relates to where the "limitations" were in applying the graph Laplacian.
About the "power":
Quote (1)
"The Laplacian makes
sense in graph theory, and in fact it is a basic tool. Moreover, the study of the discrete and
continuous versions interact in a variety of ways, so that the use of one or the other is almost
a matter of convenience in some cases. (...) Colin de Verdière’s invariant created much interest among graph theorists, because of its
surprisingly nice graph-theoretic properties. (...) Moreover, planarity of graphs can be
characterized by this invariant: $\mu(G) \leq 3$ if and only if G is planar.
Colin de Verdière’s original proof of the “if” part of this fact was most unusual in graph
theory: basically, reversing the above procedure, he showed how to reconstruct a sphere and a
positive elliptic partial differential operator $P$ on it so that $\mu(G)$ is bounded by the dimension
of the null space of $P$, and then invoked a theorem of Cheng (...) asserting that this dimension is at most $3$.
About the "limitations":
Quote (2)
"Later Van der Holst (...) found a combinatorial proof of this fact [$\mu(G) \leq 3$ if and only if G is planar]. While this may seem as a step backward (after all, it eliminated the necessity of the only application of partial differential equations in graph theory I know of), it did open up the possibility of characterizing the next case. Verifying a conjecture of Robertson, Seymour, and Thomas, it was shown by Lovász and Schrijver (...) that $\mu(G) \leq 4$ if and only if G is linklessly embedable in $\mathbb R^3$."
I did not know this paper. Thank you for this reference. Very interested in these quotes, especially in the second one. I have to see this article, and the combinatorial proof of Van der Holst. Great additional perspective! Very grateful for that!
The link just got invalid. Here is the link from springer, https://link.springer.com/chapter/10.1007/978-3-0346-0422-2_13
This is not really about the connection with graph theory, a topic I am rather ignorant of, but rather the connection to continuum notions, all of which I learned from this paper.
Consider a simplicial complex in 3 dimensions for simplicity of visualization. The 0-simplexes are vertices $(i)$, the 1-simplexes are bonds $(ij)$, 2-simplexes are triangles $(ijk)$, 3-simplexes are tetrahedra $(ijkl)$. Each simplex has an orientation and under permutation of vertices acquires a sign change of +1 or -1 if the permutation is even or odd respectively.
Now we can define functions ($p$-chains) on our simplicial complex,
$$\phi = \sum_i \phi_i (i)$$
$$\alpha = \sum_{[ij]} \alpha_{ij} (ij)$$
$$\beta = \sum_{[ijk]} \beta_{ijk} (ijk)$$
$$\gamma = \sum_{[ijkl]} \gamma_{ijkl} (ijkl)$$
where the $\alpha_{ij}$ etc. are fully anti-symmetric and the sum is over equivalence classes of simplexes (i.e. we pick one representative for each simplex from its possible permutations).
Now we define a boundary operator $\partial_p$ on $p$-simplexes. On a 0-simplex, we have $\partial_0(i) = 0$. For a 1-simplex we have
$$\partial_1(ij) = (j) - (i)$$
and we generalize this,
$$\partial_p(i_0 \cdots i_{p-1}) = \sum_n (-1)^n (i_0 \cdots \hat{i}_n \cdots i_{p-1})$$
where the hat means that vertex is removed. This is equivalent to saying that the boundary of a $p$-simplex is the sum of the $p-1$-simplices which bound it, each oriented such that their "edges" are oppositely oriented. Thus for a triangle we find
$$\partial_2(ijk) = (jk) + (ki) + (ij)$$
while for a tetrahedron we have
$$\partial_3(ijkl) = (jkl) + (kli) + (lij) + (ijk)$$
This construction automatically satisfies $\partial_{p-1} \partial_{p} = 0$ due to the "oppositely oriented edges" condition above.
Next, define the coboundary operator $\partial_p^\dagger$ which takes $p$-chains to $p+1$-chains. The definition says
$$\partial_p^\dagger (i_1 \cdots i_{p}) = \sum_{i_0@[i_1 \cdots i_{p}]}
(i_0 \cdots i_{p})$$
where $@$ means "adjacent to". Thus for a 0-simplex,
$$\partial_0^\dagger (j) = \sum_{i@j} (ij)$$
Note that the sum is over oriented 1-simplices which "point towards $(j)$". For a 1-simplex $(ij)$, $\partial_1^\dagger(ij)$ is the sum is over all triangles $(i_0 i_1 i_2)$ such that $\partial_2(i_0 i_1 i_2)$ contains $+(ij)$, and so on. This operator also satisfies $ \partial_{p+1}^\dagger \partial_p^\dagger = 0$ by construction.
The boundary and co-boundary operators act on $p$-chains linearly. We can draw an analogy with differential geometry --- in particular, the co-boundary operator is analogous to the exterior derivative, and $p$-chains are akin to exterior $p$-forms. As shown in the above-linked paper, we can think of $0$-chains as scalar fields, $1$-chains as vector fields, $2$-chains as pseudo-vector fields, and $3$-chains as pseudo-scalar fields. The properties of the boundary operators are then summed up in this figure (their $d$ is my $\partial^\dagger$):
Note that the correspondence is not an approximation (see the text for details), although one can make a connection with the continuum differential operators via a Taylor-expansion approximation in the continuum limit as the lattice spacing goes to zero.
One can now define certain vector-product operations, demonstrate Stoke's theorem, etc. utilizing this construction. In particular, we can define the Laplacian for $p$-chains as
$$\Delta_p = - (\partial_{p+1}\partial_{p}^\dagger + \partial_{p-1}^\dagger \partial_p)$$
then from the figure we find the correspondence
$$\Delta_0 \sim \mathrm{div}\,\mathrm{grad} $$
$$\Delta_1 \sim \mathrm{grad}\,\mathrm{div} - \mathrm{curl}\,\mathrm{curl}$$
$$\Delta_2 \sim \mathrm{grad}\,\mathrm{div} - \mathrm{curl}\,\mathrm{curl}$$
$$\Delta_3 \sim \mathrm{div}\, \mathrm{grad}$$
In particular, $\Delta_0 = -\partial_1 \partial_0^\dagger$ is the usual graph Laplacian, and one can show (with appropriate choice of representatives in the summations above), that
$$\Delta_0 = A - D$$
where $A$ is the adjacency matrix and $D$ is the incidence matrix of the graph (see here). In coordinate notation, it looks like
$$\Delta_0 \phi = - \partial_1 \partial_0^\dagger \sum_i \phi_i (i)$$
$$ = - \partial_1\sum_{i} \phi_i \sum_{j@i} (ji)$$
$$ = - \sum_{i} \phi_i \sum_{j@i} [(i) - (j)]$$
$$ = - \sum_{i} (i) \sum_{j@i} (\phi_i - \phi_j)$$
from which it is easy to see that the above expression is correct:
$$
\Delta_0 \phi = \sum_{i} (i) \sum_{j@i} \phi_j - \sum_{i} (i) \sum_{j@i} \phi_i \\
= \sum_i (i) \sum_j (A_{ij} - D_{ij}) \phi_j
$$
where $D_{ij} = \delta_{ij} z_i$ with $z_i$ being the coordination number of vertex $i$ and $A_{ij} = \delta_{i@j}$. The higher-order Laplacian operators are then related to the graph structure of certain bond/face/body-duals of the original graph.
There is a further connection to various topics such as de Rham cohomology, the Hodge decomposition and harmonic forms. In particular, we can decompose any $p$-chain into
$$\sigma^p = \partial_{p-1}^\dagger \alpha^{p-1} + \partial_{p+1} \beta^{p+1} + \gamma^{p}$$
where $\gamma^{p}$ is a "harmonic chain" and satisfies $\Delta_p \gamma^{p} = 0$, and corresponds to a contribution which "winds around" the lattice topologically, i.e. $\gamma^{p} \in H_p$, the $p$'th homology group of the complex. I have not seen that made more explicit anywhere yet and don't know enough about the topics myself to really comment further.
thank you very much for this extra perspective on this topic!
I wrote a blog post a while ago* on different ways of interpreting the graph laplacian from the perspectives of functional analysis, probability, statistics, differential equations, and topology, and how they connect. Some of these perspectives are covered in more detail by the other responses, but I don't necessarily think one view has primacy over the others. I think the perspective that helps connect these views is that while, as described above, it can be seen as a differential operator applied to the graph, the structure induced by that operator, in particular by its eigendecomposition, is intimately linked to the structure of the space on which it is operating, which is why so many properties can be "read off" of the Laplacian.
More recently, this lecture from Keenan Crane's discrete differential geometry class focuses largely on the differential geometric interpretation, but links to the graph perspective through the triangulation of the surface.
*Since I wrote it in 2015, there has been substantial work on the topological perspective, e.g., and graph neural networks have become ubiquitous.
Thank you very much for these useful links!
Here is another interpretation of the Laplacian (for this answer I use the notation of this answer to a similar post, in particular $\nabla$ is the [graph] gradient and $\nabla^*$ is its adjoint (i.e. one of them is the incidence matrix and the other one is its transpose).
In short: the Laplacian $\nabla^* \nabla$ is the gradient (in the sense of calculus) of the energy functional.
Note: this interpretation makes it relatively clear that (on the space perpendicular to the constant function [and eventually the "alternating" function on a bipartite graph]) applying the random walk operator $P = \mathrm{Id} - \nabla^*\nabla$ will converge to the function with the smallest energy: one is following the gradient flow.
Longer version: given a function $f:X \to \mathbb{R}$ (on the vertices), look at its energy:
$$
\mathcal{E}(f) = \|\nabla f\|_{\ell^2E}^2
$$
(if $f$ were a potential for an electrical current, then this would be the power/heat produced).
On a graph the function $f$ is just a point in $n$-dimensional space (where $n = |X|$ is the number of vertices)
so the energy $\mathcal{E}$ has a gradient (in the sense of calculus; the use of gradient here is not the same as the one of $\nabla$ above).
To compute this gradient, consider any $g$ and look at
$$
\frac{\mathrm{d}}{\mathrm{d}t}\Big|_{t=0} \|\nabla(f+ tg)\|_{\ell^2}^2
$$
Since $\nabla(f+tg) = \nabla f + t \nabla g$ and $\|\nabla h\|^2 = \langle \nabla h \mid \nabla h \rangle$, one has
$$
\frac{\mathrm{d}}{\mathrm{d}t}\Big|_{t=0} \|\nabla(f+ tg)\|_{\ell^2}^2 = \langle \nabla g \mid \nabla f \rangle
$$
Using the definition of the adjoint this is equal to $\langle g \mid \nabla^* \nabla f \rangle$.
This means that $\nabla^* \nabla f$ is the gradient of $\mathcal{E}$ at the "point" $f$.
Thank you very much. An interesting view on the graph Laplacian as well!
Actually, although I remained in the "graph Laplacian" setup, the same remains true for the Laplacian on manifolds (in the right function spaces with the correct metrics; so quite messier, which is why I only talked about the graph set-up): the Laplacian of a function $f$ gives the function $\Delta f$ you should (multiply by a very small scalar and then) remove from the function $f$ in order to decrease the energy. [I used "remove" but it could be "add" depending of which sign you use for the Laplacian]
|
2025-03-21T14:48:31.755606
| 2020-08-12T14:24:14 |
368964
|
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}
|
Stack Exchange
|
Semisimple subgroup of Euclidean group
Let $G$ be a closed and connected semisimple subgroup of the Euclidean group $E(n)$ (the group of isometries of $\mathbb R^n$).
Can we prove that $G$ is conjugate to a subgroup of $O(n)$?
I think you can do it by noting that we have the quotient homomorphism $q: E(n)\to\mathrm{O}(n)$, and homomorphic images of semisimple Lie algebras are semisimple.
More directly, use the fact that $G$ is semi-simple. The restriction of the natural (linear)representation of $E(n)$ on $\mathbb{R}^{n+1}$ preserves the subrepresentation $\mathbb{R}^n$ and hence, by Weyl's theorem, it must be the direct sum of $\mathbb{R}^n$ with a $1$-dimensional complimentary subrepresentation $L\subset\mathbb{R}^{n+1}$. This subrepresentation determines a fixed point of $G$ acting on $\mathbb{R}^n$. Hence $G$ is conjugate to a subgroup of $\mathrm{O}(n)$.
|
2025-03-21T14:48:31.755722
| 2020-08-12T14:35:57 |
368966
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368966"
}
|
Stack Exchange
|
$C^j$-topology considered by Greene and Krantz
My question is about the $C^j$ topology used by Greene and Krantz in their paper "Deformations of Complex Structures, Estimates for the $\bar{\partial}$-equation, and stability of the Bergman kernel". As it is not clear to me that this topology is the same as the usual Whitney strong (or weak) $C^j$ topology and it is not explicitly defined for functions on manifolds.
First they describe the $C^j$ topology for maps $f:U \to \mathbb{C}$ for any open set $U \subset \mathbb{C}^n$. This is done in a quite standard way, for instance:
$$ ||f||_{C^j(U)}:= \sum_{|\alpha|+|\beta| \leq j} \left|\left|\left(\frac{\partial}{\partial z}\right)^{\alpha}\left(\frac{\partial}{\partial \bar{z}}\right)^{\beta}f\right|\right|_{\infty} $$
Where $\alpha$ and $\beta$ are taken as multidices and $||\cdot||_{\infty}$ denotes the supremum norm.
Inmediately after, they define another $C^\infty$-norm. And say that it extends to a smooth manifold "via a fixed coordinate atlas". How is this extension performed? I guess you have to take a locally finite coordinate atlas and sum over all charts the previously defined norm (?). Moreover, they make a remark saying that two functions defined on $U$ are $C^\infty$ close if they are $C^k$ close for $k$ big enough and they say that this remark extends trivially to the manifold case. So implicitly they are considering a $C^j$ norm on the space of $C^\infty$ functions defined on a manifold. What is this norm?
A very similar problem arises later in page 35 when they define a topology in the space of almost-complex structures of a smooth manifold. And they claim that there are neighborhoods of the form $$S_j(\prod_{1,0},\epsilon):=\{\prod_{1,0}': \text{where }\prod_{1,0} - \prod_{1,0}' \text{ is less than } \epsilon \text{ with respect to a } C^j \text{norm}\}.$$ So again, it seems that they are considering a norm on the space of $(1,1)$ tensors on manifolds (rather than open sets) and they are taking the topology induced by this norm.
My question is, what is the precise definition of this norm that induces the $C^j$ topology for complex valued smooth functions on manifolds and how does it relate to usual Whitney topologies? (references appreciated). It looks like this topology can't be the same as the Whitney topology (otherwise the Whitney topology would be usually defined using this norm rather than the more intricate usual definition). But of course this is just a moral argument.
It is pretty much as you guessed in 1.) Instead of summing, you take the maximum. Although maybe in some settings both describe equivalent norms (I am no expert). The key point here is that the target space is $\mathbb{C}$ and not a general manifold. In general the Whitney topology does not come from a norm, although it does come from a metric when $M$ is compact (The distance Greene and Krantz talk about does generalize to manifolds).
In general one can define a norm on the space of sections of a vector bundle. In this particular case you are dealing with sections of the trivial vector bundle $M \times \mathbb{C}$ where $M$ is a compact manifold. For the general case of sections of any vector bundle, look at Section 3 of "The Banach manifold $C^k(M,N)$ by Johannes Wittmann.
Fix a finite coordinate atlas $\{(U_i, \phi_i)\}_{1, \ldots, \ell}$ such that $\bar{U}_i$ is compact and is still contained in a coordinate atlas. Then define:
$$||f||_{C^j(M)}:=\max_{1\leq i \leq \ell} ||f||_{C^j(U_i)}$$
The $||\cdot||_{C^j(U)}$ norm defined in Wittmann's work is slightly different but I think they are equivalent. (Witmann does not sum the norms for all partial derivatives but rather takes the maximum of them).
Thank you. I guess the only piece missing is that the norm resulting from summing the sup of all partial derivatives up to order $k$ is equivalent to the norm taking the maximum of those supremums. Do you know a reference for that?
|
2025-03-21T14:48:31.756064
| 2020-08-12T16:07:56 |
368973
|
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|
Stack Exchange
|
Double commutant theorem when $C^*$-subalgebra does not contain identity operator $1$
Double commutant theorem: For a unital $C^*$-subalgebra $M \subset B(H)$, one has
$$\smash{\overline M}^\text{SOT}=\smash{\overline M}^\text{WOT}=M''.$$
My question:
For a $C^*$-subalgebra $M \subset B(H)$ but don't assume $M$ contains identity operator $1$, does
$$\smash{\overline M}^\text{SOT}=\smash{\overline M}^\text{WOT}=M''?$$
$M''$ is unchanged upon adjoining $1$, so it equals the strong and weak operator closures of $\mathbb C + M$. By the way, in terms of TeX, don't use $M^{''}$ $M^{''}$; instead use $M''$ $M''$ or $M^{\prime\prime}$ $M^{\prime\prime}$. I have edited accordingly.
LSpice, thanks your edit
As @LSpice points out, $M''$ always contains the identity operator $I\in B(H)$. Now think about whether every ${\rm C}^*$-subalgebra of $M_2({\bf C})$ contains the identity matrix $I_2$ ...
If $K$ is the essential space for $M$, namely $K=\overline{\text{span}}{T(\xi): T\in M, \xi\in H}$,
then $H=K\oplus K^\perp$ and $M$ acts on $H$ like $M\oplus0$, while
$$M''=\smash{\overline M}^{\text{SOT}}\oplus \mathbb{C}I_{K^\perp}.$$
|
2025-03-21T14:48:31.756198
| 2020-08-12T16:22:30 |
368975
|
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Stack Exchange
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Graph embeddings in the projective plane: for the 35 forbidden minors, do we know their Colin de Verdière numbers?
The Graph Minor Theorem of Robertson and Seymour asserts
that any minor-closed graph property is determined by a finite set
of forbidden graph minors. It is a broad generalization e.g. of the Kuratowski-Wagner theorem, which characterizes planarity in terms of two forbidden minors: the complete graph $K_5$ and the complete bipartite graph $K_{3,3}$.
Embeddability of a graph in the projective plane is such a minor-closed property as well, and it is known that there are 35 forbidden minors that characterize projective planarity. All 35 minors are known, a recent reference from 2012 is, for example, https://smartech.gatech.edu/bitstream/handle/1853/45914/Asadi-Shahmirzadi_Arash_201212_PhD.pdf.
A classical reference is Graphs on Surfaces from Mohar and Thomassen,
Johns Hopkins University Press 2001.
I am interested in the Colin de Verdière numbers for these 35 forbidden minors and have searched for them for a while now, but could not find anything.
Question: So I wondered whether the Colin de Verdière graph invariants for the whole set of these 35 forbidden minors are actually known? I would be grateful for any reference.
UPDATE:
Updating this question based on a great comment from Martin Winter. As he points out, the Colin de Verdière number $\mu$ is known and $\mu=4$ for a handful of these 35 forbidden minors, e.g. the disjoint unions of $K_5$ and $K_{3,3}$.
Interestingly, as outlined in his answer to a related question (Algebraic graph invariant $\mu(G)$ which links Four-Color-Theorem with Schrödinger operators: further topological characterizations of graphs?), it follows that the Colin de Verdière invariant cannot provide a full characterization of graph embeddings e.g. in the projective plane.
Can you say where in the linked thesis these 35 graphs are listed? I can find the first 24 graphs in section 1.7 and 1.8. The remaining 11 cases should be listed in Chapter V, but they seem a bit more hidden.
If $\mu>5$ holds not for all forbidden minors, then we can't have "projective planarity $\Leftrightarrow$ $\mu \le 5$". So the hope is that all of them have $\mu$ larger five, or am I missing something?
And correct me if this is wrong, but the thesis lists $K_5+K_5$ (disjoint union) as one of the forbidden minors, and isn't $\mu=4$ for this graph already?
@M.Winter in the thesis, you should find all 35 of them in Figure 1.1 which is on page 8 of the text body and page 14 of the pdf
$\mu(K_5+K_5)=\mu(K_5)$ is a consequence Theorem 2.5 in "The Colin de Verdière graph parameter" by van der Holst, Lovász, Schrijver. I also added an answere here.
Here's a table containing the Colin de Verdière numbers:
Name Graph6 μ Reason
K33 + K33 4 (components linklessly embeddable)
K5 + K33 4 (components linklessly embeddable)
K5 + K5 4 (components linklessly embeddable)
K33 . K33 4 (apex)
K5 . K33 4 (apex)
K5 . K5 4 (apex)
B3 G~wWw{ 4 (apex)
C2 H~wWooF 4 (apex)
C7 G~_kY{ 4 (apex)
D1 Is[CKIC[w 4 (apex)
D4 H~AyQOF 4 (apex)
D9 I]op_oFIG 4 (apex)
D12 H^oopSN 4 (apex)
D17 G~_iW{ 4 (apex)
E6 Is[BkIC?w 4 (apex)
E11 I]op_oK?w 4 (apex)
E19 H~?guOF 4 (apex)
E20 H~_gqOF 4 (apex)
E27 I]op?_NAo 4 (apex)
F4 Is[?hICOw 4 (apex)
F6 Is[@iHC?w 4 (apex)
G1 4 (apex)
K35 4 (apex)
K45-4K2 4 (apex)
K44-e 5 (Petersen family and -2 argument)
K7-C4 4 (apex)
D3 G~sghS 4 (apex)
E5 H]oxpoF 5 (Petersen family and -2 argument)
F1 H]ooXCL 4 (apex)
K1222 4 (apex)
B7 4 (apex)
C3 4 (apex)
C4 4 (apex)
D2 4 (apex)
E2 4 (apex)
Let me give justification. Graphs with $\mu \leq 3$ are planar, hence embeddable on the projective plane. So all the $35$ graphs have $\mu \geq 4$. Since apex graphs are linklessly embeddable, and linklessly embeddable graphs have $\mu \leq 4$, the apex graphs in this table have exactly $\mu = 4$. Also, a graph is linklessly embeddable iff its components are linklessly embeddable, so the first three graphs have $\mu = 4$.
The graphs in the Petersen family are not linklessly embeddable, so they have $\mu \geq 5$. $K_{4,4}-e$ is already in the Petersen family, and $\mathcal E_5$ contains $K_{3,3,1}$ as a subgraph. They both have $\mu \geq 5$.
To see they have $\mu \leq 5$, use Theorem 2.7 in [1]: If $G=(V,E)$ is a graph, and $v$ a vertex of $G$, then $\mu(G) \leq \mu(G-v)+1$. Since we can remove $2$ vertices from $K_{4,4}-e$ to make it planar (by making it $K_{3,3}-e$), it follows that $\mu(K_{4,4}-e) \leq \mu(K_{3,3}-e)+2 = 5$. Hence $\mu(K_{4,4}-e)=5$. The same line of reasoning applies to the graph $\mathcal E_5$.
[1] Van Der Holst, Hein, László Lovász, and Alexander Schrijver. "The Colin de Verdiere graph parameter." Graph Theory and Computational Biology (Balatonlelle, 1996) (1999): 29-85.
Thanks a lot for a great answer! Very clear and super helpful. Very much appreciated
|
2025-03-21T14:48:31.756628
| 2020-08-12T16:31:59 |
368976
|
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Stack Exchange
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The variety induced by an extension of a field
If $K$ is a finitely generated field extension of $k$, then there exists an irreducible affine $k$-variety with function field $K$. The idea is that if $x_1, \dots, x_n$ are generators of $K$ under $k$, i.e each elements of $K$ is a rational function in $x_1, \dots , x_n$, then the kernel of the map $k[t_1,\dots, t_n]\to K$ is a prime ideal and the induced map between their field fractions is an isomorphism:
$(k[t_1,\dots, t_n]/I)_0\cong K$
This means $Z(I)\subseteq k^n$ is the affine irreducible variety which field fraction corresponds to $K$.
Now I have the following problem:
In this case I have $k$ equal to the function field of $\mathbb{P}^2$, and $K$ equal to the finite extension $k((\frac{l_2}{l_1})^{\frac{1}{n}},\dots, , (\frac{l_k}{l_1})^{\frac{1}{n}})$. In the paper the author tells us $K$ determine an algebraic (affine?) surface $X$ with normal singularities and a natural map $\pi: X\to \mathbb{P}^2$.
I don't understand how to define this natural map $\pi$ and what is exactly this surface $X$. I think that $K$ determine an affine variety up to birational morphisms and so I don't understand how to define exactly $X$.
Can you give me an example for $n=2$ and $k=3$, please?
You can take the normalisation of $\mathbf P^2$ in the larger field.
Can you explain me please?
One place you can read about relative normalisation is Tag 035H. I don't know what your background is, so it's hard to say more.
I know the concept of normalization but I don't understand the phrase 'in the larger field'
I believe the author of the paper refers to a special realization of one element of the birational class as you pointed out. That one should be hinted by the generators of the field extension presented. For instance in the case you asked for in the end of your post one equation should be $z_1/z_0\sqrt{\ell_2/\ell_1}-z_2/z_0\sqrt{\ell_3/\ell_1}=0$. Notice that this is "above" $z_1/z_0*(\ell_2/\ell_1)-z_2/z_0*(\ell_3/\ell_1)=0$.
You have a finite field extension $K \to L$ and a normal (even smooth) variety $X$ with fraction field $K$. Then you can take the normalisation of $X$ in $L$, meaning on each affine open $U = \operatorname{Spec} A$ (so $\operatorname{Frac} A = K$) you take the integral closure of $A$ in $L$, and glue these together for the various $U \subseteq X$. (The cited tag is merely a coordinate-free way to phrase this: the pushforward of $\mathcal O_L$ is a quasi-coherent $\mathcal O_X$-module, and you take the relative Spec of the integral closure of $\mathcal O_X$ in $\mathcal O_L$.)
I decided to turn my comment into an answer not because it is complete but because I think it can be of use.
Let $z=(z_0:z_1:z_2)$ and $u=(u_1:\ldots:u_k)$ be homogeneous coordinates of $\mathbb{P}^2$ and $\mathbb{P}^{k-1}$. First notice that the surface $X_1\subset \mathbb{P}^2\times \mathbb{P}^{k-1}$ defined by the vanishing of the $2\times 2$-minors of the matrix \begin{equation*}
\begin{pmatrix}
u_{1} & u_{2} & \cdots & u_{k} \\
\ell_{1} & \ell_{2} & \cdots & \ell_{k} \\
\end{pmatrix}
\end{equation*} is the closure of the graph of the rational map $z\mapsto (\ell_1:\ldots:\ell_k)$. Restricting the projection you get a well defined map $X_1\rightarrow \mathbb{P}^2$.
On the other hand you also have a $n$-to-$1$ map $\mathbb{P}^{k-1}\rightarrow \mathbb{P}^{k-1}$ given by $\phi_n:(t_1:\ldots:t_k)\mapsto (t^n_1:\ldots:t^n_k)$. This induces $id\times\phi_n:\mathbb{P}^2\times\mathbb{P}^{k-1}\rightarrow \mathbb{P}^2\times\mathbb{P}^{k-1}$. Now you can take $X$ to be the preimage of $X_1$ by $id\times\phi_n$.
In this way you can "see" $X\subset \mathbb{P}^2\times\mathbb{P}^{k-1}$ with coordinates $(z,t)$ as the vanishing set of minors of the matrix
\begin{equation*}
\begin{pmatrix}
t^n_{1} & t^n_{2} & \cdots & t^n_{k} \\
\ell_{1} & \ell_{2} & \cdots & \ell_{k} \\
\end{pmatrix}.
\end{equation*}
Also the map $\pi:X\rightarrow \mathbb{P}^2$ is clear. The ramification locus is induced by the ramification locus of $\phi_n$.
I'm not sure about the singularities of $X$ but I think they will depend on the relative position of lines $\ell_1,\ldots,\ell_k$.
The problem is that the group $(Z/nZ)^k$ acts on $X$ and not $(Z/nZ)^{k-1}$. What is the mistake?
The group acting effectively is $(Z/nZ)^{k-1}$, because the element $(-1,\ldots,-1) \in (Z/nZ)^k$ acts trivially on the homogeneous coordinates.
Yes, ok, but this would mean simply that the elements of the type $(a,...,a)$ belongs to the stabilizer of each point, right? We are interested only of that actions that are faithful, so this mean we must consider the new faithful action $((Z/nZ)^k)/(\cap_x G_x)$, that is isomorphic exactly to (Z/nZ)^{k-1} via the isomorphism $(a_1,..,a_k)\to (a_2-a_1, \cdots , a_k-a_1)$ ?
Can you help me also for this question, please? https://math.stackexchange.com/q/3815141
Let $n=2$ and $k=3$, and suppose by the sake of simplicity that the three lines are in general position. Then, up to projective transformations, we can assume that they are the three coordinate lines $\ell_1$, $\ell_2$, $\ell_3$ given by $z_0=0$, $z_1=0$, $z_2=0$, respectively.
Then your function field is simply $\mathbb{C}(x, \, y)(\sqrt{x}, \, \sqrt{y})$, where $x=z_1/z_0$, $y=z_2/z_0$, and the affine equation of your $(\mathbb{Z}/2\mathbb{Z})^2$-cover $X \to \mathbb{P}^2$ on the chart $z_0 \neq 0$ is $$(x, \, y) \mapsto (x^2, \, y^2).$$
Note that $X$ is projective, since it is a finite covering of a projective variety; in fact, $X$ is the union of three of these affine charts, corresponding to the three standard charts for $\mathbb{P}^2$.
A moment of thought shows that $X = \mathbb{P}^2$, and that the global equation of your bi-double cover is $$\pi \colon \mathbb{P}^2 \to \mathbb{P}^2, \quad [z_0: \, z_1: \, z_2] \mapsto [z_0^2: \, z_1^2: \, z_2^2].$$
It is an instructive exercise to factor $\pi$ through the three singular double covers $$X_i \to \mathbb{P}^2, \quad i=1,\, 2, \, 3$$ corresponding to the three non-trivial involutions in the Klein group $(\mathbb{Z}/2\mathbb{Z})^2$.
But in this case $\pi$ would be a $(Z/2Z)^3$ covering and not a $(Z/2Z)^2$-covering, right?
If $k=3$ then $k-1=2$. The covering in the affine chart is clearly a bi-double cover, non tri-double. So the extension to the projective plane must be bi-double, too. Recall that $z_0$, $z_1$, $z_2$ are homogeneous coordinates. Think of the double cover given in affine coordinates by $x \mapsto x^2$: in homogeneous coordinates it becomes $[z_0 , : ,z_1] \mapsto[z_0^2 , : ,z_1^2]$.
This is double and not bi-double, for instance because $[z_0:z_1]=[-z_0: -z_1]$ and $[-z_0:z_1]=[z_0:-z_1]$.
|
2025-03-21T14:48:31.757132
| 2020-08-12T16:36:47 |
368978
|
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"A Stasinski",
"AlexIvanov",
"Asvin",
"Kimball",
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}
|
Stack Exchange
|
Representation theory of $\operatorname{GL}_2(\mathbb Z/n\mathbb Z)$
Is there a nice reference for the finite dimensional (characteristic 0) representation theory of $\operatorname{GL}_2(\mathbb Z/n\mathbb Z)$ and $\operatorname{PGL}_2(\mathbb Z/n\mathbb Z)$ for varying $n$ and in the limit, for $\operatorname{PGL}_2(\mathbb Z_\ell)$?
I would like to know things like the number of irreducible representations, their dimensions, the minimal fields they are defined over and stuff like that.
You might try to look at https://arxiv.org/pdf/0807.4684.pdf Also, it should be worth to take a look at other related articles of Stasinski. Also, you may find some relevant information in the book of Bushnell--Henniart, "The Local Langlands conjecture for $GL_2$"
The case of $n=p$ is of course well known, e.g., Piatetskii-Shapiro's book. For n squarefree, you can write GL(2,n) as a product of GL(2,p)'s. Similarly, you can reduce to the case of GL(2) over $\mathbb Z/p^r \mathbb Z$, if this helps.
And related for $\mathbb Z_\ell$: https://mathoverflow.net/q/89184/6518 (I think there are other related questions on this site but can't find them now)
Thank you, I will take a look at the references and see if it helps in my case!
You can find further references in the answers to this question: https://mathoverflow.net/q/87254/2381. For the dimensions and multiplicities for $\mathrm{GL}_2$, Onn's paper is a convenient reference. As far as I know, the case $\mathrm{PGL}_2$ has not been written down (and may require $p>2$ to be manageable). I don't think anyone has studied the minimal fields of definition.
@AStasinski Could you please tell me which paper of Onn the linked quesion refers to? The link there no longer seems to be working.
@Asvin I have fixed the links in that question.
|
2025-03-21T14:48:31.757406
| 2020-08-12T16:58:26 |
368979
|
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"Hugo10T",
"Iosif Pinelis",
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|
Stack Exchange
|
Proof of "Prove that a sub-gaussian and isotropic random vector over a finite set T implies that the set is exponentially large"
Here the original question was asked and answered. However I have a question to the solution. If I get it right they try to show $\frac 12 I_n \leq \mathbf{E} YY^T \leq I_n$ by proving
$$ \mathbf{E} \left[ \langle X,\theta \rangle^2 {\bf 1}_{\{||X|| > 4C \sqrt{n}\}} \right] \leq \left( \mathbf{E} \left[ \langle X,\theta \rangle^4 \right] \cdot \mathbf{P}(||X||^2 > 16C^2n)\right)^{1/2} \leq \sqrt{\frac{4C^2}{16C^2}} = \frac{1}{2}$$
for all unit vectors $\theta$. However I don't see how you get $\mathbf{E} Y_i Y_j=0$ for $i\neq j$ from that. I hope someone can help me here.
In the linked answer, the inequality sign $\le$ in $\frac12\,I_n\le EYY^T\le I_n$ is not meant in the sense of the entrywise comparison. Rather, it is meant in this sense: for any two symmetric matrices $A$ and $B$, we write $A\le B$ if $B-A$ is positive semidefinite.
In this case, we have $Y=X1_{\|X\|\le4C\sqrt{n}}$, where $EXX^T=I_n$ and
$$E\langle X,\theta \rangle^2 1_{\|X\|>4C\sqrt{n}}\le\tfrac12=\tfrac12\, E\langle X,\theta \rangle^2$$
for any unit vector $\theta$. So,
$$\theta^T EYY^T\theta=E\langle Y,\theta \rangle^2
=E\langle X,\theta \rangle^2 1_{\|X\|\le4C\sqrt{n}}\le E\langle X,\theta \rangle^2=\theta^T EXX^T\theta=\theta^T I_n\theta,$$
which means $EYY^T\le I_n$, and
$$E\langle Y,\theta \rangle^2=E\langle X,\theta \rangle^2-E\langle X,\theta \rangle^2 1_{\|X\|>4C\sqrt{n}}\ge \tfrac12\,E\langle X,\theta \rangle^2,$$
which similarly means $EYY^T\ge\tfrac12\,I_n$.
Thank you very much! Is it correct that the following random vector $\tilde{Y}$ from the given proof would be defined as $\tilde{Y}:=(EYY^T)^{-1/2}Y$ ?
@Hugo10T : yes, it appears so.
Thanks. I have one last question to the given proof. How can I show that $\lVert \langle \tilde{Y},\theta \rangle \rVert_{\psi_2}\leq 2C$ holds?
@Hugo10T : Your original question was fully answered, and the additional question about $\tilde Y$ was answered too. If you have any further questions, please ask them in separate posts.
|
2025-03-21T14:48:31.757570
| 2020-08-12T17:39:32 |
368982
|
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"Daniele Tampieri",
"Willie Wong",
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|
Stack Exchange
|
Analysis of coefficients for quickly vanishing analytic vector field
Let $u = (u_1, u_2, u_3): \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a divergence-free analytic vector field for $n =3$ or $n =4$, i.e., $u_i : \mathbb{R}^n \rightarrow \mathbb{R}$ are analytic functions locally given by
$$u_i(x) = \sum_{j \in \mathbb{N}_0^n} c_j^{(i)} \cdot \prod_{k=1}^n x_k^{j_k}$$
for some real $c_j^{(i)}$.
Furthermore, we request $$|\partial_x^{\alpha} u(x)| \leq C_{\alpha K}(1+|x|)^{-K}$$
on $\mathbb{R}^n$ for any $\alpha,K$.
My question is now, can we infer some requirements for the coefficients $c_j^{(i)}$, such that the above conditions are equivalent to these requirements.
An obvious requirement is implied by Cauchy-Hadamard: $$ \limsup_{j\in\Bbb N_0^n}\sqrt[|j|]{|c_j^{(I)}|}=0.$$ Probably you can see how this restriction works jointly with the derivative growth estimate you’re interested to.
A few clarifications requested: (a) when you say $u_i$ is analytic, are you assuming necessarily that it has infinite radius of convergence? (Technically $1/(1+x^2)$ is a real analytic function too.) (b) What are the "conditions" in your question? Are you assuming divergence free throughout, or are you looking for a requirement that is equal to "divergence free + rapid decay"?
I am doubtful you can see decay rates at infinity just from some size estimates on the coefficients. $\cos(x^2)$ and $e^{-x^2}$ have very similar Taylor series.
@WillieWong a) You are right, I consider all (!) analytic functions that meet the conditions from above. b) The conditions are, analytic, divergence-free everywhere, and the inequalities.
|
2025-03-21T14:48:31.757698
| 2020-08-12T18:02:04 |
368986
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368986"
}
|
Stack Exchange
|
What relationship exists between samples of a function and samples of its vector gradient field?
A real function $f(x)$ is defined on $N$-dimensional real space where $N \ge 3$. $f(x)$ is differentiable and its gradient with respect to x is $g(x)$. So $g(x)$ is a vector field.
Assume we do not know the closed form of $f(x)$ or $g(x)$. Instead we only know the value of $f(x)$ (real numbers) on one set of data samples $x_0,x_1,\ldots$, and the value of $g(x)$ (real vectors) on a (possibly separate) set of data samples $x_0',x_1',\ldots$.
Under what condition, is it possible to derive a relationship between the function samples $f(x_0), f(x_1), \ldots$ and the gradient samples $g(x_0'), g(x_1'), \ldots$?
Intuitively, the relationship between a function and its gradient is governed by vector calculus results like Green's identities. But such results are only applicable when $f(x)$ and $g(x)$ are available in closed form. It is unclear how to apply it to the case of sampling.
|
2025-03-21T14:48:31.757786
| 2020-08-12T18:16:25 |
368988
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368988"
}
|
Stack Exchange
|
Finding particular closed paths in geometric plane regions
Let $X_m$ denote a set of $m\geq 3$ lines in $\mathbb{R}^2$ that are not all parallel. Consider the problem of determining a closed path of $kn$ points in $X_m$ $k, n \in \mathbb{Z}^+$, such that the "forward orbit" of the path has angles of incidence with the lines in $X_m$ which are strictly contained in a given set $\{\theta_1, \theta_2, ..., \theta_n\}$, and further each angle must occur in a particular order when traversing the sequence of points.
In other words, if $p_1, p_2, ..., p_n, p_{n+1}$ (with $p_{n+1} = p_1$) are a set of such points in $X_m$, join points $p_i, p_{i+1}$ with a line segment, so that that line segment makes angle $\theta_i$ with line $L_j \subset X_m$, where $p_{i+1} \in L_j$. The question then is whether such a closed path can be found for a given $X_m$, set of incidence angles $\{\theta_1, \theta_2, ..., \theta_n\}$, and ordering at which the angles must appear.
This appears to be similar to a weakening of the problem of finding periodic billiard paths in polygons, however it still appears to be hard, and I am wondering if anyone is aware of a solution, or an analogous problem that is discussed in literature.
The answer, given in this paper: https://arxiv.org/abs/2112.02207, turns out to be "yes", there always exist such closed curves. The theorem stating the answer to this question is given in the introduction to the linked paper, and I restate it here for reference:
Theorem: For any space $X_m$ with labeled lines, let $\theta_1, \theta_2,...,\theta_n$, $n\geq m \geq 3$, be any sequence of acute or right angles, and let $L_{a_1}, L_{a_2}, ..., L_{a_n}$ be a sequence of line labels such that no two consecutive labels are the same, including $L_{a_n}$ and $L_{a_1}$, and each of the $m$ possible labels occur at least once in the sequence. Then there exists a closed curve $\Gamma$ over $X_m$ that admits an incidence angle sequence $\theta_1, \theta_2, ..., \theta_n$ with respect to the line sequence $L_{a_1}, L_{a_2}, ..., L_{a_n}$ when traversed in a fixed direction. Moreover, there are either uncountably many such curves, or no more than $2^n$.
|
2025-03-21T14:48:31.757947
| 2020-08-12T19:08:39 |
368990
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368990"
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|
Stack Exchange
|
Reference request: diffusion approximation to the radiative transport
I'm looking for a good, modern reference for the diffusion approximation to the radiative transport problem. I'm aware of the text of Dautray and Lions, as well as the monograph by Bensoussan, Lions, and Papanicolaou, but these are a bit dated and difficult to read. My motivation is studying the (in)stability of the inverse radiative transport problem. More specifically, I am hoping to better understand the estimate in pages 8-10 in the following paper of H.K. Zhao.
|
2025-03-21T14:48:31.758008
| 2020-08-12T19:48:09 |
368992
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368992"
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|
Stack Exchange
|
Presentations of $\mathbf{}_3(\mathbb{F}_2)$ by three involutions, 2
I am searching for a presentation of the group $\mathbf{PGL}_3(\mathbb{F}_2)$ for which the generators are involutions $a, b, c$, and such that the following relations are present [among extra relations, of course]:
REP.: $(ab)^2 = 1, (bc)^3 = 1, (ac)^6 = 1$.
Can anyone help me out here (again) ? (Thanks !!)
$G = {\rm GL}(3,2)$ can't be generated this way. First, we may note that there is no element of order $6$ in $G$, so we must have one of : $ac = 1$ or $(ac)^2 = 1$ or $(ac)^3 = 1.$
If $ac = 1$ then $a = c$ and $G = \langle a,b \rangle $ is Abelian.
If $(ac)^2 = 1$, then $a$ and $c$ commute. But also $a$ and $b$ commute since $(ab)^2 = 1$.
Then $a \in Z(G),$ contrary to the simplicity of $G$.
If $(ac)^3 = 1,$ then $G$ is a homomorphic image of $A_{4} \cong \langle x,y : x^{2} = y^3 = (xy)^3 = 1 \rangle$, since $G = \langle ab, bc , ac \rangle $ and we have
$(ab)^2 = (bc)^3 = [(ab)(bc)]^3 = 1 $.
Recall from the previous question that $\langle ab,bc,ac \rangle$ is generated by any two of the listed generators, and is normalized by each of $a,b$ and $c$, so is normal in $G$.
|
2025-03-21T14:48:31.758112
| 2020-08-12T19:55:25 |
368993
|
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"authors": [
"Andrej Bauer",
"François G. Dorais",
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"https://mathoverflow.net/users/163629",
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"https://mathoverflow.net/users/41291",
"qk11",
"მამუკა ჯიბლაძე"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368993"
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|
Stack Exchange
|
Type Theory with no Base Types
Is there any work in type theory where no base types are assumed, e.g., that there are only function types in place ($t_1 \to t_2$ is a type whenever $t_1$ and $t_2$ are types)?
If not, are there specific reasons for that?
Are you allowing the empty model where there are no types? If so, some adjustments would need to be made, akin to those for free logic.
I'm more interested in models where there's at least one type (hence infinitely many types?) -- just that no type is a base type.
That doesn't give infinitely many types (assuming only $\to$ as a type constructor): there is a model where the only type is the unit type. Whether that's the only finite model probably depends on your axioms.
System F seems to satisfy your criteria, provided you allow other type constructors. Types are constructed from $\rightarrow$ and $\forall$. There are no base types like those of, say, system T (i.e. $\mathsf{nat}$).
In many formal systems we work with metavariables, which are meta-level symbols which stand for unknown, arbitrary or yet-to-be-determined entities. For instance, in propositional calculus we say things like "$A \land B$ is equivalent to $B \land A$" -- here $A$ and $B$ are metavariables standing for any formulas.
We can do the same thing in type theory. From your examples I surmise that you have in mind simple types, so let us stay with those. It is perfectly OK, and even useful, to consider types in which metavariables occur, as then you can say things like "$\lambda (x : A \times B). (\pi_2 x, \pi_1 x)$ is an isomorphism from $A \times B$ is isomorphic to $B \times A$". In your question you yourself used the metavariables $t_1$ and $t_2$ when you spoke of the type $t_1 \to t_2$.
Without metavariables and without base types the type theory becomes trivial because we cannot form any types at all.
Don't you still have things like product or sum of an empty family of types? Not that it makes for anything nontrivial, but...
Firstly, we're in the realm of simple types so there are no families. But if we did consider dependent types, how would you get the empty family without having the empty type first?
Definitely I am in no position to say anything competent about that, but I was thinking about something less sophisticated than dependent types, so probably I should not use the word "family" but rather, say, "tuple".
Specifically I was thinking about a programming language which has possibility to produce the product type or the sum type of any tuple of types, including the empty tuple. Is not there a syntactic counterpart of such a thing?
Pardon my ignorance, but what if we keep the metavariables but posit no base types? Would it still be trivial? For example, what if we speculate that the space of types is non-empty, and the only type constructors are of the form $t_1\to t_2$, where $t_1$ and $t_2$ are metavariables? It seems to me that we haven't committed to base types here. Would this be still trivial? Or do we need dependant types to accommodate such a move?
@მამუკაჯიბლაძე: Yes, you could posit a product operation that takes a list of types, and in ML and Haskell do precisely that, so that a * b * c is a primitive ternary product, which is neither (a * b) * c nor a * (b * c). If you allow the product of an empty list that's just a roundabout way of positing a primitive type, namely the unit type. ML does not allow it, Haskell does.
@qk11: when you work with metavariables you must always keep track of which ones are currently being used. It's a bit like polynomials, you need to know in which variables the polynomials are formed. Without any primitive types and $\to$ as the only type constructor, you cannot form any type without having at least one metavariable. Note also that metavariables may be construed as "unspecified primitive types", so a lot of what we're discussing is about what point of view you want to take.
|
2025-03-21T14:48:31.758383
| 2020-08-12T20:09:13 |
368995
|
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|
Stack Exchange
|
Chevalley-Eilenberg cohomology of polynomial vector fields on $\mathbb{A}^2$
I have a question similar to one given here.
What is the cohomology of the Lie algebra of polynomial vector fields on an affine space $\mathbb{A}^2$ over a field of characteristic $0?$ (I'm interested in the complex case, but I think there is no difference).
If I'm correct the answer to this question for $\mathbb{A}^1$ is Goncarova's theorem? However, I don't know if there is an answer for $\mathbb{A}^2.$ I found a remark in Fuchs's book on cohomology of infinite-dimensional Lie algebras that one should consider homology of polynomial vector fields instead of cohomology. I presume this is because the algebra of polynomial vector fields is dual to that of formal vector fields.
Is there an answer for cohomology as well or is it completely hopeless?
|
2025-03-21T14:48:31.758462
| 2020-08-12T20:17:41 |
368997
|
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|
Stack Exchange
|
Reference Request: Is every interval-valued probability measure consistent?
Short version: Does every interval-valued probability measure contain a conventional probability measure? I have a sense that this is a basic result about an obscure topic but I am having trouble finding a reference.
Let $(X,\Omega)$ be a measurable space, so that $X$ is a set and $\Omega\subseteq P(X)$ is a $\sigma$-algebra. An ordinary probability measure $m$ on $(X,\Omega)$ would be a function $m: \Omega \to [0,1]$ such that
$m(A^c) = 1-m(A)$ for all $A\in\Omega$
$m(\bigcup_i A_i) = \sum_i m(A_i)$ for all finite or countably infinite families $(A_i)_{i\in I}$ of disjoint sets in $\Omega$.
(These imply that $m(\varnothing) = 0$, by letting $(A_i)_{i\in I}$ be the empty family, and therefore that $m(X) = m(\varnothing^c) = 1$.)
Roughly speaking, an interval-valued probability measure instead of giving a precise value for the measure of a set, gives a range of values $\mu(A) = [m^-(A), m^+(A)]$. Formally, an interval-valued probability measure is a function $\mu: \Omega\to P([0,1])$ such that
$\mu(A)$ is a closed interval in [0,1] for each $A\in \Omega$.
$\mu(A^c) = 1 - \mu(A)$ for all $A\in \Omega$
$\mu(\bigcup_i A_i) \subseteq \sum_i \mu(A_i)$ for all finite or countably infinite families $(A_i)_{i\in I}$ of disjoint sets in $\Omega$.
(Here, we are using a common abuse of notation for adding/subtracting sets, or sets and constants: "$1 - \mu(A)$" means $\{1 - x: x\in \mu(A)\}$, and "$\sum_i \mu(A_i)$" means $\{\sum_i x_i : x_i \in \mu(A_i)\}$.)
The intuition behind these axioms is that perhaps we don't know the true values of a probability measure $m$; we only know for each measurable set $A$ that $m(A)$ is somewhere in $\mu(A) = [m^-(A),m^+(A)]$. That also tells us $m(A^c)$ is somewhere in $[1-m^+(A),1-m^-(A)] = 1 - \mu(A)$. And if we know that another measurable set $B$, disjoint from $A$, has $m(B)\in[m^-(B), m^+(B)]$, then $m(A\cup B)$ definitely can't be lower than $m^-(A)+m^-(B)$ or higher than $m^+(A)+m^+(B)$. In other words, $\mu(A\cup B)\subseteq [m^-(A)+m^-(B), m^+(A)+m^+(B)] = \mu(A) + \mu(B)$, though the containment may be strict if that reflects information we have about $A\cup B$ that we don't have about $A$ or $B$ separately, e.g. if $B = A^c$.
Examples of interval-valued probability measures:
Every ordinary probability measure $m$ gives an interval-valued measure with $m^+ = m^- = m$.
More generally, if $(m_i)_{i\in I}$ are a family of probability measures on $(X,\Omega)$, then we can define an interval-valued probability measure $\mu$ by $\mu(A) = [\inf\{m_i(A) : i\in I\}, \sup\{m_i(A) : i\in I\}]$.
For any measurable space $(X, \Omega)$, there is the maximally vague interval-valued measure $\mu$ that assigns $\mu(\varnothing) = [0,0]$, $\mu(X) = [1,1]$, and $\mu(A) = [0,1]$ for every other measurable set $A\in \Omega$.
My basic question, for which I would love a reference to the literature, is:
Given an interval-valued probability measure $\mu$, must it contain an ordinary probability measure $m$, in the sense that for all measurable $A$ we have $m(A)\in \mu(A)$? Or is it possible that $\mu$ is "inconsistent," and no choice of values for $m$ from among those allowed by $\mu$ gives a valid probability measure?
(Note that for example 3 above, this question reduces to the MO question Can we put a Probability Measure on every $\sigma$-Algebra? to which the answer is Yes for trivial reasons.)
A transfinite induction attempt to construct an ordinary probability measure from an interval-valued measure by making arbitrary choices for one measurable set at a time, then propagating the new constraints to create a new interval-valued measure with narrower intervals, seems to work just fine (so far), but the process is tedious and I'd love to reference it if the argument already exists elsewhere.
Did you invent the notion of an interval-valued probability measure, or is there existing literature?
@NateEldredge They came up in relation to density of subsets of $\mathbb{N}$, but no, I got the definition from this paper by Jamison and Lodwick, which I've rewritten here in an equivalent (but hopefully clearer) form.
I this fuzzy probability as f.i. in https://pdfs.semanticscholar.org/5213/18f77cafb280fdb45c859717c1e15100a492.pdf ?
@DieterKadelka I don't think it's quite the same as the fuzzy probability developed in that reference, which has fuzzy sets as inputs but outputs ordinary numbers. This version of probability has ordinary sets as inputs but outputs intervals; in a sense, the fuzziness is "on the other end."
|
2025-03-21T14:48:31.759008
| 2020-08-12T20:24:13 |
368998
|
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|
Stack Exchange
|
Sets of residues with only a single intersection under translation
A combinatorial game I am studying has given rise to the following question. Consider the group $\Bbb Z/n\Bbb Z$. What is the largest $m$ such that there exist $k$ sets of $m$ residues such that the intersection of a translation of each of these sets has at most 1 element? That is, if the sets are $A_1, \ldots A_k$, we require that for all $(c_1, \ldots, c_k)$ the intersection of $(A_i + c_i)$ for $1 \le i \le k$ has at most one element, where $A_i + c_i$ is set obtained by adding $c_i$ to every element in $A_i$. Alternatively, if it makes the answer simpler, we can ask what the smallest $n$ is given $m$ and $k$.
For $k=2$, the answer is simple. It is possible when $n\ge m^2$ by making one set $\{0,1,\ldots, m-1\}$ and the other $\{0,m,\ldots, m(m-1)\}$. But it's not clear to me how to extend the construction to $k\ge 3$.
If there is a simple generalization to the case where the sets of residues can be different sizes, I'd be interested in that as well. That is, we are given $k$ and $(m_1, \ldots, m_k)$, where the $i$-th set must have size $m_i$, and we are to find the smallest $n$ for which the sets can have a single mutual intersection.
Thanks in advance for any help.
It is necessary and sufficient that for any nonzero $d\in\Bbb Z/n\Bbb Z$, there exists $i\in{1,2,\dots,k}$ such that $d\notin (A_i-A_i)$.
Have you tried k different coprime integers j_i?. Then for N less than the product of the j_i, the sets which are 0 mod j_i and have about N/j_i elements each should work. For N=200, the numbers 5,6, and 7 do a nice job. I'm not sure you will do much better than m=200/7 for k=3 and N=200. Gerhard "Residue Classes Can Be Shifty" Paseman, 2020.08.12.
I think the sets work for $n\ge m$. I've edited my question to fix this. Thanks for pointing it out.
@GerhardPaseman: We should have $j_i$ as divisors of $n$ as otherwise "negative" differences would not be multiples of $j_i$. Ideally $n$ should be LCM of the $j_i$. For $k=3$ and $n=200$, we have with necessity that $m\leq 34$. See my answer for details.
It is necessary and sufficient that for any nonzero $d\in\Bbb Z/n\Bbb Z$, there exists $i\in\{1,2,\dots,k\}$ such that $d\notin (A_i-A_i)$. In other words,
$$\bigcap_{i=1}^k (A_i-A_i) = \{0\}.$$
This holds even for sets of varying sizes.
Since $| A_i-A_i|\geq |A_i| = m$, we get a necessary condition: $n-1\leq k(n-m)$, that is
$$m\leq \frac{(k-1)n+1}k.$$
For varying set sizes, it is
$$n\geq \frac{m_1+\cdots+m_k-1}{k-1}.$$
Another necessary condition can be obtained from the observation that for any $a\in\Bbb Z/n\Bbb Z$, there exist $m^k$ vectors $(c_1,\dots,c_k)$ such that $a\in \bigcap (A_i+c_i)$. Since these vectors must be distinct for distinct $a$, we have $n\cdot m^k\leq n^k$, that is
$$m\leq n^{(k-1)/k}.$$
This condition implies that the given example for $k=2$ is optimal when $m^2\leq n<(m+1)^2$.
For varying set sizes, the last condition takes form:
$$n\geq (m_1\cdots m_k)^{1/(k-1)}.$$
As for the construction, the following streamlining of Gerhard's idea does the job for a given $k$, although it's not necessarily optimal.
Take any integers $0<b_1<b_2<\dots<b_k$, set $n:=\mathrm{LCM}(b_1,\dots,b_k)$, $m:=\frac{n}{b_k}$, and select $A_i$ as any subset of $b_i(\Bbb Z/n\Bbb Z)$ with $|A_i|=m$. Indeed, in this setting, for any nonzero $d\in\Bbb Z/n\Bbb Z$, there exists $i\in\{1,2,\dots,k\}$ such that $b_i\nmid d$, implying that $d\notin (A_i-A_i)$.
For Gerhard's example with $k=3$ and $b_1=5$, $b_2=6$, and $b_3=7$, we have $n=210$ and $m=30$.
If we are also given $n$, this construction becomes more tricky as we need to pick $k$ numbers $0<b_1<b_2<\dots<b_k$ with the smallest $b_k$ possible such that $n\mid \mathrm{LCM}(b_1,\dots,b_k)$. It is clear that $b_k$ cannot be smaller than the largest primepower dividing $n$. It also follows that here it does not make sense to have $k$ greater than the number of distinct primes dividing $n$, since $b_i$ being the distinct primepowers forming the prime factorization of $n$ do the job.
|
2025-03-21T14:48:31.759286
| 2020-08-12T21:11:02 |
369001
|
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|
Stack Exchange
|
Information density of proofs?
I am a CS person so please excuse the hand-waving.
Given a set of machine-represented proofs, each different (but not necessarily proving a different thing), what sort of information-theoretic statements could we make about these?
I would define the information "density" of a proof by simply losslessly compressing it using some standard means (like lz4 or rle or whatever is suited to this domain) and comparing the compressed size to the input size. The compression ratio.
For example: if we compress them all, would they all have similar compression ratios (meaning non-redundant/repeating information content)?
Given a set of different proofs, all proving the same thing (say like the prime number theorem), what is the most dense? The least dense?
Are there useful proofs that are not "information dense", meaning they have long repeating sequences or repeating structure that is easy to compress. These would look tedious to a human but would be discover-able by a machine.
Are there any papers that have looked at this? What area is this? My googling has failed me.
This is non-trivially related to the recent question https://mathoverflow.net/questions/368893/natural-metrics-for-proof-length.
First, beware: this can only be done meaningfully within a fixed "language of proof". If you try to compare across different systems, you can get wildly different results. There is a whole domain of Proof Complexity which has lots of results about "expressivity". Adding features to your language (like let x = <big-expression> in <other-expression>) can make a huge difference. Other subtler changes can still result in exponential differences in proof length.
Similarly, which compressor you use matters. Kolmogorov Complexity relies crucially on Turing-completeness to obtain a 'stable' measure of length. Normal compressors can be used to approximate it, and give surprising good results in practice. Paul Vitanyi has written a variety of papers on that topic.
Regarding a specific example, like the prime number theorem, an obvious question becomes: what results do you consider to be 'background' and which are part of the proof? The stable answer, again relying on Kolmogorov Complexity, which I explain in Understanding expression simplification [ISSAC 2004], is to in fact consider the proof length to contain all the previous results needed to express the proof as part of its length. Then, to compare two proofs, you subtract their lengths, which takes care of 'quotienting out' all of the material that is common. However, if the two proofs take wildly different routes through mathematics, you might need to use finer concepts such as relative complexity, i.e. you compress the proofs on their own (but with the full library), and concatenated. See the works of Vitanyi for details and variants.
I would say that we currently don't really too much about this topic. We're still in the stage where we're finding uniform ways of writing down mathematical proofs. While the existing libraries are getting large, they are still not quite large enough to do convincing data mining, although that hasn't stopped many from doing so already. But they are aware that their results are very system-relative.
Some interesting early results are available on decently large databases: some use the whole of the arxiv as their data source. I quite like Discovering Mathematical Objects of Interest—A Study of Mathematical Notations for example.
The Kolmogorov complexity was the concept I was looking for (versus "compression"), thanks! I have actually heard of this before in the context of cellular automata, but I forget exactly how. Now let me attempt that paper..
|
2025-03-21T14:48:31.759687
| 2020-08-12T22:01:25 |
369005
|
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|
Stack Exchange
|
The cohomology of modular curves as a module over the Galois group
Consider the modular curve $\pi: X(N) \to X(1)$ where this map has Galois group $G = PSL_2(\mathbb Z/N\mathbb Z)$. In particular, $G$ acts on the singular cohomology $H^1(X(N),\mathbb Z)\otimes \mathbb C$ or in finite characteristic, on the etale cohomology group $H^1(X(N),\mathbb Z_\ell)\otimes_{\mathbb Z_\ell}\overline{\mathbb Q_\ell}$.
Do we know which irreducible representations of $G$ appear in the cohomology and with what multiplicities. Also, we can ask how the action of $G$ interacts with the Hecke operators, for instance. This seems to me to be very classical automorphic stuff but I have no knowledge about this area of math. Are there any friendly references?
Looking at the dimensions, I don't believe it is the regular representation.
This is classical, but I don't know a reference, One way to calculate this representation is by its character. You can use the fact that the trace of a nontrivial element is $2$ minus its number of fixed points of that element on $X(N)$. For any element that does not have order $2$, order $3$, or is unipotent there are no fixed points but for these special elements there are fixed points, so it's not quite the regular representation.
@DavidLoeffler Thank you! This is exactly the kind of thing I was looking for. The proof seems to be along the same lines as Sawin's suggestion.
Comment now reposted as an answer.
You can look at Kato, $p$-adic Hodge theory and values of zeta functions of modular forms, (4.9.3) and (4.9.4) for the question of how $G$ interacts with the Hecke operators.
Jared Weinstein's PhD thesis (http://math.bu.edu/people/jsweinst/jswthesis.pdf) is an excellent reference for this kind of thing. See section 3.4 in particular, where he computes the space $S_k(\Gamma(N), \mathbb{C})$ as a $\mathbb{C}[\mathrm{SL}_2(\mathbb{Z}/N)]$-module using an equivariant version of the Riemann--Roch formula.
Just to make sure I understand things correctly: If I wanted to use a similar decomposition for etale cohomology instead, I would have to pass to an algebraic closure, use a comparison theorem with algebraic de rham which in turn by hodge theory is the sum of sheaf cohomology of $H^0(X,\omega)$ and it's dual. Since everything is functorial, this isomorphism also preserves the structure of the group action. Do I have that right?
Yes, that's right, assuming that by $\omega$ you mean the sheaf of differentials, and you take k=2 in Jared's formulae.
Yep, that's what I had in mind. Thanks a lot!
|
2025-03-21T14:48:31.759897
| 2020-08-12T22:24:21 |
369007
|
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|
Stack Exchange
|
Examples of conjectures whose direct falsity implies different consequences than indirect falsity
Mathematics several times has statements of form
$$\mathsf{Statement A}\implies\mathsf{Statement B}$$
where $\mathsf{Statement A}$ and $\mathsf{Statement B}$ are conjectures while the implication is provable.
In such cases falsity of $\mathsf{Statement B}$ implies falsity of $\mathsf{Statement A}$. However since falsity of $\mathsf{Statement A}$ does not imply falsity of $\mathsf{Statement B}$ it might be that disproving $\mathsf{Statement B}$ might be the easiest route to disproving $\mathsf{Statement A}$. However a direct disproof of $\mathsf{Statement A}$ might reveal something else not directly revealed by $\mathsf{Statement B}$ without falsifying $\mathsf{Statement B}$. Are there known good examples?
It would be nice if falsity of $\mathsf{Statement B}$ came before falsity of $\mathsf{Statement A}$ which would make for the case that falsity of $\mathsf{Statement B}$ was indeed easier.
Please rewrite the actual question so it is clearer exactly what you seek an example of. I can’t tell what aspect of your Mertens/RH example makes it a full or partial example of what you are interested in. And the title of the post is quite hard to parse.
Mertens/RH is a partial example. I am unable to come up with one concrete example even though the possibility exists.
Both statements are conjectures and the implication is known result.
Here's an example from computability theory.
Statement $A$:"The Turing degrees are linearly ordered".
Statement $B$: "The $\Sigma^0_1$ Turing degrees are linearly ordered".
Statement $A$ was refuted by Kleene and Post 1954, with a construction that foreshadows the notion of forcing made famous by Cohen and the continuum hypothesis.
Statement $B$ was refuted by Friedberg and Muchnik in 1957 using another revolutionary method, the priority argument.
Even though the method for Statement $B$ is more powerful in its context, the "easier" method for Statement $A$ turns out, when viewed in the right light, to be an early version of a perhaps even more influential method.
If $1957$ came before $1954$ this would have been a perfect example.
I wouldn't say the Kleene-Post construction foreshadows forcing, except in the sense that the Baire category theorem foreshadows forcing. The Kleene-Post construction is essentially a category argument, showing that the Turing-incomparable pairs of reals form a comeager set.
@AndreasBlass Would you say Spector's minimal degree construction foreshadows/is forcing?
My (non-expert) impression is that Spector's construction is essentially what one must add to the original forcing ideas of Cohen to produce Sacks forcing and a minimal degree of constructibility.
|
2025-03-21T14:48:31.760102
| 2020-08-12T23:37:26 |
369011
|
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|
Stack Exchange
|
Nontrivial integer homology class implies orientability
I posted this question on MSE and I would like to see if my reasoning is correct.
Let $M^3$ be a compact, connected and oriented $3$-manifold with nonempty boundary and let $\Sigma^2$ be a compact and embedded surface such that $\Sigma \cap \partial M = \partial \Sigma$. If $\Sigma$ determines a nonzero class in $H_2(M, \partial M;\mathbb{Z})$, does it follow that $\Sigma$ is orientable?
Here is what I thought. Let $\phi : (\Sigma, \partial \Sigma) \to (M, \partial M)$ be the embedding of $\Sigma$ into $M$. This induces a map $$\phi_* : H_2(\Sigma, \partial \Sigma; \mathbb{Z}) \to H_2(M, \partial M; \mathbb{Z})$$
in homology. Since $\Sigma$ determines a nonzero class in $H_2(M, \partial M; \mathbb{Z})$, this implies that there exists a nonzero class $\alpha \in H_2(\Sigma, \partial \Sigma; \mathbb{Z})$ such that $\phi_*(\alpha) \neq 0$. In particular, $H_2(\Sigma, \partial \Sigma; \mathbb{Z}) \neq 0$, which means precisely that $\Sigma$ is orientable.
$\Sigma$ does not determine a class at all (zero or otherwise) unless it's oriented; it doesn't have a fundamental class to push forward!
|
2025-03-21T14:48:31.760198
| 2020-08-12T23:52:07 |
369012
|
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|
Stack Exchange
|
Large deviation for Brownian occupation time
I am asking for reference about the large deviation principle (LDP) for the occupation time of a Brownian motion/bridge. Let $f:\mathbb{R} \to \mathbb{R}$ be smooth and compactly supported. My question is: What is the LDP of
$$\lambda^{-1} \int_0^\lambda f(B_s) ds, \quad\lambda \to \infty \
$$
Here, $B_s, s \in [0, \lambda]$ can be either a Brownian motion or a Brownian bridge with endpoints zero.
I find that Theorem 3.1 of http://www.brunoremillard.com/Papers/wiener.pdf provides a related result. However, I do not think this note is very well-written and I am asking for a better reference. Thank you!
This is precisely the Donsker-Varadhan LDP, coupled with an application of the contraction principle. Namely, the rate function is
$$I(x)=\inf\{ J(\mu): \int f d\mu =x\}$$
where $J$ is the Donsker-Varadhan rate function. Look at the series of Donsker-Varadhan papers from 1975 (#I is the one you need) and any text on large deviations theory for the contraction principle (or the wikipedia page).
This is very helpful, thank you!
|
2025-03-21T14:48:31.760295
| 2020-08-13T01:27:05 |
369017
|
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|
Stack Exchange
|
About Homotopy Transfer Lemma
If M, A are two differential graded complexes over a commutative ring R with the following data,
$$(M,d_M) \overset{\Delta}{\longrightarrow} (A,d_A)$$ $$(A,d_A) \overset{f}{\longrightarrow} (M,d_M)$$ $$A \overset{\phi}{\longrightarrow} A $$ with the following conditions, $f \Delta=1_{M}, \Delta f=1_{A}-(d_{A} \phi+\phi d_{A}) , \phi \Delta=0, f \phi=0 $. Such a data is called Strong Deformation Retract Data(SDR).
In this setup, suppose t is a perturbation of $d_{A}$ i.e ($d_A+t$ )is a new differential in A which amounts to saying t satisfies $ [d_A,t]=0 , t^{2}=0$. Then Homotopy Transfer lemma allows you to define a new differential on $(M,\Delta_{\infty})$ so that $(M,\Delta_{\infty}) , (A,d_A+t)$ are SDR pair.
Such a solution can be formally written as described in this paper
http://citeseerx.ist.psu.edu/viewdoc/download?doi=<IP_ADDRESS>.8783&rep=rep1&type=pdf
I wanted to know if there is any Homotopy Transfer Lemma without the hypothesis $f\Delta=1$. What if there is another map $\psi: M \rightarrow M $ satisfying $f\Delta=1_{M} -(d_{M}\psi+\psi d_{M})$? Can we still expect such a lemma? If yes, can we write down an explicit formula for the new differential as before? Please point me to the right reference. Thank you!
|
2025-03-21T14:48:31.760398
| 2020-08-13T01:50:39 |
369019
|
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"Mehta",
"Mikhail Borovoi",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369019"
}
|
Stack Exchange
|
Existence of inner twisting preserve the $*$ action on simple roots
I've been trying to learn the theory of inner and outer Galois twisting in algebraic groups.
Let $k$ be a field of characteristic 0, and let $k^s$ be a separable closure. Let $G$ be a connected semisimple group over $k$. Then it is an inner $k$-form of some $k$-quasisplit group, say $G'$. There is an isomorphism $f:G \rightarrow G'$ defined over $k^s$ such that for all $\gamma$ in the Galois group $\Gamma(k^s/k)$ of $k^s/k$, we have $f^{-1} \cdot\, ^\gamma\! f \in \text{Int}(G)$.
Choose a Borel subgroup $B$ of $G$ (resp. $B'$ of $G'$) and a maximal torus $T$ (resp. $T'$) of $B$ (resp. $B'$), all defined over $k^s$. Let $\Delta$ and $\Delta'$ be the set of simple roots of $(G,T)_{k^s}$ and $(G',T')_{k^s}$. The Galois group $\Gamma(k^s/k)$ acts on each of these via the $*$-action. In the case of $G'$, since $G'$ is quasisplit, the $*$-action is just the standard action on the character lattice. Let $\rho$ be a bijection from the set of orbits in $\Delta$ onto the set of orbits in $\Delta'$.
Question: Does there exist an isomorphism $f:G \rightarrow G'$ defined over $k^s$ such that for all $\gamma \in \Gamma(k^s/k)$, we have $f^{-1} \cdot\, ^\gamma\! f \in \text{Int}(G)$, and such that the induced isomorphism $X(f)$ of based root data, when restricted to $\Delta$, coincides with $\rho$?
I basically need the Galois action to be preserved by $f$. At first I thought I could just take any inner twist $f$ and compose it with an automorphism of $G'$, but I don't think that works.
In general No, because the induced isomorphism of the based root data must induce an isomorphism of the Dynkin diagrams.
What if the group is adjoint?
|
2025-03-21T14:48:31.760535
| 2020-08-13T02:32:27 |
369021
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369021"
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|
Stack Exchange
|
Mutually orthogonal Latin hypercubes
A $d$-dimensional Latin hypercube with side length $n$ is a $d$-dimensional array with $n$ symbols such that along any line parallel to an axis, each symbol appears exactly once. Let us call a $(n,d)$ mutually orthogonal Latin hypercube (MOLH) a set of $d$ $d$-dimensional Latin hypercubes of side length $n$ such that overlaying the $d$ hypercubes results in every element of the Cartesian product of the symbols appearing exactly once. For example, $(n,2)$ MOLH are the traditional Graeco-Latin squares.
I am interested in the case where $n=d+1$. $(3,2)$ MOLHs exist, because there are Graeco-Latin squares of side length 3. But are there $(d+1,d)$ MOLHs for $d\ge3$? I haven't been able to find this answered in the literature on the subject.
Thanks for any help.
0123 1032 2310 3201
1032 2310 3201 0123
2310 3201 0123 1032
3201 0123 1032 2310
----------------------
0231 3102 1320 2013
1320 2013 0231 3102
3102 1320 2013 0231
2013 0231 3102 1320
----------------------
3210 2301 1032 0123
1032 0123 3210 2301
0123 3210 2301 1032
2301 1032 0123 3210
answers the question in the affirmative for $d=3$. This is one of 493056 distinct (but not necessarily non-isomorphic) solutions where the first cube is one of the five representatives of the main classes and the other two are considered as an unordered pair of cubes derived from one not later than the first one in the list of representatives.
Since there are considerably more 4-dimensional Latin hypercubes of side length 5, I expect there to be correspondingly more $(5, 4)$ MOLHs, but the computational task of finding them is rather larger.
|
2025-03-21T14:48:31.760659
| 2020-08-13T04:59:45 |
369024
|
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"Eric Rowell",
"Sebastien Palcoux",
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"url": "https://mathoverflow.net/questions/369024"
}
|
Stack Exchange
|
Existence of twisted metaplectic categories
The paper Classification of metaplectic modular categories by Ardonne-Cheng-Rowell-Wang (2016) mentions (in Section 3) the Grothendieck ring for the metaplectic modular categories, i.e. $SO(N)_2$, $N>1$ odd (see below).
Now by exchanging $1$ and $Z$ in lines (2) and (3), we get a new family of fusion rings.
Question 1: Are there (unitary) fusion categories corresponding to this new family?
(which could be called twisted metaplectic)
Question 2: Can this procedure be extended to other fusion categories?
There is a way to make two new families: in the right hand-side of (2) and (3), put $$\oplus (X_1 \oplus X_2)^{\oplus \frac{N-1}{2}}$$ By this modification, you still have a fusion ring, but the objects $X_1, X_2$ are now of FPdim $N$. Again you have the usual and the twisted version. Note that after this modification, the usual case with $N=3$ corresponds to the Grothendieck ring of $\mathrm{Rep}(S_4)$.
Bonus question 1: Are there (unitary) fusion categories corresponding to these new families?
Bonus question 2: Can this procedure be extended to other fusion categories?
In https://arxiv.org/pdf/2005.05544.pdf we describe a general procedure that accomplishes (1) and (2) which we call zesting.
In this case, let $\mathcal{C}$ be an odd metaplectic category. In the notation above, let $\lambda\in Z^2(\mathbb{Z}/2,Inv(\mathcal{C}))$ be chosen with $\lambda(1,1)=Z$. Then a new tensor product defined by $U\hat{\otimes} V\cong U\otimes V\otimes \lambda(d(U),d(V))$ where $d$ is the grading degree function, utilizing the $\mathbb{Z}/2$ grading on $\mathcal{C}$. In particular $X_1\hat{\otimes} X_2\cong X_1\otimes X_2\otimes Z\cong X_1\otimes X_1$ etc. As $Z$ is a boson (we are using the braiding on $\mathcal{C}$ in our construction) Proposition 6.3 in loc. cit. says that we further twist the associativity by a $3$-cocycle.
Proposition 6.4 shows that the resulting fusion category does not admit a braiding.
One can proceed in a similar way for general $N$-metaplectic categories too (i.e. $N$ can be odd or even).
Thanks for your answer, this is exactly what I was looking for!
Do you know if this procedure keeps the unitarity?
Have you heard about the other procedure mentioned in second part of the post (i.e. the bonus questions)?
(I just fixed a typo there, it was not (1) and (2), but (2) and (3))
I believe that zesting does not ruin unitarity--the various morphisms that get changed are in the original category, just twisted by some roots of unity factors typically. While this is not a proof, I think one could make it rigorous.
The other fusion rules could potentially be obtained as a Z2-equivariantization of the near-group categories of type Z/N +(N-1) in the Evans-Gannon notation. Just a guess, but the numerology seems to work out.
|
2025-03-21T14:48:31.760860
| 2020-08-13T05:40:00 |
369026
|
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"Dan Turetsky",
"Noah Schweber",
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|
Stack Exchange
|
Number of models vs. complexity for SOL theories
This was previously asked at MSE without success.
Suppose $T$ is a complete first-order theory with continuum-many countable models up to isomorphism. We define two sets of Turing degrees associated to $T$ via second-order logic:
$SecTh(T)$ is the set of Turing degrees of second-order theories of models of $T$.
$SecTh_0(T)$ is the set of Turing degrees of second-order theories of countable models of $T$.
I'm interested in how simple these sets can be. Specifically:
Is there a $T$ such that $SecTh(T)$ is not cofinal in the Turing degrees? If not, what about $SecTh_0(T)$?
Recall that $X$ is cofinal in the Turing degrees if every Turing degree is below some element of $X$.
Here's what I've been able to figure out already:
Assuming $V=L$, the answer is negative for $SecTh_0$ (and hence $SecTh$ a fortiori). The key point is that under $V=L$ the set of pointwise-definable levels of $L$ is unbounded in $\omega_1$. Given a countable $\mathcal{A}\models T$, let $\alpha_\mathcal{A}$ be the least index of a pointwise-definable level of $L$ containing an isomorphic copy of $\mathcal{A}$. The second-order theory of $\mathcal{A}$ computes the first-order theory of $L_{\alpha_\mathcal{A}}$, which in turn computes every real in $L_{\alpha_\mathcal{A}}$ by pointwise definability. Now simply use the fact that $\{\alpha_\mathcal{A}:\mathcal{A}\models T\}$ is unbounded in $\omega_1$. (Note that this is a "naive" version of the idea behind mastercodes.)
I don't actually see an immediate argument that $SecTh(T)$ need be uncountable! A priori, $Th_2(\mathcal{A})$ may not be enough to build a concrete copy of $\mathcal{A}$ in any sense that I can see. It is consistent that $SecTh_0(T)$ (and hence $SecTh(T)$) is always uncountable, since it's consistent that no two countable structures are second-order elementarily equivalent (see Marek's theorem mentioned here), but that's the best I know.
Under sufficiently strong large cardinal hypotheses, $SecTh_0(T)$ is subject to Martin's Cone Theorem, and hence for an affirmative answer for the second question it would be enough to find a $T$ such that $SecTh_0(T)$ does not contain an upper cone. Under such hypotheses and replacing $\mathsf{ZFC}$ with $\mathsf{ZF}$ + Turing Determinacy the same would hold for $SecTh(T)$, but I don't immediately see how to get this stronger result without determinacy.
I'm not practiced at working with 2nd order theories, but my first thought is the theory of a discrete linear order without endpoints, i.e. T = Th($\mathbb{Z}$). The models are linear orders of the form $\mathbb{Z}\cdot L$, but it's not clear to me the second order theory of a model will let you recover $L$, or have any particular computational power.
@DanTuretsky I don't think that will work. In a second-order way we can indeed recover $A$ from $\mathbb{Z}\cdot A$ (I want to reserve "$L$" for the constructible hierarchy): it's just $\mathbb{Z}\cdot A$ modulo the finite distance relation, and the latter is second-order definable. And this lets us get a lot of power out of the second-order theories of models of $Th(\mathbb{Z},<)$, since in particular whenever $\alpha$ is a countable ordinal we'll be able to compute $L_\alpha$ from $Th_2(\mathbb{Z}\cdot\alpha,<)$.
|
2025-03-21T14:48:31.761095
| 2020-08-13T08:06:27 |
369029
|
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"Giorgio Metafune",
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|
Stack Exchange
|
Is the maximum of derivatives of a function in (s,2)-Sobolev space (an RKHS) bounded by their norms?
Let $f(x) \in W^{s,2}(\Omega) \equiv H^s$, where $\Omega \subseteq \mathbb{R}^d$, $s > d/2$ and $W^{s,2}$ is a $(s,2)$-Sobolev space. Clearly, $W^{s,2}$ is an Reproducing Kernel Hilbert Space (RKHS) and therefore $|f(x)| \le M\left\|f\right\|_{H^s}$ for $M > 0$ holds.
The question is: does this also hold for derivatives of $f$ (they're in $L^2(\Omega)$, but does $|\nabla_x f(x)| \le M'\left\|\nabla_x f\right\|_{L^2}$ hold for some $M' > 0$?)?
Thank you
Update
Following @NateEldredge counter-example, if we tighten the requirements s.t. $s \ge 2$ then it'd seem that a bound on $|\nabla_x f(x)|$ in terms of norms does exist, albeit with a different norm.
Consider the case $d = 1$. As stated above, we assume that $f \in H^s(\Omega)$ and $s > d/2$. By requiring that $s \ge 2$, it follows that $g(x) \equiv f'(x) \in H^{s-1}(\Omega)$. Therefore, $g(x)$ is in an RKHS (albeit a different from $f(x)$'s one). Thus, $|g(x)| = |f'(x)| \le M_g \left\| f' \right\|_{H^{s-1}(\Omega)}, M_g>0$.
A $H^s$ norm verifies $\left\| f \right\|_{H^s}^2 \doteq \sum_{|\alpha|_1 \le s} \left\| D^\alpha f \right\|_{L^2(\mathbb{R}^d)}^2$. Therefore,
$$
|f'(x)|^2 \le M_g^2(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\
$$
This process can be used with any $s \ge 2$.
Is this reasoning correct?
Thx
Many things are not clear: what is a RKHS? If $f$ is a fixed function, where a linear operator appears? Is $x$ fixed? Please, edit.
RKHS - Reproducing Kernel Hilbert Space; $f$ is not fixed, but has an argument $x$;
$x \in \Omega \subseteq \mathbb{R}^d$
$x\mapsto f(x)$ is not linear. I do not understand what you are talking about.
Fixed the question. $f$ is not bounded, the evaluation functional is. But I'm asking about the implication of it being one - does this hold for derivatives of $f$ too?
I think you need $s > d/2$ in order to have an RKHS; otherwise the elements of $H^s$ are not continuous and evaluation at a point is not well-defined and does not give a bounded functional.
But I think your question has an immediate counterexample: take $d=1$, $\Omega=(0,1)$, $s=1$. Then $f'$ can be any $L^2$ function, but clearly you cannot control the pointwise values of an $L^2$ function in terms of its $L^2$ norm (they are not even well defined).
@NateEldredge OK, I've seen such a requirement ($s > d/2$), e.g. in https://arxiv.org/pdf/1709.02568.pdf. I can live with it. The question still stands though when the restriction is imposed.
In your last sentence, can you clarify what norm you want to apply to $\nabla_x f$?
@NateEldredge I'd say $L^2$?
I think that you may want to study the standard Sobolev inequalities and embedding theorems, as they give clear answers as to exactly what results of this kind are true. For example, they indicate why $s>d/2$ is necessary to get a RKHS. They are essentially the first thing that anybody working with Sobolev spaces needs to learn.
Consider $d=1$, $s=1$, $\Omega=(-1,1) \subset \mathbb{R}^1$. We can find a function $g \in L^2(-1,1)$ (or even continuous) for which $g(0)$ is arbitrarily large but $\|g\|_{L^2(-1,1)}$ is arbitrarily small. Set $f(x) = \int_{-1}^x g(t)\,dt$; then clearly $f \in H^1(-1,1)$ with $f'=g$, and we can violate any proposed bound of the form $|f'(0)| \le M' \|f'\|_{L^2(-1,1)}$.
Thx. A question: Does this change if $s > d/2$?
@qwer1304: My example does have $s>d/2$. You can modify it to use $s$ as large as you wish, if you take $g$ to be a smooth function instead.
OK, clearly this cannot work with $\left|\cdot\right|_{L^2}$ since $L^2$ is not an RKHS. Can some reasonable norm be defined s.t. the bound does work?
I think that you will have an inequality like $|\partial^\alpha f(x)| \le M'(x, \alpha) |f|_{H^s}$ under an assumption like $s > \frac{p}{2} + |\alpha|$. Again, it would come from classical Sobolev inequalities.
|
2025-03-21T14:48:31.761373
| 2020-08-13T09:25:12 |
369035
|
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|
Stack Exchange
|
Probability of complex eigenvalues
I find this is the best site to post this question, even though I considered cs.
It is a Monte Carlo experiment over the set of 10.000 n×n matrices.
If a single matrix eigenvalue is complex then python numpy package will return all the eigenvalues as numpy.complex128 type, else it will return all eigenvalues as numpy.float64 type.
Here is the algorithm:
# Monte carlo experiment in numpy
import numpy as np
from numpy import linalg as LA
e=10000 # examples
# n - matrix rank
for n in range(1,10):
cc=0 # complex counter
for i in range (e):
m=np.random.randn(n,n)# m=np.random.random((n,n))
w, _ = LA.eig(m)
if (type(w[0]).__name__ == "complex128"):
cc+=1
print(f'matrix:{n}x{n}, total cases: {e}, complex cases: {cc}, ratio: {cc/e}')
The output of this algorithm is as follows:
matrix:1x1, total cases: 10000, complex cases: 0, ratio: 0.0
matrix:2x2, total cases: 10000, complex cases: 2851, ratio: 0.2851
matrix:3x3, total cases: 10000, complex cases: 6481, ratio: 0.6481
matrix:4x4, total cases: 10000, complex cases: 8782, ratio: 0.8782
matrix:5x5, total cases: 10000, complex cases: 9674, ratio: 0.9674
matrix:6x6, total cases: 10000, complex cases: 9944, ratio: 0.9944
matrix:7x7, total cases: 10000, complex cases: 9998, ratio: 0.9998
matrix:8x8, total cases: 10000, complex cases: 9999, ratio: 0.9999
matrix:9x9, total cases: 10000, complex cases: 10000, ratio: 1.0
It is very easy to understand the 1x1 case, since we take values from floating point numbers (set $\mathbb Q$) the eigenvalue will be the number itself (also in $\mathbb Q$).
For the case 2x2 and later, we may get complex eigenvalues ($\mathbb Z$).
I am counting the complex cases (where at least one eigenvalue is complex) and providing the ratio.
I would like to get some thoughts about this ratio possible the formula how to calculate one analytically.
My thought is that the ratio is not dependent on floating point arithmetic.
I used the normal distribution $\sim \mathcal N(0,1)$, but one may test also the uniform distribution I commented.
To get a first guess of the functional dependence of the ration with respect to $n$: I would try first if f.i. this may be a logistic function. You can estimate the parameters of this function with programs as R or the python module pandas.
You may be interested in the paper "How Many Eigenvalues of a Random Matrix are Real?" by Alan Edelman, Eric Kostlan and Michael Shub
The fraction of eigenvalues that are strictly real decreases as $1/\sqrt{n}$
The probability that a $n\times n$ real matrix (with elements that are independent random variables with standard normal distributions) has only real eigenvalues is given by
$$ 2^{-n(n-1)/4}$$
Reference: A. Edelman, The Probability that a Random Real Gaussian Matrix has $k$ Real Eigenvalues, Related Distributions, and the Circular Law. Journal of Multivariate Analysis 60, 203-232 (1997).
$1-2^{- \frac{n(n-1)}{4}}$ fits nice to what I got with MC. Thanks.
@Ben Crowell There are $n^2$ independent random variables involved, namely the entries of the matrix... for this reason, the rate $n^2$ is typical of large deviations for the empirical measure of eigenvalues of random matrices.
|
2025-03-21T14:48:31.761728
| 2020-08-13T11:08:04 |
369044
|
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|
Stack Exchange
|
Is every countable discrete group a subgroup of a non discrete Lie group?
1)Let $G$ be a countable discrete group. Can $G$ be embbeded in a locally connected Lie group?
2)let $G$ be a countable discrete group with a prescribed generating set and corresponding word metric. Can $G$ be embbeded isometrically in a locally connected Lie group with its left invariant metric?
Remark: We emphasis on the word "Locally connected" because of the example $G\subset \mathbb{R}\setminus\{0 \}$ with $G=\{\pm e^n\mid n\in \mathbb{Z} \}$ with multiplication.
I assume you want more conditions, or else you can just take $G \times \mathbb R$?
@KevinCasto Thank you for your attention. To be honnest I was considering $\mathbb{Z}\subset \mathbb{R}$ or $\mathbb{Z}/2\mathbb{Z} \subset S^1$. Could you please help me to impose some extra conditions to have a minimal number of connect compenent. I had intention to ask "A connected Lie group". I am not sure if there is a connected Lie group containing the example in the "Remark".
@KevinCasto Infact one feel that $\mathbb{R}$ intrincically contains $\mathbb{Z}$. The same for$S^1$ containing finite cyclic group. So accoring to your interesting comment one impose the following question: Let G be a countable discrete group. Is $G$ a subgroup of a connected Lie group?
In that case there is a huge literature on this subject. See e.g. this question: https://mathoverflow.net/questions/110208/understanding-groups-that-are-not-linear I suppose technically you need to worry about nonlinear Lie groups like $\widetilde{SL_2}(\mathbb R)$ but I suspect the properties listed in the first answer to that question also apply to discrete subgroups of this.
@KevinCasto Thank you for this link. Befor I start reading this link, do you say that the link contains an answer to the question that every countable group is (isometrically) a subgroup of a connected Lie group?
Every Lie group is locally connected, so the answer to your question is obviously yes since every countable discrete group is a Lie group. On the other hand, many countable discrete groups do not embed (even algebraically) in connected Lie groups.
@MoisheKohanThank you. By locally connected I was (somehow) meant : non discrete.May be some example for your last sentemse?this is what I search for.
If you mean nondiscrete, then the firs comment by Kevin answers the question. As for examples, take any centerless nonresidually finite group such as BS(2,3), Baumslag Solitar group. Or take any group with undecidable word problem if you want a more exotic example.
@MoisheKohan So I realize that the ocean of Lie groups is wider and deeper than I imagined. Among various example you and Kevin pointed ot, what is a precise example of a countable group not embaddable in a connected lie group of dimension at least 1.
@KevinCasto Please read my previous comment.
$BS(2,3)=<a,b| ab^2a^{-1}=b^3>$ does not embed in a connected Lie group.
@MoisheKohan Ok thanks,I try to understand the reason.
Malcev proved that every finitely-generated matrix group $\Gamma$ (over any field) is residually finite, i.e. the intersection of all finite-index subgroups of $\Gamma$ is $\{1\}$. Baumslag-Solitar groups, such as $BS(2,3)= \langle a, b | ab^2 a^{-1} =b^3\rangle$, are among simplest examples of finitely generated groups which are not residually finite. A connected Lie group $G$ need not be linear (the universal covering group of $SL(2, {\mathbb R})$ is a standard example). However, the kernel of the adjoint representation $Ad_G$ of a connected Lie group $G$ is always contained in the center of $G$. Thus, if $\Gamma< G$ is a centerless subgroup, then the restriction of the adjoint representation $Ad_G$ to $\Gamma$ is faithful and, hence, $\Gamma$ is isomorphic to a matrix group. It is not hard to see that $BS(2,3)$ has trivial center. Thus, this group is not isomorphic to a subgroup of any connected Lie group. The same proof shows that $BS(2,3)$ is not isomorphic to a subgroup of a Lie group with finitely many components.
Remark. The standard definition of "locally connected" in topology is that every point should have a neighborhood basis consisting of connected subsets. Hence, each manifold (in particular, each Lie group) is, by definition, locally connected. Given your example, it seems that what you really had in mind is that a Lie group $G$ should have Alexandroff compactification $G\cup \{\infty\}$, such that $\infty$ admits a neighbourhood basis $U_i$ satisfying the condition that $U_i\cap G$ is connected. It is easy to see that this requirement is equivalent to the condition that $G$ is connected and 1-ended (equivalently, is neither compact nor a product of compact group with ${\mathbb R}$). I am not sure what to call this property, let's name it ($*$). Then every discrete countable group $\Gamma$ embeds in a Lie group with property ($*$), e.g., $G=\Gamma \times {\mathbb R}^2$.
Hi Moishe. This isn't really appropriate for the MO comment format, but I would like to say I have learned a lot from you and your perspective over the years. Thank you for your contributions.
@MikeMiller: Sure, you are welcome!
I would have posted this as a comment, but I have not yet unlocked that privilege. For a simpler example than Moishe Kohan's, the group $\bigoplus_{\mathbb{N}} \mathbb{Z}/2\mathbb{Z}$ does not embed in any connected Lie group, since a torsion subgroup of a connected Lie group $G$ is contained in a maximal compact subgroup of $G$ (this is a result of D. H. Lee), and it is easy to see that $\bigoplus_{\mathbb{N}} \mathbb{Z}/2\mathbb{Z}$ does not embed in $\mathrm{U}(n)$ for any $n$. Also, it follows from the Tits alternative for linear groups that any group that is not virtually solvable and does not contain a nonabelian free subgroup does not embed in a connected Lie group (or a Lie group with finitely many connected components).
|
2025-03-21T14:48:31.762109
| 2020-08-13T11:19:42 |
369047
|
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|
Stack Exchange
|
How do I fit flow values to connections in a known network?
This is not my area and I'm new to its terminology, and am posting my problem in the hope that someone will be able to direct me to where it has been solved, or who has written about it.
I have a flow network - I'm imagining intersections connected by tubes. I have several sources and sinks. I know the total volume moving through the network, and values for the total flow into and out of each intersection. I am trying to calculate the distribution of 'flow' through each tube given these conditions. The tubes, we imagine, have infinite capacity, and it 'costs' nothing to push flow through them. Is this a solved problem? What is it called?
Thanks very much for your help.
This is called the feasible flow problem. You can use network simplex or any linear programming algorithm to find the flow along each arc subject to a supply/demand constraint at each node.
Thank you so much!
Glad to help. Please mark my answer as accepted.
|
2025-03-21T14:48:31.762247
| 2020-08-13T11:35:59 |
369049
|
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|
Stack Exchange
|
Emergence of the discrete from the continuum
An almost eternal theme in Mathematics is the approximation of the Continuum by the Discrete. This core idea goes back at least to Archimedes, and remains active to these very days (and quite likely for the next thousand years) .
You have a continuous structure, say a sphere, and you approximate it as the limit of a series of discrete objects, say polyhedra.
I do not need to provide more examples, I am sure you all have plenty.
But, there is also, though by no means so prominent, a reverse direction.
I am not able to date precisely when it made its debut in the history of mathematics, but it most certainly appears in the revolutionary work of Fourier: take a discrete function, say the Heaviside step function, and approximate it via series of trigonometric functions:
So, here a discrete object is realized as a limit of continuous ones (in this case smooth functions).
I am especially intrigued by this possibility, which, pushed to the extreme, will depict a mathematical world where the Discrete is an emergent phenomenon, out of a Continuum
Thus I ask to everybody:
can you list active research in the way of approximating discrete structures via smooth ones?
For instance, a polyhedron via a series of smooth manifolds, or examples in analytic number theory, or patterns in finite combinatorics out of .....(fill the dots) .
Any thoughtful and possibly well documented answer will get my vote, regardless the domain chosen (in fact, the more examples I will harvest from heterogeneous disciplines, the happier I shall be).
On the other hand, to get the GREEN the stakes are higher: rather than single examples, a sketch of a general perspective on the Discrete as emerging from the Continuum
ADDENDUM: As pointed out by Andreas Blass, I have implicitly conflated two themes here:
discrete as limit of the continuum
emergence of the discrete from some background
Point 2) does not seem to necessarily imply point 1) and probably the same applies the other way around. Which one I am interested in? Easy: BOTH OF THEM.
But now that Andreas has already marked this point, the GREEN ANSWER would be, perhaps, a clarification on the relationship between 1 and 2 (inter alia)
The last line of your question, "the discrete as emerging from the continuum," could also include things like Thom's catastrophe theory, which don't fit the "approximation" theme of the earlier parts of the question.
Central limit theorem, anyone?
@AndreasBlass absolutely ! Yes, I conflated two things which are not necessarily entangled: 1) discrete as limit of the continuum and 2) emergence of the discrete. So, which one? Answer: I am interested in BOTH. Thom Morphogenesis is very welcome
signal processing, lossy data compression of audio and video represented by discrete samples using psychoacoustic and psychovisual models... ask Siri or Alexa...
David, I need to clarify a bit: I am not interested in discretizing the continuum here, rather the opposite: the only reason why I have used signal processing above is because it afford me to use an example of a discrete signal which can be approximated by smooth ones. I hope that someone will find something similar in geometry, combinatorics, etc
Related: When has discrete understanding preceded continuous?. Cited there: László Lovász: "Discrete and Continuous: Two sides of the same?" He looks at e.g., the Laplacian, the Colin de Verdière’s invariant, and many other examples. PDF download.
@JosephO'Rourke fantastic link!
Someone - ideally someone who understands this better than me - should write an answer about Morse theory (and its complex counterpart - Bialynicki-Birula theory)
I just stumbled onto this question googling “discrete continuum.” My guess is they are co-evolving with no obvious best solution. There’s constructor theory that’s trying to bring the best ideas from information technology into physics proper, for example.
In combinatorics, there is no shortage of functions that are hard to compute. Here are two famous examples whose input is a simple graph $G=(V,E)$:
$\alpha(G):=$ size of the largest independent set $I\subseteq V$
$\operatorname{MaxCut}(G):=$ size of the largest cut-set $C\subseteq E$
Both of these discrete functions are $\mathsf{NP}$-hard to compute. However, there are polynomial time–computable functions $\{\alpha'_k\}_{k=1}^\infty$ and $\{\operatorname{MaxCut}'_k\}_{k=1}^\infty$ that map simple graphs to real numbers (i.e., continuous quantities) in a way that monotonically decreases and converges pointwise to the desired functions.
Warning: While the runtime of the $k$th function is polynomial for a fixed precision (at least under certain models), the exponent increases with $k$. See this thread for more details on runtime considerations for semidefinite programs.
To be explicit, $\alpha'_1$ denotes Schrijver's strengthening of Lovász's semidefinite relaxation of the independence number, $\operatorname{MaxCut}'_1$ denotes the Goemans–Williamson semidefinite relaxation of maximum cut, and the sequences arise from sum-of-squares hierarchies that strengthen these relaxations. The claimed pointwise convergence is a consequence of the identities
$$\alpha'_{\alpha(G)}(G)=\alpha(G),\qquad \operatorname{MaxCut}'_{\lceil |V|/2\rceil}(G)=\operatorname{MaxCut}(G),$$
which follow from results in [Lasserre 2002] and [Fawzi-Saunderson-Parrilo 2016], respectively.
absolutely great answer! I had no idea about Lovasz number. You suggestion would be even more captivating if there was underneath something in the way of a smooth approximation of the embedded graph (something like a weird smooth hypersurfaces that approximated the embedded graph asymptotically).
@MircoA.Mannucci - You may appreciate this geometric perspective: both problems can be viewed as maximizing a linear function over a finite collection of points. One can reformulate this as a linear program over the convex hull of these points, but there are exponentially many facets in the resulting polytope. Instead, a semidefinite relaxation considers a spectrahedron that contains this polytope. The SOS hierarchy amounts to a sequence of spectrahedra that converges to the polytope. See this paper for more information in the context of MaxCut: https://arxiv.org/abs/1812.11583
very very very intriguing: thanks again Dustin! This goes a long way toward what I ultimately have in mind, namely showing that virtually any discrete geometric object can be approximated by a suitable collection of smooth ones. Plus. it is actually intriguing to me as a data scientist, but that is another story
The emergence of particles as excited states (quanta) of underlying continuous fields in quantum field theory might fit the bill. See The Unquantum Quantum by David Tong, in the section titled Emergent Integers:
Erwin Schrödinger developed an alternative approach to quantum theory
based on the idea of waves in 1925. The equation that he formulated to
describe how these waves evolve contains only continuous
quantities--no integers. Yet when you solve the Schrödinger equation
for a specific system, a little bit of mathematical magic happens.
Take the hydrogen atom: the electron orbits the proton at very
specific distances. These fixed orbits translate into the spectrum of
the atom. The atom is analogous to an organ pipe, which produces a
discrete series of notes even though the air movement is continuous.
At least as far as the atom is concerned, the lesson is clear: God did
not make the integers. He made continuous numbers, and the rest is the
work of the Schrödinger equation.
In other words, integers are not inputs of the theory, as Bohr
thought. They are outputs. The integers are an example of what
physicists call an emergent quantity. In this view, the term "quantum
mechanics" is a misnomer. Deep down, the theory is not quantum. In
systems such as the hydrogen atom, the processes described by the
theory mold discreteness from underlying continuity.
Perhaps more surprisingly, the existence of atoms, or indeed of any
elementary particle, is also not an input of our theories. Physicists
routinely teach that the building blocks of nature are discrete
particles such as the electron or quark. That is a lie. The building
blocks of our theories are not particles but fields: continuous,
fluidlike objects spread throughout space. The electric and magnetic
fields are familiar examples, but there are also an electron field, a
quark field, a Higgs field, and several more. The objects that we call
fundamental particles are not fundamental. Instead they are ripples of
continuous fields.
See Reason for the discreteness arising in quantum mechanics? for more information along these lines. In particular, note that compact operators (or more generally, operators with compact resolvents) have discrete spectra (see here and here). As this answer says:
There are several forms of discreteness in quantum theory. The
simplest one is the discreteness of eigenvalues and the associated
countable eigenstates. Those arise similarly to the discrete standing
waves on a guitar string. The boundary conditions only allow certain
standing waves that nicely fit into the enforced region in space. Even
though the string is a continuous object, its spectrum becomes
discontinuous and is naturally labeled with natural numbers.
Another reason for discreteness comes in with multi-particle systems.
Quantum theory requires that a system that is realized in space-time
contains a unitary representation of the symmetry group of space-time,
the Lorentz group. In fact, you can define a particle in quantum
theory as a subsystem that contains such a group representation. And
because you can't have any non integer fraction of a unitary group
representation, you need to have an integer number of them in your
total system. So the number of particles is also an (expected)
discrete feature, and it plays a role when you talk about single
photons for example, that are either absorbed completely or not at
all.
Yes this is a perspective which definitely goes along the lines sketched above. You got my vote. Yet, would be interesting to see what you said from a mathematical standpoint: how the discrete particles emerge from a quantized field rather than from a classical one?
@MircoA.Mannucci I added a bit more.
Mirco, regarding your second question (emergence from the continuum):
My view would be that the FUNDAMENTAL GROUP of topological objects is an example. It captures discrete properties of a continuous object, e.g. a sphere. Or seen as the deck transformation group, it expresses the discrete structure of the universal cover of a continuous object like torus, sphere, and so on.
The question is where do you draw the line in your OP. For example, one could argue the set of zeros of the sin function is also a discrete set emerging from a continous object. But I imagine this is not what you had in mind?
Interesting thought @MaryS. As for what I have in mind, yes, that too. But it is a bit ad hoc: rather, I would be interested in seeing entire discrete structures as emerging from continuous ones
Bill Lawvere has, for a long time now, promoted the idea that the discrete emerges as a limiting case by abstracting from the continuous. At the most general level, this manifests in the distinction between cohesive and constant (i.e. abstract) toposes of sets:
Quantifiers and Sheaves, Continuously Variable Sets, Variable Quantities and Variable Structures in Topoi, Toward the Description in a Smooth Topos of the Dynamically Possible Motions and Deformations of a Continuous Body, Cohesive Toposes and Cantor's 'lauter Einsen', etc...
This line of thought is worked out more precisely with the concept of Axiomatic Cohesion wherein, roughly speaking, a topos $\mathscr{E}$ is exhibited as cohesive relative to a base topos $\mathscr{S}$ via a string of adjoints between them describing how the discrete spaces $X \in \mathscr{S}$ sit inside the larger topos $\mathscr{E}$ of more general (cohesive, combinatorial, etc...) spaces. Lawvere gave some lectures on this topic in Como in 2008 and there are accompanying lecture notes. The nLab page for cohesive toposes is also quite helpful.
To bring things back down to earth a little bit, in Left and Right Adjoint Operations on Spaces and Data Types, he descirbes, in the last section, the following situation.
Suppose we have, in a cartesian closed category $\mathscr{C}$, a commutative ring object $R$ which we regard as the 'one-dimensional continuum'. We can form the correspinding 'complex numbers' ring $C = R[i]$ by defining complex multiplication on $R^2$ in the usual way. Inside of $C$ sits the multiplicative subgroup $S^1$ corresponding to the 'circle'. Finally, since we are in a cartesian closed category, we can extract from the map space $(S^1)^{(S^1)}$ the subspace $Z$ of those endomorphisms of $S^1$ which are group homomorphisms. This $Z$ can therefore be regarded as the 'integers' although it is important to note that it will not necessarily be the usual integers object $N[-1]$ derived from a natural numbers object $N$ in $\mathscr{C}$.
you got my vote, and you are dangerously closed to get the GREEN. I was aware of Lawvere's effort in this direction, though I have to say I did not have him in mind when I wrote the question. So, what is GREEN? If you read the previous answers, they all have something to do with the fundamental issue. Now, although I am pretty sure that Lawvere 's approach goes a very long way toward creating a fraework for the general theory of apporximating the discrete by the continuum I am not equally sure that it is comprehensive enough? Any thoughts? Anyways, KUDOS!
(Answer related to the one of Dustin Mixon.)
This appears as relaxation in discrete optimization: if you have an optimization problem $\min f(x)$ with a vector $x$ and the constraint that $x_i\in\{0,1\}$ for all entries $x_i$ of the vector $x$, this often becomes very hard. A common approach is to relax the constraint to $x_i\in [0,1]$. This is a continuous problem which is often easier to solve. There are several instances of the phenomenon of exact relaxation where you can prove that the relaxed problem happens to have a solution which is binary "by accident/magic".
very interesting! Can you elaborate more? Is there any partial result in the way of justifying the "magic"? I would think that the magic has to do with the fact that {0, 1} is the BOUNDARY of the unit interval, and that there is a convergent series of solutions which becomes close to 0 in all interior points. Am I off track?
Typing "exact relaxation optimization" and "exact relaxation minimization" leads you to a few interesting papers.
That answer is totally unimodular,man.
|
2025-03-21T14:48:31.763318
| 2020-08-13T11:42:42 |
369050
|
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|
Stack Exchange
|
Cone of morphism in families
I am working in derived category $D^b(X)$ of coherent sheaves on a smooth projective varitey.
Let $E,F$ be two sheaves on $X$, with $\mathrm{R}Hom(E,F)=k\oplus k[-1]$, I consider the following triangle: $$\mathrm{RHom}(E,F)\otimes E\rightarrow F\rightarrow G$$. We know that in Grothendieck group, $[F]=[G]$ since $[\mathrm{RHom}(E,F)\otimes E]=0$. Now I consider relative situtaion. Let $X\times\mathbb{P}^1\xrightarrow{\pi}\mathbb{P}^1$ and $X\times\mathbb{P}^1\xrightarrow{q} X$ to be two projection maps. Consider the relative version of the triangle above:
$$\pi^*R\pi_*\mathrm{R}\mathcal{H}om(q^*E,\mathcal{F})\otimes q^*E\rightarrow\mathcal{F}\rightarrow \mathcal{G}$$ where $\mathcal{F}$ is a coherent sheaf on $X\times\mathbb{P}^1$, $i_s^*\mathcal{F}=F$ for each $s\in\mathbb{P}^1$.
I expect in $K_0(X\times\mathbb{P}^1)$, $[\pi^*R\pi_*\mathrm{R}\mathcal{H}om(q^*E,\mathcal{F})\otimes q^*E]$ is also 0. Is this true? How do I show this if it is true?
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2025-03-21T14:48:31.763413
| 2020-08-13T11:47:34 |
369051
|
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|
Stack Exchange
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Hyper-degree sequences: How to count them and how to construct hyper-graphs from them?
From an answer to this question I have learned how to ask this question properly.
Consider a $k$-uniform hypergraph on $n$ nodes, i.e. a family of $k$-subsets of $[n]= \{1,2,\dots,n\}$ (the hyperedges).
Consider a sequence $\langle a_1, a_2, \dots a_n \rangle$ giving the numbers of hyperedges that node $i\in [n]$ is contained in. In the case of $k=2$ this is the classical degree sequence. So let me call the sequence a hyper-degree sequence when $k\leq n$ is arbitrary.
It obviously holds that $a_i \leq \binom{n-1}{k-1}$.
For $k=2$ we know by the handshaking lemma that $\sum_i a_i = 0 \text{ mod } 2$, and I assume that this holds for all $k$: $\sum_i a_i = 0 \text{ mod } k$.
My question is fourfold:
What's the best known algorithm (probably not "efficient") to check if a given sequence $\langle a_1, a_2, \dots a_n \rangle$ with $a_i \leq \binom{n-1}{k-1}$ and $\sum_i a_i = 0 \text{ mod } k$ is the hyper-degree sequence of some $k$-uniform hypergraph on $n$ nodes?
Even though it may be hard to tell exactly how many of such sequences are hyper-degree sequences, there may be a definite fraction for $n \rightarrow \infty$. How could this fraction be calculated?
Before delving into this: Are there further simple necessary conditions for a sequence to be a hyper-degree sequence? For example, for $k=2$ there must be at least $\alpha$ nodes $i \neq 1$ with $a_i \geq 1$ when $a_1 = \alpha$.
Finally: How do I construct a $k$-uniform hypergraph for a given hyper-degree sequence?
For counting them do you want your degree sequences sorted nonincreasing? It seems now the degree sequence of ${12, 23}$ would be $(1,2,1)$ as defined in the question.
@JohnMachacek: You are right, I should have been explicit about sorting, and yes, I want my degree sequences sorted.
Alexander Dewdney generalized the Havel--Hakimi algorithm for standard graph degree sequences to hypergraphs in 1975: Degree seqences in complexes and hypergraphs, Proc. Amer. Math. Soc. 53, 535-540.
You might also look at Sarah Behrens et al., New results of degree sequences of uniform hypergraphs, Electron. J. Combin. 20 (2013) P14.
How do you propose turning Dewdney's result into an algorithm without a brute-force search for $(k-1)$-graphic sequences that satisfy the theorem?
This paper seems relevant https://arxiv.org/abs/1706.03951
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2025-03-21T14:48:31.763593
| 2020-08-13T12:18:53 |
369052
|
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|
Stack Exchange
|
On Sums of powers II
In my previous question, I asked about nontrivial sums of four powers $a^m+b^n+c^k=d^l$, and whether the nature of the solutions depend on whether $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}$ is bigger, smaller, or equal to 1.
This heuristic (based on the Fermat Catalan conjecture, an analogous conjecture on three powers), was shown to have some rather glaring holes, with values of the reciprocal sum being arbitrarily close to 1/2 while still containing solutions. Therefore I propose a new question to repair the hole.
My new question is: Is it hard to find nontrivial integer coprime solutions to $\pm a^m\pm b^n\pm c^k=d^{lcm(m,l,k)}$ when $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{lcm(m,l,k)}\le1$? Are there $m,n,k$ for which it is known there are no solutions?
Also, I would like to see parametric solutions for the cases where we do know solutions.
This is one case where Roman numerals work better in a title. Gerhard "Reads 'Powers Of Two' Instead" Paseman, 2020.08.13.
Fair enough. I've changed it. The point of this question is to avoid the cases that caused an issue last time.
Setting $n=3$ in Will's example with exponents $(4,n+3,3n+3,4)$ gives a parametric solution for $(m,n,k)=(4,4,6)$ in the venue of the current question.
Let $L:=\mathrm{lcm}(m,n,)$. Then there is always a series of solutions: $$(u^n)^m-(u^m)^n + (v^{L/k})^k = v^L$$ parameterized by integer $u,v$. So, it seems that the question lacks some important restriction.
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2025-03-21T14:48:31.763836
| 2020-08-13T13:03:03 |
369054
|
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Stack Exchange
|
Periodics of Coxeter matrices for truncated Nakayama algebras
For $n \geq 3$ and $r \geq 3$ let $C_{n,r}=(c_{i,j})$ denote the $n \times n$-matrix where $c_{i,j}=1$ for $j=i,\dots,i+r-1$ (we only do this until $i+r-1>n$).
So for example for $n=7$ and $r=3$ we obtain the matrix
\begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
Define $M_{n,r}:=- C_{n,r}^{-1} C_{n,r}^T$.
Recall that a matrix $M$ is called periodic in case $M^q$ is the identity matrix for some $q \geq 1$. The smallest such $q$ is called the period in case it exists.
Question 1: Given $r \geq 3$. For which $n$ are the $M_{n,r}$ periodic and what is the period in case they are periodic? What is the maximal finite period for a given $n$ or at least a good bound?
(a more general question was asked in What are the periodic Dyck paths? )
Background: The matrices $M_{n,r}$ are the Coxeter matrices of linear Nakayama algebras of the form $A_n/J^r$, where $A_n$ is the hereditary Nakayama algebra with $n$ simples and $J$ its Jacobson radical.
The periods are derived invariants of those algebras.
It seems to be an open problem to determine the periods, see for example page 10 in https://www.math.uni-bielefeld.de/icra2012/presentations/icra2012_lenzing.pdf for a large table and section 6 for some background of this problem https://www.sciencedirect.com/science/article/pii/S0001870813000182 .
More generally one can define the Coxeter matrix of linear Nakayama algebras as in What are the periodic Dyck paths? .
Question 2: What is the maximal period of a Coxeter matrix of a linear Nakayama algebra?
for $n=3,\dots,9$ the sequence starts with 4,6,8,12,18,30,16.
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2025-03-21T14:48:31.763985
| 2020-08-13T15:29:44 |
369067
|
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}
|
Stack Exchange
|
Presentations of $\mathbf{PGL}_3(\mathbb{F}_q)$ by three involutions
If I am not mistaken, the group $\mathbf{PGL}_3(\mathbb{F}_q)$, with $\mathbb{F}_q$ the finite field with $q$ elements, can be generated by three involutions.
Where could I find such representations ?
(In the end, I am looking for surjective morphisms from planar hyperbolic reflection groups to $\mathbf{PGL}_3(\mathbb{F}_q)$.)
${\rm PSL}(3,q)$ can always be generated by three involutions. You may be right about ${\rm PGL}(3,q)$, but can you give some justification for that assertion?
@GeoffRobinson: I think I have read that somewhere in a paper (but without detailed references). For my purposes, such presentations of $\mathbf{PSL}_3(q)$ would also be very handful, so I would be very glad to see such a representation.
It is known that every non-Abelian finite simple group $G$ other than $PSU(3,3)$ is generated by three involutions. (maybe Guralnick and Malle). Most finite simple groups are known to be generated by an element of order $2$ and an element of order $3$, with the exception of a few families. If a simple group $G$ is generated by $x$ and $y$ of respective orders $2$ and $3$, then $G$ is generated by the three involutions $x,yxy^{-1}, y^{2}xy^{-2}$.
@GeoffRobinson : thanks ! I would need to know explicit presentations/relations though (of $\mathbf{PSL}_3(\mathbb{F}_q)$) in order to construct the appropriate reflection groups. But at the moment I cannot find any paper which contains such presentations ...
In cases when ${\rm PGL}(3,q) \ne {\rm PSL}(3,q)$ (i.e. when $q \equiv 1 \bmod 3$), we have $|{\rm PGL}(3,q): {\rm PSL}(3,q)| = 3$, so ${\rm PGL}(3,q)$ cannot be generated by any number of involutions!
|
2025-03-21T14:48:31.764125
| 2020-08-13T15:41:32 |
369071
|
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"Adam Chalcraft",
"Andrej Bauer",
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|
Stack Exchange
|
Is there a point in 6-dimensional space satisfying these polynomial inequalities?
I would like to know if there is a point $(a, b, p, q, x, y) \in [0,1]^6$ satisfying the following collection of inequalities.
$b \ge a$
$q \ge p$
$y \ge x$
$a \ge p \ge a^2$
$b \ge q \ge b^2$
$p \ge x \ge a p$
$q \ge y \ge bq$
$x \ge 2p - a$
$y \ge 2q - b$
$a^3 -3ap + 2x \ge 0$
$b^3- 3 b q +2 y \ge 0$
$(a/2+b/2)^3-3 (a/2+b/2) (p/2+q/2)+x+y < 0$
This seems like the sort of question that should be possible to answer, if not by hand, then with some help from a computer. Unfortunately, I am not an expert, so my attempts to work this out with Mathematica have not been successful. Any help would be greatly appreciated!
I tried calling the FindInstance function with these constraints, but it doesn't return anything in a few hours. I might try a simple discretisation approach and see if an exhaustive search over the resulting 6-D grid can find such a point. Of course, I would think that this is not the most efficient way to approach something like this!
All the conditions hold for
$$(a, b, p, q, x, y)=\left(\frac{211}{500},\frac{531}{1000},\frac{96106069}{341750000},\frac{281961}{1000000},\frac{23996819}{170875000},\frac{149721291}{1000000000}\right).$$
This result was obtained by solving the system of all the (linear) inequalities containing $x$ but not $y$ and thus getting $p\ge x\ge \max(a p,2p-a,(3a p-a^3)/2)$ (which implies $p\le a$). Then piecewise expand $\max(a p,2p-a,(3a p-a^3)/2)$; note the three corresponding cases -- call them the $x$-cases. Similarly do with the roles of $x$ and $y$ interchanged, to get the three corresponding $y$-cases. In each of the 9 combinations of the $x$- and $y$-cases, minimize the left-hand side of the last inequality in $x,y$, then in $p,q$, and finally in $a,b$. Actually, it is the combination of the $x$-case when $\max(a p,2p-a,(3a p-a^3)/2)=(3a p-a^3)/2$ and the similar $y$-case that gives the above result.
Below is the image of a piece of a Mathematica notebook providing a verification of the above statement (click on the image to enlarge it):
Would you be so kind as to reveal to the rest of us how you found these numbers?
@AndrejBauer : I have now done this.
Thanks so much, this should be very helpful when I need to check such things in the future!
@BPN - Would you be so kind as to reveal to the rest of us where this problem comes from? Unless I’ve misunderstood, it’s remarkable that the prime 5 does not appear in the question but<PHONE_NUMBER> appears as a denominator in the answer.
@AdamChalcraft : the powers of $10$ arose just because of rounding certain irrational algebraic numbers, which latter were minimizing values of $a,b$ in the described above minimization.
|
2025-03-21T14:48:31.764342
| 2020-08-13T15:50:22 |
369073
|
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|
Stack Exchange
|
Are invariant forms on homogeneous spaces necessarily closed?
Take a compact homogeneous space $G/K$, and a left $G$-invariant differential $k$-form $\omega \in \Omega^k(G/K)$. Will $\omega$ necessarily be closed? Might it even be harmonic when $G/K$ is endowed with a Riemannian metric?
No. See Bredon's book "geometry and topology" for a good discussion of the complex of invariant differential forms. For $K = 1$ you recover the Chevalley-Eilenberg complex of $\mathfrak g$.
They don't have to be closed, but each closed form differs from a closed invariant form by an exact (not necessarily invariant) form, so the de Rham cohomology of forms is the same as that of invariant forms, assuming $G$ is compact.
Note that the answer depends on the pair $(G,K)$.
For example, if $K=\{e\}$, then one is asking whether the ring of left-invariant forms on $G$ consists only of closed forms. This only happens when $G$ is abelian.
On the other hand, if $M=G/K$ is a compact Riemannian symmetric space and $G$ is the identity component of the isometry group of $M$, then, indeed, every $G$-invariant form is closed and, in fact, the ring of $G$-invariant forms on $M$ is equal to the space of harmonic forms on $M$. This is a well-known result, but for a short proof, one can consult this note by Michael E. Taylor.
For example, when $M=\mathbb{CP}^n$ endowed with its Fubini-Study metric, one has $G = \mathrm{SU}(n{+}1)/\mathbb{Z}_{n+1}$, and the only $G$-invariant forms are (linear combinations of) powers of the Kähler form $\omega$.
As another example, if $K$ is compact and $M = (K\times K)/\Delta$, where $\Delta = \{ (k,k)\ |\ k\in K \}$, then
the $(K\times K)$-invariant forms on $M$ are simply the bi-invariant forms on $K$, which are all closed.
|
2025-03-21T14:48:31.764493
| 2020-08-13T15:56:28 |
369074
|
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|
Stack Exchange
|
When every closed and connected subset is path connected
Let $X$ be a compact $T_0$ topological space such that its closed and connected subsets are path connected. Is there any characterization for such a space?
Any characterization is helpful for me. Is there any (NICE)?
Not an answer. Consider the set $C$ in the plane defined as the union of the vertical segment $S:={0}\times [0,1]$ with the graph of the function $$f:(0,1]\to\mathbb{R},$$ $f(x)=\sin(1/x)$. The set $C$ is closed, connected and I do not believe that it is path connected (Better check my claim). If my claim is correct, then the spaces you are enquiring are pretty weird: no closed subset of such a space can be homeomorphic to a closed cube of dimension $\geq 2$. With the caveat that my claim could be horribly wrong.
@LiviuNicolaescu Your space is not path connected.
@PiotrHajlasz Thanks for confirming. Finite CW complexes are excluded. Semi-algebraic or subanalytic sets are excluded. Discrete spaces have the property in the question. A nondiscrete space is ${0}\cup{1/n;;;n\in \mathbb{Z}_>0}$. I cannot think of a nontrivial connected space with the property in the question.
@LiviuNicolaescu Unless I'm mistaken, the interval $[0,1]$ has this property. Of course every compact totally disconnected space is also an example (for trivial reasons).
Wouldn't 1-dimensional CW complexes work likewise?
@DenisNardin More generally, every hereditarily locally connected (i.e. every connected subspace is locally connected) Polish space has this property.
Yes, 1-dimensional CW complexes do work.
@D.S.Lipham does such space have to be 1-dimensional?
@erz: Yes, in this case dimension is $\le 1$. Moreover, such a continuum is rational: Every point has a basis consisting of neighborhoods with countable boundary (which implies that the inductive dimension is $\le 1$). It appears that among continua, this property is exactly hlc, equivalently, rationality.
@MoisheKohan Rational is too weak to imply every closed subset is path-connected. Consider the topologists sine curve. But rim-finite implies hlc implies the property.
@D.S.Lipham: I should have said Peano continuum rather than continuum.
|
2025-03-21T14:48:31.764928
| 2020-08-13T15:58:49 |
369076
|
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|
Stack Exchange
|
Prove convergence of the integral $\int_0^{\infty} xe^{-x^6\sin^2x}dx$
I would perhaps to this post add problems of a similar kind.
Problem 1. Prove convergence of the integral $\int_0^{\infty} xe^{-x^6\sin^2x}dx$,
From
https://www.desmos.com/calculator/sjizhbtbhp
we see the function can be best interpreted as follows: $(\frac{1}{e})^{\sin^2x}$ is a periodic function, with 'peak' 1 at $\sin x=0$, 'bottom' 1/e at $\sin x=1$. The wave is greatly shrinked, or each peak becomes greatly narrowed down by taking $x^6$-th power of it (and the greater x is, the narrower each peak becomes, that's key to the integrability); though the peak value is retained for it's 1. The narrowed wave is amplified by x, but that doesn't change the integrability. We need only to show the sum of all narrow peak near x=n$\pi$ converges (for in the tail (x near $\infty$), anywhere else is almost 0 as the result of $x^6$(huge)-th power of a number less than unity.
One possible way is to realize the behavior and size of $\sin(n\pi+h)$ (h small) is exactly like that of $\sin(0+h)$, so we can (by changing independent variable) move the curve near $n\pi$ to near 0. Similar idea may be applied to another problem $\int_{\pi}^\infty \frac{dx}{x^2(\sin x)^{2/3}}$
Proof: $\sin^2(n\pi+h/2)
=\frac{1-\cos{(2n\pi+h)}}{2}=\frac{\frac{h^2}{2!}-\frac{h^4}{4!}+\dots}{2}>\frac{h^2}{2\cdot2!}$
When x is near $n\pi$ but no less than h/2 away from it, $\exp\{-\sin^2(x)x^6\}<\exp\{-\sin^2(n\pi-\pi/2)x^6\}<\exp\{-\frac{h^2}{2\cdot2!} (n\pi+h/2)^6\}$,
Let $h<\frac{1}{n^{2+\lambda}}$, where 0<$\lambda<1/2$, then $\sum_n \int_{n\pi-h/2}^{n\pi+h/2} x\exp\{-\sin^2(x)x^6\}<\sum_n h\cdot 1$, which converges (to less than $\frac{\pi^2}{6}$).
$\sum_n(\int_{n\pi-\pi/2}^{n\pi-h/2}+ \int_{n\pi+h/2}^{n\pi+\pi/2}) \exp\{-\sin^2(x)x^6\}<\sum_n\pi \exp\{-\frac{h^2}{2\cdot2!} (n\pi+h/2)^6\}=\sum_n\pi \exp\{-\frac{(\frac{1}{n^{2+\lambda}})^2}{2\cdot2!} (n\pi+h/2)^6\}=\sum_n\pi \exp\{-Kn^{2-2\lambda}\}$, where K is a positive constant, roughly $\frac{\pi^6}{4}$. The series is less than $\sum_n\pi (\frac{1}{e})^{K{n}}$, a geometric series converging to $\frac{\pi}{1-\frac{1}{e^K}}$(possibly plus a constant $C$ representing sum of the beginning iterms).
So the integral is at least bounded to a small neighbourhood of
$$l<\frac{\pi^2}{6}+\frac{\pi}{1-\frac{1}{e^{[\frac{\pi^6}{4}]}}}+C.\blacksquare$$
My questions:
How can one proceed from that to prove the convergence of the integral?
Is there a neater way to prove it?
Why $x^6$? ....
I think this question would be more appropriate for math.stackexchange.com. I believe there is some simplification of the answer possible, but the main idea you have (which is to give upper bounds for the integral on the interval $[(n-\frac 12)\pi,(n+\frac 12)\pi]$) is fine.
What’s the difference between the two sites?
Call $I$ the integral, then $$I=\sum_{k} \int_{k\pi}^{(k+1)\pi} xe^{-x^6 \sin^2x}dx=\sum_{k} \int_{0}^{\pi} (z+k\pi)e^{-(z+k\pi)^6 \sin^2z}dz=:\sum_k I_k.$$ Then
$$
I_k \le (k+1)\pi \int_0^\pi e^{-k^6 \pi^6 \sin^2 z}dz=2(k+1)\pi\int_0^{\pi/2} e^{-k^6 \pi^6 \sin^2 z}dz \le 2(k+1)\pi\int_0^{\pi/2} e^{- 4 k^6 \pi^4 z^2}dz,
$$
since $\sin z \ge (2/\pi)z$ in $[0, \pi/2]$. Finally, for $k \ge 1$
$$
\int_0^{\pi/2} e^{- 4 k^6 \pi^4 z^2}dz \le \int_0^{\infty} e^{- 4 k^6 \pi^4 z^2}dz=C k^{-3}
$$ which gives the convergence of $\sum_k I_k$.
I see it seems that we connect sin0 and sin\pi/2 and get sin z greater than linear function of z which Simplified the problem
Yes, only the behavior near the zeros of $\sin$ plays a role.
If the question isn't right for the site, it's best not to answer it...
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2025-03-21T14:48:31.765182
| 2020-08-13T17:08:02 |
369082
|
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|
Stack Exchange
|
Prove that a Boolean two-valued topos in which supports split is well-pointed
In Lawvere and Rosebrugh's Sets for Mathematics, they write
It is a theorem [MM92] that a topos is well-pointed if and only if it is Boolean, two-valued, and supports split.
[MM92] is a reference to Mac Lane and Moerdijk's Sheaves in Geometry and Logic. I have found the proof that a well-pointed topos is Boolean, two-valued and supports split (Propositions VI.7 and VI.8), but I cannot find the proof of the converse, there or anywhere else. Can anyone help me?
Given a Boolean, two-valued topos in which supports split, we want to prove it's well-pointed, meaning that, if $f,g:A\to B$ are distinct, then there is a point $p:1\to A$ such that $fp\neq gp$. Since $f$ and $g$ are distinct, their equalizer $e:E\to A$ is a proper subobject of $A$. (I'm ignoring the distinction between a subobject and a monomorphism representing it.) By "Boolean", $E$ has a non-zero complement $c:C\to A$. Since $C\neq0$, the support of $C$ is a non-zero subobject of $1$. By "two-valued", the support of $C$ is all of $1$. By "supports split" we have a morphism $p:1\to C$. I claim this point in $C$, considered as a point in $A$ (strictly speaking, the point $cp:1\to A$) does the job. Indeed, if $fcp$ were equal to $gcp$, then $cp$ would factor through the equalizer $E$ as well as its complement $C$, hence would factor through 0. But then we'd have a morphism $1\to0$, and the topos would be trivial, contrary to the assumption $f\neq g$
|
2025-03-21T14:48:31.765310
| 2020-08-13T17:18:37 |
369083
|
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"Eduardo Longa",
"Rohil Prasad",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369083"
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|
Stack Exchange
|
Behaviour of mass for currents with disjoint supports
I am sorry if this is a basic question, but I don't think in MSE I will receive any answers.
Let $(M^3,g)$ be a compact and oriented Riemannian $3$-manifold. Let $\alpha$ and $\beta$ be the integral currents supported on the compact, connected, oriented embedded surfaces $\Sigma_1$ and $\Sigma_2$ with multiplicities $\theta_1$ and $\theta_2$, that is:
$$ \alpha(\omega) = \int_{\Sigma_1} \langle \omega(x), \tau_1 (x) \rangle \theta_1(x) d \mathcal{H}^2(x), \quad \omega \in \Omega_2(M), $$
and similarly for $\beta$. Here, $\tau_1$ is a $2$-plane field that orients $\Sigma_1$ and $\theta_1$ is an integrable positive integer-valued function on $\Sigma_1$.
Assume that $\Sigma_1$ and $\Sigma_2$ are disjoint. Is it true that the mass of $\alpha$ is smaller than the mass of the sum $\alpha + \beta$?
Yes, this is true, from the fact that the supports of these currents are disjoint. In fact, I think the same argument should work for two general currents, as long as their supports have disjoint neighborhoods.
Write $M(\alpha)$ and $M(\alpha + \beta)$ for the masses of $\alpha$ and $\alpha + \beta$, respectively.
Then there is a sequence of 2-forms $\omega_k$ such that $|\omega_k|_{C^0(M)} = 1$ for every $k$ and
$$|\alpha(\omega_k)| \to M(\alpha).$$
Let $\chi: M \to \mathbb{R}$ be a smooth, non-negative cutoff function that is equal to $1$ in a neighborhood of $\Sigma_1$ and $0$ in a neighborhood of $\Sigma_2$. Then it follows that
$$\alpha(\omega_k) = \alpha(\chi\omega_k)$$
for every $k$, and moreover
$$\beta(\chi\omega_k) = 0$$
for every $k$.
It follows that
$$|(\alpha + \beta)(\chi\omega_k)| \to M(\alpha).$$
However, by definition,
$$|(\alpha + \beta)(\chi\omega_k)| \leq M(\alpha + \beta)$$
for every $k$.
It follows from the above two statements that we must have the inequality
$$M(\alpha) \leq M(\alpha + \beta).$$
Nice argument. In the particular case I described, the mass of $\alpha$ is stricly smaller than that of the sum, right?
Oh, I wasn't very careful about that. I think my argument can be fixed to show that, by picking some two-form $\eta$ supported disjointly from $\chi\omega_k$ such that $\beta(\eta) > 0$. Then, if we assume $\alpha(\chi\omega_k) > 0$ for every $k$ as well, you get $\lim_{k \to \infty} |(\alpha + \beta)(\chi\omega_k + \eta)|$ is equal to $\lim_{k \to \infty} |\alpha(\chi\omega_k)| + |\beta(\eta)|$. This is strictly greater than $M(\alpha)$, but bounded above by $M(\alpha + \beta)$ which gives you the strict inequality.
|
2025-03-21T14:48:31.765481
| 2020-08-13T17:31:32 |
369084
|
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|
Stack Exchange
|
Koszul-Malgrange Holomorphic structure on a pullback bundle
I'm finding myself a little confused about Koszul-Malgrange holomorphic structures in a certain context.
Suppose $M$ is a complex manifold, $N$ is a smooth manifold with a smooth complex vector bundle $V$ and bundle connection $\nabla$, and $f:M\to N$ is a smooth map. We form the pullback bundle $f^*V$ over $M$ and give it the pullback connection $f^*\nabla$. Let $P: f^*V \to V$ be the associated smooth projection map. Assuming $f^*\nabla^{0,1}\cdot f^*\nabla^{0,1}=0$, a theorem of Koszul-Malgrange says there is a holomorphic structure on $f^*V$ that turns it into a holomorphic vector bundle over $M$.
My question is about the following. Suppose we have a (possibly locally defined) holomorphic section $s$ of $f^*V$. In a suitable trivialization over some open set $\Omega$, so that $f^*V$ splits as $\Omega\times \mathbb{C}^n$, we can write $$s(z) = (z, h(z))$$ where $h:\Omega\to \mathbb{C}^n$ is holomorphic.
Now, consider the map $$P\circ s: \Omega \to V$$ I'm basically wondering if it makes sense to write $P\circ s =(f(z), h(z))$ for some choice of trivialization of $V$. The sort-of roadblock that I'm hitting is that the trivialization of the pullback we used should come from the Koszul-Malgrange theorem, rather than be pullback of a trivialization of $V$. So, I'm finding it hard to fiddle the definition of the pullback bundle.
I'm mainly interested in the case where $\dim M \leq \dim N$ and $f$ is regular everywhere. Then in my trivizalization over $\Omega$, the map $f_*$ should be an isomorphism of $f^*V|_{\Omega}\to V|_{f(\Omega)}$ that acts by $(z,v)\mapsto (f(z),v)$. Of course, this induces a trivialization over $f(\Omega)$. Then we can find an open set $U$ containing $f(\Omega)$ in which we have a smooth trivialization that extends our trivialization over $f(\Omega)$. I think that in this trivialization, $P\circ s$ should assume the desired form.
Okay I thought about it a bit more, and here is what I can say in general. We can find a neighbourhood $U\subset N$ such that $V|_U$ splits diffeomorphically as $U\times \mathbb{C}^n$ and the section $s$ takes the form $s(z) = (f(z), A(z)\circ h(z))$, where $A(z)$ is a smoothly varying family of invertible $n\times n$ complex matrices.
Let $(U,\varphi)$ be a trivialization of $V$ over some open set $U\subset N$ and $\Omega\subset M$ such that $f(\Omega) \subset U$. In $\Omega$ we have the pullback trivialization $\psi$, and in these trivializations the map $P$ takes $\Omega\times \mathbb{C}^n\to U\times \mathbb{C}^n$ via $(z,v)\mapsto (f(z),v)$. Meanwhile, according to Koszul-Malgrange (by the way, the easiest exposition I've found is in here http://gtnmk.droppages.com/2019/250B-2019-set.pdf), then for $p\in \Omega$ we can find $\Omega_p\subset \Omega$ and a holomorphic trivialization $k:f^*V|_{\Omega_p}\to \Omega_p \times \mathbb{C}^n$. Let us just restrict $\Omega$ so that $\Omega_p=\Omega$. This should make the notation easier.
If $s(z) = (z,b(z))$ with respect to the trivialization $\psi$, then $P\circ s= (f(z),b(z))$ with respect to the trivializations $\psi,\varphi$. Under $k$, $s$ assumes the form $s(z) = (z,h(z))$, with $h$ holomorphic. The relation between $b$ and $h$ is as follows. $\varphi$ and $k$ are trivializations, so in coordinates, we can write $$\varphi\circ k^{-1}(z,v) = (z, A(z)v)$$ for some smoothly varying family of invertible matrices $A(z)$. $A(z)$ then satisfies $$b(z) = A(z) \circ h(z)$$ which gives the result.
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2025-03-21T14:48:31.765708
| 2020-08-13T17:51:58 |
369088
|
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|
Stack Exchange
|
Permutation representation of a finite $p$-group
In the (xi) group of the classification of groups of order $p^4$ given by W.Burnside in his book," Theory of Groups Of Finite Order". The group ($\mathbb{Z_{p^{2}}}\rtimes \mathbb{Z_{p^{}}}) \rtimes_{\phi}\mathbb{Z_{p^{}}} $, have presentation
$$<a,b,c : a^{p^{2}}=b^p=c^p=e, ab=ba^{1+p},ac=cab,bc=cb>$$
I was trying to find the embedding of the above group into a symmetric group of order $p^4$, which exists by Cayley's Theorem. Is it possible to explicitly find the map ? Kindly see it.
Using the relations, you can represent every element uniquely as $c^k b^l a^m$ with $0\leq k < p$, $0\leq l < p$, $0\leq m< p^2$. Now you can work out how left multiplication with $a, b, c$ acts on the set of such representatives by again using the relations, this is easy to do explicitly. The corresponding action on the set of tuples $(k, l, m) $ as above is the desired embedding.
Actually I have some problem in understanding your answer. For example, the left action of $a$ on $c^kb^la^m$ is given by
$$a.\left( {{c^k}{b^l}{a^m}} \right) = {c^k}{b^k}{a^{1 + kp}}{b^l}{a^m} = {c^k}{b^{k + l}}{a^{1 + (k + l)p + m}}$$ how to get embedding from here ?
Well, this tells you that $a$ goes to the permutation (on the set of triples $(k, l, m)$ with the bounds mentioned in the answer) which sends $(k, l, m) \mapsto (k, k+l, 1+(k+l)p+m)$.
I get your point. Thanks, your answer is really helpful for me.
If I want to embed this image of $G$ in $S_{p^4}$ into $GL(p^4,p)$. Is there any way ? I mean, I know the concept of permutation matrices, but how to write it. I am unable to understand. Please see.
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2025-03-21T14:48:31.765847
| 2020-08-13T18:23:19 |
369092
|
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Stack Exchange
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Natural density of set of numbers not divisible by any prime in an infinite subset
Suppose $S$ is a subset of the primes with natural density $0 < \alpha < 1$ within the primes. If
$$D(X) := \{n \leq X \mid p \not \mid n \text{ for all } p \in S \}$$
(so $D(X)$ is numbers at most $X$ not divisible by any $p \in S$),
then is there a nice form for the asymptotic value of $D(X)$ as a function of $X$?
I'm guessing this might be a standard result in analytic number theory but I unfortunately don't yet know much about the area.
At least when $S$ is the inverse image of a union of Frobenius classes of $\text{Gal}(K/\mathbb{Q})$ for some number field $K$ under the Artin map, I have reason to believe that the answer is $\frac{X}{\log^{\alpha} X}$, but I don't know how to prove this either. I'm guessing the result holds in more generality though, hence the more general form of the question above - this specific case below the line reduces to the general question via Chebotarev density.
The recent paper
Matomäki, Kaisa; Shao, Xuancheng, When the sieve works. II, ZBL07207214.
gives a fairly satisfactory answer to this question in the setting of arbitrary $S$. General sieve theory gives the upper bound
$$ |D(X)| \ll X \prod_{p \in S; p \leq X} (1 - \frac{1}{p})$$
(which also matches what naive probabilistic heuristics would predict) but the matching lower bound
$$ |D(X)| \gg X \prod_{p \in S; p \leq X} (1 - \frac{1}{p}) \quad (1)$$
is not always true. For instance if $S$ consists of all the primes between $X^{1/v}$ and $X$ for some fixed $v>1$ then the RHS of (1) is about $X/v$ but the LHS is instead $v^{-v+o(v)} X$ (smooth numbers are significantly rarer than naive heuristics would predict). However in this paper (resolving a previous conjecture of Granville, Koukoulopoulos and Matomaki in the reference at the end of this answer) they show (roughly speaking) that the lower bound (1) holds if one has an inequality of the form
$$ \sum_{X^{1/v} \leq p \leq X^{1/u}: p \not \in S} \frac{1}{p} \geq \frac{1+\varepsilon}{u}$$
for some $v > u > 1$ that are not too large and some $\varepsilon > 0$ (there are examples that show that this condition is close to best possible); here the implied constant in (1) are allowed to depend on $u,v,\varepsilon$ and is basically of the form $v^{-e^{-1/u} v}$. Basically this condition is asserting that $S$ doesn't end up containing the majority of all primes between $X^{1/v}$ and $X^{1/u}$ for some bounded $u,v$, as this can lead to dramatic reductions in the size of $D(X)$.
In the situation where $S$ has natural density $\alpha < 1$ relative to the primes, summation by parts will give an asymptotic $\sum_{X^{1/v} \leq p \leq X^{1/u}; p \not \in S} \frac{1}{p} = (1-\alpha) \log(v/u) + o(1)$ as $X \to \infty$ keeping $u,v$ fixed, and so by choosing $u,v$ appropriately one can invoke the Matomaki-Shao theorem and obtain the matching lower bound (1) (presumably this can also be established by earlier results than the Matomaki-Shao paper). This already implies that $|D(X)| = X / \log^{\alpha+o(1)} X$ and presumably for your specific set $S$ you may be able to sharpen the $o(1)$ error term here using more quantitative versions of the Chebotarev density theorem.
For arbitrary $S$, one has a logarithmic version
$$ \sum_{n \in D(X)} \frac{1}{n} \asymp \log X \prod_{p \in S; p \leq X} (1-\frac{1}{p});$$
see Lemma 2.1 of
Granville, Andrew; Koukoulopoulos, Dimitris; Matomäki, Kaisa, When the sieve works, Duke Math. J. 164, No. 10, 1935-1969 (2015). ZBL1326.11055.
The estimates here are uniform in $S$, so can handle both the cases when $S$ is independent of $X$ and when $S$ depends on $X$. (But for a given $X$ the behaviour is largely driven by the intersection of $S$ with $[X^\varepsilon,X]$, so the distinction between fixed $S$ and variable $S$ is not so great in practice.)
Call a set $S$ of primes "Frobenian" if there is a finite Galois extension $K/\mathbf Q$ and a union of conjugacy classes $H$ in ${\rm Gal}(K/\mathbf Q)$ such that $S$ is equal to the set of primes $p$ that are unramified in $K$ for which the Frobenius conjugacy class of $p$ in ${\rm Gal}(K/\mathbf Q)$ lies in $H$ or, more generally, $S$ is that set of primes with finitely many exceptions. An example could be the set of all primes, with finitely many exceptions, that belong to a union of arithmetic progressions modulo $m$ that are all relatively prime to $m$, such as all primes that are $1 \bmod 4$ without the primes $5$ and $29$ or the set of all primes that are $1 \bmod 4$ together with the prime $7$.
From the MO question here I was led to a paper by Serre here or here that addresses your question in Theorem 2.4(a). Suppose the Frobenian set of primes $S$ has natural density $\alpha$ where $\alpha > 0$, so it also has "regular density $\alpha$ in the sense of Delange" (Serre's defintion (1.3)). The finitely many exceptional primes allowed in the definition of Frobenian primes above corresponds to Serre's "sufficiently large" comment in his property (1.4)(c).
Your $D(X)$ is the complement of the set $E(X) = \{n \leq X : n \text{ has some prime factor } p \in S\}$. Let $E$ be the set of all positive integers with a prime factor in $S$, so $E(X) = \{n \leq X : n \in E\}$. The set $E$ is "multiplicative": a product of relatively prime positive integers is in $E$ if and only if one of the two integers is in $E$. Then $E$ fits the hypothesis of Theorem 2.4 and your $D(X)$ is what Serre calls $E'(X)$, so Theorem 2.4(a) says for $0 < \alpha < 1$ that $D(X) = E'(X) \sim cX/(\log X)^\alpha$ for some $c > 0$, and a formula for $c$ is in equation (2.6) of Serre's paper. If $\alpha = 1$ then Theorem 2.4(b) says $D(X) = E'(X) = O(X^{1-\delta})$ for some $\delta > 0$.
Isn't it the set associated to the complement of $S$?
@WillSawin I am getting myself confused between sets and their complements. I have found a precise statement in a paper of Serre that addresses exactly the OP's question so I will edit my answer to describe that.
Under a very slight strengthening of the hypothesis, this can be done by elementary methods, namely the "Wirsing–Odoni method". The following version is Proposition 4 in this paper of mine with Finch and Sebah (I'm simplifying the hypotheses for clarity here):
Let $f$ be a multiplicative function satisfying $0\le f(n)\le 1$ for all $n$.
Suppose that there exist real numbers $\xi>0$ and $0<\beta<1$ such that
$$
\sum_{p<P}f(p)=\xi \frac P{\log P}+O\bigg( \frac P{(\log P)^{1+\beta}}\bigg) \tag{1}
$$
as $P\rightarrow \infty$. Then the product over all primes
$$
C_f=\frac1{\Gamma(\xi)} \prod_{p} \bigg( 1+\frac{f(p)}p+\frac{f(p^2)}{p^2}+\frac{f(p^3)}{p^3}+\cdots \bigg) \bigg( 1-\frac1p \bigg)^\xi
$$
converges (hence is positive), and
$$
\sum_{n<N}f(n)=C_fN(\log N)^{\xi -1}+O_f\big( N(\log N)^{\xi -1-\beta}\big)
$$
as $N\rightarrow \infty $.
In this case, $f(p^r)$ equals $1$ if $p\notin S$ and equals $0$ if $p\in S$; then $\xi=1-\alpha$ by the assumption of the natural relative density of $S$ (we see that we need a slightly quantitative statement of that density in terms of the error term in $(1)$). In this case
$$
C_f=\frac1{\Gamma(1-\alpha)} \prod_{p} \bigg( 1-\frac{1-1_S(p)}p\bigg)^{-1} \bigg( 1-\frac1p \bigg)^{1-\alpha},
$$
where $1_S(p)$ equals $1$ if $p\in S$ and equals $0$ if $p\notin S$. Notice that this is completely independent of any algebraic properties of $S$.
|
2025-03-21T14:48:31.766463
| 2020-08-13T19:31:16 |
369097
|
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Stack Exchange
|
On some prerequisites for J-holomorphic curves and Gromov-Witten invariants
I'm currently reading through J-holomorphic curves and Quantum cohomology by McDuff and Salamon, and I've been facing some unfamiliarity issues with respect to the PDE and functional analysis tools being used in there.
Say for example in understanding the local behaviour of J-holomorphic curves, we use the generalised Cauchy-Riemann equation to establish a few results about how "good" co-ordinates could be chosen, and some theorems akin to unique continuation, a variant of identity theorem (lemma 2.2.3 in the book) etc., So mostly I've been able to get along with some intuition borrowed from how Holomorphic curves are supposed to behave.
Now, my primary question is, how do I get a better intuition for geometric behaviour of J-holomorphic curves?
Also in a more broader sense are there any other fields which are a little more visual(familiar) in flavour, like say Riemannian geometry or complex geometry from where I could borrow some more intuition, especially in the context of say when I'm faced by a PDE describing a curve/surface I need some pointers on what are some primary questions about the behaviour of curve that I should start asking, (I do understand questions that one asks become very PDE specific, but I'm looking for any class of reasonable questions that one tries to ask with PDE that encounters in these areas of geometry)
Now I do understand that we get into the business of Moduli spaces of J-holo curves, in the pursuit of studying global invariants of symplectic manifolds, and things quickly tend to get more operator theoretic and all, and so essentially Riemannian geometry which uses curvature as a primary tool loses its relevance, but even then, say if there is some intuition from Riemannian geometry that I could use eventually, what kind of results in Riemannian geometry should I be looking at?
Also to get to understanding computation of Gromov-Witten invariants and proofs of results like Non-squeezing theorem (which for example uses theory of Minimal surfaces, but in a very basic sense), what is the amount of Riemannian geometry/complex geometry/variational principles (geometric analysis) do I need in my toolbox ? Would be helpful if someone could suggest a few references for the same.
I did do variational principles at a very basic level, as much as suggested in Ana Cannas Da Silva.
I hope my question doesn't come off as too vague, would be happy to clarify more specifically.
The essential idea being what can I borrow from Riemannian/Complex geometry/ Geometric analysis be it in terms of direct results or plain intuition, that would help me pursue this book in a more meaningful way.
I never got further than visualizing pseudoholomorphic curves as just plane algebraic curves, drawing their real points, but then sometimes correcting a little by remembering their complex points as much as I can. You can also work out how the Fubini-Study metric on the projective plane restricts to a quadric curve. As that curve bubbles into a pair of lines, the quadric gets a thin neck near that point where the bubbling is happening, and away from there builds up two bubbles. Gauss--Bonnet convinces you that, since the two bubbles are becoming two unit spheres, the curvature is building up near the neck into large negative curvature.
Thank you for this, at least this provides with a non trivial model example that I could work with.
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2025-03-21T14:48:31.766715
| 2020-08-13T20:11:39 |
369100
|
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Stack Exchange
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can there be a function $f:\mathbb Q_{+}^{*}\longmapsto\mathbb Q_{+}^{*}$ such that $f(xf(y))=\frac{f(f(x))}{y}$?
Problem:
Can an $f$ function be created where:$$f\colon\mathbb Q_{+}^{*}\to \mathbb Q_{+}^{*}$$
The function is defined on the set of fully positive rational numbers and is achieved:
$\forall(x,y)\in \mathbb Q_{+}^{*}\times\mathbb Q_{+}^{*},f(xf(y))=\frac{f(f(x))}{y}$
This question is similar to one of the Olympiad questions that I was very passionate about and used several ideas to solve this problem, but I did not arrive at any result from one of them by using the basic theorem in arithmetic that states that there is a corresponding application between $(\mathbb Q_{+}^{*})$and $(\mathbb Z^{\mathbb N})$ where:
$$\left\{\mathbb Z^{\mathbb N} =\text{ A set of stable sequences whose values are set in} \quad\mathbb Z\right\}$$
This app is defined like this
$$\varphi\colon\mathbb Z^{\mathbb N}\to \mathbb Q_{+}^{*} ,(\alpha_n)_{n\in\mathbb N}\longmapsto \prod_{n\in\mathbb N} P_n^{\alpha_n}$$
Where:$$\mathbb P=\left\{P_k:k\in\mathbb N\right\}\text{ is the set of prime numbers} $$
And put $x=\prod_{n\in\mathbb N}P_n^{\alpha_n},\quad y=\prod_{n\in\mathbb N }P_n^{\beta_n},\text{and}\quad $
$$f(\prod_{n\in\mathbb N}P_n^{\alpha_n})=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)$$
.
\begin{align*}
xf(y)&=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n}^{\beta_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\beta_{2n}}\right)\\
&=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n}+\beta_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{\alpha_{2n+1}-\beta_{2n}}\right)\\
\end{align*}
$\implies$
\begin{align*}
f(xf(y))&=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}-\beta_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}-\beta_{2n+1}}\right)\\
&=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n}^{-\beta_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\beta_{2n+1}}\right)\\
&=\frac{\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)}{\left(\prod_{n\in\mathbb N}P_{n}^{\beta_{n}}\right)}\\
&=\frac{f(x)}{y}\\
\end{align*}
However, this did not help me create this method
I need an idea or suggestion to solve this problem if possible and thank you for your help
Note: $(\alpha_n)_{n\in\mathbb N}\quad \text{is a stable sequence}\leftrightarrow \forall n\in\mathbb N ,\exists n_0\in\mathbb N :\left( n\geq n_0 \quad \alpha_{n}=0\right) $
Cross-posted on Math SE at Can there be a function $f\colon\mathbb Q_{+}^{}\to \mathbb Q_{+}^{}$ such that $f(xf(y))=\frac{f(f(x))}{y}$?.
Yes, this question was put by me, but .. I found that it needs strong evidence that you can check it well and wait for your suggestion @john Omielan
For $x=1$, $f(f(y))=\frac{a}{y}, a=f(f(1))$. $f'(x)=\lim \limits_{h \to 0} \frac{f(x+h)-f(x)}{h}, h \in \mathbb Q$. Then we get, $f'(f(y))f'(y)=\frac{-a}{y^2} <0$. Let, $f'(y)>0$, then $f'(f(y))<0$. And, obviously the function can't be always positive or negative. If we increase $y$, $f(y)$ increase as $f'(y)>0$. But, $f(f(y))$ decrease as $f'(f(y)<0$. Now, if $f'(y)$ changes sign then, $f(y)$ start decreasing. Hence, this becomes $f'(y)<0$ but $f'(f(y))<0$, contradiction.
There is no function $f\colon Q\to Q$ such that
$$f(xf(y))=\frac{f(f(x))}y \tag{1}$$
for all $x$ and $y$ (in $Q$), where $Q:=\mathbb Q_{+}^{*}$.
Indeed, for $x=1$ equality (1) is
$$f(f(y))=\frac{f(b)}y,$$
where $b:=f(1)$. Replacing here $y$ by $x$, from (1) we get
$$f(xf(y))=\frac{f(b)}{xy}.$$
This with $y=1$ yields
$$f(xb)=\frac{f(b)}{x},$$
or
$$f(z)=\frac cz$$
for $c:=bf(b)$ and all $z\in Q$. Now (1) becomes $y/x=x/y$ for all $x,y$ in $Q$, which is clearly false.
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2025-03-21T14:48:31.766944
| 2020-08-13T21:00:08 |
369106
|
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Stack Exchange
|
Is there a finitely generated group with the same structure as ZFC?
Is there a finitely generated computably presentable group $G$ on generator set $A$ and a computable function $f$ from first-order formulas to words on $A$ such that $\mathsf{ZFC}\vdash\sigma\leftrightarrow\tau$ iff $f(\sigma)$ and $f(\tau)$ represent the same element in $G$?
Note that is migrated from a question arising from a post on math stack exchange, as recommended by Noah Schweber https://math.stackexchange.com/q/3776818/155079
As a response to your comment "Because neither NAND or NOR are associative I was fiddling with trying to come up with a more than two parameter Boolean universal gate that is associative, and considering going into multivalued or ternary logic to achieve that", the group $A_5$ works -- see this discussion of Barrington's theorem: https://crypto.stanford.edu/~dabo/pubs/papers/barrington.html
Maybe look at https://www.sciencedirect.com/science/article/pii/S0049237X08719141 I think it tries something more logic oriented for the word problem
@BenjaminSteinberg better to also give a human-readable reference in case that url breaks: Ralph McKenzie and Richard J. Thompson, An Elementary Construction of Unsolvable Word Problems in Group Theory, Studies in Logic and the Foundations of Mathematics 71 (1973) 457-478 and here's a stable doi link: https://doi.org/10.1016/S0049-237X(08)71914-1
Wouldn't this contradict any incompleteness phenomena??
I guess that the question should include "does there exist a model of ZFC and..."?
@JohannesSchürz Which ones?
Isn't it enough to have a finitely generated computably presentable group whose word problem is $\Sigma^0_1$-universal? This paper constructs such a group: https://arxiv.org/pdf/1609.03371.pdf
@YCor No more so than "Does there exist a model of ZFC satisfying 'there are infinitely many twin primes?'" is preferable to "Are there infinitely many twin primes?" - as usual, we either assume a "naive realist" stance or implicitly embed the question itself inside ZFC.
The relation $\text{ZFC}\vdash\varphi\leftrightarrow \psi$ is a $\Sigma_1^0$-definable equivalence relation on the set $\mathcal L$ of formulas in the language of set theory. It is a corollary of Theorem 3.2 of Neis-Sorbi's "Calibrating word problems of groups via the complexity of equivalence relations" that there is a finitely generated computably presentable group with generator set $A$ whose word problem, viewed as an equivalence relation $\sim$ on the set $W$ of words on $A$, is $\Sigma_1^0$-universal. As a consequence there is a reduction from the former equivalence relation to the latter, and this just means that there is a computable function $f : \mathcal{L}\to W$ such that $f(\varphi) \sim f(\psi)$ if and only if $\text{ZFC}\vdash\varphi\leftrightarrow\psi$, which is what you want.
This is impressively non-constructive. Thanks, this is great and what I was looking for so I’ll accept it, but I am curious if there are any more natural groups that are \Sigma^0_1 Universal, as that one seemed fairly messy. This links me to the rabbit hole of papers in the domain I was looking for (and did not know existed), so it’s helpful regardless
I don't know anything about this topic, I was just feeling lucky on Google. But someone asked a related question here. It seems to me that no one was able to provide a natural group whose word problem is undecidable...
@Phylliida A minor quibble: it's totally constructive, it's just messy. In fact it's a bit better than merely constructive in this particular instance: the original paper of Miller they build on gives a uniform construction for turning a ceer into a group of the appropriate nature.
@NoahSchweber yes I realized that after I made my comment and did more reading, oops. I’ve been going through the construction to get exactly what the function and group is for my other question.
|
2025-03-21T14:48:31.767232
| 2020-08-13T21:20:58 |
369109
|
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|
Stack Exchange
|
Rigorous results on chaos in a driven damped pendulum
The harmonically driven damped pendulum is often used as a simple example of a chaotic system, the equation is just $\ddot{\phi}+\frac1q\dot{\phi}+\sin\phi=A\cos(\omega t)$. As long as $A$ and $\omega$ are small it behaves like a driven harmonic oscillator, and asymptotically settles into regular oscillations with a fixed period. However, as $A$ (or $\omega$) are increased, with the rest of parameters fixed, the system undergoes a cascade of period doubling bifurcations leading to chaotic behavior, which then gives way to regular oscillations again when it is increased further. For example, when $q=2$ and $\omega=2/3$ the first period doubling ("symmetry breaking") occurs at $A\approx1.07$ and the first chaos at $A\approx1.08$.
Unfortunately, these results seem to be obtained by numerical simulations. I am actually interested in situations where chaos does not occur. Are there known rigorous conditions on $A,\omega$ and $q$ that put the system below the first period doubling?
The most common references are to Baker-Gollub's Chaotic Dynamics and Baker-Blackburn's The Pendulum that have a lot of phase diagrams with attractors, but no theorems. From what I understand there is Melnikov's method for detecting homoclinic bifurcations rigorously (e.g. Wiggins's Global Bifurcations and Chaos), but I could not find it used to obtain this kind of result for the pendulum. I searched for papers on driven damped pendulum on MathSciNet, but they seem to use physics-style approach and/or numerical simulations. A paper that gives more of an analytic insight by perturbation methods (treating $A$ as a small parameter) is Miles, Resonance and symmetry breaking for the pendulum, Physica D: Nonlinear Phenomena, Volume 31, Issue 2, June 1988, Pages 252-268.
Do you want the exact threshold or just not too weak sufficient condition that ensures the good behavior?
@fedja Just a lower estimate would be good. I am doing a control problem where one needs to select parameter values that induce stable regular oscillations, and it reduces to DDP.
If you want to control a real system, why are simulations not good enough for you? You might also want to look into the shadowing lemma.
@Wrzlprmft Because it is for a class of systems over a range of parameters. We want to induce stable oscillations with desired period and amplitude by specifying $A,\omega$ and $q$, and to know when it is possible we need to estimate the single period region of the parameter values and how they determine the induced values there. Thanks for the tip.
|
2025-03-21T14:48:31.767421
| 2020-08-13T22:02:52 |
369111
|
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|
Stack Exchange
|
Separating a certain planar region with an open set
I have a fairly specific question related to plane separation properties. I couldn't quite see how to use Phragmen–Brouwer properties to answer it because those kind of results generally apply to closed sets.
Let $U$ be an open connected subset of the closed unit square $[0,1]^2$ such that $U\cap \partial I^2=(0,1)\times \{0\}$. Suppose $V\subseteq U$ is open (but not necessarily connected), $V\cap \partial I^2=(1/3,2/3)\times\{0\}$, and that $V$ separates $a=(1/4,0)$ and $b=(3/4,0)$ in $U$. The last condition means that $a$ and $b$ lie in distinct connected components of $U\backslash V$.
I'd basically like to know if the component of $V$ that intersects the x-axis (call it $V_0$) limits on $[0,1]^2\backslash U$ and, alone, does the job of separating $a$ and $b$ in $U$.
Question 1: If $V_0$ is the connected component of $V$ that contains $(1/3,2/3)\times\{0\}$, must $V_0$ separate $a$ and $b$ in $U$?
Question 2: Must $\overline{V_0}$ meet $[0,1]^2\backslash U$?
are you sure that by separating you mean "no path"? then take $U$ to be the open square with an extra segment, $V$ to be a thinner rectangle, but pierced by a topologist's sine curve. Perhaps a more interesting question is when separated means "no connected set"
@erz whoops! You're right. That's the standard definition of "separated by" that I meant. Thanks.
I guess then you have to show that the outer boundary of $V_0$ is connected
|
2025-03-21T14:48:31.767543
| 2020-08-13T22:13:46 |
369112
|
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|
Stack Exchange
|
Graphs all of whose cuts are positive
Let $(V, E, w)$ a weighted graph, with vertices $V$, edges $E$, and signed weight $w:E\to \mathbb R$.
I am interested to know other popular properties that are known to imply, or are equivalent to, the requirement that all cuts of the graph are positive. This part in boldface means that every time $V$ is partitioned in two nonempty sets $V=V_1 \cup V_2$, the sum of $w$ over the edges with one end in $V_1$ and the other in $V_2$, needs to be $>0$.
And by any chance, does this property have a fancy name?
This is almost obvious, but a necessary condition is that the sum of all the edge weights must be positive (since the sum of weights incident to every vertex must be positive).
true, but I think there must be something more general, an analogue of that for all cuts ?
|
2025-03-21T14:48:31.767629
| 2020-08-13T22:16:18 |
369113
|
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|
Stack Exchange
|
Who introduced the heart ($\mathcal{C}^\heartsuit$) notation in the context of $t$-structures on triangulated categories?
In the context of $t$-structures
([Wikipedia],
[nLab],
[Notes I],
[Notes II],
[HA, Definition <IP_ADDRESS>)],
[BBD, Définition 1.3.1]),
one often writes $\mathcal{C}^\heartsuit$ for the heart of a category $\mathcal{C}$ equipped with a $t$-structure $(\mathcal{C}_{\leq0},\mathcal{C}_{\geq0})$.
While I've seen this notation used in many modern references, such as [HA, Definition <IP_ADDRESS>], it seems it was not introduced together with the notion of heart in [BBD, Définition 1.3.1], at least in print.
Where did this notation first appear? Was it already commonplace in handwritten sources or chalk talks at the time of [BBD]?
|
2025-03-21T14:48:31.767704
| 2020-08-13T22:37:43 |
369114
|
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|
Stack Exchange
|
Elementary curiosity on the dual of $H^{-1}(\Omega)$
Out of curiosity, I was wondering why in the analysis of elliptic PDEs with "source" in $H^{-1}(\Omega)$ everyone writes the right-hand side in the form $$g_0- \operatorname{div} g$$ with $g_0,g\in L^2(\Omega)$ instead of the (equivalent but apparently) more precise $$f-\operatorname{div} (\nabla f)$$ with $f$ in $H_0^1(\Omega)$.
Any real advantage behind the first choice?
By Riesz representation theorem, both expressions span the whole of $H^{-1}$. There is no difference between the two in terms of generality. They give two equivalent representations of $H^{-1}$. The second one is exactly the one that one gets by Riesz representation theorem.
If you use the second representation, it gets awkward to prove that a distribution is in $H^{-1}$. The first one tells you that any derivative of an $L^2$ function is in $H^{-1}$, no special structure required
Note sure to get your point. Let me try to explain better what I meant with the post. If one is interested in the properties of equations of the type $L u = f$ with $L$ the classical 2nd order linear elliptic operator then, from my point of view, the first RHS tells that the source is the sum of an $L^2$ function and the divergence of an $L^2$ function, while the second tells that it is the sum of a "more regular" $H_0^1$ function with the divergence of an $L^2$ function which one knows being "regular" because is the gradient of an $H_0^1$ function.
So, it seems like the 2nd one reveals more structure. Also, it is the gradient of the same function $f$ so that even from the notational point of view, one does not need to introduce two functions. If one decides to write a book, I cannot find good reasons to use the first one instead of the second one, other than the willingness of following the tradition.
Maybe, an advantage is that if one derives estimates in terms of $g_0$ and $g$, then one can immediately understand what happens for "source" in the more regular space $L^2$ by simply forgetting about $g$.
@WillieWong we are not writing $g$ as the gradient of a function but as the inverse image of the operator $I-\Delta$, which is nothing but the Riesz isomorphism in $H^{-1}$.
|
2025-03-21T14:48:31.767871
| 2020-08-13T23:02:46 |
369117
|
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|
Stack Exchange
|
Matrix groups with two generators
Given two matrices $A,B\in{\rm{SL}}_2(\Bbb{R})$, is there any criterion guaranteeing that the subgroup they generate is discrete? What if one puts restrictions on $A,B$ e.g. they are both elliptic? What could the topological closure of $\langle A,B\rangle$ be if it is not discrete?
Does this answer your question: https://mathoverflow.net/questions/109967/algorithm-to-test-for-discrete-or-quasi-fuchsian-subgroups-of-psl2-c?rq=1 ?
Consider for instance the group generated by two upper triangular matrices with 1 one the diagonal, where the top right entry in the first matrix is 1 and in the second matrix is $\sqrt{2}$. The topological closure is isomorphic to $\mathbb{R}$. You can do the same trick be taking diagonal matrices with $e^1, e^{-1}$ and $e^{\sqrt{2}}, e^{-\sqrt{2}}$ and similarly with rotation matrices (the topological closure will be the circle).
@darkl: These are "elementary" subgroups and usually are excluded from the discussion of discreteness algorithms. In many ways, they are quite different from the nonelementary subgroups for which algorithms work.
Here is a survey on this question: MR0974424
Gehring, F. W.; Martin, G. J.
Iteration theory and inequalities for Kleinian groups.
Bull. Amer. Math. Soc. (N.S.) 21 (1989), no. 1, 57–63.
|
2025-03-21T14:48:31.768001
| 2020-08-13T23:51:24 |
369120
|
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|
Stack Exchange
|
Rational characters of a number field are powers of norm
Consider a number field $K/\mathbb{Q}$ and the embedding of $K^* \hookrightarrow GL_n(\mathbb{Q})$. This is the set of rational points of a $\mathbb{Q}$-algebraic group $G \subseteq GL_n(\mathbb{C})$. Then is it true that any $\mathbb{Q}$-characters of $G$ will look like $g \mapsto \det(g)^k$ for some $k \in \mathbb{Z}$. That is, on $G_\mathbb{Q} = K^*$ the character will look like $x \mapsto N_\mathbb{Q}^K(x^k) $.
I heard someone make this remark in a discussion that all the rational characters on a number field are powers of norm. I have not been able to find a reference (or other non-norm characters!).
If $H$ denotes the multiplicative group defined over $K$, then $G=\mathrm{Res}_{K/\mathbb{Q}}H$. By Section 2.61 of Milne's "Algebraic Groups - The Theory of Group Schemes of Finite Type over a Field", the group $G_{\overline{\mathbb{Q}}}$ obtained from $G$ by extension of scalars is isomorphic to the product of $H_\sigma$, where $\sigma$ runs through the embeddings $K\hookrightarrow\overline{\mathbb{Q}}$. It follows that the characters of $G$ defined over $\overline{\mathbb{Q}}$ are the maps $\prod_\sigma f_\sigma^\sigma$, where each $f_\sigma$ is a character of $H$ defined over $\overline{\mathbb{Q}}$. The characters of $G$ defined over $\mathbb{Q}$ are those maps $\prod_\sigma f_\sigma^\sigma$, which are fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. The condition "fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$" means that $f_\sigma$ is independent of $\sigma$, and it is defined over $K$. In other words, $f_\sigma$ is the map $x\mapsto x^k$ on $K^\times$ with some $k\in\mathbb{Z}$ independent of $\sigma$, which means that
$$\prod_\sigma f_\sigma^\sigma(x)=\prod_\sigma(x^k)^\sigma=N_{K/\mathbb{Q}}(x^k).$$
|
2025-03-21T14:48:31.768248
| 2020-08-13T23:58:54 |
369121
|
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|
Stack Exchange
|
Proving positive invariance
I need to prove that set $D$(A picture for Set $D$) given by
$$D=\{(x,y):0\leq x\leq L_0,~0\leq y\leq X_0,~0\leq x+y \leq R_0\}\subseteq \mathbb{R}_+^2$$ of the system:
$$\dot{x}=k_1(R_0-x-y)(L_0-x)-k_{-1}x,~\dot{y}=k_2(R_0-x-y)(X_0-y)-k_{-2}y$$
is positively invariant. I thought of checking the direction of the vector fields on the boundary of $D$ (set $D$ is a trapezium as shown in the picture) and I showed that all the vector fields point inwards except at the corner (denoted $C$) where $$\dot{x}=0,~\dot{y}<0, \text{for } x=0,~y=R_0 $$. Here, the vector field points downwards tangential to $D$. To determine what happens at this point, I thought of checking the sign of the dot product between the vector field and the normal's field. I chose the normal such that it points inside $D$ by considering $\vec{n}_C=(1,-1)$ such that:
$$(1)\cdot\dot{x}+(-1)\cdot\dot{y}$$
and by substituting $x=0,~y=R_0$, I got $$(1)\cdot\dot{x}+(-1)\cdot\dot{y}=k_{-2}y>0$$ which is positive. The positive value means that the vector fields and the normals' point in the same direct.
Is my approach fine? If it is correct, I would like to back my reasoning with some literature/theorem. Can someone help me which theorem can I use to back up my reasoning? Is there another way in which I can show that the vector fields point inwards? I appreciate in advance.
Your set $D$ is closed. So as long as your vector field does not actually point outward (it can be tangent at some points), $D$ is invariant. // That said, what you did with taking the inner product does not actually show what you want, precisely because $C$ is a corner: there exists vectors at $C$ pointing out of $D$ that has positive inner product with the particular normal you choose.
Thank you Willie Wong for that positive response. Is there a specific theorem that I can use to conclude that since D is closed then even if the vector field is tangential to it, it is invariant?
As for the second approach suggested (i,e, the existence of vectors pointing outwards with the positive inner product), I'm not really sure how I can make a conclusion about the vector field pointing inwards. Would kindly highlight that, please! Thank you.
Since all other points on the boundary except $C$ have vectors that point inward, the only point that gives us trouble (as you've noticed) is $C$. So we only need to focus on $C$ to prove that $D$ is a trapping region. Then suppose that a solution with initial condition at $C$ eventually leaves $D$. You should be able to formulate a contradiction.
|
2025-03-21T14:48:31.768434
| 2020-08-14T00:22:39 |
369122
|
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|
Stack Exchange
|
Polynomials that arises from sums of first $n$ $k$-th powers
Define
$$\displaystyle P_k(n) = \sum_{m = 1}^n m^k,$$
which is a polynomial in $n$. Indeed, we have the following list which should be familiar to most mathematicians:
$$\displaystyle P_1(n) = \frac{n(n+1)}{2}, P_2(n) = \frac{n(n+1)(2n+1)}{6}, P_3(n) = \left(\frac{n(n+1)}{2} \right)^2,$$
etc.
Curiously, we see that $P_3(n)$ is the square of a polynomial, namely $P_3(n) = P_1(n)^2$. It can be proved, using a cheeky argument that involves Catalan's conjecture (now a theorem), that this is the only time $P_k(n)$ is a perfect power: suppose otherwise, and that there exists $s \geq 2$ and $k \geq 4$ such that $P_k(n) = Q(n)^s$ for some polynomial $Q$ such that $Q(m) \in \mathbb{Z}$ whenever $m \in \mathbb{Z}$. Then in particular we have $P_k(2) = Q(2)^s$, so that
$$\displaystyle 1 + 2^k = Q(2)^s,$$
and since $k \geq 4$ this violates Catalan's conjecture that the only solution to $x^n - y^m = 1$ is $(x,y) = (3,2)$ and $(n,m) = (2,3)$. Since Catalan's conjecture was only proved in 2002 by Mihăilescu, this seems like overkill. Is there a simpler proof of this fact?
Also, regarding the polynomials $P_k(n)$, it seems that for small $k$ at least that they are all reducible. One sees that $P_k(0) = 0$ for all $k$ and, so $P_k(x)$ is divisible by $x$ as a polynomial. Moreover, considering the congruence $P_k(n) \pmod{n+1}$ reveals that the left hand side is always zero, hence $x+1 | P_k(x)$ as well. Put $Q_k(x) = \dfrac{P_k(x)}{x(x+1)}$ and
$$\displaystyle S(X) = |\{k \in \mathbb{N} : k \leq X, Q_k \text{ is reducible over } \mathbb{Q}\}|.$$
Does the limit $\displaystyle \lim_{X \rightarrow \infty} \frac{S(X)}{X}$ exist, and is it positive?
Another related question is how often the polynomials $P_k$ are square-free, that is, indivisible by the square of any polynomial over $\mathbb{C}$, say. Put
$$\displaystyle T(X) = |\{k \in \mathbb{N} : k \leq X, P_k \text{ is square-free as a polynomial}\}.$$
Does $\displaystyle \lim_{X \rightarrow \infty} \frac{T(X)}{X}$ exist, and is it positive?
Maybe including the keyword "Bernoulli numbers" somewhere would be good.
You can show that $P_k(x)$ is always divisible by $x(x+1)$, so in that sense they will always be reducible. Perhaps it is more interesting to look at $P_k(x)/(x^2+x)$?
Also, solving the diophantine equation $1+2^k=r^s$ is very easy without appealing to Mihăilescu's theorem. Just note that $2^k=(r-1)(r^{s-1}+\cdots+1)$ therefore $r=2^{t}+1$ for some $t<k$. Next divide both sides by $2^t$ and take mod $2$ to conclude that $s$ must be even. This tells you that $r+1$ also divides $2^k$, but the only way $r\pm1$ are both powers of $2$ is if $r=3$ and you get the solution $1+8=9$.
These are Faulhaber polynomials and the Wikipedia page has some good information.
@GjergjiZaimi thanks for your observation! I will update the question to ask for the reducibility of $P_k(x)/(x(x+1))$ instead.
According to wikipedia (@John Machaek's link), $P_k(x)$ is always divisible by $x^2(x+1)^2$ for odd $k\ge 3$. Moreover, from the first polynomials, it is tempting to conjecture that for even $k$ the polynomial is divisible by $(2x+1)$, which would mean that your polynomials are always reducible for $k\ge 3$
|
2025-03-21T14:48:31.768663
| 2020-08-14T00:39:53 |
369123
|
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|
Stack Exchange
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Reference for rigorous second quantization
I'd like to get acquainted with the basics of mathematical rigorous second quantization, so I'm looking for an adequate reference on this subject. I've a background in quantum mechanics, but I don't know much on quantum field thery (I don't know to what extent it is necessary to understand second quantization). The point is: my research area is rigorous statistical mechanics, so I don't intend to become an expert on the subject, but I need to understand at least the basics. The only reference I know is The Method of second quantization by F. Berezin. So far, I've just skimmed the book and it seems incredibly well-written, but it is obviously an old book and I feel it is not very modern concerning its mathematics: I'd be happier to find a more technical exposition, using modern functional analysis. However, I don't know if this reasoning is just a reflection of the fact that I know hardly nothing on the subject. So I ask: is Berezin's book the best place to start? Are there other (more technical, maybe) references on the subject? Thanks in advance!
§X.7 of Reed & Simon's Methods of Modern Mathematical Physics v.2 (Academic Press, 1975) starts out with the epigraph First quantization is a mystery, but second quantization is a functor. E. Nelson. It's pretty rigorous.
Great! I love these collection and I pretty much like everything Reed and Simon write. This sounds perfect!
One book which may be useful to you is by Edson De Faria's: Mathematical aspects of QFT.
The tract has been written by someone who definitely has a strong background in physics but is nevertheless a mathematician (former student of the famous geometer Dennis Purnell Sullivan). As you may know, there are different ways to introduce QFT, one being via Lagrangian densities, another via the Fock space (and there are others, for instance the categorical approach). This text does not cover all of them, but it is definitely something a math-oriented fellow can enjoy
Great answer! Thank you so much! I'll definitely read Edson's book. Btw, what's your opinion on Dimock's book "Quantum Mechanics and Quantum Field Theory"? It seems very good too.
simple answer: I have no opinion, because I have not read it ! But now I am going to check it out... as an aside, I know what I would LOVE to see in a book on QFT: some articulated discussion on how the different presentations are related to one another. Never seen anything like it..
PS I knew Edson quite well: we were writing our dissertations in the same crammed room some centuries ago. If you read the book and wish to reach out to him, do it: Edson is a very nice fellow,
Mirco, that's really nice. I took a course many years ago with him as well. Very nice professor, indeed.
the world is small indeed.... if you catch up with Edson send him the best from the crazy italian :)
Nice! I'll keep that in mind!
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2025-03-21T14:48:31.768881
| 2020-08-14T01:08:13 |
369125
|
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|
Stack Exchange
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When does an infinite product of random matrices have finite expected norm?
Given a finite set of matrices $A_i$, sample $n$ matrices uniformly with replacement and compute $f_n=\|A_1 A_2\cdots A_n\|^2$. When is the following limit finite?
$$\lim_{n\to \infty} E[f_n]$$
I'm especially interested in the case when $A_i=I-a x_i x_i^\intercal$ for some vector $x_i$ and positive scalar $a$.
A paper by Deffosez formulated a sufficient condition for this to be finite, and conjectured it to also be a necessary condition, Lemma 1 of "Averaged Least-Mean-Squares: Bias-Variance Trade-offs and Optimal Sampling Distributions."
For scalar $x$, their formula recovers a standard stability result:
$$a\le \frac{2 E[x^2]}{E[x^4]}$$
For vector $x$ in $d$ dimensions, second moment tensor $X^2$, fourth moment tensor $X^4$ and Einstein summation notation:
$$a\le \text{sup}_{A\in \mathcal{S}(\mathbb{R}^d)}\frac{2 A_{ij}X^2_{jk} A_{ki}}{A_{ij}X^4_{ijkl}A_{kl}}$$
However, this formulation is difficult to apply in practice -- optimization over the space of symmetric matrices where each step of optimization takes $O(d^4)$ operations. It's also hard to interpret -- which properties of the set of $A_i$'s are most responsible for divergent behavior?
Is the conjecture true?
Any suggestions for how to approximate this quantity, or obtain a "nicer" pair of necessary/sufficient conditions?
This doesn't directly answer your question, but there is a classic paper of Furstenberg and Kesten showing that the limit $\lim_{n \to \infty} \frac{1}{n} E[\log f_n]$ exists under mild hypotheses (and there is also a central limit theorem): https://mathscinet.ams.org/mathscinet-getitem?mr=121828
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2025-03-21T14:48:31.769016
| 2020-08-14T01:29:31 |
369128
|
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|
Stack Exchange
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On smoothness and roughness of a number related to triangular numbers
Define $\triangle_n$ to be the $n$th triangular number.
Define $$M_n=(2\triangle_n-1)2\triangle_n(2\triangle_n+1)=2\triangle_n(4\triangle_n^2-1).$$
Define $(\ell,k)$-smough numbers to be numbers that have all prime divisors $p$ of the form of either $p<\ell$ or $k<p$.
Is there an $\ell_0\geq6$ such that for every $\ell>\ell_0$ there is a $k_\ell$ such that for every $k\geq k_\ell$ there is always an $n$ such that $M_n$ is $(\ell,k)$-smough?
Given an $\ell$ and a $k$, how small can $n$ be such that $M_n$ is $(\ell,k)$-smough?
$\triangle$ is standard for triangular numbers. $\square$ represents sum of cubes (three dimensional squares).
$\triangle$ was used by Gauss.
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2025-03-21T14:48:31.769088
| 2020-08-14T01:38:43 |
369130
|
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Stack Exchange
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Fuchsian groups and Eichler's result
Let $G$ be a Fuchsian group of first kind contained in $\text{PSL}_2(\mathbb{R})$. A result of Eichler says, there exists a finite set $S\subset G$ such that any $\gamma$ in $G$ can be written as a product $\prod_{i=1}^{k} \gamma_i,$ where each $\gamma_i$ are either in $S,$ or power of some parabolic element coming from $S,$ for some $k\ll \log ||\gamma||.$ Howerver, the original Eichler's paper is in German, (http://matwbn.icm.edu.pl/ksiazki/aa/aa11/aa11111.pdf) and this is giving me trouble to understand the proof. Does anyone know if I can find the proof written in English somewhere else ?
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2025-03-21T14:48:31.769164
| 2020-08-14T02:02:29 |
369131
|
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Stack Exchange
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Fibrations in complex geometry
Let $X^n$ be a compact Kähler manifold with $K_X$ semi-ample, i.e., a sufficiently high power of $K_X$ is basepoint free. The associated pluricanonical system $| K_X^{\ell} |$ furnishes a birational map $$f : X \dashrightarrow \mathbb{P}^{\dim H^0(X, K_X^{\ell})-1}$$ onto some normal projective variety $Y \subset \mathbb{P}^{\dim H^0(X, K_X^{\ell})-1}$ of dimension $\kappa(X)$. Here, $\kappa(X)$ denotes the Kodaira dimension of $X$. The semi-ampleness of $K_X$ further implies that $K_X^{\ell} \simeq f^{\ast} \mathcal{O}(1)$. In particular, for every $y \in Y$ that is not contained in the discriminant locus of $f$, $K_X^{\ell} \vert_{f^{-1}(y)} \simeq \mathcal{O}_{X_y}$. Since $f$ is a submersion near $f^{-1}(y)$, we have the adjunction-type relation $K_{f^{-1}(y)} \simeq K_X \vert_{f^{-1}(y)}$ and therefore the fibres of $f$ are Calabi--Yau manifolds of dimension $n- \kappa(X)$.
In the Kähler geometry literature, it is common to refer to this map $f$ as a Calabi--Yau fibration. My question may be extremely obvious, but nevertheless:
Question: Is this a fibration in the sense of homotopy theory, i.e., does this map satisfy the homotopy lifting property?
I would say that the answer is in general no.
Think of an elliptic surface $X$ with Kodaira dimension $1$ and whose elliptic fibration contains a cuspidal curve. Then the general fibre is not homotopically equivalent to the special one (the former is homeomorphic to $S^1 \times S^1$, the latter to $S^1$, in particular their fundamental groups are different), whereas all the fibres of a Hurewicz fibration have the same homotopy type.
Edit. Actually, any elliptic surface $X$ with Kodaira dimension $1$ and whose elliptic fibration contains a nodal curve also provides a counterexample. In fact, a nodal cubic is homeomorphic to a torus "with one cycle shrunk away”; in particular, it has the homotopy type of $S^1 \vee S^2$ and the previous argument applies.
Thankyou! Do you have a reference for these homeomorphisms? Admittedly, I have never thought about homotopy types of the cuspoidal/nodal curves.
It is rather elementary. The cuspidal cubic has affine equation $y^2=x^3$, and the projection $(x, , y) \mapsto y$ induces a homeomorphism onto $\mathbb{P}^1$, which is topologically $S^2$. Regarding the nodal cubic, in can be seen as $\mathbb{P}^1$ with two points identified. Hence, topologically, we are identifying two points on $S^2$; this is clearly the same thing as shrinking a cycle on a torus, and the corresponding quotient space is homotopically equivalent to a sphere with a circle attached at a point.
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2025-03-21T14:48:31.769374
| 2020-08-14T02:47:49 |
369133
|
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|
Stack Exchange
|
A question on Grothendieck space
A Banach space $X$ is said to be Grothendieck if the weak and the weak* convergence of sequences in $X^{*}$ coincide. I have the following two questions.
Question 1. A Banach space $X$ is Grothendieck if and only if every weak*-Cauchy sequence in $X^{*}$ is weakly Cauchy?
Question 2. If $(x^{*}_{n})_{n}$ is a weak Cauchy sequence and a weak*-null sequence in $X^{*}$, is $(x^{*}_{n})_{n}$ a weak-null sequence?
Thank you!
If $(x_n^)$ is $w^$ Cauchy, then for every $x \in X$, $x_n^(x) $ has a limit, call it $f(x)$. Then $f$ is linear. Moreover, $|x_n^| \le C$, by uniform boundedness, hence $f$ is bounded, too. Then $w^$ Cauchy means $w^$ convergent.
Thanks, Giorgio. But it seems that you do not answer my questions.
The argument applies both to $w$ and $w^$ convergence, so Cauchy means convergent in both topologies and Q1 is equivalent to the definition. Q2 follows similarly, if $x_n^ \to 0$ $w^$ and is $w$-Cauchy, then it converges weak to some $z^$ and then $z^=0$ since $w$ imples $w^$. Did I overlook some point?
If $x^{}_{n}\rightarrow 0$ $ w^{}$ and is $w$-Cauchy, then it converges weak to some $x^{}\in X^{}$, not $z^{}\in X^{}$. I think that you overlook this point.
No, everything happens in $X^*$.
Maybe I can not understand your answer. Could you provide a detailed proof?
Ah, true!. See the answer below by @Dirk Werner.
I find the following criterion useful: A sequence $(x_n)$ is Cauchy iff for all subsequences $(x_{n_{k+1}}-x_{n_k})$ tends to $0$. This works for the norm topology, the weak topology and the weak$^*$ topology. This answers Q1 in the positive.
As for Q2, if $(x_n^*)$ is weakly Cauchy and weak$^*$ null, it has a limit $x^{***}$ for the weak$^*$ topology of $X^{***}$; decompose $x^{***}=x^* + x_s^{***}$, where $x_s^{***}$ is the ``singular part'' in the annihilator of $X$ in $X^{***}$. By, the assumption of Q2, $x^*=0$; i.e., $x^{***}$ is singular. This seems to be as good as it gets in a general Banach space.
Since $(x^{}_{n})$ is weak-null, $x^{}(x)=0$ for all $x\in X$,i.e.,$x^{}$ is in the annihilator of $X$ in $X^{}$. But this does not necessarily imply that $x^{}=0$. We have to prove that $x^{***}=0$.
Using the criterion you mentioned, it seems that we can only prove the necessary part of Q1, but can not prove the sufficient part.
@Dongyang: You are right!
Call the condition in Q1 Cauchy Grothendieck. Let $X$ have this property. If $(x_n^)$ is w$^$ null, it has a limit $x^{}\in X^{}$. To show that it is $0$, consider the w$^$ null sequence $(x_1^, 0, x_2^, 0, \dots)$ interlacing the given sequence with $0$. It has a limit $y^{}\in X^{}$ since the space is Cauchy Grothendieck. Now along the odd integers, the new sequence tends tends to $x^{}$, along the even integers it tends to $0$. Hence $x^{*}=0$.
@ Dirk Werner Nice argument! Q2 seems to be false since $B_{X^}$ is $w$-dense in $B_{X^{}}$ but we need an $X$ where this density is realized through sequences. Maybe some separabilty of $X^{}$ suffices and the space should not have the Grothendieck property.
@Giorgio: Yes, separability of the bidual is good enough, e.g., the bidual of the James space is separable. And a separable Grothendieck space is reflexive. -- Q2 is true in a Grothendieck space (or any other space with a wsc dual); but I read the question as about general Banach spaces.
@DirkWerner: So your answer and remark show that $X$ has the property in Q2 if and only if $X^*$ is weakly sequentially complete.
Q1 is already answered by Prof. Dirk Werner above. I simply list a number of equivalent conditions that seems to be related, but not the same as the Grothendieck property.
The following are indeed equivalent:
every weak*-null sequence in $X^{\ast}$ has a weakly Cauchy subsequence.
every bounded weak*-sequentially compact subset of $X^{\ast}$ is weakly precompact (every sequence has a weakly Cauchy subsequence).
for every bounded $T:X\to c_0$, the adjoint $T^{\ast}:\ell^1\to X^{\ast}$ is weakly precompact (i.e., $T^{\ast}$ maps bounded sets onto weakly precompact sets).
for every bounded $T:X\to Y$, where $Y$ is another Banach space with weak* sequentially compact dual ball, the adjoint $T^{\ast}$ is weakly precompact.
no weak*-null sequence in the unit ball of $X^{\ast}$ contains an $\ell^1$-subsequence.
there is no surjective bounded $T:X\to c_0$.
It is well known that $X$ is a Grothendieck space iff $X^{\ast}$ is weakly sequentially complete and (6)
While your post might contain useful information (although no proof or reference is given), I don't think that it answers the question. It looks more like a (long) comment.
This is essentially Theorem 3.1.1 in https://arxiv.org/pdf/2102.03838.pdf
@Tomasz Kania: Thank you for the reference. I was not aware that there were so many open problems pertaining to Grothendieck spaces. It will be a pleasure to read through the paper, and it sure will take some time for me.
@AlexM. I think there is a place on MO for answers in the "too long to be a comment" category which provide useful information and context for a question. I agree it would be nice to have some references.
AlexM @TimCampion: Sorry folks, I didn't include a reference since I do not have a good reference for the whole equivalence except my notes. If I included a proof myself, the post would be longer than the question and the answer. However, as Tomasz Kania graciously pointed out, the proof is almost verbatim similar to Theorem 3.1.1 in the reference he's given.
My purpose was actually to point out the difference between the 4 statements that are seemingly the same :
every weak*-null sequence in the dual is weakly null
every weak*-null sequence in the dual has a weakly null subsequence
every weak*-null sequence in the dual is weakly Cauchy
every weak*-null sequence in $X^{\ast}$ has weakly Cauchy subsequence
The first three of those is equivalent to X being a Grothendieck space. Only the 4th is weaker, as written above.
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2025-03-21T14:48:31.769771
| 2020-08-14T03:12:23 |
369135
|
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|
Stack Exchange
|
Algebraic spaces in the étale topology (proof from Stacks project)
I have a question about the proof of Lemma 78.12.1 from Stacks Project. The aim of the last paragraph of the proof is to verify that the map of sheaves in the étale topology $F \to U/R$ is an isomorphism. By Lemma 7.11.2 our job is to show that it's surjective and injective. The proof that $F \to U/R$ is injective I not understand. The used argument is:
On the other hand, the map $F \to U/R$ is injective (as a map of presheaves) since $R=U \times_{U/R} U$ again by Spaces, Theorem 63.10.5.
Why $R=U \times_{U/R} U$ imply that the map is injective?
Recall from the beginning of the proof of the lemma that $R$ is defined to be $U \times_F U$, so the surjections $U \to F \to U/R$ induce a canonical étale sheaf map $U \times_{U/R} U \to U \times_F U$ that was shown to be an isomorphism.
Let's see what this means for a test object $S$. Let $A = U(S)$, $B = F(S)$, $C = (U/R)(S)$, and let $f: A \to B$ and $g: B \to C$ be induced by the maps $U \to F \to U/R$ given before. Then we are basically saying that, if we replace $S$ with a suitably fine étale cover, then the induced set map $A \times_C A \to A \times_B A$ is an isomorphism. Here, $A \times_C A = \{ (x,y) \in A \times A | gf(x) = gf(y) \}$ and $A \times_B A = \{ (x,y) \in A \times A | f(x) = f(y) \}$. That is to say, $gf(x) = gf(y)$ if and only if $f(x) = f(y)$, or $g$ is injective on the image of $f$.
Because $U \to F$ is a surjection of sheaves, we may assume $f$ is surjective after passing to a suitable cover. We conclude that $g$ is injective, and hence $F \to U/R$ is injective.
everything is clear now except one step: Why is the induced $U \times_{U/R} U \to U \times_F U$ is an isomorphism?
@katalaveino That is the "used argument" you quoted, i.e., https://stacks.math.columbia.edu/tag/02WW
yes, I think I understand. Theorem 63.10.5 implies that since $R := U \times_F U \to U \times_S U$ is etale equiv relation, the data $(U,R, U \to U/R)$ is representation of $U/R$. But beeing represented by this data implies for $U/R$ tautologically that $R=U \times_{U/R} U$ holds? that's the point, right?
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