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https://mathoverflow.net/questions/430157 | 2 | This is a question related to
[Semistable curves of genus $g\geq 2$ form an Artin algebraic stack in the etale topology?](https://mathoverflow.net/questions/429754/semistable-curves-of-genus-g-geq-2-form-an-artin-algebraic-stack-in-the-etale)
Let $\mathcal C\rightarrow \mathcal M^{ss}\_g$ be the universal curve over the Artin stack of semistable curves of a fixed genus $g\geq 2$. Is the morphism $\mathcal C\rightarrow \mathcal M^{ss}\_g$ representable by a scheme ?
It is well-known that the fibres of the morphisms are schemes because algebraic space curves are actually schemes. But it does not mean that the total space $\mathcal C\times\_{\mathcal M^{ss}\_g} T$ is a scheme for any scheme $T$ over $\mathcal M^{ss}\_g$. The total space can be easily seen to be an algebraic space even for the stack of presetable curves. But is it possible that for the sub functor of semistable curves, the morphism $\mathcal C\rightarrow \mathcal M^{ss}\_g$ is actually represented by a scheme?
| https://mathoverflow.net/users/152042 | On the stack of semistable curves | I am posting an answer to correct my wrong comments above. The answer is based on the comment by user @johan.
For every $g\geq 0$,
denote by $\mathfrak{M}\_g$
the stack of genus-$g$ curves that are proper, geometrically connected, reduced, and at-worst-nodal. Denote the universal curve over this stack by the $1$-morphism, $$\pi\_g:\mathfrak{M}\_{g,1}\to \mathfrak{M}\_g.$$
**Theorem** [Kai Behrend, perhaps earlier sources]. These stacks are algebraic stacks (i.e., Artin stacks).
If we mark enough points on the irreducible components, the curves are projective (appropriate linear combinations of the marked points give ample divisors). Thus, the $1$-morphism $\pi\_g$ is projective after an étale base change of the target. Hence the $1$-morphism $\pi\_g$ is representable by algebraic spaces.
There is an explicit analysis by Johan de Jong and myself of the open substack of $\mathfrak{M}\_0$ parameterizing curves with at most two irreducible components, and we use this to prove many divisor class relations on the stack $\mathfrak{M}\_0$. If you restrict $\pi\_g$ over this open substack, it is projective (the dual of the relative dualizing sheaf is relatively ample). However, the next open substack is more complicated.
**Theorem** [Damiano Fulghesu]. For every $g\geq 0$, the $1$-morphism $\pi\_g$ is not representable by schemes. In fact, this already fails over the open substack of $\mathfrak{M}\_0$ parameterizing curves with at most three irreducible components.
The basic idea of the proof is Hironaka's example. Alternatively, consider the versal deformation space of a genus-$0$ curve with precisely three irreducible components. There is an extra $\mathbb{Z}/2\mathbb{Z}$-symmetry to this deformation space coming from the symmetry of the curve that interchanges the two "leaves" of the curve and maps the "middle" component back to itself. You can use that $\mathbb{Z}/2\mathbb{Z}$ symmetry to form a quotient of the family which then gives a versal family of curves whose total space is an algebraic space that is not a scheme.
Contrary to what I wrote above, we can mark points on this family. Once we mark points on the family, we can glue higher genus curves to those marked points to get similar families where the arithmetic genus is as large as we like.
It is interesting to think about what goes wrong with the suggestion in my comment. The exceptional locus for the retraction (with its reduced structure) is a Cartier divisor in $\mathfrak{M}\_g$. The image in the pullback over $\mathfrak{M}\_g$ of the universal curve does give a closed subscheme. The blowing up of this ideal sheaf is representable by projective morphisms. Fulghesu's example shows that this blowing up is not the same as $\mathfrak{M}\_{g,1}$.
| 3 | https://mathoverflow.net/users/13265 | 430215 | 174,279 |
https://mathoverflow.net/questions/429909 | 7 | The following problem is presented in the paper *Recent development of chaos theory in topological dynamics - by Jian Li and Xiangdong Ye* :
"If a dynamical system [$(X,f)$, $X$ metric space, $f$ continuous] is Li-Yorke chaotic, does there exist a Cantor scrambled set?"
I would like to know the current status of the problem. Some information is given in the paper, but it's a bit old (2015).
Thanks
| https://mathoverflow.net/users/490774 | State of the art on: "If a dynamical system is Li-Yorke chaotic, does there exist a Cantor scrambled set?" | This question was answered recently by Geschke, Grebík, and Miller:
S. Geschke, J. Grebík, and B. D. Miller, "Scrambled Cantor sets," *Proceedings of the AMS* **149** ([link](https://arxiv.org/abs/2006.08277) to the arxiv version)
They show that if $X$ is analytic (i.e., the continuous image of a Polish space), and if $(X,f)$ is a dynamical system with Li-Yorke chaos, then $(X,f)$ contains a scrambled Cantor set.
Let me point out that some assumption on $X$ (such as "analytic") is necessary here. For example, this is easiest to see if CH fails. Why? Let $(Y,f)$ be any dynamical system with Li-Yorke chaos. This means that it contains a scrambled set of size $\aleph\_1$. Let $X$ denote the closure of this set under $f$, and observe that $(X,f)$ is then a dynamical system with Li-Yorke chaos, but its cardinality is smaller than that of the Cantor space! If CH holds, one can construct such things via transfinite recursion.
| 3 | https://mathoverflow.net/users/70618 | 430218 | 174,280 |
https://mathoverflow.net/questions/429630 | 4 | I am looking for a reference or proof of a lemma (if it's true) or a counter-example otherwise. It goes as follows:
Let $B\_1$ and $B\_2$ are two concentric balls of radius $1$ and $2$ in some $n$-dimensional Euclidean space. Then for any $f$ and $F$, there exists a $C>0$ such that
$$
\text{if }\;\Delta f=\text{div}F\; \text{ weakly, then }\;
\|f\|\_{L^p\_1(B\_1)}\leq C\big(\|F\|\_{L^p(B\_2)}+\|f\|\_{L^p(B\_2)}\big).
$$
| https://mathoverflow.net/users/131004 | Reference or proof of a lemma in PDE | I would call this estimate "the classical Calderon-Zygmund estimate" but indeed it is hard to track down a statement for the right-side in divergence form. Usually it is stated as an estimate from $L^p$ to $W^{2,p}$, when what you want is $W^{-1,p}$ to $W^{1,p}$. This is unfortunate, because the latter is the more natural and useful/important estimate! These statements are of course related, and you can get from one to the other.
However, for a direct proof of the statement you want, the only place I know to find it is in [our book](https://link.springer.com/book/10.1007/978-3-030-15545-2): see Proposition 7.3 (which actually proves a slightly more general statement than the one you want). The arxiv version is [here](https://arxiv.org/pdf/1705.05300.pdf).
Note that the statement is of course only true for $1<p<\infty$, there are counterexamples given in our book as well.
| 5 | https://mathoverflow.net/users/5678 | 430219 | 174,281 |
https://mathoverflow.net/questions/430214 | 1 | Define the Fourier transform for a suitable function $f\in L^1(\Bbb R)$ by $\widehat{f}(\xi)=\int\_{\Bbb R}f(x)e^{-ix\xi} dx$.
Assume the condition $$\int\_{\Bbb R}\int\_{\Bbb R}|\widehat{f}(\xi)f(x)|^2e^{2|x\xi|} d\xi dx<\infty.$$
Put $f(x)=P(x) e^{-t x^2}$ where $t>0$ and $P$ is a
polynomial and suppose $f$ verify the above condition.
My question why $P=0$.
Thank you in advance.
| https://mathoverflow.net/users/172078 | Why we have $f=0$ | Fourier transform of $f(x)=e^{-tx^2}$ is $\hat{f}(\xi)=ce^{-\xi^2/(4t)}$.
Multiplication of $f$ by a polynomial results in applying
a differential operator with constant coefficients to $\hat{f}$, and for our $\hat{f}$ this is equivalent to multiplication on some polynomial. These polynomials play no
role in convergence (unless they are both zero), since
the expression under the exponent in the product $f(x)\hat{f}(\xi)e^{2|x\xi|}$ is
$$-tx^2+2|x\xi|-\xi^2/(4t).$$
Since this quadratic expression is sometimes positive in
the first quadrant, the integral diverges, unless $P=0$.
| 3 | https://mathoverflow.net/users/25510 | 430223 | 174,282 |
https://mathoverflow.net/questions/380958 | 1 | How can I prove the following Liouville theorem without using the mean value property?
>
> If $u$ is harmonic on $\mathbb{R}^n$ and $\int\_{\mathbb{R}^n}|\nabla u|^2 dx \leq C$ for some $C > 0$, then $u$ is constant.
>
>
>
The proof that I know indeed uses the mean value property for harmonic functions.
---
From the comments: is it rigorous to do it like this: $-\Delta u = 0 \implies \int\_{\mathbb R^n} |\nabla u|^2 = 0$ (integrating by parts, hence $u$ is constant? It seems to easy, probably I'm missing something.
| https://mathoverflow.net/users/143757 | Prove Liouville theorem without using mean value property | Below we will prove, by purely energy methods, the following sharper statement: Let $u$ be a harmonic function which satisfies $$\liminf\_{r \to \infty} \frac1{r^2} \frac{1}{|B\_r|}\int\_{B\_r} |u|^2 = 0.$$ Then $u$ is constant.
All Liouville theorems of this sort are soft/qualitative versions of harder/more quantitative regularity statements. In this case, the Liouville statement follows from the interior gradient $L^\infty$ bound for harmonic functions (as we will see below). So another version of your question could be: Is there an energy methods proof of the interior $L^\infty$ bound for harmonic functions which in particular does not use the mean value property?
The classical energy methods proof of the pointwise estimates for harmonic functions uses only two ingredients: the Caccioppoli inequality (the most basic energy estimate), and the Sobolev inequality (because you have to get pointwise bounds from $L^2$ bounds somehow!).
The Caccioppoli inequality says that
\begin{equation\*}
\int\_{B\_r}
|\nabla u|^2
\leq
\frac{C}{r^2} \int\_{B\_{2r}} u^2 \,.
\end{equation\*}
You get this by testing the equation with $\varphi^2u$ where $\varphi$ is an appropriate cutoff function.
Iterating the Caccioppoli inequality many times, we get
\begin{equation\*}
\int\_{B\_r} |\nabla^k u|^2
\leq
\frac{C}{r^{2k}} \int\_{B\_{2^kr}} u^2 \,.
\end{equation\*}
If $k > 1+\frac d2$, then we can the Sobolev inequality to get
\begin{equation\*}
\| \nabla u \|\_{L^\infty(B\_r)}^2
\leq
\frac{C}{r^2}
\frac{1}{|B\_r|}
\int\_{B\_{2^kr}} u^2\,.
\end{equation\*}
Now send $r\to \infty$ along a good subsequence for the liminf, and you discover that $\nabla u=0$, which means $u$ is constant.
| 3 | https://mathoverflow.net/users/5678 | 430233 | 174,285 |
https://mathoverflow.net/questions/430246 | 4 | Let
$$
Lu=-a\_{ij}(x)\partial\_{ij}u+b\_i(x)\partial\_i u
$$
be a uniformly elliptic operator, with $A(x)=(a\_{ij}(x))$ positive-definite.
Here I'm only considering smooth coefficients, and the domain $\Omega\subset \mathbb R^d$ is as smooth as needed (but bounded).
In one of its elementary versions, the classical [Hopf boundary point lemma](https://en.wikipedia.org/wiki/Hopf_lemma) states that if $u\in C^2(\Omega)\cap C^1(\bar\Omega)$ is a supersolution
$$
Lu\geq 0
$$
and attains a minimum point at $x\_0\in \partial\Omega$ then
$$
\partial\_\nu u(x\_0)<0,
$$
where $\nu=\nu(x\_0)$ is the outer normal to $\partial\Omega$ at $x\_0$.
In other words, the supersolution must grow linearly inside the domain close to a boundary minimum point.
There are various possible extensions, in particular $\nu$ can be any outward pointing direction, and $\Omega$ can have corners.
**Question:**
>
> Can one say anything about the behaviour of $u$ at any such boundary minimum point, assuming only that $u\in C^2(\Omega)\cap C(\bar\Omega)$?
>
>
>
The key point here being the lack of $C^1$ regularity up to the boudnary.
I need this typically for a singular eigenvalue problem, where in fact I am trying to retrieve some information on a principal eigenpair $(\lambda\_0,u\_0(x))$ to a singular problem of the form
$$
Lu\_0=\frac{1}{\theta(x)}\lambda\_0 u\_0.
$$
My weight $\theta(x)$ is a given, smooth function that is positive inside $\Omega$ but vanishes linearly, typically $\theta(x)=\operatorname {dist}\_{\partial\Omega}(x)$.
Some probabilistic arguments tell me for free that $u\_0\geq 0$ is nontrivial and $\lambda\_0>0$, with $u\_0$ being moreover $C^2$ in the interior and continuously vanishing at the boundary.
I need to show that the vanishing is linear.
Of course Hopf's lemma immediately pops up to mind, but I really cannot guarantee the $C^1$ regularity up to the boundary (and all my attempts in that direction have failed so far).
Actually for my purposes it would suffice to get
$$
c\operatorname{dist}\_{\partial\Omega}(x)\leq u\_0(x) \leq C\operatorname{dist}\_{\partial\Omega}(x)
$$
in a neighborhood of $x\_0$.
Has anyone ever heard of something like that? For example, sandwiching $u\_0$ between two local lower/upper barriers vanishing linearly would do the trick, but I dont' really know how to do that.
*Final kinky comment:* to tell the truth, really, I'm working in the one-dimensional interval $\Omega=(0,1)$. But the question is so natural that I felt I had to write it in a slightly more general framework. So, if anyone has a specific 1D trick I'll be happy too!
| https://mathoverflow.net/users/33741 | Nonsmooth version of Hopf boundary point lemma | I think this is just the comment following Lemma 3.4 of Gilbarg and Trudinger (specifically equation 3.11).
I should add that lowering the regularity of the boundary seems like a harder problem (and is I think false in the case of a square). GT require an interior sphere condition. L. Rosales (see for instance [Generalizing Hopf’s Boundary Point Lemma](https://doi.org/10.4153/CMB-2017-074-6)) seems to have the best results in this direction.
| 7 | https://mathoverflow.net/users/127803 | 430250 | 174,290 |
https://mathoverflow.net/questions/430273 | 8 | Many geometries (Riemannian, symplectic, complex, Kähler, Calabi-Yau) can be defined as categories of G-structures on manifolds with the first integrability condition (zeroing of torsion of G-structure), and structure-preserving morphisms are defined naturally if $G \to GL\_n$ is monomorphism. Is it possible to define contact geometry in such terms?
| https://mathoverflow.net/users/148161 | Is it possible to define contact manifolds as manifolds with a G-structure? | A contact structure on $M^{2n+1}$ defines a $G$-structure (actually, it defines more than one, but there is a 'minimal' $G$-structure that is preserved by all contact transformations, and that is the one that people usually consider). Conversely, a $G$-structure on $M^{2n+1}$ comes from a contact structure provided that its intrinsic torsion satisfies the appropriate identities.
For example, a contact structure on $M$ is defined by a $2n$-plane field $D\subset TM$ (satisfying some nondegeneracy property). So if $G\_0\subset\mathrm{GL}(2n{+}1,\mathbb{R})$ is the linear subgroup that preserves the subspace $\mathbb{R}^{2n}\subset\mathbb{R}^{2n{+}1}$, then a contact structure defines a $G\_0$-structure on $M$ and, conversely, that $G\_0$-structure contains the information needed to recover $D$. In that sense, the $G\_0$-structure and the $2n$-plane field $D$ are equivalent information.
However, you will note that not every $G\_0$-structure on $M$ defines a contact structure, because the corresponding hyperplane field defined by $G\_0$ might be integrable, say.
Meanwhile, via the Lie bracket, a contact structure $D\subset TM$ defines a linear bundle map $B:\Lambda^2(D)\to TM/D$ that is non-degenerate. Correspondingly, there is a subgroup $G\_1\subset G\_0$ that is the subgroup that preserves a non-degenerate mapping $b:\Lambda^2(\mathbb{R}^{2n})\to\mathbb{R}^{2n+1}/\mathbb{R}^{2n}$. Hence, a contact structure $D$ defines a canonical $G\_1$-structure on $M$ in the obvious way. Conversely, a $G\_1$-structure on $M$ whose associated induced mapping $B:\Lambda^2(D)\to TM/D$ coincides with the one induced by the Lie bracket on $M$ contains all the information in the contact structure and nothing else; automorphisms of the $G\_1$-structure are exactly the contact transformations. (The condition that the $B$ associated to the $G\_1$-structure coincide with the one induced by the Lie bracket is the analog of 'integrability' in this case. However, this is not the same as the $G\_1$-structure being 'torsion-free'.)
Moreover, there is no proper subgroup $G\_2\subset G\_1$ for which contact transformations of $M$ preserve some $G\_2$-structure on $M$. In this sense, the $G\_1$-structure defined above is the 'minimal' $G$-structure defined by a contact structure.
For $n$ odd, the only proper subgroup $G\subset\mathrm{GL}(2n{+}1,\mathbb{R})$ such that the contact transformations of $\mathbb{R}^{2n{+}1}$ preserve a *torsion-free* $G$-structure is the index $2$ subgroup $G=\mathrm{GL}^+(2n{+}1,\mathbb{R})$ while, for $n$ even, there is no proper subgroup with this property.
| 11 | https://mathoverflow.net/users/13972 | 430275 | 174,295 |
https://mathoverflow.net/questions/430242 | 6 | The question is to describe ALL integer solutions to the equation in the title. Of course, polynomial parametrization of all solutions would be ideal, but answers in many other formats are possible. For example, answer to famous Markoff equation $x^2+y^2+z^2=3xyz$ is given by Markoff tree. See also this previous question [Solve in integers: $y(x^2+1)=z^2+1$](https://mathoverflow.net/questions/414877) for some other examples of formats of acceptable answers. In general, just give as nice description of the integer solution set as you can.
If we consider the equation as quadratic in $z$ and its solutions are $z\_1,z\_2$, then $z\_1+z\_2=y^2$ while $z\_1z\_2=x^3$, so the question is equivalent to describing all pairs of integers such that their sum is a perfect square while their product is a perfect cube.
An additional motivation is that, together with a similar equation $xz^2-y^2z+x^2=0$, this equation is the smallest $3$-monomial equation for which I do not know how to describe all integer solutions. Here, the "smallest" is in the sense of question [What is the smallest unsolved Diophantine equation?](https://mathoverflow.net/questions/316708), see also [Can you solve the listed smallest open Diophantine equations?](https://mathoverflow.net/questions/400714) .
| https://mathoverflow.net/users/89064 | What are the integer solutions to $z^2-y^2z+x^3=0$? | We get $z(y^2-z)=x^3$. Thus $z=ab^2c^3$, $y^2-z=ba^2d^3$ for certain integers $a, b, c, d$ (that is easy to see considering the prime factorization). So $ab(bc^3+ad^3)=y^2$. Denote $a=TA$, $b=TB$ (each pair $(a, b) $ corresponds to at least one triple $(A, B, T)$, but possibly to several triples). You get $T^3AB(Bc^3+Ad^3)=y^2$.Thus $T$ divides $y$, say $y=TY$. We get $Y^2=TAB(Bc^3+Ad^3)$.
So, all solutions are obtained as follows: start with arbitrary $A, B, c, d$ and choose any $Y$ which square is divisible by $AB(Bc^3+Ad^3)$, the ratio is denoted by $T$ (if both $Y$ and $AB(Bc^3+Ad^3)$ are equal to 0, take arbitrary $T$).
| 7 | https://mathoverflow.net/users/4312 | 430278 | 174,296 |
https://mathoverflow.net/questions/430268 | 21 | For $\gamma$ an ordinal, let “$H\_\gamma$” be the statement:
>
> For all ordinals $\alpha$, we have $2^{\aleph\_\alpha} = \aleph\_{\alpha+\gamma}$.
>
>
>
So clearly $H\_0$ is false, and so is $H\_\omega$; in fact, $H\_\gamma$ implies that $\gamma$ is successor (because otherwise $\operatorname{cf}\gamma = \omega\_\alpha$, say, and $\aleph\_{\alpha+\gamma} = 2^{\aleph\_\alpha} = (2^{\aleph\_\alpha})^{\aleph\_\alpha} = (\aleph\_{\alpha+\gamma})^{\aleph\_\alpha} = (\aleph\_{\alpha+\gamma})^{\operatorname{cf}\aleph\_{\alpha+\gamma}}$ gives a contradiction).
On the other hand, $H\_1$ is precisely the generalized continuum hypothesis (GCH) and it is consistent relative to ZFC.
I understand from [this question](https://mathoverflow.net/questions/79920/failure-of-the-gch) and [this one](https://mathoverflow.net/questions/138308/woodins-unpublished-proof-of-the-global-failure-of-gch) that $H\_2$ is known to be consistent relative to certain large cardinal assumptions, and perhaps even $H\_k$ for any concrete $k < \omega$.
What else, if anything, is known about the consistency of the various $H\_\gamma$? Might we perhaps construct¹ a successor ordinal $\gamma$ for which $H\_\gamma$ is demonstrably false?
1. Admittedly, I don't know how to phrase this question properly, because clearly “the smallest successor ordinal for which $H\_\gamma$ is false” is a definable successor ordinal for which $H\_\gamma$ is provably false, which is clearly not what I'm asking about. But a proof in ZFC that $H\_{\omega+1}$ is false (say) would be a good answer to my question.
| https://mathoverflow.net/users/17064 | What is known about the consistency of $2^{\aleph_\alpha} = \aleph_{\alpha+\gamma}$ for all $\alpha$? | 1. By a result of Patai, $\gamma$ should be finite (this is exercise 5.15 in Jech's book).
2. For any finite $n>0, H\_n$ is consistent, see Merimovich's paper [*A power function with a fixed finite gap everywhere*](https://arxiv.org/abs/math/0005179).
---
For completeness and since not everyone has Jech's book at hand, here's how the proof of (1) goes (note that Jech uses "$\beta$" in place of "$\gamma$"). Suppose $\gamma\ge \omega$ is such that $2^{\aleph\_\eta}=\aleph\_{\eta+\gamma}$ for all $\eta$. Letting $\alpha$ be minimal such that $\alpha+\gamma>\gamma$ we have $0<\alpha\le\gamma$ and $\alpha$ is a limit. Consider $\kappa=\aleph\_{\alpha\cdot 2}$. This $\kappa$ is obviously singular and by choice of $\alpha$ we have $$2^{\aleph\_{\alpha+\xi}}=\aleph\_{\alpha+\xi+\gamma}=\aleph\_{\alpha+\gamma}$$ for each $\xi<\alpha$. From this we get $$2^\kappa=\aleph\_{\alpha+\gamma}$$ (more generally, if $\lambda$ is singular and $\mu=\max\{2^\theta:\theta<\lambda\}$ exists then $2^\lambda=\mu$). On the other hand, we have $$2^\kappa=\aleph\_{\alpha\cdot 2+\gamma}=\aleph\_{\alpha+(\alpha+\gamma)}>\aleph\_{\alpha+\gamma}$$ again by choice of $\alpha$. So we have a contradiction.
| 21 | https://mathoverflow.net/users/11115 | 430293 | 174,300 |
https://mathoverflow.net/questions/430234 | 5 | Let $\mathcal{G}\_n = \{ N(\mu,\Sigma) ; \mu \in \mathbb{R}^n, \Sigma > 0\}$ be the collection of Gaussian distributions on $\mathbb{R}^n$ with full support.
If $f : \mathbb{R}^n \to \mathbb{R}^k$ is measurable, $k\leq n$, and $f \# \gamma \in \mathcal{G}\_k$ for all $\gamma \in \mathcal{G}\_n$, is it true that $f$ is affine a.e.? Here $f\#\gamma$ denotes the pushforward of $\gamma$ under $f$.
In other words, if a function sends all Gaussian inputs to a Gaussian outputs, must that function be (essentially) affine?
If it is true, I guess it should be a classically known fact. But, a quick search didn't turn up any definitive answers. Related questions can be found on CrossValidated at [Gaussian-to-gaussian transformations](https://stats.stackexchange.com/questions/392601/gaussian-to-gaussian-transformations) and on MathOverflow at [Non-affine smooth transformation of Gaussian is Gaussian](https://mathoverflow.net/questions/420620/non-affine-smooth-transformation-of-gaussian-is-gaussian), but these only address the case for which $f$ sends a single Gaussian input to a Gaussian output (in which case there are obvious non-affine choices of $f$ that work).
It seems one should be able to expand $f$ in the Hermite basis (since $f\in L^2(N(0,I))$ follows from the hypothesis), and show coefficients for Hermite polynomials of degree two or higher must vanish by considering, e.g., the system of equations that arise by looking at moments of $f\#\gamma$, using the assumption that $f\#\gamma$ is Gaussian (so its higher moments are expressible in terms of first two moments). However, the computations get complicated….
A reference or counterexample would be appreciated.
| https://mathoverflow.net/users/99418 | Gaussian-to-Gaussian transformations are affine a.e.? | The result for $k=n$ with proof may be found in Theorem 2 of
*Nabeya, Seiji; Kariya, Takeaki*, [**Transformations preserving normality and Wishart-ness**](http://dx.doi.org/10.1016/0047-259X(86)90081-3), J. Multivariate Anal. 20, 251-264 (1986). [ZBL0602.62037](https://zbmath.org/?q=an:0602.62037).
For an interesting extension see
*Parthasarathy, K. R.*, [**Two remarks on normality preserving Borel automorphisms of (\mathbb{R}^{n})**](http://dx.doi.org/10.1007/s12044-013-0113-z), Proc. Indian Acad. Sci., Math. Sci. 123, No. 1, 75-84 (2013). [ZBL1416.60014](https://zbmath.org/?q=an:1416.60014).
| 1 | https://mathoverflow.net/users/64449 | 430309 | 174,306 |
https://mathoverflow.net/questions/430301 | 3 | I would like to ask the following question.
I am searching for a reference for the following statement:
Suppose $k$ is a perfect field. Let $A$ be a (symmetric) $k$-algebra and let $M$ be a finitely generated $A$-module. Then the following assertions are equivalent.
$\bullet$ The module $M$ is absolutely indecomposable, i.e. $M$ stays indecomposable under any ground field extension.
$\bullet$ There is an isomorphism $\text{End}\_A(M)/J(\text{End}\_A(M)) \cong k$.
A reference to a textbook would be cool.
| https://mathoverflow.net/users/91107 | Reference request for equivalent formulations of being absolutely indecomposable | This is Theorem 30.29 in
*Curtis, Charles W.; Reiner, Irving*, Methods of representation theory, with applications to finite groups and orders. Vol. I, Pure and Applied Mathematics. A Wiley-Interscience Publication. New York etc.: John Wiley & Sons. XXI, 819 p. \textsterling 40.70 (1981). [ZBL0469.20001](https://zbmath.org/?q=an:0469.20001).
$A$ doesn't need to be symmetric.
| 6 | https://mathoverflow.net/users/22989 | 430314 | 174,307 |
https://mathoverflow.net/questions/430156 | 2 | Let $X$ be a Tychonoff space, let $A,B\subset X$ be closed. Let $J\_A$ be the set of all continuous on $X$ real-valued functions which vanish on $A$.
>
> For which $X$'s is it true that $J\_A+J\_B=J\_{A\cap B}$?
>
>
>
I can prove this if $X$ is hereditary normal, as well as in the case when $C(X)$ is complete and sequential in the compact open topology (in particular if $X$ is hemi-compact compactly generated). However, in the second case the proof is somewhat indirect, so as a side here is another question:
>
> Is there a constructive-ish proof of the property in question for compact $X$'s?
>
>
>
It would be particularly nice if there was a pair of monotone maps $\varphi,\psi$ from $J\_{A\cap B}$ into $J\_A$ and $J\_B$ such that $\varphi(f)+\psi(f)=f$, for every $f\in J\_{A\cap B}$.
| https://mathoverflow.net/users/53155 | On the equality $\{f\in C(X), f|_A=0\}+\{f\in C(X), f|_B=0\}=\{f\in C(X), f|_{A\cap B}=0\}$ | This characterizes normality.
That it implies normality was observed above by Remy.
Conversely, assume $f$ vanishes on $A\cap B$. Define $h:A\cup B\to\mathbb{R}$ by $h(x)=f(x)$ if $x\in A$ and $h(x)=0$ if $x\in B$. By the Tieze-Urysohn theorem $h$ has a continuous extension $H:X\to\mathbb{R}$. That extension belongs to $J\_B$. But then $G=f-H$ belongs to $J\_A$ because $f(x)=H(x)$ on $A$. We have $f=G+H$.
| 7 | https://mathoverflow.net/users/5903 | 430316 | 174,308 |
https://mathoverflow.net/questions/430317 | 4 | Let $M$ be closed orientable $2n$-manifold, where $n$ is odd. It is well known that the $\mathbb Z$-module $H^\bullet(M;\mathbb Z)$ has graded-commutative multiplication and $H^{2n}(M;\mathbb Z)\simeq\mathbb Z$. So (by Poincaré duality) there is a skew-symmetric quadratic form $[-]\smile[-]$ on $H\_n(M;\mathbb Z)$.
Clearly, for any closed orientable $n$-submanifold $S\subset M$ we have $[S]\smile[S]=0$.
On the over hand, this can be computed as the Euler class of the normal bundle $NS$.
Indeed, this Euler class is equal to the sum of intersection points $S\cap S'$, where $S'$ is “another copy of $S$ in general position with the first one”.
**Question:** *how can we see the fact $[S]\smile[S]=0$ geometrically in terms of the normal bundle?*
It seems that, vice versa, any orientable rank $n$ vector bundle $E$ over a closed orientable $n$-manifold $S$ can be used to make closed orientable $2n$-manifold $M$, so that $E\subset M$ becomes a tubular neighborhood of its zero section $S\subset E$.
*Is this correct?*
*Do we always have the property $e(E)=0$ for $n$, $E$, $S$ as above?*
I think I can prove just that $w\_n(E)=0$ using splitting principle (recall that $e\underset2\equiv w\_n$). Also the fact $e(TS)=0$ is well-known, but how can we handle the case of arbitrary $E$?
| https://mathoverflow.net/users/76500 | Intersection form of $2n$-manifold for odd $n$ | The following argument can be phrased in terms of cohomology (it amounts to a proof that the Euler class of an odd-rank bundle is 2-torsion) but here is a purely intersection-theoretic phrasing.
It suffices to show that if $E \to S$ is any vector bundle with $\text{rank}(E) = \dim S = 2k+1$, then a generic section has no zeroes when counted with sign. As you know (and by a standard argument), the signed count of such zeroes is independent of the choice of generic section. If $\phi$ is such a section, write the oriented zero set as $Z(\phi)$.
Pick a generic section $\phi$. Then $Z(-\phi)$ can be identified with $Z(\phi)$ **with the opposite orientation**, because the negation map $\Bbb R^{2k+1} \to \Bbb R^{2k+1}$ is orientation-reversing. Thus $$\# Z(\phi) = \# Z(-\phi) = -\# Z(\phi),$$ the first equality because the count is independent of the choice of generic section, the second by the observation about orientations.
Thus $\# [S \cap S] = \# Z(\phi)$ is an integer equal to its negative, thus zero.
---
As for your follow-up questions: if $E \to S$ is any vector bundle over closed base, it may be realized as the normal bundle of an embedding of $S$ into a closed manifold by taking the fiberwise one-point compactification of $E$ into a sphere bundle over $S$.
In general, if $E$ has rank $2k+1$, the class $e(E) \in H^{2k+1}(S;\Bbb Z)$ is 2-torsion. If $E$ has rank equal to the dimension of $S$ and $S$ is oriented, this implies $e(E) = 0$. If $E$ has smaller rank than the dimension of $S$, or if $S$ is non-orientable, it is possible for this class to be non-zero.
| 9 | https://mathoverflow.net/users/40804 | 430319 | 174,309 |
https://mathoverflow.net/questions/430283 | 2 | I have a two matrices $A$ and $B$ in $\mathbb{R}^{m \times n }$ ($m \gg $ n) such that there exists an orthonormal matrix $X \in \mathbb{R}^{n \times n }$, such that:
$$AX = B$$
Given that $X$ is orthonormal this is also true:
$$A = BX^T$$
**How to find $X$?**
I tried to use [Moore-Penrose inversion](https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Linear_least-squares) $A^+$ and got non-orthonormal result $Y = A^+B$, that works only one way:
$$AY = B$$
but not in another. The problem is that both matrices $A$ and $B$ are not absolutely accurate (obtained in numerical calculations). So the non-orthonormal solution $Y$ arises. It is slightly more accurate than the exact solution $X$: $\|AY - B\| \lt \|AX - B\| < 0.001 $. But in the inverse case of course it doesn't work at all: $\|BY^T - A$$\| > 100.0$. Whilst the exact solution is good enough in both ways: $\|BX^T - A\| < 0.001$.
$A$ is full-rank, *i.e.* $A^+A = I$ **(Not really! See P.S.)**
So the question is how to find orthonormal solution of the overdetermined linear equations system?
P.S.
It turned out that my problems were caused by presence small of singular singular values of A and B. So the overall problem was ill-conditioned. I have accepted @Federico-Poloni's answer because it directly addresses the original question and has a reference. James's answer seems to be working either, but I have to choose only one.
| https://mathoverflow.net/users/491098 | Orthonormal solution of overdetermined linear equations | You can formulate this problem as an [orthogonal Procustes problem](https://en.wikipedia.org/wiki/Orthogonal_Procrustes_problem):
$$
\min\_{X \text{ orthogonal}} \|AX-B\|\_F.
$$
With a transpose you can convert from the notation in the Wikipedia page to this form: the solution is $X=UV^T$, where $A^TB = U\Sigma V^T$ is an SVD, and a proof follows from manipulating the expression using the Frobenius (trace) inner product.
| 3 | https://mathoverflow.net/users/1898 | 430322 | 174,311 |
https://mathoverflow.net/questions/430302 | 2 | Let $\mathcal L$ be the space of lattices in $\mathbb R^d$. The **$i$-th successive minimum** of $L\in \mathcal L$, denoted $\lambda\_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$.
Let $v\_1$ denote any vector in $L$ with $\lVert v\_1\rVert=\lambda\_1(L)$. Now project $L$ along $v\_1$ onto its orthogonal complement $v\_1^{\perp}$. It is easy to see that the projection image of $L$, denoted $L\_1$, is again a lattice in the hyperplane $v\_1^{\perp}$.
>
> I speculate that $\lambda\_i(L\_1) \asymp\_d \lambda\_{i+1}(L)$.
>
>
>
It is easy to see $\lambda\_i(L\_1) \le \lambda\_{i+1}(L)$ but what about the other direction?
| https://mathoverflow.net/users/489992 | Successive minima of a lattice and projection along the the shortest nonzero vector | By taking a shortest lift from $L\_1$ to $L$ you obtain that $\lambda\_{i+1}(L)\le \lambda\_i(L\_1)+\frac{1}{2}\lambda\_1(L) \le \lambda\_i(L\_1)+\frac{1}{2}\lambda\_{i+1}(L)$. This gives $\frac{1}{2}\lambda\_{i+1}(L) \le \lambda\_i(L\_1)$, i.e. $\lambda\_{i+1}(L) \le 2\lambda\_i(L\_1)$.
| 2 | https://mathoverflow.net/users/40821 | 430323 | 174,312 |
https://mathoverflow.net/questions/430305 | 2 | I'm trying to understand whether I can use the following equality in my application -- for $u,v,w \in \mathbb{R}^d$:
$$\cos(u,w)\approx \cos(u,v)\cos(v,w)$$
Where $\cos(x,y)$ gives cosine of the angle between vectors $x$,$y$
$$\cos(x,y)=\frac{\langle x, y\rangle}{\|x\| \|y\|}$$
In simulations, I'm finding it becomes a near perfect fit for $d>100$ when $v=f\_1(u), w=f\_2(v)$ where $f\_i(x)$ is a random perturbation of $x$. For instance, $f\_i(x)$ could be:
* performing random simple rotation of $x$ with angle $\le \frac{\pi}{4}$ ([details](https://math.stackexchange.com/questions/4526950/showing-that-cosu-1-u-3-cosu-1-u-2-cosu-2-u-3-for-nearby-u-i))
* adding IID standard normal random variable to each entry of $x\_i$ ([details](https://stats.stackexchange.com/questions/588349/additivity-of-angles-in-higher-dimensions))
In both cases, this equation gives a good fit even though individual cosines are far from $1$.
Questions:
1. What is a high-level explanation of this behavior?
2. What restrictions on randomly sampled $f\_i$ will let me justify using triangle equality for cosine similarity in high dimensions?
$$\cos[u,f\_2(f\_1((u))]\overset{P}{=} \cos[u,f\_1(u)]\cos[f\_1(u),f\_2(f\_1(u))]+O(d^{-\frac{1}{2}})$$
(note, there's a dimension-free bound on approximation error of this formula [here](https://arxiv.org/abs/2107.04071), making it applicable in my setting requires turning it into probabilistic bound, and removing "dimension-free" part)
| https://mathoverflow.net/users/7655 | Triangle equality for cosine similarity in high dimensions | $\newcommand{\R}{\mathbb R}$Here is a straightforward explanation in the case of adding iid standard normal random variables (r.v.'s). Here we have random vectors
\begin{equation}
U:=X,\quad V:=X+Y,\quad W:=X+Y+Z,
\end{equation}
where $X,Y,Z$ are independent standard normal random vectors in $\R^d$. Letting $\cdot$ denote the dot product, we have
\begin{equation}
EU\cdot V=EX\cdot X+EX\cdot Y=d+0=d,
\end{equation}
\begin{equation}
EV\cdot W=EX\cdot X+EY\cdot Y+2EX\cdot Y+EX\cdot Z+EY\cdot Z \\
=d+d+2\times0+0+0=2d,
\end{equation}
\begin{equation}
EU\cdot W=EX\cdot X+EX\cdot Y+EX\cdot Z=d+0+0=d.
\end{equation}
Also, since each of the dot products $U\cdot V,V\cdot W,U\cdot W$ is the sum of $d$ iid r.v.'s with finite second moments, we have
\begin{equation}
Var\,U\cdot V+Var\,V\cdot W+Var\,U\cdot W=Cd
\end{equation}
for some universal real constant $C>0$.
So, by Chebyshev's inequality,
\begin{equation}
U\cdot V\sim\_P d,\quad V\cdot W\sim\_P 2d,\quad U\cdot W\sim\_P d,
\end{equation}
where $A\sim\_P B$ means that $A/B\to1$ in probability (as $d\to\infty$).
Similarly,
\begin{equation}
U\cdot U\sim\_P d,\quad V\cdot V\sim\_P 2d,\quad W\cdot W\sim\_P 3d.
\end{equation}
So,
\begin{equation}
\cos(U,V)=\frac{U\cdot V}{\sqrt{U\cdot U}\sqrt{V\cdot V}}\sim\_P\frac1{\sqrt2},
\end{equation}
\begin{equation}
\cos(V,W)=\frac{V\cdot W}{\sqrt{V\cdot V}\sqrt{W\cdot W}}\sim\_P\frac2{\sqrt6},
\end{equation}
\begin{equation}
\cos(U,W)=\frac{U\cdot W}{\sqrt{U\cdot U}\sqrt{W\cdot W}}\sim\_P\frac1{\sqrt3}.
\end{equation}
We conclude that indeed
\begin{equation}
\cos(U,V)\cos(V,W)\sim\_P \cos(U,W).
\end{equation}
---
The similar conclusion for other functions $f\_i$ is perhaps due to linearization.
| 2 | https://mathoverflow.net/users/36721 | 430325 | 174,313 |
https://mathoverflow.net/questions/430318 | 2 | Is there any reference to the proof of following: let $T$ denote the [Lannes functor.](https://encyclopediaofmath.org/wiki/Lannes-T-functor) Then (see the link above for more details) for any finite $E$-complex $X$ (where $E$ is finite-dimensional $\mathbb F\_p$-vector space), one should have $T\_EH\_E^\*(X) = H^\*BE \otimes H^\*(X^E)$?
Wilkerson and Dwyer (”Smith theory and the functor T”, p. 2) give a reference to the unpublished manuscript “Cohomology of groups and function spaces” by Lannes. But I can not find it anywhere.
| https://mathoverflow.net/users/477793 | Fixed points cohomology via Lannes T-functor | There is something wrong with your assertion, which, when $X$ is specialized to a point, says that $$T\_EH^\*(BE) = H^\*(BE).$$ But this is **not** true. (I can't access the Dywer-Wilkerson paper you mention right at the moment.)
Lannes' famous 1992 paper in Pub. IHES (en francais!) has a functor he calls $Fix$, and $Fix(H\_E^\*(X)) \simeq H^\*(X^E)$ for $X$ a finite $E$--complex. $Fix$ is a variant of $T\_E$. Maybe this will help you.
| 1 | https://mathoverflow.net/users/102519 | 430335 | 174,317 |
https://mathoverflow.net/questions/430332 | 4 | In the theory of Sobolev space, we have the following chain rule:
* For a uniformly Lipschitz function $F : \mathbf{R}\to \mathbf{R}$ such that $F(0)=0$,
and $u\in W^{1,1}(\mathbf{R}^n)$, then we have the following chain rule:
$\partial\_j F(u)=F'(u)\circ \partial\_ju$.
But how to define the function $F'(u)$? It seems that we can't define $F'(u(.))$ a.e., and it may be not a measurable function.
| https://mathoverflow.net/users/166368 | Chain rule in Sobolev space | The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^{-1}(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one defines the product $F'(u)\nabla u$ to be 0, irrespective of the fact that $F'(u)$ is defined or not a such points. See e.g. Leoni, Morini: JEMS 9 pp 219-252
| 11 | https://mathoverflow.net/users/7294 | 430339 | 174,318 |
https://mathoverflow.net/questions/430333 | 3 | This question is motivated by the construction of the Kuznetsov component on a prime Fano threefold $X$ of index 1 (say genus $g \geq 6$, $g \neq 7, 9$):
$$
D^b(X) = \langle Ku(X), E, \mathcal{O}\_X \rangle
$$
where $E$ is the pullback of the rank 2 tautological subbundle on $Gr(2, g/2 + 2)$.
In [1], the authors construct explicit formulas for the gluing data of $E$ to $Ku(X)$ by considering $D := \langle \mathcal{O}\_X \rangle^\perp = \langle Ku(X), E \rangle$ along with inclusion functor $i : Ku(X) \hookrightarrow D$ and go on to state the left adjoint is $i^{\ast} = \mathbf{L}\_E$.
Now consider the inclusion $j : D \hookrightarrow D^b(X)$. Clearly $j \circ i$ is the inclusion of $Ku(X)$ into $D^b(X)$, so that the left adjoint $( j \circ i )^\ast$ is the usual projection functor defined for semi-orthogonal decompositions. What would the left adjoint $j^\ast$ be in this case?
[1] Jacovskis, Liu, Zhang. *Brill-Noether Theory for Kuznetsov Components and Refined Categorical Torelli Theorems for Index One Fano Threefolds*, 2022
| https://mathoverflow.net/users/458355 | Left adjoint for nested admissible categories | If $\mathcal{A} \subset \mathcal{T}$ is a right admissible subcategory and
$$
\mathcal{T} = \langle \mathcal{A}^\perp, \mathcal{A} \rangle
$$
is the corresponding semiorthogonal decomposition, the left mutation functor $\mathbf{L}\_{\mathcal{A}}$ is **defined** as the left adjoint functor of the embedding $\mathcal{A}^\perp \hookrightarrow \mathcal{T}$.
In particular, in your case $j^\* = \mathbf{L}\_{\mathcal{O}\_X}$.
| 3 | https://mathoverflow.net/users/4428 | 430361 | 174,325 |
https://mathoverflow.net/questions/430347 | 3 | Let be lipschitz $f$ on $[0,1]$ and everywhere derivable. Is it true that $f\in C^1([0,1])$ ?
| https://mathoverflow.net/users/110301 | Regularity of lipschitz and derivable function | I claim that a function with these properties need not be $C^1$.
We start with the function $f: t \in (-1,1)\setminus \{ 0 \} \mapsto \operatorname{sin}(1/t)$, and we also set $f(0) = 0$. The antiderivative of $f$, namely $F: x \in (-1,1) \mapsto \int\_0^x \operatorname{sin}(1/t) \mathrm{d} t$ is Lipschitz, but not $C^1$ because of the discontinuity of $f$ at the origin.
It remains to check that $F$ is indeed differentiable at $x = 0$. This is just an integral evaluation: specifically we need only show that
$F(x)/x = \frac{1}{x} \int\_0^x \operatorname{sin}(1/t) \mathrm{d}t \to 0$ as $x \to 0$.
For the sake of completeness, this would go as follows. Ignore the factor $1/x$ for now, and do the change of variable $\theta = 1/t$ in the integral. This becomes $-\int\_{1/x}^{\infty} \operatorname{sin}(\theta)/\theta^2 \mathrm{d}\theta = [\operatorname{cos}(\theta)/\theta^2]\_{1/x}^\infty + \int\_{1/x}^\infty \operatorname{cos}(\theta)/\theta^3 \mathrm{d}t$. After taking absolute values, one gets $\lvert F(x) \rvert \leq 2 x^2$, whence the conclusion follows after dividing by $x$.
| 6 | https://mathoverflow.net/users/103792 | 430362 | 174,326 |
https://mathoverflow.net/questions/430363 | 2 | For $G = GL\_n$, it is known that the generic fibers of the Hitchin fibration are the Picard stacks of line bundles on the corresponding spectral curves and the duality of Hitchin fibrations in this case amounts to the self-duality of the Jacobian of an algebraic curve. Note however that these statements are valid on an open subset of the Hitchin base. My question is about this open subset, See S4.2, S4.4 and Cor 4.5 of Bezrukavnikov-Braverman for an overview (and all notation that is used in the next paragraph): <https://arxiv.org/pdf/math/0602255.pdf>
Prop 4.3 of that paper states that there is a non-empty open subset $Hitch\_n^0$ of $Hitch\_n$ over which $pr\_1$ is smooth. How can we explicitly write down the open subset $Hitch\_n^0$ in this context? The proof of Prop 4.3 is unclear to me at this stage, so any clarifications would be appreciated (in particular, I'm looking for an explicit description of this open subset). Note that $Hitch\_n$ is explicitly defined in S4.1.
| https://mathoverflow.net/users/2623 | Duality of Hitchin fibrations in type A | The relevant open subset is simply the open subset where the spectral curve is smooth. Thus it is the nonvanishing locus of some discriminant polynomial.
The Hitchin fiber is the moduli space of bundles on the spectral curve, generically free of rank one, whose pushforward to the base curve is locally free of rank $n$.
For a smooth curve, the only way a generically free bundle can fail to be locally free is if it contains torsion, in which case its pushforward is never locally free, so the Hitchin fiber is simply the space of line bundles on the spectral curve generically free of rank one, i.e. the Jacobian, and thus is smooth.
However, for a non-smooth curves, one will need to compactify the Jacobian to get the Hitchin fiber, and this will introduce singularities.
| 6 | https://mathoverflow.net/users/18060 | 430364 | 174,327 |
https://mathoverflow.net/questions/430365 | 56 | Quadratic forms play a huge role in math. This leads one to wonder: Is there a theory of cubic forms, quartic forms, quintic forms and so on? I have failed to discover any. Is there any such theory? If not, is it because:
* It is not as interesting as quadratic forms?
* It is so hard that no-one has yet written about such a theory?
* It is already deeply infiltrated in math and only some smart people know about it?
| https://mathoverflow.net/users/173315 | There is a nice theory of quadratic forms. How about cubic forms, quartic forms, quintic forms, ...? | I once asked André Weil the same question.
When I was college, taking a course that discussed quadratic forms, Weil gave a guest lecture to the students about that topic. After the talk, I raised my hand and asked him why there was such a big deal in math about quadratic forms while it seemed there was nothing comparable for higher-degree forms. Weil gave an answer, but to my regret I could not understand it (difficulty hearing him) and I did not ask him later to repeat what he had said. Now many years later, I can offer an answer that I think my former student self would have found satisfactory.
Before I begin, let me point out that we all know one important higher-degree form: the determinant form of degree $n$. So it is reasonable to ask what kind of general theory there could be for higher degree forms.
First let's see that the bijection between quadratic forms and symmetric bilinear forms generalizes to higher degree.
Recall for a field $F$ not of characteristic $2$, there is a bijection between quadratic forms $Q : F^n \to F$ and symmetric bilinear forms $B : F^n \times F^n \to F$ by $Q(\mathbf x) = B(\mathbf x,\mathbf x)$ and
$$
B(\mathbf x,\mathbf y) = \frac{1}{2}(Q(\mathbf x + \mathbf y) - Q(\mathbf x) - Q(\mathbf y)).
$$ Replacing $2$ by a degree $d \geq 1$, for a field $F$ where $d! \not= 0$ (meaning $F$ has characteristic $0$ or characteristic $p$ for $p > d$) there is a bijection between forms $f : F^n \to F$ of degree $d$ and symmetric $d$-multilinear maps $\Phi : \underbrace{F^n \times \cdots \times F^n}\_{d \ {\sf copies}} \to F$ where $f(\mathbf x) = \Phi(\mathbf x,\ldots,\mathbf x)$ and
$$
\Phi(\mathbf x\_1,\ldots,\mathbf x\_d) =
\frac{1}{d!}\sum\_{\substack{J \subset \{1,\ldots, d\} \\ J \not= \emptyset}} (-1)^{d - |J|}f\left(\sum\_{j \in J} \mathbf x\_j\right),
$$
(You could include $J = \emptyset$ in the sum by the usual convention that an empty sum is $\mathbf 0$, since $f(\mathbf 0) = 0$.) For example, when $f$ is a cubic form ($d = 3$, $n$ arbitrary), the associated symmetric trilinear form is
$$
\Phi(\mathbf x,\mathbf y,\mathbf z) = \frac{1}{6}(f(\mathbf x + \mathbf y + \mathbf z) - f(\mathbf x + \mathbf y) - f(\mathbf x + \mathbf z) - f(\mathbf y + \mathbf z) + f(\mathbf x) + f(\mathbf y) + f(\mathbf z)).
$$
For example, if $f : F^3 \to F$ by $f(x\_1,x\_2,x\_3) = x\_1^3+x\_2^3+x\_3^3$ then $\Phi(\mathbf x,\mathbf y,\mathbf z) = x\_1y\_1z\_1 + x\_2y\_2z\_2+x\_3y\_3z\_3$.
The general formula for $\Phi$ in terms of $f$
shows why we want $d! \not= 0$ in $F$.
Over fields of characteristic $0$, I think this bijection is due to Weyl.
Using this bijection, we call a form $f$ of degree $d$ *nondegenerate* if, for the corresponding symmetric multilinear form $\Phi$, we have
$\Phi(\mathbf x,\mathbf y, \ldots, \mathbf y) = 0$ for all $\mathbf y$ in $F^n$ only when $\mathbf x = \mathbf 0$. (Equivalently,
we have $\Phi(\mathbf x,\mathbf x\_2, \ldots, \mathbf x\_d) = 0$ for all $\mathbf x\_2, \ldots, \mathbf x\_d$ in $F^n$ only when $\mathbf x = \mathbf 0$.) When $d = 2$ (the case of quadratic forms), this is the usual notion of a nondegenerate quadratic form (or nondegenerate symmetric bilinear form).
That the bijection between quadratic forms and symmetric bilinear forms can be extended to higher degrees suggests there might be general theory in higher degree that's just like the quadratic case, but it turns out there really are significant differences between quadratic forms and forms of higher degree. Here are two of them.
1. Diagonalizability.
Outside characteristic 2, a quadratic form can be diagonalized after a linear change of variables, but for $n \geq 3$, a form of degree $n$ might not be diagonalizable after any linear change of variables.
While any nondegenerate *binary* cubic form over $\mathbf C$ can be diagonalized (see the start of the proof of Lemma 1.7 [here](https://arxiv.org/pdf/2103.16691.pdf); in the binary case, nondegeneracy of a cubic form is equivalent to the dehomogenization being a cubic polynomial with nonzero discriminant), nondegenerate cubic forms over $\mathbf C$ in more than two variables need not be diagonalizable. For example,
the three-variable cubic form $x^3 - y^2z - xz^2$ is nondegenerate and can't be diagonalized over $\mathbf C$ for a reason related to elliptic curves: see my comments on the MO pages [here](https://mathoverflow.net/questions/238901/cubic-forms-and-finiteness-of-k-k3) and [here](https://mathoverflow.net/questions/77702/transformation-of-a-cubic-form).
For each $d \geq 3$ and $n \geq 2$ *except* for $d=3$ and $n = 2$, there are nondegenerate forms of degree $d$ in $n$ variables over $\mathbf C$ that are smooth away from $(0,0,\ldots,0)$ and are not diagonalizable. Note the diagonal form $x\_1^d + \cdots + x\_n^d$ is smooth away from the origin.
2. Group theory.
For a form $f(x\_1,\ldots,x\_n)$ over a field $F$, its *orthogonal group* is the linear changes of variables on $F^n$ that preserve it:
$$
O(f) = \{A \in {\rm GL}\_n(F) : f(A\mathbf v) = f(\mathbf v) \ {\rm for \ all } \ \mathbf v \in F^n\}.
$$
Nondegenerate quadratic forms have a rich orthogonal group (many reflections) and some higher-degree forms have a large orthogonal group: if $f$ is the determinant form of degree $n$ then its orthogonal group is ${\rm SL}\_n(F)$.
But for a form $f$ of degree $d \geq 3$ over an algebraically closed field of characteristic $0$, $O(f)$ is sometimes a *finite* group. This happens if the corresponding symmetric $d$-multilinear form $\Phi$ satisfies $\Phi(\mathbf x, \ldots, \mathbf x,\mathbf y) = 0$ for all $\mathbf y$ in $F^n$ only when $\mathbf x = 0$. When $d \geq 3$ this condition is different from nondegeneracy as defined above. Let's say such $f$ and $\Phi$ are *nonsingular*. That nonsingular forms of degree $d$ have a finite orthogonal group over $\mathbf C$ is due to Jordan.
It also holds over algebraically closed fields of characteristic $p$ when $p > d$ (so $d! \not= 0$ in the field).
As an example, the orthogonal group of $x\_1^d + \cdots + x\_n^d$ over $\mathbf C$ when $d \geq 3$ has order $d^n n!$: it contains only the compositions of $n!$ coordinate permutations and scaling of each of the $n$ coordinates by $d$th roots of unity. Taking $n = 2$, this reveals a basic difference between the concrete binary forms $x^2 + y^2$ and $x^d + y^d$ for $d \geq 3$ that you can tell anyone who asks you in the future how higher degree forms are different from quadratic forms.
In retrospect, the label used for the second topic ("Group theory") really applies to both topics. For a field $F$,
the group ${\rm GL}\_n(F)$ acts on the forms of degree $d$ in $n$ variables with coefficients in $F$, and the first topic is about the orbit of $x\_1^d + \cdots + x\_n^d$ under this action while the second topic is about the stabilizer of $f(x\_1,\ldots,x\_n)$ under this action.
Concerning papers and books, I'll just mention one of each. There is Harrison's paper "A Grothendieck ring of higher degree forms" in J. Algebra 35 (1978), 123-138 [here](https://core.ac.uk/download/pdf/82089961.pdf) and Manin’s book *Cubic forms: algebra, geometry, arithmetic*. Manin mentioned a recurring nightmare he had about this book, soon after he finished it, in an interview with Eisenbud [here](https://www.youtube.com/watch?v=2wGRUBlsizE?t=177).
| 71 | https://mathoverflow.net/users/3272 | 430367 | 174,329 |
https://mathoverflow.net/questions/429873 | 5 | Given compact Kähler manifolds $X$ and $X'$ deformation equivalent over the unit disk $\Delta \subset \mathbb{C}$. More precisely, there is a proper holomorphic surjective map
\begin{align\*}
\pi\colon \mathcal{X}\to \Delta
\end{align\*}
and $t,t' \in \Delta$ such that $X$ and $X'$ are biholomorphic to the fibers $\pi^{-1}(t), \pi^{-1}(t')$ respectively.
Is there a deformation of $X$ and $X'$ over $\Delta$ such that every fiber is Kähler?
I am specially interested in the case where $X$ and $X'$ are of hyperkähler type, i.e. irreducible holomorphic symplectic. In other words, simply connected and admitting a unique holomorphic symplectic form. I know that there are (large) deformations where the deformed space is not Kähler.
| https://mathoverflow.net/users/141157 | Can deformation equivalent Kähler manifolds always be obtained by a deformation where all the fibers are Kähler? | I don't think this is known. For hyperkahler manifolds, conjecturally,
all smooth complex deformations are class C and birational to hyperkahler.
If this is true, your conjecture would follow automatically. The only
relevant publication that I am aware of is
<https://arxiv.org/abs/1703.02001>
*Arvid Perego*
Kählerness of moduli spaces of stable sheaves
over non-projective K3 surfaces
We show that a moduli space of slope-stable sheaves over a K3 surface is an irreducible hyperkähler manifold if and only if its second Betti number is the sum of its Hodge numbers h2,0, h1,1 and h0,2.
---
Perego proves that a (smooth)
limit of hyperkahler manifolds is Fujiki class
C if $b\_2=h^{2,0}+ h^{1,1} + h^{0,2}$.
This is a bit weaker than what you need, of course.
All the best
Misha
| 1 | https://mathoverflow.net/users/3377 | 430386 | 174,336 |
https://mathoverflow.net/questions/430255 | 7 | Is there a theory T such that:
* T includes all the axioms of [CZF](https://en.wikipedia.org/wiki/Constructive_set_theory).
* T includes the Idealization, Standardization, and Transfer schemas from [IST](https://en.wikipedia.org/wiki/Internal_set_theory#Formal_axioms_for_IST).
* Every axiom of T is a theorem of IST.
* T has [Church's rule](https://en.wikipedia.org/wiki/Disjunction_and_existence_properties#Related_properties). Explicitly, for every formula $\phi$ in IST's language, if $T \vdash \forall^{st} x \in \omega. \exists^{st} y \in \omega. \phi(x, y)$, then there is a computable function $f$ such that $T \vdash \forall^{st} x \in \omega. \phi(x, f(x))$.
(If not, is there a nice way to adjust IST schemas that makes the answer yes (but such that adding LEM to T still results in the full IST)?)
EDIT: Thinking about it more, the Idealization schema probably needs revised to be the universal closure of $$(\forall^{st} z. z \text{ finite} \land \forall x \in z. \psi(x) \implies \exists y. \forall x \in z. \phi(x, y)) \iff (\exists y. \forall^{st} x. \psi(x) \implies \phi(x, y))$$ to be useful. That way, when constructing the $y$ on the left side you actually have some information about $z$ to work with.
---
A first candidate I considered was $\{\phi : \exists t. IST \vdash t \Vdash\_{tr} \phi\}$ where $\Vdash\_{tr}$ is from Definition 5.2 in *[CZF has the disjunction and numerical existence property](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.103.2816)* but $t \Vdash\_{tr} \phi$ is only defined for $\phi$ in the language of first order set theory (which does not include the IST axioms). The obvious generalization leads to things like $\exists N \in \omega. \forall^{st} m \in \omega. N > m$ requiring nonstandard realizers, and permitting those seems to cause problems.
The *reduction algorithm* in *[Internal set theory: A new approach to nonstandard analysis](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.139.8758)* has a way to convert formulas in IST's language to first order set theory, and all of the IST axioms get translated to theorems of ZFC. So the next candidate is the set of statements that reduce to an axiom of CZF (or any set theory containing CZF which obeys Church's rule, such as IZF). The problem is that the reduction algorithm involves a conversion to prenex normal form, I can't figure out how to adapt the reduction algorithm to be compatible with intuitionistic logic.
| https://mathoverflow.net/users/65915 | Is there a constructive version of internal set theory? | As you have proven (by a well-known construction), one cannot expect to have full Transfer in constructive NSA. For different but related reasons, full Standardisation is off the table, though its restriction to reals seems semi-constructive. In fact, one can have Idealisation and weak versions of the other IST axioms, working in constructive/intuitionistic finite type arithmetic. There is even an associated proof translation (and reduction algorithm), which may well be adaptable to CZF.
In particular, you should take a look at the following paper by Benno van den Berg et al:
<https://arxiv.org/abs/1109.3103>
They consider Heyting arithmetic in all finite types (with function extensionalty). This is only the starting point, there are a number of follow-up papers (see google/mathscinet).
| 8 | https://mathoverflow.net/users/33505 | 430387 | 174,337 |
https://mathoverflow.net/questions/363882 | 6 | The Monster group is the largest of the sporadic simple groups, and has been proven by Wilson to also be a Hurwitz group. It has a presentation in terms of Coxeter groups, specifically Y443 along with the "spider" relator, and quotienting out by the center. However, I am interested in a presentation as a Hurwitz group.
Specifically, what is a presentation of the monster in terms of two elements a and b such that a has order 2, b has order 3, and ab has order 7? Also, what is the smallest possible order of the commutator of a and b in such a presentation?
| https://mathoverflow.net/users/38744 | Presentation of the Monster as a Hurwitz group | I have computed two pairs of generators $(a,b)$ of the Monster satisfying
the relations $a^2 = b^3 = (ab)^7 = 1$ using [1]. In both cases $a$ is of
class 2B, $b$ is of class 3B in the Monster, and the commutator $[a, b]$
has order 39. This gives an upper bound for the minimal order of that commutator.
More details of the computation are given in subdirectory
`applications/Hurwitz` of [1].
[1] Martin Seysen, the python mmgroup package,
<https://github.com/Martin-Seysen/mmgroup>
| 8 | https://mathoverflow.net/users/105705 | 430393 | 174,340 |
https://mathoverflow.net/questions/430337 | 1 | Consider a $k$ regular graph of $n$ vertices, where $3 \leq k \leq (n-1)$. Is there any upper or lower bound, in the worst case, known for either the tree-width or the clique width of each $k$ regular family?
| https://mathoverflow.net/users/166840 | Tree width and clique width of regular graphs | For each $k$-regular family, the treewidth and cliquewidth can be both $\Theta(n)$, due to the existence of expanders.
By [On Balanced Separators, Treewidth, and Cycle Rank](https://arxiv.org/abs/1012.1344) Thm. 2.1, $tw(G) \geq \tilde s(G) -1$, and by the definition of the strict balanced separator number, $\tilde s(G)$ is at least the size of a balanced separator of a graph. A random regular graph is an expander, which in turn makes the size of balanced separators $\Theta(n)$. Thus $tw(G)=\Theta(n)$.
By [The Tree-Width of Clique-Width Bounded Graphs without $K\_{n,n}$](https://link.springer.com/chapter/10.1007/3-540-40064-8_19), every graph of clique-width $k$ which does not
contain the complete bipartite graph $K\_{n,n}$ for some $n > 1$ as a subgraph has tree-width at most $3k(n − 1) − 1$. It's possible to take $n=3$ for random regular graphs, thus the clique-width is also $\Theta(n)$.
| 3 | https://mathoverflow.net/users/125498 | 430403 | 174,343 |
https://mathoverflow.net/questions/430336 | -1 | Let $G$ be a finite group with no abelian subfactor in its composition series.
Is $G$ obtained from simple groups by iterating semidirect products?
(Initially it was asked whether $G$ is a direct product of simple groups, but $A\_5\wr A\_5$ was mentioned as an immediate counterexample.)
| https://mathoverflow.net/users/475097 | Splitting of a finite group with no abelian subfactor in composition series | There are groups that look like wreath products, but where the base group has no complement, so they are not semidirect products. The theory is described in an old paper of mine (and probably elsewhere):
D. F. Holt, Embeddings of group extensions into Wreath products, Quar. J. Math. (Oxford) 29 (1978), 463--468.
It can be seen as a non-abelian analogue of Shapiro's lemma in cohomology.
Roughly speaking, let $D$ be a group extension of $N$ by $H$ (meaning $N$ is the normal subgroup), suppose that the group $G$ contains $H$ as a subgroup, and let $\Omega$ be the set of (right or left) cosets of $H$ in $G$ with the transitive (right or left) action of $G$ on $\Omega$.
Then there is corresponding wreath product like extension $E$ of $N^\Omega$ by $G$, where the factors of $N^\Omega$ are permuted under conjugation corresponding to the action of $G$ on $\Omega$, and the normalizer in $E$ of a factor of the base group modulo the other factors of the base group is naturally isomorphic to $D$.
Furthermore, the extension $E$ is non-split if and only if $D$ is.
So, for example, we could take $D = {\rm Aut}(A\_6)$ (as in spin's answer) with $N = {\rm Inn}(A\_6) \cong A\_6$, and $H = C\_2^2$, and $G = A\_5$, so $|\Omega| = 15$. This results in a non-split extension of $A\_6^{15}$ by $A\_5$, which cannot be constructed from simple groups by taking iterated semidirect products.
| 8 | https://mathoverflow.net/users/35840 | 430413 | 174,345 |
https://mathoverflow.net/questions/415409 | 4 | Let $\mathcal{V}$ be a semi-monoidal category, meaning it satisfies the axioms of a monoidal category except missing a unit and the unit axiom. One could then still go about defining a $\mathcal{V}$-category by dropping the requirement of having unit morphisms.
One concern is that without unit morphisms there is no way to define an underlying category $\mathcal{C}\_0$ associated to a $\mathcal{V}$-category $\mathcal{C}$. However other than that it seems to me that other parts of the theory makes sense.
Has such categories been considered previously anywhere in literature? I can find several mentions of semi-monoidal category e.g. <https://arxiv.org/abs/math/0507349>
but no mentions of semi-enriched categories. Of particular interest to me would be if there is a version of the enriched Yoneda lemma for such categories.
| https://mathoverflow.net/users/160045 | Enriched categories over a semi-monoidal category | Expanding upon my comment: categories without units are called [semicategories](https://ncatlab.org/nlab/show/semicategory). You can enrich a semicategory in a semigroupal category, which is what you describe. The Yoneda lemma is subtle with semicategories, but see [On regular presheaves and regular semi-categories](http://www.numdam.org/item/CTGDC_2002__43_3_163_0/). However, note that the authors work with semicategories enriched in a monoidal category: this is because, despite the definition of enriched semicategory and semifunctor not needing a unit in $\mathcal V$, a unit is necessary to define an enriched notion of natural transformation between semifunctors. I am not aware of a reference that explicitly develops the theory of semicategories enriched in semigroupal categories.
| 4 | https://mathoverflow.net/users/152679 | 430422 | 174,348 |
https://mathoverflow.net/questions/430424 | 6 | Let $M^8 \subset B^9 \setminus \{ 0 \} \subset \mathbf{R}^9$ be a properly embedded, stable minimal hypersurface. Suppose that $0 \in \overline{M}$ is an isolated singularity of the surface.
**Question.**
(i) Are the tangent cones at the origin regular, that is $\operatorname{sing} \mathbf{C} = \{ 0 \}?$ (ii) If *one* cone in $\operatorname{VarTan}(\lvert M \rvert,0)$ is regular, is there a 'direct' way of showing that they must *all* be regular? (By 'direct' I would mean without going via uniqueness of tangent cones à la Lojasiewicz–Simon.)
I believe (i) used to be an open question, but I am curious about its current status or partial progress. The Schoen–Simon regularity theory implies that the cones are either regular or cylindrical of the form $\mathbf{C}^7 \times \mathbf{R}$.
*Correction.* Otis Chodosh pointed out in his answer that this is incorrect. The aforementioned regularity theory of Schoen–Simon shows that the singular set of every tangent cone $\mathbf{C} \in \operatorname{VarTan}(\lvert M \rvert,0)$ is one-dimensional, but it can consist of several rays, one for each singularity in the link of the cone. It's the tangent cones taken at these rays that are cylindrical, of the form described above.
| https://mathoverflow.net/users/103792 | What is the current status on bad tangent cones at isolated singularities? | (i) This used to be a wide open area, but recently there has been some progress: Gabor Székelyhidi has constructed an example of an isolated singularity with a cylindrical tangent cone here: <https://arxiv.org/pdf/2107.14786.pdf> . If you are willing to change the metric from the flat metric to some other Riemannian metric, Leon Simon has constructed examples with cylindrical tangent cones but wild singular set <https://arxiv.org/pdf/2101.06401.pdf> (it's not known if this can be done in an analytic background metric).
(ii) I don't think there is any (known) way to prove that if one tangent cone is regular then they all are (without proving uniqueness of tangent cones). The basic problem is that the closure of the set of regular minimizing (or stable, etc) cones could contain singular cones. So it's not clear how to rule out a singular point having a "nearly" singular tangent cone AND a fully singular tangent cone (besides proving that the nearly singular cone is actually unique).
---
Finally, I will note that
>
> The Schoen–Simon regularity theory implies that the cones are either regular or cylindrical of the form $\mathbf{C}^7\times \mathbf{R}$
>
>
>
is not correct. In general, a tangent cone to a stable (or minimizing) hypersurface in $\mathbf{R}^9$ is $\mathbf{C}(\Sigma)$ where $\Sigma\subset\mathbf{S}^8$ is minimal with isolated singularities (this could all occur with multiplicity in the stable case, but Schoen--Simon basically says that this is not an issue). There's no reason that the singularities of $\Sigma$ have to be perfectly aligned along a spine, like with the link of (Simons cone) $\times \mathbf{R}$.
Note that in the usual "dimension reduction" arguement, you say: if $M^8\subset \mathbf{R}^9$ is stable minimal with singularity at $0$, then either (i) the tangent cone is smooth or (ii) you can take an ITERATED tangent cone (tangent cone to the tangent cone) and find something of the form $\mathbf{C}^7\times \mathbf{R}$. Note that there exists $x\_i \in M$ and $\lambda\_i\to\infty$ so that $\lambda\_i(M-x\_i)$ converges to this iterated tangent cone (so it's a blowup limit) but there need not be any actual tangent cone of this form (a priori, I am not sure that such an example has actually been constructed, but it seems very likely to exist).
| 6 | https://mathoverflow.net/users/1540 | 430437 | 174,354 |
https://mathoverflow.net/questions/430434 | 2 | In Angella and Tomassini's [paper](https://link.springer.com/article/10.1007/s00222-012-0406-3) p.75, there is an exact sequence:
>
> $\cdots\to B^{\bullet,\bullet}\to H\_{\bar\partial}^{\bullet,\bullet}\to H\_A^{\bullet,\bullet}\to \cdots$
>
>
>
where $B^{\bullet,\bullet}:=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$, and recall that Dolbeault cohomology $H\_{\bar\partial}^{\bullet,\bullet}:=\frac{\ker\bar\partial}{\text{im }\bar\partial}$, and Aeppli cohomology $H\_A^{\bullet,\bullet}:=\frac{\ker\partial\bar\partial}{\text{im }\partial+\text{im }\bar\partial}$, obviously, there are natural maps of $f:B^{\bullet,\bullet}\to H\_{\bar\partial}^{\bullet,\bullet}$ and $g:H\_{\bar\partial}^{\bullet,\bullet}\to H\_A^{\bullet,\bullet}$, but I wonder how do we get $\text{im } f=\ker g$?
My thinking process is like that:
In order to get the expression of $B^{\bullet,\bullet}$, we should compute the kernel of the map $g$, for a $\bar\partial$-closed form $\alpha$, let $g(\alpha)=0\in H\_A^{\bullet,\bullet}$, then we get $\alpha=\partial\beta+\bar\partial\gamma$, since $\alpha$ is $\bar\partial$-closed, we get $\bar\partial\partial\beta=0$, thus $\alpha\in\ker\bar\partial\cap\text{im }\partial+\text{im }\bar\partial$, then I get stuck, can anyone help me to get $B^{\bullet,\bullet}=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$?
| https://mathoverflow.net/users/99826 | The kernel of $H^{\bullet,\bullet}_{\bar\partial}(X)\to H_A^{\bullet,\bullet}(X)$ | Let $\alpha$ be a $\bar{\partial}$-closed form. Denote its Dolbeault cohomology class by $[\alpha]\_{\bar{\partial}}$ and its Aeppli cohomology class by $[\alpha]\_A$; note that the map $g$ is given by $g([\alpha]\_{\bar{\partial}}) = [\alpha]\_A$. Likewise, if $\alpha' \in \ker\bar{\partial}\cap\operatorname{im}\partial$, denote the corresponding element in $B^{\bullet,\bullet}$ by $[\alpha']\_B$; note that the map $f$ is given by $f([\alpha']\_B) = [\alpha']\_{\bar{\partial}}$.
As you noted, if $[\alpha]\_{\bar{\partial}} \in \ker g$, then $\alpha = \partial\beta + \bar{\partial}\gamma$, so $[\alpha]\_{\bar{\partial}} = [\partial\beta + \bar{\partial}\gamma]\_{\bar{\partial}} = [\partial\beta]\_{\bar{\partial}}$. Moreover, since $\bar{\partial}\alpha = 0$, we see that $\bar{\partial}\partial\beta = 0$ and hence $\partial\beta \in \ker\bar{\partial}\cap\operatorname{im}\partial$. Therefore, we can form the element $[\partial\beta]\_B$ which satisfies $f([\partial\beta]\_B) = [\partial\beta]\_{\bar{\partial}} = [\alpha]\_{\bar{\partial}}$, so $[\alpha]\_{\bar{\partial}} \in \operatorname{im}f$ and hence $\ker g \subseteq \operatorname{im}f$.
Suppose now that $[\alpha]\_{\bar{\partial}} \in \operatorname{im}f$, then there is $\alpha' \in \ker\bar{\partial}\cap\operatorname{im}\partial$ with $[\alpha']\_{\bar{\partial}} = [\alpha]\_{\bar{\partial}}$. As $\alpha' \in \operatorname{im}\partial + \operatorname{im}\bar{\partial}$, we see that $g([\alpha]\_{\bar{\partial}}) = g([\alpha']\_{\bar{\partial}}) = [\alpha']\_A = 0$, so $[\alpha]\_{\bar{\partial}} \in \ker g$ and hence $\operatorname{im}f \subseteq \ker g$.
| 5 | https://mathoverflow.net/users/21564 | 430439 | 174,355 |
https://mathoverflow.net/questions/430391 | 5 | Background: A referee has suggested a shorter proof of one of my results, but I'm having trouble justifying one of their assertions. The setting is that $A$ is a commutative ring, and the referee's suggestion is to reduce to the case that $A$ is local and finite (as a set) in the following way:
* First assume that $A$ is finitely generated as a $\mathbb{Z}$-algebra.
* Then form the set of ideals $\mathcal{J} = \{I \subseteq A\text{ ideal}: A/I\text{ is finite and local}\}$.
* Claim that $\bigcap\_{I\in\mathcal{J}} I = 0$.
* Deduce that the canonical homomorphism $A \to \prod\_{I\in\mathcal{J}} A/I$ is injective.
* The result for $A$ follows from the result for each finite, local $A/I$.
I'm fine with each of these steps except for the middle one, that the intersection of all the ideals must be $0$. I can tell that $\mathcal{J}$ contains every maximal ideal, even every power of every maximal ideal, so $\bigcap \mathcal J$ must be contained in the Jacobson radical of $A$. Conversely, each ideal $I\in \mathcal{J}$ must be contained in a unique prime $\mathfrak{m}\_I$, which must then be maximal since $A/I$ is finite, hence Artinian and dimension $0$. For every example I cook up, $\bigcap \mathcal J$ is zero, even when the Jacobson radical is nonzero, but I can't see why it must happen in general (or if it doesn't).
---
Precise question:
>
> Let $A$ be a commutative, unital ring that is finitely generated as a $\mathbb{Z}$-algebra. Let $\mathcal{J} = \{I \subseteq A\text{ ideal}: A/I\text{ is finite as a set and local}\}$. Must it be the case that $\bigcap\_{I\in\mathcal{J}} I = 0$?
>
>
>
| https://mathoverflow.net/users/1474 | For a finite-type $\mathbb{Z}$-algebra $A$, is the intersection of all ideals $I$ such that $A/I$ is finite and local necessarily zero? | A proof of the proposed result that is similar (if not identical) to [Peter Kropholler's](https://mathoverflow.net/a/430392/84349) and YCor's proofs can be derived from two *well-known* results, namely **[Lemma 2](https://mathoverflow.net/questions/57515/a-finitely-generated-mathbbz-algebra-that-is-a-field-has-to-be-finite)** and **Theorem 3** below, together with the **[Artin-Rees lemma](https://en.wikipedia.org/wiki/Artin%E2%80%93Rees_lemma)** (alternatively [Theorem 18.4.v, 3]; see proof of Claim 5 and subsequent note).
We shall establish:
>
> **Claim 1.** Let $R$ be a commutative unital ring. Let $\mathcal{I}$ be intersection of all ideals $I$ of $R$ such that $R/I$ is a [local ring](https://en.wikipedia.org/wiki/Local_ring) of finite cardinal. If $R$ is a finitely generated $\mathbb{Z}$-[algebra](https://en.wikipedia.org/wiki/Algebra_over_a_field#Generalization:_algebra_over_a_ring), then $\mathcal{I}$ is $\{0\}$.
>
>
>
The main results we need are:
>
> **[Lemma 2](https://mathoverflow.net/questions/57515/a-finitely-generated-mathbbz-algebra-that-is-a-field-has-to-be-finite).** [Lemma 4.8, 1]. A field which is finitely generated as a ring is finite.
>
>
>
>
> **Theorem 3.** [Theorem 4.19 (Nullstellensatz, General form), 2].
> Let $R$ be a [Jacobson ring](https://en.wikipedia.org/wiki/Jacobson_ring) and let $S$ be a finitely generated $R$-[algebra](https://en.wikipedia.org/wiki/Algebra_over_a_field#Generalization:_algebra_over_a_ring).
> Then $S$ is a Jacobson ring.
>
>
>
As an intermediate step, we shall prove:
>
> **Claim 4.** Let $R$ be a finitely generated $\mathbb{Z}$-algebra. If $R$ is local, then $R$ is a finite ring.
>
>
>
>
> *Proof.* Since $R$ is Noetherian, its unique maximal ideal $\mathfrak{m}$ is finitely generated. As $R$ is Jacobson by Theorem 2, the ideal $\mathfrak{m}$ is also the [nilradical](https://en.wikipedia.org/wiki/Nilradical_of_a_ring) of $R$. Consequently, there is $n \ge 1$ such that $\mathfrak{m}^n = 0$, which shows in particular that $R$ is Artinian ($R$ is zero-dimensional and Noetherian). To conclude, it only remains to show that the residual field $R/\mathfrak{m}$ of $R$ is finite, which is given by Lemma 1.
>
>
>
The following result mentioned by YCor is instrumental.
It can be proved by means of the [Artin-Rees lemma](https://en.wikipedia.org/wiki/Artin%E2%80%93Rees_lemma) as indicated by Peter Kropholler, or alternatively by using a result of Matlis'theory of injective modules of Noetherian rings [Theorem 18.4.v, 3].
>
> **Claim 5.** Let $R$ be a commutative unital Noetherian ring. Then $R$ is residually local and Artinian, i.e.,
> for every non-zero $x \in R$ there is an ideal $I$ of $R$ such that $x \notin I$ and $R/I$ is local and Artinian. (In other words, the intersection of all ideals $I$ such that $R/I$ is local and Artinian, results in the null ideal.)
>
>
>
>
> *Proof.* Let $x \in R \setminus \{0\}$ and let $I$ be an ideal of $R$ maximal among the ideals of $R$ not containing $x$. Such an $I$ exists by Zorn's lemma. We shall prove that $\overline{R} = R/I$ is local. Let $\overline{x} = x + I$. By construction, we know that $\overline{x}$ is contained in every non-zero ideal of $\overline{R}$. It also follows from our assumptions on $x$ and $I$ that $\overline{R}\overline{x}$ is a simple $\overline{R}$-module, so that the annihilator $M$ of $\overline{x}$ is a maximal ideal of $\overline{R}$. We claim that there is $n \ge 1$ such that $M^n = \{0\}$. If the claim holds true, then any prime ideal of $\overline{R}$ contains a power of $M$ and hence is equal to $M$, which shows that $\overline{R}$ is local and Artinian.
> Reasoning by way of contradiction, we assume that $M^n \neq \{0\}$ for every $n \ge 1$. As $\overline{R}$ is Noetherian, we can apply the Artin-Rees lemma [Theorem 8.5, 3]. This lemma yields a positive integer $c$ such that $M^n \cap \overline{R} \overline{x} = M^{n - c}(M^c \cap \overline{R} \overline{x})$ for every $n > c$. Taking $n = c + 1$, we obtain that $\overline{R} \overline{x} = M \overline{R} \overline{x} = \{0\}$, which is the desired contradiction. Observe indeed that $M^c$ and $M^{c + 1}$ are non-zero by assumption, so that both ideals contain $\overline{x}$.
>
>
>
**Note.** In order to prove of Claim 5, we can use a result from the Matlis's theory of [injetive modules](https://en.wikipedia.org/wiki/Injective_module) over Noetherian rings instead of the Artin-Rees lemma. It goes as follows.
>
> By construction, the ideal $\overline{R}\overline{x} \simeq \overline{R}/M$ is an essential $\overline{R}$-submodule of $\overline{R}$. Therefore the [injective hull](https://en.wikipedia.org/wiki/Injective_hull) $E(\overline{R}\overline{x}) \simeq E(\overline{R} / M)$ contains $\overline{R}$. In particular, $1 \in E(\overline{R}\overline{x})$, so that $M^n \cdot 1 = \{0\}$ for some $n \ge 1$ by [Theorem 18.4.v, 3]. Thus $M^n = \{0\}$, which implies that $\overline{R}$ is local and Artinian, as desired.
>
>
>
Now we are in position to prove Claim 1.
>
> *Proof of Claim 1.* Combine Claims 4 and 5.
>
>
>
---
[1] R. Swan, "Excision in algebraic K-theory", 1971.
[2] D. Eisenbud, "Commutative Algebra with a View Towards Algebraic Geometry", 1995.
[3] H. Matsumura, "Commutative Ring Theory", 1989.
| 2 | https://mathoverflow.net/users/84349 | 430446 | 174,357 |
https://mathoverflow.net/questions/430445 | 1 | Let $X$ be a smooth projective surface. Let $p$ be a closed point of $X$. Let $k(p)$ be the corresponding skyscraper sheaf, then actually we could use Grothendieck-Riemann-Roch to calculate the Chern of the coherent sheaf $k(p)$.
However, now let $U\cong \mathrm{Spec}A$ be a open subset containing $p$, $m$ be the maximal idea of $A$ corresponding $p$. Then we have a closed embedding $$i: Y:=\mathrm{Spec}(A/m^2)\to X$$
Now, how do we calculate the Chern class of $i\_{\*}(\mathcal{O}\_{Y})$.
Since $Y$ is not reduced, it seems we could not use GRR. Could you give me some ideas about it? Thanks a lot.
| https://mathoverflow.net/users/158876 | Chern class of torsion sheaf support on a point | If you can figure out the Chern character of $i\_\*(\mathcal{O}\_Z)$ where $Z = \text{Spec}(A/m)\hookrightarrow X$ by GRR then you use the following short exact sequence:
$$0 \rightarrow m/m^2 \rightarrow \mathcal{O}\_Y\rightarrow \mathcal{O}\_Z\rightarrow 0$$
Note the pushforward is also going to be exact. The kernel $m/m^2$ is a $2$-dimensional vector space over $A/m$ so its pushforward is going to be direct sum of two $i\_\*(\mathcal{O}\_Z)$. Now you can calculate the Chern character of $i\_\*(\mathcal{O}\_Y)$.
| 3 | https://mathoverflow.net/users/127776 | 430447 | 174,358 |
https://mathoverflow.net/questions/430450 | 3 | This [paper](https://reader.elsevier.com/reader/sd/pii/0167715294901104?token=A61379C3631ADCAC256F64615D513FEE297A7F515D6873C962AA17DFBF8B8992D4BBB4DE58A9018CA4D09009E587C0E9&originRegion=us-east-1&originCreation=20220914163053) proves a probabilistic version of Taylor's theorem
\begin{equation\*}
\mathbb{E}g(X) = \sum\_{k=0}^{n-1} \frac{g^{(k)}(0)}{k!} \mathbb{E}X^k + \frac{\mathbb{E}X^n}{n!} \mathbb{E} g^{(n)}(X\_{(n)}),
\end{equation\*}
where $X\_{(n)}$ is another random variable derived from $X$ and $g^{(n)}$ is the $n$-th derivative of $g$. Suppose we take $n=4$ and we know that $g^{(4)} < 0$ (e.g. due to concavity), then it seems to follow that
\begin{equation\*}
\mathbb{E}g(X) = \sum\_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}X^k + \frac{\mathbb{E}X^4}{4!} \mathbb{E} g^{(4)}(X\_{(4)})
< \sum\_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}X^k .
\end{equation\*}
Hence we can get explicit upper bound on $\mathbb{E}g(X)$. However, a *crucial assumption* in the probabilistic Taylor theorem is that $X$ is non-negative. This means we cannot apply the result to a centered random variable like $X-\mu$, where $\mu = \mathbb{E}X$. Suppose again that $g^{(4)} < 0$. Are there any results that would allow me to conclude something like the following?
\begin{equation\*}
\mathbb{E}g(X-\mu) = \sum\_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}(X-\mu)^k + \frac{\mathbb{E}(X-\mu)^4}{4!} \mathbb{E} g^{(4)}(Y) < \sum\_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}(X-\mu)^k,
\end{equation\*}
where $Y$ would be another random variable linked to $X-\mu$.
| https://mathoverflow.net/users/159841 | Probabilistic Taylor theorem for concave functions | If $g^{(4)}\le0$, [then](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder)
$$g(x)=\sum\_{k=0}^3\frac{g^{(k)}(0)}{k!}\,x^k+\frac{x^4}4\,
\int\_0^1g^{(4)}(sx)(1-s)^3\,ds
\le\sum\_{k=0}^3\frac{g^{(k)}(0)}{k!}\,x^k$$
for real $x$.
Replacing here $x$ by $X-\mu$ and assuming that $E|X|^3<\infty$, we get
$$Eg(X-\mu)\le\sum\_{k=0}^3\frac{g^{(k)}(0)}{k!}\,E(X-\mu)^k,\tag{1}\label{1}$$
as desired.
---
(The strict inequality $<$ in \eqref{1} will not hold in general, even if $g^{(4)}<0$. In particular, the strict inequality in \eqref{1} will not hold if $P(X=\mu)=1$.)
| 3 | https://mathoverflow.net/users/36721 | 430456 | 174,362 |
https://mathoverflow.net/questions/430440 | 13 | Let $f\in C([0,1],[0,1])$ be such that:
$$\forall x\in [0,1], \; \exists k\in \mathbb N, \; f^{\circ k}(x)=0.$$
Is it true that $f$ is nilpotent (i.e., that there is some $k$ such that $f^{\circ k}=0$)?
Here $f^{\circ k}$ denotes the $k$th iterate of $f$.
| https://mathoverflow.net/users/110301 | Does locally nilpotent imply nilpotent for continuous self-maps of intervals? | Yes, this implies that $f$ is nilpotent.
As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.
Decompose the open set $\{x: f(x)>0\}=\bigcup I\_n$ into its connected components. Clearly, since each set $[0,a]$ is invariant, the zeros of $f$ must accumulate at $0$. On the other hand, I claim that the $I\_n$ do not accumulate at $x=0$.
Indeed, if they did, we could start out with any $I\_0$ and then $f(\overline{I\_0})=[0,b\_0]$ for some $b\_0>0$. By assumption, $I\_1\subseteq [0,b\_0]$ for some $I\_1$. Let $K\_1=\{x\in \overline{I\_0}: f(x)\in \overline{I\_1}\}$. Since $0\notin\overline{I\_n}$ for all $n$, the orbit of any $x\in K\_1$ will not yet have reached zero after one iteration.
Continue in this style: $f(\overline{I\_1})=[0,b\_1]\supseteq I\_2$ for some $I\_2$. Let $K\_2 =\{x\in K\_1: f^2(x)\in \overline{I\_2}\}$. The compact sets $K\_n$ are nested, so $\bigcap K\_n\not=\emptyset$, but if $x\in K\_n$, then $f^k(x)\not= 0$ for $k\le n$, so this point never reaches zero.
It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.
| 7 | https://mathoverflow.net/users/48839 | 430458 | 174,363 |
https://mathoverflow.net/questions/430419 | 7 | I'm a science/math journalist [ger] and currently I'm working on an article about prize money for open problems (Millennium Prize Problems and such). One section will be about the history of prize money and rewards. While I know about the culture of betting and rewards for open problems in the Lwow school and Erdös' faible for betting and rewarding, I wonder if there is a reliable source on the history of such rewards. In particular I'm interested in the question since when have money rewards for open problems started being a thing in mathematics and what some of the historic examples for such rewards are.
Thank you for helping me out :-)
| https://mathoverflow.net/users/479777 | Best sources for the history of prize money for open mathematical problems | Here is a long list of [Mathematics awards](https://en.wikipedia.org/wiki/List_of_mathematics_awards) and many of them have a prize money too (you can click on each of them in order to read more and find winners data).
Another couple of important prizes to mention (without taking into account the Clay Mathematics Institute prize of $1,000,000$ USD for each of the well-known Millennium Problems) are the $1,000,000$ USD cash prize for the first person able to prove the Beal Conjecture, the $120,000,000$ JPY (i.e., more than $1,000,000$ USD) for the first mathematician who will solve the Collatz conjecture (see [Prize for the Collatz conjecture](https://www.prnewswire.com/news-releases/bakuage-offers-prize-of-120-million-jpy-to-whoever-solves-collatz-conjecture-math-problem-unsolved-for-84-years-301326629.html)).
Lastly, it is worth to mention the $2,000$ USD prize (plus a bottle of champagne) waiting for the first solver of each of the problems listed on this page: [2,000 USD prizes](https://unsolvedproblems.org/).
| 3 | https://mathoverflow.net/users/481829 | 430472 | 174,368 |
https://mathoverflow.net/questions/430401 | 3 | (The construction of matrix $\mathbf{A}$ is not difficult to be understood. You can first jump to **A Toy Example** to take a glance. Any idea or suggestion would be appealing for me.)
---
**The Original Problem:**
Given $N,D\in\mathbb{Z}^+~(D\ge N)$ and $\alpha\in\mathbb{R}^+$, the vector $\mathbf{p}$ and the matrix $\mathbf{A}\_\mathbf{p}$ are defined as follows:
* $\mathbf{p}=[p\_1,p\_2,\cdots,p\_N]$, where $p\_i$s are selected from $\{1,2,\cdots,D\}$ and satisfying the condition of $(p\_1<p\_2<\cdots<p\_N)$.
* Given $\mathbf{p}$, there is $\mathbf{A}\_\mathbf{p}=[a\_{ij}]\_{N\times N}$, where $a\_{ij}=e^{-\alpha |p\_i-p\_j|}$.
I am trying to find out the property of $\det(\mathbf{A}\_\mathbf{p})$. Based some of my findings, I am confused by the following two subproblems:
$(\text{Q}1)$ Can we conveniently calculate the value of $\det(\mathbf{A}\_\mathbf{p})=f(\mathbf{p})$ for a given $\mathbf{p}$? In other words, is there a way to explicitly unfold $\det(\mathbf{A}\_\mathbf{p})$?
$(\text{Q}2)$ Is $\mathbf{A}\_\mathbf{p}$ positive semi-definite?
$(\text{Q}3)$ Does $\det(\mathbf{A}\_\mathbf{p})$ hit its minimal value only when $(p\_{i+1}-p\_i=1)$? By the way, in this case, $\mathbf{A}\_\mathbf{p}$ will become a special symmetric Toeplitz matrix.
---
**A Toy Example:**
Given $N=3$, $D=10$ and $\alpha =1$. I construct $\mathbf{p}\_1=[3,4,5]$ and $\mathbf{p}\_2=[2,5,7]$. Then we have:
$$
\det \left( \mathbf{A}\_{\mathbf{p}\_1} \right) =\left| \begin{matrix}
1& e^{-1}& e^{-2}\\
e^{-1}& 1& e^{-1}\\
e^{-2}& e^{-1}& 1\\
\end{matrix} \right|\approx 0.748,
$$
and
$$
\det \left( \mathbf{A}\_{\mathbf{p}\_2} \right) =\left| \begin{matrix}
1& e^{-3}& e^{-5}\\
e^{-3}& 1& e^{-2}\\
e^{-5}& e^{-2}& 1\\
\end{matrix} \right|\approx 0.979 > \det \left( \mathbf{A}\_{\mathbf{p}\_1} \right) .
$$
---
**Some of My Efforts:**
I may have the following observations:
$(\text{O}1)$ The diagonal elements of $\mathbf{A}\_\mathbf{p}$ are all ones since $|p\_i-p\_i|=0$.
$(\text{O}2)$ All elements of $\mathbf{A}\_\mathbf{p}$ are in $[0,1]$.
$(\text{O}3)$ $\mathbf{A}\_\mathbf{p}$ is symmetric since $|p\_i-p\_j|=|p\_j-p\_i|$.
$(\text{O}4)$ Actually, the order of $\mathbf{p}\_i$s do not affect the value of $\det(\mathbf{A}\_\mathbf{p})$.
I guess that $\mathbf{A}\_\mathbf{p}$ has the following two properties:
$(\text{P}1)$ The answer of $(\text{Q}2)$ is "Yes", *i.e.*, $\mathbf{A}\_\mathbf{p}$ is positive semi-definite.
$(\text{P}2)$ The answer of $(\text{Q}3)$ is "Yes", *i.e.*, $\left[\det(\mathbf{A}\_\mathbf{p})=\min\left\{{\det(\mathbf{A}\_{\mathbf{p}\_k})}\right\}\right] \Leftrightarrow \left[ \forall i, ~p\_{i+1}-p\_{i}=1 \right]$.
The above conjectures of $(\text{P}1)$ and $(\text{P}2)$ is empirically presented. I write the following Python code to validate them and find that all randomly generated $\mathbf{A}\_\mathbf{p}$ satisfy $(\text{P}1)$ and $(\text{P}2)$:
```
import numpy as np
import random
from scipy import spatial
alpha = 1
N = 10
X = np.arange(N).reshape(N, 1)
X = np.exp(-alpha * spatial.distance.cdist(X, X))
X_det = np.linalg.det(X)
for D in range(N, 1000):
for i in range(100):
p = np.array(random.sample(range(1, D + 1), N)).reshape(N, 1)
A = np.exp(-alpha * spatial.distance.cdist(p, p))
A_det = np.linalg.det(A)
if A_det <= 0:
print(A_det, p.reshape(N,)) # det(A) <= 0
exit(0)
if A_det < X_det and abs(A_det - X_det) > 1e-8:
print(p, p.reshape(N,)) # det(A) < det(X) with numerical tolerance
exit(0)
print('Done.')
```
I test many combinations of $\{\alpha, N, D\}$. I see that there is no any case satisfy the conditions of `A_det <= 0` and `A_det < X_det`.
---
**Why I Try to Study $\det(\mathbf{A}\_\mathbf{p})$?**
I study the entropy of multivariate Gaussian distributions with some special covariance matrices (*i.e.*, the above defined $\mathbf{A}\_\mathbf{p}$). The entropy value is related to $\det(\mathbf{A}\_\mathbf{p})$ (you can see more details from my previous problems below).
I have made efforts and spent more than 14 days on it. Specifically, I read some textbooks, papers and blogs related to it. Here are some previous problems posted by me: [Problem 1](https://math.stackexchange.com/questions/4524167/expectations-of-multivariate-gaussian-entropy), [Problem 2](https://math.stackexchange.com/questions/4526630/prove-disprove-the-matrix-convergence-in-probability), [Problem 3](https://math.stackexchange.com/questions/4529767/prove-disprove-mathbbe-ln-det-mathbfa-rightarrow-0-when-bounded) and [Problem 4](https://math.stackexchange.com/questions/4525722/is-the-determinant-of-mathbfa-in0-1n-times-n-bounded). However, I am still stucked. Now I think that the key step is to resolve the problem I posted above.
I am sorry for occupying much public resource of this platform. But I really want to resolve the problems, especially $(\text{Q}2)$ and $(\text{Q}3)$. Could you please provide help or some tips?
| https://mathoverflow.net/users/490652 | $\min(\det(\mathbf{A}))$ for special matrix $\mathbf{A}$ | **Q1** The determinant is $\prod\_{n=1}^{N-1} (1 - e^{-2\alpha(p\_{n+1}-p\_n)})$.
**Q2** Yes, using the answer to **Q1**.
**Q3** Yes, using the answer to **Q1**.
---
---
The formula for **Q1** is proved by induction on $N$.
The base case $N=1$ is just $\det(1)\_{1\times1} = 1$.
For $N>1$, let $c = e^{-\alpha(p\_N^{\phantom.} - p\_{N-1}^{\phantom.})}$,
and let ${\bf A'\_p}$ be the matrix obtained from ${\bf A\_p}$
by subtracting $c$ times row $N-1$ from row $N$
and then subtracting $c$ times column $N-1$ from column $N$.
Then $\det{\bf A'\_p} = \det{\bf A\_p}$.
But the last row and column of ${\bf A'\_p}$ are all zero
except for the $(N,N)$ entry which is $1-c^2$.
Therefore $\det{\bf A'\_p}$ is $1-c^2$ times the determinant of
the symmetric matrix obtained from ${\bf A\_p}$ by deleting its
$N$-th row and $N$-th column -- which is just ${\bf A\_{p'}}$
where ${\bf p'} = [p\_1, p\_2, \ldots, p\_{N-1}]$.
This completes the induction step and the proof.
---
**Q2** By the product formula, $\det A$ is positive, as is the determinant of
every symmetric minor of $A$ (which is of the same form as $A$ for some
subsequence of $p\_1,\ldots,p\_N$).
---
**Q3** This too follows from the product formula:
each factor $1 - e^{-2\alpha(p\_{n+1}-p\_n)}$ is at least
$1 - e^{-2\alpha}$, with equality if and only if $p\_{n+1} - p\_n = 1$.
---
---
We can also deduce the positive answer to **Q2** by constructing
linearly independent probability distributions $\mu\_i$ on the real numbers
whose covariance matrix is proportional to ${\bf A\_p}$: the distribution $\mu\_i$
chooses integers $p \geq p\_i$ with probability
$(1-e^{-\alpha}) e^{-\alpha(p-p\_i)}$.
If the real numbers $p\_i$ are not required to be integers then we can achieve
the same goal using continuous $\mu\_i$ with distribution functions
$\alpha e^{-\alpha(p-p\_i)}$ supported on $p \geq p\_i$.
| 5 | https://mathoverflow.net/users/14830 | 430486 | 174,370 |
https://mathoverflow.net/questions/430355 | 30 | Starting with a single stick of unit length, a point $p \in (0, 1)$ is picked uniformly at random along the stick and the stick is snapped, producing two sticks of length $p$ and $1-p$.
At each next stage, a stick is picked uniformly at random, and a point is picked uniformly at random along the length of that stick, and it is snapped.
**Question:** After n snaps, what is the expected length of the longest remaining stick?
**Remarks:**
Myself and a friend of mine did some simulations and found some pretty unexpected results. The expected value after $500$ splits is approximately $0.2098$, which is massive for that many splits, at least intuitively.
On the other hand, it can be proven rather easily that the expected value does go to $0$ as $n \to \infty$. But the decay seems extremely slow.
| https://mathoverflow.net/users/173490 | Expected length of longest stick in a stick snapping process | It seems that the length of the longest stick is of order $n^{2\sqrt{2}-3} = n^{-0.171\ldots}$ as $n\to\infty$. This follows from a discrete-time analogue of the homogeneous fragmentation process, see chapter 1.5 of J. Bertoin, *Random fragmentation and coagulation processes*. Vol. 102. Cambridge University Press, 2006.
Let us denote by $X\_{n,0} \geq X\_{n,1} \geq \cdots \geq X\_{n,n}$ the ordered sizes of the sticks after $n$ snaps. The first result we need is the following.
**Lemma:** $\chi\_{n}(p) := \mathbb{E}\left[\sum\_{i=0}^n X\_{n,i}^p\right] = \frac{1}{n!}\left(\frac{2}{1+p}\right)\_n,$ where $(a)\_n=a(a+1)\cdots (a+n-1)$ is the rising Pochhammer symbol.
**Proof:** We have
\begin{align\*}
\mathbb{E}\left[\sum\_{i=0}^n X\_{n,i}^p\middle| X\_{n-1,0},\ldots\right] &= \frac{1}{n}\sum\_{k=0}^{n-1}\Big(\mathbb{E}\left[x^p + (X\_{n-1,k}-x)^p\middle|X\_{n-1,k}\right]+\sum\_{\substack{i=0\\i\neq k}}^{n-1}X\_{n-1,i}^p\Big)\\
&= \frac{1}{n}\left(\frac{2}{p+1}+n-1\right)\sum\_{i=0}^{n-1}X\_{n-1,i}^p,
\end{align\*}
where $x$ is uniform in $(0,X\_{n-1,k})$. Hence $\chi\_n(p) = \frac{1}{n}\left(\frac{2}{p+1}+n-1\right) \chi\_{n-1}(p)$.
Together with $\chi\_0(p)=1$, this gives the claimed formula for $\chi\_n(p)$. $\square$
Following Corollary 1.4 in Bertoin's book, we note that
\begin{equation}
n^{\frac{p-1}{p+1}} \chi\_n(p) = \frac{1}{\Gamma\left(\frac{2}{p+1}\right)} + O(n^{-1}).
\end{equation}
In particular it is bounded for any $p>-1$. Since $X\_{n,0}^p < \sum\_{i=0}^n X\_{n,i}^p$, we deduce that $n^{\frac{p-1}{p+1}}X\_{n,0}^p$ is bounded as $n\to\infty$. Hence
\begin{equation}
\limsup\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} \leq -\frac{1}{p}\frac{p-1}{p+1} \leq -\frac{1}{\bar{p}}\frac{\bar{p}-1}{\bar{p}+1} = 2\sqrt{2}-3,
\end{equation}
because the maximum $\bar{p}$ of $-\frac{1}{p}\frac{p-1}{p+1}$ is achieved at $\bar{p} = 1+\sqrt{2}$.
Similarly one can derive a matching lower bound by noting that
\begin{equation}
X\_{n,0}^\epsilon \geq \frac{\sum\_{i=0}^{n}X\_{n,i}^p}{\sum\_{i=0}^{n}X\_{n,i}^{p-\epsilon}}
\end{equation}
for any $\epsilon>0$, which implies
\begin{equation}
\liminf\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} \geq - \frac{\frac{p-1}{p+1} - \frac{p-\epsilon-1}{p-\epsilon+1}}{\epsilon}
\end{equation}
for any $\epsilon>0$ and $0<p<\bar{p}$. Letting $\epsilon$ approach $0$ and $p$ approach $\bar{p}$, we thus have
\begin{equation}
\liminf\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} \geq -\frac{2}{(1-\bar{p})^2}=2\sqrt{2}-3.
\end{equation}
We may therefore conclude that
\begin{equation}
\lim\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} =2\sqrt{2}-3.
\end{equation}
| 18 | https://mathoverflow.net/users/47484 | 430495 | 174,373 |
https://mathoverflow.net/questions/430492 | 7 | Let $q$ be a prime power, let $n$ be a positive integer and let $\mathbb{F}\_q$ be the finite field of cardinality $q$. I have some computational evidence that the set $$\{x^n+(-1)^nay^n:x,y\in\mathbb{F}\_q\}$$ is the whole of $\mathbb{F}\_q$, unless $q$ is small (with respect to $n$).
This does make sense because if, for instance, $q-1=n$, then the set
$\{ x^n+(-1)^nay^n:x,y\in\mathbb{F}\_q \}$ only consists of $\{0,1,a,1+a\}$ because each element of $\mathbb{F}\_q$ raised to the $n$ equals either $0$ or $1$.
Can anyone justify the computational evidence?
| https://mathoverflow.net/users/45242 | Sum of two $n$th powers in finite fields | Yes, this is true, and is proved e.g. as Corollary 3 of Small's "[Diagonal equations over large finite fields](https://doi.org/10.4153/CJM-1984-016-6)" (Can. J. Math. 1984).
Small actually gives explicit bounds on how large $q$ needs to be in terms of $n$ — in particular the equation $ax^n+by^n$ generates all of $\mathbb{F}\_q$ whenever $a,b\in\mathbb{F}\_q\setminus\{0\}$ and $q>(\delta-1)^4$, where $\delta=(n,q-1)$.
| 11 | https://mathoverflow.net/users/385 | 430496 | 174,374 |
https://mathoverflow.net/questions/430488 | 13 | Let $X$ be a reflexive Banach space. Then the convex hull of the extreme points of the unit ball is weakly dense by the Krein-Milman theorem and Kakutani's theorem. My question is, if there is an example of a reflexive Banach space $X$ whose unit ball does not equal the convex hull of its extreme points? Such an example $X$ must be infinite-dimensional and can't be strictly convex. My feeling is that this should be known, but I couldn't find anything about it. I've asked this question on StackExchange as well.
| https://mathoverflow.net/users/58366 | Is there a reflexive Banach space whose ball is not the convex hull of its extreme points? | I'd try $X:=\ell\_2$ with an equivalent but non strictly convex norm.
Let $(e\_k)\_{k\ge0}$ be the standard Hilbert basis of $\ell\_2$. Consider the sets:
* $A:=\{0\}\cup\{2^{-k}e\_k:k\ge 1\}$,
* $B$, the closed unit ball of $\ell\_2$,
* $C:=\overline{\text{co}}\, A$, a compact subset of the hyperplane $e\_0^\perp$.
* $D:=\overline{\text {co}}\big(B\cup( e\_0+C)\cup(-e\_0-C)\big)$.
Clearly, $C:=\big\{\sum\_{k\ge1}2^{-k}a\_ke\_k: a\_k \ge0 \text{ for all } k\ge1 \text{ and } \sum\_{k\ge1}a\_k\le1\big\},$ so $\text{ext}\, C=A$ and ${\text{co}}(\text{ext}\, C)={\text{co}}\, A\subsetneq C. $
The set $D$ is a bounded, closed, convex, symmetric nbd of $0$ in $\ell\_2$, so it is the closed unit ball in an equivalent norm of $\ell\_2$.
We have $D\cap (e\_0+e\_0^\perp)=e\_0+C;$ the set of extremal points of $D$ in the affine hyperplane $e\_0+e\_0 ^\perp$ is exactly the set $e\_0+A$, which implies that ${\text{co}}(\text{ext} D)\subsetneq D$ because the trace of these sets on $e\_0+ e\_0 ^\perp$ are ${\text{co}}(\text{ext} D)\cap (e\_0+ e\_0 ^\perp) =e\_0+\text{co}A\subsetneq D \cap (e\_0+ e\_0 ^\perp) $.
| 7 | https://mathoverflow.net/users/6101 | 430502 | 174,376 |
https://mathoverflow.net/questions/430462 | 1 | Let $X$ be an $n$ dimensional standard Gaussian and let $U$ be an $n \times n$ orthogonal matrix. Then, the random vector $Z = U^\top X$ is also distributed as a standard Gaussian in $R^n$ and we have $E[\prod\_{i=1}^n Z\_i^2] = 1$ by independence.
Is there a method for bounding such functions if $X$ was an $n$ dimensional vector where each coordinate is an independent Rademacher $\pm1$ random variable. For instance, for $Z = U^\top X$, the coordinates are not independent anymore. Can one show that $E[\prod\_{i=1}^n Z\_i^2]$ is polynomial in $n$ or $\exp(n)$ even in this setting?
More generally, is there a way of comparing such quantities to the Gaussian setting or universality statements that are applicable in such settings where the random vector may have some weak dependencies?
| https://mathoverflow.net/users/491250 | Comparison of Rademacher and Gaussian moments under linear transformations | By the arithmetic mean--geometric mean inequality and the condition $Z=U^\top X$ for an orthogonal matrix $U$,
$$\prod\_1^n Z\_i^2\le\Big(\frac1n\sum\_1^n Z\_i^2\Big)^n
=\Big(\frac1n\sum\_1^n X\_i^2\Big)^n=1,$$
since $X\_i=\pm1$ for all $i$. So,
$$E\prod\_1^n Z\_i^2\le1,$$
as desired.
| 2 | https://mathoverflow.net/users/36721 | 430508 | 174,379 |
https://mathoverflow.net/questions/430518 | 1 | As we know we can construct unitary matrix as $H=H\_1H\_2\dots,H\_k$, by stacking householder matrices $H\_i\in \mathbb{R}^{d\times d}$. The number of householder matrices we use, i.e., $k$, determines the expressivity of construction.
Conversely, for a given unitary matrix $H$, how can I know how many householder matrices I need at least for building this $H$? I mean how can I know the minimal $k$.
| https://mathoverflow.net/users/127318 | How many householder matrices do I need for constructing a given unitary matrix? | A Householder decomposition of a $d \times d$ unitary $H$ can be achieved with *at most* $d$ Householder matrices (see [Theorem 1](https://core.ac.uk/download/pdf/82680205.pdf)) and a tighter upper bound is $d-m$ Householder matrices where $m = \dim( \operatorname{ker}(H-I\_d) )$ (see [Theorem 2](https://core.ac.uk/download/pdf/82680205.pdf)).
*Uhlig, Frank*, [**Constructive ways for generating (generalized) real orthogonal matrices as products of (generalized) symmetries**](http://dx.doi.org/10.1016/S0024-3795(01)00296-8), Linear Algebra Appl. 332-334, No. 1-3, 459-467 (2001). [ZBL0982.65049](https://zbmath.org/?q=an:0982.65049).
| 3 | https://mathoverflow.net/users/64449 | 430525 | 174,384 |
https://mathoverflow.net/questions/430433 | 3 | I have a few related questions. First I would appreciate it if someone could provide me with a reference for the following
"Complex unitary irreducibles of virtually abelian groups have bounded degrees."
I am only concerned with complex unitary representations in this question.
Do we have a quantitative version of the above statement? For example, is the following true?
"If $G$ is a virtually abelian group with a finite index abelian subgroup H, then the degrees of irreducibles of $G$ are bounded above by $[G:H]$"
If H is additionally normal, is it true that the degrees of irreducibles of G divide $[G:H]$?
If these are not true in general, would they be true if $G$ is finitely generated in addition to being virtually abelian?
| https://mathoverflow.net/users/491235 | Irreducibles of virtually abelian finitely generated groups | I can at least explain how this bounded degree property works roughly when $G$ is a countable virtually abelian group. Kaplansky shows in his paper Groups with representations of bounded degree, Canadian J. Math vol. 1 (1949), 105-112 that the group ring of a virtually abelian group satisfies a polynomial identity.
If $K$ is any field, then the image of $KG$ under an irreducible representation is a primitive $K$-algebra satisfying this same polynomial identity. In Kaplansky, Rings with a polynomial identity, Bull. Amer. Math. Soc. 54 (1948) 575–580 that a primitive PI algebra is of the form M\_k(D) where D is the division algebra which is the commutant of the irreducible representation and that D is finite dimensional over its center. Note that k can be bounded in terms of the degree of the polynomial identity (in Passman's book you can get precise bounds on the degree) because the smallest degree polynomial identity satisfied by $M\_r(K)$ goes to infinity as $r$ goes to infinity.
If $G$ is countable, then every simple $\mathbb CG$-module is countable dimensional and hence the commutant of an irreducible representation has countable dimension over $\mathbb C$. But the only division algebra over $\mathbb C$ of countable dimension is $\mathbb C$ itself. Thus any irreducible representation of $\mathbb CG$ has image $M\_r(\mathbb C)$ where $r$ is bounded.
In any event once you know that each representation is of finite degree then you can take a normal abelian subgroup of index say $m$ and use Clifford theory to say each irreducible of $G$ has degree at most $m$.
| 4 | https://mathoverflow.net/users/15934 | 430535 | 174,386 |
https://mathoverflow.net/questions/430530 | 9 | I have already asked this at [MSE](https://math.stackexchange.com/questions/4524317/propagators-and-pdes) but did not get an answer.
In quantum field theory one encounters the retarded, advanced and Feynman propagators as certain solutions to a wave equation. Mathematically, these derivations are somewhat magical (typically one inserts an "infinitesimally small" iε
term, and then the different propagators result from different integration contours around certain poles). On the other hand, there is a mathematically rigorous theory of fundamental solutions to PDEs, but I have never seen anything analogous to these propagators in a PDE book. Can somebody recommend a source (book/lecture notes/paper), where the retarded, advanced and Feynman propagators are treated in a mathematically rigorous way, so that I can see the connection between QFT and PDE theory?
| https://mathoverflow.net/users/89014 | Propagators and PDEs | You'll find some more info about the fundamental solutions of the wave equation in chapter 5.D of [Folland](https://press.princeton.edu/books/hardcover/9780691043616/introduction-to-partial-differential-equations) and chapter I.7 of [Trèves](https://store.doverpublications.com/0486453464.html).
The trick you use is the idea that (tempered) distributions, even ones with non-trivial singular support, can have well-defined Fourier transforms.
So it'll help if you know a bit about distribution theory, but if not, the chapter on that in Folland is quite good.
I like what Folland has to say about propagators:
>
> ...there are a number of natural but quite different fundamental solutions of the wave equation. (In other words, there are a number of natural but quite different distributions that agree with the function $[4\pi^2(|\xi|^2 — \tau^2)]^{-1}$ *away from the light cone!*)
>
>
>
(Emphasis mine.)
Hopefully this helps!
| 9 | https://mathoverflow.net/users/49417 | 430536 | 174,387 |
https://mathoverflow.net/questions/430532 | 1 | Here are two different versions of Gaussian ballot theorems in the literature, each on different while similar events but the rate is quite different:
1. [P39, Probability Result 1](http://wrap.warwick.ac.uk/129549/7/WRAP-moments-functions-squareroot-cancellation-chaos-Harper-2020.pdf): For any independent sequence of Gaussian random variables $G\_m$, mean-zero and variances in between $1/20$ and $20$, and denote $S\_j:=\sum\_{m=1}^j G\_m$ we have for any $a>0$ $$\mathbb{P}(S\_j\le a,\forall j\le n)\asymp\min(1,\frac{a}{\sqrt{n}}).$$
and
2. [P19, Theorem 5](https://arxiv.org/pdf/1601.00582.pdf): For $G\_m$ as above, but now variances are the same(say $1$). Let $\delta>0$. Then there is some constant $C(\delta)>0$ such that for all $B>0$, $b\le B-\delta$ and $n\geq 1$ we have $$\mathbb{P}(S\_n\in(b,b+\delta),S\_j\le B, \forall j<n)\le C\frac{(1+B)(1+B-b)}{n^{3/2}},$$
and if $\delta<1$, $$\mathbb{P}(S\_n\in[0,\delta], S\_j\le 1,\forall 0<j<n)\geq \frac{1}{Cn^{3/2}}.$$
So I kind of don't understand, say the upper bound part, why by changing the control on last step from $\le a$ to be in some fixed range will make the rate shrink by $n$. Or the question may be phrased as, why are these two results consistent with each other?
| https://mathoverflow.net/users/174600 | What happens in the difference rate between these two versions of ballot theorem? | $\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}\newcommand\ep\varepsilon$The paper linked in your post contains a reference to a reference to an apparent proof, which I hope results in an overall valid proof.
Anyhow, we can assess the plausibility of the relations
\begin{equation\*}
P(S\_j\le a\ \forall j\le n)\asymp\min(1,a/\sqrt{n}) \tag{1}\label{1}
\end{equation\*}
and
\begin{equation\*}
P(S\_n\in(b,b+\de),S\_j\le B\ \forall j<n)\le C\frac{(1+B)(1+B-b)}{n^{3/2}} \tag{2}\label{2}
\end{equation\*}
by replacing the probabilities there by the corresponding probabilities for the standard Wiener process $(W\_t)$.
Indeed, let $M\_1:=\max\_{t\in[0,1]}W\_t$. Let $\Phi$ and $\vpi$ denote the standard normal cdf and pdf, respectively. Let $\ep:=1/\sqrt n\to0$.
Then, by the reflection principle,
\begin{equation\*}
P(M\_1\le \ep a)=2(\Phi(\ep a)-\Phi(0))\asymp\min(1,a\ep), \tag{1a}\label{1a}
\end{equation\*}
which suggests that \eqref{1} is plausible.
Now, to assess the plausibility of \eqref{2}, write
\begin{equation\*}
\begin{aligned}
&P(W\_1\in\ep(b,b+\de),M\_1\le \ep B) \\
&=P(W\_1\in\ep(b,b+\de))-P(W\_1\in\ep(b,b+\de),M\_1>\ep B) \\
&=P(W\_1\in\ep(b,b+\de))-P(W\_1\in\ep(2B-b-\de,2B-b)),
\end{aligned}
\end{equation\*}
again by the reflection principle. So,
\begin{equation\*}
P(W\_1\in\ep(b,b+\de),M\_1\le \ep B)=g(\de)-g(0)=\de\,g'(c\de), \tag{2a}\label{2a}
\end{equation\*}
where
\begin{equation\*}
g(t):=\Phi(\ep(b+t))-\Phi(\ep(2B-b-\de+t)) \tag{2b}\label{2b}
\end{equation\*}
and $c=O(1)$. Note that for $t=O(1)$
\begin{equation\*}
g'(t)=\ep\,[\vpi(\ep(b+t))-\vpi(\ep(2B-b-\de+t))] \\
=O(\ep^2\,|\vpi'(\ep c\_1)|)=O(\ep^2\,\ep |c\_1|\vpi(\ep c\_1))=O(\ep^3),
\end{equation\*}
where $c\_1=O(1)$.
So,
\begin{equation\*}
P(W\_1\in\ep(b,b+\de),M\_1\le \ep B)=O(\ep^3)=O(1/n^{3/2}), \tag{2c}\label{2c}
\end{equation\*}
which suggests that \eqref{2} is plausible as well, at least as far as the order of magnitude is concerned.
Thus, looking at \eqref{1a}, we can say that $P(M\_1\le \ep a)\asymp\ep$ for small $\ep$ because $P(M\_1\le \ep a)$ is a **first-order** difference of the values of $\Phi$ for arguments near $0$ differing by $\asymp\ep$, **and $\Phi'(0)\ne0$**.
On the other hand, looking at \eqref{2a} and \eqref{2b}, we can say that $P(W\_1\in\ep(b,b+\de),M\_1\le \ep B)=O(\ep^3)$ because $P(W\_1\in\ep(b,b+\de),M\_1\le \ep B)$ is a **second-order** difference of the values of $\Phi$ for arguments differing from $0$ by $O(\ep)$, **and $\Phi''(0)=0$** (so that $\Phi''(\ep c\_2)=O(\ep)$ if $c\_2=O(1)$). The second order of the difference in question is most clearly seen in the case when $b=B-\de$, which implies that $g(t)=\Phi(\ep(B+t-\de))-\Phi(\ep(B+t))$.
| 2 | https://mathoverflow.net/users/36721 | 430548 | 174,390 |
https://mathoverflow.net/questions/430545 | -1 | Let $a,b$ be coprime and say $0<a<b<2a$.
Consider the quadratic system:
$$\alpha\delta-\beta\gamma=1$$
$$(\alpha^2-(\alpha\delta+\beta\gamma))a^2b+\beta^2b^3+(2\alpha\beta-\beta\delta)ab^2-\alpha\gamma a^3=0$$
$$(-\alpha\gamma+2\gamma\delta)a^2b+(\delta^2-(\alpha\delta+\beta\gamma))ab^2-\beta\delta b^3+\gamma^2a^3=0$$
1. Does it admit an integral solution in $\alpha,\beta,\gamma,\delta$? Is so, then is there a polynomial time algorithm in $\log|ab|$?
I think there are no integral solutions. We get the criteria
$$\beta^2\equiv\beta\delta\equiv 0\bmod a$$
$$\gamma^2\equiv\alpha\gamma\equiv 0\bmod b$$
This constraint seems very restrictive with respect to $\alpha\delta-\beta\gamma=1$.
2. If not, then does it admit a rational solution in $\alpha,\beta,\gamma,\delta$? Is so, then is there a polynomial time algorithm in $\log|ab|$?
| https://mathoverflow.net/users/10035 | Does this quadratic system admit an integral or a rational solution? | In my comments I employed Maple, which uses tools like Grobner bases to solve polynomial equations. But now I'll try to do it by hand. Let $E\_1,E\_2,E\_3$ be the three equations. A rational solution of $\{E\_1,E\_2,E\_3\}$ can be turned into an integer solution of $\{E\_2,E\_3\}$ by multiplying by a common denominator, so I'll first look for integer solutions of $\{E\_2,E\_3\}$. Coprimality of $a,b$ is silently assumed throughout and variables are integer unless specified otherwise.
From $E\_2$, $\beta$ is a multiple of $a$, while from $E\_3$, $\gamma$ is a multiple of $b$. Put $\beta=ax, \gamma=by$. Now (using $a,b\ne 0$), the equations factor:
\begin{align}
(bx+\alpha)(ay-bx-\alpha+\delta)&=0.\tag{$E\_{2a}$} \\
(ay+\delta)(ay-bx-\alpha+\delta)&=0.\tag{$E\_{3a}$}
\end{align}
The first case is $\alpha=-bx$, which turns $E\_{3a}$ into $(ay+\delta)^2=0$ so $\delta=-ay$. So this integer
solution of $\{E\_2,E\_3\}$ is
$$ \alpha=-bx, \beta=ax, \gamma=by, \delta=-ay.$$
However, this gives $\alpha\delta-\beta\gamma=0$ so we can't
satisfy $E\_1$ even by scaling. So this solution is discarded.
The second case is $ay-bx-\alpha+\delta=0$, which is a factor
of both $E\_{2a}$ and $E\_{3a}$. So we can choose $\delta$ arbitrarily and set $\alpha=ay-bx+\delta$. So this integer solution of $\{E\_2,E\_3\}$ is
$$ \alpha = ay-bx+\delta, \beta=ax, \gamma=by. \tag{S} $$
We can locate all **rational solutions of $\{E\_1,E\_2,E\_3\}$**
by uniformly scaling $\alpha,\beta,\gamma,\delta$ to satisfy $E\_1$. However, the scale factor might be irrational so we must be careful. What we need is that $\alpha\delta-\beta\gamma=\rho^2$ where $\rho$ is some rational number. Solving that for $x$, substituting and scaling down by $\rho$,
we find that all rational solutions of $\{E\_1,E\_2,E\_3\}$ have the form
$$ \alpha=\frac{a^2y^2+awy+\rho^2}{(ay+w)\rho},
\beta=\frac{(w^2+awy-\rho^2)a}{b(ay+w)\rho},
\gamma=\frac{by}{\rho}, \delta=\frac{w}{\rho} $$
for arbitrary integer $y,w$ and rational $\rho$.
If we allow $y,w$ to be rational, we can dispense with $\rho$:
$$ \alpha=\frac{a^2y^2+awy+1}{ay+w},
\beta=\frac{(w^2+awy-1)a}{b(ay+w)},
\gamma=by, \delta=w. $$
Now looking for **integer solutions of $\{E\_1,E\_2,E\_3\}$** we must satisfy $E\_1$.
Substituting (S) into $E\_1$ we get
$$ (bx-\delta)(ay+\delta)=-1. \tag{$E\_1b$} $$
So we have $bx-\delta=1,ay+\delta=-1$ or vice-versa.
Both options give $(bx-\delta)^2=1$, so $\delta=bx\pm 1$,
and also $ay+bx=0$.
Since $a,b$ are coprime, we can satisfy $ay+bx=0$ only with $x=az, y=-bz$.
In summary, the integer solutions of $\{E\_1,E\_2,E\_3\}$ are
\begin{align\*}
\alpha&=1-abz,\beta=a^2z,\gamma=-b^2z,\delta=1+abz \\
\alpha&=-1-abz,\beta=a^2z,\gamma=-b^2z,\delta=-1+abz.
\end{align\*}
for arbitrary integer $z$.
| 2 | https://mathoverflow.net/users/9025 | 430559 | 174,393 |
https://mathoverflow.net/questions/430543 | 6 | $\DeclareMathOperator\Int{Int}\DeclareMathOperator\Ext{Ext}$Suppose $E\subset\mathbb{R}^n$ is a set of finite perimeter and suppose that the measure theoretic boundary $\partial^\*E=\mathbb{R}^n\setminus(\Int(E)\cup \Ext(E))$ is closed, where $\Int(E)=\{x\in\mathbb{R}^n\,:\, \lim\_{r\to 0}\frac{\mathscr{L}^n(E\cap B\_r(x))}{\omega\_n r^n}=1\}$ and $\Ext(E)=\{x\in\mathbb{R}^n\,:\,\lim\_{r\to 0}\frac{\mathscr{L}^n(E\cap B\_r(x))}{\omega\_n r^n}=0\}$. Is it true that then there is a Borel set $\tilde{E}$ with $\mathscr{L}^n(\tilde{E}\mathbin\Delta E)=0$ and $\partial^\*E=\partial \tilde{E}$?
| https://mathoverflow.net/users/351083 | If the measure theoretic boundary is closed must it coincide with the topological boundary? | This is indeed true; you can find the answer to your question in Francesco Maggi's book for example, by combining Proposition 12.9 and Remark 15.3. I'll paraphrase the argument.
Let $E \subset \mathbf{R}^n$ be a Caccioppoli set, and $\mu\_E$ be its Gauss–Green measure. Then $\partial^\* E \subset \operatorname{spt} \mu\_E$, and as the support is closed, one has
\begin{equation}
\overline{\partial^\* E} = \operatorname{spt} \mu\_E.
\end{equation}
Now, said proposition gives the existence of a Borel set $\tilde{E}$ so that $\mathcal{H}^n(E \triangle \tilde{E}) = 0$ and $\partial \tilde{E} = \operatorname{spt} \mu\_{\tilde{E}}$. But then
\begin{equation}
\partial \tilde{E} = \operatorname{spt} \mu\_{\tilde{E}} = \operatorname{spt} \mu\_E = \overline{\partial^\* E} = \partial^\* E.
\end{equation}
| 4 | https://mathoverflow.net/users/103792 | 430577 | 174,399 |
https://mathoverflow.net/questions/336219 | 0 | I want to know what is following sum coefficient looks like. We sum over all integers $p$, $q$ also we put the condition that $q$ is even. Also, it should depend on the parity of $k$
$$\bar{S}(k)=\sum\_{p+q=k}[p]^{2m+1}q $$
The box symbol over $p$ denote that when p=0 it should be treated as $[0]=1$
I have seen the degree of the polynomial is 2m+3 in $k$. Can we claim that all the coefficient of the polynomial $S(k)$ is positive?
For example when $m=0$ and $k$ is even we have
$\bar{S}(k)=4 \binom{k/2+1}{3}+k$ hence all coefficient is positive.
Is it true in general all coefficient would be positive for $\bar{S}
(k)?$
What about the sum below where r is even depending on the parity is all the coefficient is non nengative or it has general closed formula in $k$?
$$\bar{R}(k)=\sum\_{p+q+r=k}[p]^{2m+1}[q]^{2m'+1}r.$$
| https://mathoverflow.net/users/45170 | If the coefficient of the polynomial positive | When $m=1$ and $k$ is odd, I compute that $\bar{S}(k) = (3k^5-20k^3+17k)/120$, which has a negative coefficient.
| 1 | https://mathoverflow.net/users/2807 | 430586 | 174,401 |
https://mathoverflow.net/questions/430547 | 3 | In [this paper](https://arxiv.org/abs/2003.10470), Danny Calegari shows that taut foliations in (let's say closed for simplicity) 3-manifolds are precisely those which admit a map $f: M \to S^2$ which restricts to a branched cover on each leaf. I'm trying to unpack what this means for the topology of $M$.
The TL;DR is my understanding is that this implies that $M$ admits a foliation by circles. But this implies $M$ is Seifert fibered, which is a contradiction because $M$ could be, say, hyperbolic.
I'm not sure where the following argument goes wrong. Calegari's argument implies that we can choose $f$ so that a given total transversal $\gamma$ of our foliation is precisely the set of critical points of $f$. Moreover, we can assume that these branch points are all order 2. Let's say that $\gamma = f^{-1}(0)$.
Away from $0$, $f$ is a submersion and so $\ker f$ is a 1-dimensional distribution. At a point in $\gamma$, our assumptions imply that there are coordinates where $f$ looks like
$$(x, y, z) \mapsto (x^2 - y^2, x^2 + y^2).$$ Thus we see that the distribution $D = \ker f$ extends over $\gamma$, by being generated in these coordinates by the vector field $\frac{\partial}{\partial z}$. So we have a global 1-dimensional distribution $D$, which is tangent to each of our fibers of $f$ and moreover tangent to $\gamma$. So it integrates to a foliation which is precisely this union of circles.
| https://mathoverflow.net/users/314845 | Do taut foliations leafwise branch covering S^2 yield foliations by circles? | To see what is going on, I think it is helpful to carefully work out an explicit example. In Section 2.1 of Calegari's paper, he explains how to deal with 3-manifolds that fiber over the circle. Here is a brief description of what he does.
I will start by setting up notation (and I won't follow Calegari here since I prefer to write it a little differently). Let $\Sigma\_g$ be a compact oriented genus $g$ surface and let $\psi\colon \Sigma\_g \rightarrow \Sigma\_g$ be an orientation-preserving diffeomorphism. Let $M\_{\psi}$ be the mapping torus of $\psi$, so $M\_{\psi} = \Sigma\_g \times [0,1] / \sim$ where $(p,1) \sim (\psi(x),0)$ for all $x \in \Sigma\_g$. The fibers of the map $M\_{\psi} \rightarrow S^1$ are all genus-$g$ surfaces, and form a taut foliation.
This points to one place where I think you slightly misunderstand what Calegari does. It's easy to find a transversal to the above taut foliation that hits each fiber exactly once. But you can't ensure that this is the entire branch locus and that the branching is simple since there might not be a branched cover $\Sigma\_g \rightarrow S^2$ with a single branch point of order $2$. For what he is doing, you'll have to take a more complicated transversal with several components.
Let $P \subset S^2$ be a set of $4g$ distinct points. We can then easily construct a map $\pi\colon \Sigma\_g \rightarrow S^2$ that is a degree $(g+1)$ branched cover, where the branch locus is exactly $P$ and each point of $p$ is a simple branch point (i.e., has degree $2$).
Now comes the key step in Calegari's argument. He argues that we can find an orientation-preserving diffeomorphism $\phi\colon S^2 \rightarrow S^2$ taking $P$ to $P$ such that $\psi\colon \Sigma\_g \rightarrow \Sigma\_g$ is a lift of $\phi$, i.e., such that
$$\pi(\psi(x)) = \phi(\pi(x)) \quad \quad \text{for all $x \in \Sigma\_g$}.$$
Letting $M\_{\phi}$ be the mapping torus of $\phi$, we thus get a map
$F\colon M\_{\psi} \rightarrow M\_{\phi}$ that is a fiberwise branched cover.
Calegari stops the argument here, and omits the final step of why this gives a map $f\colon M\_{\psi} \rightarrow S^2$ that is a branched cover on each fiber. Here is what I think he has in mind. The mapping class group of $S^2$ is trivial, so the map $\phi\colon S^2 \rightarrow S^2$ is isotopic to the identity. Of course, this isotopy moves the points in $P$. This implies that there is a fiber-preserving diffeomorphism $\lambda\colon M\_{\phi} \rightarrow M\_{\text{id}} = S^2 \times S^1$. The map $f\colon M\_{\psi} \rightarrow S^2$ is then
$$M\_{\psi} \stackrel{F}{\longrightarrow} M\_{\phi} \stackrel{\lambda}{\longrightarrow} S^2 \times S^1 \stackrel{\text{proj}}{\longrightarrow} S^2.$$
One feature of this is that it is not the case that the branch points in the fibers all map to the same points in $S^2$; indeed, as you go around the circle of fibers the images of branch points in $S^2$ trace out a braid. I think it would hard to avoid this.
Since the composition $\lambda \circ F$ is fiber-preserving, your $1$-dimensional foliation I think will be exactly the standard foliation of the mapping torus $M\_{\psi}$ whose leaves are unions of the images of sets $x \times [0,1]$ in $M\_{\psi}$. If $x$ is a periodic point of $\psi$, then this leaf will be a closed circle, but for non-periodic points it will be $\mathbb{R}$.
| 3 | https://mathoverflow.net/users/317 | 430587 | 174,402 |
https://mathoverflow.net/questions/429842 | 13 | The physicist [Yoichiro Nambu](https://en.wikipedia.org/wiki/Yoichiro_Nambu) introduced in a 1950 paper [A Note on the Eigenvalue Problem in Crystal Statistics](http://www.totoha.net/archiv/3675.pdf) the notion of an "*eigenoperator*" (page 12, see [Nambu and the Ising model](https://arxiv.org/abs/2209.01122) for a recent discussion of this early work by the 2008 Nobel laureate).
Given a self-adjoint operator $H$, the eigenoperator $X$ satisfies
$$HX-XH=\lambda X,\;\;\lambda\in\mathbb{R}.$$
**Q:** A Google search for "eigenoperator" does not return much, has this notion found its place in the mathematical literature, perhaps under a different name?
---
I append a screenshot of the relevant paragraph from Nambu's paper:
Footnote 6) refers to Nambu's paper [On the Method of the Third Quantization](https://www.semanticscholar.org/paper/On-the-Method-of-the-Third-Quantization.-Nambu/6f3f8b38cf6c6963627cfc883afc580d347b4e95), where $X$ is called an "eigenmatrix".
| https://mathoverflow.net/users/11260 | Has Nambu's notion of an "eigenoperator" found a place in the mathematical literature? | Such an $X$ is an *eigenvector of $\,\operatorname{ad}(H)$*. Joint eigenspace decompositions of several $\operatorname{ad}(H\_i)$ are commonplace in math since the work of Lie, Killing, Cartan, with the joint eigenvectors called *root vectors*. So I would say that the notion “had a place” already before Nambu.
| 10 | https://mathoverflow.net/users/19276 | 430589 | 174,403 |
https://mathoverflow.net/questions/430596 | 9 | My first idea on how to do this would be:
1. $0\in\mathbf Z$
2. $\forall x\in\mathbf Z,Sx\in\mathbf Z\land Px\in\mathbf Z$
3. $P$ and $S$ are injective
4. $\forall x\in\mathbf Z,PSx=SPx=x$
5. some induction axiom that holds for decreasing sequences as well (intuitively, if $\phi$ holds for $x\_0\in\mathbf Z$ and $\forall x\in\mathbf Z,\phi(x)\to\phi(Px)$ then $\phi$ holds for all numbers less than $x\_0$)
6. addition (with the two standard axioms $\forall x\in\mathbf Z,x+0=x$ and $\forall x,y\in\mathbf Z,Sx+y=S(x+y)$ and an extra axiom $\forall x,y\in\mathbf Z,Px+y=P(x+y)$)
7. multiplication
defining $-1=P0$, $-2=PP0$, etc.
Is this the canonical way to do it (if so, what is it called) or is there a problem I haven't noticed (if so, what would be the "right way")?
| https://mathoverflow.net/users/nan | What is the canonical way to extend Peano's axioms to the set of all integers? | The canonical way to extend Peano arithmetic to the integers is not by changing the language or axioms, but instead by treating integers as equivalence classes of pairs of natural numbers.
We think of an integer $x$ as any pair $(a,b)$ of natural numbers with $a-b=x$. More formally, we translate arithmetic statements about integers by first replacing the integers with pairs of natural numbers, and then transforming the results by
\begin{align}
0\_Z& \to (0,0)\\
S(a,b)& \to (Sa,b)\\
(a,b)+(c,d)& \to (a+c,b+d)\\
(a,b)(c,d)& \to (ac+bd,ad+bc)\\
(a,b)=(c,d)& \to a+d=b+c
\end{align}
This process turns every arithmetic statement about integers into an arithmetic statement about natural numbers instead. Eg $$(\exists x,y,z\in\mathbb{Z})x^3+y^3+z^3=30$$
can be translated idiomatically (i.e. after performing the mechanical translation and then some of the usual simplifications) as
\begin{align}(\exists a,b,c,d,e,f)& a^3+3ab^2+c^3+3cd^2+e^3+3ef^2=\\
&b^3+3ba^2+d^3+3dc^2+f^3+3fe^2+30\end{align}
| 21 | https://mathoverflow.net/users/nan | 430600 | 174,407 |
https://mathoverflow.net/questions/430606 | 3 | Let $\Gamma=\left<\mathcal{S}\,|\,\mathcal{R}\right>$ be a group defined by the presentation, where each relator $r\in\mathcal{R}$ is a reduced word of length 3 consisting of three different symbols only in $\mathcal{S}$, not in $\mathcal{S}^{-1}$.
Take a proper subset $\mathcal{S'}$ of $\mathcal{S}$, and let $\Gamma'$ be the subgroup of $\Gamma$ generated by $\mathcal{S'}$. Define $\mathcal{R'}$ be a set consisting of all of the relators in $\mathcal{R}$ whose three symbols are in $\mathcal{S'}$. Let us formalize my question:
(Question) Does the natural surjection (generator to generator) $\left<\mathcal{S'}\,|\,\mathcal{R'}\right> \twoheadrightarrow \Gamma'$ have trivial kernel?
Surely, there are easy counterexamples. For $G=\left<a,b,c\,|\,abc,bac\right>$, take a subset $\{a,b\}$ of the generating set. In this case, we see $\mathcal{R'}$ is empty. To avoid these trivial cases, we need the following property:
(C1) If $c\in\mathcal{S}$ and $c\in\left<\mathcal{S'}\right>$($c$ is in the subgroup of $\Gamma$ generated by $\mathcal{S'}$), then $c\in\mathcal{S'}$.
Now, after choosing $\mathcal{S'}$ satisfying (C1), can we answer (Question) positively? I think there are counterexamples, but it seems a bit complicated to me. Could you recommend some references on these groups?
Revision: the answer to the original question above is negative, as Roland Bacher neatly demonstrated. If we refine the question, adding the condition $\Gamma$ satisfies (C2) below, would the question be still meaningful?
(C2) For any two different symbols $a,b\in\mathcal{S}$, $a,a^{-1},b,b^{-1}$ are four different elements in $\Gamma$ as group elements.
| https://mathoverflow.net/users/490332 | Subgroup of a group with length 3 relators | Here is an explicit example showing that assumptions (C1) and (C2) are not sufficient for your question. Let
$$G = \langle a,b,c,d \mid abc, bdc, dac, adb \rangle.$$
A quick computer calculation shows that $|G|=24$ and in fact $G \cong {\rm SL}(2,3)$ with generator images
$$ \left( \begin{array}{cc}1&2\\1&0\end{array} \right),
\left( \begin{array}{cc}0&2\\1&1\end{array} \right),
\left( \begin{array}{cc}2&2\\0&2\end{array} \right),
\left( \begin{array}{cc}2&0\\1&2\end{array} \right).$$
So all of the generators represent distinct elements of $G$ (in fact they all have order 6) , and there are no relators involving fewer than three generators.
| 5 | https://mathoverflow.net/users/35840 | 430621 | 174,411 |
https://mathoverflow.net/questions/430591 | 1 | Assume you have $n$ independent binary variables $\{x\_1,\dots,x\_n\}$ and for each variable $x\_i$ you know that its value is equal to $1$ with a probability $p\_i$. I would like to enumerate the joint assignments of these variables in such a way that in $k$ steps I capture as much volume of the joint probability mass function as possible. Does there exist some approximation algorithm which runs faster than the algorithm which enumerates the assignments according to their joint probability $\Pi p\_i$. Or something related?
| https://mathoverflow.net/users/491356 | Choosing $k$ different assignments of binary variables in order to capture the largest volume of the joint probability distribution | You want to find a $k$-subset $S\subseteq\{0,1\}^n$ to maximize
$$\sum\_{(y\_1,\dots,y\_n)\in S} \prod\_{i=1}^n \left(p\_{i}y\_i+(1-p\_i)(1-y\_i)\right).$$
Equivalently, find the $k$ largest (with multiplicity)
$$\prod\_{i=1}^n \left(p\_{i}y\_i+(1-p\_i)(1-y\_i)\right).$$
Equivalently, find the $k$ largest (with multiplicity)
$$\sum\_{i=1}^n \left((\log p\_{i})y\_i+(\log(1-p\_i))(1-y\_i)\right).$$
Some integer linear programming (ILP) solvers have an option to find the $k$ best solutions. An alternative approach to get the $k$ best solutions is to call an ILP solver $k$ times. After each solve, yielding optimal solution $\hat{y}\in\{0,1\}^n$, impose an additional no-good cut
$$\sum\_{i:\hat{y}\_i = 0} y\_i + \sum\_{i:\hat{y}\_i = 1} (1 - y\_i) \ge 1$$
that excludes only that solution.
| 0 | https://mathoverflow.net/users/141766 | 430651 | 174,421 |
https://mathoverflow.net/questions/430646 | 2 | I'd be applying for a Ph.D. at various grad schools in the U.S. in the coming months and while I know I'd like to pursue research in the field of Algebraic Topology, I am not knowledgeable enough yet to figure out the exact subfield that would suit me best. I would like to know the best and quickest way to get a brief overview of the major active research areas in Topology (especially Algebraic Topology) so that I can start reading up in the areas that interest me and get in touch with the relevant professors in that area, while also meeting the application submission deadlines.
I've taken an introductory course in Homotopy Theory and Fundamental Groups and another in Simplicial Homology Theory in my Masters, besides a basic and advanced courses in General Topology. I thoroughly enjoyed my General Topology courses, especially the problems on compactness, connectedness, the separation axioms etc. I also liked the concept of Homotopy more than simplicial Homology, mostly because the construction of the simplicial complex seemed too geometric in nature. The book we used was mostly this: [Introduction to Topology by Tej Bahadur Singh](https://rads.stackoverflow.com/amzn/click/com/B07ZS1D3H8) and [Topology by Munkres](https://rads.stackoverflow.com/amzn/click/com/0131816292).
However, while browsing through the profiles of professors, I see their areas of interests mentioned as symplectic topology, stable homotopy theory, Floer homology etc. most of which I am unfamiliar with and would like to know which amongst these would most align with my interests. Also, almost nobody seems to mention General Topology as their broad area of research which makes me wonder if it is not an active area of research and that Algebraic Topology is the natural progression that everyone moves on to.
While consulting my past professors would be ideal for this and I am trying that as well, they seem simply too busy for elaborate discussions.
I've gone through the suggestions [here](https://math.stackexchange.com/q/158171/116321) but the plan laid out there is too long for my situation.
| https://mathoverflow.net/users/124001 | Most efficient way of getting a brief overview of the current active research areas in Algebraic Topology | Algebraic topology is a research area where often even the statements (and certainly the proofs) of active research areas require a lot of background to understand. For that reason, the answer to the linked stack exchange question from 2012 is really spot on. The best way to decide which area you might want to research is to read graduate textbooks in that area (or talk to experts, or go to seminars, but it sounds like this is more difficult for the OP) and see if you like the types of objects being studied, enjoy the nature of the proofs, etc.
For example, you said you prefer homotopy over more geometric arguments. If you start reading a book in symplectic topology or knot theory, you might find you don't like it because it's too geometric. Others might start reading graduate textbooks (or survey papers or research papers, as Tyler Lawson recommended in the linked thread) in homotopy theory and not like when it starts to draw tools from category theory. This kind of trial and error is the best way to settle on a research discipline where you actually enjoy working.
That said, it's also wise early in your career to think a bit about which areas are "hot." As someone who loved learning General Topology from the Munkres book, I was sad to learn that point-set topology is not anywhere near as active now as it was 50 years ago. There are very few departments doing point-set topology research, and it's difficult to get a job if your expertise is point-set topology. Careerwise, it would be better to be in a "hotter" area. I think there are several such areas that might be of interest based on what you've written about your background so far, including homotopy theory, knot theory, symplectic geometry, topological data analysis, differential topology, differential geometry/analysis, etc.
In US PhD programs, you are not generally expected to know what you will research before you show up. Your first year is usually spent taking graduate courses to get a firm foundation across pure mathematics (think: algebra, analysis, and topology) and then pass your qualifying exams. After that, you take more advanced courses and reading courses with potential advisors, where you get closer to the edge of human knowledge. In those courses, you learn the necessary background to understand modern proofs and the statements of interesting open problems. There's a lot of trial and error. In the first two years of my PhD program, I considered going into research in algebraic geometry, knot theory, ergodic theory, and homotopy theory, eventually settling on the latter because I enjoyed proofs and conferences in that area the most. Breadth is a good thing, because it makes you a better mathematician, expands what you can teach, and you never know when you'll find a surprising connection (see, e.g., recent work of Blumberg and Abouzaid on "homotopy Floer homology").
So, even though you are looking for the "most efficient" way, I think the best way is to peruse the books recommended in the link you already had. Even just reading the table of contents, preface, and first chapter of each of those would give you a sense of which you liked and which you didn't. It would help you a bit to have a rough idea of which type of topology you find most interesting so far, so that you can apply to graduate programs with that specialty, but don't expect to learn in the next few months what thing you'll be researching for your whole career. Expect that you might change your mind once you're already in grad school or even later: lots of people have switched into topological data analysis from other things, for example, even years after their PhD.
Most of all, try to pick a PhD program where you think you have the best chance of being successful and happy. When you look at a program, in addition to asking yourself if they have research in the subspecialty of topology that you think you want to study, also ask: do they nurture their students? Do lots of people drop out? Are students spending huge amounts of time teaching and getting no attention from faculty? Are their students getting the types of jobs afterwards that you currently think you might want? Is it in a place you can see yourself living? Etc. Think about the culture. To do good work, it helps to be in a good working environment, and leading a happy life as much as possible.
| 3 | https://mathoverflow.net/users/11540 | 430659 | 174,424 |
https://mathoverflow.net/questions/430552 | 0 | [Legendre's formula](https://mathworld.wolfram.com/LegendresFormula.html) can be very easily be generalised as mentioned [here](https://projecteuler.net/action=redirect;post_id=111677) (visible after login) which is like this
${\pi}(v,p)={\pi}(v,p-1)-1.[{\pi}(v/p,p-1)-{\pi}(p-1,p-1)]$
${ \big\downarrow}$
$S(v,p)=S(v,p-1)-p[S(v/p,p-1)-S(p-1,p-1)]$
This is still $O(n^{3/4})$ algorithm
I have been trying to achieve the same with [Lehmer's formula](https://mathworld.wolfram.com/LehmersFormula.html)
$\pi(n)=φ(n, a)+\frac{1}{2}(b+a-2)(b-a+1)-\sum\_{i=a}^{b}{\pi}(\frac{x}{p\_i})-\sum\_{i=a+1}^c \sum\_{j=i}^{b\_{i}}[{\pi}(\frac{x}{p\_ip\_j})-(j-1)]$
where $b = n^{1/2}, c = \pi(n^{1/3}), a = \pi(n^{1/4}), φ(n,a)-$ **number of integers** in [1;n] such that they are not divisible by any prime among first $a$ primes.
${ \big\downarrow}$
$S(n)=\Phi(n, a)+\frac{1}{2}(\sum b+\sum a-2)(\sum b-\sum a+1)-\sum\_{i=a}^{b}p\_i{S}(\frac{x}{p\_i})-\sum\_{i=a+1}^c \sum\_{j=i}^{b\_{i}}[p\_ip\_j{S}(\frac{x}{p\_ip\_j})-\sum(j-1)]$
The above expression works fine till $n=8$; after that it doesn't.
I am not exactly sure what to do with $\frac{1}{2}(b+a-2)(b-a+1)$ It should not be left as is.
Adding $\sum$ may not be the right way to go.
Then I realised
$\frac{1}{2}(b+a-2)(b-a+1)=a+(a+1)+(a+2)+.......for (b-a) terms$
My required sum should be $\sum a+\sum (a+1)+\sum (a+2)+.......for (b-a) terms$ but even that gives an expression $(b-a)[(a^2+ab+b^2)-1]/6$ which is also giving the correct answer for $n \le 8$.
I have successfully transitioned to from $φ(n, a)$ to $\Phi(n, a)$ where $\Phi(n, a)$ - **sum of integers** in [1;n] such that they are not divisible by any prime among the first $a$ primes.
I guess the equivalent expression of $\frac{1}{2}(b+a-2)(b-a+1)$ should be $s'(a)+s'(a+1)+s'(a+2)+.......for (b-a)$
note: $s'(x)$ is the sum of first $x$ primes; $S(x)$ is the sum of all primes up to $x$.
PS: I am not looking for any suggestion of alternate $O(n^{2/3})$ or similar algos.
For that I have already gone through [Fastest Algorithm to Compute the Sum of Primes?](https://mathoverflow.net/questions/81443/fastest-algorithm-to-compute-the-sum-of-primes) and more similar literature.
As a part of this question I am only looking to generalise the Meissel-Lehmer.
| https://mathoverflow.net/users/483720 | How can I convert Meissel's/Lehmer's formula for prime counting to get sum of primes | This might be a "shameless plug" but i did recently "generalize Meissel-Lehmer to count sum of the powers of primes". This was supposed to give an exposition, so it might be of help
<https://arxiv.org/abs/2111.15545>
| 2 | https://mathoverflow.net/users/483197 | 430663 | 174,426 |
https://mathoverflow.net/questions/430665 | 4 | Let $G$ be an almost $k$-simple group that is also simply connected (so that $G(k)^{+}=G(k)$). For opposite parabolic subgroups $P$ and $P^{-}$, it is known that $G(k)^{+}$ is generated by the unipotent radicals $R\_u(P)(k)$ and $R\_u(P^{-})(k)$ (Prop 1.5.4 in Margulis's book "Discrete Subgroups of Semisimple Lie groups).
Can this generation be extended to bounded generation? That is, is $G(k)=G(k)^{+}$ boundedly generated by $R\_u(P)(k)$ and $R\_u(P^{-})(k)$?
| https://mathoverflow.net/users/15311 | Bounded generation of group by unipotent radicals of opposite parabolic subgroups | I hope you will permit me to write $P^+$ in place of $P$. Put $U^\pm = R\_u(P^\pm)$.
Yes, at least in the split case.
Suppose first that $P^+$ and $P^-$ are minimal. Put $T = P \cap P^-$. By working in $\operatorname{SL}\_2(k)$ or $\operatorname{PGL}\_2(k)$, you see that, for all $t \in k^\times$ and all roots $\alpha$ of $T$ in $G$, you can write $\alpha^\vee(t)$ as a product of $6$ elements of $U^+(k)$ and $U^-(k)$. This covers $T(k)$, using $6r$ elements, where $r$ is the semisimple rank of the group. Once you have written, for each element $w$ of the (finite!) Weyl group of $T$ in $G$, a representative of $w$ as a product of elements of $U^+(k)$ and $U^-(k)$ (which, again by a rank-$1$ computation, can be done using at most $3\ell$ elements, where $\ell$ is the length of a minimal expression for $w$ as a product of reflections), you only need $2$ more elements to generate the corresponding Bruhat cell.
Now continue to suppose that $G$ is split, but drop the assumption that $P^+$ and $P^-$ are minimal. Put $M = P^+ \cap P^-$, and let $T$ be a split maximal torus in $M$.
Let $\alpha$ be a root of $T$ in $M$. Since we (should) have assumed that $P^+$ does not contain any isotropic factor of $G$, there is some root $\beta$ of $T$ in $U^+$ such that $\alpha + \beta$ is also a root of $T$ in $U^+$. Then, for a suitable Chevalley–Steinberg system $(u\_r : \operatorname{Add} \to U\_r)\_{\text{$r$ a root}}$, we have for all $t \in k$ that $[u\_{-\beta}(t), u\_{\alpha + \beta}(1)]$ lies in $u\_\alpha(t)U^-(k)$. That is, each element of $U\_\alpha(k)$ is a product of $4$ elements of $U\_{-\beta}(k) \subseteq U^-(k)$ and $U\_{\alpha + \beta}(k) \subseteq U^+(k)$ with an element of $U^-(k)$.
This shows that the group of $k$-rational points of every root subgroup of $M$ is boundedly generated by $U^+(k)$ and $U^-(k)$; so, if $B\_M^\pm$ are opposite Borel subgroups of $M$ containing $T$, then $R\_u(B\_M^\pm)(k)$ are boundedly generated by $U^+(k)$ and $U^-(k)$; so $R\_u(B\_M^\pm)(k)U^\pm(k) = R\_u(B\_M^\pm\cdot U^\pm)(k)$ is boundedly generated by $U^+(k)$ and $U^-(k)$. Since $B\_M^+\cdot U^+$ and $B\_M^-\cdot U^-$ are opposite Borel subgroups of $G$, we have reduced to the previous case.
| 6 | https://mathoverflow.net/users/2383 | 430666 | 174,428 |
https://mathoverflow.net/questions/428326 | 4 | I believe this should be a well known result, but I wasn’t able to prove or find a good reference for it.
Let $E$ and $F$ be $n$-regular, respectively $m$-regular vector bundles in the sense of Castelnuovo-Mumford, the is it true that $E\otimes F$ is at most $n+m$-regular?
The result is true for modules, let’s say $M$ and $N$ provided that $Tor^1(M,N)$ is zero, due to a result by Caviglia. This is the case here locally, so I would like to claim it for the vector bundle case, but I am not sure if this holds, because of the difference in the definition of the regularity for modules and vector bundles.
Any help or reference will be appreciated.
| https://mathoverflow.net/users/43027 | Castelnuovo-Mumford regularity for tensor products of vector bundles | Yes, furthermore, your statement holds even if $E$ is locally free and $F$ is coherent.
You can apply the following fact to prove it:
>
> Suppose a coherent sheaf $\mathscr F$ on $\mathbf P$ is resolved by a
> long exact sequence $$\cdots \rightarrow \mathscr F\_2\rightarrow
> \mathscr F\_1\rightarrow \mathscr F\_0\rightarrow \mathscr F\rightarrow
> 0$$ of coherent sheaves on $\mathbf P$. If $\mathscr F\_i$ is
> $(m+i)-$regular for every $i\geq0$, then $\mathscr F$ is $m-$regular.
>
>
>
You can prove this fact by chasing through the long exact sequence.
| 2 | https://mathoverflow.net/users/491416 | 430671 | 174,430 |
https://mathoverflow.net/questions/430627 | 5 | I have been looking into the Fermat quintic equation $a^5+b^5+c^5+d^5=0$. To exclude the trivial cases (e.g. $c=-a,d=-b$), I will take $a+b+c+d$ to be nonzero for the rest of the question. It can be shown that if $a+b+c+d=0$, then the only solutions in the rational numbers (or even the real numbers) are trivial.
I had the idea of representing the values $a, b, c, d$ as roots of a quartic polynomial, and then trying to force the Galois group to have low order. As the Fermat quintic is symmetric, it yields an equation in the coefficients of the quartic, which are satisfied by the quartic equations: $5x^4-5x^3+5qx^2-5rx+5(q-1)(q-r)+1=0$, so clearly the most general case of Galois group $S\_4$ is attainable. One could also fix one or two of the roots, obtaining a smaller Galois group.
However, what I have yet to discover, is a quartic in the family above which has a square discriminant, forcing the Galois group to be a subgroup of the alternating group $A\_4$. Is there an example of such a quartic? Is there one with Galois group a proper subgroup of $A\_4$? Better yet, are there infinite families of such quartics?
| https://mathoverflow.net/users/38744 | Small Galois group solution to Fermat quintic | I have answers to your first two questions, and some insight into the third.
First, there is a quartic in the family above with Galois group contained in $A\_{4}$. One example is found by taking $q = 1/5$ and $r = 0$. This gives $5x^{4} - 5x^{3} + x^{2} + 1/5$, which has Galois group $A\_{4}$. A quartic has Galois group contained in $A\_{4}$ if and only if the discriminant $D(q,r)$ is a square, and the discriminant of your quartic family is degree $4$ in $r$. This means that if you find a single $A\_{4}$ polynomial in the family, this gives a rational point on $y^{2} = D(q,r)$. This is a genus $1$ curve, and there's a good chance that this gives you an elliptic curve with rank $\geq 1$ and hence infinitely many rational $r$ with the same $q$. In particular, for $q = 1/5$, the elliptic curve is $y^{2} + xy = x^{3} + x^{2} - 12642x - 612239$ and has rank $2$.
Second, there are quartics in your family with Galois group a proper subgroup of $A\_{4}$, namely $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. I know of three examples: $(q,r) = (49/120,1/12)$, $(791/1560,17/120)$, and $(1949/3680,189/736)$. In the first case, the splitting field is $\mathbb{Q}(\sqrt{-3},\sqrt{10})$. In the second case, the splitting field is $\mathbb{Q}(\sqrt{10}, \sqrt{-1599})$. In the third case, the splitting field is $\mathbb{Q}(\sqrt{-230},\sqrt{410})$.
Third, I do not know if there are infinite families of such quartics. I focused on the case that the Galois group is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. This occurs when the resolvent cubic has a rational root, and when the discriminant is a square. The resolvent cubic in your family is a cubic curve in $x$, $q$ and $r$ and a projectivization of this polynomial defines a cubic surface in $\mathbb{P}^{3}$. This surface is rational (and so there is a two parameter subfamily of your family where the Galois group is contained in the dihedral group of order $8$). Imposing the restriction that the discriminant is a square leads to a surface of the form $X : y^{2} = f(a,b,c)$, where $f$ is a homogeneous degree $8$ polynomial. There are a couple of rational curves on $X$ that arise from the points at infinity on the projectivization of the resolvent cubic, which do not correspond to pairs $(q,r)$. I have not found any other curves of genus $0$ or $1$ on $X$. A computation with Magma indicates that $X$ is a K3 surface, and it is conjectured that there is a number field $K$ for which $X(K)$ is Zariski dense. It is plausible that $X$ has many rational points.
| 2 | https://mathoverflow.net/users/48142 | 430677 | 174,433 |
https://mathoverflow.net/questions/430616 | 3 | Let $A$ be a set of $n$ elements. Let $S\_1,\dots,S\_n$ be independent $k$-element random subsets of $A$. What is the probability that $S\_1,\dots, S\_n$ evenly cover $A$, i.e. each element of $A$ belongs to exactly $k$ random subsets?
| https://mathoverflow.net/users/97209 | Random covering of a set | It follows immediately from the special case (with $m=n$ and $s=t=k$) of Theorem 1 of [Canfield and Mckay](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v12i1r29) that the probability -- say $p\_{n,k}$ -- in question is
$$\sim\frac1{\sqrt e}\dfrac{\displaystyle{\binom{n}{k}^n}}{\displaystyle{\binom{n^2}{k n}}} \tag{1}\label{1}$$
if $n\to\infty$ and either $k=o(n^{1/2})$ or $k\asymp n-k\asymp n$. (Indeed, writing $M\_{ij}:=1(j\in S\_i)$ for all $i$ and $j$ in $[n]:=\{1,\dots,n\}$, we have a bijection between the set of all "even" $k$-coverings $(S\_1,\dots,S\_n)$ of the set $[n]$ and the set of all $0$-$1$ "incidence" matrices $M=(M\_{ij}\colon(i,j)\in[n]^2)$ with all row sums and all column sums equal $k$.)
According to Conjecture 1 of [Canfield and Mckay](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v12i1r29), \eqref{1} holds uniformly over all $k\in\{1,\dots,n-1\}$ as $n\to\infty$.
A prehistory of this problem is discussed in [Canfield and Mckay](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v12i1r29) as well.
---
For $k=2,3,4$,
$$p\_{n,k}=C\_{n,k}\Big/\binom nk^n,$$
where $(C\_{n,2})$, $(C\_{n,3})$, and $(C\_{n,3})$ are [A001499](https://oeis.org/A001499), [A001501](https://oeis.org/A001501), and [A058528](https://oeis.org/A058528), respectively.
| 4 | https://mathoverflow.net/users/36721 | 430683 | 174,436 |
https://mathoverflow.net/questions/430631 | 3 | Let $W$ be a standard Brownian motion, and $\mathcal F\_t$ it’s natural filtration. Let $H$ be a continuous process, adapted to $\mathcal F\_t$ and integrable with respect to $W$.
**Question:** Is it true that for all a.s. finite $\mathcal F\_t$-stopping times $\sigma$ we have
$$\lim\_{h \to 0+} \frac{1}{W\_{\sigma + h} - W\_{(\sigma - h) \vee 0}} \int\_{(\sigma - h)\vee 0}^{\sigma + h} H\_s \, dW\_s = H\_\sigma$$
in probability?
*Note: By convention we set $\frac{1}{0} = \infty.$*
| https://mathoverflow.net/users/173490 | Lebesgue differentiation theorem at a stopping time | Yes. Let $h \in (0,1)$, $Q\_h = \frac{1}{W\_{\sigma + h} - W\_{(\sigma - h) \vee 0}} \int\_{(\sigma - h)\vee 0}^{\sigma + h} (H\_s - H\_{(\sigma - h) \vee 0}) \, dW\_s$ and set $M\_h = \sqrt{-\log(h)}$. Because $H$ is continuous, and hence, $\lim\_{h \to 0} H\_{(\sigma - h) \vee 0} = H\_{\sigma} $ almost surely, the claim holds if we can prove that $Q\_h$ converges to zero in probability.
For arbitrary $\epsilon>0$ \begin{align\*}
& P( |Q\_h| > \epsilon) = A\_h + B\_h \quad \text{where} \\
& A\_h := P( |Q\_h| > \epsilon , |W\_{\sigma + h} - W\_{(\sigma - h) \vee 0} | > M\_h \sqrt{\sigma + h - (\sigma - h) \vee 0 } )\\
& B\_h := P( |Q\_h| > \epsilon , |W\_{\sigma + h} - W\_{(\sigma - h) \vee 0} | < M\_h \sqrt{\sigma + h - (\sigma - h) \vee 0} )
\end{align\*}
By the strong Markov property of BM (given $\mathcal{F}\_{(\sigma-h) \vee 0}$), $$
B\_h \le P(|W\_{\sigma + h} - W\_{(\sigma - h) \vee 0} | > M\_h \sqrt{\sigma + h - (\sigma - h) \vee 0} ) = \mathcal{N}(0,1)[M\_h, \infty] \le e^{-M\_h^2 } \;.
$$
By Markov's inequality, the strong Markov property of BM (given $\mathcal{F}\_{(\sigma-h) \vee 0}$), and Itô isometry, \begin{align\*}
& A\_h \le P\left(\left| \frac{1}{\sqrt{\sigma + h - (\sigma - h) \vee 0}} \int\_{(\sigma - h)\vee 0}^{\sigma + h} (H\_s - H\_{(\sigma - h) \vee 0}) \, dW\_s \right| > M\_h \epsilon \right) \\
& \le E\left(\left| \frac{1}{\sqrt{\sigma + h - (\sigma - h) \vee 0}} \int\_{(\sigma - h)\vee 0}^{\sigma + h} (H\_s - H\_{(\sigma - h) \vee 0}) \, dW\_s \right|^2 \right) M\_h^{-2} \epsilon^{-2} \\
& \le E\left( \frac{1}{\sigma + h - (\sigma - h) \vee 0} \int\_{(\sigma - h)\vee 0}^{\sigma + h} (H\_s - H\_{(\sigma - h) \vee 0})^2 ds \right) M\_h^{-2} \epsilon^{-2}
\end{align\*}
By continuity of $H$, we see that $A\_h\searrow 0$ and $B\_h \searrow 0$, and in turn, $P(|Q\_h| > \epsilon) \searrow 0$ as $h \searrow 0$.
Thus, $Q\_h$ converges to zero in probability, as required.
| 4 | https://mathoverflow.net/users/64449 | 430684 | 174,437 |
https://mathoverflow.net/questions/430571 | 3 | **Motivation/Hand-Wavy Question:**
In [this post](https://mathoverflow.net/questions/335736/low-degree-polynomial-approximation-of-the-piecewise-linear-function-x-mapsto), it was asked what the best local approximation of $f(x):=\max\{0,x\}$ is by a *polynomial* of a given degree; with the answer provided by Chebyshev's [Equioscillation theorem](https://en.wikipedia.org/wiki/Equioscillation_theorem). Instead, I'm looking for a smooth (not necessarily analytic) global approximation of $f$ (which should be possible since I can now leverage smooth bump functions) such that $f$'s Lipschitz constant is also approximated by the smooth function's $C^k$-norms:
$$
\|f\|\_k\,:=\,\max\_{0\le j \le k} \sup\_{x\in\mathbb{R}} \left |\partial^{j} f (x) \right |.
$$
---
**Rigerous Question:**
For every $\delta\_1,\, \delta\_2$ and each $k\in \mathbb{N}$, I'm looking for a smooth (not necessarily analytic) approximation $f\_{k,\delta}\in C^{k,1}(\mathbb{R})$ satisfying
* $\sup\_{x\in \mathbb{R}}\, \big|f(x)-f\_{k,\delta}(x)\big|<\delta\_1$,
* $\big|\|f\_{k,\delta}\|\_{C^{k}} - \operatorname{Lip}(f)\big| = \big|\|f\_{k,\delta}\|\_{C^{k}} - 1\big| <\delta\_2$.
Is such an approximation possible? Atleast can I get $\delta\_1$ arbitrarily small even if I cannot get $\delta\_2$ to approach 1?
---
If no, what what is the largest $k$ for which this is possible. If yes, can we also require that $f\_{k,\delta}$ belong to the Sobolev spaces $W^{\tilde{k},2}(\mathbb{R})$ for appropriately large $\tilde{k}$ (given by the Sobolev embedding theorem)?
| https://mathoverflow.net/users/36886 | Smooth approximation of the $\max\{0,x\}$ function with controlled derivatives | Let us consider $[-1,1]$ instead.
**Lemma 1**. For any $\epsilon>0$, there exists $f\_\epsilon\in C^{\infty}\left(\left[0,1\right]\right)$,
a bijection from $\left[0,1\right]$ onto itself, such that
\begin{align\*}
f\_\epsilon(0) & =0,\\
f\_\epsilon (1) & =1,\\
\forall k\in\mathbb{N},k\geq1,f\_\epsilon^{(k)}\left(0\right)=f\_\epsilon^{(k)}\left(1\right) & =0,\\
\sup\_{\left[0,1\right]}\left|f\_\epsilon^{\prime}\left(x\right)\right| & <1+\epsilon.
\end{align\*}
A constructive proof is [here](https://math.stackexchange.com/questions/1736090/cut-off-function-construction/4365567#4365567). With this lemma, you see that for any $\epsilon$, define $f\_{1,\delta} = \begin{cases} 0 &\textrm{if } x<0 \\ f\_\delta &\text{otherwise } \end{cases}$ then $f\_{1,\delta}\in C^\infty[-1,1]$ and fits the bill for $k=1$ (and $\delta\_1=\delta\_2=\delta$ wlog).
For $k\geq2$ that isn't possible. Pietro Majer in the comments to your question has given a nice counter example, and Giorgio Metafune has explained why this cannot be for the same reason even if you only ask the second derivative to be bounded by an arbitrary constant ($\delta\_1\to0$, $\delta\_2=10$ for example). By Arzela-Ascoli, the sequence of first derivative would be equicontinuous and (up to extraction of a subsequence) converge uniformly in $C^1$ to $x^+$ which is not in $C^1$.
| 2 | https://mathoverflow.net/users/40120 | 430694 | 174,442 |
https://mathoverflow.net/questions/430674 | 4 | $\DeclareMathOperator\RoT{RoT}$I'm interested in the following ring. Fix a (Noetherian?) base ring $R$, and consider the category of finitely generated projective $R$-modules equipped with endomorphisms up to isomorphism. Since $\otimes$ distributes over $\oplus$, you can view this as a commutative semiring with those two operations. Then you can quotient by the equivalence that $(V,f) \sim (W,g)$ if there exist $(V^\prime,f^\prime)$ and $(W^\prime,g^\prime)$ such that $(V,f) \oplus (V^\prime,f^\prime) \cong (W,g) \oplus (W^\prime,g^\prime)$ and $f^\prime$ and $g^\prime$ are both commutators $f^\prime = AB - BA$ and $g^\prime = XY - YX$. We get a ring, which I've been calling the "ring of traces" over $R$ or $\RoT(R)$.
When all projective modules over $R$ are free (e.g. $R$ is a field or a PID), then $\RoT(R) \cong R$, and the isomorphism is realized by the trace map (which is why I'm calling it the ring of traces). This is true because all endomorphisms are represented by matrices with entries in $R$, and the commutator subalgebra of $\mathfrak{gl}\_n(R)$ is $\mathfrak{sl}\_n(R)$, the algebra of traceless matrices — any commutator has trace zero and any trace zero matrix is a commutator. You can also show that if $R$ is a Dedekind domain, the ring of traces over $R$ is isomorphic to $R$, though it's a little more involved, and still involves writing endomorphisms in terms of matrices.
**THE QUESTION**: is there a ring $R$ for which this "ring of traces" $\RoT(R)$ is not isomorphic to $R$? Such an $R$ which is Noetherian, and for which $\operatorname{Spec}R$ is smooth (or at least normal) would be ideal, but I can't find one even without those restrictions. If no, can we patch together a scheme which would give an affirmative answer to the analogous question for schemes/vector bundles/endomorphisms?
**EDIT:** In response to the comment below, I should add the following: Note that $(V,f)\oplus(V,g) \sim (V,f+g)$, due to the following calculation involving matrices with entries in $\operatorname{End}(V)$.
$$ \begin{pmatrix} 0 & f & 0 \\ 0 & 0 & g \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 0 \\ I\_V & 0 & 0 \\ 0 & -I\_V & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 & 0 \\ I\_V & 0 & 0 \\ 0 & -I\_V & 0 \end{pmatrix}\begin{pmatrix} 0 & f & 0 \\ 0 & 0 & g \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} f & 0 & 0 \\ 0 & -f-g & 0 \\ 0 & 0 & g \end{pmatrix} $$
So $(V,f) \oplus (V,-f-g) \oplus (V,g)$ is a commutator.
$$ \begin{pmatrix} 0 & f+g \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ I\_V & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ I\_V & 0 \end{pmatrix}\begin{pmatrix} 0 & f+g \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} f+g & 0 \\ 0 & -f-g \end{pmatrix} $$
So $(V,f+g) \oplus (V,-f-g)$ is a commutator too.
Adding the second to $(V,f) \oplus (V,g)$, we get
$$ (V,f) \oplus (V,g) \sim (V,f) \oplus (V,g) \oplus (V,f+g) \oplus (V,-f-g). $$
Adding the first to $(V,f+g)$, we get
$$ (V,f+g) \sim (V,f+g) \oplus (V,f) \oplus (V,-f-g) \oplus (V,g). $$
The right hand sides of the two equivalences are isomorphic to each other (permute the factors), so the left hand sides are equivalent, and $(V,f) \oplus (V,g) \sim (V,f+g)$. Which is a very natural thing to want from something that is supposed to be a trace!
| https://mathoverflow.net/users/141571 | "Ring of traces" over a ring R | $\newcommand\HH{\mathit{HH}}$I write $[V,f]$ for the class of $(V,f)$ in your ring.
Let me prove the following property of $[V,f]$ : if $f: V\to W, g: W\to V$, then $[V,gf] = [W,fg]$.
Indeed, $[V,gf] = [V\oplus W, (gf, 0)]$, because $(W,0)$ is a commutator.
Further, by your trick about sums, $[P,-h] = -[P,h]$ so that $[V,gf] - [W,fg] = [V\oplus W, (gf,0)]+[V\oplus W, (0,-fg)] $ which, by using your trick again, is $[V\oplus W, (gf, -fg)]$. Now I claim that this is a commutator of the obvious things, namely $(v,w)\mapsto (0,f(v))$ and $(v,w)\mapsto (g(w),0)$.
This is a simple computation: applying the first composite to $(v,w)$ gives $(gf(v),0)$ and the second $(0,fg(w))$.
This proves the cyclic invariance claim. Now, by definition more or less, $\HH\_0(R)$ is the universal recipient of a cyclically invariant map $(V,f)\mapsto “\operatorname{tr}(V,f)”$. The trick here is to use cyclic invariance or summation with $(Q,0)$ to reduce from arbitrary projective to free, and then from free to “free on 1 generator” by using the standard basis of $\operatorname{End}(R^n)$.
Conversely, sending $(V,f)$ to its Hattori–Stallings trace in $\HH\_0(R)$ satisfies your relation that commutators are sent to $0$.
So they are the same. For a commutative ring, $\HH\_0(R) = R$, as pointed out in the [comments](https://mathoverflow.net/questions/430674/ring-of-traces-over-a-ring-r#comment1108227_430674).
| 6 | https://mathoverflow.net/users/102343 | 430700 | 174,443 |
https://mathoverflow.net/questions/430396 | 6 | Suppose $G \subset \text{Iso}(M)$ is a Lie group acting smoothly on a (pseudo-)Riemannian manifold $(M, g)$. Then $G$ induces a Lie algebra of Killing vectors on $M$. In [this](https://www.semanticscholar.org/paper/Einstein-Tensor-and-3%E2%80%90Parameter-Groups-of-with-Goenner-Stachel/9917663943e0aef06647d2dabe5891667393538b) paper by Goenner and Stachel (in particular in appendix B), they mention how to find a coordinate representation for the metric out of the Killing vectors and the Casimir invariant. However, I do not understand what they mean by "solving the structure equations for the Killing vector fields". Certainly one cannot find unique solutions to the structure equations since adding multiples of commuting vector fields yields a new solution. The system of first order PDE's that one obtains from the coordinate representations of the Lie brackets are unwieldy at best, so it seems one needs to incorporate the Killing equation in order to work through the problem. Yet, the Killing equation depends on the metric, which is precisely what we are trying to find. What exactly are the authors trying to say?
| https://mathoverflow.net/users/491183 | Does the isometry group determine the Riemannian metric? | I think that you are missing some hypotheses on the action of $G$, otherwise there are trivial counterexamples. For example, if $G\subset\mathrm{Iso}(M,g)$ is the trivial group, then one clearly cannot reconstruct $g$ from just knowing $G$ as a subgroup of $\mathrm{Diff}(M)$. Beyond that, the most you can hope for is to construct $g$ up to constant multiples, since $\mathrm{Iso}(M,cg) = \mathrm{Iso}(M,g)$ for any nonzero constant $c$.
A more reasonable question is to ask how to tell whether, given a (connected) Lie subgroup $G\subset\mathrm{Diff}(M)$ or, equivalently, its Lie algebra ${\frak{g}}\subset\mathrm{Vect}(M)$, one can tell whether or not there is a (pseudo-Riemannian) metric $g$ such that $G\subset\mathrm{Iso}(M,g)$ and, if so, provide an algorithm for finding $g$.
Some hint that this might be tricky can be seen by considering the torus $M = \mathbb{R}^2/\mathbb{Z}^2$ and letting the group be $G\simeq\mathbb{R}$ acting as $t\cdot[x,y] = [x {+} t, y {+} at]$ for some irrational constant $a$. The only $G$-invariant metrics are of the form $g = E\,\mathrm{d}x^2 + 2F\,\mathrm{d}x\,\mathrm{d}y+G\,\mathrm{d}y^2$ where $E$, $F$, and $G$ are constants with $EG-F^2\not=0$. On the other hand, if $a$ were rational, then $G$ would be isomorphic to $S^1 = \mathbb{R}/\mathbb{Z}$ and there would be arbitrary functions worth of $G$-invariant metrics.
In general, you don't expect $G$ to determine a unique metric up to multiples unless $G$ acts transitively on $M$, and, even then, the metric may not be unique up to constant multiples. A good example, is $M = G$ where $G$ acts on itself by left translations. Then any left-invariant metric on $G$ will do (and each of them is completely determined by its value at the identity of $G$).
Meanwhile, in the transitive case, it's not hard to determine whether there is an invariant metric and whether or not it is unique. You just have to look at the (closed) subgroup $G\_p\subset G$ consisting of the elements of $G$ that fix some given $p\in M$. That group acts linearly on $T\_pM$, and you just need to know whether that linear representation of $G\_p$ fixes a non-degenerate inner product on $T\_pM$. If $G\_p$ is connected, this is a purely linear algebra problem, since one can easily compute the Lie algebra of $G\_p$ as represented in $T\_pM$ and then, by solving a linear system of equations, find the subspace $S\_p$ of $G\_p$-invariant quadratic forms on $T\_pM$. Then, assuming that this subspace contains a non-degenerate quadratic form, you are done. If $G\_p$ is not connected, you first find the corresponding subspace $S^\circ\_p$ for $G\_p^\circ$, the identity component of $G\_p$, and then deal with the problem of finding the $G\_p/G\_p^\circ$-invariant elements of $S\_p^\circ$, again, an algebra problem.
| 5 | https://mathoverflow.net/users/13972 | 430707 | 174,445 |
https://mathoverflow.net/questions/430708 | 2 | Let $\Omega=[0,1]\times [0,1]$ be the square. We say a function $f\in H^1(\Omega)$ is periodic on $\Omega$ if $f(x,0)=f(x,1)$ and $f(0,y)=f(1,y)$ (in the sense of traces of course). Now consider the problem
$$\begin{cases}-\Delta u+au=f,\\ u,u\_x,u\_y\text{ are periodic on }\Omega.\end{cases} $$
It is easily seen that the variational formulation of the above equation is
$$(\nabla u,\nabla v)\_{L^2}+(u,v)\_{L^2}=(f,v)\_{L^2},\;\;\;\;\text{ for all } v\in H^1\_{per}(\Omega):=\{v\in H^1(\Omega):v\text{ is periodic on }\Omega\}. $$
* When $a>0$, we have the ideal situation where the solution operator $T\_a:f\mapsto u$ is well defined (Lax-Milgram on the Hilbert space $H^1\_{per}$), self-adjoint (since $(\nabla u,\nabla v)\_{L^2}+a(u,v)\_{L^2}$ is an inner product) and compact on $L^2(\Omega)$ (by Sobolev embeddings). Thus one can find a Hilbert basis for $L^2(\Omega)$ composed from eigenfunctions $\{\phi\_n\}$ of $T\_a$. Thus we have that $u=\sum\_n (\lambda\_n+a)^{-1}(f,\phi\_n)\_{L^2}\phi\_n$ where the $\lambda\_n$'s are the eigenvalues of $-\Delta$.
* When $a\leq 0$, we have two situations given by the Fredholm alternative applied to the inverse of the operator of $L\_\epsilon u=-\Delta u+au+\epsilon u$ (where $\epsilon$ is chosen so that the problem is coercive).
1. If $-a$ is not an eigenvalue of $-\Delta$, then a unique solution $u$ exists for the given $f$. Thus we can also define a solution operator $T\_a:f\mapsto u$ which is not necessarily continuous from $L^2$ to $H^1$ (a priori). Do the eigenvalues of $T\_a$ still form a basis for $L^2(\Omega)$? If not how do we represent a solution in this case?
2. When $-a$ is an eigenvalue, either we have no solutions or $f \perp V$ where $V$ is a subspace of dimension $n\in\mathbb{N}$. (I believe it's some eigenspace). How to represent solution in this case?
| https://mathoverflow.net/users/345667 | Representing solutions of $-\Delta u+au=f$ when $a\leq 0$ | Take the eigenbasis $\phi\_n$ you exhibited for $a=0$. It is still an eigenbasis for $-\Delta + a$, with shifted eigenvalues $\lambda\_n + a$.
Write
$$
E\_a=\{n\in\mathbb N : \lambda\_n =- a\},
$$
which is either an empty set or a finite set. If $E\_a=\emptyset$ (case 1.) then
$$
u = \sum\_{n\in\mathbb N} \frac{1}{\lambda\_n + a} \langle\phi\_n,f\rangle \phi\_n,
$$
and this sum is well defined. Note that only finitely many terms correspond to the case $\lambda\_n+a<0$, since $\lambda\_n\to\infty$.
If $E\_a\neq\emptyset$ (case 2.) note that if $\langle\phi\_n,f\rangle\neq0$ for some $n\in E\_a$, there isn't a solution, as an integration by parts against $\phi\_n$ will show.
If $\langle\phi\_n,f\rangle=0$ for all $n\in E\_a$, then any solution writes
$$
u=\sum\_{n\in E\_a} \alpha\_n \phi\_n + \sum\_{n\not\in E\_a} \frac{1}{\lambda\_n + a} \langle\phi\_n,f\rangle \phi\_n,
$$
where the $\alpha\_n$ are arbitrary. So to obtain uniqueness, you need to set $\langle\phi\_n,u\rangle$ for all $n \in E\_a$. When $a=0$, there is a unique eigensolution, corresponding to $\lambda=0$, which is the constant eigensolution $\phi\_0=1$. So usually one sets $\langle\phi\_0,u\rangle=\int\_{[0,1]^2} u dxdy=0$ to ensure uniqueness (but you could choose another constant if you so wished).
| 1 | https://mathoverflow.net/users/40120 | 430711 | 174,446 |
https://mathoverflow.net/questions/430476 | 3 | Let $\operatorname{wt}(n)$ be [A000120](https://oeis.org/A000120), i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $a(n,m)$ be the sequence of numbers $k$ such that $\operatorname{wt}(k)=m$. In other words, $a(n,m)$ is the $n$-th number with binary weight equals $m$.
I conjecture that
$$a(1,2^{n-1}+n)+\sum\limits\_{i=1}^{2^{n-1}}a(i+2,2^{n-1}+n)=(2^{n+1}+1)2^{2^{n-1}+n-1}-1$$
I also conjecture that numbers of the form $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ have only $2$ partitions into parts with binary weight equals $2^{n-1}+n$, exactly the number itself and the sum above.
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$ | Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have
$$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$
Then the sum in question can be easily computed:
\begin{split}
& a(1,2^{n-1}+n)+\sum\_{i=1}^{2^{n-1}}a(i+2,2^{n-1}+n)\\
&= \sum\_{i=0}^{2^{n-1}+1} a(i+1,2^{n-1}+n) - (2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-1}) \\
&=(2^{n-1}+1)(2^{2^{n-1}+n+1} - 1) - 2^{n-1}(2^{2^{n-1}+2}-1) + 2^{2^{n-1}+n-1} \\
&= (2^{n+1}+1)2^{2^{n-1}+n-1} - 1.
\end{split}
| 2 | https://mathoverflow.net/users/7076 | 430713 | 174,448 |
https://mathoverflow.net/questions/430657 | 2 | Let $(\Theta, H, \mu)$ be an abstract Wiener space, i.e. let $(\Theta, \lVert \cdot \rVert\_{\Theta})$ be a separable Banach space, let $(H, \langle \cdot, \cdot \rangle\_{H})$ be a separable Hilbert space densely embedded in $\Theta$, and let $\mu$ be a Gaussian measure on $\Theta$ with characteristic functional $\exp( - \frac{1}{2} \Vert f \Vert\_H^2 ), f \in \Theta^{\ast}$. Let $(e\_i)\_{i = 1}^{\infty} \subseteq \Theta^{\ast}$ be an orthonormal basis of $H$ contained in $\Theta^{\ast}$.
Let $\Psi: \Theta \rightarrow \mathbb{R}$ be a measurable function contained in the $d$-th inhomogeneous Wiener–Itô Chaos $\mathcal{H}^{(\leq d)}(\Theta, \mu)$, i.e. the $L^2(\Theta, \mu)$ closure of the linear span of
$$
\Bigl\{ \prod\_{\alpha\_i \in \alpha} h\_{\alpha\_i}(e\_i) : \alpha = (\alpha\_1, \alpha\_2, \dotsc) \in \mathbb{N}^{\mathbb{N}}, \lvert \alpha \rvert \leq d \Bigr\}
$$
where $h\_1, h\_2, \dotsc$ are the usual Hermite polynomials.
**Question:** In the situation above, does there necessarily exist a measurable $\Phi: \Theta \rightarrow \mathbb{R}$ which equals $\Psi$ for $\mu$-almost every $\theta \in \Theta$ and is continuous on a neighbourhood of $H$?
**Remark:** I'm actually interested in the case where $\mathbb{R}$ is replaced by another separable Banach space $E$, but I'm more than happy to understand the special case first, and I don't think the difference is too drastic for this purpose anyway.
**Special Cases:** In the case where $\dim \Theta < \infty$ (and $\mu$ is non-degenerate), this is of course true. Since in that case the ONB $(e\_i)\_{i = 1}^{\infty}$ is finite, the (inhomogeneous) Wiener–Itô Chaos is finite-dimensional and spanned by polynomials, $\Psi$ itself is continuous on all of $\Theta$, which constitutes a neighbourhood of $H$.
**(Edit) Remark:** Since on $\mathbb{R}$ with the standard normal distribution $\gamma\_1$, an indicator function like $1\_{[0, \infty)}$ does not have a continuous version, I tried finding a counterexample to my question by looking for a Bernoulli random variable defined on $\Theta$, and then modifying it to something like $1\_{\{e\_1(\cdot) > 0\}}$. However, one can show that random variables in finite Wiener–Itô Chaos are Malliavin differentiable, but a (non-trivial) Bernoulli random variable is not. So maybe the answer to my question is positive after all.
---
**Motivation**: The motivation for this comes from trying to show a large deviation principle (LDP) for the measures $(\Psi^{\ast} \mu\_{\varepsilon})\_{\varepsilon>0}$ on $\mathbb{R}$ (but generically $E$), where $(\mu\_{\varepsilon})\_{\varepsilon>0}$ is the family of measures on $\Theta$ defined by $\mu\_{\varepsilon}(\cdot) = \mu(\varepsilon^{-1/2}( \cdot))$. Those of course satisfy an LDP with rate
$$ I(x) = \frac{1}{2} \lVert x \rVert\_H^2 , ~~x \in H $$
and $I(x) = \infty$ for $x \notin H$. Now, if $\Psi: \Theta \rightarrow \mathbb{R}$ were continuous, then the contraction principle would assert that the rate function of $(\Psi^{\ast} \mu\_{\varepsilon})\_{\varepsilon>0}$ is
$$ I(s) = \inf \Bigl\{ \frac{1}{2} \lVert h \rVert\_H^2 : \Psi(h) = s \Bigr\}, ~~ s \in \mathbb{R}.$$
However, since $\mu(H) = 0$, the image of $H$ under $\Psi$, which controls the rate function, could be just about anything. However, if there was a continuous $\Phi$ which equals $\Psi$ $\mu$-almost surely and is continuous on a neighbourhood of $H$, then this gives a canonical candidate for the rate function.
Alternatively, one might use an extended version of the contraction principle to get an LDP here. But that version requires that there exist measurable functions $\Psi\_N: \Theta \rightarrow \mathbb{R}$ which are continuous on neighborhoods of $H$ and converge uniformly to $\Psi$ on that neighborhood, implying that $\Psi$ had to be continuous that neighborhood.
| https://mathoverflow.net/users/117692 | $\Psi$ in finite Wiener–Itô Chaos implies existence of continuous representative on neighborhood of Cameron–Martin space? | The answer to the question is no, even in the case $d=1$. Take for example $H = \ell^2$, $\Theta = \{\xi\,:\, \|\xi\| = \sup\_{n\ge 1} |\xi\_n|/(1+\log n) < \infty\}$, $\mu$ the law of i.i.d. normals, and $\Psi(\xi) = \sum\_n \xi\_n/n^{3/4}$. Assume by contradiction that a version of $\Psi$ exists that is continuous at $0$, so that there exist $K < \infty$ and $\epsilon > 0$ such that $|\Psi(\xi)| \le K$ for almost every $\|\xi\| \le 2\epsilon$. In particular, it is the case that, for every $h$ with $\|h\| \le \epsilon$, $|\Psi(\xi + h)| \le K$ on the ball of radius $\epsilon$. On the other hand, if $h \in H$, then the Cameron-Martin theorem tells us that the law of $\Psi(\cdot + h)1\_{\|\cdot\| \le \epsilon}$ is equivalent to that of $\Psi(\cdot)1\_{\|\cdot\| \le \epsilon}$ and that, writing $f\_n = n^{-3/4}$, we have $\Psi(\cdot+h) = \Psi(\cdot) + \langle{h,f}\rangle$ almost surely. Since we can easily find an element $h \in H$ such that $\|h\|\le \epsilon$ but $\langle{h,f}\rangle \ge 2K$, we have a contradiction.
Regarding the motivation though, there is of course an LDP for elements in Wiener chaoses of finite order and the LDP functional is the one you would expect, see for example Michel Ledoux's "A note on large deviations for Wiener chaos".
| 4 | https://mathoverflow.net/users/38566 | 430714 | 174,449 |
https://mathoverflow.net/questions/430705 | 4 | Let $ f $ be a function on $ \mathbb{T}=[0,1] $ ($ 1 $-periodic) with bounded variation. Prove that if $ \widehat{f}(k)=\int\_0^1f(x)e^{-2\pi ikx}dx=o(1/|k|) $, then $ f\in C(\mathbb{T}) $. I do not know how to use the assumption that $ f $ is of bounded variation. Can you give me some hints?
| https://mathoverflow.net/users/241460 | The decay of Fourier coefficients and the continuity of functions | The (distributional) derivative $\mu=f'$ is a measure, and by assumption $\widehat{\mu\_n}=o(1)$. By Wiener's theorem, this implies that $\mu$ does not have a point part, so $\mu$ is a continuous measure and thus $f$ is a continuous function. Compare Corollary 13.11 of my lecture notes [here](https://math.ou.edu/%7Ecremling/teaching/lecturenotes/fa-new/ln13.pdf).
| 4 | https://mathoverflow.net/users/48839 | 430715 | 174,450 |
https://mathoverflow.net/questions/430581 | 0 | Let $(X,d)$ be a complete separable geodesic *(thus connected and path connected)* metric space of non-positive curvature (in the sense of [Ballmann](https://people.mpim-bonn.mpg.de/hwbllmnn/archiv/NPC0606.pdf#page=3)) and fix some $x\_0\in X$. Let $C\_0(I,X)$ denote the set of continuous functions $f:[0,1]\rightarrow X$ satisfying
$$
f(0) = x\_0 \mbox{ and } \sup\_{x\in X}\, d(f(x),x\_0) < \infty.
$$
Define the metric $D$ on $C\_0(X)$ by
$$
D(f,g)\,:=\, \sup\_{x\in X}\, d(f(x),g(x))
,
$$
and note that $D$ is always finite.
Is $C\_0(I,X)$ a complete and separable space of non-positive curvature? I mean this is true if $X$ is Banachian but I don't know about the general case...
| https://mathoverflow.net/users/491352 | Curvature of space of paths into a NPC space | Maybe I'm missing something. But it looks like if you take $X = \mathbb{R}$ than $C\_0(I, X)$ is a set of continuous functions on $[0,1]$ with $0$ at $0$. And than it's universal aka (almost) contains an isometric copy of every finite metric space. And NPC is very restrictive in terms of what $4$-point subsets are allowed.
PS: maybe you can consider an $L\_2$-metric
$$D\_{L\_2}(f,g)\,:= \sqrt{\int\_{[0,1]} \big(d(f(x),g(x))\big)^2}$$ and than is will be NPC. But it will not be complete.
| 1 | https://mathoverflow.net/users/32454 | 430723 | 174,453 |
https://mathoverflow.net/questions/430726 | 6 | **"Real-life" motivation.** The German satirical magazine *Der Postillon* [suggested](https://www.der-postillon.com/2016/08/7-wege-rauchen.html) a few measures for deterring smokers from their bad habit. I especially liked the idea of inserting one "prank cigarette" per pack, giving the smoker a reminder in form of a (mild) explosion after lighting it. This motivated the following problem.
**Problem statement.** Suppose we are given $n$ empty packs of cigarettes, each to be filled with $n$ cigarettes. Amongst the $n^2$ cigarettes in total to be distributed to the $n$ empty packs, there are $n$ "prank cigarettes" to be distributed in a uniform manner into the $n$ packs.
**Questions.**
1. Let $M\_n$ be the expected value of the maximum number of prank cigarettes any pack receives. Is there an explicit formula for $M\_n$? If not, do we have $\lim\sup\_{n\to\infty}M\_n/n > 0$ or $\lim\sup\_{n\to\infty}M\_n/\log(n) > 0$?
2. Let $E\_n$ be the expected value of packs without any prank cigarettes. Is there an explicit formula for $E\_n$? If not, do we have $\lim\sup\_{n\to\infty}E\_n/n > 0$ or $\lim\sup\_{n\to\infty}E\_n/\log(n) > 0$?
| https://mathoverflow.net/users/8628 | Expected maximum number of "prank cigarettes" in an average pack | $\newcommand{\Si}{\Sigma}$Let $X\_n$ be the maximum number of prank cigarettes any pack receives, so that $M\_n=EX\_n$. Note that $X\_n=\max(N\_1,\dots,N\_n)$, where $(N\_1,\dots,N\_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of [Raab and Steger](https://link.springer.com/chapter/10.1007/3-540-49543-6_13),
\begin{equation\*}
X\_n\sim r\_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1}
\end{equation\*}
in probability (as $n\to\infty$).
We will also prove
>
> **Proposition 1:**
> \begin{equation\*}
> EX\_n^2\ll r\_n^2.
> \end{equation\*}
>
>
>
(As usual, we write $A\ll B$ to mean $A=O(B)$.)
By Proposition 1 and the [de la Vallée-Poussin theorem](https://en.wikipedia.org/wiki/Uniform_integrability#Relevant_theorems), $X\_n/r\_n$ is uniformly integrable, which implies
>
> \begin{equation\*}
> M\_n=EX\_n\sim r\_n
> \end{equation\*}
>
>
>
(which agrees with Qiaochu Yuan's heuristics/conjecture).
*Proof of Proposition 1:*
Let $[n]:=\{1,\dots,n\}$. Note that
\begin{equation\*}
EX\_n^2\ll\sum\_{m\in[n]}mP(X\_n\ge m)
=\Si\_1+\Si\_2, \tag{3}\label{3}
\end{equation\*}
where
\begin{equation\*}
\Si\_1:=\sum\_{1\le m\le 4r\_n}mP(X\_n\ge m),\quad \Si\_2:=\sum\_{4r\_n<m\le n}mP(X\_n\ge m).
\tag{4}\label{4}
\end{equation\*}
It is easy to bound $\Si\_1$:
\begin{equation}
\Si\_1\le\sum\_{1\le m\le 4r\_n}m\ll r\_n^2. \tag{5}\label{5}
\end{equation}
Let us now bound $\Si\_2$.
Recall that $X\_n=\max(N\_1,\dots,N\_n)$, where $(N\_1,\dots,N\_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$
\begin{equation\*}
P(X\_n\ge m)\le\sum\_{i\in[n]}P(N\_i\ge m)
=nP(N\_1\ge m). \tag{7}\label{7}
\end{equation\*}
Next, $N\_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence
\begin{equation\*}
\begin{aligned}
P(N\_1\ge m)&=\sum\_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\
&\le\sum\_{k=m}^n\binom nk\frac1{n^k}\le\sum\_{k=m}^n\frac1{k!}\ll\frac1{m!}.
\end{aligned}
\tag{11}\label{11}
\end{equation\*}
Further, for $m>4r\_n$, eventually (that is, for all large enough $n$),
\begin{equation\*}
m!\ge(m/e)^m=\exp(m\ln(m/e)) \\
\ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13}
\end{equation\*}
So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13},
\begin{equation\*}
\Si\_2\le\sum\_{4r\_n<m\le n}mnP(N\_1\ge m) \\
\ll\sum\_{4r\_n<m\le n}mn\frac1{n^3}\le1\ll r\_n^2. \tag{15}\label{15}
\end{equation\*}
Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$
---
Quite similarly one can show that
\begin{equation\*}
EX\_n^p\sim r\_n^p
\end{equation\*}
for each real $p>0$.
| 10 | https://mathoverflow.net/users/36721 | 430731 | 174,456 |
https://mathoverflow.net/questions/430669 | 6 | Let $(\mathbb{S}^2,g)$ be a Besse sphere, that is, a Riemannian sphere all of whose geodesics are closed. By a result of Gromoll and Grove, all the geodesics are simple (no self-intersections) and have the same length $L > 0$, say.
**Question 1**: in the celebrated book of Besse entitled "Manifolds all of whose geodesics are closed", the author asserts that every minimizing geodesic $\gamma$ from $p \in \mathbb{S}^2$ to an arbitrary point $q \in \mathbb{S}^2$ has length less than $L/2$, and thus the diameter of $(\mathbb{S}^2,g)$ is less than $L$. How does one prove these claims?
**Question 2**: In the paper by Bott entitled "On manifolds all of whose geodesics are closed", the author defines the index of $(\mathbb{S}^2,g)$ (or for any manifold of higher dimension with this property on the geodesics) as the index of any closed geodesic in the surface. Why is this number independent of the geodesic and of its parametrization?
| https://mathoverflow.net/users/85934 | On properties of Besse spheres | It so happens I recently read this same passage in Besse's book. I am no expert, but here is how I understand it.
**Question 1.** Let $p,q \in \mathbf{S}^2$ be two arbitrary points, and $\eta: [0,L] \to \mathbf{S}^2$ be a closed geodesic containing both points, fixing $\eta(0) = p$ for example. Going along $\eta$ either in the given direction $\eta'(0)$, or in the opposite direction $-\eta'(0)$, one encounters $q$ after a distance $L/2$ at the most. Therefore $d(p,q) \leq L/2$, and as $p,q \in \mathbf{S}^2$ were arbitrary, the diameter of $\mathbf{S}^2$ is also at most $L/2$.
*Remark.* Besse gets a bound of $L$ for the diameter, because the situation there is slightly different: they work with manifolds for which all geodesics issued from a (fixed) point $p$ are loops with length $L$.
**Question 2.** The second question is a bit trickier; to answer it I will still use Besse's book. There the constancy of the index is stated as a part of Theorem 7.23, the Bott–Samelson theorem. Note that technically Bott states that all geodesics issued *from the same point $p$* have the same index; this is also how it's stated in Besse's book, so I'll focus on this claim.
*Claim.* The geodesics of length $L$ issued from $p \in \mathbf{S}^2$ have the same index.
We write $\lambda(\gamma) \in \mathbf{Z}$ be the index of any closed geodesic $\gamma: [0,L] \to \mathbf{S}^2$ with $\gamma(0) = p = \gamma(L)$. (We consider the index as a geodesic arc with fixed endpoints, rather than as a geodesic loop.) The Morse index theorem states that $\lambda(\gamma)$ equals the number of conjugate points $\gamma(t)$ for all times $0 < t < L$, counted with multiplicity.
Let $(\gamma\_s \mid -\delta < s < \delta)$ be some continuous variation of $\gamma\_0 := \gamma$ through geodesic loops at $p$. The times at which one encounters the first, second, $\dots,n$th conjugate points along $\gamma\_s$ vary continuously with $s$. The only way the index $\lambda(\gamma\_s)$ could change would be if conjugate points either appeared in the arc $\gamma\_s$ or disappeared from it. In either case, some conjugate points would need to cross the base point $p$, adding to the multiplicity of $p$ along the arc $\gamma\_{s^\*}$ for some $s^\* \in (- \delta,\delta)$.
This is impossible because the multiplicity of $p$ to itself along any geodesic loop is already 'maxed out': because the Jacobi fields are $L$-periodic, the point $p$ is conjugate to itself with multiplicity one. (This is the largest possible multiplicity on a two-dimensional surface.) Therefore the loops issued from $p$ all have the same index, which finishes the proof of the claim.
| 2 | https://mathoverflow.net/users/103792 | 430733 | 174,457 |
https://mathoverflow.net/questions/390478 | 4 | What is the Hausdorff dimension of the subset $S\_c \subset [0,1]$ of points such that the [critical exponent](https://en.wikipedia.org/wiki/Critical_exponent_of_a_word) of their binary expansion is $c$? It's clear that $\dim\_H S\_{\infty}=1$, but what can be said for $c<\infty$?. Also, what is $\dim\_H \cup\_{c\in [2,\infty)} S\_c$? Are there techniques for the evaluation of the Hausdorff dimension which are applicable to these cases?
The text by Falconer [1] doesn't seem to cover this kind of questions.
**Update**: it occurred to me that even the estimate of the (upper) box dimension of $S\_c$ is not trivial. Already for integer $c$, it leads to a tricky combinatorial problem, that is, how many distinct binary words of length $n$ are there having no factor of the form $w^c$ ($w \in \{0,1\}^\*$)?
It's also not easy for me to guess whether the Hausdorff and box dimensions could disagree.
[1]: Falconer, K. (2004). Fractal geometry: mathematical foundations and applications. John Wiley & Sons.
| https://mathoverflow.net/users/167834 | Hausdorff dimension and critical exponent of words | Here are a few observations: (the last one is maybe the main point of interest)
$\bullet$ Your $S\_c$ is a closed subset of $[0,1]$ invariant under $x \mapsto 2x \pmod 1$ (usually there's a subtlety about points with multiple expansions, but none of those are in $S\_c$ for $c < \infty$), and so I think a result of Furstenberg ([1]) shows that it has equal Hausdorff and box dimension. They should both be equal to the topological entropy $h(S\_c)$ of $S\_c$ viewed as a subshift on $\{0,1\}$ (using $\log\_2$ in definition of entropy).
$\bullet$ In [1], Kolpakov proves that $dim(S\_3) = h(S\_3) \geq 0.5435$ (I've converted to $\log\_2$, he used $\ln$)
$\bullet$ for all $c < \infty$, $dim(S\_c) = h(S\_c) < 1$, since at the least you forbid the word $0^{\lceil c \rceil}$, and all proper subshifts of $\{0,1\}^{\mathbb{N}}$ have entropy less than $1$.
$\bullet$ (this is quite long and probably hard to read, sorry) There are probably better methods, but you can use the techniques of my paper [3] to bound the entropy (and therefore dimension) of $S\_c$ from below. Let's restrict to integer $c$ for now. Then $S\_c$ is a subshift defined by forbidden list $\{w^c \ : \ w \in \{0,1\}^\*\}$, where $w^c$ is the $c$th power of $w$.
The technique from that paper to bound entropy from below is to rescale at some length $j$, and then view your subshift $X$ (here $S\_c$) as a new subshift whose alphabet is $A^{(j)} = L\_j(X)$ (the set of $j$-letter words in $X$) and whose forbidden list is $F^{(j)}$, a set of forbidden concatenations of `letters' $A^{(j)}$ which induces the same forbidden words as those for $X$. More formally,
$a\_1 \ldots a\_i$ is in $F^{(j)}$ if the $ij$-letter concatenation $a\_1 \ldots a\_i$ contains a forbidden word for $X$, but neither $a\_1 \ldots a\_i-1$ nor $a\_2 \ldots a\_i$ contains a forbidden word for $X$. This forbidden list $F^{(j)}$ on $A^{(j)}$ induces a subshift $X^{(j)}$ where each point is a sequence of $c$-letter blocks whose concatenation is in $X$. (Formally, $X^{(j)}$ is isomorphic to $X$ under the $c$th power of the shift.)
Here, if we use $X = S\_c$ and set $j = c$, then it's easy to bound from above the number of $i$-'letter' words in $F^{(c)}$. There are none for $i = 1$ by definition. For larger $i$, if $a\_1 \ldots a\_i \in F^{(c)}$, then either $a\_1 \ldots a\_i$ is a $c$th power of an $i$-letter word (less than $2^i$ possibilities) or contains a $c$th power of an $(i-1)$-letter word ($c$ choices for where in $a\_1$ the power starts, less than $2^{i-1}$ possibilities for the base of the power, less than $2^c$ choices for the remaining letters). If only shorter $c$th powers were in $a\_1 \ldots a\_i$, then one of $a\_1 \ldots a\_{i-1}$ or $a\_2 \ldots a\_i$ would contain a $c$th power, disallowed by definition. So, the number of $i$-letter forbidden words in $F^{(c)}$ is less than $2^i + c2^c 2^{i-1} \leq c 2^{c+i}$.
Now, Theorem 4.1 from [3] gives a lower bound for entropy of a subshift. It states that if $X$ is a subshift with alphabet $A$ and an infinite forbidden list $F$ with $F\_n$ words of length $n$ for every $n$, and if
$\sum F\_i x^i < x(|A| - k + 1) - 1$ for some $x > 1/|A|$ and $k < |A|$, then
$h(X) \geq \log k$. This is hideous, but we can make it easier.
We'll apply to the $c$-letter rescaling $S\_c^{(c)}$ of $S\_c$ defined above, and assume that $c \leq 10$. Then $|A^{(c)}| = 2^c - 2$ (the only illegal $c$-letter words in $S\_c$ are $0^c$ and $1^c$). Let's take $k = 2^{c-1} - 1$ and $x = 4/2^c$. Then $x(|A^{(c)}| - k + 1) = 1$, so we need only verify that
$\sum (F^{(c)})\_i x^i < 1$; recall that $(F^{(c)})\_i \leq c 2^{c+i}$. Finally,
$\sum\_{i = 2}^\infty (F^{(c)})\_i x^i \leq \sum\_{i=2}^\infty c 2^{c+i} (4/2^c)^i =
c 2^c \sum\_{i=2}^\infty (8/2^c)^i = c 2^c \frac{64/2^{2c}}{1 - (8/2^c)}
= \frac{64c}{2^c-8}$.
This is less than $1$ for $c \leq 10$, so Theorem 4.1 shows that in that case,
$h(S\_c^{(c)}) \geq \log k = \log(2^{c-1} - 1)$. But $h(S\_c^{(c)}) = c h(S\_c)$ (since the former is isomorphic to the latter under the $c$th power of the shift). So, at long last, we get $h(S\_c) \geq \frac{\log(2^{c-1} - 1)}{c}$.
In particular, this approaches $1$ as $c \rightarrow \infty$, so this should answer one of your questions: $dim(\bigcup S\_c) = 1$.
I should say that I found several papers estimating entropies of $S\_c$ from below, and would not at all be surprised if one gives a much easier proof that $dim(\bigcup S\_c) = 1$. But all I found were for specific values of $c$.
[1] H. Furstenberg, Disjointness in ergodic theory, minimal sets, and a problem in Diophantine approximation. Math. Systems Theory, 1 (1967)
[2] R. Kolpakov, Efficient lower bounds on the number of repetition-free words. J. Integer Seq. 10 (2007), no. 3
[3] R. Pavlov, Ronnie, On subshifts with slow forbidden word growth.
Ergodic Theory Dynam. Systems 42 (2022), no. 4
| 6 | https://mathoverflow.net/users/116357 | 430734 | 174,458 |
https://mathoverflow.net/questions/429702 | 5 | This is a follow-up to the question of Joseph O'Rourke [Which metric spaces have this superposition property?](https://mathoverflow.net/q/118008/)
A metric space $X$ will be called all-set-homogeneous if for any subset $A\subset X$ any distance-preserving map $A\to X$ can be extended to an isometry $X\to X$.
*Classical examples* include Euclidean spaces, Lobachevsky spaces, and spheres (all with canonical metrics and up to rescaling).
>
> Is it true that complete all-set-homogeneous *length*-metric spaces include only the classical examples *plus* the so-called [universal $\mathbb{R}$-trees](https://arxiv.org/abs/math/9904133) of finite valence?
>
>
>
We have a partial answer, which requires the following definition.
Given a metric space $X$, consider all pseudometrics induced on $n$ points $x\_1,\dots, x\_n\in X$.
Such a metric is completely described by $n{\cdot}(n-1)/2$ real numbers $|x\_i-x\_j|\_X$, so it can be encoded by a point in $\mathbb{R}^{n{\cdot}(n-1)/2}$.
The set $F\_n(M)\subset \mathbb{R}^{n{\cdot}(n-1)/2}$ of all these points is called the $n^\text{th}$ fingerprint of $X$.
We know that all-set-homogeneous *length*-metric spaces with closed fingerprints include only the basic examples.
(We do not assume local compactness.)
The proof is very short, you may check it [here](https://arxiv.org/abs/2211.09671), but likely it is not new.
>
> Is it new?
>
>
>
**Remarks**
* Locally compact all-set-homogeneous length spaces include only basic examples; in fact, in this case, 3-point-homogeneity is sufficient. It follows from the result of J. Tits ["Sur certaines classes d'espaces homogènes de groupes de Lie" (1955)];
* It seems that condition that the space is length can not be seriously connected to all-set-homogeneity. So we expect that there are counterexamples to the first question.
* It was shown by Birkhoff that all-set-homogeneous geodesic space with local uniqueness of geodesics has to be classical [See his *Metric foundations of geometry. I*].
| https://mathoverflow.net/users/1441 | All-set-homogeneous spaces | This is more of an observation and a long comment than an answer.
The observation is that a fairly standard application of the Erdős–Rado theorem implies that in any metric space with more than $2^{2^{\aleph\_0}}$ many elements, there is an infinite sequence $(a\_i)\_{i<\omega}$ of distinct elements satisfying $d(a\_i,a\_j) = d(a\_0,a\_1)$ for any $i<j<\omega$. By the argument in your note this is enough to imply that any all-set-homogeneous metric space (with no assumption of completeness or length-ness) has cardinality at most $2^{2^{\aleph\_0}}$.
The comment regards connections between your question and concepts that have been studied in model theory, particularly in [the model theory of metric structures](https://faculty.math.illinois.edu/%7Ehenson/cfo/mtfms.pdf).
In your note, you mention a variation of this question in which you only require homogeneity over finite sets. In discrete structures, this kind of homogeneity is studied in model theory and is referred to as *ultrahomogeneity* (as opposed to ordinary homogeneity, which requires a stronger relationship between the two finite sets of elements to get an automorphism). The relevant concept here is that of a [*Fraïssé class*](https://en.wikipedia.org/wiki/Fra%C3%AFss%C3%A9_limit), which is a class of finite structures satisfying the properties of the class of finite substructures of an ultrahomogeneous structure. Fraïssé's theorem is the statement that it is possible to construct the structure itself from the class in a particular way. In particular, countable ultrahomogeneous structures (in a finite relational language) are precisely characterized as the Fraïssé limits of Fraïssé classes (although this doesn't necessarily help that much with actually characterizing such structures precisely).
In my subfield of model theory, known as *continuous logic*, we apply model-theoretic techniques to structures with underlying metrics, such as metric spaces, Banach spaces, $\mathbb{R}$-trees, and so on. [Fraïssé limits have been studied in the context of continuous logic by Ben Yaacov](https://www.jstor.org/stable/43864204), although frankly the necessary formalism is quite technical when compared to the discrete version. As is often the case in continuous logic, some of the defining properties of Fraïssé classes need to be modified to only hold in some approximate sense. Here the natural notion is that of approximate ultrahomogeneity. In particular, a metric space $X$ is *approximately ultrahomogeneous* if for any finite $A \subseteq X$, any distance-preserving map $f : A \to X$, and any $\varepsilon > 0$, there is an isometry $g : X \to X$ such that $d(f(a),g(a)) < \varepsilon$ for every $a \in A$.
So, using Ben Yaacov's machinery, I am able to give a technical and possibly useless answer to a modification of a question you did not ask here, which is that the separable approximately ultrahomogeneous metric spaces are precisely the limits of Fraïssé classes (in the sense of Ben Yaacov) of finite metric spaces.
Unfortunately, the all-set-homogeneity property doesn't have analogs that have been studied by model theorists to my knowledge. The issue is that, as we already saw, properties of this sort tend to limit the size of the structure involved, and this makes it a little awkward to use model-theoretic tools.
Finally, I can comment on the axiomatizability of the class of length spaces in continuous logic. A fairly well-known fact is that complete length spaces can be characterized as those complete metric spaces that admit approximate midpoints (i.e., for any $a,b \in X$ and any $\varepsilon > 0$, there is a $c \in X$ such that $d(a,c)$ and $d(c,b)$ are both within $\varepsilon$ of $\frac{1}{2}d(a,b)$. This condition can be easily expressed in continuous logic by the closed condition $\sup\_{xy}\inf\_{z} \max(|d(x,z)-\frac{1}{2}d(x,y)|,|d(z,y)-\frac{1}{2}d(x,y)|) = 0$, so for any $r > 0$, the class of (complete) length spaces of diameter at most $r$ is elementary (in the sense of continuous logic, which by default only deals with complete metric structures). Dealing with unbounded metric spaces, while doable in continuous logic, is a bit trickier. It can be accomplished using Ben Yaacov's [extension of continuous logic to unbounded metric structures](https://doi.org/10.1142/S0219061308000737). In this situation too, it is possible to axiomatize the class of length spaces, but it's a little bit more complicated to state how precisely.
| 2 | https://mathoverflow.net/users/83901 | 430738 | 174,459 |
https://mathoverflow.net/questions/430742 | 7 | Several years ago, I mentioned offhandedly to a colleague that I had noticed that, if you extend the $\mathsf E\_n$ series downwards in the natural way, by removing nodes from the long arm of $\mathsf E\_8$, then the fundamental group of $\mathsf E\_n$ (i.e., the quotient of the weight lattice by the root lattice) is cyclic of order $9 - n$ for $3 \le n \le 8$. (Specifically, $\mathsf E\_5 = \mathsf D\_5$, $\mathsf E\_4 = \mathsf A\_4$, and $\mathsf E\_3 = \mathsf A\_2 + \mathsf A\_1$.) I thought that this was just a funny coincidence, but my colleague said that there was actually more to it, and gave me a beautiful explanation of the deeper reason.
Unfortunately, as is the way of such things, I did not write down the beautiful explanation and so forgot it; and, upon reaching out to my colleague several years later, I find that they can no longer re-construct their beautiful explanation. Can you provide any explanation other than "that's how it works out"?
| https://mathoverflow.net/users/2383 | Why is the fundamental group of $\mathsf E_n$ cyclic of order $9 - n$? | If we define the $E\_n$ lattice in the algebraic geometer's way as the orthogonal complement of $(1,\dots, 1, 3)$ in the Lorentzian unimodular lattice $I\_{n,1}$, i.e. in $\mathbb Z^{n+1}$ with intersection form $$\langle(x\_1,\dots, x\_{n+1}), (y\_1,\dots, y\_{n+1})\rangle =x\_1y\_1 + x\_2 y\_2 + \dots + x\_n y\_n - x\_{n+1} y\_{n+1}, $$
then there is an immediate proof: because $(1,\dots, 1, 3)$ is a primitive vector of norm $9-n$ and the complement of a primitive vector of norm $k$ in a unimodular lattice always has fundamental group of norm $k$ (with the isomorphism of the dual lattice modulo the lattice to $\mathbb Z/k$ on the dual lattice given by the pairing with that vector).
One can check that the lattice given by this construction has the $E\_n$ Dynkin diagram for all $n\geq 3$ using the roots given by all vectors with a series of $0$s, then a $1$, then a $-1$, then a series of at least one $0$, and the vector $(0,\dots, 0,1,1,1,1)$.
---
Another perspective:
For any series of Dynkin diagrams where we add one vertex after a number in a line, a simple recurrence shows the determinant of the Cartan matrix (= order of the fundamental group) of the $n$th member of the series must be a linear function of $n$. Finding the right linear function for $E\_n$ then requires computing two examples.
Given that the fundamental group has order $9-n$ for $n\geq 3$, the fact that it is cyclic is nontrivial only for $n=5$, so not much of a coincidence.
---
A third perspective: A $k$-torsion element of the dual lattice modulo the lattice is the same thing as an element of the lattice tensor $\mathbb Z/k$ with zero dot product with every vector of the lattice, i.e. a $\mathbb Z/k$-linear combination of the simple roots with zero dot product with any root. These can be calculated directly from the Dynkin diagram.
On an $E\_n$-type lattice, if the value at the vertices near the end are
$$ \begin{array}{ccc} a & b & c \\ d & & \end{array}$$
then $a=2d$ and $a+c=2b$ and $b=2c$ so $a=3c$ so because $2d=3c$ we have $c=2(d-c)$, $b= 4(d-c)$, $a=6(d-c)$, and $c=3(d-c)$, and then, continuing left, every vertex is a multiple of $d-c$. This shows every element of the fundamental group is a multiple of
$$ \begin{array}{ccc} \dots & 2 & 3 & 4 & 5 & 6 & 4 & 2 \\&&&&& 3 & & \end{array}$$
showing the fundamental group is cyclic, and this element is $9-n$-torsion since it is only satisfies the orthogonality condition in $\mathbb Z/k$ if the label $9-n$ of the unique vertex of $E\_{n+1}$ that's missing in $E\_n$ is divisible by $k$, showing the fundamental group is cyclic of order $9-k$.
This proof, despite appearances, works for $n=3$, where we just have the additional relation $a=0$ to work with.
---
All three arguments work for $n>9$ as well as $n\leq 8$, with the first and third arguments still determining the fundamental group and the second argument determining its order.
| 8 | https://mathoverflow.net/users/18060 | 430743 | 174,461 |
https://mathoverflow.net/questions/430706 | 14 | The well-known Sylvester–Gallai Theorem states that a set of $n>2$ points in $R^2$ not all on a line contains two points such that the line passing through these two points does not contain a third point in the set.
One of the reasons this is somewhat tricky to prove is that it is false in some geometries, including finite field geometry. This can be seen by considering, for example, all $p^2$ points in $F\_p^2$ a plane over a finite field.
My question is the following:
>
> Let $n$ be a set of (say) $n=p^{1/100}$ points in $F\_p^2$ a large finite field of prime order. Then does the conclusion of the Sylvester-Gallai theorem hold?
>
>
>
I am restricting the question to prime order fields to (1) avoid issues with subfields, and to (2) try to prevent the analog of the Hesse Configuration, a counterexample to the corresponding statement over the complex numbers.
Many of the proofs of the real case use the ordering of the field, so will run into trouble in a finite field. On the other hand, once the sets are really really small (much smaller than what is being considered here) there might be hope of formally translating the problem to the complexes and using the characterizations of possible counter-examples [see: [https://arxiv.org/pdf/math/0403023.pdf]](https://arxiv.org/pdf/math/0403023.pdf).
I assume this question would been studied before, but I can't find a reference. I'm mostly interested in a positive result, so if there is a counter-example, I'd be interested to know if there is an understanding of counterexamples, which there is in the complex case.
| https://mathoverflow.net/users/630 | Sylvester–Gallai theorem for small sets in a finite field | The answer is no, basically thanks to [Menelaus's theorem](https://en.wikipedia.org/wiki/Menelaus%27s_theorem). First observe as a routine application of the [Bombieri-Vinogradov theorem](https://en.wikipedia.org/wiki/Bombieri%E2%80%93Vinogradov_theorem) that there are infinitely many primes $p$ such that $p-1$ contains a factor $d \sim p^{1/100}$, so in particular the multiplicative group ${\bf F}\_p^\times$ contains a subgroup $H$ of order $d$. Now let $ABC$ be any triangle in the plane ${\bf P F}\_p^2$ and consider the set of all points $E$ on the line $AC$ such that the ratio $CE/EA$ (defined in the obvious fashion) lies in $H$, together with the set of all points $D$ on the line $CB$ such that $BD/DC$ lies in $H$, and the set of all points $F$ on the line $AB$ such that
$-AF/FB$ lies in $H$. Menelaus's theorem tells us that this is a Sylvester-Gallai configuration of order $3d \sim p^{1/100}$ which is not all contained in one line. It contains some points at infinity, but one can apply a generic projective transformation to create an affine configuration (or alternatively one can scale the three copies of $H$ here appropriately).
This "Menelaus configuration" also basically appears (in the context of the p-adics) in [this answer](https://mathoverflow.net/a/258473/766) of David Speyer to a related MathOverflow question; Ben Green and I also encountered similar configurations in [our paper on this topic](https://arxiv.org/abs/1208.4714). In the spirit of that paper (which conveys the moral that sets with few ordinary lines tend to arise from cubic curves, although we could only make this heuristic rigorous over the reals), one could imagine that these Menelaus configurations are essentially the *only* Sylvester-Gallai configurations in this regime (the other cubic curves beyond triangles not giving good examples due to the presence of tangent lines, or (in the case of three concurrent lines) because the underlying group is of prime order), though that would seem rather beyond the reach of our current technology to prove. (In high dimensions, there are however some results constraining the rank of Sylvester-Gallai configurations over finite fields; see Section 5.3 of [this survey of Dvir](https://arxiv.org/pdf/1208.5073.pdf).)
| 12 | https://mathoverflow.net/users/766 | 430748 | 174,462 |
https://mathoverflow.net/questions/430744 | 4 | In 1979, Hakimi and Schmeichel [1] initiated such a study by determining the maximum number of triangles and 4-cycles possible in an $n$-vertex planar graph (see also [2] for a small correction).
* [1] S. Hakimi, E. Schmeichel. On the Number of Cycles of Length k in a Maximal Planar Graph. J. Graph Theory 3 (1979): 69–86.
* [2] A. Alameddine. On the Number of Cycles of Length 4 in a Maximal Planar Graph. J. GraphTheory, 4(1980): 417–422.
In 2019, Győri et al. [3] studied the maximum number of pentagons on planar graphs.
* [3] Győri E, Paulos A, Salia N, et al. The maximum number of pentagons in a planar graph[J]. arXiv preprint arXiv:1909.13532, 2019.
I have a problem about whether there is a similar result for planar bipartite graphs. For example, an **upper bound** on the number of **4 cycles** (or any other **even cycles**). I haven't found it yet. If not, it is not clear what the difficulty is.
I once asked a similar question at Math. Stack and did not seem to see any results.
* [The upper bound on the number of four cycles of 3-connected quadrangulation graphs.](https://math.stackexchange.com/questions/4423841/the-upper-bound-on-the-number-of-four-cycles-of-3-connected-quadrangulation-grap)
| https://mathoverflow.net/users/171032 | Is there any study on the bounds on the number of even cycles for planar bipartite graphs? | Every $n$-vertex planar graph has at most $O(n^k)$ copies of $C\_{2k}$. Note that the bipartite assumption is not needed. A more general result is proven in my paper [Subgraph densities in a surface](https://www.cambridge.org/core/journals/combinatorics-probability-and-computing/article/subgraph-densities-in-a-surface/D2A8D2326DB06835D536A303DAE30338) with Gwenaël Joret and David Wood, where we prove that the same bound holds for graphs embeddable in any fixed surface. In fact, we determine the maximum number of copies of $H$ in an $n$-vertex graph embeddable in a surface of Euler genus $g$, for *every* fixed graph $H$ (up to a multiplicative constant).
There is a matching lowerbound, and in the case of an even cycle, the construction is bipartite. Thus, the answer to your question is $\Theta(n^k)$ for $C\_{2k}$. For the construction, take $C\_{2k}$ and blow-up every other vertex into a stable set of size around $n/k$. This is a bipartite planar graph with $\Omega(n^k)$ copies of $C\_{2k}$.
| 2 | https://mathoverflow.net/users/2233 | 430764 | 174,467 |
https://mathoverflow.net/questions/430749 | 3 | I'm helping in the translation of an article on code theory and I got the following: $t$-frameproof and $t$-wise intersecting, but I don't know what its correct translation into Spanish would be. Can you help me please?
the definition goes like this:
>
> we say that a code $\mathcal{C}$ is $t$-frameproof, or $t$-wise intersecting, if the supports of any $t$
> nonzero codewords have a nonempty common intersection.
>
>
>
| https://mathoverflow.net/users/218991 | What would be the correct Spanish translation? | These articles on encoding [[1](https://upcommons.upc.edu/bitstream/handle/2117/193058/Fernandez-Soriano-Domingo-Seb%C3%A9.pdf?sequence=1&isAllowed=y) and [2](https://web.ua.es/en/recsi2014/documentos/papers/codigos-con-propiedades-de-localizacion-basados-en-matrices-de-bajo-sesgo.pdf)] in Spanish gives some suggestions for the translation of technical words in that field. Many words are kept in the original English.
The word "frameproof" is translated as "código a prueba de incriminaciones", abbreviated CI, so that $t$-frameproof could be rendered as $t$-CI (or you might just write $t$-frameproof and note the Spanish translation of "frameproof").
A $t$-wise intersection is "una intersección $t$-tuplas".
| 1 | https://mathoverflow.net/users/11260 | 430765 | 174,468 |
https://mathoverflow.net/questions/430775 | 3 | A graph is [planar](https://en.wikipedia.org/wiki/Planar_graph) if it can be drawn on the plane in such a way that its edges do not cross each other.
A graph is [$k$-planar](https://en.wikipedia.org/wiki/1-planar_graph) if it can be drawn on the plane in such a way that each of its edges is crossed at most $k$ times.
There is also a concept of [almost planar graphs](https://mathworld.wolfram.com/AlmostPlanarGraph.html) that relies on edge deletions or contractions.
I am dealing with another kind of graphs: they can be drawn in the plane in such a way that at most $k$ pairs of edges intersect.
Is there a terminology for such graphs?
| https://mathoverflow.net/users/158328 | Planar graphs - more or less | These are the graphs with *pairwise crossing number* or *pair-crossing number* at most $k$. Note that it is an open problem whether the pair-crossing number is actually equal to the usual [crossing number](https://en.wikipedia.org/wiki/Crossing_number_(graph_theory)) of a graph. See [Crossing number, pair-crossing number, and expansion](https://www.sciencedirect.com/science/article/pii/S0095895604000437) and [The Graph Crossing Number and its Variants: A Survey](https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS21) for more info.
| 5 | https://mathoverflow.net/users/2233 | 430777 | 174,470 |
https://mathoverflow.net/questions/430785 | 6 | It is "well-known" (e.g. stated [here](https://math.stackexchange.com/a/56117/27659) without proof and sketched [here](https://math.stackexchange.com/a/123660/27659)) that $\mathrm{Z\_2}$ proves $\mathrm{Con(PA)}$ using the "usual" model-theoretic proof, that is one can build a notion of model $\mathbb{N}$ and satisfaction $\models$ such that $\mathrm{Z\_2}$ proves $\mathbb{N}\models\mathrm{PA}$.
I'm struggling to reproduce the proof, with the main difficulty being defining the satisfaction relation $\models$; the issue being how to define it so that for any (code) of a predicate $\phi\rightarrow\psi$, one has
$$\mathbb{N}\models \phi\rightarrow\psi\quad \mbox{iff}\quad \mathbb{N}\models\phi\ \mbox{implies}\ \mathbb{N}\models\psi $$
which must use the fact that $\phi$ and $\psi$ are structurally smaller than $\phi\rightarrow\psi$.
Does anyone know a detailed reference for this folklore result?
| https://mathoverflow.net/users/36103 | How does one prove the consistency of $\mathrm{PA}$ in $\mathrm{Z_2}$? | I don't know a detailed reference, but here's a more detailed proof sketch.
The key point is that the "naive" understanding of $\models$ admits a $\Sigma^1\_1$ definition, which $\mathsf{Z\_2}$ can unproblematically implement. (There are various ways to sharpen this of course.)
One way to do this is via **Skolemization**: given a first-order-arithmetic sentence $\varphi$ in prenex normal form, let $\hat{\varphi}$ be its Skolemization. This $\hat{\varphi}$ has the form $\exists \overline{F}\forall\overline{x}\theta$ for some tuple of function **symbols** $\overline{F}$, some tuple of variables $\overline{x}$, and some first-order quantifier-free formula $\theta$ in the language of arithmetic + the $\overline{F}$s. The transition from $\varphi$ to $\hat{\varphi}$ is appropriately simple, and the property "The tuple of **functions** $\overline{f}$ witness $\hat{\varphi}$ in $\mathbb{N}$" is expressible by a universal formula in the language of $\mathsf{Z}\_2$, so we get the desired $\Sigma^1\_1$ definition.
(Another approach, which I actually prefer since for more complicated structures than $\mathbb{N}$ it avoids choice, is via *trees*. Basically, each $\varphi$ has a corresponding "big syntax tree" $T\_\varphi$, with the truth of $\varphi$ in $\mathbb{N}$ corresponding to the existence of a subtree of $T\_\varphi$ satisfying a certain $\Pi^0\_1$ property. Again, we get a $\Sigma^1\_1$ definition. Here the syntactic structure of $\varphi$ is more visible: it corresponds roughly to the height of $T\_\varphi$. For example, if $\varphi\equiv \theta\vee\psi$ - and re: your quesiton, note that implications are just disjunctions in classical logic - then $T\_\varphi$ consists of a root node labelled "$\vee$," copy of $T\_\theta$ on the "left" of the root, and a copy of $T\_\psi$ on the "right" of the root.)
Call this $\Sigma^1\_1$ property "witnessability." We can prove in $\mathsf{RCA\_0}$ alone that for every sentence $\varphi$ *at most one of* $\varphi$ and $\neg\varphi$ is witnessable, and that the inference rules of first-order logic preserve witnessability. More substantively, we can prove in $\mathsf{RCA\_0}$ + "For all sets $A$ and numbers $n$, the $n$th jump of $A$ exists" (this is actually a bit stronger than $\mathsf{ACA\_0}$) that each $\mathsf{PA}$-axiom is witnessable and that if $\varphi$ is not witnessable then $\neg\varphi$ is. Finally, using $\Sigma^1\_1$ comprehension you can prove that the set of witnessable sentences exists, and thus we have a negation-complete consistent theory extending $\mathsf{PA}$, so $\mathsf{PA}$ is consistent.
(In fact, the appeal to $\Sigma^1\_1$ comprehension is totally unnecessary here, but I think it helps make things more intuitive.)
| 4 | https://mathoverflow.net/users/8133 | 430786 | 174,471 |
https://mathoverflow.net/questions/430759 | 5 | Pythagorean number of a totally real field $\mathbb{K}$ is the minimal number $N$ of
squares $t\_k^2$ required to represent a totally positive $0\leq x\in \mathbb{K}$ as $x=\sum\_{k=1}^N t\_k^2$, where $t\_k\in\mathbb{K}$, $1\leq k\leq N$. E.g. for $\mathbb{K=Q}$ one has $N=4$, by [Lagrange 4-squares theorem](https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem), and
for $\mathbb{K=R}$ one has $N=1$.
What is known about Pythagorean numbers of real cyclotomic fields, i.e. fields $\mathbb{Q}[\zeta\_n+\zeta\_n^{-1}]$, with $\zeta\_n$ a primitive $n$-th root of unity?
EDIT: $N\leq 4$ by Hilbert-Landau-Siegel Theorem (page 78 in Hilbert's [book](https://math.berkeley.edu/%7Ewodzicki/160/Hilbert.pdf)), see also [sum of squares in ring of integers](https://mathoverflow.net/q/14456). So the question is whether $N=4$ is the case for every $n$, or we can do better.
---
NB. Not all elements $x>0$ in $\mathbb{K}$ need be totally positive, so the question is about only these $x$ which are totally positive.
| https://mathoverflow.net/users/11100 | Pythagorean numbers of real cyclotomic fields | **Content warning**: This answer assumes knowledge of Hasse-Minkowski and mild (local) class field theory. The latter could be eliminated with some more explicit computations in local fields. References to all required facts can be found in Milne's notes or Cassels-Frohlich, as well as Serre's a course in arithmetic for the basics of quadratic forms. (Updated to correct the case of $d=2$ but the final answer remains the same.)
Start with the case of finite extensions $K$ of $\mathbf{Q}\_p$.
**Claim:** All elements of $K$ are sums of two squares if and only if $i \in K$. If $p = 2$, all elements of $K$ are sums of three squares if and only if $[K:\mathbf{Q}\_2]$ is even.
**Proof:** If $i \in K$ then
$$n = \left(\frac{n+1}{2}\right)^2 + \left(i \cdot \frac{n-1}{2}\right)^2.$$
If $i \notin K$ and $L = K(i)$ then $L/K$ is cyclic of order two. Thus $K^{\times}/N\_{L/K}(L^{\times})$ also has order two by local class field theory. But the norms are percisely the elements which are sums of two squares, so "half" the elements of $K^{\times}$ are not sums of two squares.
Let $D/\mathbf{Q}\_2$ denote the unique non-split quaternion algebra. This is just the Hamilton quaternions over $\mathbf{Q}\_2$ and corresponds to the Hilbert symbol $(-1,-1)\_{\mathbf{Q}\_2}$. That is, $D = \mathbf{Q}\_2[i,j]$ with $i^2=j^2=-1$ and $ij = -j i$. By local class field theory, $D$ splits over $K$ if and only if $[K:\mathbf{Q}\_2]$ is even. If $D$ splits over $K$ then the Hilbert symbol $(-1,-1)\_{K}$ is trivial and there exists $x,y \in K$ with
$$-1 = x^2 + y^2.$$
But now
$$n = \left(\frac{n+1}{2}\right)^2 + \left(x \cdot \frac{n-1}{2}\right)^2
+ \left(y \cdot \frac{n-1}{2}\right)^2$$
is the sum of three squares. Conversely, if every element of $K$ is the sum of three squares then $-1 = x^2 + y^2 + z^2$ and the quaternion $1 + xi+ yj+zij \in D\_K$ has norm zero, which implies that $D\_K$ has zero divisors and thus splits. This completes the proof.
If $F/\mathbf{Q}$ is any global field, then $\alpha \in F$ is a sum of $d$ squares if and only if
$$x^2\_1 + x^2\_2 + \ldots + x^2\_d - \alpha x^2\_{d+1} = 0$$
has a non-trivial solution. By Hasse-Minkowski, this can be checked locally at each place.
1. If $\alpha \in F$ is totally positive then it has a real solution as soon as $d \ge 1$.
2. If $v$ is a finite place of odd residue characteristic then it has a solution as soon as $d \ge 3$ because $x^2\_1+x^2\_2+x^2\_3=0$ already has a non-trivial solution since it is a smooth conic and thus it is isomorphic to $\mathbf{P}^1$ over $F\_v$.
3. Let $v$ have residue characteristic $2$. There is a non-trivial solution as soon as $d \ge 4$ for any $\alpha$ over a local field. For $d=2$ or $d=3$, we can determine whether it has a solution by the local result above. (By weak approximation if $F\_v = K$ and not every element of $K$ is a sum of $d$ squares then not every element of $F$ is a sum of $d$ squares; being a sum of $n$ squares in $K$ is an open condition.)
So the answer for a general number field $F$ is:
1. $d = 2$ if and only if the localizations $F\_v$ for all $v$ contain a square root of $-1$. But this implies that the only primes in $F$ which can split completely are $1 \bmod 4$ and so by Cebotarev $F$ contains $\mathbf{Q}(i)$. If $i \in F$ then every element is a sum of two squares.
2. $d \le 3$ if and only if the localizations $F\_v$ for all $v$ of residue characteristic $2$ have even degree over $\mathbf{Q}\_2$.
3. $d \le 4$.
For example, given a quadratic field $F = \mathbf{Q}(\sqrt{D})$ of discriminant $D$, then $d = 2$ if $D = -4$, and $d \le 3$ if $D \not\equiv 1 \bmod 8$.
For the specific fields $F = \mathbf{Q}(\zeta + \zeta^{-1})$ where $\zeta$ is a primitive root of unity of order $n$, write $n = 2^k m$. The Galois group of $\mathbf{Q}(\zeta)$ is
$$(\mathbf{Z}/2^k \mathbf{Z})^{\times} \oplus (\mathbf{Z}/m \mathbf{Z})^{\times},$$
the Galois group of $F$ is the quotient of this group by $(-1,-1)$. Tthe decomposition group $C$ of $v|2$ in $\mathbf{Q}(\zeta)$ is
$$C:=(\mathbf{Z}/2^k \mathbf{Z})^{\times} \oplus \langle 2 \rangle,$$
and the decomposition group $D$ of $v|2$ in $F$ is $C/C \cap \langle -1,-1 \rangle$. Certainly $i \notin F$ so $d > 2$. Hence $3$ squares are always required for these fields $F$ for any $n$. To see when $d=3$ it suffices to compute whether the decomposition group has even order or not. We find:
1. If $k=0$ or $k=1$, then the order of $D$ is the order of $2 \bmod n$. Hence $d=3$ when $2$ either has order divisible by $4$, or has order $2r$ with $r$ odd and in addition $2^r \not\equiv -1 \bmod n$.
2. If $k=2$ and $2$ has even order modulo $n$ then $4 | |D|$ and so $2 | |C|$. If $2$ has odd order modulo $n$ then $(-1,-1)$ does not lie in $C$, and so the quotient $|D|$ still has even order. Thus $d = 3$ in all cases.
3. If $k \ge 3$ then $4 | |C|$ and the decomposition group of $v|2$ of $F$ always has even order, so $d=3$.
For example, if $n=p > 2$ is prime, then $d=3$ if and only if the multiplicative order of $2 \bmod p$ is divisible by $4$, otherwise $d=4$. The few $n$ for which $d=3$ are $n = 5, 8, 12, 13, \ldots $
| 6 | https://mathoverflow.net/users/488927 | 430788 | 174,473 |
https://mathoverflow.net/questions/430771 | 5 | $\DeclareMathOperator{\Res}{Res}
\DeclareMathOperator{\Cor}{Cor}$
This question was [asked in MSE](https://math.stackexchange.com/q/4533187/37763).
It got no answers or comments, and so I post it here.
Let $H$ be a subgroup of a finite group $G$, and let $M$ be a $G$-module,
that is, an abelian group on which $G$ acts. Write $n=[G:H]$.
Consider the restriction and corestriction homomorphisms in Tate cohomology:
\begin{align\*}
\Res\colon\, &H^{-1}(G,M) \to H^{-1}(H,M),\\
\Cor\colon\, &H^{-1}(H,M) \to H^{-1}(G,M).
\end{align\*}
Since
$$ \Cor \circ \Res =n,$$
we know that for $\xi \in H^{-1}(G, M)$,
$$\text{if}\ \, \Res \xi=0,\ \text{then}\ \, n\xi=0.$$
>
> **Question:**
> What is an example of $(G,H,M,\xi)$ such that
> $$n\xi=0,\ \ \text{but}\ \ \Res \xi \neq 0\ ?$$
>
>
>
| https://mathoverflow.net/users/4149 | Restriction vs. multiplication by $n$ in Tate cohomology | Let $G = \Bbb Z/4$, acting on the Gaussian integers $M = \Bbb Z[i]$ via multiplication by $i$. The transfer $M\_G \to M^G$ is given by multiplication by $1 + i + (-1) + (-i) = 0$, so
$$H^{-1}(G,M) = M\_G = \Bbb Z[i] / (1-i) \cong \Bbb Z/2.$$
In particular, all elements are 2-torsion.
Similarly, if $H$ is the index-2 subgroup $\Bbb Z/2$, then
$$H^{-1} = M\_H = \Bbb Z[i] / (2).$$
Under this identification, the restriction $\Bbb Z[i]/(1-i) \to \Bbb Z[i]/(2)$ is multiplication by $(1+i)$. The restriction is therefore injective.
| 7 | https://mathoverflow.net/users/360 | 430789 | 174,474 |
https://mathoverflow.net/questions/430781 | 1 | Suppose $x\_i$ come from 2-d standard Normal centered at 0. What is the range of $a$ for which the following iteration converges almost surely?
$$w\_{i+1} = w\_i-a x\_i \langle w\_i, x\_i \rangle$$
For 1-d, one can write $w\_{i+1}^2$ as $w\_i^2=w\_0^2\prod\_i (1-\alpha x\_i^2)$, take log and apply central limit theorem. For 2-D, one could use a [similar approach](https://stats.stackexchange.com/a/586427/511), but using matrix logarithm instead of regular logarithm. However, this side-steps some technical details (we have a product of rank-1 matrices, how can you take matrix logarithm of that?) so I'm wondering if the resulting estimate is valid.
In particular, it predicts that iteration converges iff $\alpha \in (0.12131, 0.91861)$, while in [simulations](https://www.wolframcloud.com/obj/yaroslavvb/newton/forum-largest-a.nb), I'm seeing convergence for $\alpha=0.93$ (but divergence for $\alpha=0.94$)
The question is equivalent to finding $\alpha$ such that $C\_i$ converges almost surely where
$$C\_i=w\_i w\_i^T$$
$$C\_{i+1}=(1-\alpha x\_i x\_i^T)C\_i(1-\alpha x\_i x\_i^T)$$
| https://mathoverflow.net/users/7655 | Range of $a$ such that $w \leftarrow w-a x \langle w, x \rangle$ converges almost surely? | As you say, the threshold is around $a=0.937087$.
The exact condition for convergence is
$$
\int\_{0}^{2\pi}\int\_0^\infty\log(1-2ar^2\cos^2\theta+a^2r^4\cos^2\theta)re^{-r^2/2}\,dr\,d\theta<0.
$$
I don't think there will be a clean formula for $a$.
The derivation of the expression is as follows:
Let $r\_j=\|w\_j\|$. The entire problem is rotationally symmetric. Additionally, the operation is linear. Combining these two facts, we have $r\_{j+1}=Z\_jr\_j$ where $(Z\_j)$ is an i.i.d sequence of random variables. Taking logarithms, we have that $r\_j\to 0$ if $\mathbb E \log Z<0$ and $r\_j\to\infty$ if $\mathbb E \log Z>0$.
To compute $\mathbb E\log Z$, suppose without loss of generality that $w\_j=(1,0)$. Write $x\_{j+1}=(r\cos\theta,r\sin\theta)$ where $r$ is distributed as a Rayleigh random variable and $\theta$ is uniform on $[0,2\pi]$. Then
$$
w\_{j+1}=(1,0)-a(r\cos\theta,r\sin\theta)r\cos\theta,
$$
so that $Z\_{j+1}=\|w\_{j+1}\|$. The expression for $\mathbb E\log Z\_{j+1}$ is then as displayed in the first display equation.
| 5 | https://mathoverflow.net/users/11054 | 430790 | 174,475 |
https://mathoverflow.net/questions/430756 | 1 | Let $S$ be a reducible compact complex analytic space, thus we have the decomposition $S=\bigcup\_{i=1}^n {V\_i}$ where $V\_i$ is the irreducible component of $S$. Let $L$ be a line bundle on $S$, I wonder if we can give some descriptions about $H^0(S,L)$ and $H^0(V\_i,L)$.
I have seen this question [Globally generated line bundle on reducible curve](https://math.stackexchange.com/questions/3375341/globally-generated-line-bundle-on-reducible-curve "Globally generated line bundle on reducible curve"). For the special case that a reducible variety with two components $Y$ and $Z$ intersecting at a simple node $p$, then one can say the global sections of $L$ is the codimension one subspace of $H^0(L|\_Y) \oplus H^0(L|\_Z)$ consisting of sections with the same value at $p$. But this case is too peculiar.
So, in general case, can we say something about $H^0(S,L)$ and $H^0(V\_i,L)$ or the relationships of their dimensions? One motivation lies in the following theorem by Grauert.
>
> Let $f: X \rightarrow Y$ be a proper morphism of complex spaces and $\mathcal{S}$ a coherent analytic sheaf on $X$ which is flat with respect to $Y$ (or $f$), which means that the $\mathcal{O}\_{f(x)}$-modules $\mathcal{S}\_{x}$ are flat for all $x \in X .$ Set $\mathcal{S}(y)$ as the analytic inverse image with respect to the embedding $X\_{y}$ of in $X$. Then for any integers $i, d \geq 0$, the set
> $$
> \left\{y \in Y \,\, |\,\, \dim\_{\mathbb{C}} H^{i}\left(X\_{y}, \mathcal{S}(y)\right) \geq d\right\}
> $$
> is an analytic subset of $Y$.
>
>
>
The fibre $X\_y$ surely can be reducible, we write the decomposition $X\_y=\bigcup\_{i=1}^n {V\_i}$. So, now if we know the dimension of $H^{i}\left(X\_{y}, \mathcal{S}(y)\right) $, can we infer whether there exists some $i$, such that we can say something about $H^{i}\left(V\_i, \mathcal{S}|{V\_i}\right)? $
| https://mathoverflow.net/users/141609 | Global sections of a line bundle on a reducible complex space | In any case $H^0(S,L)$ is a subspace in $\oplus H^0(V\_i,L)$, but the way it sits there depends on the way the components $V\_i$ are glued together. In the simplest case where they for a simple normal crossing configuration, there is an exact sequence
$$
0 \to H^0(S,L) \to
\bigoplus H^0(V\_i,L) \to
\bigoplus\_{i < j} H^0(V\_i \cap V\_j,L),
$$
where the last map is given by the differences of the restriction maps.
| 1 | https://mathoverflow.net/users/4428 | 430792 | 174,476 |
https://mathoverflow.net/questions/430796 | 4 | Let $[n] := \{1,\dots,n\}$, for some large integer $n$, and let $\mathcal{F}$ be a family of 2-element subsets of $[n]$.
The famous [Erdös-Ko-Rado (EKR) theorem](https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Ko%E2%80%93Rado_theorem) says that if $|\mathcal{F}| > {n - 1 \choose 1} = n-1$, then $\mathcal{F}$ must contain (at least) two disjoint subsets.
**Question:**
* Assuming that $|\mathcal{F}|$ is significantly larger than the bound required by the EKR theorem, is there a result that gives a lower bound on the number of *pairwise* disjoint 2-element subsets that $\mathcal{F}$ must contain?
* More generally, does there exist such a lower bound on the number of disjoint $k$-subsets if we consider families of $k$-element subsets that are much larger than ${n -1 \choose k-1}$?
| https://mathoverflow.net/users/491512 | On the number of disjoint subsets of a large set families | Your first question is simply asking what is the minimum number of edges an $n$-vertex graph must have to force a [matching](https://en.wikipedia.org/wiki/Matching_(graph_theory)) of size $m$. This number was determined exactly by a classic result of [Erdős and Gallai](https://citeseerx.ist.psu.edu/viewdoc/summary;jsessionid=1D0279A99A414F91F8003CD4304965E9?doi=10.1.1.210.7468). They proved that if the maximum size of a matching of an $n$-vertex graph $G$ is $m$, then $G$ has at most $\max\{\binom{2m+1}{2}, \binom{m}{2}+(n-m)m\}$ edges. Moreover, this bound is tight for all $n,m \geq 1$.
Your second question was asked by Erdős in 1965 and has received considerable attention. I believe the state of the art is the paper [The Erdős Matching Conjecture and concentration inequalities](https://arxiv.org/abs/1806.08855) by Frankl and Kupavskii. See the references therein for more information.
| 1 | https://mathoverflow.net/users/2233 | 430808 | 174,478 |
https://mathoverflow.net/questions/430809 | 3 | This question is similar to [a question I asked last year](https://mathoverflow.net/questions/402333), but I'm not asking for the same thing.
---
Let $S$ be a set of ordinals, and consider the Levy collapse $\operatorname{Col}(\omega,S)$. Let $G$ be $\operatorname{Col}(\omega,S)$-generic. Must every unbounded subset of $\omega$ in $V[G]$ contain an unbounded further subset in the ground model $V$? I believe the answer is yes, with a proof similar to the linked question, but I am unsure (the proof cannot be brought over verbatim).
I am also interested in the more general question: Is it possible to characterise posets $\mathbb{P}$ such that for all filters $G$ which are $\mathbb{P}$-generic, every subset of $\omega$ in $V[G]$ has an unbounded further subset in $V$?
| https://mathoverflow.net/users/146831 | Unbounded subset of $\omega$ in $V[G]$ has an unbounded subset in $V$? | No, this does not hold.
Given an infinite set $A$, let $\hat{A}=\{\langle n, a\_1,...,a\_n\rangle: n\in\mathbb{N}\}$ where $\langle\rangle$ is an appropriate "tupling" operation and $a\_i$ is the $i$th smallest element of $A$. Note that any infinite subset of $\hat{A}$ lets us reconstruct all of $\hat{A}$ itself. This means in particular that if $X\in\mathcal{P}^{V[G]}(\omega)\setminus\mathcal{P}^V(\omega)$ then $\hat{X}$ is an infinite set with no infinite subset in $V$. So indeed **any** forcing which adds a real - including $Col(\omega,S)$ whenever $S$ is uncountable - will add infinite subsets of $\omega$ not containing any infinite subset from the ground model.
In fact, an even stronger phenomenon can occur. Identify an (infinite) set of natural numbers $A$ with its **principal function** $p\_A$ which on input $i$ gives the $i$th smallest element of $A$. Note that if $A\subseteq B$ then $p\_A$ grows at least as fast as $p\_B$. Now $Col(\omega,2^{\aleph\_0})$ adds a function dominating all ground model functions, which is a massive strengthening of "adding a real with no infinite ground model subset." This raises an interesting follow-up question: must $Col(\omega,\omega\_1)$ add a function not escaped by any function in the ground model? This takes us into the realm of **cardinal chracteristics of the continuum**, and see in particular [this old question of mine](https://mathoverflow.net/questions/77891/can-we-collapse-omega-1-to-omega-without-adding-a-dominating-real).
| 5 | https://mathoverflow.net/users/8133 | 430810 | 174,479 |
https://mathoverflow.net/questions/430800 | 9 | For positive integer $n$ the following value of a hypergeometric function
$$\_4F\_3(n,n,n,2n,1+n,1+n,1+n,-1)$$
based on the first few terms looks like
$$ R\_1(n) + R\_2(n) \pi^2$$
where $R\_{1,2}(n)$ are rational numbers. Is there a way to express $R\_{1,2}(n)$ in a closed form using “simple” operations like power, ratio, factorial, etc?
| https://mathoverflow.net/users/41312 | Closed form for ₄F₃(n,n,n,2n;1+n,1+n,1+n;−1) | Fiddling with Maple, I get: if $n$ is a positive real number, then
$$
{{\_4\mathrm F\_3}(n,n,n,2\,n;\,n+1,n+1,n+1;\,-1)}={\frac {{n}^{2}
\sqrt {\pi}\,\Gamma(n+1)\,\psi^{(1)}(n)}{{4}^{n}\,\Gamma \left( n
+{\frac{1}{2}}\right)}. }
$$
Here, $\psi^{(1)}$ is the [trigamma function](https://en.wikipedia.org/wiki/Trigamma_function),
$\psi^{(1)}(z) = \frac{d^2}{dz^2}\log(\Gamma(z))$.
---
It seems, for positive integer $n$, we have that $\psi^{(1)}(n)$ has the form $\pi^2/6 - \text{rational}$.
Francois pointed out the recurrence, so that
$$
\psi^{(1)}(n) = \frac{\pi^2}{6}-\sum\_{k=1}^{n-1}\frac{1}{k^2}
$$
| 13 | https://mathoverflow.net/users/454 | 430811 | 174,480 |
https://mathoverflow.net/questions/430797 | 5 | Let $P \in \mathbb Z[X]$ be monic, separable, of degree $d$, $K$ its splitting field over $\mathbb Q$ and $G$ the Galois group of $K$ over $\mathbb Q$.
Now, let $p$ be a prime number unramified in $K$. If $P \text{ mod } p$ factorizes as a product of $n\_1$ linear factors, $n\_2$ quadratic irreducible factors, $\dots, n\_d$ irreducible factors of degree $d$, then we know that $G$ contains a permutation of cycle type $(n\_1, n\_2, \dots, n\_d)$ by lifting the Frobenius automorphism of the extension $(\mathcal O\_K/\mathfrak p)/\mathbb F\_p$ to $G$, where $\mathfrak p$ is any prime ideal of $\mathcal O\_K$ above $p$.
My question is : what happens at ramified primes $p$ ? What kind of cycle type can we get from factorizations of $P \text{ mod } p$ with multiplicities ?
I know every lift of the Frobenius is of the form $\sigma \tau$ where $\sigma$ is a fixed lift and $\tau \in I\_p$, the inertia subgroup at $p$, but that doesn't tell me much about its cycle type. I don't know what to expect from an irreducible factor of degree $k$ with exponent $m$ in $P \text{ mod } p$. Does it imply there is a permutation with a corresponding sequence of $m$ distinct $k$-cycles ?
| https://mathoverflow.net/users/133679 | Cycle type in Galois group from ramified primes | It does *not* imply there is such a permutation in the Galois group, and your question has an interesting history.
First of all, I would say the question you ask is arguably the wrong one: the more natural object to focus on is a prime ideal factorization of $p$, not a factorization of a polynomial mod $p$.
Let $F = \mathbf Q(\alpha)$ where $\alpha$ has minimal polynomial $\varphi(x)$ over $\mathbf Q$ of degree $n$ and $L$ is the Galois closure of $F$ over $\mathbf Q$. For a prime number $p$, let the ideal
$p\mathcal O\_F$ have $g$ different prime ideal factors with ramification indices and residue field degrees $e\_i$ and $f\_i$ for $i = 1, \ldots, g$. Frobenius asked if there is an element of ${\rm Gal}(L/\mathbf Q)$ that permutes the $n$ roots of $\varphi(x)$ with $e\_i$ disjoint cycles of length $f\_i$ for $i = 1, \ldots, g$. This is a better way to pose the question you're asking. Frobenius (and Dedekind) knew the answer is yes when $p$ is unramified in $F$. What if $p$ is ramified in $F$? Frobenius conjectured in 1882 that the answer is still yes.
However, there are counterexamples when $p$ is ramified. None were known until the 1990s, when Thomas Hawkins asked Jean-Pierre Serre about the conjecture of Frobenius. (The 100+ year wait for counterexamples was simply because the conjecture had been forgotten, not that it really needed 100 years of progress in math.) Serre provided counterexamples, first with $n = 10$ and later with $n = 6$. This is discussed in Hawkins' book *The Mathematics of Frobenius in Context*. The conjecture is Conjecture 9.17 on page 323 and the counterexample of degree $6$ is built up in Section 9.3.4. I'll now show you what that counterexample is.
Take $\varphi(x) = x^6 - 3x^5 - 5x^4 + 15x^3 - 12x^2 + 4x-4$, which is irreducible over $\mathbf Q$ with Galois group over $\mathbf Q$ being isomorphic to $S\_4$. (In the book, Hawkins gives some motivation for this: Serre did not just create it out of nothing.) The Galois group acts on the $6$ roots of $\varphi(x)$ as even permutations. For a root $\alpha$ of $\varphi(x)$, $F := \mathbf Q(\alpha)$ has degree $6$ over $\mathbf Q$ and $3\mathcal O\_F = \mathfrak p^2\mathfrak q$, where $\mathfrak p$ and $\mathfrak q$ have residue field degree $2$ ($6 = 2 \cdot 2 + 1 \cdot 2$). But no element of ${\rm Gal}(L/\mathbf Q)$ acts on the $6$ roots as a permutation with cycle type $(2,2,2)$ since such permutations are odd permutations of the roots.
In this example, $\varphi(x) \equiv x^4 (x^2+2x+2) \bmod 3$, so the way $3\mathcal O\_F$ factors is not reflected in the way $\varphi(x) \bmod 3$ factors. That is no surprise: the Dedekind-Kummer theorem relating the factorization of a polynomial mod $p$ and the prime ideal factorization of $p\mathcal O\_F$ need not be valid at primes dividing the discriminant of the polynomial. (Note in the example above that $\mathcal O\_F \not= \mathbf Z[\alpha]$.)
In the MAA review of Hawkins' book by David Roberts [here](https://www.maa.org/press/maa-reviews/the-mathematics-of-frobenius-in-context-a-journey-through-18th-to-20th-century-mathematics), a different counterexample of degree $6$ is given: $\Phi(x) = x^6−6x^4+6x^2−6+2$. Like $\varphi(x)$, $\Phi(x)$ is irreducible over $\mathbf Q$ and its Galois group over $\mathbf Q$ is $S\_4$ acting as even permutations of the $6$ roots of $\Phi(x)$. When $F = \mathbf Q(\alpha)$ for $\alpha$ a root of $\Phi(x)$, $5$ is ramified with decomposition $\mathfrak p^2\mathfrak q$ where the prime ideals have residue field degree $2$. As in the previous example, no element of the Galois group of $\Phi(x)$ can act on the roots of $\Phi(x)$ as a permutation with cycle type $(2,2,2)$ since that would be an odd permutation. This time the gods are shining down on us, since the factorization of $\Phi(x) \bmod 5$ matches the factorization of $5\mathcal O\_F$: $\Phi(x) \equiv (x^2+x+2)^2(x^2+3x+3) \bmod 5$. So this is a counterexample to the exact question you asked even though I first said it was not really the right way to ask the question.
Ultimately this false conjecture of Frobenius is not important, since the way we really use results like this (turning the shape of a prime ideal factorization into the shape of a permutation on the roots in a Galois group) is in Chebotarev-type applications, for which a finite set like the set of ramified primes is negligible. Still, it is a question you can ask and it's nice to have it settled.
| 10 | https://mathoverflow.net/users/3272 | 430814 | 174,481 |
https://mathoverflow.net/questions/430701 | 1 | Let $(X,\tau,d)$ be a space where $\tau$ is a topology and $d$ is a metric, where the topology $\tau$ is not necessarily compatible with $d$.
Is there a canonical name for such a structure (maybe under some extra assumption on the relation between the metric and the topology)?
For example, assuming that all closed balls $\overline{B}(x,\epsilon)=\{y\in X\mid d(x,y)\leq \epsilon\}$ define with the metric are also closed in the topology $(X,\tau)$.
**Edit:** I will give some more context accordingly to the comments, since the question as stated right now might be a bit too general.
Spaces with two topologies $(X,\tau,\sigma)$ are called bitopological spaces, and there is a lot of theory behind them.
I recall from a conversation that if the second topology $\sigma$ comes from a metric $d$ and the metric interacts "nicely" with the topology, this structure $(X,\tau,d)$ goes under a different name and that this notion has a wide literature as well.
I was hoping to get some information about this type of structure.
| https://mathoverflow.net/users/121875 | Name of a space with both a topology and a metric that are not compatible? | A "topometric space" is a triple $(X,\tau,\rho)$ where $\tau$ is a topology on $X$ and $\rho$ is a metric on $X$ which is lower semi-continuous with respect to the topology. These are useful in continuous logic. More details can be found in papers by Ben Yaacov.
| 3 | https://mathoverflow.net/users/128723 | 430820 | 174,482 |
https://mathoverflow.net/questions/430825 | 7 | I need the dimensions of the Atkin-Lehner eigenspace for the paper I'm writing.
As is well known, the cuspidal space $S\_{k}(\Gamma\_{0}(N))$ can be decomposed by Atkin Lehner involution. For example, when $N=21=3\times7$, we have the decomposition
$$
S\_{k}(\Gamma\_{0}(N))=S^{(++)} \oplus S^{(+-)} \oplus S^{(-+)} \oplus S^{(--)}
$$
where $S^{(++)}$, $S^{(+-)}$, $S^{(-+)}$, and $S^{(--)}$ are the subspaces of $S\_{k}(\Gamma\_{0}(N))$ for which pairs of eigenvalues for $W\_3$, $W\_7$ are (+1,+1), (+1,-1), (-1,+1), (-1,-1) respectively.
In addition, dimensions of each summand spaces are 0,0,1,0.
This is from [Table 5 of Antwerp IV](https://wstein.org/Tables/antwerp/table5/).
For k=2, the dimension of each Atkin-Lehner eigenspace has already been given by Table 5 of Antwerp IV and David Kohel.
And for any weight k, I found [the following documentation](https://trac.sagemath.org/ticket/9455).
However, using this, it is difficult for me to find the dimensions of the eigenspaces.
So, is there any idea or data?
| https://mathoverflow.net/users/491555 | How to get the dimension of Atkin-Lehner eigenspace or do you have any data already obtained? | You can compute these dimensions using *modular symbols* (an auxiliary space which has the same Hecke action as modular forms, but is easier to compute). Here's a Sage example for weight 4 cusp forms of level Gamma0(17):
```
sage: S=ModularSymbols(Gamma0(17), weight=4, sign=1).cuspidal_submodule()
sage: S.atkin_lehner_operator().charpoly().factor()
(x + 17) * (x - 17)^3
```
Sage normalises the Atkin–Lehner operator so that its square is multiplication by $(-N)^{k-2}$, since that normalisation works better for odd weights (it avoids introducing square roots). If you want to normalise so that the operator becomes an involution, as is usual for $\Gamma\_0$ levels, you need to scale by $N^{(k - 2) / 2}$. So, in my example, the scaling factor is 17, and the output means that the +1 eigenspace has dimension 3, and the -1 eigenspace has dimension 1.
You can compute the local AL operators for each prime $p \mid N$ similarly, and get the simultaneous eigenspaces by taking intersections of kernels – here's a code snippet which does this:
```
sage: def AL_eigenspace_dims(N, k):
....: facs = [p^r for (p, r) in N.factor()]
....: S = ModularSymbols(Gamma0(N), sign=1, weight=k).cuspidal_submodule()
....: Wmats = [S.atkin_lehner_operator(f).matrix() for f in facs]
....: for s in cartesian_product([ [-1, 1] for f in facs ]):
....: Vs = [(Wmats[i] - s[i]*facs[i]**((k-2)/2)).kernel() for i in range(len(facs))]
....: V = reduce(lambda u,v: u.intersection(v), Vs)
....: print(s, V.dimension())
....:
sage: AL_eigenspace_dims(21, 2)
(-1, -1) 0
(-1, 1) 1
(1, -1) 0
(1, 1) 0
sage: AL_eigenspace_dims(21, 12)
(-1, -1) 7
(-1, 1) 6
(1, -1) 7
(1, 1) 8
```
This modular symbol method will be slow if k or N is large, because it involves linear algebra with matrices having about $kN$ rows and columns. (For $k N$ around 500, it takes roughly ten seconds on my machine). In contrast, the algorithm in Lloyd Kilford's draft code from your link uses the Riemann–Roch theorem, and the running time is dominated by factorising N, so it should be practical for N up to about 100 digits; but his implementation only covers the case of weight 2. In principle it should be possible to extend the Riemann–Roch method to higher weights, but I don't know if anyone has implemented this in Sage.
| 10 | https://mathoverflow.net/users/2481 | 430826 | 174,484 |
https://mathoverflow.net/questions/430836 | 5 | Might there be a research team that has formalised the Riemann Hypothesis? So far I have encountered two related questions:
1. [Is there a formulation of the Riemann Hypothesis in first-order arithmetic?](https://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence)
2. [Can the Riemann Hypothesis be undecidable?](https://mathoverflow.net/questions/79685/can-the-riemann-hypothesis-be-undecidable)
A Google search reveals that the Prime Number Theorem has been formalised in HOL-light
so it is reasonable to infer that the Riemann Hypothesis has been formalised [3]. If so, might there be a publication that analyses the different trade-offs that were involved?
References:
-----------
1. Marc Larsson et al. Coqtail. 2022. Github repository. <https://github.com/coq-community/coqtail-math>
2. Sylvie Boldo, Catherine Lelay, and Guillaume Melquiond. Coqueliquot: A user-friendly Library for Real Analysis for Coq. 2015.
3. John Harrison. HOL light: an overview. 2009. [https://www.cl.cam.ac.uk/~jrh13/slides/tphols-18aug09/slides.pdf](https://www.cl.cam.ac.uk/%7Ejrh13/slides/tphols-18aug09/slides.pdf)
| https://mathoverflow.net/users/56328 | Formalisation of the Riemann Hypothesis | I just learned of the following formalisation led by Brandon Gomes and Alex Kontorovich from Andrej Bauer(via email):
>
> Brandon Gomes & Alex Kontorovich. Formalization of the Riemann
> Hypothesis in the Lean Theorem Prover. Github repository. 2020.
> <https://github.com/bhgomes/lean-riemann-hypothesis>
>
>
>
| 8 | https://mathoverflow.net/users/56328 | 430854 | 174,490 |
https://mathoverflow.net/questions/430835 | 1 | $\DeclareMathOperator\SU{SU}$I am looking for a (generalized) Euler angles decomposition for $\SU(N)\ (N>1)$ in the following fashion:
$$
\SU(N)\ni m = a\, u \, b
$$
where $a,b$ are independent diagonal $\SU(N)$-matrices each of which accounts for $N-1$ parameters
while $u\in \SU(N)$ is parametrized by the remain $(N-1)^2$ parameters. For instance the matrices $u$ might form a $U(N)$-isomorphic subgroup of $\SU(N)$. Notice that in the case $N=2$ this decomposition reduces to the known Euler's one $a(b)= \exp(i \alpha(\beta) \sigma\_3)$ and $U = \exp(i \gamma \sigma\_2)$ where $\sigma\_j$ is a Pauli matrix and $\alpha$, $\beta$, $\gamma$ are the angles.
I found in literature [Bertini, Cacciatori, and Cerchiai - On the Euler angles for $\operatorname{SU}(N)$](https://doi.org/10.1063/1.2190898) and [Tilma and Sudarshan - Generalized Euler angle parametrization for $\operatorname{SU}(N)$](https://iopscience.iop.org/article/10.1088/0305-4470/35/48/316/meta) other kinds of generalizations of Euler angles decompositions, however they differ in form from the one above.
Can you point me to some relevant literature? Can you show me a working parametrization of $u$ such that the whole $\SU(N)$ is covered?
| https://mathoverflow.net/users/145660 | On Euler angles decomposition of $\mathrm{SU}(N)$ | Building on the paper [Idel and Wolf - Sinkhorn normal form for unitary matrices](https://arxiv.org/abs/1408.5728) Colin McQuillan [suggested](https://mathoverflow.net/questions/430835/on-euler-angles-decomposition-of-mathrmsun#comment1108718_430835), it is easy to see that every $\operatorname{SU}(N)$ matrix $m$ can be decomposed as
$$
m = a \,u\, b
$$
with $a,b \in \operatorname{SU}(N)$ and diagonal, and
$$
u = F \,\left(\frac{1}{\mathrm{det}(V)} \oplus V\right) F^\dagger \in \operatorname{SU}(N)
$$
where $F\_{jk} = \frac{1}{\sqrt{N}}e^{\frac{2\pi i}{N}jk}$ is the Fourier transform unitary matrix and $V\in \operatorname{U}(N-1)$.
Notice that the number of parameters of this factorization matches the dimension of $\operatorname{SU}(N)$.
| 1 | https://mathoverflow.net/users/145660 | 430858 | 174,493 |
https://mathoverflow.net/questions/422450 | 0 | In the post (cross-posted in Mathematics Stack Exchange with identificator MSE [**4244256**](https://math.stackexchange.com/questions/4244256/conjectures-inspired-in-the-context-of-casas-alvero-conjecture-via-the-logarith) and same title) we assume that $P(x)=a\_0+a\_{1}x+\ldots+a\_{n-1}x^{n-1}+a\_{n}x^n$ is a polynomial of degree $1<\deg(P)=d=n$ defined over a field $K$ of characteristic zero. We denote its derivatives as $P^{(i)}(x)$ (writting $P^{(0)}(x)=P(x)$), and $a\_n$ denotes the leading coefficient of $P(x)$.
I've stated two conjectures (as speculations from the fact) inspired in that I've proven inductively that each polynomial $p(x)$ of degree $1<\deg(p)$ (and with corresponding $a\_n\neq 0$) satisfies $$p(x)=a\_n\cdot\left(\frac{n-l}{\frac{d}{dx}\log p^{(l)}(x)} \right)^n\tag{1}$$
for each integer $l$ with $0\leq l<n-1$.
**Conjecture 1.** *Let* $P(x)$ *a polynomial of degree* $1<n$, *thus we assume* $P^{(n)}(0)\neq 0$. *If the equation* $$P(x)=\frac{P^{(n)}(0)}{n!}\cdot\left(\frac{n-l}{\frac{d}{dx}\log P^{(l)}(x)} \right)^n$$ *holds for each integer* $0\leq l<n-1$, *then* $P(x)$ *has the form* $$P(x)=\frac{P^{(n)}(0)}{n!}\cdot(x-\alpha)^n$$ *for some element* $\alpha\in K$.
I know the statement of the called Casas-Alvero conjecture from Wikipedia [*Casas-Alvero conjecture*](https://en.wikipedia.org/wiki/Casas-Alvero_conjecture). I've speculated (while I don't know if this has a good mathematical content, or if these ideas are in the literature in some way more or less explicit) if from previous simple idea one can to state an equivalent form of Casas-Alvero conjecture.
**Conjecture 2.** *Let* $P(x)$ *a polynomial of degree* $1<n$ *and leading coefficient* $a\_n\neq 0$. *The Casas-Alvero conjecture is equivalent to that the equation* $$P(x)\cdot\left(P^{(l+1)}(x)\right)^n=a\_n\left((n-1)P^{(l)}(x)\right)^n,\tag{2}$$
*holds for each integer* $0\leq l<n-1$.
>
> **Question (Updated, considering the kindly advice of moderator in comments).** I would like to know **what work can be done** about the veracity of Conjecture 1, can you prove or refute Conjecture 1? **Many thanks.**
>
>
>
I don't know if these conjectures Conjecture 1 and Conjecture 2 are in the literature, my idea was very simple: that is to study the logarithmic derivative of derivatives of a given polynomial of the cited form, if it is in the literature please refer it in a comment or in your answer and I try to search and study this from the corresponding articles.
Remark: recently a professor solve a related question [1].
References:
-----------
[1] *Iterated derivatives and polynomials that are the power of a linear polynomial*, Mathematics Stack Exchange (Sep 12, 2021), post with identificator MSE [**4248459**](https://math.stackexchange.com/questions/4248459/iterated-derivatives-and-polynomials-that-are-the-power-of-a-linear-polynomial)
| https://mathoverflow.net/users/142929 | Conjectures inspired in the context of Casas-Alvero conjecture, via the logarithmic derivative of derivatives of a polynomial | I'm probably making a stupid mistake, but is Conjecture 1 possibly rather easy? At $\ell=0$ we're assuming
$$ p(x) = \frac{a\_n}{n!} \left( \frac{n}{\frac{d}{dx} \log p(x)} \right)^n $$
Write $\frac{d}{dx} \log p(x) = \dfrac{q(x)}{r(x)}$, a reduced rational expression. Then the right hand side of the hypothesis is
$$ \frac{a\_n}{n!} \left( \frac{n r(x)}{q(x)} \right)^n, $$
still a reduced rational expression, and this is equal to $p(x)$, a polynomial. So $q(x)$ is a constant.
Factor $p(x) = \frac{a\_n}{n!} (x-r\_1)^{e\_1} \dotsm (x-r\_k)^{e\_k}$ (with all $e\_i > 0$ and $r\_i$ pairwise distinct). Then the logarithmic derivative is
$$ \frac{q(x)}{r(x)} = \frac{d}{dx} \log p(x) = \sum \frac{e\_i}{x-r\_i} . $$
The numerator of this sum is
$$ \sum\_{i=1}^k e\_i \prod\_{j \neq i} x-r\_j . $$
This is a nonzero polynomial (leading coefficient $\sum e\_i = n = \deg p$) of degree $k-1$ (the number of distinct roots of $p$) and it's easy to see that its value at each $r\_i$ is nonzero, so there will be no cancellation with the denominator. So, this is $q(x)$. It must be that this is constant. So the number of distinct roots of $p$ must be $k=1$, $p$ is a $k$th power.
This was only using the hypothesis at $\ell=0$, not all $0 \leq \ell < n-1$, so perhaps you can tell me if I've made some silly error or misunderstanding.
| 1 | https://mathoverflow.net/users/88133 | 430859 | 174,494 |
https://mathoverflow.net/questions/430868 | 8 | Let $P\in \Bbb{Z}[X]$ be a polynomial with degree $d>1$.
It is conjectured that for all such $P$, their range for integer inputs $R\_P:=P(\Bbb{Z})$ has finite intersection with the set of factorials $\{n!:n\ge 0\}$.
We say that $P$ is “good” if there does not exist some $Q\in \Bbb{Z}[X]\setminus \{X\}$ such that $P \mid P\circ Q$. Examples: if $P=X^2$ then $Q=2X$ shows $P$ isn’t good; if $P=X^2-1$, then $Q=X^2$ shows $P$ isn’t good.
I was curious if there are any counter-examples to the following stronger claim:
For all such good $P$, $R\_P$ does not contain an infinite sequence $a\_1<a\_2<\dots$ where $a\_i \mid a\_{i+1}$ for $i\ge 1$. Or even stronger, there exists a constant $C=C\_P$ so that $R\_P$ does not contain divisibility chains longer than $C$.
Also, is there a nice characterization for when $P$ is good?
| https://mathoverflow.net/users/130484 | Divisibility chains and polynomials | Every $P$ is a counterexample. Indeed, given a polynomial $P$ consider the recursive sequence $b\_{n+1}=f(b\_n)$ where I take $f(x)=x+P(x)$, say. Then $P(b\_{n+1}) = P(b\_n + P(b\_n)) \equiv P(b\_n) \equiv 0 \bmod P(b\_n)$ since $x-y \mid P(x)-P(y)$ in general. Setting $a\_n=P(b\_n) \in R\_P$ this says that $a\_{n+1}$ is divisible by $a\_n$.
One can replace $f$ by more general polynomials. An important property of $f$ is that it permutes the roots of $P$.
| 8 | https://mathoverflow.net/users/31469 | 430870 | 174,497 |
https://mathoverflow.net/questions/430866 | 5 | Consider a [set theory](https://en.wikipedia.org/wiki/Alternative_set_theory) with the following axioms:
1. [separation](https://en.wikipedia.org/wiki/Axiom_schema_of_specification): $\exists y \forall x (x \in y \leftrightarrow \phi \land x \in a)$ where $y$ is not free in $\phi$
2. [reflection](https://en.wikipedia.org/wiki/Reflection_principle): $\phi \to \exists u \phi^u$
where $\phi^u$ bounds all *unbounded* quantifiers in $\phi$ to $u$ (see [this question](https://math.stackexchange.com/questions/4528459/does-relativization-of-formulas-replace-bounded-quantifiers-or-not)). This theory proves:
| Result | Parameters | Formula $\phi$ |
| --- | --- | --- |
| existence | | $\top$ |
| [pairing](https://en.wikipedia.org/wiki/Axiom_of_pairing) | $a,b$ | $\exists x (x = a) \land \exists x (x = b)$ |
| [union](https://en.wikipedia.org/wiki/Axiom_of_union) | $a$ | $\forall x\_{\in a} \forall y\_{\in x} \exists z (z = y)$ |
| [infinity](https://en.wikipedia.org/wiki/Axiom_of_infinity) | $a$ | $\exists x (x = a) \land \forall x \exists y (x \in y \land \forall z\_{\in y} (z = x))$ |
| [collection](https://en.wikipedia.org/wiki/Axiom_schema_of_replacement#Collection) for a [$\Delta\_0$](https://en.wikipedia.org/wiki/L%C3%A9vy_hierarchy#Definitions) formula $\psi$ | $a$ | $\forall x\_{\in a} \exists y \psi$ |
| [transitive model](https://en.wikipedia.org/wiki/Transitive_model) for a $\Delta\_0$ formula $\psi$ | | $\psi \land \forall x \forall y\_{\in x} \exists z (z = y)$ |
What is the consistency strength and interpretability strength of this theory? Can it prove full collection? Has it been studied in the literature?
| https://mathoverflow.net/users/74578 | How strong is separation + reflection of unbounded quantifiers? | This theory is mutually interpretable with second-order arithmetic $\mathsf{Z}\_2$ and $\mathsf{ZFC}-\mathsf{PowerSet}$ (and hence equiconsistent with them). Note that the mentioned theories are well-known to be mutually interpretable: the interpretation of $\mathsf{ZFC}-\mathsf{PowerSet}$ in $\mathsf{Z}\_2$ is achieved by carriying out the construction of $L$.
Trivially your theory is a subtheory of $\mathsf{ZFC}-\mathsf{PowerSet}$ and hence is interpretable there. The non-trivial part of interpreting $\mathsf{Z}\_2$ in your theory is to show there that there is a model of second-order Peano arithmetic (in the signature with just the successor).
For that you simply take any set $A$ such that it contains an empty set and for any $x\in A$ there is a set of the form $\{x\}\cup x\in A$, i.e. it satisfies $$(\exists x\in A)(\forall y\in x) y\ne y \land (\forall x\in A)(\exists y\in A)((\forall z\in x)z\in y \land (\forall z\in y)(z=x\lor z\in x)).$$ Then you consider (set-encoded) binary relations $E$ on $A$ s.t. $x\_1 \mathrel{E} x\_2$, for all empty $x\_1,x\_2\in A$ and for every $x\_1 \mathrel{E} x\_2$ and $y\_1,y\_2$ of the shapes $\{x\_1\}\cup x\_1$, $\{x\_2\}\cup x\_2$ we have $y\_1\mathrel{E} y\_2$. Then you construct a least binary relation $E\_0$ like this. For this you, starting from a Cartesian square $U$ of $A$ (a set containing at least one presentation of pair $(x,y)$, for all $x,y\in A$), construct the set $E\_0\subseteq U$ that consists of all $z\in U$ of the form $(x,y)$ such that some presentation of $(x,y)$ is in every binary relation of the considered form. You put $N$ to be the subset of $A$ consisting only of $E\_0$-reflexive points. Finally you define the successor relation $x \mathsf{R} y$ to be $(\exists x',y'\in N)(x\mathrel{E} x'\land y\mathrel{E} y'\land \text{$y'$ is of the shape $\{x'\}\cup x'$})$. The structure $(N,E,S)$, where $E$ serves as equality clearly will be a model of second-order Peano arithmetic.
Your theory doesn't prove collection. In fact even the extension of Zermelo set theory with choice $\mathsf{ZC}$ by your reflection principle doesn't prove collection. For this let me show that $\mathsf{ZFC}$ proves consistency of this theory. We cosider an $\omega$-sequence of ordinals $\omega=\alpha\_0<\alpha\_1<\ldots$ such that each $\alpha\_{n+1}$ is least such that for every transitive model $M$, there is a transitive model $M'\in V\_{\alpha\_{n+1}}$ for which $M\cap V\_{\alpha\_n}=M'\cap V\_{\alpha\_n}$ and $M$ and $M'$ satisfy the same first-order formulas with parameters from $M\cap V\_{\alpha\_n}$. It is easy to see that for $\alpha\_\omega=\lim\_{n<\omega} \alpha\_n$ the model $V\_{\alpha\_\omega}$ is a model of $\mathsf{ZC}$ togethe r with reflection.
Don't know if this theory was studied.
| 7 | https://mathoverflow.net/users/36385 | 430883 | 174,501 |
https://mathoverflow.net/questions/430887 | 14 | A [Seifert surface](https://en.wikipedia.org/wiki/Seifert_surface) of a knot is a surface whose boundary is the knot. The genus of a knot is the minimal genus among all the Seifert surfaces of the knot. My question is, is any algorithm known to find the genus of a knot?
Note that it’s been known since the 1980’s that Seifert’s algorithm for finding a Seifert surface does not suffice to find the genus of a knot. To wit, there even exist knots for which Seifert’s algorithm never produces a minimal-genus Seifert surface no matter what diagram of the knot you take.
In any case, is there some diagrammatic knot invariant that allows you to calculate the genus?
| https://mathoverflow.net/users/5017 | Is there an algorithm for the genus of a knot? | Jaco and Oertel's paper *[An algorithm to decide if a three-manifold is a Haken manifold](https://www.sciencedirect.com/science/article/pii/0040938384900399)* [1984], plus a bit of work, gives a doubly exponential time algorithm to compute the Seifert genus. (In practice their algorithm lies in exp-poly.)
Agol, Hass, and Thurston's paper *[The computational complexity of knot genus and spanning area](https://www.ams.org/journals/tran/2006-358-09/S0002-9947-05-03919-X/S0002-9947-05-03919-X.pdf)* reduces the time required to exp-poly (and proves that "genus less than $g$" lies in NP).
Recently announced work of Lackenby *[Unknot recognition in quasi-polynomial time](http://people.maths.ox.ac.uk/lackenby/quasipolynomial-talk.pdf)* [2021] gives a Haken hierarchy of the initial knot complement. If this can be improved to be a taut sutured manifold hierarchy, then that will reduce the time required (for the genus problem of a knot in the three-sphere given as a diagram) to quasi-polynomial.
---
The above (and the other answers) answer the question as asked in its first paragraph. However, its final paragraph asks if the genus can be computed from a "diagrammatic knot invariant". The answer there seems to be "not in general", but this is not a theorem. In particular there are families of knots (fibered, alternating) where the spans of certain polynomial invariants (Alexander, Jones) record the genus.
| 12 | https://mathoverflow.net/users/1650 | 430895 | 174,507 |
https://mathoverflow.net/questions/430906 | 2 | Let $\lambda$ be a limit ordinal and let $T$ be a tree such that for every element $t \in T$ and every $\beta < \lambda$, there is a branch of length at least $\beta$ that contains $t$. Does it follow that $T$ has a branch of length $\lambda$?
| https://mathoverflow.net/users/5199 | Existence of a branch of limit ordinal length | No. Aronszajn trees are the classical example here. The formal definition of an Aronszajn tree is simply a tree of height $\omega\_1$ where every level is countable. This tells you nothing about your requirement. However, the standard construction, indeed the one due to Aronszajn, is the "model tree", in a sense.
This is a tree of height $\omega\_1$ (with countable levels), such that any node $t$ we can find a branch, in the form of a maximal chain, of height $\beta$, for any limit ordinal $\operatorname{ht}(t)<\beta<\omega\_1$ which contains $t$. However, there are no branches of length $\omega\_1$ in an Aronszajn tree.
In most, if not all, investigations of these sort of trees we tend to begin by making these assumptions: every point is a splitting point; every point has extensions arbitrarily high; any point in a limit level is the limit of a unique branch.
Aronszajn trees have the stricter condition that each level is countable, as well. We can generalise these to $\kappa>\omega\_1$ which involves interesting combinatorics and large cardinal axioms.
| 4 | https://mathoverflow.net/users/7206 | 430907 | 174,509 |
https://mathoverflow.net/questions/430860 | 4 | Let $\text{Mod}\_g$ be the mapping class group of a closed oriented genus-$g$ surface $\Sigma\_g$ and let $H = H\_1(\Sigma\_g;\mathbb{Q})$. Fix some $r \geq 0$. It is known that the cohomology group $H^k(\text{Mod}\_g;H^{\otimes r})$ is independent of $g$ once $g$ is sufficiently large relative to $g$ and $r$. Does anyone know a concrete description of it?
For $r=0$, this is just the Madsen-Weiss theorem. For $r \geq 1$, this could be extracted from the paper
E. Looijenga,
Stable cohomology of the mapping class group with symplectic coefficients and of the universal Abel-Jacobi map,
J. Algebraic Geom. 5 (1996), no. 1, 135-150.
However, this paper really answers a much more complicated question where you look at the cohomology in an irreducible algebraic representation of $\text{Sp}(2g,\mathbb{Q})$. You could assemble this to get information about $H^{\otimes r}$, but given how complicated Looijenga's answer is this would lead to something terrible. I'm hoping there is a reasonable closed-form answer for these specific representations.
I've worked through Looijenga's argument and extracted the following special case: if $k$ is even, then $H^k(\text{Mod}\_g;H) = 0$, while if $k$ is odd of the form $k = 2n-1$, then
$$H^k(\text{Mod}\_g;H) = \bigoplus\_{i=0}^{n-2} H^{2i}(\text{Mod}\_g;\mathbb{Q}).$$
Thanks to Dan Petersen in the comments for pointing out that I had originally screwed this up, as well as a related calculation for $r=2$.
| https://mathoverflow.net/users/491583 | Stable cohomology of mapping class group with coefficients in $H^{\otimes n}$ | Appendix B of Randal-Williams' ["Cohomology of automorphism groups of free groups with twisted coefficients"](https://link.springer.com/article/10.1007/s00029-017-0311-0) gives a stable description of the graded $\mathbb{Q}[\Sigma\_q]$-module $H^\*(\Gamma\_g;H^{\otimes q})$.
| 4 | https://mathoverflow.net/users/32022 | 430914 | 174,510 |
https://mathoverflow.net/questions/430632 | 6 | I am currently dealing with discrete Fourier transform and correlation technique to construct the spectrum of a broad band signal. It's already known that if I have enough observations of the signal, in frequency domain it's a Gaussian spectrum.
I want to explore more on the frequency spectrum estimation of this kind of signal when there are not enough samples or there are samples with gaps in between sets of observation or random sampling cases. In these cases, the existing DFT or correlation techniques don't work very well.
I have come across many frequency estimation techniques such as [1]. The techniques like this can estimate the discrete frequencies present in a signal when limited data is available. However, for a continuous broad band spectrum, these techniques can fail as the computational complexity is extremely high.
I have 2 questions.
1. How many coherent samples in observation is enough to characterize a Gaussian distribution in frequency? Should it be estimator dependent (I suppose not). If it's not estimator dependent, how to find this number?
2. Is it useful to get a broader perspective on harmonic analysis based on some functional analysis? I'm very curious to get a broader perspective to such estimators. Will it be helpful for my current problem?
[1] C. Andrieu and A. Doucet, “Joint bayesian model selection and estimation of noisy sinusoids via reversible jump MCMC,” IEEE Transactions on Signal Processing, vol. 61, no. 14, pp. 3653–3655, 2013.
| https://mathoverflow.net/users/489481 | Harmonic analysis for a beginner | ***Q1.*** Your question 1 does not have a definite answer, this will very much depend on the uniformity of your data and on the noise level. To characterize your spectrum, the method of spectral estimation with a Gaussian process prior seems well suited. The paper [Bayesian Nonparametric Spectral Estimation](https://arxiv.org/abs/1809.02196) contains both a comprehensive description of the method and a computer code you can use.
***Q2.*** For question 2, [harmonic analysis](https://en.wikipedia.org/wiki/Harmonic_analysis) and [spectral analysis](https://en.wikipedia.org/wiki/Spectral_density_estimation) refer to very different fields of research, the former is not particularly relevant for the data analysis problem that you describe. Some texts on spectral analysis that you could look into: [Spectral Analysis for Physical Applications](https://books.google.nl/books/about/Spectral_Analysis_for_Physical_Applicati.html?id=FubniGJ0ECQC) and [Spectral Analysis of Signals.](https://www.maths.lu.se/fileadmin/maths/personal_staff/Andreas_Jakobsson/StoicaM05.pdf)
| 1 | https://mathoverflow.net/users/11260 | 430916 | 174,511 |
https://mathoverflow.net/questions/418235 | 13 | Let $M$ be a compact oriented $n$-manifold. Denote $\Omega^k := {\bigwedge}^k T^\*M$ the vector bundle of differential $k$-forms, and let $\Omega^{\text{odd}} := \bigoplus\_{\text{$k$ odd}} \Omega^k$ and $\Omega^{\text{even}} := \bigoplus\_{\text{$k$ even}} \Omega^k$ be the bundles of odd and even differential forms, respectively.
When are $\Omega^{\text{odd}}$ and $\Omega^{\text{even}}$ isomorphic as vector bundles over $M$?
---
A few observations:
* If $n$ is odd they are isomorphic, $\Omega^{\text{odd}} \simeq \Omega^{\text{even}}$, as can be seen e.g. by introducing a Riemannian metric $g$ on $M$, and observing that the Hodge star $\star: \Omega^k \to \Omega^{n - k}$ is an isomorphism for eack $k$ and hence also between $\Omega^{\text{odd}}$ and $\Omega^{\text{even}}$.
* More generally, if $\chi(M) = 0$, then $\Omega^{\text{odd}} \simeq \Omega^{\text{even}}$. Proof: there is a non-vanishing section $s$ of $\Omega^1 \simeq TM$. Write $\Omega^1 = \mathbb{R}s \oplus V$ for some vector bundle $V$. Then $$\Omega^{\text{odd}} = s \wedge \Omega^{\text{even}}(V) \oplus \Omega^{\text{odd}}(V) \simeq \Omega(V) \simeq s \wedge \Omega^{\text{odd}}(V) \oplus \Omega^{\text{even}}(V)
= \Omega^{\text{even}},$$
where $\Omega(V)$ is the full exterior bundle of $V$, and $\Omega^{\text{odd/even}}(V)$ are the odd/even parts of $\Omega(V)$.
* However, if $n = 2$ they are not isomorphic unless $M$ is the torus: the Euler class of $\Omega^{\text{odd}} = \Omega^1$ is then non-zero, while $\Omega^{\text{even}} = \Omega^0 \oplus \Omega^2 \simeq \mathbb{R}^2$ has trivial Euler class.
* If $n = 2d$, $\Omega^{\text{even}}$ and $\Omega^{\text{odd}}$ can be seen to have identical Chern classes $c\_i$, except possibly the top one $c\_d$. (I define the Chern class of a real vector bundle to be the Chern class of its complexification.) Namely, if $SM \subset TM$ denotes the unit sphere bundle, and $\pi: SM \to M$ is the footpoint projection, then the pullbacks $\pi^\*\Omega^{\text{odd}} \simeq \pi^\*\Omega^{\text{even}}$ are isomorphic. (Proof: since there is a non-vanishing tautological section $\tau(x, v) := g\_x(v, \bullet)$ of $\Omega^1$, we may repeat the argument in the second point.) By Gysin sequence, $\pi^\*$ is an isomorphism on $H^\bullet(M)$ for $\bullet \leq n - 1$, so $c\_i(\Omega^{\text{odd}}) = c\_i(\Omega^{\text{even}})$ for $i \leq d - 1$, and $c\_d(\Omega^{\text{odd}}) - c\_d(\Omega^{\text{even}})$ is a multiple of $\chi(M)$ in $H^{2d}(M, \mathbb{Z})$.
* If $n = 4$, my computations suggest that $\Omega^{\text{even}}$ and $\Omega^{\text{odd}}$ have equal all Chern, and Stiefel–Whitney characteristic classes. One needs to use that $\Omega^2 = \Omega^1 \wedge \Omega^1$ and the formula for $c\_2$ of a wedge product. In particular, $c\_2(\Omega^{\text{odd}}) - c\_2(\Omega^{\text{even}}) = 0$ is the *zero multiple* of the Euler characteristic.
* My guess after all of this is that $\Omega^{\text{even}} \not \simeq \Omega^{\text{odd}}$ if $n$ is even and $\chi(M) \neq 0$, but I was not able to prove this.
| https://mathoverflow.net/users/84963 | When are bundles of odd and even differential forms isomorphic? | I will explain that as long as $n>2$ the real vector bundles $\Omega^{even}$ and $\Omega^{odd}$ over $M$ are isomorphic.
If $n>2$ then $dim(\Omega^{even}) = dim(\Omega^{odd}) = 2^{n-1} > n$ and so $\Omega^{even}$ and $\Omega^{odd}$ are isomorphic if and only if they are stably isomorphic. This follows from obstruction theory, applied to the map $BO(2^{n-1}) \to BO$. We are therefore left with analysing the real $K$-theory class
$$\Omega^{even} - \Omega^{odd} \in KO^0(M).$$
As Meier's comment points out, the complex version of this would be the complex $K$-theory Euler class of $TM$, and we can take inspiration from that construction as follows.
For a real vector space $V$ consider the chain complex of vector bundles
$$V \times \Lambda^0 V \overset{(v,w) \mapsto (v, v \wedge w)}\longrightarrow V \times \Lambda^1 V \overset{(v,w) \mapsto (v, v \wedge w)}\longrightarrow V \times \Lambda^2 V \longrightarrow \cdots,$$
which is exact over every point other than $0 \in V$. This defines (see e.g. Definition 9.23 of Spin Geometry) a compactly-supported real $K$-theory class
$$v\_V \in KO^0\_c(V).$$
This construction can be done fibrewise to any vector bundle, so in particular gives a class
$$v\_{TM} \in KO^0\_c(TM),$$
which we can consider as a class in $\widetilde{KO}^0(Th(TM))$, the reduced real $K$-theory of the Thom space of $TM$. Pulling $v\_{TM}$ back along the zero-section $s\_0 : M \to Th(TM)$ gives the complex of vector bundles
$$\Omega^0 \overset{0}\longrightarrow \Omega^1 \overset{0}\longrightarrow \Omega^2 \overset{0}\longrightarrow \cdots$$
over $M$, which represents the class $\Omega^{even} - \Omega^{odd} \in KO^0(M)$ that we are interested in.
On the other hand, the inclusion of a tangent fibre $S^n \to Th(TM)$ is $n$-connected, so there is a factorisation up to homotopy
$$s\_0 : M \overset{q}\longrightarrow S^n \longrightarrow Th(TM)$$
(the map $q$ has degree $\chi(M)$, but that will not matter for the argument). We obtain the equation
$$\Omega^{even} - \Omega^{odd} = q^\*(v\_{\mathbb{R}^n}) \in KO^0(M).\tag{1}\label{eq}$$
We now turn to determining the class $v\_{\mathbb{R}^n} \in KO^0\_c(\mathbb{R}^n) = \widetilde{KO}^0(S^n) = KO^{-n}$. It is easy to see from the construction in terms of exterior powers that
$$v\_{\mathbb{R}^n} = (v\_{\mathbb{R}^1})^n \in KO^{-n},$$
and not hard to see that $v\_{\mathbb{R}^1} = \eta \in KO^{-1} = \mathbb{Z}/2\{\eta\}$. It follows that $v\_{\mathbb{R}^n}=0$ for $n > 2$, as $\eta^3 \in KO^{-3} = 0$, and combining this with \eqref{eq} proves the claimed result.
**Addendum**. It is interesting to see what happens for $n=2$. As explained in the question, over an orientable surface the Euler class distinguishes $\Omega^{even}$ and $\Omega^{odd}$, but one can still wonder about them as stable vector bundles. The analysis above, including the fact that $q$ has mod 2 degree $\chi(M)$, shows that
$$\Omega^{even} - \Omega^{odd} = \chi(M) \cdot p^\*(\eta^2) \in KO^0(M),$$
where $p : M \to S^2$ is the map that collapses the complement of a ball. As $\eta^2$ has order 2, this shows that these bundles are stably isomorphic if $\chi(M)$ is even.
On the other hand taking $M = \mathbb{RP}^2$ we have $\Omega^{even} = \mathbb{R} \oplus L$, where $L$ is the tautological line bundle, and $\Omega^{odd} = T\mathbb{RP}^2 \cong\_{stable} L^{\oplus 3} - \mathbb{R}$. These are not stably isomorphic, as they have different second Stiefel--Whitney classes: this implies that $\chi(\mathbb{RP}^2) \cdot p^\*(\eta^2) = p^\*(\eta^2) \neq 0 \in KO^0(\mathbb{RP}^2)$, which is indeed the case.
| 3 | https://mathoverflow.net/users/318 | 430918 | 174,512 |
https://mathoverflow.net/questions/430831 | 2 | **Motivation**: I'm trying to understand the proof of Theorem 3.1 in [Antonelli, Saut, and Sparber - Well-Posedness and averaging of NLS with time-periodic dispersion management](https://arxiv.org/abs/1204.4468). Though in the following I'm raising kind of abstract question.
Consider non-linear Schrödinger equation (NLS):
$$i\partial\_tu +\Delta u = |u|^{2}u, \quad u(t\_0, x)= \varphi(x).$$
Suppose $X$ (space of functions on $\mathbb R^d$) is Banach space and assume that
$$\|u\|\_{L\_I^{\infty}X} \lesssim \|\varphi\|\_{X} + |I|\|u\|^3\_{L\_{I}^{\infty}X}$$
here $I$ is small time interval.
**Questions**: (1) How to use the standard continuity argument to say that: there exists a solution $u$ to NLS in $I \times \mathbb R^d$ such that
$$\|u\|\_{L^{\infty}\_{I} X} \leq C \|\varphi \|\_{X}$$
for sufficiently small $I$? (2) What is standard continuity argument?
| https://mathoverflow.net/users/173418 | What is standard continuity argument for well-posedness? | I understand continuity argument a bit differently than the other post. Suppose you've shown the inequality
$$\|u\|\_{L^\infty(I,X)} \leq C(\|\varphi\|\_{X} + |I|\|u\|\_{L^\infty(I,X)}^3) \tag{1},$$
for some absolute constant $C>0$ and all intervals $I$ with $|I|\leq\delta$.
Since $u(t\_0)=\varphi$ and your solution $t\mapsto u(t)$ is continuous with values in $X$, there exists a $t\_\*>t\_0$ such that for $t\in [t\_0,t\_\*]$, $\|u(t)\|\_{X}\leq \frac{3C}{2}\|\varphi\|\_{X}$. Let $T\_\*>t\_0$ be the maximal such $t\_\*$, that is
$$\forall t\_0\leq t\leq T\_\*, \enspace \|u(t)\|\_{X}\leq \frac{3C}{2}\|\varphi\|\_{X}$$
and there exists a sequence $t\_n\rightarrow T\_\*^{+}$ such that $\|u(t\_n)\|\_{X}>\frac{3C}{2}\|\varphi\|\_{X}$. Note this implies $\|u(T\_\*)\|\_{X} = \frac{3C}{2}\|\varphi\|\_{X}$. If no such $T\_\*$ exists (i.e., $\|u(t)\|\_{L^\infty([t\_0,\infty),X)}\leq \frac{3C}{2}\|\varphi\|\_{X}$), then there is nothing to prove, so we may assume otherwise.
We can use inequality (1) to get a lower bound for $T\_\*-t\_0$. Indeed, setting $I=[t\_0,T\_\*]$, if $|I|>\delta$, then there is nothing to prove. If $|I|\leq \delta$, then
$$\|u\|\_{L^\infty(I,X)} \leq C\|\varphi\|\_{X} + \frac{27C^4(T\_\*-t\_0)}{8}\|\varphi\|\_{X}^3.$$
This implies that $\frac{27C^3(T\_\*-t\_0)}{8}\|\varphi\|\_{X}^2 \geq \frac{1}{2}$, otherwise we would have
$$\|u\|\_{L^\infty(I,X)} < \frac{3C}{2}\|\varphi\|\_{X} \Longrightarrow \|u(T\_\*)\|\_{X} < \frac{3C}{2}\|\varphi\|\_{X},$$
which contradicts our choice of $T\_\*$.
| 3 | https://mathoverflow.net/users/54316 | 430919 | 174,513 |
https://mathoverflow.net/questions/430924 | 2 | I want to find a Banach space $E$ and a compact operator $K:[0,1]\times E \rightarrow E$ (that is, $K$ maps every bounded sequence onto a sequence that converges up to a subsequence) satisfying the following conditions:
1. $K(0,\cdot) = 0$
2. There is a $r>0$ and a sequence $(\lambda\_n,u\_n)\in [0,1]\times \overline{B}\_r(0)$ such that $\lambda\_n\rightarrow 0$ but, for each $N\in \mathbb{N}$, it is possible to find $n>N$ such that $K(\lambda\_n,u\_n)\not\in B\_r(0)$.
My attempt: let $E=c\_0$ endowed with the maximum norm, where $c\_0$ is the Banach space of the sequences that converges to zero. Consider the operator $K:[0,1]\times c\_0\rightarrow c\_0$ defined by
$$K(\lambda,u)=2(\lambda u\_1,\lambda^{1/2} u\_2^2,\ldots,\lambda^{1/n} u\_n^n).$$
If we take $r=1$, then we have
$$ K(1/n,e\_n) = \frac{2}{n^{1/n}}\rightarrow 2>1=r.$$
The problem with my attempt is that apparently the operator $K$ is not compact.
| https://mathoverflow.net/users/173595 | Example of a compact operator that is not uniformly continuous | Your idea almost works.
Let $E = \ell\_1$ be the space of absolutely summable sequences.
Let
$$ K(\lambda, (u\_k)\_{k\in\mathbb{N}}) = (2\sum \lambda^{1/k} |u\_k|, 0, 0, \ldots) $$
Given any bounded set in $E$, its image is (essentially) a bounded set in $\mathbb{R}$, and so is pre-compact.
$K(0,\cdot)$ is obviously $0$.
Letting $e\_n$ be the element of $\ell\_1$ that has a $1$ on the $n$th spot and zero otherwise, you get
$$ K(1/n, e\_n) = (2 (1/n)^{1/n}, 0, 0, 0, \ldots) $$
which is eventually outside $B\_1(0)$.
| 3 | https://mathoverflow.net/users/3948 | 430932 | 174,517 |
https://mathoverflow.net/questions/430926 | 1 | Let $G$ be a compact connected Lie group. We denote by $\mathfrak{g}$ the Lie algebra of $G$ and by $\mathfrak{g}^\*$ the dual space of $\mathfrak{g}$. Let $\mathcal{O}\_r: = G\cdot r$ be a generic coadjoint orbit of $G$.
The coadjoint orbit $\mathcal{O}\_r$ endowed with the Kirillov–Kostant–Souriau $\omega$ is a symplectic manifold. I've read that it is also a Kähler manifold; meaning that there exists a unique almost complex structure $J$ on $\mathcal{O}\_r$ which is compatible with $\omega$ and such that the form $g(\cdot,\cdot):= \omega(\cdot,J\cdot )$ is a Riemannian metric on $\mathcal{O}\_r$.
>
> Given an element $\beta \in \mathcal{O}\_r $, then the tangent space of $\mathcal{O}\_r$ at $\beta$ is $T\_\beta \mathcal{O}\_r = \lbrace \xi\_{\mathcal{O}\_r}(\beta), \xi \in \mathfrak{g}\rbrace$ , where $\xi\_{\mathcal{O}\_r}(\beta) = \frac{d}{dt}\rvert\_ {t=0} e^{-t \xi}\cdot\beta$.
> What is $J(\xi\_{\mathcal{O}\_r}(\beta) )$, $\xi \in \mathfrak{g} $ ?
>
>
>
| https://mathoverflow.net/users/172459 | Question about the Kähler structure on generic coadjoint orbits | Put $T = G\_r$. We may, and do, assume that $\beta = r$, and simply describe a $T$-invariant complex structure on $\operatorname T\_{\mathcal O\_r}(r)$.
Instead of having one 1-dimensional subspace of $\mathfrak g$ for every root $\alpha$, we get a $2$-dimensional subspace $\{X\_\alpha + \overline{X\_\alpha} \mathrel: X\_\alpha \in (\mathfrak g\_{\mathbb C})\_\alpha\}$ for every pair of roots $\{\alpha, \overline\alpha = -\alpha\}$.
Our Kähler structure treats this $2$-dimensional space, which I will provocatively call $\mathfrak g\_{\pm\alpha}$ since its complexification is $(\mathfrak g\_{\mathbb C})\_\alpha \oplus (\mathfrak g\_{\mathbb C})\_{-\alpha}$, as a $\mathbb C$-vector space *via* the (isomorphic) projection to $(\mathfrak g\_{\mathbb C})\_\alpha$, and then rotates by $i$—but we must choose $i$ appropriately to get a negative definite metric. After our discussion in the comments, I think I have finally cleaned up the relevant signs.
Fix a root $\alpha$ of $T$ in $\mathfrak g\_{\mathbb C}$. Put $i\_\alpha = -\lambda\lvert\lambda\rvert^{-1}$, where $\lambda = r(\mathrm d\alpha^\vee(1))$ ($H\_\alpha \mathrel{:=} \mathrm d\alpha^\vee(1)$ is sometimes called the coroot, but I prefer to reserve that terminology for $\alpha^\vee$ itself), so that $i\_\alpha$ is a square root of $-1$. Then $J$ carries $\xi\_{\mathcal O\_r}(r)$, where $\xi = X\_\alpha + \overline{X\_\alpha}$, to $\xi'\_{\mathcal O\_r}(r)$, where $\xi' = i\_\alpha(X\_\alpha - \overline{X\_\alpha})$, for every $X\_\alpha \in (\mathfrak g\_{\mathbb C})\_\alpha$.
| 2 | https://mathoverflow.net/users/2383 | 430935 | 174,518 |
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