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https://mathoverflow.net/questions/390812
|
4
|
By implementing the techniques described in and similar to
>
> A. Dujella, J. C. Peral, *Elliptic curves with torsion group Z/8Z or Z/2Z x Z/6Z*, arXiv, Number Theory [math.NT] (2013), arXiv:[1306.0027v1](https://arxiv.org/pdf/1306.0027v1.pdf)
>
>
>
>
> A. J. MacLeod, *A Simple Method for High-Rank Families of
> Elliptic Curves with Specified Torsion*, arXiv, Number Theory [math.NT] (2014), arXiv:[1410.1662v1](https://arxiv.org/abs/1410.1662v1)
>
>
>
we found two $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$ elliptic curves with the suspected rank $5$. Magma or mwrank can uncover $4$ generators for each curve.
```
[1,0,0,-14578233419504842866768626074144915305633167440,677262813789165866285590491715440177836227774714086370178932369721600]
(-1632905513070906593633395/16 : 2124657596658008539974203142878778065/64 : 1)
(10622742675684263293819723408153580/141909670681 : 124725874840414877086253409903032599643909082039040/53458650132568829 : 1)
(-97457668455374061775698849271818397251850996320108070/699579678211711463845331372281 : 1251876399993507238708312434030333353053490882764701323776880839986798621571460/585134598109537035210690491594808002277939571 : 1)
(49337296288524457271016713026090973088733461258708686528/509102607359595399318340821876529 : 151796511564094491944492436957734460721901141802177579465531284308649800667607130768/11487035994099737573838589665510598244834003265833 : 1)
```
```
[1,0,1,-737894644327026219218841358570509913846652580014,190908291625785611127850212005200136993025927512858679851032108028536736]
Generator 1 is [-238715637021392602565547:-594518321028603238275438200237984267:1];
Generator 2 is [31360030622781021392151674068077015054:10671104634442837701763318571303490297384474168604:41410962494173];
Generator 3 is [1458463852616891911932692956390480548:-2127910767977542086809771217107258154280506806417:8268141253861];
Generator 4 is [-1269285019645294204766876727074196:-936846131419637244134703316207143480358743016:1672446203];
```
Each curve has $7$ isogenous curves: three $2$-isogenous $\mathbb{Z}/6\mathbb{Z}$, one $3$-isogenous $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, and three $3$-isogenous $\mathbb{Z}/2\mathbb{Z}$ curves each.
```
[ 1, 0, 0, -14578233419504842866768626074144915305633167440, 677262813789165866285590491715440177836227774714086370178932369721600 ]
[ 1, 0, 0, -16788121482662999494757620303458923799396647440, 458347722412840961402654372801843479314855058739320714200591470465600 ]
[ 1, 0, 0, -12390456997202053746439995758428243843549687440, 887546443368809714827401664475279559448327008402628774422894004977600 ]
[ 1, 0, 0, -14576851441951382397539853329545081741153167440, 677397680379739007790920569085466233004214499673105145976879825721600 ]
[ 1, 0, 0, -41980864276959314492810225577242671428565265440, -2468181464491007069038907228919975481228430981974833640494773900794000 ]
[ 1, 0, 0, -623018003626406527989379968810953241503944571740, -189260266529123210476476660352196009349286414306954926139458416124443260 ]
[ 1, 0, 0, 103607077681372989451472258964699809067497272860, -15814718849970915410189591344868440743288201743513473235175995725111140 ]
[ 1, 0, 0, -14765289440014632645792271909007165829857967440, 658984511291994677032788803916334626722274364249125813187527019161600 ]
```
```
[ 1, 0, 1, -737894644327026219218841358570509913846652580014, 190908291625785611127850212005200136993025927512858679851032108028536736 ]
[ 1, 0, 1, -241916501236723307413266334879564479597172572014, -43233279468511210455812058234371488153394628551933612572678486665396064 ]
[ 1, 0, 1, 1666422596383330350336832332119338002274496477386, 1176171118768273493852310932836981453901128598294298129440036617833140096 ]
[ 1, 0, 1, -11077862174482229377663715428315484777959481765414, 14190706014518294309757776770418575190667486067178220709161242656555664576 ]
[ 1, 0, 1, -19291288676001541738319227851451141863129028466429, -32597782170371077908961912239683117512235494535789630982779370974269425448 ]
[ 1, 0, 1, -19288797687595273046405834919359334932574148466429, -32606625123070618542609678065286321263560745277882760553634793344061425448 ]
[ 1, 0, 1, -15915357490227593061488748204997524479146121906429, -44371276360828247024588958692803998413327921261654610033506325574902129448 ]
[ 1, 0, 1, -22707075676275789485763994411373670135990015026429, -20258339007143308239877852937993242901252252662392631030077662487428721448 ]
```
All the attempts to uncover the fifth missing generators for the curves in question and all their respective isogenous curves are unsuccessful yet.
The methods that were tried so far:
1. A full run of `MordellWeilShaInformation` in the properly installed Magma V2.26-2 in Ubuntu 20.04 Linux with 16 Gb of RAM. (Note: a $\mathbb{Z}/2\mathbb{Z}$ curve `[ 1, 0, 0, -623018..., -189260... ]` crashes Magma currently, a bug report is filed).
2. Mwrank (Jul 28 2020) runs with bounds $-b14$.
3. Magma script searches on the $2$- and $4$-coverings with the bounds $2^{40}$. Only the $2$- and $4$-coverings where all known generators were excluded, were considered (as described in the question [here](https://mathoverflow.net/questions/384358/z-8z-elliptic-curve-with-a-missing-generator)).
4. Magma script searches on the $6$- and $12$-coverings. Only the $6$- and $12$-coverings where all known generators were excluded, were considered (as described in the answer [here](https://mathoverflow.net/questions/385814/3-6-12-descent-for-z2xz6-elliptic-curves)). (Note: a random $3$-covering for the element `S3toA(0)`$=<1,1,1>$ produced by `ThreeDescentCubic(E, <1,1,1>)` was used to create $6$- and $12$-coverings, a full run of `ThreeSelmerGroup(E)` was not performed as it would take too long).
```
// S3, S3toA := ThreeSelmerGroup(E); S3; S3toA; S3toA(0);
threecovers_1, maps3toE := ThreeDescentCubic(E, <1,1,1>); threecovers_1; maps3toE;
```
>
> **Question.** We would greatly appreciate any hint leading to the discovery of the fifth generator(s) for the curve(s) in question or the respective isogenous curves.
>
>
>
Your name will be published at the bottom of Andrej Dujella's [page](https://web.math.pmf.unizg.hr/%7Eduje/tors/z2z6old5.html) when the fifth generator is uncovered for the curve(s).
|
https://mathoverflow.net/users/95511
|
Z2xZ6 elliptic curves with missing generators
|
By implementing the techniques described in
>
> T. Fisher, [Finding rational points on elliptic curves using 6-descent and 12-descent](https://www.dpmms.cam.ac.uk/%7Etaf1000/papers/sixandtwelve.pdf), *Journal of Algebra* **320** (2008), no. 2, 853-884,
>
>
>
Tom Fisher himself used $12$-descent to find the missing generators for both mentioned curves. The Magma files are available for download: [example2.m](https://mega.nz/file/HsxiyZCD#v27IEn9KMvlWuXNnT4sOxMpq6ZBZZavhh_04dgXYLpY) (33 kB) and [example1.m](https://mega.nz/file/zgp2SBSK#N3AiSr-vO8BXvmM-U6gpzjpBLT03ZFHZg2IgPR7Cg0M) (30 kB). The fifth generators have heights $446.24$ and $375.99$.
He added, "As I indicate in the attached files, it turns out that these points could in principle have been found using $4$-descent, but not in practice since the product of Tamagawa numbers is so large. This is typically the case for curves in these torsion families.
The reason I can use $12$-descent on these curves is that the rational
$3$-torsion point makes the $3$-descent practical (although not using Magma's `ThreeSelmerGroup`, which was never designed to take advantage of this extra
structure)."
**EDIT:** Both curves were added to Andrej Dujella's $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$ [rank $5$ page](https://web.math.pmf.unizg.hr/%7Eduje/tors/z2z6old5.html).
|
5
|
https://mathoverflow.net/users/95511
|
410743
| 168,038 |
https://mathoverflow.net/questions/410712
|
6
|
$\DeclareMathOperator\ddiv{div}\DeclareMathOperator\tr{tr}\newcommand{\conf}{\mathrm{conf}}$Consider this PDE on a symmetric tensor $h$ on $S^2$:
$$\Delta \text{tr}(h) - \ddiv(\ddiv(h)) + \tr(h) = f$$
where $f \in L^2(S^2)$ and $\Delta$, $\ddiv$ and $\tr$ are with respect to the round metric on $S^2$.
I wish to show that there exists at least one solution to this.
If we assume for simplicity that $h = \frac{1}{2} \tr(h) g\_{S^2}$, then the PDE becomes
$$\Delta \tr(h) + 2\tr(h) = 2f$$
which doesn't have a solution for every $f$ since $-2$ is an eigenvalue of $\Delta$ (I am assuming that $\Delta + \lambda$ is not surjective if $-\lambda$ is an eigenvalue; is this correct?). I am not sure how to approach this.
One approach is decomposing $h$ into its trace part and a conformal Lie derivative of a vector field $X$: $h = \frac{1}{2} \tr(h) g\_{S^2} + \mathcal{L}\_{\conf}X$. Then the PDE becomes:
$$\frac{1}{2}\Delta \tr(h) - \ddiv(\Delta\_{\conf}X) + \text{tr}(h) = f$$
where $\Delta\_{\conf}$ is the conformal laplacian on vector fields.
I am not able to continue.
Any help is appreciated.
|
https://mathoverflow.net/users/138705
|
Solving $\Delta \text{tr}(h) - \mathrm{div}(\mathrm{div}(h)) + \text{tr}(h) = f$ on $S^2$
|
Your second-order differential operator appears when one takes the variation of the scalar curvature of the sphere by a symmetric tensor $h\mapsto \frac{d}{dt}|\_{t=0}\mathrm{Sc}\_{g+th}$. In constant sectional curvature, such operators were studied for instance by Calabi (60'), and then in constant scalar curvature by Ebin (69').
Edit: I have taken another look at Ebin's paper and figured out a similar, yet simpler argument than my original one (you can still find it below). I think this argument provides an answer to your question in any geometry: let $(M,g)$ be an arbitrary closed Riemmanian manfiold. Define a linear second order operator $\gamma:S^{2}(T^{\*}M)\rightarrow C^{\infty}(M)$ by $\gamma h=-\Delta\_{g}tr\_{g}h+\mathrm{div}\mathrm{div}h-(h,Ric\_{g})\_{g}$. Consider its $L^{2}$-dual $\gamma^{\*}:C^{\infty}(M)\rightarrow S^{2}(T^{\*}M)$. It can be checked to assume the form $\gamma^{\*}f=\mathrm{Hess}\_{g}f-fRic\_{g}-\Delta\_{g}fg$. The symbol of $\gamma$ is $|\xi|^{2}tr+i\_{\xi}i\_{\xi}$ while the symbol of $\gamma^{\*}$ is $\xi\otimes\xi+|\xi|^{2}I$. Thus the symbol of $\gamma\gamma^{\*}$ is a constant multiplied by $|\xi|^{4}$, so $\gamma\gamma^{\*}$ is an elliptic fourth order operator (a "bilaplacian"). Therefore, if you want to solve $\gamma h=f$ then replace $h=\gamma^{\*}g$ and solve $\gamma\gamma^{\*} g=f$. This has a solution if and only if $f$ satisfies a compatibility condition: it must be $L^{2}$-orthogonal to the finite dimensional kernel of $\gamma\gamma^{\*}$, which in this case is equal to the kernel of $\gamma^{\*}$.
In your original case of the sphere, $Ric\_{g}=g$ and so $(h,Ric\_{g})\_{g}=tr\_{g}h$.
My original answer is below:
Here is what I suggest to solve your question in the special case of $S^{2}$.
If you have a Riemmanian manifold with constant sectional curvature, $(M,g)$, consider the second order differential operator $H\_{g}=\frac{1}{2}{(d^{\nabla^{g}}d^{\nabla^{g}}\_{V}+d^{\nabla^{g}}\_{V}d^{\nabla^{g}}})-\frac{1}{2}\kappa g\wedge$.
$\kappa$ here is the sectional curvature of the space, so in your case it can be taken to be $\kappa=1$. The wedge product here is known as the Kulkarni-Numizu product of symmetric tensor fields (there is a more generlized version of this wedge product, operating on "double forms", also adressed by Kulkarni, but this notion will do). $d^{\nabla^{g}}$ and $d^{\nabla^{g}}\_{V}$ here are the exterior covariant derviatvie of the first and second index of a symmetric tensor, respectively, where we think of symmetric tensors as examples of vector-valued differential forms $\Omega^{1}(M ; T^{\*}M)$. The image of $H\_{g}$ lies in $\Omega^{2}(M;\Lambda^{2}T^{\*}M)$, and in the case where $M$ is two-dimensional every element in this space is fully determined by its "scalar curvature", namely if $\sigma\in\Omega^{2}(M;\Lambda^{2}T^{\*}M)$ then $\sigma=\frac{1}{4}(tr\_{g}tr\_{g}\sigma) g\wedge g$.
In the case where $\kappa=1$, a direct calcultion shows that taking the trace twice from $H\_{g}$ yields the operator $tr\_{g}tr\_{g}H\_{g}=-\Delta\_{g}tr\_{g}+\mathrm{div}\mathrm{div}-tr\_{g}$, so solving your equation is equivelent of solving $H\_{g}h=\frac{1}{4} f g\wedge g$, where $g$ is the metric of the sphere. Replacing $h=H^{\*}\_{g}\psi$ for $\psi\in\Omega^2(M;\Lambda^{2}T^{\*}M)$ where $H\_{g}^{\*}$ is the formal $L^{2}$ dual of $H\_{g}$ yields the equation $H\_{g}H^{\*}\_{g}\psi=\frac{1}{4} f g\wedge g$. Note how since $H\_{g}$ operates on symemtric tensors, the image of $H^{\*}\_{g}$ is a symmetric tensor in $\Omega^{1}(M;T^{\*}M)$.
Another calculation then shows that the principle symbol of $H\_{g}H^{\*}\_{g}$ in the case where $M$ is two dimensional is $|\xi|^{4}$. Thus this is an elliptic fourth order differential opertor (a "bilaplcian"), and so the equation $H\_{g}H^{\*}\_{g}\psi=\frac{1}{4} f g\wedge g$ is solvable for $\psi$ if and only if $ \frac{1}{4} f g\wedge g$ is orthogonal to the kernel of $H\_{g}H^{\*}\_{g}$. By duality, this kernel is equal to $\mathrm{ker} H^{\*}\_{g}$. I am not sure if this kernel is trivial when $M$ is simply connected. If not, then this orthogonality yields a compatability condition which $f$ must satisfy in order for the equation to have a solution.
|
4
|
https://mathoverflow.net/users/144247
|
410745
| 168,039 |
https://mathoverflow.net/questions/410744
|
1
|
Let $H$ be a finite simple graph, with $v\_H\ge3$ vertices and $e\_H\ge3$ edges. Say $H$ is strict 2-balanced if $\frac{e\_H-1}{v\_H-2}\gt \frac{e\_K-1}{v\_K-2}$ for all proper subgraphs $K$ with $v\_K\ge3$, and say $H$ is strict balanced if $\frac{e\_H}{v\_H}\gt \frac{e\_K}{v\_K}$ for all proper subgraphs $K$ with $v\_K\ge1$.
Does a strict 2-balanced graph must to be strict balanced? (May add conditions such that the density of the graph is not so small, bigger than 2 etc.)
|
https://mathoverflow.net/users/160959
|
Does a strict 2-balanced graph must to be strict balanced?
|
This is true. Indeed, provided $H$ has a connected component with at least two edges, then just being $2$-balanced, that is, knowing that $\frac{e\_H-1}{v\_H-2} \ge \frac{e\_K-1}{v\_K-2}$ for all $K \subset H$ is enough to imply that $H$ is strictly balanced. To prove this, suppose that $H$ is not strictly balanced and let $K \subset H$ be a subgraph with $\frac{e\_K}{v\_K} \ge\frac{e\_H}{v\_H}$. That is, $e\_K v\_H \geq e\_H v\_K$. Since $H$ is $2$-balanced, we have $\frac{e\_H-1}{v\_H-2} \ge \frac{e\_K-1}{v\_K-2}$. Multiplying out gives
$$e\_H v\_K - 2e\_H - v\_K + 2 \geq e\_K v\_H - 2e\_K - v\_H + 2.$$
Substituting in $e\_K v\_H \geq e\_H v\_K$ on the right-hand side, we get that
$$e\_H v\_K - 2e\_H - v\_K + 2 \geq e\_H v\_K - 2e\_K - v\_H + 2.$$
Cancelling the like terms gives $-2e\_H - v\_K \geq -2e\_K - v\_H$, which in turn implies that $2(e\_K - 1) - (v\_K - 2) \geq 2(e\_H - 1) - (v\_H - 2)$ and this can be rewritten as
$$\left(2\frac{e\_K-1}{v\_K-2} - 1\right)(v\_K - 2) \geq \left(2\frac{e\_H-1}{v\_H-2} - 1\right)(v\_H - 2).$$
However, this is a contradiction, since $2\frac{e\_H-1}{v\_H-2} - 1 \geq 2\frac{e\_K-1}{v\_K-2} - 1$ by assumption, $\frac{e\_H-1}{v\_H-2} > 1/2$ for all $H$ with a connected component with at least $2$ edges and $v\_H - 2 > v\_K - 2$. The remaining case, where $H$ is a matching is not strictly $2$-balanced unless $H$ is a single edge, so the required result also holds in this case.
|
3
|
https://mathoverflow.net/users/66275
|
410750
| 168,042 |
https://mathoverflow.net/questions/410703
|
5
|
In [these slides of a talk](http://www.giovannicuri.com/Talks/Slides_Kanazawa2010.pdf) Giovanni Curi shows that the generalized uniformity principle follows from Troesltra’s uniformity principle and from the subcountability of all sets, which are both claimed to be consistent with CZF. Subcountability’s consistency with CZF is not surprising in light of counterintuitive results like that subsets of finite sets aren’t necessarily finite, but it seems to have a different flavor.
What are the intuitions or motivations for subcountability?
What references prove that subcountability is consistent with CZF?
|
https://mathoverflow.net/users/312621
|
Subcountability
|
An intuition for ESC (every set is subcountable, i.e., a subquotient of the natural numbers) in a predicative framework is that everything is built up from below starting with natural numbers, so we may assume that every set can be represented as a set of codes (natural numbers) quotiented out by an equivalence relation (denoting equality of whatever the codes represent).
For the consistency of CZF + ESC (indeed, CZF + REA + ESC), see Michael Rathjen's *Choice principles in constructive and classical set theories*, Thm. 8.3:
[http://www1.maths.leeds.ac.uk/~rathjen/acend.pdf](http://www1.maths.leeds.ac.uk/%7Erathjen/acend.pdf)
|
6
|
https://mathoverflow.net/users/2004
|
410755
| 168,045 |
https://mathoverflow.net/questions/410069
|
5
|
Consider the stochastic differential equation as follows:
$$X\_t=X\_0+t+\int\_0^t\frac{dW\_s}{1+m(s)},\quad \forall t\ge 0,~~~~~~~~~~~~~~~(\ast)$$
where $X\_0>0$ is square integrable and $m(t)=\mathbb P[\inf\_{0\le s\le t}X\_s>0]$ for all $t\ge 0$. Can we prove that $(\ast)$ admits at most one solution $(X,m)$ (assuming its existence)?
Any answer, comments or references are highly appreciated.
|
https://mathoverflow.net/users/261243
|
Uniqueness of the solution to some SDE
|
$\newcommand{\F}{\mathcal{F}}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$According to a comment by the OP, $X\_0$ and $(W\_t)$ are independent. So, without loss of generality (wlog), $X\_0$ is a real number $x\_0>0$.
Let $\F$ denote the set of all nonincreasing functions from $[0,\infty)$ to $[0,1]$. Define the (nonlinear) operator $F$ from $\F$ to $\F$ as follows: for each $f\in\F$ and each real $t\ge0$,
\begin{equation\*}
F(f)(t):=P(\inf\_{s\in[0,t]}X^f\_s>0), \tag{1}
\end{equation\*}
where
\begin{equation\*}
X^f\_t:=x\_0+t+\int\_0^t\frac{dW\_s}{1+f(s)}. \tag{2}
\end{equation\*}
We have to show that the equation
\begin{equation\*}
F(f)=f \tag{3}
\end{equation\*}
has at most one solution $f$ in $\F$.
Before doing this, let us prove, for completeness, the existence of a solution of (3). Here we just detail the argument in the comment by user GJC20 on this [previous answer](https://mathoverflow.net/a/410689/36721).
Let $f\_0:=0$ and $f\_n:=F(f\_{n-1})$ for all natural $n$. Then $f\_1=F(f\_0)\ge0=f\_0$ and hence, by that previous answer, $f\_n\ge f\_{n-1}$ for all natural $n$. So, the uniformly bounded sequence $(f\_n)$ converges (as $n\to\infty$) pointwise to some $\bar f\in\F$, which implies $f\_{n+1}=F(f\_n)\to F(\bar f)$ pointwise. So, $F(\bar f)=\bar f$, that is, $\bar f$ is a solution of (3).
To prove that (3) has at most one solution in $\F$, suppose that functions $f\_1$ and $f\_2$ in $\F$ are solutions of (3). Let
\begin{equation\*}
t\_0:=\sup\{t\in[0,\infty)\colon f\_1(s)=f\_2(s)\ \forall s\in[0,t)\}. \tag{4}
\end{equation\*}
Then wlog $t\_0<\infty$. Take any $h\in(0,1)$. To obtain a contradiction, suppose that
\begin{equation\*}
\ep:=\|f\_1-f\_2\|>0, \tag{5}
\end{equation\*}
where $\|g\|:=\sup\{|g(t)|\colon t\in[0,t\_0+h]\}$.
The crucial point is to use the same kind of time change as in the mentioned previous answer: The process
\begin{equation\*}
\text{$(X^{f\_i}\_t)$ equals the process $(x\_0+t+W\_{\tau\_i(t)})$ in distribution,} \tag{\*}
\end{equation\*}
where
\begin{equation\*}
\tau\_i(t):=\int\_0^t\frac{ds}{(1+f\_i(s))^2}; \tag{6}
\end{equation\*}
here and in what follows, $i\in\{1,2\}$.
Note that $\frac1{(1+z)^2}$ is $2$-Lipschitz in $z\ge0$. Therefore and in view of (6), (4), and (5),
\begin{equation\*}
|\tau\_1(t)-\tau\_2(t)|\le\int\_{t\_0}^{\max(t\_0,t)} ds\,\Big|\frac1{(1+f\_1(s))^2}-\frac1{(1+f\_2(s))^2}\Big|
\le 2h\ep \tag{7}
\end{equation\*}
for all $t\in[0,t\_0+h]$.
Since $\frac14\le\frac1{(1+f\_i)^2}\le1$, the functions $\tau\_i$ are Lipschitz-continuous and strictly increasing on $[0,\infty)$ from $\tau\_i(0)=0$ to $\tau\_i(\infty-)=\infty$, and the inverse functions $\tau\_i^{-1}$ are defined, strictly increasing, and $4$-Lipschitz on $[0,\infty)$. It follows that for all $u\in[0,\tau\_2(t\_0+h)]$
\begin{equation\*}
|\tau\_1^{-1}(u)-\tau\_1^{-1}(u)|\le4\sup\_{t\in[0,t\_0+h]}|\tau\_1(t)-\tau\_2(t)|\le8h\ep, \tag{8}
\end{equation\*}
in view of (7).
By (\*), for all real $t\ge0$,
\begin{equation\*}
F(f\_1)(t)-F(f\_2)(t)=D\_1(t)+D\_2(t), \tag{9}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
D\_1(t)&:=P(\inf\_{u\in[0,\tau\_1(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\
&-P(\inf\_{u\in[0,\tau\_2(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0)
\end{aligned}
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
D\_2(t)&:=P(\inf\_{u\in[0,\tau\_2(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\
&-P(\inf\_{u\in[0,\tau\_2(t)]}(x\_0+\tau\_2^{-1}(u)+W\_u)>0).
\end{aligned}
\end{equation\*}
Using e.g. (as an overkill) [Lemma 8, p. 407](https://epubs.siam.org/doi/10.1137/1115047) (or its Russian original, [Lemma 8, p. 423](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=tvp&paperid=1854&option_lang=eng)) and (8), we get
\begin{equation\*}
\|D\_2\|\le C\Big(1+\frac1{\sqrt{t\_0+h}}\Big)h\ep\le2C\sqrt h\,\ep, \tag{10}
\end{equation\*}
where $C$ is some universal positive real constant.
Let $\tau\_{\min}(t):=\min(\tau\_1(t),\tau\_2(t))$ and $\tau\_{\max}(t):=\max(\tau\_1(t),\tau\_2(t))$.
Take any $t\in[0,t\_0+h]$ and then let $u\_1:=\tau\_{\min}(t)$, $\de:=\tau\_{\max}(t)-\tau\_{\min}(t)$, $x\_1:=x\_0+\tau\_1^{-1}(u\_1)$, and $G:=1-\Phi$, where $\Phi$ is the standard normal cdf. Let $G\big(\frac{x\_1}{\sqrt{u\_1}}\big):=0$ if $u\_1=0$.
Then
\begin{equation\*}
\begin{aligned}
&|D\_1(t)| \\
&=P(\inf\_{u\in[0,\tau\_{\min}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\
&-P(\inf\_{u\in[0,\tau\_{\max}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\
&=P(\inf\_{u\in[0,\tau\_{\min}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0, \\
&\inf\_{u\in(\tau\_{\min}(t),\tau\_{\max}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)\le0) \\
&\le P(\inf\_{u\in[0,u\_1]}W\_u>-x\_1,\inf\_{u\in(u\_1,u\_1+\de]}W\_u\le-x\_1) \\
&=P(\inf\_{u\in[0,u\_1+\de]}W\_u\le-x\_1)
-P(\inf\_{u\in[0,u\_1]}W\_u\le-x\_1) \\
&=2G\Big(\frac{x\_1}{\sqrt{u\_1+\de}}\Big)-2G\Big(\frac{x\_1}{\sqrt{u\_1}}\Big) \\
&\le\frac\de{x\_1^2}\le\frac\de{x\_0^2}\le\frac{2h\ep}{x\_0^2}.
\end{aligned}
\tag{11}
\end{equation\*}
The first inequality in (11) follows because the function $\tau\_1^{-1}$ is increasing and $x\_0+\tau\_1^{-1}(u\_1)=x\_1$.
The fourth, last equality in (11) follows by the reflection principle.
The second inequality there follows because $\frac{\partial}{\partial u}G\big(\frac x{\sqrt u}\big)\le\frac1{2x^2}$ all real $x>0$ and $u>0$.
The last inequality in (11) follows by (7), since $\de=\tau\_{\max}(t)-\tau\_{\min}(t)$.
Collecting (5), (9), (11), and (10), for any real $h\in(0,1)$ such that $\frac{2h}{x\_0^2}+2C\sqrt h<1$, we have
\begin{equation\*}
\ep=\|f\_1-f\_2\|=\|F(f\_1)-F(f\_2)\|\le\|D\_1\|+\|D\_2\|\le\Big(\frac{2h}{x\_0^2}+2C\sqrt h\Big)\ep<\ep,
\end{equation\*}
which is the mentioned desired contradiction with (5).
So, $\ep=0$, which means that $f\_1(s)=f\_2(s)\ \forall s\in[0,t\_0+h)$, which contradicts (4). This final contradiction shows that $t\_0=\infty$ in (4), and we are done.
|
4
|
https://mathoverflow.net/users/36721
|
410768
| 168,050 |
https://mathoverflow.net/questions/410767
|
1
|
Given a Hilbert space (separable) $\mathcal{H}$ with an orthonormal basis $\{e\_i\}\_{i=1}^{\infty}$, define an operator $T$ with domain $\mathcal{D}(T)$ equal to the span of $\{e\_i\}$ by $Te\_i:=\lambda\_ie\_i$, for some $λ\_i∈R$.
**My question:** Is $T$ necessarily self-adjoint (i.e. $T=T^\*$)? It is easy to see that $T$ is a symmetric operator and therefore $T \subset T^\*$. I understand that I can make use of the definition of adjoint $T^\*$ and $T$ to show that $T$ and $T^\*$ agree on all basis vectors and therefore any finite linear combinations of them? But, how can one show that that they agree on the closure of the span? In other words, how do I show (if it is possible in the first place) that $T^\* \subset T$?
Any help will be truly appreciated. Thanks.
|
https://mathoverflow.net/users/nan
|
A question on the self-adjointness of an operator
|
No. For example, if the $\lambda\_i$ are bounded, we see that $T$ can be extended to an operator $S$ defined everywhere. Then $\text{dom}(T^\*)\supseteq \text{dom}(S^\*) = \mathcal H$, so $T^\*$ is defined everywhere as well.
So can we find the domain of $T^\*$ in the general case? Recall that the domain of $T^\*$ consists of all $v = \sum\_iv\_ie\_i\in\mathcal H$ for which there is $w\in\mathcal H$ such that for all vectors $u\in\mathcal H$ we have $\langle w,u\rangle = \langle v,Tu\rangle$. Applying this on $u=e\_i$ gives $\langle w, e\_i \rangle = \lambda\_iv\_i$, so we get $w = \sum\_i\lambda\_iv\_i$.
Conversely, if $\sum\_i\lambda\_iv\_i \in\mathcal H$ it is easy to check that $w = \sum\_i\lambda\_iv\_i$ satisfies the above condition.
So $v\in\text{dom}(T^\*)$ if and only if $\sum\_i|\lambda\_i|^2|v\_i|^2 < \infty$.
There are always such vectors $v$ that are not in the linear span of the basis vectors, for example we can take $|v\_i| = \min(\frac1{i^2},\frac1{i^2|\lambda\_i|^2})$. So $T$ is never self-adjoint.
|
3
|
https://mathoverflow.net/users/470870
|
410770
| 168,051 |
https://mathoverflow.net/questions/410734
|
1
|
The Chebyshev polynomials $(T\_k)\_{k \in \mathbb{N}\_0}$ are defined recursively by
$$
T\_0(x)=1 , \ \ T\_1(x)=x, \ \ T\_{k+1}(x)=2x\,T\_k(x)-T\_{k-1}(x) \ .
$$
With this one can find the explicit formulas
\begin{align}
& T\_{2l}(x) = l \sum\limits\_{j=0}^{l} (-1)^{l-j} \frac{(l+j-1)!}{(l-j)!(2j)!} \, (2x)^{2j}\\
& T\_{2l+1}(x) = \frac{2l+1}{2} \sum\limits\_{j=0}^l (-1)^{l-j} \frac{(l+j)!}{(l-j)!(2j+1)!} \, (2x)^{2j+1}
\end{align}
for all $l \in \mathbb{N}$. However I needed to normalize these polynomials into
\begin{align\*}
& \widetilde{T}\_{2l}(x) = l \sum\limits\_{j=0}^{l} (-1)^{l-j} \frac{(l+j-1)!}{(l-j)!(2j)!} \left( x^{2j} - \frac{1}{j+1} {2j \choose j} \right)\\
& \widetilde{T}\_{2l+1}(x) = \frac{2l+1}{2} \sum\limits\_{j=0}^l (-1)^{l-j} \frac{(l+j)!}{(l-j)!(2j+1)!} \, x^{2j+1}
\end{align\*}
for $l \in \mathbb{N}$ and $\widetilde{T}\_0(x)=0, \, \widetilde{T}\_1(x) = \frac{x}{2}$. I'm pretty sure any link to the recursion formular of the (non-normalized) Chebyshev polynomials is gone. Can anyone help me find out if there is a new recursion formula for $\widetilde{T}\_{k+1}$, depending on $\widetilde{T}\_0,...,\widetilde{T}\_{k}$? I think it might have some interesting applications for a certain graph theoretical setting.
Any help is much appreciated.
|
https://mathoverflow.net/users/409412
|
Recursive formula from given explicit formula for normalized Chebyshev polynomials
|
If I read your formulas correctly you define $\tilde T\_{2l}(x)=T\_{2l}(x/2)-C\_l$ and $\tilde T\_{2l+1}(x)=T\_{2l+1}(x/2)$, where
$$C\_l=l\sum\_{j=0}^l (-1)^{l-j}\frac{(l+j-1)!}{(l-j)!(2j)!}\frac 1{j+1}\binom{2j}j.$$
This can be rewritten
$$C\_l=(-1)^l l!\sum\_{j=0}^l\frac{(-l)\_j(l)\_j}{j!(2)\_j},$$
where $(a)\_j=a(a+1)\dotsm(a+j-1)$. By the Chu-Vandermonde summation formula,
$$C\_l=(-1)^ll!\frac{(2-l)\_l}{(2)\_l}=\begin{cases}1, & l=0,\\
-1/2, & l=1,\\ 0, & l\geq 2.\end{cases}$$
So your polynomials are given by
$$\tilde T\_k(x)=T\_k(x/2)-\delta\_{0,k}-\frac 12\delta\_{2,k}.$$
They satisfy the recursion
$$\tilde T\_{k+1}(x)=x\tilde T\_k(x)-\tilde T\_{k-1}(x),\qquad k\geq 4.$$
|
1
|
https://mathoverflow.net/users/10846
|
410784
| 168,054 |
https://mathoverflow.net/questions/410550
|
1
|
Consider $P(X,Y)$ discrete and $Z = f(Y)$ with $f$ deterministic. The function $f$ identifies a partition of the elements of the alphabet $\mathcal{Y}$ of $Y$. Each outcome $z \in \mathcal{Z}$ is a subset $z \subseteq \mathcal{Y}$. Let $Z' = f'(Y)$ be identical to $Z$ except for two elements $z\_1', z\_2' \in \mathcal{Z}'$ that are derived from $z\_1, z\_2 \in Z$ by "moving" one element $\bar{y} \in z\_1$ from $z\_1$ to $z\_2$. This means that $z\_1' = z\_1 \setminus \{\bar{y}\}$ and $f'(\bar{y}) = z\_2' = z\_2 \cup \{\bar{y}\}$. Does the following relationship hold?
$$
P(X=x, Y=y \mid Z=z\_1) = P(X=x, Y=y \mid Z'=z\_1') \cdot \frac{P(Z=z\_1)}{P(Z'=z\_1')}
\quad \text{for} \quad
y \ne \bar{y}
$$
If correct, does it also imply $P(X=x, Y=y, Z=z\_1) = P(X=x, Y=y, Z=z\_1')$ for $y \ne \bar{y}$? If not, please, provide a counterexample.
|
https://mathoverflow.net/users/101100
|
Identity for special case of Markov chain
|
$\newcommand{\Y}{\mathcal{Y}}\newcommand{\ZZ}{\mathcal{Z}}$It appears that for all $u\in\Y$ and all $z\in\ZZ$ we have
\begin{equation\*}
f(u)=z\iff u\in z \tag{1}
\end{equation\*}
and
\begin{equation\*}
f'(u)=\begin{cases}
z\_1'&\text{ if }u\in z\_1'=z\_1\setminus\{\bar y\}, \\
z\_2'&\text{ if }u\in z\_2'\cup\{\bar y\}, \\
f(u)&\text{ if }u\notin z\_1'\cup z\_2'=z\_1\cup z\_2.
\end{cases}
\tag{2}
\end{equation\*}
Then the identity
\begin{equation\*}
P(X=x,Y=y|Z=z\_1)=P(X=x,Y=y|Z'=z\_1')\frac{P(Z=z\_1)}{P(Z'=z\_1')} \tag{3}
\end{equation\*}
does not hold in general, even if $y\ne\bar y$. Indeed (assuming $P(Z=z\_1)\ne0$ and $P(Z'=z\_1')\ne0$), we can rewrite (3) as
\begin{equation\*}
P(X=x,Y=y,Z=z\_1)=P(X=x,Y=y,Z'=z\_1')\frac{P(Z=z\_1)^2}{P(Z'=z\_1')^2}. \tag{5}
\end{equation\*}
However,
\begin{equation\*}
P(Z=z\_1)=P(f(Y)=z\_1)=P(Y\in z\_1)=P(Y\in z\_1'\cup\{\bar y\})
=P(Y\in z\_1')+P(Y=\bar y)
\end{equation\*}
and
\begin{equation\*}
P(Z'=z\_1')=P(f'(Y)=z\_1')=P(Y\in z\_1'),
\end{equation\*}
so that
\begin{equation\*}
\text{$P(Z'=z\_1')=P(Z=z\_1)$ iff $P(Y=\bar y)=0$. }\tag{6}
\end{equation\*}
Also,
\begin{multline\*}
P(X=x,Y=y,Z=z\_1)=P(X=x,Y=y,f(y)=z\_1) \\
=P(X=x,Y=y)1(f(y)=z\_1)=P(X=x,Y=y)\,1(y\in z\_1)
\end{multline\*}
and
\begin{multline\*}
P(X=x,Y=y,Z'=z\_1')=P(X=x,Y=y,f'(y)=z\_1') \\
=P(X=x,Y=y)\,1(f'(y)=z\_1')=P(X=x,Y=y)\,1(y\in z\_1'),
\end{multline\*}
so that
\begin{equation\*}
P(X=x,Y=y,Z=z\_1)=P(X=x,Y=y,Z'=z\_1') \tag{7}
\end{equation\*}
if $y\ne\bar y$.
Thus, (5) does not hold for all $y\ne\bar y$ unless $P(Y=\bar y)=0$, and hence (3) does not hold for all $y\ne\bar y$ unless $P(Y=\bar y)=0$.
On the other hand, the identity
\begin{equation\*}
P(X=x,Y=y|Z=z\_1)=P(X=x,Y=y|Z'=z\_1')\frac{P(Z'=z\_1')}{P(Z=z\_1)},
\end{equation\*}
which is equivalent to (7) (which also was in question), will always hold for all $y\ne\bar y$.
|
1
|
https://mathoverflow.net/users/36721
|
410785
| 168,055 |
https://mathoverflow.net/questions/410782
|
1
|
Let $\mathcal{G}$ be a simple (no self-edges) undirected graph with $N$ vertices, and denote $\mathbf{A}$ its adjacency matrix: $A\_{ij}=1$ if there exists an edge between vertex $i$ and vertex $j$.
$\left[\mathbf{A}^k\right]\_{ab}$ will represents the number of walks of length $k$ between vertices $a$ and $b$, in other words, the number of ways to go from $a$ to $b$ in $k$ steps.
Question
--------
Is there a simple way to count the number of walks of length $k$ starting from $a$ that passes via vertex $b$? (the step where it reaches $b$ is irrelevant).
A related question would be, given a random walk starting at $a$ what is the probability that it passes through $b$ at least once in $k$ steps?
I am afraid that these are hard questions to answer if we do not add details about $\mathcal{G}$. If this is the case, I would already find it interesting to know the answer of these questions in the case of $r-$regular graphs: All vertices possess $r$ edges.
|
https://mathoverflow.net/users/420641
|
Number of walks on a graph passing through a specific vertex
|
Let matrix $B$ be obtained from $A$ by removing the row and column corresponding to $b$. Then the number of walks of length $k$ starting from $a$ and passing vertex $b$ is given by
$$\mathrm{rowsum}\_a(A^k) - \mathrm{rowsum}\_a(B^k),$$
where $\mathrm{rowsum}\_a$ is the sum of the row corresponding to $a$.
|
1
|
https://mathoverflow.net/users/7076
|
410786
| 168,056 |
https://mathoverflow.net/questions/410661
|
54
|
Let $G$ be a finite group. Let $r\_2\colon G \to \mathbb{N}$ be the square-root counting function, assigning to each $g\in G$ the number of $x\in G$ with $x^2=g$. Perhaps surprisingly, $r\_2$ does not necessarily attain its maximum at the identity for general groups, see [Square roots of elements in a finite group and representation theory](https://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory).
I'm interested in whether $r\_2(g)$ can attain a value above $0.999|G|$ for some non-identity element $g\in G$.
**Update:** Thanks to everybody who participated in the discussion. The lemma proved here influenced greatly the statement of Theorem 4.2 in <https://arxiv.org/pdf/2204.09666.pdf> . Proposition 3.12 in this same paper is essentially the answer posted by GH from MO.
|
https://mathoverflow.net/users/160715
|
How many square roots can a non-identity element in a group have?
|
Here is a streamlined and simplified version of the posts by Saúl Rodríguez Martín and Emil Jeřábek.
**Theorem.** Assume that $G$ is a finite group, and $r\_2(g)>(3/4)|G|$ holds for some $g\in G$. Then $G$ is an elementary abelian $2$-group, and $g$ is the identity element.
*Proof.* Fix any element $y\in G$, and consider the sets
$$S=\{x\in G: x^2=g\},\qquad T=\{x\in S:xy\in S\}.$$
By the union bound,
$$|G\setminus T|\,\leq\, 2|G\setminus S|<|G|/2,$$
hence $|T|>|G|/2$. For any $x\in T$, we have $(xy)^2=x^2$,
which implies that
$$xyx^{-1}=(xy)x^{-1}=(xy)^{-1}x=y^{-1}.$$
So $\{x\in G:xyx^{-1}=y^{-1}\}$ contains more than half of the elements of $G$, whence it contains all elements of $G$. In particular, $y=y^{-1}$, which shows that $G$ is an elementary abelian $2$-group. Moreover, $g$ is the identity element, since the identity element is the only square in $G$.
|
24
|
https://mathoverflow.net/users/11919
|
410789
| 168,057 |
https://mathoverflow.net/questions/410793
|
7
|
Here is a question that occurred to me while learning about neural networks. For $t\in\mathbb{R}$ put $t\_+=\max(0,t)$, so $t\_+=t$ if $t\geq 0$ and $t\_+=0$ if $t\leq 0$. (This is RELU=rectified linear unit in neural network language.) Note that $t=t\_+-(-t)\_+$ and $|t|=t\_++(-t)\_+$.
For the maximum of two variables we have
\begin{align\*}
\max(x,y) &= x + (y-x)\_+ \\
&= \left(x+y+(x-y)\_++(y-x)\_+\right) /2 \\
&= \left((x+y)\_+-(-x-y)\_++(x-y)\_++(y-x)\_+\right)/2
\end{align\*}
Is there a similar expression for $\max(x,y,z)$ as a linear combination of terms $\phi(x,y,z)\_+$ with $\phi$ linear? I suspect not, but I have not found a proof. More generally, is there a nice characterisation of the functions of several variables that do admit such a representation?
|
https://mathoverflow.net/users/10366
|
RELU representation of $\max(x,y,z)$
|
There's indeed no such way to write $\max(x,y,z)$.
**Lemma.** Consider on $\mathbf{R}^n$ a function $f(x)=\sum\_{i\in I}^m t\_iL\_i(x)\_+$ where the $L\_i$ are nonzero, pairwise non-positively-collinear linear forms, $t\_i$ are nonzero scalars. Suppose that
for some $i$, $f$ is locally linear at $x\_0$ with $x\_0\in\mathrm{Ker}(L\_i)$. Then there exists $j$ (unique) such that $t\_iL\_i-t\_jL\_j=0$.
[So $L\_j=\frac{t\_i}{t\_j}L\_i$; since these are not positively collinear, necessarily $t\_i$ and $t\_j$ have distinct signs and in particular $j\neq i$. This also proves uniqueness. Thus the contribution $t\_iL\_i(x)\_++t\_jL\_j(x)\_+$ is linear, namely equal to $t\_iL\_i(x)$.]
*Proof.* Induction on $n$. For $n=1$, necessarily $m\le 2$. We can suppose $i=1$. We can rescale to assume $L\_1(x)=x$. So $\mathrm{Ker}(L\_1)=\{0\}$. So $f(x)=x\_++t(ax)\_+$ with $a\le 0$, for all $x$. The only possibility to get $f$ smooth at $0$ is to have $a=-1$.
For $n=2$, if no $L\_j$, $j\neq i$, is collinear to $L\_i$, then all such $L\_j$ are smooth at $x\_0$, and we get a contradiction. So for some $j$, $L\_j$ is collinear to $L\_i$, and we see that the only possibility for scalars is $t\_iL\_i=t\_jL\_j$.
For $n\ge 3$. Fix $i$ as in the lemma; up to rescale, we can suppose $t\_i=1$. Let $P$ be any hyperplane passing through $x\_0$. By induction, there exist $j\_P$ such that in restriction to $P$ we have $L\_i=t\_{j\_P}L\_{j\_P}$.
Suppose by contradiction that $L\_i-t\_jL\_j$ is nonzero for all $j\neq i$. Then there exists $x$ such that $L\_i(x)-t\_jL\_j(x)\neq 0$ for all $j$. Let $P$ be a hyperplane passing through $x\_0$ and $x$. Then we get a contradiction.
**Corollary.** There is no function of the form $x\mapsto\sum\_j t\_jL\_j(x)\_+$ ($L\_j$ linear) on $\mathbf{R}^3$ that equals $f:(x,y,z)\mapsto \max(x,y,z)$.
*Proof.* Write it as $x\mapsto\sum\_k L\_k(x)\_++a\_kL\_k(-x)\_+$, where the $L\_k$ are nonzero and pairwise non-collinear. When $\mathrm{Ker}(L\_k)$ is not one of the hyperplanes $x\_i=x\_j$, there is an element in this kernel at which $f$ is locally linear (namely any point with three distinct coordinates). Then the lemma ensures that $a\_k=-1$, so that the contribution $L\_k(x)\_+-L\_k(-x)\_+$ is linear. So we can rewrite the given map as
$$f(x)(=\max(x\_1,x\_2,x\_3))=L(x)+ \sum\_{1\le i\neq j\le 3}a\_{ij}(x\_i-x\_j)\_+$$
with $L$ linear. Write $L(x)=\sum\_{i=1}^3a\_ix\_i$.
Since $f$ is symmetric, we have $f(x\_1,x\_2,x\_3)=\frac16\sum\_{\sigma\in\mathfrak{S}\_3}f(x\_{\sigma(1)},x\_{\sigma(2)},x\_{\sigma(3)})$. Hence we can suppose that the $a\_i$ and $a\_{ij}$ are symmetric too, which means that $a\_i=b$ for some $b$ and $a\_{ij}=c$ for some $c$. Thus
$$f(x)=b(x\_1+x\_2+x\_3)+c\left(\sum\_{1\le i\neq j\le 3}(x\_i-x\_j)\_+\right)$$
for all $x$.
Evaluation at $x=(1,1,1)$ yields $b=1/3$. Evaluation at $x=(0,0,1)$ yields $1=b+2c$, so $c=1/3$, and evaluation at $x=(0,0,-1)$ yields $0=-b+4c$, and this is a contradiction.
|
4
|
https://mathoverflow.net/users/14094
|
410805
| 168,060 |
https://mathoverflow.net/questions/410798
|
48
|
Briefly, I was wondering if someone can suggest an angle for introducing the gist of Galois groups of polynomials to (advanced) high school students who are already familiar with polynomials (factorisation via Horner, polynomial division, discriminant and Vieta's formulas for quadratic equations).
I am struggling to find a coherent approach that makes use of their current understanding, e.g., whether to give an intro on group theory, or to describe permutation operations on roots. This is all intended for a short course introduction and not an extended program. Any advice would be quite helpful.
|
https://mathoverflow.net/users/115841
|
Ideas for introducing Galois theory to advanced high school students
|
I have now twice taught Galois theory to advanced high school students at [PROMYS](https://promys.org/). This is a six week course, meeting four times a week, for students who already are comfortable with proofs and, in particular, have seen basic number theory. The second time, I taught the course as an IBL course, and you can read my worksheets [here](http://www.math.lsa.umich.edu/%7Espeyer/PROMYSGaloisTheory2021/). Here is what I have done, and some thoughts about ways to do less.
---
**Showing that there is no universal quintic formula:** Both times, I started with a weak version of unsolvability of the quintic. This would make a very natural stopping point for a less ambitious course.
Consider the general degree $n$ polynomial
$$x^n + c\_{n-1} x^{n-1} + \cdots + c\_1 x + c\_0 = (x-r\_1) (x-r\_2) \cdots (x-r\_n).$$
Point out that $c\_1$, ..., $c\_n$ are symmetric polynomials in the roots $r\_1$, ..., $r\_n$. Exhibit the quadratic, cubic and (optionally) quartic formulas and point out that they compute they express the roots $r\_i$ in terms of the coefficients $c\_i$ **while staying entirely inside the polynomials**. For example, the quadratic formula is $r\_1 = \tfrac{-c\_1 + \sqrt{c\_1^2-4c\_2}}{2}$, and $c\_1^2-4c\_2 = (r\_1-r\_2)^2$, so you can take the square root without leaving the world of polynomials. (By prepared for discussions about what you mean by square root, since teachers have taught them that square roots are always positive; what you mean is any expression whose square is $c\_1^2 - 4 c\_2$.)
Announce that your goal is to show, for $n \geq 5$, that there is no expression for $r\_1$, ..., $r\_n$ in terms of $c\_1$, ..., $c\_n$, using the operators $+$, $-$, $\times$, $\div$, $\sqrt[n]{\ }$ such that every $n$-th root stays inside the symmetric rational functions. It is worth taking some time to get some buy in that students understand the goal and realize it is nontrivial and a reasonable approximation to what we might informally state as "there is no quintic formula".
With this as the goal, define the symmetric group $S\_n$ and note how it acts on formulas in $r\_1$, ..., $r\_n$. Define the sign homomorphism $S\_n \to \pm 1$ by the action of $S\_n$ on $\prod\_{i<j} (r\_i - r\_j)$ and define the alternating permutations $A\_n$ as the permutations with sign $1$. Show that, for $n=3$ or $4$, there are polynomials which define homomorphisms $A\_n \to \mathbb{C}^{\ast}$. Prove that, for $n \geq 5$, there are no homomorphisms $A\_n \to \mathbb{C}^{\ast}$.
Now, define $F$ to be the set of all rational functions invariant for $A\_n$. Clearly, $F$ is closed under $+$, $-$, $\times$, $\div$. Now the key Lemma: If $f \in F$ is nonzero, and $f = g^n$ for some $g$ in $\mathbb{C}(r\_1, \dots, r\_n)$, then $\sigma : \tfrac{\sigma(g)}{g}$ would be a group homomorphism $A\_n \to \mathbb{C}^{\ast}$. But we showed (for $n \geq 5$) that there are no such homomorphisms! So $g$ must also be in $F$. Thus, our operations can never get us outside $F$, and in particular (for $n \geq 5$) we cannot get to $r\_1$, $r\_2$, ..., $r\_n$. $\square$
I really like this approach because it introduces so many key concepts -- symmetries, characters, a set closed under the field operations, defining a subfield by its symmetries -- without ever needing to define a field or a group as an abstract object. The first time, I did the above argument in a week of lectures and then went back to point out the key abstractions of "field", "group" and "character" hiding in the proof. The second time, I did it in 3 weeks of IBL and introduced the group theory language explicitly as I went, but held back the definition of a field until we had completed the proof. [Worksheet 9](http://www.math.lsa.umich.edu/%7Espeyer/PROMYSGaloisTheory2021/9_PROMYSWorksheets2021.pdf) is the climax.
As a side note, I never needed to prove the fundamental theorem of symmetric functions, though I assigned it as homework, and it is well-motivated by these results. The proof only needs the easy containment $\mathbb{C}(c\_1, \ldots, c\_n) \subseteq \mathbb{C}(r\_1, \ldots, r\_n)^{S\_n}$, not the equality.
---
**Getting to abstract fields** If you want to do anything harder than this, I think you need to define fields.
Here, the fact that my students have already seen basic number theory is a huge advantage: They already know that, for $p$ a prime, every nonzero element of $\mathbb{Z}/p \mathbb{Z}$ is a unit, so they find it very quick to believe and prove the same thing about $k[x]/p(x) k[x]$ for $p(x)$ an irreducible polynomial.
There is a conceptual obstacle, though, which is convincing my students that they really can treat $\mathbb{Q}[\sqrt[3]{2}]$ as the same thing as $\mathbb{Q}[x]/(x^3-2) \mathbb{Q}[x]$. It takes them a long time to believe that, for example, knowing that $x^2+x+1$ is a unit in the ring $\mathbb{Q}[x]/(x^3-2) \mathbb{Q}[x]$ really proves that $\tfrac{1}{\sqrt[3]{2}^2+\sqrt[3]{2}+1}$ is in $\mathbb{Q}[\sqrt[3]{2}]$. I think it is worth spending time to make them sit with this discomfort until they resolve it.
There is a reasonable half way goal to aim for here -- that $\sqrt[3]{2}$ can't be computed with $+$, $-$, $\times$, $\div$, $\sqrt{\ }$. That has the advantage of using field extensions, and the multiplicativity of degree, but not splitting fields or Galois groups. I talked about that more [here](https://matheducators.stackexchange.com/questions/9784/impossibility-of-trisecting-the-angle-doubling-the-cube-and-alike-what-are-rea/9799#9799).
To my surprise, in summer 2021, I never actually wound up needing to prove that the degree of a field extension was well defined! I talked about bases and spanning sets, and showed that $1$, $x$, ..., $x^{\deg(f)-1}$ was a basis for $k[x]/f(x) k[x]$, but I never proved that two bases of a field had the same cardinality or that degree was multiplicative! This was quite a relief; in summer 2018, I took a week away from the main material to do a crash course on linear algebra, and I lost a lot of momentum there.
---
**If you are going for the full unsolvability of the quintic**
You need to introduce splitting fields and their automorphisms. At first, I found students swallow these with no hesitation, but they don't really realize what they've said yes to. When you tell them that there is an automorphism of $\mathbb{Q}[\sqrt[3]{2}, \omega]$ which maps $\sqrt[3]{2}$ to $\omega \sqrt[3]{2}$, the students who don't just say yes to everything will start to feel very concerned. I think this is probably the point where it pays off to have really gotten them used to the notion that computations with polynomials really do prove things about concrete subfields of $\mathbb{C}$.
The **key lemma**, from this perspective, is that if $f(x)$ is an irreducible polynomial in $F[x]$, and $K$ is a normal extension of $F$ in which $f$ has a root, then $f$ splits in $K$ and $\text{Aut}(K/F)$ acts transitively on the roots of $f$ in $K$. From this, deduce:
**Theorem** Suppose that $f(x)$ is an irreducible polynomial of degree $n \geq 5$ over $F\_0$, let $K$ be a field in which $f$ splits and suppose that $\text{Aut}(K/F\_0)$ acts on the roots of $f$ by $S\_n$. Then there is no tower of radical extensions $F\_0 \subset F\_1 \subset \cdots \subset F\_r$ in which $f(x)$ has a root.
It's worth taking the time to convince students that this really is a rigorous and powerful formalization of "you can't solve the quintic with radicals".
The key proof is to embed everything into a single splitting field $L$; let $G = \text{Aut}(L/F\_0)$. The details are on [Worksheet 17](http://www.math.lsa.umich.edu/%7Espeyer/PROMYSGaloisTheory2021/17_PROMYSWorksheets2021.pdf), but the key point is that, on the one hand, we have a surjection $G \to S\_n$ but, on the other hand, we have a sequence of subgroups $G \trianglerighteq G\_0 \trianglerighteq G\_1 \trianglerighteq \cdots \trianglerighteq G\_r = \{ e \}$ with each $G\_{i+1}$ the kernel of a character $G\_i \to \mathbb{C}^{\ast}$, and this is impossible. (The warm-up version that I start the course with is easier because $G$ is $S\_n$, so we are building the composition series in a concrete group we know, rather than a very abstract group where all we can say is that it surjects to $S\_n$.)
There are a surprising number of things I didn't need to prove! I never introduced the notion of separability; the result is correct as stated for a normal inseparable extension. I also never proved the fundamental theorem of Galois theory! This proof starts with a chain of subfields and uses it to construct a chain of subgroups, but we never need the reverse construction! In particular, if there were two different subgroups that stabilized the same subfield, it would not effect the proof in any way.
I also never needed the abstract notion of a quotient group! I always had a concrete homomorphism $G \to H$, and talked about its image and kernel; I never needed to show that every normal subgroup was the kernel of a homomorphism. I put this on the homework, but it was never used in class.
---
Once you get here, a final issue is to construct **an explicit example of a polynomial where the Galois group is $S\_n$**. The easy root is to take the polynomial $x^n + c\_{n-1} x^{n-1} + \cdots + c\_1 x + c\_0$ with coefficients in $\mathbb{C}(c\_1, \ldots, c\_n)$, since it is easy to see that $\text{Aut}{\big(} \mathbb{C}(r\_1, \ldots, r\_n)/\mathbb{C}(c\_1, \ldots, c\_n){\big)}$ is $S\_n$.
If you want to give an actual quintic with coefficients in $\mathbb{Q}$ whose Galois group is $S\_5$, there are various hacks to do this. What I did was to tell my students that, in 10 years, this would no longer be an issue: We will just write down a random polynomial $x^5 + c\_4 x^4 + \cdots + c\_0$ with integer coefficients and roots $(r\_1, \ldots, r\_5)$, compute the degree $120$ polynomial $g(x):= \prod\_{\sigma \in S\_5} {\big( }x-r\_{\sigma(1)} - 2 r\_{\sigma(2)} - \cdots - 5 r\_{\sigma(5)} {\big)}$, and check that the result is irreducible; this will prove that the polynomial has Galois group $S\_5$. The need for a clever proof is simply because modern computers can't\* compute and factor such large polynomials. I then did show them the clever proof that an irreducible quintic with two complex roots has Galois group $S\_5$, which was the last result of the course.
\* I might be wrong about this! I just attempted the case of a random integer monic quintic on my laptop. The first time, I didn't use enough working precision in computing the roots, and the degree $120$ polynomial didn't even even have real coefficients. But I went back and told Mathematica to use 100 digits for every floating point computation, and it got the polynomial in quite reasonable time, with every coefficient within $10^{-90}$ of an integer. This could make a genuine in class demo! Of course, I would need to know whether I am actually computing the degree $120$ polynomial correctly, but it seems unlikely that floating point errors would come out so near integers every time. I'm not sure whether I can trust Mathematica's integer polynomial factorization for such large polynomials, but I know that Mathematica's routines for characteristic $p$ factorization are very good, and factoring my polynomial modulo the first $10$ primes finds a a degree $100$ factor modulo $3$; this is already enough to prove irreducibility. The future is now!
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68
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https://mathoverflow.net/users/297
|
410811
| 168,062 |
https://mathoverflow.net/questions/410759
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2
|
After QR decomposition of a matrix, $M$, the columns of Q are orthonormal. Is it possible after obtaining *Q*, we recover unnormalized column vectors from $Q$? For example, the matrix M has the following $Q$ matrix.
$M=
\begin{bmatrix}1&-1&4\\1&4&-2\\1&4&2\\1&-1&0
\end{bmatrix}$
The $Q$ matrix is
$Q=\begin{bmatrix}0.5& -0.5& 0.5\\0.5& 0.5& -0.5\\0.5& 0.5& 0.5\\0.5&-0.5&-0.5 \end{bmatrix}$
This is for a certain application in spectroscopy. The process is as follows using Gram-Schmidt method.
The raw data is an experimental spectroscopy data. In general, we will have background signals, call them v1, v2, v3 as column vectors. Now measure the spectrum of the molecule of interest. This vector is m. Orthonormalize v1,v2 and v3 to obtain e1, e2 and e3. The magnitude of e1,e2,and e3 is 1. I am following the notation of Wikipedia's article on Gram-Schmidt [Wiki](https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process#The_Gram%E2%80%93Schmidt_process).
Then here comes the key step: After an orthonormal basis set of e1, e2, and e3 has been obtained, from the background spectra, measure the spectrum of molecule of interest. This vector is $m$.
Now orthogonalize $m$, with respect to e1,e2, and e3 to obtain another vector $u$. In other words, the dot product u.e1=u.e2=u.e3 = 0.
Since $u$ has not been normalized, i.e., it has a length given by sqrt($u.u$). The length of $u$ is of interest for measurement purposes.
I wanted to know if this is possible by QR method or not?
|
https://mathoverflow.net/users/142414
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Is it possible to obtain orthogonal (but not normalized) vectors after QR factorization?
|
It looks like you want $m - QQ^Tm$, is that correct?
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3
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https://mathoverflow.net/users/1898
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410813
| 168,063 |
https://mathoverflow.net/questions/410808
|
3
|
I was reading a problem list by Erdos ([doi](https://doi.org/10.1111/j.1749-6632.1989.tb16392.x)). On page 144 (which is the 12-th page of the pdf), a problem stuck out to me.
For positive integer $n$, let $h(n)$ be the smallest $k$ such that $[n] := \{1,\dots,n\}$ can be partitioned into $k$ parts $A\_1,\dots,A\_k$ so that $A\_i \cup A\_j$ lacks arithmetic progressions of length four for each $i,j \in [k]$. I was wondering if this quantity $h(n)$ has been studied, and if so what bounds are known.
**Observations:** A simple upper bound is the following. Let $c:[k] \to [n]$ be a coloring (which obviously corresponds to a partition of $[n]$ into $k$ parts). If there exists a monochromatic arithmetic progression of length 3, $P = \{n,n+d,n+2d\} \subset [1,2n/3]$, then we have $d<n/3$ implying $n+3d \in [n]$ meaning there exists an arithmetic progression of length four.
Hence we must have $W\_{h(n)}(3) > 2n/3$. I am not too familiar with the bounds for the multicolor van der Waerden numbers, but by Bloom and Sisak's density result this implies $h(n) \gg \log^{1+c}(n)$ for some $c>0$.
Also, I'm pretty sure a bound of $h(n)\ll n^{1-\epsilon}$ for some $\epsilon > 0$ should follow by appropriately modifying the work of Tomon and Pach on rainbow arithmetic progressions.
|
https://mathoverflow.net/users/130484
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What are bounds on this van der Waerden-esque problem?
|
The bound $h(n) \ll n^{2/3}$ is achievable by the following twisted cubic construction: Let $X=\mathbb{F}\_p^3$ ($p\geq5$) and $Y\_{ij}=\{(t,t^2+i,t^3+j): t \in \mathbb{F}\_p\}$. The sets $Y\_{ij}$ partition $X$ and $Y\_{ij} \cup Y\_{kl}$ never contains any 4-AP.
If there were a 4-AP, $\{a,b,c,d\}$, there must be two elements belonging to $Y\_{ij}$ and two belonging to $Y\_{kl}$, since none of $Y\_{ij}$ contains a 3-AP. We may assume $\{a,b\} \in Y\_{ij}$ and $\{c,d\} \in Y\_{kl}$. In this case, the vectors $a-b$ and $c-d$ would be parallel, but a twisted cubic does not contain any pair of secants that are parallel.
We can "project" $\mathbb{F}\_p^3$ on $\mathbb{Z}$ by sending $(a,b,c)$ (taking values in $0 ... p-1$) to $4ap^2+2bp+c$. In this projection, if a subset of $\mathbb{F}\_p^3$ does not contain a 4-AP, the image would not contain a 4-AP either. So we can project the $Y\_{ij}$s on $\mathbb Z$ without creating any 4-APs. By translating the projected $Y\_{ij}$s by $4xp^3+2yp^2+zp$ $(x,y,z \in \{0,1\})$ and renaming the variables, we can partition $0 ... 8p^3-1$ by $8p^2$ sets.
Thus the bound $h(n) \ll n^{2/3}$ is achievable.
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3
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https://mathoverflow.net/users/125498
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410815
| 168,065 |
https://mathoverflow.net/questions/410821
|
1
|
By the thick subcategory theorem, if $X, Y$ are finite $p$-local spectra of type $m,n$ respectively, then $Y$ can be built from $Y$ in a finite number of "steps" iff $n \geq m$. Here, a "step" can be given by taking a cofiber, a direct sum, or a retract of previously constructed spectra.
Moreover, by the nilpotence theorem, it's always possible to pass from height $n$ to height $n+1$ in *one* step: if $X$ has type $n$, then it admits a $v\_n$-self map, whose cofiber is of type $n+1$.
**Question 1:** Fix a prime $p$ and $n \in \mathbb N$. Does there exist a(n $\infty$-)category $J$, and an $\infty$-functor $F$ from $J$ to the category of type $n$ spectra, whose colimit $\varinjlim F$ exists and is a nonzero finite spectrum $Y$ of type $\geq n+2$?
**Question 2:** Same question, but we only require the nonzero finite spectrum $Y$ of type $\geq n+2$ to be a *retract* of $\varinjlim F$.
I think Question 2 is probably the better formulation. Note that by compactness, in (2) we may assume that $J$ is a finite ($\infty$-)category, and thus we may assume without loss of generality that $J = \Delta^{op}\_{\leq N}$ is a truncated simplex category.
|
https://mathoverflow.net/users/2362
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Can a finite, type $n+k$ spectrum be a (non-iterated) colimit of finite, type $n$ spectra for $k \geq 2$?
|
Yes, in a silly way : let $X,Y$ be nonzero, $X$ has type $n$ and $Y$ $n+2$.
Then $X\oplus Y$ has type $n$, and has $Y$ as a retract, so it solves Question 2 with $J= \*$.
But in fact, taking the cofiber of $id\_X\oplus 0$ on this spectrum yields $Y\oplus \Sigma Y$, which also has type $n+2$, so it solves Question 1 too.
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4
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https://mathoverflow.net/users/102343
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410826
| 168,071 |
https://mathoverflow.net/questions/410657
|
2
|
In this question, I follow the book "An invitation to quantum groups and duality" by Timmerman, p259.
Let $G$ be a locally compact group and $C$ be a $C^\*$-algebra. Assume an action
$$\alpha: G \to \operatorname{Aut}(C)$$
is given. Define a $\*$-homomorphism
$\delta\_0: C \to C\_b(G,C)$ by $\delta\_0(c)(g) = \alpha\_g(c)$. I try to understand the proof of theorem 9.2.4. In particular, I want to prove that
$\delta\_0(C)C\_0(G)$ is dense in $C\_0(G,C).$
Note that we can embed $C\_0(G)\hookrightarrow C\_0(G, M(C))$ via $f \mapsto f(-)1\_{M(C)}$ where $M(C)$ denotes the multiplier algebra of $C$. Then $\delta\_0(C)C\_0(G)$ is a multiplication in $C\_0(G,M(C))$ which returns an element of $C\_0(G,C)$ since $C$ is an ideal in $M(C)$.
Why is the density claim true? Timmerman writes:
>
> The equation $\delta\_0(\alpha\_{g^{-1}}(c))(g) = \alpha\_g(\alpha\_{g^{-1}}(c)) = c$ shows that for every $g \in G$ and every $c \in C$ there is $f \in \delta\_0(C)$ with $f(g) = c$. Now the density follows from a standard argument.
>
>
>
What is this 'standard argument'?
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https://mathoverflow.net/users/216007
|
Action of a group $G$ induces a coaction on $C_0(G)$
|
The key ideas here are to exploit compactness, and to use a [Partition of unity](https://en.wikipedia.org/wiki/Partition_of_unity). In fact, I think the most useful formulation of this, for locally compact spaces, can be found in Rudin, "Real and Complex Analysis", Theorem 2.13:
>
> Let $G$ be a locally compact space, $K\subseteq G$ compact, and let $K\subseteq U\_1\cup\cdots\cup U\_n$ be a finite open cover. There are continuous functions $f\_i:G\rightarrow [0,1]$ with compact support contained in $U\_i$, and with $\sum\_i f\_i(x) = 1$ for each $x\in K$.
>
>
>
We call the $(f\_i)$ a *partition of unity subordinate to $(U\_i)$*.
Let $X$ be the closed linear span $\delta\_0(C) C\_0(G)$, a $C\_0(G)$-submodule of $C\_0(G,C)$. What Timmermann shows is that for each $g\in G,c\in C$ there is some $f = f\_{g,c}\in X$ with $f(g)=c$. Here is a sketch of how to proceed:
* Let $K\subseteq G$ be compact and fix $c\in C$. For each $g\in K$, the function $f\_{g,c}$ is continuous. Thus, for $\epsilon>0$, there is an open cover $(U\_g)\_{g\in K}$ of $K$ with $h\in U\_g \implies \|f\_{g,c}(h) - c\|<\epsilon$.
* By compactness, find finitely many points $(g\_i)$ with $(U\_{g\_i})$ a cover of $K$.
* Pick a partition of unity subordinate to $(U\_{g\_i})$, say $(g\_i)$.
* Set $f = \sum g\_i f\_{g\_i,c}$
* For $h\in K$, if $h\in U\_{g\_i}$ then $\|f\_{g\_i,c}(h)-c\|<\epsilon$, while if $h\not\in U\_{g\_i}$ then $g\_i(h)=0$. Conclude
$$ f(h) = \sum\_i g\_i(h) f\_{g\_i,c}(h) = \sum\_{h\in U\_i} g\_i(h) f\_{g\_i,c}(h)
\approx\_\epsilon c, $$
using that $\sum\_i g\_i(h)=1$.
Identify $C\_0(G) \odot C$ with a subspace of $C\_0(G,C)$. We have shown that if $g\in C\_0(G)$ has compact support and $c\in C$ then $g\otimes c\in X$. Taking the closure, conclude that $C\_0(G)\otimes C\subseteq X$. So now take your favourite proof that $C\_0(G)\otimes C \cong C\_0(G,C)$ to conclude that $X = C\_0(G,C)$. (Your "favourite proof" will also use a partition of unity argument).
---
Actually, I think there is a much nicer, and well-explained, but still elementary argument given by Soltan in "Examples of non-compact quantum group actions", [doi.org/10.1016/j.jmaa.2010.06.045](https://doi.org/10.1016/j.jmaa.2010.06.045), also [arXiv:1001.0520](https://arxiv.org/abs/1001.0520). See Proposition 2.1. In fact, in Section 2 of this paper is developed some general theory, and Proposition 2.3 gives a somewhat more "conceptual" (and easier) proof of the fact you are after, using some general theory.
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2
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https://mathoverflow.net/users/406
|
410827
| 168,072 |
https://mathoverflow.net/questions/49035
|
4
|
$\DeclareMathOperator\lcm{lcm}$This is a rather severe revision of [a question I asked recently](https://mathoverflow.net/questions/48888/). We know over the integers that $\gcd(a^2,b^2)=\gcd(a,b)^2$. We might prove this via unique factorization. In building the theory of prime factorization we use the fact that $\gcd(a,b)$ exist. This fact is sometimes proved with the slick (to a beginner) move that there must be a minimal element in $\lbrace d\mid d>0 \text{ and }d=as+bt\rbrace$ and it must be a common divisor and indeed the greatest such. This also shows that there is a linear combination $\gcd(a,b)=as+bt$. I'm sure standard terminology exists but allow me to call any such pair $(s,t)$ *Bézout cofactors* for the pair $(a,b)$. Of course with the extra condition $\lvert s\rvert<b$ we also have $\lvert \rvert|<a$ and two cofactor pairs, one with $s<0<t$ and one with $t<0<s$. The same things are true in more general settings such as polynomials. The slick argument above gives no indication how to find $s$, $t$ but we do know the extended Euclidean algorithm. To hone in on my question I'll stick to the case that $\gcd(a,b)=1$: Given that $a,b$ are integers with cofactors $s$, $t$ such that $as+bt=1$, there are also cofactors $s'$, $t'$ for $a^2$, $b^2$. One proof would be that evidently $\gcd(a,b)=1$ so also $\gcd(a^2,b^2)=1$ (via a small bit of theory) and hence the slick argument gives that $s'$, $t'$ exist (and the Euclidean algorithm will only take about twice as long to find $s'$, $t'$ as it will for $s$, $t$). However it is easy to check that $a^2\cdot s^2(as+3bt)+b^2\cdot t^2(3as+bt)=1$. And this establishes the existence of $s'$, $t'$ constructively given only the fact that $a$, $b$, $s$, $t$ belong to some ring, pairwise commute, and satisfy $as+bt-1=0$. (An aside: I believe I can prove that no $s'$ and $t'$ cubic in $a,b,s,t$ make $a^2s'+b^2t'=1$.) My question has to do with similar translations to polynomial identities. I'll start with a specific instance based on $ab=\gcd(a,b)\lcm(a,b)$ and then attempt to state my general question.
>
> The following is true over the integers: if $\gcd(u,v)=1$ and $au=bv$ then there is a $w$ with $a=vw$. Is the following true as well: if $A,B,U,V,S,T$ are commuting variables and we are given the expressions $US+VT-1$ and $AU-BV$, is there an (explicit) expression $W=W(A,B,S,T,U,V)$ such that $A-VW$ is in the ideal of $\mathbb{Z}(A,B,U,V,S,T)$ generated by $AS+BT-1$ and $AU-BV$?
>
>
>
Note that $au=bv$ would be $\lcm(a,b)$, also $b=uw$ and $w=\gcd(a,b)$.
>
> Consider theorems of integer divisibility whose premises and conclusions can be written as multinomial equations ($d \mid a$ becomes $a-da'=0$ , $\gcd(a,b)=1$ becomes $as+bt-1=0$ etc.). Is there always (or when is there) a derivation of the conclusion purely from manipulation of $\mathbb{Z}$ multinomials?
>
>
>
This is partly idle curiosity, but I also find that sometimes an explicit constructive solution is very useful to improve results.
|
https://mathoverflow.net/users/8008
|
Explicit Bézout cofactors
|
The answer to the question posed by Aaron Meyerowitz to darij grinberg in the [comments](https://mathoverflow.net/questions/49035/explicit-b%c3%a9zout-cofactors#comment119719_49035) is unfortunately negative, even in the integers, by taking $a=c=u=v=w=0$ but $b=d=1$. However, it has a positive answer when $u\neq 0$.
Assume $ab=cd$, $ax+cy=u$, $uv=a$, and $uw=c$. Using the latter two equalities, we can solve for $a$ and $c$. The first two equalities now become
$$
uvb=uwd\qquad \text{ and }\qquad uvx+uwy=u.
$$
As $u\neq 0$, and we are working in an integral domain, we obtain
$$
vb=wd\qquad\text{ and }\qquad vx+wy=1.
$$
We find
$$
b=b(vx+wy)=(vb)x+bwy=(wd)x+bwy=w(dx+by).
$$
Similarly, $d=v(dx+by)$. Thus, taking $t:=dx+by$ works.
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2
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https://mathoverflow.net/users/3199
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410832
| 168,075 |
https://mathoverflow.net/questions/410802
|
17
|
**Question:** Let $p$ be an odd prime. Does there exist a closed manifold $M$ with $\widetilde H^\ast(M; \mathbb Q) = 0$ but $\widetilde H^\ast(M; \mathbb F\_p) \neq 0$?
When $p = 2$, an example is given by $\mathbb R \mathbb P^2$.
After discussion, this question turns out to be equivalent to [this other question](https://mathoverflow.net/questions/410577/for-which-n-does-there-exist-a-closed-manifold-of-chromatic-type-n). That is, over at that question Saal Hardali explained that if $M$ is a closed manifold, then the chromatic type of $M$ at a prime $p$ is either 0 or 1. Both possibilities are realized at $p=2$; the question is whether chromatic type 1 is realized at odd primes. Chromatic type 0 just means having nonvanishing rational (co)homology. So the question is whether there exists a closed manifold $M$ which is rationally contractible but whose $p$-localization is nontrivial for an odd $p$. This turned out to be mistaken, thanks to Ben Wieland for pointing this out at the other question.
**Side Question:** When $p=2$, what are some other examples of $M$ with $\widetilde H^\ast(M;\mathbb Q) = 0$ but $\widetilde H^\ast(M;\mathbb F\_2) \neq 0$ besides $M = \mathbb R \mathbb P^{2n}$ and products thereof?
|
https://mathoverflow.net/users/2362
|
Does there exist a closed manifold with vanishing reduced rational cohomology but nonvanishing odd torsion cohomology?
|
As mme noted in the comments, such examples cannot exist in odd dimensions, for Euler characteristic reasons. They can't exist in dimension 2 either, by classification. I claim that in all other dimensions $2n > 2$ we have (plenty of) examples.
Let $N$ be a rational homology $2n$-sphere, that is a $2n$-manifold with $H\_\*(N; \mathbb{Q}) = H\_\*(S^{2n}; \mathbb{Q})$. For every prime $p$ there exists a rational homology $2n$-sphere with $\dim\_{\mathbb{F}\_p} H\_\*(N;\mathbb{F}\_p) > 2$. For instance, you can take a spun lens space (any spun rational homology $2n-1$-sphere would do).
Now, the integral homology of $M = \mathbb{RP}^{2n} \# N$ splits as a direct sum of that of the two summands in all dimensions strictly between 0 and 2n, and it vanishes in dimension 2n (because $M$ is non-orientable) and it is $\mathbb{Z}$ in dimension 0 (because $M$ is connected). (This is Exercise 6 in Section 3.3 of Hatcher's *Algebraic topology*.) That is, $M$ is a rational homology ball, and its homology has as much $p$-torsion as that of $N$.
For the side question, if we choose $N$ to have no 2-torsion in its homology (e.g. spinning an odd lens space should do the trick), this gives plenty of examples.
|
20
|
https://mathoverflow.net/users/13119
|
410833
| 168,076 |
https://mathoverflow.net/questions/410831
|
2
|
Let $G$ and $G'$ be [quivers](https://ncatlab.org/nlab/show/quiver). If their [path categories](https://ncatlab.org/nlab/show/path+category#free_category_on_a_directed_graph) $Path[G]$ and $Path[G']$ are isomorphic, does is follow that $G$ is isomorphic to $G'$?
|
https://mathoverflow.net/users/471475
|
Does the path category of a quiver determine the quiver up to isomorphism?
|
Yes. Given a Quiver $G$, you can identify $G$ as the subquiver of $Path[G]$ of arrows that are not identity and cannot be written as composite of non-identity arrows. So any isomorphism between $Path[G]$ and $Path[G']$ send elements of $G$ to elements of $G'$ and restrict to an isomorphism between $G$ and $G'$.
|
7
|
https://mathoverflow.net/users/22131
|
410834
| 168,077 |
https://mathoverflow.net/questions/410773
|
1
|
In $\mathbb{P}^1\times\mathbb{P}^2$ take a general divisor $X$ of type $(0,2)$. Consider two general divisors $H\_1,H\_2$ of type $(2,1)$ and set $Y = X\cap H\_1\cap H\_2$.
Let $Z$ be the blow-up of $X$ along $Y$. I would like to ask whether there is a rank $3$ vector bundle $\mathbb{E}$ on $\mathbb{P}^1$ such that $Z$ can be embedded as a divisor in $\mathbb{P}(\mathbb{E})$ and if so how the splitting type of $\mathbb{E}$ can be computed.
Thank you very much.
|
https://mathoverflow.net/users/14514
|
Embedding of a blow-up
|
The map $Z \to \mathbb{P}^1$ is a conic bundle, so to understand the vector bundle $\mathbb{E}$ it is enough to compute the pushforward of the anticanonical class.
Now, the anticanonical class of $Z$ can be written as $-K\_X - E$, where $E$ is the exceptional divisor of the blowup, so its pushforward to $X$ is isomorphic to $I\_Y(-K\_X)$. If you use the identification of $X$ with $\mathbb{P}^1 \times \mathbb{P}^1$, this is $I\_Y(2,2)$. Using the Koszul resolution for $I\_Y$, one obtains
$$
0 \to \mathcal{O}(-2,-2) \to \mathcal{O}^{\oplus 2} \to I\_Y(2,2) \to 0.
$$
Pushing this forward to the first factor, one obtains
$$
0 \to \mathcal{O}^{\oplus 2} \to p\_{1\*}(I\_Y(2,2)) \to \mathcal{O}(-2) \to 0.
$$
Therefore, $\mathbb{E}$ has splitting type $(0,0,-2)$ or $(0,0,2)$, depending on which convention about the projective bundle you use.
|
3
|
https://mathoverflow.net/users/4428
|
410838
| 168,078 |
https://mathoverflow.net/questions/410777
|
5
|
Let $\mathfrak{X}\_{CK}^{\perp}$ be the space of vector fields on $S^2$ that are $L^2$-orthogonal to conformal Killing vector fields. Let $\mathfrak{X}\_{CK}$ be the 6-dimensional space of conformal Killing vector fields on $S^2$.
Can we find a vector field $Y \in \mathfrak{X}\_{CK}^{\perp}$ and a vector field $W \in \mathfrak{X}\_{CK}$ that is not Killing such that
$$\mathrm{div}(Y) = \mathrm{div}(W)$$
|
https://mathoverflow.net/users/138705
|
Finding vector fields on $S^2$ with equal divergence
|
I think that this is not possible:
Per my comment on [Divergence of conformal Killing vector fields on $S^2$ and the spherical harmonics](https://mathoverflow.net/questions/410824/divergence-of-conformal-killing-vector-fields-on-s2-and-the-spherical-harmoni?noredirect=1#comment1054264_410824) you want to solve
$$
\textrm{div} (Y) = -2a\cdot x
$$
for $Y$ orthogonal to conformal-KVF's and $a \in \mathbb{R}^3$ fixed (nonzero). Suppose you can do this. Then, we find that
$$
Y = a^T + W
$$
for $W$ divergence free. From this question <https://math.stackexchange.com/questions/1898371/divergence-free-vector-field-on-a-2-sphere> you can see that $W = J \nabla f$ for some function $f \in C^\infty(S^2)$ (for $J$ the complex structure on $S^2$). Now, by assumption, $Y$ is orthogonal to $a^T$:
\begin{align\*}
0 & = \int\_{S^2} Y\cdot a^T \\
& = \int\_{S^2}|a^T|^2 + (J \nabla f) \cdot a^T \\
& = \int\_{S^2}|a^T|^2 - \nabla f \cdot J a^T \\
& = \int\_{S^2}|a^T|^2 + f \textrm{div}(J a^T).
\end{align\*}
However, note that $Ja^T$ is a KVF and thus divergence free. This implies that $\int\_{S^2}|a^T|^2 = 0$. This is a contradiction.
|
5
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https://mathoverflow.net/users/1540
|
410839
| 168,079 |
https://mathoverflow.net/questions/410837
|
4
|
I am curios where in the literature was the first time written the following conjecture.
Say we have we have an elliptic curve $E$ given by the Weierstrass equation $y^2=x^3+AX+B$ with $A,B\in \mathbb{Z}$. Then the number of integral points should satisfy $E(\mathbb{Z})<<\_{\varepsilon} |\Delta|^{\varepsilon}$ for any $\varepsilon>0$.
|
https://mathoverflow.net/users/41010
|
integral points on elliptic curves in terms of discriminant
|
Not quite what you asked, but too long for a comment. In a book in 1978 Lang conjectured that on a (quasi)minimal Weierstrass equation, we have
$$\bigl|E(\mathbb{Z})\bigr|\le{C}^{\operatorname{rank}E(\mathbb{Q})},$$ where $C$ is an absolute constant. And assuming "standard conjectures", we have $$\operatorname{rank}E(\mathbb{Q})\ll\log{N\_E}/\log\log{N\_E}.$$ Since the conductor is smaller than the discriminant, combining these gives the conjecture that you quote, in slightly stronger form that one can take $\epsilon=c/\log\log\Delta$ for an absolute constant $c$.
|
8
|
https://mathoverflow.net/users/11926
|
410845
| 168,080 |
https://mathoverflow.net/questions/410840
|
4
|
From the definition of $\zeta(z):= \sum\_{k=1}^\infty \tfrac{1}{k^z}$ for $\mathrm{Re}(z)>1$ it is obvious that $\zeta(2k)\downarrow 1$ as $k \rightarrow \infty$. I am interested in the "true" speed of this convergence.
I know that e.g. $\sum\_{k=1}^\infty (\zeta(2k)-1) = \tfrac{3}{4}$ holds (Use the definition and switch up the order of summation). So the convergence speed must be higher than that of $\tfrac{1}{k}\downarrow 0$.
The software Mathematica even evaluates the sum $\sum\_{k=1}^\infty k^2(\zeta(2k)-1)$ to be $\tfrac{\pi^2}{8}$, but I don`t know how to prove this result or whether to trust it. This would mean that the true convergence speed is higher than that of $\tfrac{1}{k^3}\downarrow 0$.
Anyways, are there theorems in the literature that yield this convergence speed? Or even better: Inequalities of the form
\begin{equation}
\zeta(2k)-1 \leq \frac{C\_\ell}{k^\ell} \qquad \text{ for } k \in \mathbb{N}
\end{equation}
for some explicit constant $C\_\ell$ depending only on $\ell\in \mathbb{N}$? I'm not proficient in number theory and might have looked in the wrong places (such as Abramowitz and Stegun so far, which only contains the series that yields the $\tfrac{3}{4}$).
|
https://mathoverflow.net/users/471478
|
Speed of convergence of $\zeta(2k)\to 1$?
|
Here is an explicit bound. The sum $\sum\_{n > N} n^{-s}$ for real $s > 1$ is bounded by the integral
$$\int\_N^\infty x^{-s} = N^{1-s} / (s-1).$$
Therefore for any $N$ you have
$$0 < \zeta(s) - (1 + 2^{-s} + \cdots + N^{-s}) < N^{1-s} / (s-1).$$
E.g., with $N = 3$ you get
$$0 < \zeta(s) - 1 - 2^{-s} - 3^{-s} < 3^{1-s}/(s-1).$$
|
13
|
https://mathoverflow.net/users/20598
|
410846
| 168,081 |
https://mathoverflow.net/questions/410757
|
15
|
In my analysis research, I came across the following problem. Given $n$ positive real numbers $x\_1,\dots,x\_n$, consider the $n$-many power sums
$$ p\_3 = x\_1^3 + x\_2^3 + \dots + x\_n^3 , $$
$$ p\_5 = x\_1^5 + x\_2^5 + \dots + x\_n^5 , $$
$$ \vdots $$
$$ p\_{2n+1} = x\_1^{2n+1} + x\_2^{2n+1} + \dots + x\_n^{2n+1} . $$
Do the values of the power sums $p\_3,p\_5,\dots,p\_{2n+1}$ uniquely determine $x\_1,\dots,x\_n$ (up to reordering)?
I was wondering if this problem exists in the literature? I know the answer to this problem is "yes" in the case of the first $n$ power sums $p\_1,p\_2,\dots,p\_{n}$ by [Newton's identities](https://en.wikipedia.org/wiki/Newton%27s_identities): the power sums $p\_1,p\_2,\dots,p\_{n}$ determine the elementary symmetric polynomials $e\_1,\dots,e\_n$ via explicit formulas, from which we can construct a degree-$n$ polynomial with roots $x\_1,\dots,x\_n$ and appeal to the fundamental theorem of algebra.
I believe the answer is yes, and I've checked some special cases using a computer. Using analysis techniques I can easily get a local uniqueness statement. Indeed, the function $f:(x\_1,\dots,x\_n) \mapsto (p\_3,p\_5,\dots,p\_{2n+1})$ has a Jacobian matrix $Df$ that looks like a Vandermonde matrix, and this makes it easy to compute its determinant (and minors). In particular, on the simplex $\{ 0 < x\_1 < x\_2 < \dots < x\_n \} \subset\mathbb{R}^n$ the Jacobian matrix $Df$ has nonzero determinant, and so the inverse function theorem tells us that $f$ is locally injective. Moreover, this argument applies to all minors of $Df$, and so $Df$ is a [strictly totally positive](https://en.wikipedia.org/wiki/Totally_positive_matrix) matrix. Such matrices turn out to be diagonalizable with distinct positive eigenvalues, but I haven't been able to conclude that $f$ is globally injective.
|
https://mathoverflow.net/users/471391
|
Do power sums determine the variables?
|
Let $\gamma (x) = (x^3, x^5, \dots, x^{2n+1})$. By rearranging both sides (using that all the polynomials have odd degree and therefore are antisymmetric), the problem can be restated as:
>
> Let $0\le a,b$, $a+b \le 2n$ integers, and $x\_1, \dots x\_a,y\_1, \dots y\_b \in \mathbb{R}\_{\ge 0}$. Then
>
>
> $$ \sum\_{i=1}^a \gamma(x\_i) = \sum\_{i=1}^b \gamma(y\_i) $$
>
>
> has no nontrival solutions.
>
>
>
From what I can see from the [MathSciNet review](https://mathscinet.ams.org/mathscinet-getitem?mr=309867), that is precisely the content of [1]. I haven't been able to find that paper, but I found it as a reference in [2], which also contains the idea of the proof (in a slightly more general setting) in the second half of Section 3. The key idea is using that the Jacobian matrix is a strictly totally positive matrix, indeed!
[1] J. Steinig, *On some rules of Laguerre's, and systems of equal sums of like powers*. Rend. Mat. (6) 4 (1971), 629–644 (1972).
[2] [S. W. Drury and B. P. Marshall, *Fourier restriction theorems for degenerate curves.* Mathematical Proceedings of the Cambridge Philosophical Society, 101(3), 541-553.](https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/fourier-restriction-theorems-for-degenerate-curves/7E17A1219062870146EA1D31A0DD211A)
|
10
|
https://mathoverflow.net/users/165826
|
410848
| 168,082 |
https://mathoverflow.net/questions/410844
|
1
|
Let $3 \leq k < n \in \mathbb{N}$. By $[n]^k$ we denote the collection of the subsets of $n = \{0,\ldots,n-1\}$ that have size $k$. We say that a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(n,E)$ is $k$-*uniform* if $E\subseteq [n]^k$. Moreover, $H=(n, E)$ is *linear* if $|e\_1\cap e\_2| \leq 1$ for $e\_1\neq e\_2\in E$, and it is *maximal linear* if $E\subseteq E'\subseteq [n]^k$ and $E\neq E'$ imply that $(n, E')$ is no longer linear.
**Question.** Are there integers $k < n$ with $k\geq 3$ and maximal linear hypergraphs $H\_i = (n, E\_i)$ for $i = 1,2$ such that $|E\_1| \neq |E\_2|$? (If yes, it would also be interesting to know how big the difference of the edge sets can become in terms of $n$, but this information is not needed for acceptance of answer.)
|
https://mathoverflow.net/users/8628
|
Number of edges in $k$-uniform linear hypergraph
|
For the case $k=3$ we have *partial Steiner triple systems* as a design theory name for $3$-uniform linear hypergraphs. The *spectrum* $S^{(3)}(v)$ consists of the sizes of maximal partial Steiner triple systems taking triples from a $v$-element set. The paper below gives the final steps in determining $S^{(3)}(v)$ for each $v$.
I can access the paper through my university, but I did not find a freely available version.
[The spectrum of maximal partial steiner triple systems](https://doi.org/10.1007/BF01388482)
However, the main result (and other references) can be found in
[Maximal designs and configurations - a survey](https://actamath.savbb.sk/pdf/acta2302.pdf)
looking in Section 9 about partial Steiner triple systems. In particular, you do have multiple sizes in the spectrum. If $v = 12k + r$, then the smallest is around $12k^2$ while the largest is around $24k^2$ (of course with linear and constant terms depending on cases $\pmod{12}$ as well as other caveats).
|
4
|
https://mathoverflow.net/users/51668
|
410851
| 168,085 |
https://mathoverflow.net/questions/410704
|
11
|
Here [A Question on a second countable $T\_2$ space](https://math.stackexchange.com/questions/1215409/a-question-on-a-second-countable-t-2-space), Paul asked if every second countable Hausdorff space has a $G\_\delta$-diagonal. In the comments Brian M. Scott answered that, at the time (2015), the answer to that question might not be known. Interestingly enough, here [Does second countable and functionally Hausdorff imply submetrizable?](https://mathoverflow.net/questions/280359/does-second-countable-and-functionally-hausdorff-imply-submetrizable), Taras Banakh showed two years later that, as a consequence of a more general result, every second countable functionally Hausdorff space is submetrizable (in particular, every such space has a $G\_\delta$-diagonal).
My question is whether the answer to Paul's original question is already known. I couldn't find anything definitive about it.
|
https://mathoverflow.net/users/146942
|
Second countable vs. $G_\delta$-diagonal
|
There exists a counterexample to this question of Paul.
It suffices to find a second-countable Hausdorff space $X$ that has two properties:
(1) the space $X\times X$ is Baire;
(2) for any nonempty open sets $U,V\subseteq X$ we have $\overline U\cap\overline V\ne\emptyset$.
Such a space $X$ cannot have $G\_\delta$-diagonal. Indeed, assuming that the diagonal $\Delta\_X$ is of type $G\_\delta$ in $X\times X$, we can write $(X\times X)\setminus \Delta\_X$ as the union $\bigcup\_{n\in\omega}F\_n$ of closed subsets $F\_n$ of $X\times X$. Property (2) implies that $X$ has no isolated points and hence the diagonal $\Delta\_X$ is nowhere dense in $X\times X$. Since the space $X\times X$ is Baire, there exists $n\in\omega$ such that $F\_n$ has non-empty interior in $X\times X$. Then there are nonempty open sets $U,V$ in $X$ such that $U\times V\subseteq F\_n$ and hence $\overline U\times\overline V\subseteq F\_n$. Then $(\overline U\times \overline V)\cap\Delta\_X=\emptyset$ and hence $\overline U\cap\overline V=\emptyset$, which contradicts the property (2).
For the space $X$ one can take [the projective space of the countable product of lines](https://mathoverflow.net/q/46986/61536) $\mathbb R^\omega$. So, $X$ is the quotient space of $\mathbb R^\omega\setminus\{0\}^\omega$ by the equivalence relation $x\sim y$ iff $\mathbb Rx=\mathbb Ry$. Observe that the quotient map $q:\mathbb R^\omega\setminus\{0\}^\omega\to X$ is open. This fact can be used to show that $X$ is Hausdorff and its square $X\times X$ is Baire. Moreover, for any nonempty open set $U\subseteq X$ the closure $\overline{U}$ contains the image $q[(\{0\}^n\times \mathbb R^{\omega\setminus n})\setminus\{0\}^\omega]$ for some $n\in\omega$, which implies that the condition (2) is satisfied.
|
10
|
https://mathoverflow.net/users/61536
|
410853
| 168,086 |
https://mathoverflow.net/questions/410847
|
6
|
We first consider the sheaf of holomorphic functions $\mathcal{O}(\mathbb{C}^n)$ on $\mathbb{C}^n$. By [Oka coherence theorem](https://en.wikipedia.org/wiki/Oka_coherence_theorem), $\mathcal{O}(\mathbb{C}^n)$ is coherent over itself.
Now we consider a finite group $G$ acting on $\mathbb{C}^n$ and let $\pi: \mathcal{C}^n\to \mathbb{C}^n/G$ be the projection. We define a sheaf $\bar{\mathcal{O}}$ on $\mathbb{C}^n/G$ as follows: for any open subset $U\subset \mathbb{C}^n/G$, we define
$$
\bar{\mathcal{O}}(U):=\{f\in \mathcal{O}(\pi^{-1}(U))|f \text{ is }G-\text{invariant.}\}
$$
>
>
> >
> > My question is: is this sheaf $\bar{\mathcal{O}}$ coherent over itself too?
> >
> >
> >
>
>
>
|
https://mathoverflow.net/users/24965
|
Do we have the Oka coherence theorem for finite group actions?
|
This would be true. You need two facts:
1. Grauert's theorem that coherent sheaves are preserved by proper direct images. This implies
$\pi\_\*\mathcal{O}\_{\mathbb{C}^n}$ is coherent.
2. Sub modules of coherent sheaves are coherent. Therefore
$$\tilde{\mathcal{O}} = \pi\_\*\mathcal{O}\_{\mathbb{C}^n}^G\subset \pi\_\*\mathcal{O}\_{\mathbb{C}^n}$$
is coherent
|
8
|
https://mathoverflow.net/users/4144
|
410854
| 168,087 |
https://mathoverflow.net/questions/411851
|
6
|
By definition, an $n$-dimensional Delzant polytope $P$ is not necessarily a lattice polytope. But
is there a natural way (or operations) to turn $P$ into a lattice polytope using the fact that the edge vectors incident to any vertex of $P$ form an integral basis of $\mathbb{Z}^n$? And what do these operations mean in symplectic geometry/topology?
The background of this question is: when people explain the equivalence between symplectic toric manifold and smooth projective toric variety, they often assume the moment polytope to be both lattice and Delzant. So I feel there must be a reason for this assumption.
|
https://mathoverflow.net/users/471224
|
From Delzant polytope to lattice polytope
|
There is a way to turn a Delzant polytope into a lattice polytope, but there is no natural or canonical way. Note that the Delzant condition implies that the normal vector to each facet (codimension 1 face) is integral. By letting the defining equation for the hyperplane containing each facet rational (by slightly translating hyperplanes), you can make the coordinates of all vertices rational without changing the combinatorial structure. Now the polytope is integral after scaling. As you see, this is far from being canonical.
The above operation changes the cohomology class represented by the symplectic form. So you can think of it as choosing a different symplectic form on the same smooth manifold. If you view a symplectic toric manifold as a symplectic reduction, you are taking a different regular value of the moment map.
Some people regard symplectic toric manifolds as smooth projective varieties, but strictly speaking, that is wrong. Any symplectic toric manifold is diffeomorphic to a smooth projective variety, but it may not be symplecally embedded into a projective space equipped with the Fubini-Study form $\omega\_{FS}$. In fact, by Kodaira embedding theorem, a symplectic toric manifold $(M, \omega)$ can be (symplectically) embedded into $\left(\mathbb{P}^N, \omega\_{FS}\right)$ for some $N$ if and only if $[\omega] \in H^2(M, \mathbb{Z})$. Even if we allow scalar multiples of $\omega\_{FS}$, the class $[\omega]$ can only be a multiple of an integral one.
For an easy example, consider $(S^2 \times S^2, \sigma \oplus \lambda\sigma)$ where $\sigma$ is a volume form on $S^2$ and $\lambda>0$ is irrational. It can never be embedded into a projective space preserving the symplectic structure. You can take rational $\lambda$ to find an embedding, but then you are dealing with a different symplectic manifold.
|
6
|
https://mathoverflow.net/users/11846
|
411855
| 168,090 |
https://mathoverflow.net/questions/411858
|
5
|
This is a reference request, since the answer is probably well known, but I could not find it.
Given a finitely generated group $\Gamma$ with a generating set $S$, define the word norm $l = l\_S : \Gamma \rightarrow \mathbb{N}$ to be
$$l (g) = \min \lbrace k : \exists s\_1,...,s\_k \in S, g = s\_1 ... s\_k \rbrace ,$$
i.e., $l(g)$ is the distance between $e$ and $g$ in the Cayley graph of $\Gamma$ (w.r.t the generating set $S$).
My question: Let $\Gamma = \rm SL\_3 (\mathbb{Z})$ (with some finite generating set). For $1 \leq i,j, \leq 3, i \neq j$ and $m \in \mathbb{Z}$, denote $e\_{i,j} (m)$ to be the elementary matrix with $1$'s along the main diagonal, $m$ in the $(i,j)$-entry and $0$ in all other entries. What can one say about the growth rate of $l (e\_{i,j} (m))$?
My naive attempt for an answer gives me $l (e\_{i,j} (m)) = O (\log^3 (m))$:
1. For convenience, we fix the generating set $S = \lbrace e\_{i,j} (\pm 1), e\_{i,j} (\pm 2) : 1 \leq i,j, \leq 3, i \neq j \rbrace$.
2. Using commutator (Steinberg) relations
$$ e\_{i,j} (2^{2^{r+1}}) = [e\_{i,k} (2^{2^{r}}), e\_{k,j} ( 2^{2^{r}})]$$
it is not hard to show by induction on $r$ that $l (e\_{i,j} (2^{2^r})) \leq 4^{r} $.
3. Again by the commutator relations, it follows that for every $2^r \leq d < 2^{r+1}$ it holds that $l (e\_{i,j} (2^{d})) \leq 4^{r+1} \leq 4 d^2$.
4. Thus it follows that for every $2^d \leq m < 2^{d+1}$,
$$l (e\_{i,j} (m)) \leq \sum\_{t =0}^d l (e\_{i,j} (2^t)) \leq 4 \sum\_{t =0}^d t^2 = 4 \frac{d (d+1) (2d +1)}{6} = O (\log^3 (m)).$$
As noted above - this computation is quite naive. Are there better known results?
|
https://mathoverflow.net/users/3461
|
Growth of the word norm for elementary matrices in $\rm SL_3 (\mathbb{Z})$
|
The answer is that this is $\simeq\log(m)$. Where $f\sim g$ means that $f\preceq g\preceq f$ and $f\preceq g$ means that eventually $f\le cg$ for some $c>0$.
This is a particular case of a result of Lubotzky, Mozes and Raghunathan. But it's easy to prove directly.
First since the matrix norm of $e\_{ij}(1)^m$ grows linearly and the matrix norm is submultiplicative, one immediately sees that $|e\_{ij}(1)^m|\succeq \log(m)$.
Now we use that $m\ge 3$, so we can suppose $(i,j)=(1,3)$. Then consider the subgroup $\Gamma$ of matrices that are identity, except the entries 13,23 that are arbitrary, and the block 11 12 21 22 which is an arbitrary integral power of $\begin{pmatrix}2&1\\1& 1\end{pmatrix}$. Then $\Gamma$ is a cocompact lattice in the 3-dimensional group SOL. It is immediate that in the group SOL, if $v$ is in the normal abelian subgroup (written additively), then the word length of $v$ (with respect to any compact generating subset) is $\simeq \log(\|v\|)$. In particular, for fixed nonzero $v$ the word length of $mv$ (=$v^m$, switching back to multiplicative notation) is $\simeq\log(m)$. Since cocompact lattices are undistorted, we deduce that the same holds in $\Gamma$. Hence $|e\_{13}(1)^m|\preceq \log(m)$ in $\Gamma$, and hence in the larger group $\mathrm{SL}\_n(\mathbf{Z})$.
|
8
|
https://mathoverflow.net/users/14094
|
411861
| 168,091 |
https://mathoverflow.net/questions/409314
|
2
|
Let's consider the parabolic system
$$
\begin{cases}
u\_t - \Delta u -a\Delta(uv) = 0 \\
v\_t - \Delta v - b\Delta(uv) = 0
\end{cases}
$$
with $a,b >0$. What is the name of this system? Are there known results about existence and uniqueness?
|
https://mathoverflow.net/users/nan
|
Parabolic system with coupling in the diffusion
|
Such systems belong to the family of **cross diffusion systems**.
For $W^{1,p}$ data ($p$ greater than the dimension), local existence and uniqueness stems from Amann's theorem (look for *Dynamic theory of quasilinear parabolic systems paper* from 1989). For weak solutions (in bounded domains), you can look at the work of Chen and Jungel *Analysis of a parabolic cross-diffusion population model without self-diffusion* in *J. Diff. Eq.* (and authors citing it) based on the remark that and adapted LlogL entropy is dissipated along positive solutions. In population dynamics they have been introduced by Shigesada Kawasaki and Terramoto (*Spatial segregation of interacting species*, *J. Theor. Biol* 1979).
I am far from being exhaustive, there exists a huge literature on those systems (and more complicated versions).
|
2
|
https://mathoverflow.net/users/161129
|
411872
| 168,093 |
https://mathoverflow.net/questions/410841
|
8
|
Recall that a subset $A \subset \mathbb Z\_+$ of positive integers *syndetic* if there exists a $d>0$ such that every positive integer has distance at most $d$ to an element of $A$. It is called *piecewise syndetic* if it is the intersection of a syndetic set with a subset of $[0,\infty)$ containing arbitrarily long intervals.
Let us call $n \in \mathbb Z\_+$ *powerful* if for every prime $p$, the multiplicity $\nu\_p(n) \neq 1$. (Or, equivalently: $n$ can be expressed as a product of a square and a cube.)
Here is the question: is the set of powerful numbers piecewise syndetic?
|
https://mathoverflow.net/users/14233
|
Is the set of powerful numbers piecewise syndetic?
|
The answer is no. A set $S$ to be piecewise syndetic iff there is an integer $d$ such that there exist intervals $I$ of arbitrary length such that distances between elements of $S\cap I$ are bounded by $d$. In particular, $|S\cap I|\geq\frac{1}{d}|I|$. I will show no such $d$ exists.
For any prime $p$, the fraction of those which are either not divisible by $p$ or divisible by $p^2$ is equal to $1-\frac{p-1}{p^2}$. Further, these conditions are independent - formally, if we consider take $P=p\_1\dots p\_k$, the product of first $k$ primes, then from the Chinese remainder theorem, in any interval $I$ of length $P^2$ the fraction of integers $n$ in $I$ such that $v\_{p\_i}(n)\neq 1$ for all $i$ is equal to
$$C:=\prod\_{i=1}^k\left(1-\frac{p-1}{p^2}\right).$$
This gives an upper bound of $CP^2$ for the number of powerful numbers an interval of length $P^2$. Now, as $k$ tends to infinity, then $C$ tends to zero (this essentially follows from the fact sum of reciprocals of primes diverges), in particular for large enough $k$ it becomes smaller than $\frac{1}{d}$, so no interval $I$ of length $P^2$ contains more than $\frac{1}{d}|I|$ powerful numbers.
|
14
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https://mathoverflow.net/users/30186
|
411875
| 168,095 |
https://mathoverflow.net/questions/411868
|
10
|
Laver showed in 1995 that the period of the first row of certain [Laver tables](https://en.wikipedia.org/wiki/Laver_table) is unbounded, assuming that a rank-into-rank cardinal exists.
The most accessible proof of his result that I was able to find is in chapter 12 of Patrick Dehornoy's Braids and Self-Distributivity ([Springer 2000](https://link.springer.com/book/10.1007/978-3-0348-8442-6)). The proof is quite technical. Laver defines an algebra on the models of set-theory - technically, on certain elementary embeddings of huge sets. He defines two operations on such elementary embeddings, quotients out certain large infinities to get finite results, and investigates their properties.
My question is: why should elementary embeddings of ZFC have anything to do with the periodicity of these finite tables?
More generally, is there some guiding intuition I can use to make sense of Laver's very complex construction? I find it difficult to understand how Laver could have plowed through this weird and intricate technical construction without having some reason to think it would lead to some kind of specific finitary result.
|
https://mathoverflow.net/users/472518
|
Motivation for Laver's use of large cardinals to show finite combinatorial properties of Laver tables
|
As was already mentioned in the comments, the premise of the question is somewhat backwards. Indeed, looking at [Laver's paper](https://www.sciencedirect.com/science/article/pii/S0001870885710146?via%3Dihub), the combinatorial structures now known as Laver tables were not at all his initial motivation. Instead, from the start and throughout Laver was interested in elementary embeddings of some $V\_\lambda$ and the algebras they produce under certain operations, one of which is "applying" one elementary embedding to another. This operation satisfies a relation $a(bc)=(ab)(ac)$ (and thus defines what is sometimes known as a [shelf](https://ncatlab.org/nlab/show/shelf)). For instance, Laver has shown that the shelf generated by one elementary embedding $j$, which he denotes $\mathscr A\_j$, is free.
Now, $\mathscr A\_j$ is a large and complicated algebra. One of Laver's results (essentially Theorem 11 of the paper) is that the study of $\mathscr A\_j$ can be reduced to studying certain finite algebras, $\mathscr A\_{j,n}$, which have $2^n$ elements. These algebras are again defined using elementary embeddings, but as Laver explains after the theorem, these algebras also have a completely explicit description not referring to any large cardinals - these are the Laver tables that you are familiar with.
Therefore Laver's motivation was studying certain algebras which arise naturally if you are interested in large cardinals. Any relation to finitary structures was an afterthought to him.
|
14
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https://mathoverflow.net/users/30186
|
411883
| 168,098 |
https://mathoverflow.net/questions/411885
|
1
|
For $d, m \in\mathbb{N}$ fixed, let $P\equiv P(x) := \sum\_{|\alpha|\leq m} c\_\alpha\cdot x^\alpha$ be a real polynomial in $d$ variables of (total) degree $m$. (That is, the above sum ranges over all multiindices $\alpha=(i\_1, \ldots, i\_d)\in\mathbb{N}\_0^{\times d}$ of length $|\alpha|\equiv i\_1+\ldots + i\_d$ less than $m$.)
Is it possible to estimate the maximum coefficient $\|c\_\alpha\|\_\infty := \max\_{|\alpha|\leq m}|c\_\alpha|$ of $P$ against its uniform norm $\|P\|\_{\infty;K}:= \sup\_{x\in K}|P(x)|$, for $K$ some compact set in $\mathbb{R}^d$?
That is, does there exist a compact set $K$ in $\mathbb{R}^d$ together with a constant $\kappa \equiv \kappa(m,d,K)>0$ such that
$$\tag{1}\|c\_\alpha\|\_\infty \ \leq \ \kappa\cdot \|P\|\_{\infty; K} \qquad \text{ for each } \ P \ \text{ as above}?$$
Any references are welcome.
**Edit:** Do you know if $\kappa$ can be chosen independent of $m$, or at least such that the sequence $(\kappa(m,d,K))\_{m\geq 0}$ is bounded (for $K$ and $d$ fixed)?
|
https://mathoverflow.net/users/472548
|
Inequality between coefficients of a polynomial and its supremum
|
Yes, such a $\kappa$ exists for every compact set $K$ with non-empty interior. Here is an abstract linear-algebra argument.
Let $V$ be the real vector space spanned by the multi-indices $\alpha$ with length at most $m$.
We have a linear map $A\colon V \to \mathbb R^K$, sending $(c\_\alpha)$ to the function $x\to \sum\_\alpha c\_\alpha x^\alpha$. Since different polynomials cannot agree on a set with non-empty interior, this map is injective. There is then a finite subset $F\subset K$ such that the composition $V\xrightarrow A\mathbb R^K\to\mathbb R^F$ is a linear isomorphism (for example, pick the elements of $F$ one by one, making sure that the rank of the composition increases by 1 each time).
Now $\kappa$ exists by the fact that all linear isomorphisms between finite-dimensional vector spaces are bicontinuous.
If you want an explicit value for $\kappa$, you can use discrete derivatives (see <https://en.wikipedia.org/wiki/Finite_difference>) to write $c\_\alpha$ as a linear combination of some values of $P$.
|
2
|
https://mathoverflow.net/users/470870
|
411886
| 168,099 |
https://mathoverflow.net/questions/410548
|
8
|
The irreducibility of the commuting variety $\{(A,B) \in \mathcal{M}\_{n}(\mathbb{C})^2, \ AB = BA \}$ is well-known (see for instance [On Dominance and Varieties of Commuting Matrices](https://doi.org/10.2307/1970336) by Gerstenhaber). I am interested in the irreducibility of some special linear sections of the commuting varieties. Namely, let $W\_1$, $W\_2$ be two $k$-dimensional subspaces of $\mathbb{C}^n$ ($n$ and $k$ are fixed). Is the variety:
$$ \{(A,B)\in \mathcal{M}\_{n}(\mathbb{C})^2, \ AB = BA, \ W\_1 \subset \operatorname{Ker}(A), \ W\_2 \subset \operatorname{Ker}(B) \}$$
known to be irreducible? Or perhaps are there examples where it is not?
|
https://mathoverflow.net/users/37214
|
Irreducibility of linear sections of the commuting variety
|
It is already reducible in the toy case $n=2, k=1$ where $W\_1, W\_2 \subseteq \mathbb{C}^2$ are two distinct lines. Without loss of generality, have them be spanned by the standard basis vectors respectively, so that
$$A = \begin{bmatrix}0 & a\_{12} \\ 0 & a\_{22} \end{bmatrix}, \quad
B = \begin{bmatrix}b\_{11} & 0 \\ b\_{21} & 0 \end{bmatrix}.$$
The equation $AB = BA$ now reads
$$\begin{bmatrix}a\_{12}b\_{21} & 0 \\ a\_{22}b\_{21} & 0 \end{bmatrix} = \begin{bmatrix}0 & b\_{11}a\_{12} \\ 0 & b\_{21}a\_{12} \end{bmatrix}.$$
(As you can see, $AB=BA=0$.) But we have
$$(a\_{22}b\_{21}, a\_{12}b\_{21}, a\_{12}b\_{11}) = (a\_{22},a\_{12}) \cap (a\_{12},b\_{21}) \cap (b\_{11},b\_{21})$$
of ideals in four variables, so there are three irreducible components of dimension $2$ here.
In contrast, in the case $W\_1 = W\_2$, the equation $AB=BA$ instead has the form of the single equation $ab'=a'b$, which is not only irreducible, but of a different dimension and degree.
So I imagine it's complicated in general, depending on $n,k$ and $\dim(W\_1 \cap W\_2)$. I wouldn't be surprised if there's a story here similar to Springer fibers, which vary in dimension and reducibility, but in a very nice way.
**edit**:
Investigations in Macaulay2 for the case $n=4, k=2$ give the following:
* If $W\_1 + W\_2 = \mathbb{C}^4$, then the variety has six irreducible components (two of dimension 9, four of dimension 8).
* If $\dim(W\_1 \cap W\_2) = 1$, then the variety has eight irreducible components (three of dimension 9 and five of dimension 8).
* If $W\_1 = W\_2$, then the variety has two irreducible components, both of dimension 10, but with different degrees of generators.
|
8
|
https://mathoverflow.net/users/45505
|
411896
| 168,102 |
https://mathoverflow.net/questions/410776
|
1
|
Here is the goal sum where the [Pochhammer Symbol](https://mathworld.wolfram.com/PochhammerSymbol.html) with the [Incomplete Beta function series](https://functions.wolfram.com/GammaBetaErf/Beta3/06/01/03/01/01/0003/)
$$\sum\_{m=0}^\infty \frac{\text B\_z(m+a,b-m)x^m}{m!}=\sum\_{m=0}^\infty z^{m+a}\sum\_{n=0}^\infty\frac{(1-(b-m))\_nz^n x^m}{(m+a+n)m!n!}=\frac{z^a}a\sum\_{m,n\ge0}\frac{(a)\_{m+n} (b-1)\_{m+n}(xz)^mz^n}{(a+1)\_{m+n}(b-1)\_m m! n!}$$
Now use the [Kampé de Fériet](https://www.researchgate.net/publication/341441364_Novel_reductions_of_Kampe_de_Feriet_function) function also [found on Wolfram Mathworld](https://mathworld.wolfram.com/KampedeFerietFunction.html):
$$\text F^{p,r,u}\_{q,s,v}\left(^{a\_1,…,a\_p;c\_1,…,c\_r;f\_1,…,f\_u}\_{b\_1,…,b\_q;d\_1,…,d\_s;g\_1,…,g\_v}\ x,y\right)\mathop=^\text{def}\sum\_{m=0}^\infty\sum\_{n=0}^\infty\frac{\prod\limits\_{j=1}^p(a\_j)\_{m+n} \prod\limits\_{j=1}^r(c\_j)\_m \prod\limits\_{j=1}^u (f\_j)\_n x^my^n}{\prod\limits\_{j=1}^q (b\_j)\_{m+n} \prod\limits\_{j=1}^s(d\_j)\_m \prod\limits\_{j=1}^v(g\_j)\_n m!n!}$$
Therefore:
$$\sum\_{m=0}^\infty \frac{\text B\_z(m+a,b-m)x^m}{m!}\mathop= ^{|z|,-a\not\in\Bbb N}\frac{z^a}a \text F^{2,0,0}\_{1,1,0}\left(^{a,b-1;;}\_{a+1;b-1;}\ xz,z\right)$$
The second sum of interest is:
$$\sum\_{m=0}^\infty \text B\_z(m+a,b-m)x^m\mathop= ^{|z|,-a\not\in\Bbb N}\frac{z^a}a \text F^{2,1,0}\_{1,1,0}\left(^{a,b-1;1;}\_{a+1;b-1;}\ xz,z\right)$$
Is there a simpler closed form or decomposition formula for this case of the Kampé de Fériet function? A solution verification is also needed, but the derivation should be correct. Please give feedback.
|
https://mathoverflow.net/users/245836
|
Simplification of $\sum_{m=0}^\infty \text B_z(m+a,b-m)x^m,\sum_{m=0}^\infty \frac{\text B_z(m+a,b-m)x^m}{m!}$ in terms of Kampé de Fériet function
|
I'm not sure if this is considered simpler, but still:
\begin{split}
\sum\_{m=0}^\infty \frac{\text B\_z(m+a,b-m)x^m}{m!} &= \sum\_{m=0}^\infty \frac{x^m}{m!}\int\_0^z y^{m+a-1}(1-y)^{b-m-1}\,{\rm d}y\\
&=\int\_0^z y^{a-1}(1-y)^{b-1} e^{\frac{xy}{1-y}}\,{\rm d}y.
\end{split}
Similarly,
$$\sum\_{m=0}^\infty \text B\_z(m+a,b-m)x^m =\int\_0^z \frac{y^{a-1}(1-y)^b}{1-y-xy}\,{\rm d}y.$$
|
1
|
https://mathoverflow.net/users/7076
|
411904
| 168,104 |
https://mathoverflow.net/questions/411901
|
4
|
In a [paper](http://www.numdam.org/item/?id=AIF_1977__27_2_1_0) by Saffari and Vaughan there appears a complicated-looking double sum
$$\Sigma\_1=\sum\_{\rho\_1}\sum\_{\rho\_2}\frac{(1+\theta)^{\rho\_1}-1}{\rho\_1}\cdot \frac{(1+\theta)^{\bar{\rho\_2}}-1}{\bar{\rho\_2}} \cdot \frac{2^{1+\rho\_1+\bar{\rho\_2}}-2^{-1-\rho\_1-\bar{\rho\_2}}}{1+\rho\_1+\bar{\rho\_2}} \cdot \frac{2^{2+\rho\_1+\bar{\rho\_2}}-1}{2+\rho\_1+\bar{\rho\_2}} \cdot x^{1+\rho\_1+\bar{\rho\_2}},$$
where both sums are over all non-trivial zeros $\rho\_k=\beta\_k+i\gamma\_k$ of the Riemann zeta-function, and $\theta \in (0,1]$ is a parameter. Using the inequality $|z\_1z\_2|\leq\frac{1}{2}(|z\_1|^2+|z\_2|^2)$, they state that one can bound $\Sigma\_1$ by
$$\Sigma\_1\ll \sum\_{\rho\_1}\sum\_{\rho\_2} x^{1+2\beta\_1} \min(\theta^2, \gamma\_1^{-2}) (1+|\gamma\_1-\gamma\_2|)^{-2}.$$
I can see where the final factor $(1+|\gamma\_1-\gamma\_2|)^{-2}$ comes from, however the other two factors confuse me. The Riemann hypothesis is assumed in the final result, but does not appear to have been used here. Even if it was, I can only see how this implies the bound $x^{1+\rho\_1+\bar{\rho\_2}} \ll x^{1+2\beta\_1}.$ What I mainly can't understand is how they bound
$$\frac{(1+\theta)^{\rho\_1}-1}{\rho\_1}\cdot \frac{(1+\theta)^{\bar{\rho\_2}}-1}{\bar{\rho\_2}} \ll \min(\theta^2, \gamma\_1^{-2}).$$
Unconditionally, I can show that $$\frac{(1+\theta)^{\rho\_1}-1}{\rho\_1}\cdot \frac{(1+\theta)^{\bar{\rho\_2}}-1}{\bar{\rho\_2}} \ll \theta^2 \left(\frac{1}{\gamma\_1^2} + \frac{1}{\gamma\_2^2} \right).$$
But, to get the original bound, it seems like you would need to bound $\gamma\_2$ with $\gamma\_1$, which sounds unlikely at best. Am I missing any magic tricks to do with double-sums/convergence/Riemann zeros?
|
https://mathoverflow.net/users/151669
|
Double sum over zeros of Riemann zeta-function
|
Using $\theta\in[0,1]$ and $\mathrm{Re}(\rho)\leq 1$, we see that
$$\frac{(1+\theta)^\rho-1}{\rho}=\int\_1^{1+\theta}t^{\rho-1}\,dt$$
has absolute value at most $\min(\theta,3|\rho|^{-1})$. Therefore,
$$\Sigma\_1\ll x\sum\_{\rho\_1}\sum\_{\rho\_2}x^{\beta\_1+\beta\_2}\min(\theta,|\gamma\_1|^{-1})\min(\theta,|\gamma\_2|^{-1})(1+|\gamma\_1-\gamma\_2|)^{-2}.$$
Now we apply
$$x^{\beta\_1+\beta\_2}\min(\theta,|\gamma\_1|^{-1})\min(\theta,|\gamma\_2|^{-1})\ll x^{2\beta\_1}\min(\theta^2,\gamma\_1^{-2})+x^{2\beta\_2}\min(\theta^2,\gamma\_2^{-2})$$
and the symmetry $\rho\_1\leftrightarrow\rho\_2$ to arrive at
$$\Sigma\_1\ll x\sum\_{\rho\_1}\sum\_{\rho\_2}x^{2\beta\_1}\min(\theta^2,\gamma\_1^{-2})(1+|\gamma\_1-\gamma\_2|)^{-2}.$$
|
4
|
https://mathoverflow.net/users/11919
|
411916
| 168,106 |
https://mathoverflow.net/questions/411910
|
11
|
Does there exist a finitely generated group $G$ and an automorphism $\Phi\colon G \to G$ such that there are $\Phi$-periodic elements with unbounded period?
If $G$ is merely countable and not finitely generated there are easy examples: consider a free or free abelian group of countably infinite rank, let $x\_1,x\_2,\ldots$ be a standard generating set, and consider the automorphism that cyclically permutes the first $2$ generators, the next $3$ generators, the next $4$ generators and so on.
|
https://mathoverflow.net/users/135175
|
Does there exist an automorphism of a finitely generated group with periodic points of unbounded period?
|
1. Yes, and even with an inner automorphism.
Namely, start from your example of a countable group $G$ with elements $x\_n$ ($n$ ranging over some arbitrary subset) and automorphism $f$ such that $x\_n$ lies in an $n$-cycle for $f$. Let $G'$ be the semidirect product $G\rtimes\langle f\rangle$. Finally, embed $G'$ into a finitely generated group $H$ (every countable group embeds into a f.g. group, an old result of Higman-Neumann-Neumann). Then in $H$, for the inner automorphism defined by $f$, the element $x\_n$ lies in an $n$-cycle. [NB: as $G$ can be chosen to be abelian and hence $G'$ metabelian, $H$ can be chosen to be 4-step solvable, by the Neumann-Neumann proof of the previous embedding theorem.]
2. Note: there are no examples among linear groups (= subgroup of $\mathrm{GL}\_d$ over a field) with an inner automorphism. Indeed, for the inner automorphism $i\_g$ defined by an element $g$, the centralizer of $g^{n!}$ is an increasing sequence of subgroups. But in a matrix group the centralizers in the algebra of matrices are subspaces. So there is no strictly infinite sequence of centralizers. Hence the finite $i\_g$-cycles have bounded length.
3. Here's another example, due to B.H. Neumann 1937, with the additional feature of being residually finite. Namely, consider a set $X$ consisting of the infinite disjoint union of sets $X\_n$ ($n$ ranging over an infinite set $I$ of integers $\ge 5$), each $X\_n$ being in bijection with $\{1,\dots,n\}$. Let $c$ be the element acting as the $n$ cycle $n\mapsto 1\mapsto 2\mapsto\dots$ on each $X\_n$, and $t$ the element acting as the transposition $1\mapsto 2\mapsto 1$ on each $X\_n$. Let $G\_I$ be the subgroup of permutations of $X$ generated by $\{t,c\}$. Then $G\_I$ is residually finite (since it acts faithfully with finite orbits). Also $G\_I$ contains, for each $n\in I$, as normal subgroup, the alternating group $\mathrm{Alt}\_n$ of $X\_n$ (acting trivially elsewhere). Let $s\_n$ be the 3-cycle $1\mapsto 2\mapsto 3\mapsto 1$ acting on $X\_n$, identity elsewhere. Then we see that $s\_n\in G\_I$, and the inner automorphism defined by $c$ admits $s\_n$ as element of period exactly $n$.
|
11
|
https://mathoverflow.net/users/14094
|
411929
| 168,115 |
https://mathoverflow.net/questions/411939
|
5
|
I just started my first year of university and because I'm visually impared I have trouble seeing what's written on the chalkboard.
I've partially solved this problem by purchasing chalk from hagoromo and asking my professors to use it. (Leaves a wider and nicer line and erases more cleanly).
But the chalkboard still gets dirty from chalk which reduces the contrast between the board and the chalk, which in turn makes it harder for me to see.
My university supplies what look to be whiteboard erasers for the blackboard which I suspect isn't a good idea.
I'm looking for suggestions for premium chalkboard eraser brands or DIY options.
Thanks
EDIT: My university washes all the boards regardless of department every day at the end of the day.
(At least in my campus).
|
https://mathoverflow.net/users/472573
|
Chalkboard eraser
|
The Hagoromo chalk, now produced in South Korea, is well accompanied by a professional Korean microfibre eraser (600,000 fibers per square inch), as explained by professor Bayer on his [Chalk page.](https://www.math.columbia.edu/~bayer/LinearAlgebra/Video/Chalk.php)
For more on chalk, there is this older [MO post.](https://mathoverflow.net/q/26267/11260)
|
12
|
https://mathoverflow.net/users/11260
|
411940
| 168,118 |
https://mathoverflow.net/questions/411918
|
10
|
Let $R$ be the hyperfinite $II\_1$-factor. We know that $R$ is isomorphic to $R\otimes R$. So, $L\_\infty(0,1) \otimes R$ is a von Neumann subalgebra of $R$.
I am not sure whether it is sure for any type $II\_1$ von Neumann algebra $M$, i.e.,
is $L\_\infty(0,1) \otimes M$ a von Neumann subalgebra of $M$?
|
https://mathoverflow.net/users/91769
|
Tensor product of a von Neumann algebra and $L_\infty $
|
If $\mathbb F\_I$ denotes the free group on $I$ generators with $\lvert I \rvert > 1$, then $L^\infty(0, 1) \overline \otimes L \mathbb F\_I$ is not isomorphic to a von Neumann subalgebra of $L \mathbb F\_I$. For $\lvert I \rvert > \aleph\_0$ this is Corollary 6.4 in [S. Popa: Orthogonal pairs of subalgebras in finite von Neumann algebras, J. Op. Th. 9(1983), 253-268]. The general case $\lvert I \rvert > 1$ is Theorem 1 in [[N. Ozawa: Solid von Neumann algebras, Acta Math. 192 (2004), 111-117](https://doi.org/10.1007/BF02441087)].
|
11
|
https://mathoverflow.net/users/6460
|
411955
| 168,121 |
https://mathoverflow.net/questions/411960
|
2
|
$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Gal{Gal}$Let $ L $ be a cyclic Galois extension of $ \mathbb{Q} $ of degree $ 6 $. So $ G = \Gal(L/\mathbb{Q}) $ is a cyclic group of order $ 6 $. Then we have a homomorphism $ \phi : G \rightarrow \GL (L) $ defined by $ \phi(\sigma)(g) = \sigma(g) $ for all $ \sigma \in G $. Thus by representation theory we can consider $ L $ as a $\mathbb{Q}G $ module. Then $ L $ can be written as a direct-sum decomposition of $ r $ distinct irreducible submodules. What is the value of $ r $? And what are the irreducible components?
|
https://mathoverflow.net/users/215016
|
Irreducible components of a cyclic extension over $ \mathbb{Q} $
|
By the [normal basis theorem](https://en.wikipedia.org/wiki/Normal_basis#Group_representation_point_of_view), $L$ is the regular representation of $G$, i.e., $r = 6$ and the components are the $6$ distinct characters of $G$.
EDIT: As pointed out in the comments, particularly by [@FrançoisBrunault](https://mathoverflow.net/questions/411960/irreducible-components-of-a-cyclic-extension-over-mathbbq/411962#comment1055624_411962), that is $L \otimes\_{\mathbb Q} \mathbb C$ as a $\mathbb C[G]$-module; $L$ itself as a $\mathbb Q[G]$-module has $4$ summands, corresponding to the trivial and sign characters, and the two other pairs of complex conjugate characters.
|
3
|
https://mathoverflow.net/users/2383
|
411962
| 168,122 |
https://mathoverflow.net/questions/411967
|
3
|
We have the $j$-invariant defined as
I have that
$$
j(\tau)=\frac{1}{q}+\sum\_{k\geq 0}c\_kq^k,
$$
where $q=e^{-2\pi t}$ ($\tau=it$) and $c\_k\sim e^{4\pi\sqrt{k}}/(k^{3/4}\sqrt{2})$.
The inversion formula for the $j$-invariant is
$$
q=j^{-1}+\sum\_{k\geq 2}d\_kj^{-k}.
$$
Thus, I would like to know some upper bound or asymptotic formula for $d\_k$.
Any hint or reference?
|
https://mathoverflow.net/users/120084
|
Growth of the coefficients of the inversion of the $j$-invariant function
|
It's in the OEIS: <https://oeis.org/A066396>
There's a formula there that gives an approximation of the form (in your notation)
$$
d\_k \sim A \cdot (-1)^{k+1}\cdot B^k / k^{3/2}
$$
where $A\approx1943.54943\dots$
and $B\approx2311.3945621\dots\,$.
|
4
|
https://mathoverflow.net/users/11926
|
411969
| 168,127 |
https://mathoverflow.net/questions/411982
|
7
|
Given prime $p\ge 11$, $S$ is a subset of $\mathbb{Z}\_p\times\mathbb{Z}\_p$ with $3p-3$ elements. Prove: $S$ has a subset $T$ with $p-1$ elements, such that$\sum\_{x\in T}x\equiv (0,0)\pmod{p}$.
|
https://mathoverflow.net/users/472630
|
$p-1$ elements in $\mathbb{Z}_p\times\mathbb{Z}_p$ with a sum $(0,0)$
|
In this post a sum over sets means the additive-combinatorial sum, i.e. $\sum A\_i=\{a\_1+...+a\_i : a\_1 \in A\_1, ..., a\_i \in A\_i\}$.
**Lemma.** Let $(a\_1,a\_2,... ,a\_{2p−2})$ be a sequence of $2p−2$
elements of $\mathbb Z\_p$, where $p$ is a prime. Then either
* there exists a subsequence $A$ of length $p-1$, in which the sum of the elements equals zero, or
* $p$ elements take the same value in the sequence.
**Proof:** We may assume $a\_1 \leq a\_2 \leq ... \leq a\_{2p-2}$. If there are $p$ consecutive values, then we are in the second case, otherwise $a\_1 \neq a\_{p}$, $a\_2 \neq a\_{p+1}$, $...$, $a\_{p-1} \neq a\_{2p-2}$, and we can define the sets $B\_i=\{a\_i,a\_{i+p-1}\}$. By the Cauchy-Davenport theorem, $|\sum B\_i|-1 \geq \sum (|B\_i|-1)$ if $\sum B\_i$ is not the whole $\mathbb Z\_p$, but this is impossible because $\sum (|B\_i|-1) = p-1$. So $0\in ∑B\_i$ and it's possible to pick one element in each $B\_i$ to get a zero sum.
Now we prove the main theorem.
Let $X$ be a sequence where the number of occurences of $x$ is one less than the number of occurences of $x$ in $S$ as the first index, i.e. $|\{i:X\_i=x\}|=|\{y: (x,y)\in S\}|-1$. The sequence has length $2p-3$ if every element appears in $S$ as the first index, and at least $2p-2$ otherwise.
If $X$ has length $2p-3$, we may assume that the elements we choose have first indices $1,2,3...(p-1)$, and we try to find appropriate second indices. Let $Y\_i$ be the set of second indices of the elements of $S$ having first index $i$ ($i \neq 0$). We may assume $\sum |Y\_i| \geq 2p-2$, for otherwise we can choose all the elements $(0,x)$ ($x \neq 0$). Thus $\sum (|Y\_i|-1) \geq p-1$. By the Cauchy-Davenport theorem, $|\sum Y\_i|-1 \geq \sum (|Y\_i|-1)$ if $\sum Y\_i$ is not the whole $\mathbb Z\_p$, but this is impossible because $\sum (|Y\_i|-1) \geq p-1$. So $0\in ∑Y\_i$ and there is an appropriate choice of second indices.
If $X$ has length at least $2p-2$, we may find a $(p-1)$-subsequence $A$ in $X$ that sum to zero, or there are $p$ elements with the same first index (this case is trivial). The sequence $A$ is our choice of first indices, and the point is again trying to find appropriate second indices. Let $Y\_i$ be the set of **k-sums** of second indices of the elements of $S$ having first index $i$ ($i \in A$), where $k$ is the number of occurences of $i$ in $A$. As $k$ is smaller than the number of elements in $S$ having first index $i$ (by the definition of $X$), the set $Y\_i$ has size at least (number of occurences of $i$ in $A$)+1. Now $|\sum Y\_i|-1 \geq \sum (|Y\_i|-1)$ if $\sum Y\_i$ is not the whole $\mathbb{Z} \_p$, but this is impossible because $\sum (|Y\_i|-1) \geq |A| = p-1$. So $0\in ∑Y\_i$ and there is an appropriate choice of second indices.
|
5
|
https://mathoverflow.net/users/125498
|
411988
| 168,133 |
https://mathoverflow.net/questions/410855
|
1
|
I am working with an integral within the context of a Carleman estimate, and am trying to manifest its reality (with the later goal of finding a lower bound for $-S$ in the $L^2$ sense) but am having trouble. Although I believe the operator is symmetric from my calculations, there might be small errors, so I wanted to first ask the question of symmetricity of the operator $S$ that I will implicitly define below. If the answer to the question is yes, I am interested in how to manifest the reality of its corresponding $L^2$ weighted norm. For context, let $ f\in C\_0^\infty(\mathbb{R}\times[0,1])$ take values in $\mathbb{C}$, $\alpha\in\mathbb{R}$, $\phi\in C^\infty([0,1])$, take values in $\mathbb{R}$ and define the function $\psi(x,t)=x+\phi(t)$. I am interested in the following integral
$$
\begin{split}
\int f^\dagger Sf &:= \alpha \int 192\alpha^4 \psi^3 f^\dagger \partial\_x f + \alpha^248\psi^4 f^\dagger \partial\_ {xx}f+6i\alpha^2 \psi^2 \phi'(t)f^\dagger \partial\_x f - 48\alpha^2 \
\psi f^\dagger \partial\_{xxx}f\\
&\quad-12 \alpha^2 \psi^2 f^\dagger \partial\_{xxxx} f -12 \alpha^2 f^\dagger \partial\_{xx}f+(1/2)i\phi'(t)f^\dagger \partial\_{xxx}f + f^\dagger \partial^6\_{x} f \\
&\quad+ \frac{1}{4}f^\dagger f\left(48\alpha^2 \psi^2 - 256 \alpha^6 \psi^6 +24i \alpha^2 \psi \phi'(t) -\frac{1}{4}\psi\phi'' - \frac{1}{4} \phi \right),
\end{split}$$
where the integrals are computed over $[0,1]\times \mathbb{R}$, and the implicitly defined operator $S$ should be symmetric according to my calculations.
My questions are:
1. Is the operator $S$ defined above indeed symmetric?
2. If yes, how to manifest the reality of the integral? In other words, how to show that the terms containing the imaginary unit $i$ either, *e.g*., disappear or turn into the real/imaginary part of some expression.
Edit 1: I added a pre-factor of $\alpha$ inside the integral
Edit 2: I added a forgotten factor of $\alpha^2$ to the term $48\psi^4 f^\dagger \partial\_ {xx}f$
|
https://mathoverflow.net/users/152473
|
Is this operator symmetric and, if so, how to manifest the reality of its $L^2$ weighted norm?
|
*The small print refers to the original expression of the OP,
without the subsequent edits.*
Let me try something simple: $\alpha=0$, $\phi(t)\equiv 0$, then
$$I=\int\_{-\infty}^\infty dx\int\_0^1 dt\, f^\dagger Sf =\int\_{-\infty}^\infty dx\int\_0^1 dt\,\left(48x^4f^\dagger\frac{\partial^2 f}{\partial x^2}+f^\dagger\frac{\partial^6 f}{\partial x^6}\right),$$
and the question is whether this integral is real for a complex valued function $f(x,t)$.
The answer is no, for example, take $f(x,t)=e^{-x^2}(1+ix^2)$, then
$I=-\left(\frac{87}{4}+72 \,i\right) \sqrt{\frac{\pi }{2}}.$
---
Let me now consider the revised integral of the OP,
\begin{split}
I=\int f^\dagger Sf &:= \alpha \int 192\alpha^4 \psi^3 f^\dagger \partial\_x f + \alpha^2 48\psi^4 f^\dagger \partial\_ {xx}f+6i\alpha^2 \psi^2 \phi'(t)f^\dagger \partial\_x f - 48\alpha^2 \
\psi f^\dagger \partial\_{xxx}f\\
&\quad-12 \alpha^2 \psi^2 f^\dagger \partial\_{xxxx} f -12 \alpha^2 f^\dagger \partial\_{xx}f+(1/2)i\phi'(t)f^\dagger \partial\_{xxx}f + f^\dagger \partial^6\_{x} f \\
&\quad+ \frac{1}{4}f^\dagger f\left(48\alpha^2 \psi^2 - 256 \alpha^6 \psi^6 +24i \alpha^2 \psi \phi'(t) -\frac{1}{4}\psi\phi'' - \frac{1}{4} \phi \right),
\end{split}
To check whether this is real I consider order by order in $\alpha$,
$$I=\alpha I\_1 + \alpha^3 I\_3+\alpha^5 I\_5+\alpha^7 I\_7.$$
Each $\alpha$-independent term $I\_p$ should be separately real.
$\bullet$ Start with $I\_1$,
$$I\_1=\int\_{-\infty}^\infty dx\int\_0^1 dt\,\left(\tfrac{1}{2}i\phi'f^\dagger \frac{\partial^3 f}{\partial x^3} + f^\dagger \frac{\partial^6 f}{\partial x^6} -\tfrac{1}{16}(x+\phi)\phi''f^\dagger f - \tfrac{1}{16} \phi f^\dagger f\right).$$
Perform a partial integration with respect to $x$, keeping in mind that $\phi$ depends on $t$ only,
$$I\_1=\int\_{-\infty}^\infty dx\int\_0^1 dt\,\left(-\tfrac{1}{2}i\phi'f \frac{\partial^3 f^\dagger}{\partial x^3} + f \frac{\partial^6 f^\dagger}{\partial x^6} -\tfrac{1}{16}(x+\phi)\phi''f^\dagger f - \tfrac{1}{16} \phi f^\dagger f\right).$$
The second expression equals the complex conjugate of the first, hence $I\_1$ is real.
$\bullet$ Continue with $I\_3$,
$$I\_3=\int\_{-\infty}^\infty dx\int\_0^1 dt\,\left(12\left\{4(x+\phi)^4 f^\dagger \frac{\partial^2 f}{\partial x^2} - 4 \
(x+\phi) f^\dagger \frac{\partial^3 f}{\partial x^3}- (x+\phi)^2 f^\dagger \frac{\partial^4 f}{\partial x^4}\right\} +6\phi'\left[i (x+\phi)^2 f^\dagger \frac{\partial f}{\partial x}+ i(x+\phi) f^\dagger f\right]-12 f^\dagger \frac{\partial^2 f}{\partial x^2}+12 (x+\phi)^2f^\dagger f
\right).$$
If you now take the complex conjugate and do partial integrations with respect to $x$, you see that the last two terms are separately real, the sum of the two terms between square brackets is real, but the sum of the three terms between curly brackets is *not* real.
$\bullet$ Next is $I\_5$,
$$I\_5=192\int\_{-\infty}^\infty dx\int\_0^1 dt\, (x+\phi)^3 f^\dagger \frac{\partial f}{\partial x}.$$
Its complex conjugate is
$$\bar{I}\_5=-192\int\_{-\infty}^\infty dx\int\_0^1 dt\, \left[(x+\phi)^3 f^\dagger \frac{\partial f}{\partial x}+3(x+\phi)^2f^\dagger f\right].$$
This will in general be different from $I\_5$, so it is complex. The remaining term $I\_7$ has a real integrand $\propto (x+\phi)^6 f^\dagger f$, so it is real.
**Conclusion:** the corrected expression in the OP is still not real. One way to fix this is to add $I\_5$ to $I\_3$, so if instead of $192\alpha^4$ one would write $192\alpha^2 $ in the very first term of the integral.
|
1
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https://mathoverflow.net/users/11260
|
412001
| 168,137 |
https://mathoverflow.net/questions/411876
|
5
|
A Tzitzeica surface has the property that the ratio of the surface’s Gaussian curvature and the fourth power of the distance from the origin to the tangent plane at any arbitrary point of the surface is constant.
My question is: are there Tzitzeica surfaces with constant negative Gaussian curvature?
|
https://mathoverflow.net/users/111304
|
Tzitzeica surface
|
The answer is 'no'.
Suppose that $M\subset\mathbb{E}^3$ is a smooth connected surface. If the ratio of the Gauss curvature $K$ and $p^4$ is constant (where $p(x)$ is the distance from $T\_xM$ to the origin is constant) and $K$ is constant and nonzero, then $p$ is also constant. However, if $p$, which is known as the *support function* of the surface, is constant and the Gauss curvature $K$ is nonzero, then the surface is a portion of a sphere centered at the origin, and hence $K$ is a positive constant.
|
4
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https://mathoverflow.net/users/13972
|
412021
| 168,142 |
https://mathoverflow.net/questions/355732
|
3
|
Let's call a sequence $a\_1, \ldots, a\_n$ *suitable* if for any positive integer $d$ there is at most one index $i$ such that $a\_i = a\_{i + d}$ and all elements $a\_{i + 1}, \ldots, a\_{i + d - 1}$ are not equal to $a\_i$.
For each $k$, I'm interested in longest suitable sequences with all elements in $\{0, \ldots, k - 1\}$. There is a suitable sequence of length $3k - 1$: start with numbers $0, \ldots, k - 1$ in order, followed by first $2k - 1$ elements of [A025480](http://oeis.org/A025480). E.g., for $k = 3$ this sequence would look as follows: $0, 1, 2, 0, 0, 1, 0, 2$. It isn't difficult to prove that this pattern works for any $k$.
With brute-force I've discovered a few curious observations:
* $3k - 1$ appears to be the maximum length of a suitable sequence with elements in $\{0, \ldots, k - 1\}$;
* The number of longest suitable sequences appears to be $k! \times $[A002047](http://oeis.org/A002047)$[k]$.
How can this be explained?
|
https://mathoverflow.net/users/106512
|
Distinct distances between adjacent equal elements
|
This can be explained as follows.
Assume that $(a\_1,\ldots,a\_{n})$ is a suitable sequence.
For every $j\in \{0,\ldots,k-1\}$ denote $A(j)=\{i:a(i)=j\}$ and denote by $m(i)$ and $M(i)$ the minimal and maximal, respectively, elements of $A(j)$. Then $$\sum\_{j=0}^{k-1} \left(M(i)-m(i)\right)\geqslant 1+2+\ldots+(n-k),$$
since LHS equals to the sum of distances between adjacent equal elements, and there are at least $n-k$ such distances (exactly $n-k$ if $A(j)\ne \emptyset$ for all $j$). Since all guys $m(j)$ are distinct, their sum is at least $1+2+\ldots+k$, analogously the sum of $M(j)$ is at most $(n-k+1)+\ldots+n$, thus $$
\sum\_{j=0}^{k-1} M(i)-m(i)\leqslant (n-k+1)+\ldots+n-(1+\ldots+k)=k(n-k),$$
and we get $k(n-k)\geqslant 1+\ldots+n-k=(n-k)(n-k+1)/2$, or $n\leqslant 3k-1$.
Also we get that if $n=3k-1$ is the length of a suitable sequence $(a\_1,\ldots,a\_{3k-1})$, then this may be only possible if $a\_1,\ldots,a\_k$ are all distinct, and so are $a\_{2k-1},\ldots,a\_{3k-1}$. For $s=1,2,\ldots,2k-1$ denote by $f(i)$ the minimal positive number such that $a\_i=a\_{i+f(i)}$. Then $f(1),\ldots,f(2k-1)$ are well-defined and form a permutation of $1,\ldots,2k-1$. Also the numbers $i+f(i)$ must form a permutation of $k+1,\ldots,3k-1$. And if this all happens, we get $k!$ corresponding suitable sequences. It remains to note that then $(i-k,f(i)-k,2k-i-f(i))$ for $i=1,2,\ldots,2k-1$ are the columns of a $3\times (2k-1)$ zero-sum array (see definition at[A002074](http://oeis.org/A002047)), and viceversa.
|
1
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https://mathoverflow.net/users/4312
|
412027
| 168,144 |
https://mathoverflow.net/questions/411987
|
8
|
Suppose $U$ is a (possibly singular) scheme and $X$ is a compactification (potentially unnecessary at least in characteristic $0$). Let $\pi:X\to \*$ be the map to the point (though one can consider more general maps as well). There is a classically known pro-algebraic "compactly supported global sections" functor defined by Deligne in the appendix to Hartshorne's "Residues and Duality". Namely, one has a functor $$\pi\_!: \mathrm{Coh}\_U\to D^b \mathrm{ProFinVect},$$ given informally by taking a sheaf $F$ to the fiber of the map $$\Gamma(U, F)\to \Gamma(\mathring{\delta}\_X, F),$$
where $\mathring{\delta}\_X: = \widehat{X\setminus U}\cap U$ is the punctured formal boundary. (More explicitly, one defines a functor on coherent sheaves on $X$ and applies it to any continuation of $F$, or equivalently, to $j\_\*F$ as an ind-object. It's easy to see that this can be done fully inside the $\infty$-category of complexes of pro-coherent sheaves on $X$.)
This functor is described in the condensed language in Lecture 11 of Scholze and Clausen's [Lectures](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf), though here my understanding is limited (and the lecture works under a smoothness condition, which is surely unnecessary in defining $\pi\_!$).
Since $\pi\_!$ is valued in complexes of pro-vector spaces, one can define the dual contravariant functor $(\pi\_!)^\*:D^b Coh(U)\to D^b Vect$ (as a functor of $\infty$-categories). As $U$ varies, this forms a presheaf of "distributions on $F$" (which I think is a sheaf in general) which, for $X$ smooth, agrees with the Serre dualizing sheaf. Write $S\_X$ for the sheaf of distributions on the constant sheaf, i.e. (the sheafification of) $$S\_X:U\mapsto (\pi\_!^U)^\*(\mathcal{O}\_X).$$
I have three questions about this construction.
1. What is the name of the resulting complex for a general proper (and arbitrarily singular) $X$? I want to call it the dualizing complex, but have only seen that defined under some homological singularity restrictions on $X$ like the Gorenstein property.
2. Is there a way to interpret this construction in the condensed language (e.g., is this what would be called $\pi^!\mathcal{O}$ in the Clausen-Scholze [lectures](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf)? Is it clear that their construction works in the non-smooth context?)
3. The construction in the Deligne appendix only defines $\pi\_! F$ as a complex of pro-vector spaces, not pro-finite vector spaces. It is very easy to make it pro-finite-dimensional (and thus with an ind-finite dimensional dual) by rewriting his construction $\infty$-categorically, but I have never seen this done. Is there a reference to this?
|
https://mathoverflow.net/users/7108
|
Compactly supported sections of coherent sheaves and the dualizing complex
|
Isn't the dualizing complex defined in general in the proper case by taking applying the right adjoint of $\pi\_\ast$ to $k$? That's what I'll take as the definition anyways. The Gorenstein property just means that the dualizing complex is invertible.
It is true that this can be computed by the formula you write down, and one way to see this is as in our lecture notes. (We do define compactly supported cohomology without the smoothness assumption.) So yes, this can be interpreted in our setting.
I know only few references on those compactly supported cohomology groups, and I'm not aware of any that work $\infty$-categorically.
And as you guessed, you actually don't have to sheafify this construction, it's already a sheaf.
|
4
|
https://mathoverflow.net/users/6074
|
412028
| 168,145 |
https://mathoverflow.net/questions/412022
|
3
|
Let $P\equiv P(x) := \sum\_{|\alpha|\leq m} c\_\alpha\cdot x^\alpha$ be a real polynomial in $d$ variables of (total) degree $m$, where $d, m \in\mathbb{N}$ are fixed.
(I.e., the above sum ranges over all multiindices $\alpha=(i\_1, \ldots, i\_d)\in\mathbb{N}\_0^{\times d}$ of length $|\alpha|\equiv i\_1+\ldots + i\_d$ less than $m$.)
Denote by $P\_k = \sum\_{|\alpha|=m}c\_\alpha\cdot x^\alpha$, $ 0\leq k\leq m$, the $k$-th homogeneous component of $P$.
I was wondering the following: Given a compact subset $K$ in $\mathbb{R}^d$, is it possible to for $\varphi\_K(P):=\max\_{0\leq k\leq m}\|P\_k\|\_{\infty; K}$ (or indeed for any $\ell\_p$-norm of $(\|P\_k\|\_{\infty;K})\_{k\geq 0}$ with $1\leq p \leq \infty$) find a constant $\kappa=\kappa(d,K)$ such that
$$\tag{1} \varphi\_K(P) \ \leq \ \kappa\cdot \|P\|\_{\infty; K} \qquad \text{ for each } \ P \ \text{ as above} \ ?$$
(Here, $\|f\|\_{\infty;K}:=\sup\_{x\in K}|f(x)|$ is the uniform norm over $K$.) Any references are welcome.
|
https://mathoverflow.net/users/472548
|
Estimate the homogeneous components of a polynomial against its maximum
|
The answer is no. E.g., let $d=1$, $K=[0,1]$, and, for $x\in K$,
$$P(x):=T\_n(x):=n\sum\_{0\le k\le n/2}\frac{(-1)^k}{n-k}\binom{n-k}k2^{n-2k-1}x^{n-2k}
=\cos(n\arccos x),$$
the $n$th [Chebyshev polynomial](https://en.wikipedia.org/wiki/Chebyshev_polynomials#Explicit_expressions).
Then $\|P\|\_{\infty;K}\le1$, whereas (say) for $k=0$ we have $\|P\_k\|\_{\infty;K}=2^{n-1}\to\infty$ as $n\to\infty$.
|
4
|
https://mathoverflow.net/users/36721
|
412035
| 168,148 |
https://mathoverflow.net/questions/412053
|
7
|
I'm concerned about the group structure on $[X,S^n]$, i.e. the set of homotopy classes of continuous maps from $X$ to $S^n$.
On the one hand, $[X,Y]$ has a group structure that is natural with respect to $X$ if and only if $Y$ is an H-space. The naturality is in the sense that $f:X'\to X$ induces a homomorphism $f\_{\star}:[X,Y]\to[X',Y]$. It's known that $S^n$ is an H-space only for $n=0,1,3,7$. Thus, for a general $X$, there's no natural group structure on $[X,S^n]$ when $n$ takes other values.
On the other hand, when $X$ is a closed smooth $d$-manifold, the Pontryagin-Thom construction establishes a bijection between $[X,S^n]$ and $\Omega\_{d-n}^{fr}(X)$, i.e. the (unstable) $(d-n)$th framed bordism set of $X$. When $d<2n-1$, two transverse $(d-n)$-submanifolds in $X$ have no intersection and the disjoint union induces a group structure on $\Omega\_{d-n}^{fr}(X)$. Then we can make $[X,S^n]$ into a group via the Pontryagin-Thom bijection.
Hence $[X,S^n]$ is a group when $X$ is a closed smooth manifold whose dimension $<2n-1$. I would like to know how "natural" or how "unnatural" this group structure is: (1) if $X'$ is a closed smooth manifold whose dimension $<2n-1$, does a smooth map $f:X'\to X$ induce a group homomorphism? (2) When $n=0,1,3,7$, does this group structure coincide with the H-space induced group structure?
**Edit 1** The question has been edited to correct a mistake pointed out by Gregory.
**Edit 2** As mentioned by Tyrone, $[X,S^n]$ is a cohomotopy group when $X$ is a complex of dimension $<2n-1$, with the multiplication induced by the folding map. It reduces to the H-space induced group if $n=0,1,3,7$. My **Q2** essentially asks whether *cohomotopy groups* coincide with *framed bordism groups* (when the latter applies).
|
https://mathoverflow.net/users/472749
|
The group structure on $[X,S^n]$ induced by the framed bordism
|
This is an answer to question (2). Let $n=0,1,3,7$ and $i=1,2$ and $d\leq 2n-2$. Let $e=(1,0,\ldots,0)\in S^n$
Let $f\_i:X\rightarrow S^n$ be two maps representing framed submanifolds $(M\_i,\nu\_i)$. Let $T\_i$ be tubular neighborhoods of $M\_i$.
By the assumptions on the dimensions, the maps can be chosen in such a way that
* $T\_1\cap T\_2=\emptyset$.
* The value $-e\in S^n$ is regular for both $f\_1,f\_2$
* $f\_i^{-1}(\{-e\})=M\_i$ and the framing induced by the differential is $\nu\_i$.
* $f\_i(X\setminus T\_i)=e$.
Let $g:X\rightarrow S^n$ be the product map $g(x)=f\_1(x)f\_2(x)$. By the choices made above $-e$ is a regular value and $g^{-1}(\{-e\})=M\_1\cup M\_2$. The induced framing on $M\_1\cup M\_2$ is $\nu\_1,\nu\_2$ on the respective components. This shows that the group structures coincide.
|
9
|
https://mathoverflow.net/users/12156
|
412056
| 168,155 |
https://mathoverflow.net/questions/412059
|
8
|
Recall a topological space is called planar if it can be embedded in $S^2$. I'm interested in understanding hyperbolic groups with planar boundaries.
In [1], it is shown that if a one-ended hyperbolic group has 1-dimensional planar boundary, then this boundary is either a circle or a Sierpinski carpet. It is also well known that there exists hyperbolic groups with boundary homeomorphic to $S^2$, and moreover the only 0-dimensional such space is the Cantor set.
My question is: What other planar spaces arise as boundaries of hyperbolic groups? Does a nice "list" of these spaces exist, or is this a hopeless problem? What if we restrict ourselves to just one-ended groups?
Any references would be appreciated. Thanks!
---
**References**
[1] Kapovich, Michael, and Bruce Kleiner. "Hyperbolic groups with low-dimensional boundary." Annales scientifiques de l'Ecole normale supérieure. Vol. 33. No. 5. 2000.
|
https://mathoverflow.net/users/135406
|
For which planar topological spaces $Z$ does there exist a hyperbolic group $\Gamma$ with $\partial \Gamma \cong Z$?
|
There are many further examples with local cut points, which can be obtained by amalgams over $\mathbb{Z}$, as @YCor suggests in his comment.
Perhaps the easiest example is obtained by gluing *three* one-holed tori along their boundary. The resulting group $\Gamma$ is hyperbolic, and its boundary cannot be a Sierpinski carpet or $S^2$, since it has a local cut point, coming from either of the limit points of the amalgamating copy of $\mathbb{Z}$). On the other hand, that local cut point has valence 3, so the boundary also cannot be a circle.
An even richer family of examples can be constructed via the following construction. Take any simple closed curve $\gamma$ on the boundary of a handlebody $U$, and take the double
$\Gamma=\pi\_1(U)\*\_{\langle\gamma\rangle} \pi\_1(U)$ .
This will be a 3-manifold group, and hence its boundary will be planar.
By the Convergence Group Theorem of Tukia--Casson--Jungreis--Gabai, the boundary of $\Gamma$ is only a circle if $\Gamma$ is a surface group, which in turn only occurs if $\gamma$ was a ``surface element" of $\pi\_1(U)$. And the boundary won't be a Sierpinski carpet, because it has a local cut point.
It's complicated to describe the $\partial\Gamma$ for this construction in general, but you can make a start by looking at the papers of [Cashen--Macura](https://arxiv.org/abs/1006.2123) and [Cashen](https://arxiv.org/abs/1009.2492), who explain how to construct the *decomposition space* $D(\gamma)$. The cut-point structure of $D(\gamma)$ gives information about the cut-point structure of $\partial\Gamma$.
As far as I know, the problem of characterising all planar compacta that arise as boundaries hasn't been worked out, but I suspect it is tractable, and would make a very nice thesis problem, say. One would also want to make use of Bowditch's theorem, which explains how local cut points in the boundary correspond to cyclic splittings, and the Strong Accessibiltiy Theorem of Louder--Touikan, which asserts (under mild hypotheses) that hierarchies of cyclic splittings terminate.
|
8
|
https://mathoverflow.net/users/1463
|
412085
| 168,159 |
https://mathoverflow.net/questions/412091
|
11
|
Consider the generating function
$$
G\_n(x\_1,x\_2,\ldots,x\_n, t\_1,t\_2,\ldots,t\_n) =\sum\_{\lambda}s\_{\lambda}(x\_1,x\_2,\ldots, x\_n) t\_1^{\lambda\_1}t\_2^{\lambda\_2} \cdots t\_n^{\lambda\_n},
$$
where the sum is over all partitions $\lambda=(\lambda\_1, \lambda\_2,\ldots, \lambda\_n)$ and $s\_\lambda$ is a Schur polynomial.
For small $n$ such generating function is easy to find, for example for $n=2$ by direct calculation we have
$$
G\_2(x\_1,x\_2, t\_1,t\_2)=\sum\_{\lambda}s\_{\lambda}(x\_1,x\_2) t\_1^{\lambda\_1}t\_2^{\lambda\_2}=\frac{1}{(1-x\_1 t\_1)(1-x\_2 t\_1)(1-x\_1 x\_2 t\_1 t\_2)}.
$$
If we put $t\_1=t\_2=1$ then we come to well-known Littlewood identity
$$
\sum\_{\lambda}s\_{\lambda}(x\_1,x\_2)=\frac{1}{(1-x\_1)(1-x\_2)(1-x\_1 x\_2)}.
$$
**Question.** Is there any close expression for the generating function $G\_n$ for arbitrary $n?$
|
https://mathoverflow.net/users/40637
|
Generating function for Schur polynomials
|
This is done in my paper [The character generator of SU(*n*)](https://klein.mit.edu/~rstan/pubs/pubfiles/44.pdf). I believe there was an essentially the same previous MO question, but I am unable to find it.
|
15
|
https://mathoverflow.net/users/2807
|
412095
| 168,162 |
https://mathoverflow.net/questions/412070
|
1
|
If $H=(V,E)$ is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) and $\kappa\neq \emptyset$ is a cardinal, then a map $c:V\to \kappa$ is said to be a *colouring* if for every edge $e\in E$ with $|e|\geq 2$ the restriction $c\restriction\_e: e\to \kappa$ is not constant. The smallest cardinal $\kappa$ for which there is a colouring $c:V \to \kappa$ is said to be the *chromatic number* of $H$ and is denoted by $\chi(H)$.
The *dual* of the hypergraph $H=(V,E)$ is $H^\partial = (E, E\_V)$ where $$E\_V = \big\{S\subseteq E: \text{there is }v\_0\in V \text{ such that } S = \{e\in E: v\_0\in e\}\big\}.$$
**Question.** Given cardinals $\kappa, \lambda \geq 2$, is there always a hypergraph $H$ with $\chi(H) = \kappa$ and $\chi(H^\partial) = \lambda$?
|
https://mathoverflow.net/users/8628
|
Chromatic number and taking duals of hypergraphs
|
If I understand correctly, $\chi(H^\partial)$ is the least number of colours which suffice to colour the edges of $H$ so that each vertex is incident with edges of at least two different colours. And it seems to me that $H$ is isomorphic to $(H^\partial)^\partial$ provided that, for any vertices $x,y\in V(H)$, there is an edge $e\in E(H)$ such that $|e\cap\{x,y\}|=1$. I believe the answer is affirmative for all $\kappa,\lambda\ge2$. Each of the example hypergraphs is either an ordinary graph, or the dual of a graph, or the disjoint union of a graph and the dual of a graph.
If $\kappa\ge4$ and $\lambda=2$, let $H=K\_\kappa$ (the complete *graph* of order $\kappa$).
If $\kappa=2$ and $\lambda\ge4$, let $H=K\_\lambda^\partial$.
If $\kappa\ge4$ and $\lambda\ge4$, let $H=K\_\kappa\cup K\_\lambda^\partial$ (disjoint union).
If $\kappa=2$ and $\lambda=2$, let $H=C\_4$.
If $\kappa=3$ and $\lambda=3$, let $H=K\_3$.
If $\kappa=3$ and $\lambda=2$, let $H=K\_4-e$.
If $\kappa=2$ and $\lambda=3$, let $H=(K\_4-e)^\partial$.
If $\kappa\ge4$ and $\lambda=3$, let $H=K\_\kappa\cup K\_3$.
If $\kappa=3$ and $\lambda\ge4$, let $H=K\_3\cup K\_\lambda^\partial$.
|
2
|
https://mathoverflow.net/users/43266
|
412111
| 168,167 |
https://mathoverflow.net/questions/412113
|
9
|
$\DeclareMathOperator\CT{CT}$
Let $\CT\_t(f(t))$ denote the constant term of the Laurent polynomial of $f(t)$.
Define the two functions $F(x\_1,\dots,x\_n)$ and $G(y)$ by
$$F:=\prod\_{i=1}^nx\_i^{-1}(1-x\_i)^{-2}\prod\_{1\leq i<j\leq n}
(1-x\_i-x\_j)^{-1} \qquad \text{and} \qquad
G:=n!\cdot y^{-n}e^{(n+1)y+y^2/2}.$$
I like to ask:
>
> **QUESTION.** Is the following true? It would be great if there is a direct way to compare these two.
> $$\CT\_{x\_1}\CT\_{x\_2}\cdots\CT\_{x\_n}\left(F(x\_1,\dots,x\_n)\right)=\CT\_y\left(G(y)\right).$$
>
>
>
**NOTE 1.** The sequence on the right-hand side is available at the [OEIS as A301741](https://oeis.org/A301741) with an explicit evaluation.
**NOTE 2.** Incidentally, we also have (a consequence of Han's formula [proved here](https://arxiv.org/pdf/0808.0928.pdf))
$$\CT\_{x\_1}\CT\_{x\_2}\cdots\CT\_{x\_n}\left(F(x\_1,\dots,x\_n)\right)=
\sum\_{\lambda\vdash n}f^{\lambda}\prod\_{u\in\lambda}
\frac{(n+2)^{h\_u}+n^{h\_u}}{(n+2)^{h\_u}-n^{h\_u}};$$
where $h\_u$ is the *hook-length* of cell $u$ (in the Young diagram of $\lambda$) and $f^{\lambda}$ is the *number of Standard Young Tableau of shape $\lambda$* ([given by the hook-length formula](https://en.wikipedia.org/wiki/Hook_length_formula)).
**NOTE 3.** A cute analogue: let $f:=\prod\_{i=1}^nx\_i^{-1}(1-x\_i)^{-1}\prod\_{1\leq i<j\leq n}(1+x\_i+x\_j)$ and $g:=n!\cdot y^{-n}e^{ny-y^2/2}$. Then,
$$\CT\_{x\_1}\CT\_{x\_2}\cdots\CT\_{x\_n}\left(f(x\_1,\dots,x\_n)\right)
=\CT\_y(g(y)).$$
**Proof.** Fedor's reasoning applies (it'd be nice to employ Richard's too)
\begin{align\*}
{\rm CT}\, f&=
[x\_1\ldots x\_n] \prod\_i (1-x\_i)^{-1}\prod\_{i<j}(1+x\_i+x\_j) \\
&=[x\_1\ldots x\_n]\prod\_i\exp(x\_i)\prod\_{i<j}\exp(x\_i+x\_j-x\_ix\_j)\\
&=[x\_1\ldots x\_n] \exp\left(\sum x\_i+\sum\_{i<j}(x\_i+x\_j-x\_ix\_j)\right)\\
&=[x\_1\ldots x\_n]\exp\left(n\cdot S-S^2/2\right).
\end{align\*}
|
https://mathoverflow.net/users/66131
|
Extracting constant terms: is there a direct way?
|
For power series $u(x\_1,\ldots,x\_n),v(x\_1,\ldots,x\_n)$ call $u,v$ similar and write $u\sim v$ if all monomials $\prod x\_i^{c\_i}$ with $c\_i\in \{0,1\}$ have equal coefficients in $u,v$. In other words, if $u$ is congruent to $v$ modulo the ideal generated by $x\_i^2$'s. Note that this similarity respects addition and multiplucation, and that $(1-x\_i)^{-1}\sim \exp(x\_i)$ and $(1-x\_i-x\_j)^{-1}\sim 1+x\_i+x\_j+2x\_ix\_j\sim\exp(x\_i+x\_j+x\_ix\_j)$. Thus
\begin{align\*}
{\rm CT}\, F&=
[x\_1\ldots x\_n] \prod\_i (1-x\_i)^{-2}\prod\_{i<j}(1-x\_i-x\_j)^{-1}\\&=
[x\_1\ldots x\_n]\prod\_i\exp(2x\_i)\prod\_{i<j}\exp(x\_i+x\_j+x\_ix\_j)\\&=[x\_1\ldots x\_n] \exp\left(
\sum 2x\_i+\sum\_{i<j}(x\_i+x\_j+x\_ix\_j)
\right)\\
&=[x\_1\ldots x\_n]\exp\left((n+1)S+S^2/2\right),
\end{align\*}
where $S=x\_1+\ldots+x\_n$ (since $S^2/2\sim \sum\_{i<j} x\_ix\_j$). Now if we expand $\exp\left((n+1)S+S^2/2\right)$ as a power series in $S$, we get $[x\_1\ldots x\_n]S^n=n!$ and $[x\_1\ldots x\_n]S^k=0$ for $k\ne n$, thus your identity.
|
16
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https://mathoverflow.net/users/4312
|
412119
| 168,169 |
https://mathoverflow.net/questions/412123
|
2
|
Consider the adjacency matrix $\mathbf{A}$ of a one dimensional lattice of size $N$. That is, $A$ is a $N\times N$ matrix with $A\_{ij}=1$ if vertex $i$ adjacent to vertex $j$ (there exists an edge between $i$ and $j$). Whether there are boundary conditions or not (1D line or a closed ring), the spetrum of $A$ is very well known: there exists an analytic formula for both its eigenvalues and eigenvectors for any finite $N$.
Furthermore, let $\mathbf{w}$ be a random matrix of size $N\times N$ such that $\left[\mathbf{w}\right]\_{ij}=w\_{ij}$ is normally distributed with mean $0$ and variance $1/N$. As $N\to\infty$, the spectrum of $\mathbf{w}$ is also well known: if $\mathbf{w}$ has symmetric entries then its eigenvalues will follow the wigner semi-circle distribution. I am not sure about the distribution of the eigenvectors.
Now, let $\mathbf{B}$ be a $N\times N$ matrix such that $B\_{ij}=w\_{ij}A\_{ij}$, i.e $\mathbf{B}$ represents the 1D lattice with random weights between links.
Here are a few questions I had:
1. Is the distribution of the eigenvalues known?
2. What will be the joint distribution of eigenvalues?
3. What will be the joint distribution of eigenvectors?
4. In the wigner case, eigenvectors and eigenvalues are independent, is it still the case for $\mathbf{B}$?
|
https://mathoverflow.net/users/420641
|
Are the eigenvalues of the 1D lattice with random weights known?
|
I presume this will depend on the connectivity of the 1D lattice. Let me consider the simple case of nearest neighbor connections, when the adjacency matrix $A$ is tridiagonal with the same values on each diagonal. The statistics of the matrix $B$ when the nonzero elements are i.i.d. random variables was studied in [General Tridiagonal Random Matrix Models, Limiting Distributions and Fluctuations](https://arxiv.org/abs/math/0610827) (2006).
|
1
|
https://mathoverflow.net/users/11260
|
412126
| 168,171 |
https://mathoverflow.net/questions/411947
|
2
|
My question arose from the proof of Proposition 31.7 of "The book of involutions."
It says "… is the Tits algebra of the quasisplit group $(G\_{\nu\_G})\_{F\_{\chi}}$, hence it is trivial." I understood every part of the proof but this sentence. I guess this sentence is true due to the positive answer to the following question.
Let $G$ be a quasi-split simple connected semisimple algebraic group over a field $F$ and $\Gamma=\operatorname{Gal}(F\_\text{sep}/F)$. If $\chi$ is a $\Gamma$-invariant character of the center of $G$, is the minimal Tits algebra $A\_{\chi}$ of $G$ split?
I didn't look closely at the construction of Tits algebra but I'm just taking the notion of Tits algebra as a Galois descent of $\operatorname{End}\_L(V)$, where $L$ is a splitting field of $G$ and $V$ is the minimal representation of $G\_L$ corresponding to $\chi$. From this, however, I cannot see why the Tits algebra of a quasisplit group is split. Is it true? If so, how can I show this?
|
https://mathoverflow.net/users/304053
|
Tits algebra of the quasi-split semisimple algebraic groups
|
I found and read the original paper ***<Représentations linéaires irréductibles d'un groupe réductif sur un corps
quelconque>*** by Jacques Tits.
In Theorem 3.3 of this paper, it is shown that the Tits algebra of a quasi-split semisimple algebraic group $G$ is split.
|
2
|
https://mathoverflow.net/users/304053
|
412138
| 168,175 |
https://mathoverflow.net/questions/412148
|
2
|
Given matrix $X \in \mathbb{R}^{m\times n}$ and sequence $\left\{X^k\right\}\_k$ converges to $X$ according to the Frobenius norm. I wonder that $\sigma\_i(X^k)$ converge $\sigma\_i(X)$ or not (where $\sigma\_i(X)$ is singular values of $X$)?
|
https://mathoverflow.net/users/472818
|
Limitation through the singular values
|
Yes, one can prove that
$$
|\sigma\_i(A+E) - \sigma\_i(A)| \leq \|E\| \quad \quad \forall i
$$
which implies that the singular values are continuous. This follows, for instance, from Weyl's inequalities.
See also <https://math.stackexchange.com/q/2783345/65548> , which links to a proof of these inequalities by Qiaochu Yuan.
|
2
|
https://mathoverflow.net/users/1898
|
412150
| 168,178 |
https://mathoverflow.net/questions/412103
|
13
|
Suppose $G$ is a group. Consider the set $G^G$ of all functions $G \to G$, which forms a group under elementwise multiplication. Now, for all $g \in G$ let’s define $c\_g \in G^G$ as the constant function $c\_g(x) \equiv g$, and $id \in G^G$, as the identity map $id(x) = x$. Now, consider the subgroup $E(G) = \langle \{c\_g | g \in G\} \cup \{id\} \rangle$.
$E(G)$ preserves several “finiteness” properties of $G$:
>
> If $G$ is finite then $E(G)$ is also finite.
>
>
>
*Proof*: $|E(G)| \leq |G^G| = |G|^{|G|}$
>
> If $G$ is finitely generated then $E(G)$ is also finitely generated.
>
>
>
*Proof*: If $A$ is a generating set of $G$, then $\{c\_g | g \in A\} \cup \{id\}$ is a generating set of $E(G)$.
>
> If $G$ is residually finite then $E(G)$ is also residually finite.
>
>
>
*Proof*: Consider the following class of maps $\pi\_g: E(G) \to G, f \mapsto f(g)$ for all $g \in G$. All $\pi\_g$ are homomorphisms and each non-trivial element of $E(G)$, maps to a non-zero element of $G$ under some of $\pi\_g$. The rest follows from residual finiteness of $G$.
However, there is also a fourth “finiteness” property I am interested in but do not know how to deal with:
>
> If $G$ is finitely presented, does that mean that $E(G)$ is also finitely presented?
>
>
>
I suspect, it should be, but have no idea how to prove it.
[This question on MSE](https://math.stackexchange.com/q/4330034/407165)
|
https://mathoverflow.net/users/110691
|
Does this group construction preserve finite presentability?
|
It seems the answer is no, $E(G)$ can fail to be finitely presented even if $G$ is finitely presented.
I claim that a counterexample is given by $G=F\_2\times F\_2$.
First, as @SeanEberhard explains in the comments, $E(F\_2\times F\_2)$ is isomorphic to the subgroup $H$ of $F\_3\times F\_3$ generated by $\{(x,1),(y,1),(1,x),(1,y),(z,z)\}$, where $\{x,y,z\}$ is a generating set for $F\_3$. Now I claim that $H$ is not finitely presented.
First I claim that $H$ contains the commutator subgroup of $F\_3\times F\_3$. Any simple commutator in $F\_3\times F\_3$ is an ordered pair of simple commutators in $F\_3$, so it suffices to show that any $(g,1)$ for $g$ a simple commutator in $F\_3$ lies in $H$ (and by a parallel argument also $(1,g)$). This is easy to see because given any word in $(x,1),(y,1),(z,1)$ (and their inverses) representing an element of $[F\_3,F\_3]\times\{1\}$, we can replace each instance of $(z,1)$ with $(z,z)$ (and $(z^{-1},1)$ with $(z^{-1},z^{-1})$) and get a word representing the same element of $[F\_3,F\_3]\times\{1\}$ but now lying in $H$.
Now that we know $H$ contains the commutator subgroup, we can use the Bieri-Neumann-Strebel-Renz invariant $\Sigma^2(F\_3\times F\_3)$ to conclude that $H$ is not finitely presented. Indeed, $H$ lies in the kernel of the homomorphism $\chi\colon F\_3\times F\_3 \to \mathbb{Z}$ sending $(z,1)$ to $1$, $(1,z)$ to $-1$, and all of $(x,1),(y,1),(1,x),(1,y)$ to $0$ (I guess it actually equals this kernel), and the computation of $\Sigma^2(F\_3\times F\_3)$ (see, e.g., [Bux-Gonzalez](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/S0024610799007991)) reveals that $[\chi]$ is not in $\Sigma^2(F\_3\times F\_3)$, since the "dead" edge from $(x,1)$ to $(1,x)$ in the defining flag complex has empty "living link". (Actually, I think $\Sigma^2(F\_3\times F\_3)$ is empty, so probably we didn't even need to isolate a particular $\chi$, and in fact every infinite index subgroup of $F\_3\times F\_3$ containing the commutator subgroup fails to be finitely presented, but I'll just leave this analysis as-is.)
---
Edit: By request and for the sake of being self-contained, here is the proof that $E(F\_2\times F\_2)\cong H$, compiled from @SeanEberhard's comments. Here $\{x,y,z\}$ is a free basis of $F\_3$ and $H$ is the subgroup of $F\_3\times F\_3$ generated by $\{(x,1),(y,1),(1,x),(1,y),(z,z)\}$.
Let $\{a,b\}$ be a free basis of $F\_2$. We have an epimorphism $\phi\colon F\_3\to E(F\_2)$ given by sending $x$ to $c\_a$, $y$ to $c\_b$, and $z$ to $id\_{F\_2}$, and we claim it is injective (hence an isomorphism). Indeed, given any non-empty reduced word in $c\_a$, $c\_b$, and $id\_{F\_2}$ (and their inverses) representing an element $f$ of $E(F\_2)$, since $F\_2$ is mixed identity-free, we can evaluate $f$ at some element of $F\_2$ to get a non-trivial element of $F\_2$, which means $f$ is a non-trivial element of $E(F\_2)$. Hence $F\_3 \cong E(F\_2)$. This also shows $F\_3\times F\_3 \cong E(F\_2)\times E(F\_2)$. Let us identify $H$ with its isomorphic image in $E(F\_2)\times E(F\_2)$, generated by $\{(c\_a,c\_1),(c\_b,c\_1),(c\_1,c\_a),(c\_1,c\_b),(id\_{F\_2},id\_{F\_2})\}$.
Now consider the monomorphism $\psi \colon F\_2^{F\_2} \times F\_2^{F\_2} \to (F\_2\times F\_2)^{F\_2\times F\_2}$ given by sending $(f,g)$ to the function $f\times g$ defined by $(f\times g)(w,v):=(f(w),g(v))$. The restriction of $\psi$ to the subgroup $H$ is an isomorphism onto its image, which is generated by $c\_{(a,1)}$, $c\_{(b,1)}$, $c\_{(1,a)}$, $c\_{(1,b)}$, and $id\_{F\_2\times F\_2}$, hence is $E(F\_2\times F\_2)$ as desired.
|
13
|
https://mathoverflow.net/users/164670
|
412154
| 168,180 |
https://mathoverflow.net/questions/412167
|
8
|
When reading through [Loregian and Riehl - Categorical notions of fibration](https://arxiv.org/abs/1806.06129), on p. 3 there is a remark that confuses me about notation. Given a $2$-category $\mathcal C$ one usually defines $\mathcal C^\text{op}$ to be the category $\mathcal C$ but with inverted $1$-cells and $\mathcal C^\text{co}$ to be the category $\mathcal C$ but with inverted $2$-cells.
Many constructions like for example limits and colimits seem to be named according to the co-notation but are mathematically op-dual. Similarly in the example above.
Hence the question is, whether there are any interesting constructions that are named “correctly” and also whether I am even correct about this feeling about things being switched up here.
|
https://mathoverflow.net/users/145805
|
Why aren‘t op and co switched?
|
The problem is that for a long time there were only 1-categories and hence only one kind of duality, and sometimes people called it "op" and sometimes "co". Colimits and cofibrations were therefore the only possible notion of dual for limits and fibrations, and take place in the opposite category, which was also the only notion of dual for a category, and denoted "op" as being short for "opposite". Sometimes both "op" and "co" terminologies are in use for the same thing, e.g. "oplax" and "colax".
One should not attempt to remedy this by departing from the standard definitions of $C^{\rm op}$ and $C^{\rm co}$ for 2-categories. Among other reasons, they align with standard notions of duality for enriched categories: for a general $V$ there is only one notion of opposite $V$-category, denoted $C^{\rm op}$, and when a 2-category is regarded as a $\rm Cat$-enriched category this produces the 1-cell dual.
I think the most satisfying perspective on this (though not fully satisfying) is that the 2-cell dual *of* a 2-category, $C^{\rm co}$, corresponds to classical dual constructions (often denoted by "co") *on* its objects, regarded as generalized categories. For instance, while a comonad *on* a category $A$ is equivalently a monad on the category $A^{\rm op}$, a comonad *in* a 2-category $C$ (on an object $A\in C$) is equivalently a monad in $C^{\rm co}$. (A monad in $C^{\rm op}$ is just the same as a monad in $C$.)
|
16
|
https://mathoverflow.net/users/49
|
412169
| 168,182 |
https://mathoverflow.net/questions/412137
|
3
|
**EDIT**: The definition of a Suslin measurable set I wrote here is incorrect. It should be that $\mathcal{S}$ contains the [*field*](https://en.wikipedia.org/wiki/Field_of_sets) (or algebra) of open subsets of ${}^\omega\omega$ (or, in other words, it contains all open and closed subsets of ${}^\omega\omega$). I do not intend to re-ask this question, so the question and answers here refer to the unchanged definition below.
---
Recall that a *Suslin scheme* is a set of subsets of reals ${}^\omega\omega$ of the form:
$$
\langle X\_s : s \in {}^{<\omega}\omega\rangle
$$
and that the *Suslin operation* $\mathcal{A}$ is an operation that takes a Suslin scheme $\mathcal{X} := \langle X\_s : s \in {}^{<\omega}\omega\rangle$ and yield:
$$
\mathcal{A}(\mathcal{X}) := \bigcap\_{a \in {}^\omega\omega}\bigcup\_{n \in \omega} X\_{a\upharpoonright n}
$$
The set of all **Suslin measurable sets**, call it $\mathcal{S}$, is the smallest set of subsets of reals such that:
* $\mathcal{S}$ contains all open subsets of ${}^\omega\omega$.
* $\mathcal{S}$ is closed under the Suslin operation (i.e. if $\mathcal{X}$ is a Suslin scheme in which $X\_s \in \mathcal{S}$ for all $s$, then $\mathcal{A}(\mathcal{X}) \in \mathcal{S}$).
---
A result of Nikodym says that the set of Baire subsets of reals is closed under the Suslin operation (Corollary 4.8 of Todorcevic's *Introduction to Ramsey spaces*). Thus, every Suslin measurable subset of reals has the Baire property. The questions I have are:
1. Can we prove, in $\mathsf{ZFC}$ and $\mathsf{ZF}$, that there exists a Baire subset of reals that is not Suslin measurable?
2. If it is not provable in $\mathsf{ZF}$, is there a well-known model of set theory in which every subset of reals is Suslin measurable?
(Side question: There seems to be very little literature that discusses Suslin measurable sets. Is there another term for such sets?)
**EDIT**: To clarify, a subset $X$ of real is Baire (or has the Baire property) if $X = U \, \triangle \, M$, where $U$ is open and $M$ is meagre.
|
https://mathoverflow.net/users/146831
|
A Baire subset of reals that is not Suslin measurable
|
While Gabe's answer is deleted ("One shouldn't try to work in ZF at 5am"), let me work in ZFC.
(a) The usual middle-thirds Cantor set $C$ is nowhere dense in $\mathbb R$. It has cardinal $\mathfrak c = 2^{\aleph\_0}$. So every subset of $C$ is again nowhere dense in $\mathbb R$. Every subset of $C$ has the property of Baire. There are at least $2^{\mathfrak c}$ sets with the property of Baire. [In fact, there are exactly $2^{\mathfrak c}$ sets with the property of Baire.]
(b) Start with any family $\mathscr U$ of sets, and define
$$
\mathcal A(\mathscr U) = \{ \mathcal A(\mathcal X) :
\mathcal{X} = \langle X\_s : s \in {}^{<\omega}\omega\rangle , X\_s \in \mathscr U \text{ for all } s \in {}^{<\omega}\omega\} .
$$
The Suslin operation is idempotent. That is, if $\mathscr U$
is any family of sets, then $\mathcal A(\mathcal A(\mathscr U)) = \mathcal A(\mathscr U)$.
The family of "Suslin measurable sets" (a.k.a. coanalytic sets) is
$\mathcal A(\mathscr G)$, where $\mathcal G$ is the family of all open subsets of $\mathbb R$.
(c) Let $\mathscr G\_0 = \{(a,b) : a,b\in\mathbb Q, a<b\}$. Then
$\mathcal A(\mathscr G) = \mathcal A(\mathscr G\_0)$. Since $\mathscr G\_0$ is countable and ${}^{<\omega}\omega$ is countable, there are at most $\aleph\_0 ^{\aleph\_0} = \mathfrak c$ Souslin schemes in $\mathscr G\_0$. We conclude that $\mathcal A(\mathscr G\_0)$ has cardinal at most $\mathfrak c$. So there are at most $\mathfrak c$ Suslin measurable sets in $\mathbb R$. [In fact, there are exactly $\mathfrak c$ Suslin measurable sets.]
(d) Conclude there is a set with the property of Baire that is not Suslin measurable.
|
7
|
https://mathoverflow.net/users/454
|
412171
| 168,184 |
https://mathoverflow.net/questions/412159
|
3
|
Let $G:=(E,V,W)$ be a weighted graph and let $d\_G$ be its graph metric, defined by on any two edges $e\_1,e\_2\in E$ by
$$
d\_G(e\_1,e\_2)\triangleq
\inf\_{\gamma}\, \sum\_{v\in \gamma} W(v),\qquad\tag{0}\label{0}
$$
where the infimum is taken over all sequences of vertices $\gamma=(v\_1,\dots,v\_t)$ connecting $e\_1$ to $e\_2$; where $W:V\rightarrow \mathbb{R}$ is the *weight function* defined on unweighted graph $(E,V)$.
In Ballmann's book [Lectures on spaces of nonpositive curvature](https://people.mpim-bonn.mpg.de/hwbllmnn/archiv/NPC0606.pdf) conditions are given which guarantee that $d\_G$ has non-positive curvature (*[in the sense of Alexandrov](https://en.wikipedia.org/wiki/Alexandrov_space)*); namely, there is some $c>0$ satisfying:
$$
\inf\_{v}\, W(v)\geq c.\tag{1}\label{1}
$$
My question is, when is $(E,d\_G)$ a $\operatorname{CAT}(\kappa)$ space, for some $\kappa\in \mathbb{R}$?
---
My hypothesis is that there exist some partition $C\_1,\dotsc,C\_N$ of $E$ into open balls such that \eqref{1} holds; where the infimum in \eqref{0} is taken over all paths $\gamma$ contained in $C\_i$; where $e\_1,e\_2\in C\_i$.
|
https://mathoverflow.net/users/469470
|
When is a graph a $\operatorname{CAT}(\kappa)$ space?
|
The generalized Cartan--Hadamard theorem states that a length space is CAT(1) if it is locally CAT(1) + any closed curve of length $<2{\cdot}\pi$ is null-homotopic in class of curves of length $<2{\cdot}\pi$.
If the space is a graph, then the latter is equivalent to saying that any cycle has length at least $2{\cdot}\pi$.
|
3
|
https://mathoverflow.net/users/1441
|
412175
| 168,185 |
https://mathoverflow.net/questions/412163
|
5
|
[This](https://ncatlab.org/nlab/show/Lawvere%27s+fixed+point+theorem#precise_statement) proof of Lawvere's fixed point theorem suggests (since it uses $\lambda$ notation) that it is written in the *internal language* of cartesian closed categories (which is the $\lambda$-calculus, as explained e.g. in Part I of Lambek and Scott's book *Introduction to higher order categorical logic*). However, one needs quantifiers in order to formulate the notion of *point-surjective* and the *existence* of a fixed point $s\colon 1\to B$. So strictly speaking, the proof can't take place in the internal language, since the $\lambda$-calculus doesn't have quantifiers and assumptions. (Higher-order logic admits quantifiers, but that is only available in elementary toposes and not in cartesian closed categories in general.)
**Question:** Is there some way of making the proof formally work in some "internal language" of cartesian closed categories? Or is the $\lambda$ notation used in the proof just an informal explanation of the proof rather than an indication for the use of the internal language?
|
https://mathoverflow.net/users/471475
|
Internal language proof of Lawvere's fixed point theorem for cartesian closed categories
|
You're right that the statement of the theorem, and the entirety of the proof, don't fit inside the internal logic of a CCC. However, once given $f:B\to B$, the definition of $q$ and the proof that it is a fixed point of $f$ can take place inside that internal language.
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6
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https://mathoverflow.net/users/49
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412177
| 168,186 |
https://mathoverflow.net/questions/412158
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4
|
The notion of a [group object](https://en.wikipedia.org/wiki/Group_object) $G$ in a category with finite products can *either* be defined with [a few commutative diagrams](https://ncatlab.org/nlab/show/group+object#in_terms_of_internal_group_objects) or via requiring that each hom set $\hom(X,G)$ is a group. There is a theorem that these two definitions are equivalent.
**Question**: Does the equivalence of these two definitions follow from [Kripke–Joyal semantics](https://ncatlab.org/nlab/show/Kripke-Joyal+semantics#properties) (Theorem 3.2)?
After all, Kripke–Joyal semantics gives a characterization of truth in the internal language of a topos using *generalized elements*. (And the second definitions renders as "for each $X$, the generalized elements with stage $X$ constitute a group".)
|
https://mathoverflow.net/users/471475
|
Group objects via diagrams or generalized elements — Kripke–Joyal?
|
Sort of. Traditional Kripke–Joyal semantics can only force the truth of predicates, not the existence of structure. So once you have multiplication, unit, and inversion maps, you could use KJ semantics to prove that the necessary diagrams commute if and only if they render each homset into a group. (At least, once you generalize KJ semantics to categories with finite products — I'm not sure I've seen it written down in more generality than categories with finite limits.)
However, Awodey, Gambino, and Hazratpour have recently generalized KJ semantics to dependent type theory ([Kripke–Joyal forcing for type theory and uniform fibrations](https://arxiv.org/abs/2110.14576)), and you might be able to use that version to deal with the algebraic structure as well.
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5
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https://mathoverflow.net/users/49
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412178
| 168,187 |
https://mathoverflow.net/questions/412157
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5
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$\DeclareMathOperator\Lip{Lip}\DeclareMathOperator\AE{AE}$**Background**
*Gelfand triples.* Let $\mathcal B$ be a Banach space, $\mathcal B^\*$ its dual space, and $\mathcal H$ a Hilbert space. The triple $(\mathcal B,\mathcal H, \mathcal B^\*)$ is called a Gelfand triple if the following embeddings are continuous
$$
\mathcal B \hookrightarrow \mathcal H \hookrightarrow \mathcal B^\*.
$$
An example that I am familiar with is the triple $(BV(\Omega), L^2(\Omega), BV^\*(\Omega))$, where $\Omega \subset \mathbb R^2$ is bounded and $BV(\Omega)$ is the space of functions of bounded variation.
*Arens-Eells space.* Let $X$ be a compact pointed metric space with base point $e$. An *elementary molecule* is defined as follows (Nik Weaver, Lipschitz Algebras, 2nd ed.)
$$
m\_{pq} := \delta\_p - \delta\_q,
$$
where $\delta\_p, \delta\_q$ are delta-functions placed at $p,q$.
The Arens-Eells space $\AE(X)$ (also known as the Lipschitz-free space) is the completion of the linear span of elementary molecules with respect to the Arens-Eells norm
$$
\|{m}\|\_{\AE} := \inf \left\{\sum\_{i=1}^n |{a\_i}| d(p\_i,q\_i) \colon m = \sum\_{i=1}^n a\_i m\_{p\_iq\_i} \right\},
$$
where $d(p,q)$ is the distance between $p,q \in X$.
The dual of the Arens-Eells space is the $\Lip\_0(X)$ space of all Lipschitz functions on $X$ vanishing at $e$, equipped with the following norm
$$
\|f\|\_{\Lip\_0} := \Lip(f),
$$
where $\Lip(f)$ denotes the Lipschitz constant.
**Question**
Is there a Hilbert space $\mathcal H$ such that $(\AE(X), \mathcal H, \Lip\_0(X))$ form a Gelfand triple? It would suffice for me to think of $X$ as the unit ball in $\mathbb R^n$ with base point $0$, equipped with the euclidean metric.
|
https://mathoverflow.net/users/160379
|
Existence of a Gelfand triple involving the Arens–Eells space (aka Lipschitz free space)
|
Your question is equivalent to asking if there is an injective bounded linear operator from $AE(X)$ into a Hilbert space when $X$ is a compact metric space. The answer is "yes" because $AE(X)$ is separable. It is elementary to construct a nuclear injective linear operator from an arbitrary separable Banach space into a Hilbert space.
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6
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https://mathoverflow.net/users/2554
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412186
| 168,192 |
https://mathoverflow.net/questions/412067
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7
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A topological space is said to be quasi-Polish if it is second-countable and completely quasi-metrizable (see for an introduction de Brecht's article: *de Brecht, Matthew*, [**Quasi-Polish spaces**](http://dx.doi.org/10.1016/j.apal.2012.11.001), Ann. Pure Appl. Logic 164, No. 3, 356-381 (2013). [ZBL1270.03086](https://zbmath.org/?q=an:1270.03086).) ).
These spaces have been introduced because, among other things, they allow the generalization of some results of classical descriptive set theory in non-Hausdorff environments. Now my question is:
* An Hausdorff quasi-Polish space needs to be Polish? Do we have a counter-example of an Hausdorff quasi-Polish space which is *not* Polish?
A strictly related [question](https://mathoverflow.net/questions/221405/hausdorff-open-image-of-a-polish-space), which asks basically the same thing, but is a bit old, so maybe there are new results that came up since then. Moreover, as the OP's comment points out, the question contained in Tkachuk's book is assuming all spaced Tychonoff and I'm not.
Thanks!
|
https://mathoverflow.net/users/141146
|
Hausdorff quasi-Polish spaces
|
I hope that the following space $P\mathbb Q^\omega$ is second-countable and quasi-Polish but not Polish.
Let $\mathbb Q$ be the field of rational numbners endowed with the discrete topology. Then its countable power $\mathbb Q^\omega$ is a Polish vector space over the field $\mathbb Q$. Let $P\mathbb Q^\omega$ be the projective space of $\mathbb Q$. So, $P\mathbb Q^\omega$ is the quotient space of $\mathbb Q^\omega\_\circ=\mathbb Q^\omega\setminus\{0\}^\omega$ by the equivalence relation $\sim$ defined as $x\sim y$ iff $\mathbb Qx=\mathbb Qy$. Is is easy to see that the quotient map $q:\mathbb Q^\omega\_\circ\to P\mathbb Q^\omega$ is open, which implies that the space $P\mathbb Q^\omega$ is second-countable. It is known that the space $P\mathbb Q^\omega$ is Hausdorff but not Urysohn (moreover, $P\mathbb Q^\omega$ is superconnected in the sense that for any nonempty open sets $U\_1,\dots,U\_n$ in $P\mathbb Q^\omega$ the intersection $\bar U\_1\cap\dots\cap\bar U\_n$ is not empty). Therefore, $P\mathbb Q^\omega$ is not metrizable and hence not Polish.
Now we show that $P\mathbb Q^\omega$ is quasi-Polish. Let $\mathbb Q^{<\omega}\_\circ=\bigcup\_{n\in\omega}(\mathbb Q^n\setminus\{0\}^n)$ be the countable set of all finite nonzero sequences of rational numbers and let $\{s\_n\}\_{n\in\omega}=\mathbb Q^{<\omega}\_\circ$ be an enumeration of the set $\mathbb Q^{<\omega}\_\circ$. For every sequence $s\in \mathbb Q^{<\omega}\_\circ$ let $\ell(s)$ be its length and ${\uparrow}s=\{x\in\mathbb Q^\omega\_\circ:x{\restriction}\_{\ell(s)}=s\}$.
Let $q[{\uparrow}s]$ be the image of ${\uparrow}s$ under the quotient map $q:\mathbb Q^\omega\_\circ\to P\mathbb Q^\omega$. It follows that $\{q[{\uparrow}s]:s\in\mathbb Q^\omega\_\circ\}$ is a countable base of the topology of $P\mathbb Q^\omega$.
For every $n\in\omega$, consider the quasi-pseudometric
$d\_n:P\mathbb Q^\omega\times P\mathbb Q^\omega\to\{0,1\}$ defined by $d\_n{-1}(1)=q[{\uparrow}s\_n]\times (P\mathbb Q^\omega\setminus q[{\uparrow}s\_n])$.
Finally consider the quasi-metric $d:P\mathbb Q^\omega\times P\mathbb Q^\omega\to[0,1]$ defined by $$d=\sum\_{n\in\omega}\tfrac1{2^{n+1}}d\_n.$$
I hope that the quasi-metric $d$ witnesses that $P\mathbb Q^\omega$ is a quasi-Polish space.
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3
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https://mathoverflow.net/users/61536
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412202
| 168,200 |
https://mathoverflow.net/questions/412110
|
5
|
Let me be precise about what I mean in the title. Let $A$ be a $C^\*$-algebra, which we identify with its image of its universal representation $(\pi, H)$, so the second dual of $A$ is canonically identified with the von Neumann algebra $A''$ generated by $\pi(A) = A$. Denote $\widetilde{A}$ the minimal unitalization of $A$ in $A''$, and $M(A)$ the maximal one, i.e. $M(A) = \{x \in A'' \mid xA \cup Ax \subset A\}$, which is of course a copy of the multiplier algebra of $A$. It is known (see e.g. Pedersen's book, Theorem 3.12.9) that $M(A)\_{\mathrm{sa}} = (\widetilde{A}\_{\mathrm{sa}})\_m \cap (\widetilde{A}\_{\mathrm{sa}})^m$, where $(\widetilde{A}\_{\mathrm{sa}})\_m$ denotes the set of strong limits in $A''$ of bounded decreasing nets in $\widetilde{A}\_{\mathrm{sa}}$, and $(\widetilde{A}\_{\mathrm{sa}})^m$ the set of strong limits of bounded increasing nets in $\widetilde{A}\_{\mathrm{sa}}$. Of course $M(A)\_{\mathrm{sa}} \subset A''\_{\mathrm{sa}}$ and the inclusion is strict in general. There are a variety of topologies on $M(A)$. Besides the ones that inherited from $B(H)$ as a subspace, one also often considers the strict topology on $M(A)$, which is the locally convex topology defined by the family of semi-norms $\|(\cdot)a\|$ and $\|a(\cdot)\|$ with $a$ running through $A$. It is also known that $M(A)$ is in fact the strict completion of $A$ as a locally convex space.
Now von Neumann's double commutation theorem says that the ultra-strong-$\*$ closure of $A$ in $B(H)$ is $A''$. On $M(A)$, denote the ultra-strong-$\*$ topology that is inherited from $H$ by $\sigma^\*$, and the strict topology by $\beta$. If $\sigma^\* = \beta$, then $M(A)$ is ultra-strongly-$\*$ complete since it is strictly complete, in particular $M(A)$ is ultra-strongly-$\*$ closed in $B(H)$, forcing $M(A) = A''$, but we know this is not always the case, although we do have $M(A) \subset A''$ by the above discussion, i.e. the $\beta$-closure of $A$ is always contained in the ultra-strong-$\*$ closure of $A$. Since finer topology gives smaller closure, my question is, is it true that we always have $\beta$ finer than $\sigma^\*$?
Here's a special case to get started. Note that if $A = K(H)$, then $A''=B(H)$. In this case, it is fairly easy to see that the strict topology on $M(A)$ is finer than the strong-$\*$ topology on $M(A)$. Indeed, take any rank-one projection $p\_\xi$ onto $\mathbb{C}\xi$ with $\xi$ being an arbitrary unit vector of $H$, one has, for any $x \in B(H) = M(K(H)) = K(H)''$, that $\|xp\_\xi\| = \|xp\_\xi x^\*\|^{1/2} = \|x\xi\|$ and $\|x^\*p\_\xi\| = \|x^\* p\_\xi x\|^{1/2} = \|x^\*\xi\|$. As $p\_\xi \in K(H)$, and $\|(\cdot)\xi\|$ together with $\|(\cdot)^\*\xi\|$ form a generating family of semi-norms of the strong-$\*$ topology, we've shown that indeed the strict topology on $M(K(H)) = B(H)$ is finer than the strong-$\*$ one. It is known that the strict topology, the strong-$\*$ topology, the ultra-strong-$\*$ topology and the Arens-Mackey topology on $B(H)$ all agree on bounded parts. So it is reasonable to compare the strict topology to other topologies that are finer than the strong-$\*$ one but still agree with it on bounded parts. Two such topologies are already mentioned, the ultra-strong-$\*$ topology and the Arens-Mackey topology. We know the dual of $B(H)$ equipped with the strict topology is exactly the predual $B(H)\_\*$ of $B(H)$, so the Arens-Mackey topology is finer than the strict one. The question remains about the comparison between the ultra-strong-$\*$ topology and the strict topology, even in this special case where we do have $M(A)=A''$.
|
https://mathoverflow.net/users/128540
|
Is the strict topology on the multiplier algebra of a $C^*$-algebra always finer than the ultrastrong-$*$ topology?
|
The answer is "yes".
Use the standard technique: if necessary, replace $H$ by $H\otimes\ell\_2$ to ensure that for each $\omega\in B(H)\_\*$ (the predual of $B(H)$, the trace-class operators on $H$) there is $\xi\in H$ so that $\omega(y) = (y(\xi)|\xi)$ for each $y\in A''$. As $A$ acts non-degenerately on $H$ (because $A''$ must be unital) and as replacing $H$ by $H\otimes\ell\_2$ does not change this, we may apply the [Cohen–Hewitt factorization theorem](https://en.wikipedia.org/wiki/Cohen%E2%80%93Hewitt_factorization_theorem). This allows us to find $a\in A, \xi'\in H$ with $a(\xi') = \xi$.
Now let $(x\_i)$ be a net in $M(A)$ which converges strictly (that is, for the $\beta$ topology) to $x\in M(A)\subseteq A''$. This means that $x\_i a \rightarrow xa$ and $a^\*x\_i\rightarrow a^\*x$ in norm, so also $x\_i^\*a\rightarrow x^\*a$, in norm. In particular,
$$ \lim\_i \omega((x\_i-x)^\*(x\_i-x))
= \lim\_i \|(x\_i-x)\xi\|^2
= \lim\_i \|(x\_i-x)a(\xi')\|^2 = 0, $$
and similarly $\omega((x\_i-x)(x\_i-x)^\*) \rightarrow 0$. As $\omega$ was arbitrary, this shows that $x\_i\rightarrow x$ in the $\sigma$-strong$^\*$-topology, that is, for the $\sigma^\*$ topology.
---
In fact, we can avoid the "standard technique". Restriction of functionals defines a quotient map $B(H)\_\* \rightarrow M\_\*$ from the trace-class operators to the (unique) predual of our von Neumann algebra $M$. Then the ultra-strong$^\*$-topology is given by the seminorms
$$ M\ni x \mapsto ( \omega(x^\*x) + \omega(xx^\*) )^{1/2} $$
as $\omega$ varies through the positive part of $M\_\*$. In this way, we see that there is no dependence on $H$. (I guess there is an extra argument needed here: that a *positive* member of $M\_\*$ lifts to a positive trace-class operator).
In our case, $M=A''\cong A^{\*\*}$ the bidual, as the representation of $A$ is universal. So $M\_\* \cong A^\*$ the dual of $A$. For $\mu\in A^\*$ and $a\in A$ define $a\cdot\mu\in A^\*$ to be the functional $A\ni b\mapsto \mu(ba)$. There are various ways to show that the linear span of $\{ a\cdot\mu : a\in A, \mu\in A^\* \}$ is norm-dense in $A^\*$. (For example, polar decomposition of functionals and the GNS construction.) We can hence directly apply Cohen--Hewitt to $A^\*$ to show that for each $\mu\in A^\*$ there is $a\in A, \mu'\in A^\*$ with $\mu = a\cdot\mu'$. Similarly we can regard $A^\*$ as a right $A$-module, and then the same argument shows that we can find $a'\in A,\mu''\in A^\*$ with $\mu' = \mu''\cdot a'$, so that $\mu = a\cdot\mu''\cdot a'$ (this being a bimodule, so the order of left/right actions doesn't matter).
Now the same argument works: if $(x\_i)$ in $M(A)\subseteq A^{\*\*}$ converges strictly to $x\in M(A)$ then $\|(x\_i-x)a\| \rightarrow 0$ and $\|a'(x\_i-x)^\*\| = \|(x\_i-x)a'^\*\| \rightarrow 0$, and so
$$ \mu((x\_i-x)^\*(x\_i-x)) = \mu''(a'(x\_i-x)^\*(x\_i-x)a) \rightarrow 0. $$
Thus we have converted the argument to not depend on $H$.
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7
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https://mathoverflow.net/users/406
|
412209
| 168,202 |
https://mathoverflow.net/questions/401347
|
2
|
I think it is standard and common to use Lax-Milgram theorem to prove the existence of solution to elliptic equation. However, can we use it to establish the existence of parabolic equation? I do not find some examples in standard PDE textbooks.
Suppose I have a parabolic equation
$$ \partial\_t u - \partial\_{x\_j}(a\_{ij} \partial\_{x\_i} u) + b\_i \partial\_{x\_i} u + c u =f(x,t)$$
on $\Omega \times [0,T]$. Then the weak formulation should be
$$ \int\_{\Omega} \partial\_{t} u \varphi + a\_{ij} \partial\_{x\_i} u \partial\_{x\_j} \varphi + b\_i \partial\_{x\_i} u \varphi + c u \varphi-f\varphi=0,$$
for all $\varphi(x) \in H^1\_0(\Omega)$ and a.e. $t\in[0,T]$. But I do not know how can we define the bilinear mapping in this way. Or, is it impossible to prove the existence via Lax-Milgram? May I get some help? Thanks!
|
https://mathoverflow.net/users/87922
|
Lax-Milgram and the existence of solution to parabolic equation
|
Finally I found the exact theorem from Chapter 3.1.2 in the book "Elliptic & Parabolic Equations" written by Zhuoqun Wu, Jingxue Yin and Chunpeng Wang.
|
1
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https://mathoverflow.net/users/87922
|
412236
| 168,210 |
https://mathoverflow.net/questions/412242
|
4
|
Let $\mathbf{a}=(a\_0,\cdots,a\_n)\in\mathbb{Z}^{n+1}$ be an array of integers, let $\mathcal{E}(\mathbf{a})=\oplus\_{i=0}^n\mathcal{O}(a\_i)$ be the vector bundle on $\mathbb{P}^1$, and let $\mathbb{P}(\mathcal{E}(\mathbf{a}))$ be its projectivization. We know that $\mathbb{P}(\mathcal{E}(\mathbf{a}))$ and $\mathbb{P}(\mathcal{E}(\mathbf{b}))$ are isomorphic iff $\mathbf{a}-\mathbf{b}=(k,\cdots,k)$ for some $k\in\mathbb{Z}$.
For $n\geq2$, is there an explicit numerical criterion telling whether $\mathbb{P}(\mathcal{E}(\mathbf{a}))$ and $\mathbb{P}(\mathcal{E}(\mathbf{b}))$ are deformation equivalent?
(In case $n=1$, we know Hirzebruch surfaces are deformation equivalent iff $a\_1-a\_0\equiv b\_1-b\_0$ mod $2$)
|
https://mathoverflow.net/users/nan
|
Deformation equivalence of $\mathbb{P}^n$ bundles over $\mathbb{P}^1$
|
The projective bundles $\mathbb{P}(a)$ and $\mathbb{P}(b)$ are deformation equivalent if and only if
$$
\sum a\_i \equiv \sum b\_i \bmod n + 1.
$$
To show this it is enough to check that $\mathbb{P}(a)$ is deformation equivalent to $\mathbb{P}(a')$, where
$$
a' = (a\_0 - 1, a\_1 + 1, a\_2, \dots, a\_n).
$$
For this assume that $a\_0 \le a\_1$. Then there is an exact sequence
$$
0 \to \mathcal{O}(a\_0 - 1) \to \mathcal{O}(a\_0) \oplus \mathcal{O}(a\_1) \to \mathcal{O}(a\_1 + 1) \to 0,
$$
hence there is a deformation equivalence between vector bundles
$$
\mathcal{O}(a\_0 - 1) \oplus \mathcal{O}(a\_1 + 1)
\quad\text{and}\quad
\mathcal{O}(a\_0) \oplus \mathcal{O}(a\_1)
$$
that corresponds to deformation to zero of the extension class of the above exact sequence. Taking the sum with the other line bundles and passing to projectivizations, one obtains the required deformation equivalence for projective bundles.
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5
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https://mathoverflow.net/users/4428
|
412246
| 168,211 |
https://mathoverflow.net/questions/412240
|
2
|
Let $f(n)$ = 1 if $n$ belongs to [A014689](https://oeis.org/A014689), $\operatorname{prime}(n)-n$, the number of nonprimes less than $\operatorname{prime}(n)$. Here $\operatorname{prime}(n)$ is the $n$-th prime number, $\operatorname{prime}(1)=2$.
Let $a(n)$ be the $n$-th composite numbers, $a(1)=4$.
Then I conjecture that
$$a(n) = 1 + a(n-1) + f(n)$$
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Difference between $n$-th and $(n-1)$-th composite numbers
|
Surely $f(n)$ is meant to be the indicator function of the range of the function $k\mapsto p(k)-k$, where $p(k)$ denotes the $k$-th prime number. With this supplemented definition, the conjecture is true.
Indeed, $n=p(k)-k$ holds if and only if there are $n-1$ composite numbers up to $p(k)$, that is, $a(n-1)<p(k)<a(n)$. Therefore, $f(n)=1$ means that there is a prime number between $a(n-1)$ and $a(n)$. So we have $a(n)=a(n-1)+2$ when $f(n)=1$, and we have $a(n)=a(n-1)+1$ when $f(n)=0$.
|
8
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https://mathoverflow.net/users/11919
|
412252
| 168,214 |
https://mathoverflow.net/questions/412229
|
2
|
$\DeclareMathOperator\MCG{MCG}$Let $\Gamma=\MCG(S\_{g, 0,0 })$ be mapping class group of closed hyperbolic surfaces. Let $V=C^0(S)$ be the set of vertices of the simplicial curve complex. We are studying the well known group action $\Gamma \times V \to V$, yet we find ourselves utterly incapable of computing and translating elements $([\phi], [\alpha]) \mapsto [\phi(\alpha)]$ where $\alpha$ represents a "simple closed curve" in $V$. In otherwords it appears that the action of the MCG on the curve complex is incomputable for closed surfaces.
I am aware of Mark C. Bell's curver program. It's documentation indicates that it uses results of Saul Schleimer relating paths in a "flip graph" with words in the pointed mapping class groups. I am not aware of any flip graph analogy for the closed mapping class groups.
***My question:** What is best computational approach to being able to solve the following problem for the group action $\Gamma \times V \to V$:*
``*Given $[\phi] \in \Gamma, [\alpha] \in V, [\beta] \in V$, determine whether $[\phi(\alpha)]=[\beta]$ is True or False in $V$.*"
Evidently the fact that we have no linear representations of $\Gamma$ is one computational obstruction, for otherwise all the quantities could essentially be represented as matrices.
In my research I have interest in constructing finite subsets $I$ for which a chain sum $\sum\_{[\phi] \in I} \sum\_{[\alpha] \in B}[\phi(\alpha)]=0$ mod 2. Here $B$ is a finite subset of $V$ based on Nathan Broaddus' homology spheres. (e.g. $B$ is the vertex set if the two dimensional spheres in genus $g=2$ discovered by Broaddus).
(\*\*) I'll accept Sam Nead's answer since it suggests that Birman's exact sequence tells us that forgetful map from pointed mapping classes to unpointed mapping classes is onto, with kernel equal to the image of $\pi\_1(S, pt)$ under the Push map. So it's obvious from Birman's sequence that Bell-Webb's curver program can be used to compute pure mapping class actions on the curve complex. Although this raises the question of whether the membership problem for image of $\pi\_1(S, pt)$ by the push map is computable in the pointed groups. Assuming it is, then to solve the subset-sum problem I'll start with the pointed groups in curver.
|
https://mathoverflow.net/users/20516
|
Is action of MCG on the curve complex computable for closed surfaces? [Yes: Birman Exact Sequence]
|
You write "it appears that the action of the MCG on the curve complex is incomputable for closed surfaces."
This is not correct. Geva Yashfe points out one approach in the comments. Here is another, with references given below.
Fix $S$ a connected, closed, oriented surface of genus $g$. Fix a one-vertex triangulation of $S$. We represent simple closed curves via "edge coordinates" (or we could use normal coordinates). Note that (as $S$ is closed) two curves can have different edge coordinates yet be isotopic.
Thus your problem reduces to determining isotopy equivalence among such representatives. (The presence of a mapping class is dealt with by understanding (say) how Dehn twists act on edge coordinates.) It is a theorem that two curves are isotopic if and only if they cobound an annulus, perhaps after performing a sequence of "bigon" moves. Detecting such annuli or bigons, and performing bigon moves, can be done computationally.
A naive implementation will be polynomial time in the edge weights. Thinking more deeply gives an algorithm that is polynomial time in the logarithm of the edge weights. This brings us to the beginnings of modern research in this area, and so answers your question (in italics).
You can find versions and discussions of this in papers such as
* Schaefer, Sedgwick, and Štefankovič [2002]
* Agol, Hass, and Thurston [2005]
* Erickson and Nayyeri [2012]
as well as work of Bell (partly with Webb).
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4
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https://mathoverflow.net/users/1650
|
412255
| 168,215 |
https://mathoverflow.net/questions/412256
|
5
|
Consider a smooth quadric surface $Q\subset\mathbb{P}^3$ over a field $k$. Are there natural hypotheses one can put on $k$ in order to ensure the existence of a line defined over $k$ on $Q$?
|
https://mathoverflow.net/users/14514
|
Lines on quadric surfaces
|
Yes, it suffices that $k$ have no nontrivial quadratic extensions.
Since the surface is geometrically $\mathbb P^1 \times \mathbb P^1$, the space of lines is geometrically a union of two copies of $\mathbb P^1$. Arithmetically, the components are defined over a quadratic extension of $k$. If there are no nontrivial quadratic extensions then they are defined over $k$.
Each component is then a form of $\mathbb P^1$. Since the anticanonical bundle has degree $2$, a section of the anticanonical bundle has zero locus a scheme of length two. Because $k$ has no nontrivial extensions of degree $2$, the zero locus consists of either one or two $k$-points, so the space parameterizing lines has a $k$-point and thus there is a rational line.
At least in characteristic not $2$, this condition is necessary, as if $k$ has a quadratic extension then it contains a nonsquare $a$ and the form $x^2 -y^2 + z^2 - a w^2$. Any line on this surface must, over $k(\sqrt{a})$, be a line on $(x+y)(x-y) + (z- \sqrt{a}w)( z+\sqrt{a}w)$, thus of the form $\alpha (x+y) + \beta (z-\sqrt{a}w) = \alpha (x-y) - \beta (z+\sqrt{a}w)=0$ or of the form $\alpha (x+y) + \beta (z+\sqrt{a}w) = \alpha (x-y) - \beta (z-\sqrt{a} w) =0$, but the Galois group exchanges these two types of lines so this is impossible.
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5
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https://mathoverflow.net/users/18060
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412259
| 168,216 |
https://mathoverflow.net/questions/412257
|
6
|
Let $X$ be a noetherian scheme and $\mathcal{I} \subset \mathcal{O}\_X$ a coherent sheaf of ideals. Suppose that $\mathcal{I}^d$ is locally-free for some power $d$. Then the blowing up $\mathrm{Bl}\_{\mathcal{I}^d} \to X$ is an isomorphism. However, $\mathrm{Bl}\_{\mathcal{I}} \cong \mathrm{Bl}\_{\mathcal{I}^d}$ as $X$-schemes so the map $\pi : \mathrm{Bl}\_{\mathcal{I}} \to X$ is also an isomorphism. By definition of blowing up this implies that the inverse image of $\mathcal{I}$ which is equal to $\mathcal{I}$ is is locally-free.
My question is about a direct algebraic proof of this fact translated into commutative algebra:
Let $A$ be a noetherian local ring and $I \subset A$ be an ideal such that $I^d = (f)$ for some non zero-divisor $f$. Then $I = (g)$ for some non zero-divisor $g$.
I can see how to prove this when $A$ is a UFD but I don't see how a direct proof would go of the general case. Is there an easy argument or is it necessary to argue with the Rees Algebra?
|
https://mathoverflow.net/users/154157
|
If power of an ideal is locally free then it is locally free
|
Here is a proof if $A$ is an integral domain.
Let $x\_1,\dots, x\_n$ generate $I$. Then $\prod\_{i=1}^n x\_i^{e\_i}$ generate $I^d =f$ for vectors $e\_i$ of nonnegative integers satisfying $\sum\_i e\_i=d$. These generators are all multiples of $f$ and can't all lie in the maximal ideal times $f$ so, since $A$ is local, one of them must be a unit times $f$. Fix such a generator $\prod\_{i=1}^n x\_i^{e\_i}$.
Without loss of generality, $e\_1>0$.
Then for all $j$ from $1$ to $n$, $(x\_j/x\_1)\prod\_{i=1}^n x\_i^{e\_i}$ is also in $I^d$, thus is a multiple of $f$, hence is a multiple of $\prod\_{i=1}^n x\_i^{e\_i}$. Dividing by $x\_1^{e\_1-1} \prod\_{i=2}^n x\_i^{e\_i}$, we see that $x\_j$ is a multiple of $x\_1$. Becuase this works for all $j$, $I=(x\_1)$.
|
7
|
https://mathoverflow.net/users/18060
|
412261
| 168,217 |
https://mathoverflow.net/questions/412274
|
2
|
Recently I have been reading the paper *The categorical origins of Lebesgue integration* by Tom Leinster (<https://arxiv.org/pdf/2011.00412.pdf>). In this paper, he said that：
>
> For $n \geq 0$, let $E\_{n}$ be the subspace of $L^{p}[0,1], (1\leq p<\infty)$ consisting of the equivalence classes of step functions constant on each of the (open) intervals $\left(\frac{i-1}{2^{n}}, \frac{i}{2^{n}}\right)$, $\left(1 \leq i \leq 2^{n}\right)$. Write $E=\bigcup\_{n \geq 0} E\_{n}$, which is the space of step functions whose points of discontinuity are dyadic rationals. Then $E$ is dense in the set of all step functions on $[0,1]$, which in turn is dense in $L^{p}[0,1]$; so $E$ is dense in $L^{p}[0,1]$. It follows that $L^{p}[0,1]$ is the colimit (direct limit) of the diagram $E\_{0} \hookrightarrow E\_{1} \hookrightarrow \cdots$ in the category $\mathbf{\text{Ban}}$ of Banach spaces with linear contractions as morphisms.
>
>
>
I know that the category $\mathbf{\text{Ban}}$ is cocomplete, i.e., the colimit of any diagram exists (because it has coproducts and coequalizers). However, I just have no idea **why the density of $E$ in $L^p[0,1]$ implies just that $L^p[0,1]$ is the colimit of the increasing sequence $E\_{0} \hookrightarrow E\_{1} \hookrightarrow \cdots$ in the category $\mathbf{\text{Ban}}$ of Banach spaces with linear contractions**? I thought about this for two days but didn't figure out anything. Can anyone give a convincing or inspiring explanation?
Any help is appreciated.
|
https://mathoverflow.net/users/154064
|
Direct limit of the sequence $E_{0} \hookrightarrow E_{1} \hookrightarrow \cdots$ in the category of Banach spaces
|
A colimit is an object $E$ together with morphisms $i\_n:E\_n\to E$ commuting with the inclusions $i\_{n,m}:E\_n\to E\_m$ (i.e., $i\_m\circ i\_{n,m}=i\_n$) such that, for every sequence of morphisms $f\_n:E\_n\to Y$ with $f\_m\circ i\_{n,m}=f\_n$, there is a unique $f:E\to Y$ with $f\_n=f\circ i\_n$ for all $n$.
You have to show that $i\_n:E\_n\hookrightarrow L^p([0,1])$ satisfy this universal property. Given $f\_n:E\_n\to Y$ as above you can define a linear contraction $\varphi: \bigcup\_{n\in\mathbb N} E\_n \to Y$ by $\varphi(x)=f\_n(x)$ for $x\in E\_n$ (which is well-defined) and, because of the density of the union in $L^p([0,1])$, define $f:L^p([0,1])\to Y$ as the unique continuous extension of $\varphi$.
|
7
|
https://mathoverflow.net/users/21051
|
412277
| 168,220 |
https://mathoverflow.net/questions/412266
|
0
|
In his YouTube video [New rank records for elliptic curves having rational torsion](https://www.youtube.com/watch?v=3kxLBpj1Mzc), Noam Elkies uses systems of equations at 6:16 and 8:38 to present $\mathbb{Z}/3\mathbb{Z}$ curves of [rank 14](https://web.math.pmf.unizg.hr/%7Eduje/tors/z3old891011121314.html) and [rank 15](https://web.math.pmf.unizg.hr/%7Eduje/tors/z3.html).
I created a similar system of equations for a $\mathbb{Z}/6\mathbb{Z}$ curve of [rank 9](https://web.math.pmf.unizg.hr/%7Eduje/tors/z6.html) (edit: removed original infographic):
$$x+y+z=–73826006279$$
$$xyz=–13211438249850179974544071008000$$
I should have negated both equations, but it's too late now, as Andrej Dujella has already provided all the [integer solutions](https://twitter.com/dujella1/status/1473410623388241920).
Now I am trying to create a similar system for a curve without $3$-torsion, specifically for a $\mathbb{Z}/8\mathbb{Z}$ curve of rank 1 mentioned on p. 20 in our recently submitted paper
>
> Halbeisen, Hungerbuehler, Voznyy, and Zargar, *A geometric approach to elliptic curves with torsion groups $\mathbb Z/10\mathbb Z$, $\mathbb Z/12\mathbb Z$, $\mathbb Z/14\mathbb Z$, and $\mathbb Z/16\mathbb Z$*, [arXiv: 2106.06861](https://arxiv.org/abs/2106.06861)
>
>
>
$$y^2 = x^3 + 10226878x^2 + 43046721x$$
>
> **Question $1$:** For the mentioned $\mathbb{Z}/8\mathbb{Z}$ curve, is it possible to keep a format similar to a $\mathbb{Z}/3\mathbb{Z}$ case (two equations, three variables, +, -, $\times$)? I would really like to avoid squaring/cubing the same variable.
>
>
>
>
> **Question $2$:** Is it possible to keep a similar format for any other curve having a $2$-torsion, but not $3$-torsion, i.e. for torsion subgroups $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z}$, $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/10\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/8\mathbb{Z}$?
>
>
>
|
https://mathoverflow.net/users/95511
|
Systems of equations for elliptic curves without $3$-torsion
|
Is this system of equations satisfactory for Question 1:
$x+y+x \times z+y \times z=6400$,
$x \times y \times z=6561$?
It seems that all curves with torsion groups containing $\mathbb{Z}/4\mathbb{Z}$ can be obtained in this way. By taking $x+y+x \times z+y \times z= d$, $x \times y \times z=-cd$, we get the elliptic curve $[1,-c/d,-c/d,0,0]$ with a point $[0,0]$ of order $4$.
|
4
|
https://mathoverflow.net/users/21337
|
412281
| 168,223 |
https://mathoverflow.net/questions/410799
|
6
|
Let $\eta(n)$ be [A006337](https://oeis.org/A006337), an "eta-sequence" defined as follows:
$$\eta(n)=\left\lfloor(n+1)\sqrt{2}\right\rfloor-\left\lfloor n\sqrt{2}\right\rfloor$$
Sequence begins
$$1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1$$
Let $a(n)$ be [A091524](https://oeis.org/A091524), $a(m)$ is the multiplier of $\sqrt{2}$ in the constant $\alpha(m) = a(m)\sqrt{2} - b(m)$, where $\alpha(m)$ is the value of the constant determined by the binary bits in the recurrence associated with the Graham-Pollak sequence.
Sequence begins
$$1, 1, 2, 2, 3, 4, 3, 5, 4, 6, 7, 5, 8, 6, 9, 7, 10, 11, 8, 12, 9, 13$$
Then we have an integer sequence given by
$$b(n)=(a(n))^2\eta(n)$$
Sequence begins
$$1, 2, 4, 8, 9, 16, 18, 25, 32, 36, 49, 50, 64, 72, 81, 98, 100, 121, 128, 144, 162, 169$$
I conjecture that $b(n)$ is a sequence of $k^2$ and $2k^2$ ordered in ascending order.
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Sequence of $k^2$ and $2k^2$ ordered in ascending order
|
Denote by $f(n)$ the sequence of squares and double squares in ascending order. We have to prove that $f(n)=b(n)=(a(n))^2\eta(n)$. Consider two cases.
1. $f(n)=k^2$. Then the number of squares and double squares not exceeding $k^2$ equals $n$, that is, $n=k+\lfloor k/\sqrt{2}\rfloor$. Therefore $n<k(1+1/\sqrt{2})$ that is equivalent (by multiplying to $2-\sqrt{2}$) to $n\sqrt{2}>2n-k$, and $\lfloor n\sqrt{2}\rfloor\geqslant 2n-k$. On the other hand, $n+1>k(1+1/\sqrt{2})$ that analogously yields $(n+1)\sqrt{2}>2(n+1)-k$ and $\lfloor (n+1)\sqrt{2} \rfloor\leqslant 2n-k+1$. Since also $\lfloor (n+1)\sqrt{2} \rfloor\geqslant \lfloor n\sqrt{2} \rfloor+1\geqslant 2n-k+1$, this implies that $\eta(n)=1$. According to [OEIS](https://oeis.org/A091524) we have $a(n)=a(\lfloor k(1+1/\sqrt{2}\rfloor)=k$, thus $(a(n))^2\eta(n)=k^2$ as needed.
2. $f(n)=2k^2$. Then the number of squares and double squares not exceeding $2k^2$ equals $n$, that is, $n=k+\lfloor k\sqrt{2}\rfloor$. So, $n<k(\sqrt{2}+1)<n+1$, and (by multiplying to $\sqrt{2}-1$) we get $n\sqrt{2}<n+k$ and $(n+1)\sqrt{2}>(n+1)+k$. This yields $\eta(n)=2$. Again by [OEIS](https://oeis.org/A091524) we get $a(n)=a(\lfloor k(1+\sqrt{2}\rfloor)=k$ and $(a(n))^2\eta(n)=2k^2$.
|
9
|
https://mathoverflow.net/users/4312
|
412287
| 168,224 |
https://mathoverflow.net/questions/412008
|
14
|
I apologise in advance if this is an elementary question more fitted for Math Stack Exchange. The reason why I have decided to post here is that the question I am used to seeing on that site are not of the open-ended format of the one I am asking.
It is now the second time I have been studying Calculus (first self-taught, now in school) and we are going over the proof of Cauchy's mean value theorem (the precursor to l'Hopital's rule). I do understand the proof, and the intuitive explanation about parametrised curves in a plane, but I still think the statement of the theorem looks relatively obscure. Why are we considering a ratio and not something else? This led me to try and generalise, and this is where we get to my question.
I first tried finding a function $h: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that for any functions $f,g: D \subset \mathbb{R} \rightarrow \mathbb{R}$ satisfying Cauchy's mean value theorem's hypotheses, for any interval $[a,b] \subset D$, there exists $x \in [a,b]$ such that $h(f'(x), g'(x))=h(f(b-a), g(b-a))$. Beyond making a few tries and finding a few counterexamples, I realised this wasn't really in the spirit of a mean value theorem: we are trying to make an analogy, if we may use this term, between $f'(x)$ and $f(b)-f(a)$, while in both Lagrange and Cauchy's mean value theorems the analogy is made between $f'(x)$ and $\frac{f(b)-f(a)}{b-a}$. So I started looking for $h$ such that there exists $x$ such that
$$
h(f'(x), g'(x))= h\left(\frac{f(b)-f(a)}{b-a}, \frac{g(b)-g(a)}{b-a}\right).
$$
But this didn't really lead me anywhere.
>
> The question I'm asking is precisely this: can we say anything more about functions $h(x,y) \neq \frac{x}{y}$ satisfying these statements? Suppose we simplify even further, and consider, for example, only the functions $h\_{\alpha,\beta}(x,y) = x^{\alpha}y^{\beta}$. Can we maybe prove that only those with $\alpha = k, \beta = -k$ for some $k$ work? (in addition, clearly, to those with $\alpha\beta = 0$) Is this even interesting to investigate?
>
>
>
Thanks in advance for helping me. My knowledge doesn't really go far beyond Calculus and Linear Algebra (say, Spivak and Axler's books) but I will try to understand your replies.
|
https://mathoverflow.net/users/472669
|
Generalisation of Cauchy's mean value theorem
|
When I reviewed this question a few days ago, I thought there was something sounding familiar in it but I did not remembered what it was: now I have remembered. This problem was fully solved by [Alessandro Faedo](https://en.wikipedia.org/wiki/Alessandro_Faedo) in paper [1]: in his ZBMath review, Peter Bullen says
>
> The author determines all the functions $F(X,Y)$ such that, for every $f(x)$ and $g(x)$ continuous in $[a, b]$ with $f^\prime(x)$ and $g^\prime(x)$ defined for $a<x<b$, there exists a point $\xi$, with $a<\xi<b$ such that
> $$
> F\left(\frac{f(b)-f(a)}{b-a}, \frac{g(b)-g(a)}{b-a}\right) = F(f^\prime (\xi), g^\prime(\xi))
> $$
> If $F(X,Y)={X/Y}$ then we have the Cauchy mean value theorem.
>
>
>
**Synopsis of Faedo's paper**
Following [Willie Wong's comment](https://mathoverflow.net/questions/412008/generalisation-of-cauchys-mean-value-theorem/412290#comment1062017_412290), I am adding here a short synopsis of the paper. Indeed, apart from being written in Italian, the exposition does not follow the now common explicit definition/explicit theorem paradigm, but the mathematics is "embedded" almost seamlessy in the prose, thus it is not easily understand for a foreigner: this is a common point of several writings by Italian mathematicians of that generation.
**§1. Basic definitions** (pp. 489-490).
In this section, the author give the basic definitions he will use throughout the paper.
Let $\Phi(X,Y)\in C^0(\Bbb R^2\setminus E)$ where $E$ is a singular set without interior points such that
$$
E= E\_1 \cup E\_2
$$
where
* $E\_1$ is the set of points of $\Bbb R^2$ such that
$$
\lim\_{(X,Y)\to (X\_o,Y\_o)} \Phi(X,Y)=\pm\infty \quad\forall (X\_o,Y\_o)\in E\_1
$$
* $E\_2$ is the set of points of $\Bbb R^2$ such that
$$
\lim\_{(X,Y)\to (X\_o,Y\_o)} \Phi(X,Y)\;\text{ does't exist }\; \forall (X\_o,Y\_o)\in E\_2
$$
**Definition**. $\Phi(X,Y)\in C^0(\Bbb R^2\setminus E)$ is a *Cauchy function* if, for all $f, g$ continuous on a given closed interval $[a,b]$ and differentiable in its interior, such that
1. the points
$$
X = \frac{f(b)-f(a)}{b-a},\; Y= \frac{g(b)-g(a)}{b-a}
$$
do not belong to $E\_1$ and
2. for each $x\in ]a,b[$ the points $X=f^\prime(x)$ and $Y=g^\prime(x)$ do not belong to $E\_2$,
there exists at least one point $\xi\in ]a,b[$ such that
$$
\Phi\left(\frac{f(b)-f(a)}{b-a}, \frac{g(b)-g(a)}{b-a}\right) = \Phi(f^\prime (\xi), g^\prime(\xi)). \label{1}\tag{1}
$$
**§2. A slight generalization Cauchy's mean value theorem, and a particular class of Cauchy functions** (pp. 490-492).
In this section the author first proves the following theorem:
**Theorem (Generalized Cauchy's mean value theorem)**. If $f, g$ are continuous on a given closed interval $[a,b]$ and differentiable in its interior, and $h, k\in\Bbb R$ are two constants such that
* $g(b)-g(a)+ k(b-a)\neq 0$
* equations $f^\prime(x)+h=0$ and $g^\prime(x)+k=0$ are never simultaneously true for every $x\in]a, b[$
then there exists at least one point $\xi\in ]a,b[$ such that
$$
\begin{split}
\left[\frac{f(b)-f(a)}{b-a} +h\right]\cdot\left[\frac{g(b)-g(a)}{b-a}\right]^{-1} & \\
\frac{f(b)-f(a) + h(b-a)}{g(b)-g(a) + k(b-a)} & = \frac{f^\prime (\xi)+h}{g^\prime(\xi)+k}.
\end{split}
$$
The proof is a straightforward application of the standard Cauchy's mean value theorem. $\blacksquare$
The above theorem shows that $$
\Phi(X,Y)=(X+h)(Y+k)^{-1}
$$
is a Cauchy function, and the author notes immediately that,
* for any given continuous function $F:\Bbb R \to \Bbb R$, the function $\Phi\_1(X,Y)=F((X+h)(Y+k)^{-1})$ is in the same way Cauchy,
* describes their simple $E\_1$ and $E\_2$ sets and,
* by abuse of notation, calls $\Phi\_1$ the class of Cauchy functions generated by $(X+h)(Y+k)^{-1}$, which is thus called the *generating function* of the class.
He then notes that the another way of representing functions of class $\Phi\_1$ is to use arbitrary zero-degree homogeneous functions: if $F^\ast(z\_1,z\_2)$ is such a function, then for each real $c\neq 0$ we have
$$
F^\ast(c z\_1, c z\_2) = F^\ast(z\_1,z\_2) = F^\ast\!\left(\frac{z\_1}{z\_2}, 1\right)
$$
and we can represent any function of the class $\Phi\_1$ as
$$
\Phi\_1(X,Y)=F^\ast(X+h,Y+k).
$$
**§3 and §4. The class $\Phi\_2$ and an example of homogeneous function which is not Cauchy** (pp. 492-493).
The Author shows that also linear functions of the form
$$
\Phi\_2(X,Y)=c\_1 X + c\_2 Y
$$
are Cauchy functions, and again he defines as $\Phi\_2$ the class generated by such function.
In paragraph 4 he shows that there exists a positively homogeneous function, precisely the function $\Phi^\ast(X,Y)=(X^2+Y^2)^{\alpha/2}$, $\alpha\neq 0$ which is not a Cauchy function, proving thus that there are not other classes of Cauchy functions that can be constructed in the same way as $\Phi\_1$.
**§5. A necessary condition for a function of the form $\Phi(X,Y)=X-\psi(Y)$ to be a Cauchy function** (pp. 493-495).
The core result of the paper is the following lemma:
**Lemma**. Let $\psi\in C^2(\Bbb R)$. A necessary condition for a function $\Phi(X,Y)= X-\psi(Y)$ to be a Cauchy function is that $\psi(Y)$ is linear.
**Proof**. The demonstration goes by contradiction: assume that $\Psi$ is a Cauchy function and suppose that there exists at least a number $Y\_0$ for which $\psi^{"}(Y\_0)\neq 0$. Moreover, without restriction to generality, let's assume that $\psi^{"}(Y\_0)>0$. Then the function
$$
F(Y)=\psi(Y)-Y\psi^\prime(Y\_0)
$$
has a local minimum in $Y=Y\_0$ since
$$
F^\prime(Y\_0)=0\; \wedge \; F^{"}(Y\_0)>0
$$
Now choose a function $g\in C^1([a,b])$ satisfying the following properties:
* its values at the endpoints of $[a,b]$ satisfy the following relation
$$
\frac{g(b)-g(a)}{b-a} =Y\_0,
$$
* $|Y\_0 -g^\prime(x)|<\delta$ for all $x\in[a,b]$ and a sufficiently small $\delta>0$,
* $Y\_0\neq g(x)$ for all $x$ belonging to a subset of $[a,b]$ of positive (Lebesgue) measure.
For example we can define an indexed family of such functions as
$$
g(x) =Y\_0 x +\varepsilon \sin\frac{2\pi x}{b-a}\quad 0<\varepsilon<\delta
$$
Finally define $f^\prime(x)=\psi(g^\prime(x))$: this implies that
$$\DeclareMathOperator{\Dm}{\operatorname{d}\!}
f(x)= f(a) + \int\limits\_a^x \psi(g^\prime(x))\Dm x \iff \frac{f(b)-f(a)}{b-a} = \frac{1}{b-a}\int\limits\_a^b \psi(g^\prime(x))\Dm x.
$$
Then, for any $g$ satisfying the above properties, we have
$$
\begin{split}
\frac{1}{b-a}\int\limits\_a^b F(g^\prime(x))\Dm x &= \frac{1}{b-a}\int\limits\_a^b \psi(g^\prime(x))\Dm x - \frac{\psi^\prime(Y\_0)}{b-a}\int\limits\_a^b g^\prime(x)\Dm x\\
& = \frac{1}{b-a}\int\limits\_a^b \psi(g^\prime(x))\Dm x - \psi^\prime(Y\_0)\frac{g(a)-g(b)}{b-a}\\
&> \frac{1}{b-a}\int\limits\_a^b F(Y\_0)\Dm x = F(Y\_0) =\psi(Y\_0) -Y\_0 \psi^\prime(Y\_0) \\
&= \psi\left(\frac{g(a)-g(b)}{b-a}\right) - \psi^\prime(Y\_0) \frac{g(a)-g(b)}{b-a}
\end{split}
$$
This implies that
$$
\begin{eqnarray}
\frac{1}{b-a}\int\limits\_a^b \psi(g^\prime(x))\Dm x & > & \psi\left(\frac{g(a)-g(b)}{b-a}\right)\\
&\Updownarrow &\\
\frac{f(b)-f(a)}{b-a} & > &\psi\left(\frac{g(a)-g(b)}{b-a}\right)\label{2}\tag{2}
\end{eqnarray}
$$
while $f^\prime(x) -\psi(g^\prime(x)) =0 $ for each $x\in [a,b]$. This finally implies, contradicting the hypothesis, that $\Phi (X,Y) =X - \psi(Y)$ does not satisfy \eqref{1} thus it is not a Cauchy function: therefore it must be $\psi^{"}(x)=0$ for all $x\in[a,b]$. $\blacksquare$
I said this is the core result of the paper since the main theorem, proved in the following section of the paper, follows from an application of the ideas developed in this lemma.
**§6. The main theorem** (pp. 496-497).
The characterization of Cauchy functions given by Faedo is expressed by the following
**Theorem**. Let $\Phi(X,Y)\in C^2(\Bbb R^2\setminus E)$ where $E$ is defined as above. Then a necessary condition for $\Phi(X,Y)$ to be a Cauchy function is that each level curve defined by the equation
$$
\Phi(X,Y)=c,\quad c=\text{const.}\label{3}\tag{3}
$$
is a piecewise linear curve $\Gamma\_c$ whose vertex are points $(X,Y)$ for which, simultaneously,
$$
\frac{\partial\Phi}{\partial X}=0\;\wedge\;\frac{\partial\Phi}{\partial Y}=0.
$$
**Proof**. Let $(X\_0,Y\_0)$ be a point in $\Bbb R^2\setminus E$ for which the partial derivatives
$$
\dfrac{\partial\Phi}{\partial X}\bigg|\_{(X\_0,Y\_0)}\text{ and }\dfrac{\partial\Phi}{\partial Y}\bigg|\_{(X\_0,Y\_0)}
$$ are not simultaneously equal to zero. For example and without restriction to generality, let
$$
\left.\frac{\partial\Phi}{\partial X}\right|\_{(X\_0,Y\_0)}\neq 0.
$$
Since $\Phi(x\_0,Y\_0)=c$, equation \eqref{3} defines a function $X=\psi(Y)$ with $X\_0=\psi(Y\_0)$ and
$$
\Phi(\psi(Y),Y)-c =0
$$
at least in a neighborhood of $Y\_0$, thus can find a $\delta >0$ such that if $|Y-y\_0| <\delta$ then also $|X\_0-\psi(Y)|<\delta$. Moreover, since $\frac{\partial\Phi}{\partial X}\neq 0$, the function $X\mapsto\Phi(X,Y\_0)$, considered in a neighborhood of $X=X\_0$, takes the value $c$ only in $X=X\_0$. From here on it is possible to proceed as in the proof of the lemma in §5: assume that $\psi^\prime(Y\_0)>0$ and put
$$
\overline X = \frac{f(b)-f(a)}{b-a},\; \overline Y= \frac{g(b)-g(a)}{b-a}.
$$
We have that
$$
\frac{g(b)-g(a)}{b-a} = Y\_0
$$
and moreover
$$
\frac{f(b)-f(a)}{b-a} = \frac{1}{b-a}\int\limits\_a^b \psi(g^\prime(x))\Dm x.
$$
with $|X\_0-\psi(g^\prime(x))|<\delta$, and thus
$$
\Bigg|\frac{f(b)-f(a)}{b-a} - X\_0\Bigg|=\Bigg| \frac{1}{b-a}\int\limits\_a^b [\psi(g^\prime(x))-X\_0]\Dm x\Bigg|<\delta.
$$
This implies $|\overline{X}-X\_0|<\delta$ and $|\overline{Y}-Y\_0|<\delta$ and due to the arbitrariness of $\delta$ we have
$$
\Phi(\overline{X},\overline{Y}) \neq \Phi(X\_0,Y\_0) = c.
$$
Equation \eqref{2} of §5 thus proves that
$$
\Phi\left(\frac{f(b)-f(a)}{b-a}, \frac{g(b)-g(a)}{b-a}\right) \neq \Phi(f^\prime (x), g^\prime(x))=c\quad\forall x\in[a,b],
$$
thus it must be $\psi^{"}(Y\_0)$ for otherwise $\Phi(X,Y)$ is not a Cauchy function. Thus $\psi(Y)$ must be a linear function, and this implies that
$$
\psi^\prime(Y) = \text{const.} = - \frac{{\partial \Phi}/{\partial Y}}{{\partial \Phi}/{\partial X}}
$$
thus ${{\partial \Phi}/{\partial X}}\neq 0$ implies ${{\partial \Phi}/{\partial Y}}\neq 0$, therefore $X=\psi(Y)$ is defined for every $X$ except those for which ${\partial \Phi}/{\partial X} = {\partial \Phi}/{\partial Y} = 0$. $\blacksquare$
**§7 and §8. Properties of the piecewise linear level curve $\Gamma\_c$ and rational Cauchy functions.**
In §7 Faedo proves that on the points $(X\_0,Y\_0)$ where the gradient of the Cauchy function $\Phi$ does not vanish, two level curves $\Gamma\_c$ and $\Gamma\_{c\_0}$ with $c\neq c\_0$ do not intersect nor self-intersect. In the last paragraph, the Author shows that the only rational Cauchy functions are those belonging to the classes $\Phi\_1$ defined in §2 and $\Phi\_2$ defined in §3.
**Reference**
[1] Sandro Faedo, "Sul teorema di Cauchy degli incrementi finiti" [On Cauchy's theorem about finite increments] (Italian), Rendiconti di Matematica, VI Serie 10 (1977), 489-499 (1978), [MR0480904](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0480904), [Zbl 0384.26002](https://www.zbmath.org/?q=an%3A0384.26002).
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21
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https://mathoverflow.net/users/113756
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412290
| 168,225 |
https://mathoverflow.net/questions/412269
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3
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What is meant by the statement and the proof of Lemma 3.2 in Chapter XII of Dehornoy's book [*Braids and Self-Distributivity*](https://link.springer.com/book/10.1007/978-3-0348-8442-6)?
That lemma states "Assume that $j\_1$ and $j\_2$ are elementary embeddings of ($R\_\lambda$ , $\in$) into itself. Then so is $j\_1[j\_2]$."
Here, $j\_1[j\_2]$ was defined in Definition 3.1, where $j\_1$ and $j\_2$ were elementary embeddings of the limit rank $R\_\lambda$ into itself (by "$R\_\lambda$" Dehornoy means what other authors call $V\_\lambda$, $\lambda$ an ordinal).
$\bigcup\_{\gamma\lt\lambda} j\_1(j\_2\big| R\_\gamma)$
The difficulty I have here is that definition 3.1 does not actually define an elementary embedding of $R\_\lambda$ into itself. $j\_2\big| R\_\gamma$ is a set of ordered pairs whose first element ranges over $R\_\gamma$, and so $j\_1$ of that is a set of ordered pairs whose first element only occurs in the image of $j\_1$. In particular, the elementary embedding being defined is not defined for elements of $R\_\lambda$ not in the range of $j\_1$.
I may have seen some other authors suggest that one can define $j\_1[j\_2]$ to be the identity on the complement of the range of $j\_1$, but Dehornoy does not do this, and this convention would not be consistent with the terminology in the proof of Lemma 3.2 besides.
Similarly, the proof of lemma 3.2 just does not make sense for the same reason.
**Edit:** Here are some notes based on the answer and comments. Perhaps they will be helpful to future readers.
Let $j$ be an elementary embedding from $R\_\lambda$ to $R\_\lambda$. Let $X$ be a proper subset of $R\_\lambda$. Let $\phi$ be the formula $x\in y$. For each $a$ in $X$, $\phi(a, X)$ holds, and so $j(a)$ is in $j(X)$. Let $Y=\lbrace j(x):x\in X\rbrace$, the pointwise application of $j$ to $X$. Then $j(X)=Y\cup K$ where $K$ does not intersect the range of $j$. If $X$ is finite then $K$ is empty. If $X\subset X'$ then $K\subset K'$ where $K'$ corresponds to $X'$ as $K$ does to $X$.
If $X$ is an ordered pair $(x,y)$ then $j(x)$ is the ordered pair $(j(x), j(y))$.
Hence, if $X$ is a function, then $j(X)$ is the function $Y\cup K$, where $Y$ is the pointwise application of $j$ to $X$ consisting of all ordered pairs $(j(x),j(y)):(x,y)\in X)$. $K$ is also a function whose domain does not intersect the range of $j$.
Now, $j\_2$ is the set of ordered pairs $(x,j\_2(x))$ for $x$ in $R\_\lambda$. Hence, $j\_1[j\_2]$ is the union of the set $\lbrace (j\_1(x), j\_1(j\_2(x)): x\in R\_\lambda$ with a function $K$ whose domain does not intersect the range of $j\_1$.
This shows at once that for $z$ in the range of $j\_1$, that $j\_1[j\_2](z)=j\_1(j\_2(j\_1^{-1}(z)))$, motivating the self-distributivity of $[]$.
As to the purpose and structure of these $K$, I have no insight.
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https://mathoverflow.net/users/472518
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Dehornoy's proof that the application of two elementary embeddings is an elementary embedding
|
As Monroe Eskew points out in comments: $j\_1(j\_2|\_{V\_\gamma})$ really means $j\_1$ *applied to* $j\_2|\_{V\_\gamma}$ (not the pointwise image of $j\_2|\_{V\_\gamma}$ under $j\_1$, as you seem to be reading it). So since $j\_2|\_{V\_\gamma}$ is a function $V\_\gamma \to V\_{j\_2(\gamma)}$, you get that $j\_1(j\_2|\_{V\_\gamma})$ is a function $V\_{j\_1(\gamma)} \to V\_{j\_1(j\_2(\gamma))}$, as required.
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7
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https://mathoverflow.net/users/2273
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412296
| 168,228 |
https://mathoverflow.net/questions/412297
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2
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Let $U$ be the forgetful functor from categories to quivers. Then the left adjoint $F$ of $U$ is the functor sending a quiver to its [path category](https://ncatlab.org/nlab/show/path+category#free_category_on_a_directed_graph). [It's a fact](https://mathoverflow.net/questions/410831/does-the-path-category-of-a-quiver-determine-the-quiver-up-to-isomorphism) that $F$ is injective up to isomorphism, i.e., if $Path[G]\cong Path[G']$, then $G\cong G'$.
**Question:** Is it usually the case that "free" functors are injective up to isomorphism? In particular, what can be said about the following setting in universal algebra: for each variety $V$ there is a forgetful functor $U\colon V\to \mathbf{Set}$, which has a left adjoint. Are there examples in which "free" functors are not injective up to isomorphism?
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https://mathoverflow.net/users/471475
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Are free functors usually injective up to isomorphism?
|
Yes, there are examples where $V$ is a variety of algebras and the left adjoint to the forgetful functor $U: V \to \mathbf{Set}$ is not injective on isomorphism classes of objects. Here are two.
* Take the algebraic theory consisting of no operations and the single equation $x = y$. Then $V$ is the category of sets with at most one element, and $U$ is the inclusion. The left adjoint $F$ maps the empty set to the empty set and every nonempty set to $1$.
* Take the algebraic theory consisting of a single constant $c$ and the equation $x = c$. Then $V$ is the terminal category, and $U$ maps its object to the one-element set $1$. The left adjoint $F$ maps everything to $1$.
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6
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https://mathoverflow.net/users/586
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412301
| 168,230 |
https://mathoverflow.net/questions/409830
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3
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Given a rooted tree $T\_r$ (up to isomorphism), define the probability $P(T\_r)$ as the probability of ending up with $T\_r$ if one starts with a single (root) vertex and incrementally connects new vertices one-by-one, such that at each step the vertex being connected to on the existing tree is chosen at random
For example: There are 4 4-vertex rooted trees (up to isomorphism).
3 have probability: $1/6$, and 1 has probability: $1/2$
$v$--$v\_r$--$v$--$v$
I came up with a [Java program](https://github.com/swamikevala/NumberTrees) for calculating these probabilities, but this is only feasible for trees of up to about 12 or 13 vertices.
The question is: Is there any formula or algorithm for calculating the probability of a tree, based on the attributes of 2 or more of its subtrees?
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https://mathoverflow.net/users/155247
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Probability calculation of rooted trees
|
We can see any tree $T$ as the Hasse diagram of a poset whose smallest element is the root. I will freely identify the tree with the corresponding poset.
If we label by $i$ the $i$'th vertex added in the process, we get a rooted tree equipped with a linear extension, and each such equipped rooted tree occur with probability $\frac{1}{(n-1)!}$ (where $n$ is the number of vertices).
The number of ways to get a given rooted tree in this way is the number of linear extensions of this tree up to automorphisms of the rooted tree.
Hence the probability you're looking for is given by
$$\frac{L(T)}{|\text{Aut}(T)| (n-1)!}$$
where $L(T)$ is the number of linear extensions of $T$ and $\text{Aut}(T)$ is the automorphism group of $T$.
There are algorithms for computing the former in $O(n^2)$ (<https://www.researchgate.net/publication/226291352_On_computing_the_number_of_linear_extensions_of_a_tree>). I am not sure for the latter, but it can probably be computed efficiently using a variant of the AHU algorithm.
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2
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https://mathoverflow.net/users/160416
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412308
| 168,233 |
https://mathoverflow.net/questions/412306
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2
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Let $X$ be a smooth complex projective variety *of general type*; in my applications, I work with a surface, but let me ask this question in full generality.
Assume that for some $m \geq 1$ the vector bundle $S^m \Omega\_X^1$ is generated by global sections, namely, the evaluation map $$H^0(X, \, S^m \Omega\_X^1) \otimes \mathcal{O}\_X \to S^m \Omega^1\_X$$ is surjective.
>
> **Question.** Is it true that $K\_X$ is ample? Otherwise, what is a counterexample?
>
>
>
I started working on these topics rather recently, so I apologize if this question turns out to be trivial for the experts. Any answer and/or reference to the relevant literature will be highly appreciated.
**Edit** (12/26/2021). Follow-up question about the base-point freeness of $|K\_X|$ asked as [MO412382](https://mathoverflow.net/questions/412382/projective-variety-of-general-type-such-that-sm-omega-x1-is-globally-genera).
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https://mathoverflow.net/users/7460
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Projective variety of general type such that $S^m \Omega_X^1$ is globally generated
|
If $X$ is a surface it is true. In general, a smooth projective variety with $S^m\Omega ^1\_X$ globally generated does not contain any smooth rational curve $C$. Indeed $\Omega ^1\_C$ is a quotient of $\Omega ^1\_X$, so $S^m\Omega ^1\_C$ is also globally generated, which of course implies $g(C)\geq 1$.
Now if $X$ is a surface, this implies that $K\_X$ is ample (in fact $K\_X$ is ample if and only if $X$ does not contain any smooth rational curve with square $\,-1$ or $-2$).
**Edit:** As pointed out by YangMills in the comments, the result holds in all dimensions: if a smooth projective variety $X$ of general type contains no smooth rational curve, $K\_X$ is ample — see Lemma 2.1 in [arxiv.1606.01381](https://arxiv.org/pdf/1606.01381.pdf).
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6
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https://mathoverflow.net/users/40297
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412311
| 168,234 |
https://mathoverflow.net/questions/412288
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14
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Let $T$ be the solid 2-torus and let $\sim$ be the equivalence relation on $T$ generated by the relation $\{(\alpha,\beta) \sim (\beta,\alpha) \mid \alpha, \beta \in S^1\}$ on the boundary $\partial T=S^1\times S^1$. What kind of space is $T/{\sim}$?
I believe that it is $S^3$, but I can't prove it.
I posted this question already here:
<https://math.stackexchange.com/questions/4339019/identifying-boundary-of-solid-torus-with-itself-by-swapping-coordinates>
Unfortunately, I didn't get a satisfying answer. Not sure if it's because it is harder than I thought or actually nobody cares. Maybe both :)
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https://mathoverflow.net/users/58211
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Quotient of solid torus by swapping coordinates on boundary
|
Your quotient manifold is homeomorphic to $S^3$. I know this because it is a closed 3-manifold and its fundamental group is trivial, so I'm quoting the Poincare conjecture/Perelmann's theorem. The fundamental group can be seen to be trivial because there is a CW-decomposition of the (2-manifold) torus with one 0-cell, three 1-cells and two 2-cells that is respected by your involution, where the 1-cells are the longitude, meridian and diagonal of the torus, and the two 2-cells are triangles. The involution swaps the longitude and meridian while fixing the diagonal pointwise, and swaps the two 2-cells. The quotient of the torus by the involution has one vertex, two 1-cells and one 2-cell. The fundamental group of this space is infinite cyclic, generated by the loop around the 1-cell that started life as both the longitude and meridian of the torus.
This quotient space can be viewed as a copy of the Mobius strip, as in Taras Banakh's comment.
But now think about how the insides fits in: a CW-structure on the solid torus can be made by adding in one more 2-cell, attached around the meridian of the torus, and one 3-cell. The extra 2-cell wraps around the generator for the fundamental group of the piece discussed in the previous paragraph. So the fundamental group of the overall space is trivial.
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12
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https://mathoverflow.net/users/124004
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412314
| 168,235 |
https://mathoverflow.net/questions/412320
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4
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A sequence $a\_n$ is called *log-convex* if $\mathcal{L}(a\_n):=a\_{n+1}a\_{n-1}-a\_n^2\geq0$ for all $n$; it is *infinitely log-convex* provided that all the iterates $\mathcal{L}^k(a\_n)$ are still log-convex, $k\geq1$. Here $\mathcal{L}^2(a\_n)=\mathcal{L}(\mathcal{L}(a\_n))$, etc.
Consider in particular the well-known sequence of Catalan numbers $C\_n=\frac1{n+1}\binom{2n}n$. It has been established that both $C\_n$ and some of its "quantum" versions are log-convex.
I would like to ask:
>
> **QUESTION.** Is it true that $C\_n$ is infintely log-convex? It appears to be so.
>
>
>
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https://mathoverflow.net/users/66131
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Is $C_n$ infinitely log-convex?
|
I think so. If $a\_n=\int\_X f^n(x)d\mu(x)$ for a certain positive function $f$ on a measure space $(X,\mu)$, then $$a\_{n-1}a\_{n+1}-a\_n^2=\int\_{X\times X} f^{n-1}(x)f^{n+1}(y)d\mu(x)d\mu(y)-\int\_{X\times X} f^{n}(x)f^{n}(y)d\mu(x)d\mu(y)\\=\frac12\int\_{X\times X} f^{n}(x)f^{n}(y)\left(\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}-2\right)d\mu(x)d\mu(y)$$
has also a similar representation.
But
$$
C\_n=4^n\frac{(2n-1)!!}{(2n)!!}\cdot \frac1{n+1}=\frac2{\pi}\int\_0^{\pi/2} (2\sin x)^{2n}dx\int\_0^{1} t^n dt.
$$
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10
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https://mathoverflow.net/users/4312
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412329
| 168,239 |
https://mathoverflow.net/questions/412337
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2
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Consider the following statement about a connected, reductive group $G$ over a field $k$:
>
> Every finite, normal subgroup $N$ of $G$ is central.
>
>
>
In characteristic $0$, this is true, and the proof is easy: since $N$ is smooth, it suffices to show that $N(\overline k)$ is contained in $Z(G)(\overline k)$; so we can consider, for $n \in N(\overline k)$, the orbit map $g \mapsto g n g^{-1}$ on $G\_{\overline k}$, whose image is a connected subscheme of $N\_{\overline k}$, hence equals the singleton $\{n\}$.
One cannot reason in the same way in positive characteristic, since not every subgroup scheme of $G$ is smooth. Is the statement nonetheless still true?
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https://mathoverflow.net/users/2383
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Finite, normal subgroups of reductive groups in positive characteristic
|
Oops, it turns out that I already knew the answer to this, in a different context.
This can fail in characteristic $2$: the kernel of the exceptional isogeny $\operatorname{SO}\_{2n + 1} \to \operatorname{Sp}\_{2n}$ is not central. I originally supposed that this sort of exceptional behavior might be peculiar to characteristic $2$ (a supposition to which @YCor responded in a [comment](https://mathoverflow.net/questions/412337/finite-normal-subgroups-of-reductive-groups-in-positive-characteristic/412340#comment1056681_412340)), but @FriedrichKnop points out in a [comment](https://mathoverflow.net/questions/412337/finite-normal-subgroups-of-reductive-groups-in-positive-characteristic/412340#comment1056714_412340) that it is not so: in any non-$0$ characteristic, the kernel of the Frobenius isogeny is not central.
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2
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https://mathoverflow.net/users/2383
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412340
| 168,242 |
https://mathoverflow.net/questions/412156
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3
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I'm reading Richard Ryan's article *"A simpler dense proof regarding the abundancy index"* and got stuck in his proof for Theorem 2. The Theorem is stated as follows:
Suppose we have a fraction of the form $\frac{2n-1}{n}$, where $2n-1$ is prime.
(i) ...
(ii) If $n$ is odd and $I(b)=\frac{2n-1}{n}$ for some $b$, then $b$ is odd; moreover, if $2n-1$ does not divide $b$, then $b(2n-1)$ is a perfect number.
Ryan's proof: Suppose that $n$ is odd and $b$ is even. Let $m$ be the greatest integer such that $2^m$ divides $b$. Once again, there is a prime factor, $q$, of $\sigma(2^m)$ that divides $b$. Thus $I(b) > I(2^mq) > 2$ and we have a contradiction. Finally, if the prime number $2n - 1$ does not divide $b$ then, since $I$ is multiplicative, $I(b(2n - 1)) = 2$.
The 'once again' comes from (i): ... let $I(b)=\frac{2n-1}{n}$ for even $b$ and $m$ be the greatest integer such that $2^m$ divides $b$. There is a prime factor, $q$, of $\sigma(2^m)$ that
also divides $b$ since $\sigma(2^m) = 2^{m+1} - 1 \neq 2n - 1$. Thus
$I(b) > I(2^mq) \geq \frac{2^{m+1}-1}{2^m}\cdot\frac{2^{m+1}}{2^{m+1}-1}=2$,
which contradicts $I(b)=\frac{2n-1}{n}$.
I see that $\sigma(2^m)=2^{m+1}-1\neq 2n-1$, because $n$ is odd and thus $2n$ is not a power of two. It is also possible to easily calculate, that $b\neq2^m$ so $b$ must have some other factor in addition to $2^m$ and that factor has a prime factor, $p$, for which $2^mp$ divides $b$. But Ryan writes that $\sigma(2^m)$ must have the prime factor $q$ and arrives at a contradiction through that. Could someone explain this to me? Also, I don't understand how $I(q)\geq\frac{2^{m+1}}{2^{m+1}-1}$
EDIT: As Jose pointed out, the answer below is incorrect. I think I made some progress in the right direction below.
$I(b)=\frac{\sigma(b)}{b}=\frac{2n-1}{n}\implies n\sigma(b)=b(2n-1)\implies n\sigma(2^m)\sigma(a)=b(2n-1)$, because $b$ is even but not a power of two. Now $\sigma(2^m)=2^{m+1}-1\neq2n-1$ from Ryan's proof makes sense: $\sigma(2^m)$ divides b so it must have the prime factor $q$, which divides $b$. Thus
$I(b)>I(2^mq)=\frac{2^{m+1}-1}{2^m}\cdot\frac{q+1}{q}$. However, I still do not understand how $I(q)=\frac{\sigma(q)}{q}=\frac{q+1}{q}\geq\frac{2^{m+1}}{2^{m+1}-1}$.
Abundancy index $I$ is defined as $I(n)=\frac{\sigma(n)}{n}$, where $\sigma$ counts the sum of divisors. Both $I$ and $\sigma$ are multiplicative.
I did not find a tag for abundancy index.
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https://mathoverflow.net/users/472825
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Help with R. Ryan's "A simpler dense proof regarding the abundancy Index."
|
**Hint:** Since $q \mid \sigma(2^m)$, then
$$q \leq 2^{m+1} - 1,$$
which implies that
$$\frac{1}{q} \geq \frac{1}{2^{m+1} - 1}.$$
Can you finish?
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1
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https://mathoverflow.net/users/10365
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412344
| 168,245 |
https://mathoverflow.net/questions/412324
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5
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The number of the so-called [Baxter permutations of length $n$](https://en.wikipedia.org/wiki/Baxter_permutation) is computed by
$$a\_n=\frac1{\binom{n+1}1\binom{n+1}2}\sum\_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}.$$
There has also been a $q$-analogue of this sequence
$$a\_n(q)=\frac1{\binom{n+1}1\_q\binom{n+1}2\_q}\sum\_{k=0}^{n-1}q^{3\binom{k+1}2}
\binom{n+1}k\_q\binom{n+1}{k+1}\_q\binom{n+1}{k+2}\_q.$$
Hence $a\_n(1)=a\_n$.
Now, if we focus on the special values at $q=-1$ then it appears that the resulting sequence does seem to exhibit small prime factors, suggesting that one might be able to find a closed formula. Notice that such is not true when $q\neq-1$. Here are the first few values of $a\_n(-1)$ for $n\geq1$: $1, 0, 0, -2, -4, 0, 0, 18, 42, 0, 0, -240, -600, 0, 0, 3850, 10010, 0, 0, -68796,\dots$.
So, I like to ask:
>
> **QUESTION.** Is there a nice formula for $a\_n(-1)$? Of course, one may restrict to the non-zero values.
>
>
>
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https://mathoverflow.net/users/66131
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Closed formula for $(-1)$-Baxter sequences
|
Note that $-1$ is a root of the polynomial $q\to {n\choose k}\_q$ of multiplicity
$$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\
0,&\text{otherwise}.\end{cases}$$
Thus ${n+1\choose 2}\_{-1}$ is always non-zero, and ${n+1\choose 1}\_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio
$$h(n,k):=\frac{{n+1\choose k}\_q {n+1\choose k+1}\_q{n+1\choose k+2}\_q}
{{n+1\choose 1}\_q{n+1\choose 2}\_q}\,\,\,\text{at}\,\,\,q=-1
$$
is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}\_q {n+1\choose k+1}\_q{n+1\choose k+2}\_q$ equals 2 in this case, while for ${n+1\choose 1}\_q {n+1\choose 2}\_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus
$$
a\_n(-1)=\sum\_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}=
\sum\_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\
=\sum\_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}.
$$
* If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a\_n(-1)=-a\_n(-1)$ and thus $a\_n(-1)=0$.
* If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a\_n(-1)=-a\_n(-1)=0$.
* Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a\_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}\_{-1}={t\choose \lfloor m/2\rfloor}$$ we get
$$
a\_{4s}(-1)=\frac2{2s}\sum\_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}
$$
(because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that
$$
\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}=
\frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit)
$$
This is similar to [Dixon's identity](https://en.wikipedia.org/wiki/Dixon%27s_identity), and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained [in this answer](https://mathoverflow.net/a/214930/4312), so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to
$$
f(x,y,z)=\prod\_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i)
$$
and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.
* For $n=4s+1$ we again get $h(n,k)=0$ for odd $k$, and for the sum over even $k=2i$ we apply the identity ${n+1\choose k+1}\_q={n+1\choose 1}\_q{n\choose k}\_q/{k+1\choose 1}\_q$. This allows to cancel out the singular term ${n+1\choose 1}\_q$, and applying ${k+1\choose 1}\_{-1}=0$ for even $k$ we get
$$
a\_{4s+1}(-1)=\frac1{2s+1}\sum\_{i=0}^{2s}(-1)^i{2s+1\choose i}{2s+1\choose i+1}{2s\choose i}.
$$
This corresponds to Dyson's conjecture (now a theorem) $${\rm CT} \prod\_{i\ne j}(1-x\_i/x\_j)^{a\_i}=\frac{(\sum a\_i)!}{\prod a\_i!}$$
for $(a\_1,a\_2,a\_3)=(s,s,s+1)$, and we get Roland Bacher's answer
$$
a\_{4s+1}(-1)=\frac{(-1)^s}{2s+1}{3s+1\choose s,s,s+1}.
$$
|
7
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https://mathoverflow.net/users/4312
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412356
| 168,247 |
https://mathoverflow.net/questions/412358
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9
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I have come across the word 'period' in several contexts and I wonder if these notions are related.
(1) The period map and domain: Let $ \pi : X \rightarrow B $ be a proper holomorphic submersion of complex manifolds. If a fiber $ X\_0 $ of $ \pi $ is (compact) Kahler, then the cohomology groups $ H^k(X\_0, \mathbb{Z}) $ (modulo torsion) are Hodge structures of weight $ k $. Now if we consider a small enough neighborhood $ U $ of $ 0 \in B $, then all fibers of U are Kahler, the Hodge numbers $ h^{p,q} $ are constant, and there is a fixed isomorphism (#) $ H^k(X\_b, \mathbb{C}) \cong H^k(X\_0, \mathbb{C}) $ from Ehresmann's theorem. So the numbers $ a\_p = \text{dim} F^pH^k(X\_b, \mathbb{C}) $ are also constant. The period map then sends $ b \in U $ to the (image under (#) of the) flag $ \{F^pH^k(X\_b, \mathbb{C})\}\_p $ in the flag variety $ G(a\_k , \ldots , a\_1, H^k(X\_0, \mathbb{C})) $. The period domain is the subset of the flag variety whose flags satisfy $ F^pH^k(X\_0, \mathbb{C}) \oplus \overline{F^{k+1-p}H^k(X\_0, \mathbb{C})} = H^k(X\_0, \mathbb{C}) $.
(2) Period: A real number is a period if it is an integral $ \int\_{p \ge 0} q(x\_1, \ldots, x\_n) dx $ where $ p $ is a polynomial with rational coefficients and $ q $ is a rational function with rational coefficients. We can allow $p,q $ to be algebraic functions, this is equivalent. So abelian integrals $ \int\_0^r \frac{dx}{\sqrt
{4x^3 - ax - b}} $ are periods ($ r,a,b \in \mathbb{Q} $).
(3) Period of a cohomology group: I came across this notion after reading Daniil Rudenko's [question](https://mathoverflow.net/questions/332387/are-we-better-in-computing-integrals-than-mathematicians-of-19th-century) on this website. This one I do not understand at all.
So my question is: How are these notions related? Surely (2) and (3) are related as you're supposedly computing an integral in (3) as well. And I suspect (1) and (3) are related as well as you're dealing with some cohomology group.
I apologize if the question is a bit too general but I'd like to understand this.
|
https://mathoverflow.net/users/152391
|
Different occurences of the word 'period' in algebraic geometry
|
The second and the third are pretty much equivalent.
Indeed, "the period" in XIX century sense is essentially
the same as the discrepancy between the branches of a
multi-valued function, obtained as an integral of an
algebraic function. If you take the Riemann surface
associated with this algebraic function (that is,
a branched covering of C where it is well-defined),
the period becomes an integral of the holomorphic
differential $fdz$ associated with this function
over a closed loop, that is, an integral of a
holomorphic 1-form over a closed cycle.
The modern definition of "periods" is, more or less,
"the pairing between holomorphic differential forms
and integral homology". However, this notion can
(and often is) extended to Hodge structures, and
this is related to the first notion you mention.
Define the Teichmuller space of a complex manifold
as the space of all complex structures (here the
complex structures are understood as endomorphisms
of $TM$ satisfying $I^2=-Id$ and the integrability
condition) up to isotopies: $Teich= \frac{Comp}{Diff\_0}$.
Never mind that this space might be horribly non-Hausdorff;
the period map is naturally defined on $Teich$, and
(if you are lucky) it would help to make sense even of the
non-Hausdorff pathologies.
The period map is the map taking $I\in Teich$
to the Hodge structure on $H^\*(M,I)$, which is
understood as a point in the appropriate flag space.
This map is holomorphic, and, if you are lucky,
defines a biholomorphism between $Teich$ and
the space of Hodge structures ("period space").
This holds, unfortunately, only for complex tori,
but weaker versions of this statement are true
for K3, hyperkahler manifolds, complex curves
and in some other cases. These results are
called "global Torelli theorems".
The local Torelli theorems are statements about
the local structure of the period map, saying
(in most cases) that is it locally an immersion;
this is true, for instance, for Calabi-Yau manifolds.
The modern definition of periods is not
entirely equivalent to the traditional, because
the Hodge structure contains more data than
just holomorphic forms; however, if you know
the Hodge structure on cohomology, you know
the pairing between the holomorphic forms
and the integer homology, hence it is an
extension of the XIX century notion.
If you restrict yourself to the first cohomology,
the Hodge structure is a flag $H^{1,0}(M)\subset H^1(M)$;
in this case the XIX century notion of "periods"
coincides with the Hodge-theoretic notion.
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9
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https://mathoverflow.net/users/3377
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412360
| 168,248 |
https://mathoverflow.net/questions/412357
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1
|
With COVID19 becoming a pandemic I saw some researchers trying to model it with fractional differential equations (FDE) instead of ordinary differential equations (ODE). From a technical standpoint I wondered what's the reason in doing so. I don't see how FDEs appear "naturally" when modelling diseases. For example lets look at a simple SIR-model
$$\frac{dS}{dt} = -\beta IS $$
$$\frac{dI}{dt}=\beta IS - \gamma I$$
$$\frac{dR}{dt} = \gamma I$$
With $S$ being the amount of susceptible persons, $I$ being the amount of infectious persons and $R$ being the amount of recovered persons. $\beta > 0$ the average number of contacts per person and $\gamma>0$ being the transition rate.
What would be the benefit of looking at it as a fractional differential equation with parameter $\alpha \in (0,1)$? (Let's say e.g. with a Caputo-Derivative)
I know, that a major difference between ODE and FDE is, that the solution of the FDE is not "local" in a sense that the solution in a point $t\_1$ depends on the values of the solution on the whole intervall $[0,t\_1]$. Whereas in ODE this is not really the case, we don't necessarily need the information about the values of the solution on the whole interval $[0,t\_1]$. Often this is interpreted as the solution having a "memory" but why is it good for modelling epidemics, that the solution, i.e. the amount of susceptible, infectious and recovered people has a "memory"?
Apart from that, are there any technical advantages from using FDE instead of ODE? I know, that we have slightly different results regarding the stability of equilibrium points. (if we look at $u'=f(u)$, with equilibrium point $u\_\*$, then by the principle of linearized stability, we know that $u\_\*$ is asymptotically stable if the Eigenvalues $\lambda$ of $f'(u\_\*)$ have negative real part. Whereas for FDE of order $\alpha$ we have asymptotic stability if $|arg(\lambda)|>\frac{\alpha \pi}{2}$). (see [this](https://arxiv.org/abs/1512.04989) paper for a proof of this statement). However I don't see how this would "help" us in the model given above.
**Question**
Why should we use FDEs instead of ODEs, when it comes to modelling infectious diseases? What are FDEs able to do, that ODEs are not? Is there a way to show that FDEs appear "naturally" in modelling diseases? (Literature hints are always appreciated!) Thanks in advance!
P.S. I already asked this question on Math stackexchange two weeks ago, however I thought mathoverflow may be better suited for this question. If its not I apologize in advance!
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https://mathoverflow.net/users/473107
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Why should we model infectious diseases with fractional differential equations?
|
There are two issues here: Firstly, fractional derivatives are non-local operators, so they can be used to model processes with a "memory", where the prior history governs the future evolution. Secondly, the exponent of the fractional derivative can be used as a fit parameter to improve the agreement with data.
Studies in the literature based on the [compartmental models](https://en.wikipedia.org/wiki/Compartmental_models_in_epidemiology) for the spread of an infectious diseases typically compare a bit better with data when the ODE is substituted by a FDE. The improvement is not large and might well be due mainly to the additional fit parameter. The memory effect might as well be included by introducing additional compartments, I don't think there is a truly compelling reason to prefer FDE over ODE.
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1
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https://mathoverflow.net/users/11260
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412361
| 168,249 |
https://mathoverflow.net/questions/412345
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2
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Let $f:[0,\infty)\to [0,\infty)$ be an increasing function satisfying
$$\int\_0^\infty f(x)\frac{dx}{1+x^2}=\infty.$$
Can we find a continuous increasing function $F$ on $[0,\infty)$ satisfying
$$\int\_0^\infty F(x)\frac{dx}{1+x^2}=\infty$$
and $F(x)\leq f(x)?$
|
https://mathoverflow.net/users/184109
|
Given an increasing function $f$, to find a continuous function satisfying properties of $f$
|
Yes. Since $f$ is increasing, it is almost everywhere continuous and in particular locally integrable. So we can define $F$ by $F(t)=0$ for $t<1$ and $F(t)=\int\_{t-1}^tf(x)dx$ for $t\geq1$. This function will be continuous and increasing and satisfies $f(x-1)\leq F(x)\leq f(x)$. From the first inequality you can shown that the integral $\int\_0^\infty F(x)\frac{dx}{1+x^2}$ diverges.
|
2
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https://mathoverflow.net/users/470870
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412367
| 168,253 |
https://mathoverflow.net/questions/412346
|
0
|
Suppose that we have a bounded polynomial defined on $[0,1]$. I think because it is just polynomial, root finding algorithms would easily and without any instability find all its roots. Am I right?
And what is the fastest and most stable root-finding algorithm for my problem?
Edit: I introduce my polynomial as:
$$ f(x) = B\_n(x) - x$$
$B\_n(x)$ is a bernstein polynomial with positive and strictly increasing coefficients $a\_k$:
$$B\_n(x) = \sum\_{k=1}^n \binom n ka\_kx^k (1-x)^{n-k}$$
$$a\_0=0 < a\_1 < a\_2 < ... < a\_n=1$$
|
https://mathoverflow.net/users/155971
|
Are root finding algorithms stable for bounded polynomials?
|
Finding **real** roots is not going to be stable, even if you assume the polynomial to be monic and have bounds on where the interesting stuff happens. As an example, consider $(x^2 + \varepsilon)(x^2 - 2x + 1 - \varepsilon)$. If $\varepsilon < 0$, the only real root is at $0$, if $\varepsilon > 0$ the only real root is at $1$.
If you want a guarantee that the algorithm will converge to a root with bounds on the convergence speed, you'd need to either promise the existence of roots with odd multiplicty, or be willing to accept complex roots as solutions.
|
2
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https://mathoverflow.net/users/15002
|
412375
| 168,256 |
https://mathoverflow.net/questions/412372
|
7
|
Let $f\colon X\to Y$ be a continuous map. Then $f$ induces a geometric morphism $f^\ast\dashv f\_\ast\colon \mathrm{Sh}(X)\leftrightarrows\mathrm{Sh}(Y)$, whose left adjoint is called *inverse image* and whose right adjoint is called *direct image*.
Why is the left adjoint called *inverse image* and why is the right adjoint called *direct image*?
Especially the following makes it confusing: The *direct image* $f\_\ast$ is defined by
$$F\in\mathrm{Sh}(X)\mapsto (U\in\mathcal O(Y)\mapsto F(f^{-1}(U))).$$
[Usually](https://en.wikipedia.org/wiki/Image_(mathematics)#Inverse_image) one calls $f^{-1}(U)$ the *inverse image* of $U$ under $f$.
|
https://mathoverflow.net/users/471475
|
Direct and inverse image terminology
|
There is a precise, almost literal, sense in which $f^\* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$ generalises the inverse image as defined in elementary set theory.
Observe that open subspaces $V \subseteq Y$ correspond to subterminal objects in $\textbf{Sh} (Y)$: the sheaf of sections of the inclusion $V \hookrightarrow Y$ is a subterminal object, and every subterminal object is isomorphic to one of this form.
Since $f^\*$ preserves finite limits, it preserves subterminal objects, and in fact it sends the sheaf corresponding to $V$ to the sheaf corresponding to $f^{-1} V$.
So we may think of $f^\* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$ as being an extension of the set-theoretic inverse image operation, and this justifies the name "inverse image functor".
Once you have an "inverse image functor", there is a powerful temptation to call its partner the "direct image functor".
My personal opinion is that the name "direct image functor" is unsuitable for the *right* adjoint of the inverse image functor.
For one thing, $f\_\* : \textbf{Sh} (X) \to \textbf{Sh} (Y)$ is not a generalisation of the set-theoretic direct image operation, as I will now explain.
Let $U$ be an open subspace of $X$.
What do you suppose $f\_\*$ applied to the subterminal object corresponding to $U$ yields?
It is a subterminal object, of course, but it is not (the subterminal object corresponding to) the direct image
$$\exists\_f U = \{ y \in Y : \exists x \in U . y = f (x) \} = \{ y \in Y : \exists x \in X . y = f (x) \land x \in U \}$$
which is not even guaranteed to be open if we do not assume $f : X \to Y$ is an open map.
Instead, it corresponds to
$$\forall\_f U = \{ y \in Y : f^{-1} \{ y \} \subseteq U \} = \{ y \in Y : \forall x \in X . y = f (x) \Rightarrow x \in U \}$$
which is always open: indeed, $\forall\_f U$ is the union of all open $V \subseteq Y$ such that $f^{-1} V \subseteq U$.
Unfortunately, the reality is that $f^\* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$ does not always have a left adjoint that generalises $\exists\_f$, so while it is $\exists\_f$ that bears the name "direct image" in elementary set theory, for want of a better name it is $f\_\*$ that gets the name in topos theory.
|
10
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https://mathoverflow.net/users/11640
|
412376
| 168,257 |
https://mathoverflow.net/questions/412365
|
12
|
Is there a geometric theory $T$ and a Grothendieck topos $\mathcal E$ such that (2) holds but (1) doesn't:
1. $\mathcal E$ 2-represents the 2-functor
$$\mathbf{GrothTop}\to\mathbf{Cat}$$
which sends a Grothendieck topos $\mathcal E$ to the category of models of $T$ in $\mathcal E$.
2. $\mathcal E$ represents the 1-functor
$$\mathrm h\mathbf{GrothTop}\to\mathrm h\mathbf{Cat}\to \mathbf{Set}$$
which can be obtained by truncating the above functor to the 1-categorical level and composing with the functor sending a category $\mathcal C$ to the set $\mathrm{Ob}(\mathcal C)/\mathord{\cong}$ of all objects of $\mathcal C$ up to isomorphism.
Conversely, (1) implies (2), right?
|
https://mathoverflow.net/users/471475
|
Definition of "classifying topos"
|
This is a bit surprising to me, but the two statements turn out to be equivalent.
A topos $\mathcal{E}$ is completely determined by the functor
$$\mathbf{Geom}(-,\mathcal{E}) : \mathbf{GrothTop}^\mathrm{op} \to \mathbf{Cat}$$ (this is some kind of 2-Yoneda Lemma). But we can also look at the 1-category $\mathrm{h}\mathbf{GrothTop}$ in which we identify geometric morphisms that are isomorphic. Then the (1-categorical) Yoneda Lemma says that $\mathcal{E}$ is completely determined by the functor
$$\mathbf{Geom}(-,\mathcal{E})/\!\cong\,\, : \mathrm{h}\mathbf{GrothTop}^\mathrm{op} \to \mathbf{Class}$$
to the category of classes (we have to work with classes rather than sets because there can be a proper class of geometric morphisms up to isomorphism between two toposes).
In particular, suppose that $\mathbf{Geom}(-,\mathcal{E})/\!\cong\,\,$ agrees with the functor that sends each Grothendieck topos $\mathcal{F}$ to the collection of $T$-models in $\mathcal{F}$ up to isomorphism, for some geometric theory $T$. Then $\mathcal{E}$ is the classifying topos of $T$. To prove this, we need to use the fact that the classifying topos of a geometric theory always exists (thanks to @Mike Shulman for pointing this out).
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10
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https://mathoverflow.net/users/37368
|
412387
| 168,260 |
https://mathoverflow.net/questions/412385
|
15
|
While exploring the Baxter sequences from [my earlier MO post](https://mathoverflow.net/questions/412324/closed-formula-for-1-baxter-sequences), I obtained a rather curious identity (not listed on OEIS either). I usually try to employ the Wilf-Zeilberger (WZ) algorithm to justify such claims. To my surprise, WZ offers **two different** recurrences for each side of this identity.
So, I would like to ask:
>
> **QUESTION.** Is there a conceptual or **combinatorial reason** for the below equality?
> $$\frac1n\sum\_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}
> =\frac2{n+2}\sum\_{k=0}^{n-1}\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}.$$
>
>
>
**Remark 1.** Of course, one gets an alternative formulation for the Baxer sequences themselves:
$$\sum\_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}}{\binom{n+1}1\binom{n+1}2}
=2\sum\_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}}{\binom{n+1}1\binom{n+2}2}.$$
**Remark 2.** Yet, here is a restatement to help with combinatorial argument:
$$\sum\_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}
=2\sum\_{k=0}^{n-1}\binom{n+1}k\binom{n}k\binom{n+1}{k+2}.$$
|
https://mathoverflow.net/users/66131
|
A rather curious identity on sums over triple binomial terms
|
Just playing around with it: The RHS multiplied by $n$ is the same as
$$2 \sum\_{k=0}^{n-1} \binom{n+1}{k} \binom{n}{k} \binom{n+1}{k+2}.$$
Subtracting this from $n$ times the LHS gives
$$\sum\_{k=0}^{n-1} \binom{n+1}{k} \binom{n+1}{k+2} \left( \binom{n}{k+1} - \binom{n}{k} \right).$$
Now you check that replacing $k$ with $n-1-k$ changes the sign of the summand, so the sum is zero.
|
22
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https://mathoverflow.net/users/20598
|
412389
| 168,261 |
https://mathoverflow.net/questions/412373
|
1
|
Let $(p\_1)^{k\_1}(p\_2)^{k\_2}\dots$ be the prime factorization of $\varphi(n)$. Assuming that we have a value of order $(p\_x)^{k\_x}$ for all $x$, can we calculate the discrete log of any value in $\mathbb{Z}\_n^\times$ efficiently? Alternatively, if we have a value of order $p\_x$ for all $x$, can we calculate the discrete log efficiently?
Here the interest is in the discrete log of any base, i.e. for the general case. There may be algorithms that work well for a specific base, but here the focus should be on algorithms for all bases.
|
https://mathoverflow.net/users/24942
|
Does having the discrete logarithm of prime factors of $n$ allow us to calculate any discrete log more efficiently?
|
There are no known general-purpose algorithms that can compute the discrete logarithm efficiently with this additional information.
A special case of your question arises when $n$ is a "[safe prime](https://en.wikipedia.org/wiki/Safe_and_Sophie_Germain_primes)," i.e. a prime number such that $p=\frac{n-1}{2}$ is also prime (a Sophie-Germain prime). In this case, $\varphi(n)=2p$. Then $-1$ has order $2$, and it is very straightforward to find an element of order $p$ (pick any $g\in(\mathbb{Z}/n\mathbb{Z})^\times$ besides $1$ and $-1$, and compute $g^p$ using square-and-multiply. If $g^p=1$, then $g$ has order $p$; otherwise $g^p=-1$ and $g^2$ has order $p$). However, safe primes are often [explicitly sought out](https://doi.org/10.1007/0-387-23483-7_367) as moduli for Diffie-hellman protocols precisely because there is no known attack on the discrete log problem for such $n$.
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2
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https://mathoverflow.net/users/404359
|
412408
| 168,267 |
https://mathoverflow.net/questions/412207
|
1
|
Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). If $\kappa>0$ is a cardinal, we say that $H$ is $\kappa$-*uniform* if $|e|=\kappa$ for all $e\in E$.
If $X$ is a non-empty set, then a map $c:V\to X$ is said to be a *colouring* if for every edge $e\in E$ with $|e|\geq 2$ the restriction $c\restriction\_e: e\to X$ is not constant. The smallest cardinal $\kappa$ for which there is a colouring $c:V \to \kappa$ is said to be the *chromatic number* of $H$ and is denoted by $\chi(H)$.
The *dual* of the hypergraph $H=(V,E)$ is $H^\partial = (E, E\_V)$ where $$E\_V = \big\{S\subseteq E: \text{there is }v\_0\in V \text{ such that } S = \{e\in E: v\_0\in e\}\big\}.$$
**Question.** Given cardinals $\alpha, \beta \geq 2$, is there an $\alpha$-uniform hypergraph $H=(V,E)$ with $\chi(H) = \beta$ and $\chi(H^\partial) = 2$?
|
https://mathoverflow.net/users/8628
|
Chromatic number of duals of uniform hypergraphs with large edges
|
**Theorem.** For any cardinals $\alpha,\beta\ge2$ there is an $\alpha$-uniform hypergraph $H$ with $\chi(H)=\beta$ and $\chi(H^\partial)=2$.
**Proof.** Let $V=\bigcup\_{\xi\in\beta}V\_\xi$ where the sets $V\_\xi$ are pairwise disjoint and $|V\_\xi|\gt\alpha\beta$. Let $E=\{e\in[V]^\alpha:|\{\xi\in\beta:e\cap V\_\xi\ne\varnothing\}|\ge2\}$.
Plainly $H=(V,E)$ is an $\alpha$-uniform hypergraph and $\chi(H)=\beta$. To see that $\chi(H^\partial)=2$ color each vertex red or blue so that each set $V\_\xi$ contains at least $\alpha$ vertices of each color. Then each vertex of $H$ is contained in both a monochromatic edge and a bichromatic edge, whence $\chi(H^\partial)=2$.
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3
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https://mathoverflow.net/users/43266
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412412
| 168,269 |
https://mathoverflow.net/questions/412369
|
3
|
I see that Lipschitz continuity is a common assumption used in optimisation, statistics, machine learning, etc.
Could you point me in the direction of some literature that discusses why Lipschitz continuity is commonly assumed? Does it naturally occur in applications? Contains an important class of functions?
EDIT: I’ve found that it’s central for ODEs because of Picard-Lindelöf theorem but I’d like to have something closer to statistics or optimization.
|
https://mathoverflow.net/users/104860
|
Reference request: importance of Lipschitz continuity
|
**In Mathematical/High Dimensional Statistics:**
One fairly amazing result is for $X=(X\_1,\dots,X\_n)$ where the coordinates are i.i.d. standard Gaussians, and $f:R^n \to R$ a $L$-Lipschitz function (w.r.t. Euclidean norm), then the random variable $f(X) - E[f(X)]$ is sub-Gaussian with sub-Gaussian norm $L$, i.e. it satisfies for any $\epsilon \ge 0$:
$$
P(|f(X)- E[f(X)]| \ge \epsilon) \le 2 e^{-\epsilon^2/2 L^2}.
$$
It is somewhat surprising (to me at least) how often this result comes in handy - particularly when working with rademacher/gaussian complexities in mathematical statistics/learning theory and also in random matrix analysis.
To give one concrete example, if $X$ is a $n \times p$ random matrix with i.i.d. standard gaussian entries, and $Y$ is another matrix, then by Weyl's inequality:
$$
\max\_{k=1,\dots, p} |\sigma\_k(X) - \sigma\_k(Y)| \le \|X-Y \|\_F,
$$
where $\sigma\_k$ is the $k$-th largest singular value. This tells us that the $k$-th singular value is a 1-Lipschitz function of the matrix, and so we immediately get
$$
P(|\sigma\_k(X) - E[\sigma\_k(X)]| \ge \epsilon) \le 2e^{-\epsilon^2/2}.
$$
I would suggest the following two books for more on this kind of usage of Lipschitz continuity:
1. High Dimensional Statistics - A non-asymptotic viewpoint by Martin J Wainwright
2. High Dimensional Probability by Roman Vershynin
**In machine learning:**
The concept of Lipschitz continuity is also important in the current machine/deep learning literature where model robustness (which they refer to as adversarial robustness) is an important issue. To understand the robustness of a complicated model (such as a neural network), a lot of work has gone into training networks that define an input-output map with small Lipschitz constant. The intuition here is that if my model is robust, it should not be too affected by perturbations in the input, $f(x+\delta x) \approx f(x)$, and this would be ensured by having $f$ be L-Lipschitz where $L$ is small. See this paper [Training robust neural networks using Lipschitz bounds](https://arxiv.org/pdf/2005.02929.pdf) and their references for more on this.
|
7
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https://mathoverflow.net/users/473170
|
412418
| 168,270 |
https://mathoverflow.net/questions/412271
|
10
|
In an [article](https://www.semanticscholar.org/paper/A-potential-analogue-of-Schinzel%27s-hypothesis-for-Bender-Wittenberg/cc4b8b7082c3ac64809adf5d333ed1c9a6d83f01), I've found the following result. Unfortunately, it was derived from a general, somewhat complicated theory, that would be cumbersome for this result alone.
>
> Assume that $\mathbb F\_p$ is the field with $p$ elements, and that we choose an integer $a$ with $1< a < 4p$, such that $a$ be coprime to $p(p-1)$ (for example, $a = 2p - 1$ is suitable). Let $t$ be transcendental over $\mathbb F\_p$, and set $K = \mathbb F\_p(t)$. Consider the polynomial $$P(X) = X^{4p} +t^a$$ in $K[X]$.
>
>
> Then for every $f\in \mathbb F\_p[t]$, $P(f)$ is reducible in $\mathbb F\_p[t]$.
>
>
>
I would like a direct proof of this result. Any idea?
|
https://mathoverflow.net/users/62826
|
Proving that polynomials belonging to a certain family are reducible
|
It follows from [this](https://mathscinet.ams.org/leavingmsn?url=http://projecteuclid.org/euclid.pjm/1103036322) result of Swan: If $t$ divides $f(t)$ there is nothing to do. So assume that this is not he case. Then with $F(t)=P(f(t))=f(t)^{4p}+t^a$, $F'(t)=at^{a-1}$ and $F(t)$ is separable. From this one easily gets that the discriminant of $F(t)$ is a square in $\mathbb F\_p$ (see the alternative formula for the discriminant at the beginning of the article of Swan). By Swan's result, the number of irreducible factors of $F(t)$ is congruent to $\deg F(t)=4p\deg f$ mod $2$, hence it is even and therefore $>1$.
*Note added later:* In fact the special case needed here is already due to Dickson, as pointed out by Swan in his paper. The argument is easy: If $F(t)$ is separable of degree $n$ and irreducible, then its Galois group contains an $n$-cycle. If $n$ is even, then this $n$-cycle is an odd permutation, so the discriminant of $F(t)$ is not a square. (Assuming that we are working over a finite field of odd characteristic.)
|
9
|
https://mathoverflow.net/users/18739
|
412419
| 168,271 |
https://mathoverflow.net/questions/412421
|
2
|
I'm wondering if the following conjecture is true:
>
> Let $\mathcal{A}$ be an isogeny class of elliptic curves over $\mathbf{Q}$. Fix an odd prime $p$ of good reduction. Then there is a curve $E \in \mathcal{A}$ such that the quantity $$\dfrac{L(E,1)}{\Omega\_E}$$ is a $p$-adic unit.
>
>
>
That is, the quantity $\dfrac{L(E,1)}{\Omega\_E}$ can always be made a $p$-adic unit after "shifting by an isogeny". I wanted to ask:
1. Is this conjecture known to be true? It seems like if this is true, this would be well-known, but I'd like to confirm.
2. If it is true, does anyone have a reference?
|
https://mathoverflow.net/users/394740
|
$p$-adic valuation of $L$ values for elliptic curves
|
The conjecture is **False**.
Let $\mathcal A$ be the isogeny class [6690j](https://www.lmfdb.org/EllipticCurve/Q/6690/j/) and $p=7$. The quantities $\dfrac{L(E,1)}{\Omega\_E}$ for the two [elliptic](https://www.lmfdb.org/EllipticCurve/Q/6690/j/2) [curves](https://www.lmfdb.org/EllipticCurve/Q/6690/j/1) are $7$ and $49$ respectively, neither a $p$-adic unit.
|
2
|
https://mathoverflow.net/users/125498
|
412423
| 168,273 |
https://mathoverflow.net/questions/406701
|
2
|
(**Preamble:** We have asked this same question in [MSE](https://math.stackexchange.com/questions/4269988) two weeks ago, without getting any answers. We have therefore cross-posted it to MO, hoping that it gets answered here.)
The topic of [odd perfect numbers](https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers) likely needs no introduction.
Denote the *classical sum of divisors* of the positive integer $x$ by $\sigma(x)=\sigma\_1(x)$.
If $n$ is odd and $\sigma(n)=2n$, then we call $n$ an *odd perfect number*. Euler proved that a hypothetical odd perfect number must necessarily have the form $n = p^k m^2$ where $p$ is the *special prime* satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Descartes, Frenicle, and subsequently [Sorli](http://hdl.handle.net/10453/20034) conjectured that $k=1$ always holds. Dris conjectured that the inequality $p^k < m$ is true in his [M. Sc.](https://arxiv.org/abs/1204.1450) [thesis](https://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html).
[Brown (2016)](https://arxiv.org/abs/1602.01591) eventually produced a proof for the weaker inequality $p < m$. (Pomerance informed Brown that our past proofs for the estimate $p < m$ may be (inherently) flawed. See this [MO question](https://mathoverflow.net/questions/443950) for the details of Pomerance's rebuttal.)
Now, [recent](https://mathoverflow.net/questions/376268) [evidence](https://mathoverflow.net/questions/393738) suggests that $p^k < m$ may in fact be false.
**THE ARGUMENT**
Let $n = p^k m^2$ be an odd perfect number with special prime $p$.
Since $p \equiv k \equiv 1 \pmod 4$ and $m$ is odd, then $m^2 - p^k \equiv 0 \pmod 4$. Moreover, $m^2 - p^k$ is **not a square** ([Dris and San Diego (2020)](http://nntdm.net/volume-26-2020/number-4/33-38/)). Note: I am already too sleepy to add the details, but Dris and San Diego's proof in NNTDM for the statement "$m^2 - p^k$ is **not a square**" is incomplete and flawed as originally published since our proof makes use of the estimate $p < m$. (Nonetheless, it is still *possible* to prove the same statement, essentially by using Acquaah and Konyagin's result in [(IJNT, 2012)](https://www.worldscientific.com/doi/abs/10.1142/s1793042112500935). Will submit a corrigendum for [(Dris and San Diego (2020))](http://nntdm.net/volume-26-2020/number-4/33-38/) to NNTDM as soon as possible.)
This implies that we may write
$$m^2 - p^k = 2^r t$$
where $r \geq 2$ and $2^r \neq t$. (I have removed the condition $\gcd(2,t)=1$ since we still have not completely ruled out the possibility that $m^2 - p^k$ may be a power of two. We do know that under the assumption that $k=1$ holds, then $m^2 - p^k$ is not a power of two when $p$ is a Fermat prime.)
It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:
$$\text{Case (1): } m > t > 2^r$$
$$\text{Case (2): } m > 2^r > t$$
$$\text{Case (3): } t > m > 2^r$$
$$\text{Case (4): } 2^r > m > t$$
$$\text{Case (5): } t > 2^r > m$$
$$\text{Case (6): } 2^r > t > m$$
Note that these six (6) cases may be summarized as:
$$\text{Case (A): } m > \max(2^r, t)$$
Cases (1) and (2) are included under this Case (A).
$$\text{Case (B): } \max(2^r, t) > m > \min(2^r, t)$$
Cases (3) and (4) are included under this Case (B).
$$\text{Case (C): } \min(2^r, t) > m$$
Cases (5) and (6) are included under this Case (C).
---
We can easily rule out Case **(5)** and Case **(6)**, as follows:
Under Case **(5)**, we have $m < t$ and $m < 2^r$, which implies that $m^2 < 2^r t$. This gives
$$5 \leq p^k = m^2 - 2^r t < 0,$$
which is a contradiction.
Under Case **(6)**, we have $m < 2^r$ and $m < t$, which implies that $m^2 < 2^r t$. This gives
$$5 \leq p^k = m^2 - 2^r t < 0,$$
which is a contradiction.
---
Under Case **(1)** and Case **(2)**, we can prove that the inequality $m < p^k$ holds, as follows:
Under Case **(1)**, we have:
$$(m - t)(m + 2^r) > 0$$
$$p^k = m^2 - 2^r t > m(t - 2^r) = m\left|2^r - t\right|.$$
Under Case **(2)**, we have:
$$(m - 2^r)(m + t) > 0$$
$$p^k = m^2 - 2^r t > m(2^r - t) = m\left|2^r - t\right|.$$
---
So we are now left with Case **(3)** and Case **(4)**.
Under Case **(3)**, we have:
$$(m + 2^r)(m - t) < 0$$
$$p^k = m^2 - 2^r t < m(t - 2^r) = m\left|2^r - t\right|.$$
Under Case **(4)**, we have:
$$(m - 2^r)(m + t) < 0$$
$$p^k = m^2 - 2^r t < m(2^r - t) = m\left|2^r - t\right|.$$
Note that, under Case **(3)** and Case **(4)**, we actually have
$$\min(2^r,t) < m < \max(2^r,t).$$
But the condition $\left|2^r - t\right|=1$ is sufficient for $p^k < m$ to hold.
Our inquiry is:
>
> **QUESTION:** Is the condition $\left|2^r - t\right|=1$ also necessary for $p^k < m$ to hold, under Case **(3)** and Case **(4)**?
>
>
>
Note that the condition $\left|2^r - t\right|=1$ contradicts the inequality
$$\min(2^r,t) < m < \max(2^r,t),$$
under the remaining Case **(3)** and Case **(4)**, and the fact that $m$ is an integer.
|
https://mathoverflow.net/users/10365
|
On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II
|
>
> **QUESTION:** Is the condition $\left|2^r - t\right|=1$ also necessary for $p^k < m$ to hold, under Case **(3)** and Case **(4)**?
>
>
>
The answer is no.
It is true that $p^k\lt m\implies |2^r-t|\not=1$.
*Proof* :
Since you ruled out (5) and (6), and showed that, under (1) or (2), $m < p^k$ holds, one can say that it is true that $p^k\lt m\implies$ (3) or (4).
Also, since, under (3) or (4), $2^r$ and $t$ cannot be consecutive integers, one can say that $|2^r-t|\not=1$.
As a result, one can say that
$$p^k\lt m\implies (3)\ \text{or}\ (4)\implies |2^r-t|\not=1.\quad\blacksquare$$
|
1
|
https://mathoverflow.net/users/34490
|
412427
| 168,274 |
https://mathoverflow.net/questions/412413
|
1
|
On the first page of the old paper [Solution of the first boundary value problem for an equation of continuity of an incompressible medium](https://www.researchgate.net/publication/281324404_Solution_of_the_first_boundary_value_problem_for_an_equation_of_continuity_of_an_incompressible_medium) of Bogovskii, the notations $W\_p^l(\Omega)$ and $L\_p^l(\Omega)$ are used without explanation.
I assume that $W\_p^l(\Omega)$ denotes the usual Sobolev space of functions in $L\_p(\Omega) := \{ f : \Omega \to \mathbb{R} : f \textrm{ is measurable and } \int\_{\Omega} |f|^p < \infty \}$ whose weak derivatives of order up to $l$ are also in $L\_p(\Omega)$.
I have not seen the notation $L\_p^l(\Omega)$ used before, but according to the first sentence of this paper it is a "Sobolev space". **What does this notation mean?**
|
https://mathoverflow.net/users/473159
|
The meaning of $L_p^l(\Omega)$ in a paper of Bogovskii on Sobolev spaces
|
In [1], chapter 1, §1.1.2 p. 2, $L\_p^l(\Omega)$ is defined as the space of distributions on $\Omega$ whose derivatives of order $l$ belong to the space $L\_p(\Omega)$ as defined above. The relation between $W\_p^l(\Omega)$ and this space is given again in [1], chapter 1, §1.1.4, p. 7 and is
$$
W\_p^l=L\_p^l(\Omega) \cap L\_p(\Omega).
$$
**Reference**
[1] Vladimir G. Maz’ya, *Sobolev spaces. With applications to elliptic partial differential equations*, translated from the Russian by T. O. Shaposhnikova, 2nd revised and augmented edition (English) Grundlehren der Mathematischen Wissenschaften 342. Berlin: Springer (ISBN 978-3-642-15563-5/hbk; 978-3-642-15564-2/ebook), pp. xxviii+866 (2011), [MR2777530](https://mathscinet.ams.org/mathscinet-getitem?mr=MR2777530), [Zbl 1217.46002](https://www.zbmath.org/?q=an%3A1217.46002).
|
2
|
https://mathoverflow.net/users/113756
|
412429
| 168,275 |
https://mathoverflow.net/questions/412104
|
1
|
A matrix $\begin{bmatrix}w&x\\y&z\end{bmatrix}\in\mathbb Z^{2\times 2}$ is unimodular if $$|wz-xy|=1$$ holds.
>
> Is there a parametrization of such matrices with $|w||z|-xy=1$
> $$w,z<0<\max(y|z|,|w|x)<\frac{|w||z|+xy}2?$$
>
>
>
Additionally I would prefer to have the constraint
$$\max(|w|,|z|,x,y)\leq2\min(|w|,|z|,x,y)$$ which I think has to hold automatically if a parametrization exists.
|
https://mathoverflow.net/users/10035
|
On parametrization of a type of unimodular $2\times2$ integral matrix
|
You put so many restraints on your variables that there are actually no matrices satisfying all the conditions you want at the same time.
In order to simplify things a bit, note that by replacing the matrix
$\begin{bmatrix}w&x\\y&z\end{bmatrix}$ by $\begin{bmatrix}-w&x\\y&-z\end{bmatrix}$ you might as well want to parametrize $x,y,w,z$ such that:
1. $w,z > 0$
2. $wz-xy = 1$
3. $0 < \max(yz,wx)<\frac{wz+xy}2$
Now according to 3 at least one of x,y has to be > 0, and since both are nonzero the equation $xy = wz-1 \geq 0$ implies that both $x,y > 0$.
From combining 2 and 3. it follows that:
4. $wx < \frac{wz+xy}2 = wz - 1/2$
5. $yz < \frac{wz+xy}2 = xy + 1/2$ hence $ yz \leq xy$
Now 4. implies $x<z$ while 5. implies $z\leq x$ these both obviously can't be satisfied at the same time.
It often helps if you have a set of constraints that you want to be satisfied, to try and find an easier looking set of constraints. The only thing I have been trying to arrive at this prove is just try to simplify things as much as possible, until I arrived at a contradiction.
I have the feeling this might have been a better suited question for stack exchange.
|
4
|
https://mathoverflow.net/users/23501
|
412447
| 168,286 |
https://mathoverflow.net/questions/412417
|
2
|
Let $\Omega \subseteq \mathbb R^n$ be a bounded open set with smooth boundary. Let $k\geq 1$, $\alpha\in(0,1)$, $a\_{ij},b\_i,f \in C^{k,\alpha}(\Omega)$ for $i,j=1,...,n$, and define the operator
$$L = \sum\_{i,j=1}^n a\_{ij} \partial\_{ij} + \sum\_{i=1}^n b\_i \partial\_i.$$
Assume further that $L$ is uniformly elliptic, i.e. $\sum\_{i,j} a\_{ij}(x) \xi\_i\xi\_j\geq \lambda \|\xi\|^2$ for all $x \in \Omega$, $\xi\in\mathbb R^n$, and some $\lambda > 0$.
Standard elliptic regularity theory ensures that the Dirichlet problem $L u =f$ in $\Omega$, with $u=0$ on $\partial\Omega$, admits a solution $u \in C^{k+2,\alpha}$,
whose Holder norm depends only on that of the above data.
I was wondering whether there are any known conditions under which the regularity of the coefficients $b\_i$ can be weakened from $C^{k,\alpha}$ to $C^{k-1,\alpha}$, while retaining the existence of a solution $u \in C^{k+2,\alpha}$. It is unclear to me whether the proof of Schauder's estimates based on reduction to constant-coefficient equations (e.g. Theorem 13.2.1. of Jost's text) can be adapted to this setting under any sensible conditions. (While my question is general, I will note that $f$ happens to be of class $ C^{k+2,\alpha}$ in my particular use case.)
|
https://mathoverflow.net/users/111925
|
Second-order elliptic regularity with rough coefficients
|
In dimension $d=1$, let's try
$$
u^{\prime\prime}+b u^\prime =0,
$$
a solution is
$$
u^\prime = \exp\left({-\int\_0^x b(t) \textrm{d} t}\right)
$$
So the regularity of $u^\prime$ is that of $b$,+1, and that of $u$ is that of $b$,+2. So you cannot get regularity of $b$+3 in general.
Based on this example, you would need truly miraculous cancelations between $b$ and $a$ for things to work out exactly the right way. In fact,
$$
a\_{ij}u\_{,ij}=-b\_{i}u\_{,i}+f
$$
means that if $u\_{,ij}$ and $a\_{ij}$ are $C^{k,\alpha}$ as well as $f$, then so is $b\_{i}u\_{,i}$. So $u\_{,i}$ should cancel at every not $C^{k,\alpha}$ point for $b\_i$, that's asking a lot. Limiting to the case of finitely many such points, your problem locally very much look like the one dimensional problem I wrote above, and the solutions will behave accordingly..so I venture that the answer is no.
|
3
|
https://mathoverflow.net/users/40120
|
412455
| 168,288 |
https://mathoverflow.net/questions/336501
|
4
|
For integers $n\geq 1$ I denote the Euler's totient function as $\varphi(n)$ and the divisor function $\sum\_{1\leq d\mid n}d$ as $\sigma(n)$, that are two well-known mulitplicative functions. We assume also the theory of odd perfect numbers, see if you want the corresponding section of the Wikipedia with title *[Perfect number](https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers)*.
It is easy to prove the following statement, on assumption that there exists an odd perfect number $x$.
>
> **Fact.** If $x$ is an odd perfect number then $$\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)\tag{1}\label{1}$$
> holds.
>
>
>
**Computational fact.** For integers $1\leq n\leq 5000$, the only solution of \eqref{1} is $n=1$. To see it, after some seconds, choose *GP* as language and evaluate next code (it is just a line written in Pari/GP) in the web page Sage Cell Server
`for (x = 1, 5*10^3,if (eulerphi(x^(sigma(x))*(sigma(x))^x)==2^(x-1)*x^(3*x-1)*eulerphi(x), print(x)))`
I believe that the following conjecture holds.
>
> **Conjecture.** *The only solution of our equation \eqref{1} is the integer $1$.*
>
>
>
**Motivation for the post.** My belief is that an interesting way (but my attempts were failed) to study the unsolved problem related to odd perfect numbers (that is if there exist any of them) should be to create intrincated/artificious equations similar than \eqref{1} involving the sum of divisors functions and the Euler's totient function with the purpose to invoke inequalitites, asymptotics, heuristics or conjectures for these arithmetic functions (my belief is that the problem of odd perfect numbers is related to the distribution of prime numbers, thus maybe in the equations *similar than* \eqref{1} that previously I've evoked should be required also that arise functions as the radical of an integer $\operatorname{rad}(x)$ or even the prime-counting function $\pi(x)$, both specialized for odd perfect numbers $x$).
>
> **Question.** What work can be done to prove of refute previous conjecture, that the only solution of $$\varphi\left(n^{\sigma(n)}\sigma(n)^n\right)=2^{n-1} n^{3n-1}\varphi(n)$$ should be $n=1$? It is welcome unconditionally statements or heuristics, but also feel free to invoke conjectures if you can get some advanced statement.
>
>
>
Thus, as how it is perceived in the title of the post, previous **Question** is also an invitation to add remarkable statements about the nature of the solutions of \eqref{1}, if we are in the situation that the Question can not be solved.
**Last remarks to to emphasize my ideas.** What is saying myself thus previous **Motivation** and **Question**? That of couse I understand that the equation/characterization for odd perfect nubmers by means the equation $\sigma(x)=2x$ for odd integers $x\geq 1$ is easiest (to understand and study it) than others involving more arithmetic functions, but in my belief is that there exists a chance to get some statement for odd perfect numbers by the method to create more intrincated/artificious equations.
I think that my question is interesting, and I think that arises in a natural way when one tries to drop solutions like $2^{2^{\lambda-1}-1}$, that is the sequence *[A058891](https://oeis.org/A058891)* from the On-Line Encyclopedia of Integer Sequences, for equations like this $$\varphi(x^x\sigma(x))=x^x\varphi(x).$$
See if you want the code
`for (x = 1, 10^4,if (eulerphi((x^x)*sigma(x))==(x^x)*eulerphi(x), print(x)))`
---
I would like to refer that certain characterizations of primes are feasibles from answers of next posts (the problem [2] remains as unsolved), these posts are not directly related to this my post in MathOverflow, but maybe can be inspiring for some user of MathOverflow since are similar problems. Thus I justify this last paragraph as a compilation of similar equations for constellations of primes: thanks to the excellence of the user who provides the answer of [1] we've the characterization of Sophie Germain primes and similarly for twin primes; thanks to the excellence of the user mathlove who provided the answers of problems [2] and [3] we've a characterization of Mersenne exponents, Fermat primes and near-square primes.
[1] [*From the equation* $\sigma(x^{\varphi(y)})=\frac{1}{\varphi(x)}(x^y-1)$ *involving arithmetic functions to a characterization of Sophie Germain primes*](https://math.stackexchange.com/questions/3578715/from-the-equation-sigmax-varphiy-frac1-varphixxy-1-involving), question **3578715** from Mathematics Stack Exchange (Mar 12 '20).
[2] [*From the equation* $\sigma(x^{\sigma(y)-1})=\frac{1}{\varphi(x)}(x^{y+1}-1)$ *involving arithmetic functions to a characterization of Mersenne exponents*](https://math.stackexchange.com/questions/3587159/from-the-equation-sigmax-sigmay-1-frac1-varphixxy1-1-invo), question **3587159** from Mathematics Stack Exchange (Mar 19 '20).
[3] [*On characterizations for near-square primes and Fermat primes in terms of equations involving arithmetic functions*](https://math.stackexchange.com/questions/3588192/on-characterizations-for-near-square-primes-and-fermat-primes-in-terms-of-equati), question **3588192** from Mathematics Stack Exchange (Mar 20 '20).
|
https://mathoverflow.net/users/142929
|
What work can be done to study the solutions of $\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)$?
|
Here is a proof that if an odd integer $x>1$ satisfies (1), then $x$ is a perfect number.
First, by using the property that $\varphi(nm)=n\varphi(m)$ whenever $\mathrm{rad}(n)\mid\mathrm{rad}(m)$, we can rewrite (1) as
$$\big(\frac{\sigma(x)}2\big)^{x-1}\cdot \frac{\varphi(x\sigma(x))}{\varphi(x)} = x^{3x-\sigma(x)}.$$
Since the fractions in the l.h.s. are integer, the r.h.s. is also integer and odd, and thus $\sigma(x)=2y$ for some odd $y$ such that $\mathrm{rad}(y)\mid \mathrm{rad}(x)$. Then we further simplify the above equation to
$$y^x = x^{3x-2y}.$$
In other words,
$$\big(\frac{y}x\big)^x = x^{2(x-y)}.$$
Now, if $x>y$, then lhs < 1 while rhs > 1.
Vice versa, if $x<y$, then lhs > 1 while rhs < 1.
Hence, $x=y$, meaning that $\sigma(x)=2x$. QED
|
4
|
https://mathoverflow.net/users/7076
|
412456
| 168,289 |
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