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https://mathoverflow.net/questions/447874 | 0 | Consider the simplest SIR model:
$$S'=-a SI$$
$$I'=a SI - b I$$
$$R'=b I$$
It is known that the waiting time of an infeccious person in the compartment $I$ follows an exponential behavior with rate $b$.
I was trying to figure out how this relates with a Poisson process, but I am not finding the "hook".
For instance, if the people are the domain of the random variables, and the random variables give us the time of infectious, I think it is a Poisson non-homogeneous process with rate $aS$.
However, once infected, I don't know how to describe the recovery. I mean, the domain $I$ is always changing, I don't know how to describe this.
Thank you.
| https://mathoverflow.net/users/172600 | Poisson Process x SIR model | The solution of this ODE defined over a compact interval $\left[0,T\right]$ for each initial condition can be formally cast as the limit (as the number of individuals $N$ scales to infinite) of a stochastic process modelling the peer-to-peer spread of a virus with Poisson clocks.
**Remark.** This answer is just intended to provide some references on the matter and break down a bit the overall *standard* path (along these references) that show how these non-stochastic ODEs emerge formally from the stochastic interacting dynamics among $N$ individuals (or nodes in a complete graph) in the *thermodynamic* limit, i.e., as $N\longrightarrow \infty$.
**In words.** i) characterize the microscopic state of the system (i.e., the state of each of the $N$ individuals as infected, susceptible or removed over time); ii) via peer-to-peer interacting rules, ascertain the dynamics, i.e., how the microscopic state will evolve over time according to neighbouring interactions (one may resort to an interacting particle system [1], where the interactions among individuals are driven by Poisson clocks); iii) Characterize the macroscopic processes, e.g., the fraction of infected individuals or the fraction of susceptible individuals over time (these processes build on the microscopic one); iv) take the limit as $N\rightarrow \infty$ and under certain conditions the macroscopic processes converge to the solution of the SIR or SIS ODE (e.g., following Kurtz approach [2-3], the macro processes are governed by the solution to the ODE up to a martingale term that vanishes as $N$ grows to infinite, yielding the concentration).
Just to depict an overall thermodynamic limit program in a nutshell, you have the following possibility.
**Microscopic process.** Consider $N$ individuals. Let $X^N\_i(t)\in\left\{-1,0,1\right\}$ to be a process representing the state of an individual $i$ at time $t\geq 0$: i) $X^N\_i(t)=0$ means that individual $i$ is susceptible; ii) $X^N\_i(t)=1$ means infected; and iii) $X^N\_i(t)=-1$ means removed at time $t$.
**Microscopic dynamics [1].** Once infected, each individual $i$ turns on two clocks: a) *Infection clock* -- after a random time $T\_I\sim {\sf Exp}(1/a)$, it randomly chooses another individual and infects him -- if the chosen individual is already infected, then nothing happens; b) *Removed clock*-- after a random time $T\_r\sim {\sf Exp}(1/b)$, it transitions to the state removed (and the clocks are turned off).
**Macroscopic process.** Define $I^N(t):= \sum\_{i=1}^N \mathbf{1}\_{\left\{X^N\_i(t)=1\right\}}/N$; $S^N(t):= \sum\_{i=1}^N \mathbf{1}\_{\left\{X^N\_i(t)=0\right\}}/N$ and $R^N(t):= \sum\_{i=1}^N \mathbf{1}\_{\left\{X^N\_i(t)=-1\right\}}/N$ as the fraction of infected, susceptible and removed individuals at time $t$ among $N$ individuals in the population.
**Macroscopic dynamics.** This characterization yields -- the macro-processes will be characterized by the integral of the Poisson processes governing the microscopic interactions and via compensating these Poisson processes, you obtain --
$$S^{N}(t) = S^{N}(0) + M\_S^N(t) - a\int\_{0}^t S^N(s-)I^N(s-)ds$$
$$I^{N}(t) = I^{N}(0) + M\_I^N(t) + a\int\_{0}^t S^N(s-)I^N(s-)ds-b\int\_{0}^t I^N(s-)ds,$$
where $M\_S^N(t)$ and $M\_I^N(t)$ are martingales that converge in probability to zero (w.r.t. the uniform norm on the space of Skorokhod paths defined over the compact interval $\left[0,T\right]$, $\mathcal{D}\left[0,T\right]$). This will further imply that the processes $\left(I^N(t)\right)\_{t\in\left[0,T\right]}$ and $\left(S^N(t)\right)\_{t\in\left[0,T\right]}$ converge weakly (w.r.t. the uniform norm on $\mathcal{D}\left[0,T\right]$) to the solution of the ODE with initial condition $I(0)$ and $S(0)$ if $I^N(0)$ and $S^N(0)$ converge in distribution to $I(0)$ and $S(0)$.
**More on the convergence.** That the martingale converges to zero is granted by Doob's inequality. This will imply that the stochastic macro-processes $\left(I^N(t)\right)\_{t\in \left[0,T\right]}$ and $\left(S^N(t)\right)\_{t\in \left[0,T\right]}$ are tight (a useful result in this part is the stochastic version of Arzelà–Ascoli theorem, e.g., in [4]). This implies that $\left(I^N(t)\right)\_{t\in \left[0,T\right]}$ and $\left(S^N(t)\right)\_{t\in \left[0,T\right]}$ admit convergent subsequences whose limit is necessarily the solution of the ODE. Since these ODE's have unique solution (vector field is Lipschitz and they live in a compact invariant set $\left[0,1\right]^2$) the sequence converges.
**Remark.** If the graph of interactions is not complete, then $\left(I^N(t)\right)\_{t\in \left[0,T\right]}$ and $\left(S^N(t)\right)\_{t\in \left[0,T\right]}$ are not Markov and the problem becomes quite intricate. I believe that there is now a growing body of literature devoted to establishing this limit over nontrivial non-complete graphs.
[1] T. M. Liggett, Interacting particle systems, Springer, New York, 1985.
[2] T. G. Kurtz, Solutions of ordinary differential equations as limits of pure jump markov processes, Journal of Applied Probability, 7 (1970), pp. 49–58.
[3] S. N. Ethier and T. G. Kurtz, Markov Processes: Characterization and Convergence, ser. Probability and Statistics. New York, NY: John Wiley & Sons, Inc., 1986.
[4] J. Jacod and A. Shiryaev, Limit Theorems for Stochastic Processes, 2nd ed., ser. Grundlehren der mathematischen
Wissenschaften. New York, NY: Springer-Verlag, 2003, vol. 288.
| 3 | https://mathoverflow.net/users/138242 | 447904 | 180,352 |
https://mathoverflow.net/questions/447892 | 0 | I am interested in the exponential sum
$$\sum\_{n=1}^X \frac{e(c\_1n^2+c\_2 n)}{1-e(c\_1n)}$$
where $c\_2$ is irrational and $e(x)=e^{2\pi i x}$. I know if the denominator is not there, this is a Weyl sum with a lot of bounds.
Is there anything that can be done to relate this sum to a Weyl sum or get some other bounds?
Edit, one thing that might be nice is to get a bound like
$$\left|\sum\_{n=1}^X \frac{e(c\_1n^2+c\_2 n)}{1-e(c\_1n)}\right|\ll \left|\sum\_{n=1}^X e(c\_1n^2+c\_2 n)
\right|\max\_{1\leq n\leq N} \frac{1}{|1-e(c\_1n)|}.$$
Is a bound like this possible?
| https://mathoverflow.net/users/479223 | Exponential sum with weight in bottom | Under your assumption that no $n$ from $1$ to $X$ has $|1 -e(c\_1n)|<\epsilon$, we have an upper bound for your sum of the form $2 \epsilon^{-1} \log X + O(X)$.
This is the "trivial bound" in that it just comes from estimating $$\sum\_{n=1}^X \frac{1}{ |1 - e(c\_1 n )|}.$$
By the pidgeonhole principle, the number of $n$ with $e( c\_1 n)$ at an angle less than $\theta$ from $1$ is at most $ 2 \theta /\epsilon$ since otherwise there would be $2$ such $n$ with angles within an interval of length less than $\epsilon$ and then the absolute value of their difference would have an angle less than $\epsilon$ and thus have $|1 -e(c\_1n) |< \epsilon$.
If $\theta\_1,\dots, \theta\_X$ are any set of angles such that the number with distance less than $\theta$ from $0$ is at most $2\theta/\epsilon$ then
$$ \sum\_{n=1}^{X} \frac{1}{ |1 -e(\theta\_n)|} \leq \sum\_{n=1}^{X} \frac{1}{ |1 -e ( n \epsilon/2))|}$$
because $|1- e(c\_1 n )|$ is an increasing function of the angular difference, and this sum is well-approximated by the integral
$$ 2 \epsilon^{-1} \int\_{\epsilon/2}^{ X \epsilon/2} \frac{ d \theta}{ | 1- e(\theta)|}$$
and since $\frac{1}{ 1- e(\theta)} = \frac{1}{ \theta} + O(1)$ , the integral is bounded by $\log X + O(X \epsilon)$, giving the claim.
Improving this bound by a significant factor seems to require estimates at many different scales, since the integral producing the $\log X$ term has equal contributions from many different scales. I suspect this can be done in such a way that the estimate at each scale is essentially a Weyl sum over a Bohr set, which probably means that cancellation can indeed be found under hypotheses on $c\_1,c\_2$, but one couldn't hope to save more than the $\log X$ factor as one is unlikely to get a bound for the overall sum greater than the bound for a single term.
| 3 | https://mathoverflow.net/users/18060 | 447907 | 180,354 |
https://mathoverflow.net/questions/447888 | 6 | We know several amazing techniques about the derived category $Perf (X)$ of a smooth projective variety such as the whole theory of Fourier-Mukai transforms. On the other hand, from a dg-categorical point of view, it is natural to work with smooth proper varieties instead of smooth projective varieties. Indeed, (dg) derived categories do not see projectivity of a variety and some natural birational geometry operations can produce proper Fourier-Mukai partners(<https://mathoverflow.net/a/369756/177839>). However, as far as I know, nearly all of the papers on derived categories of smooth varieties assume projectivity. So my question is whether there are specific reasons people do not work with proper varieties. More specifically
1. Are there any fundamental results such as ones coming from Bondal, Orlov, Kawamata, Bridgeland, Mukai etc. (I’m not trying to be exhaustive at all) that have counter-examples for proper cases?
2. Is there any fundamental result that is known to work for proper cases?
3. Are there technical/philosophical reasons why people seem to avoid proper cases?
I would really appreciate any comment to any of the questions.
| https://mathoverflow.net/users/177839 | Derived categories of smooth proper varieties? | The study of derived categories is a special case of the more general study of semiorthogonal components of derived categories. By Chow lemma for any proper variety $X$ there is a blow up $\pi \colon X' \to X$ such that $X'$ is projective, and by Orlov's blowup formula the derived category of $X$ is a semiorthogonal component of the derived category of $X'$. So, from this point of view, the story of proper varieties is a part of the story of projective varieties.
| 7 | https://mathoverflow.net/users/4428 | 447914 | 180,356 |
https://mathoverflow.net/questions/447470 | 3 | Let $P$ be a polarization of a Hilbert space $\mathcal{H}$, i.e. a bounded idempotent: consider a group $G=GL\_{res}(\mathcal{H}):=\{g \in GL(\mathcal{H}): [g,P] \in HS\}$ (where $HS$ is the set of all Hilbert Schmidt operators). For $g \in G$ let $P\_g:=gPg^{-1}$ and define $F\_{g,h}=P\_gP\_h-P\_h+I$. One can check that $F\_{g,h}$ are Fredholm: therefore one can define the so called *determinant line* $det(F\_{g,h})$ to be $\Lambda^{top}(ker(F\_{g,h})) \otimes \Lambda^{top}(coker(F\_{g,h}))^\*$. Using this data one can construct a (small) category where (the set of) objects is $G$ and for $g,h \in G$ we put $Hom(g,h)=det(F\_{g,h})$. However it is not immediate that this constitutes a category: in order to do so one have to prove some identifications:
>
> How to prove that $det(F\_{g,h} \cdot F\_{h,k}) \cong det(F\_{g,h}) \otimes det(F\_{h,k})$?
>
>
>
Of course both spaces are in fact one dimensional therefore there is an *abstract* isomorphism but I suspect that this isomorphism should be somehow *canonical* and natural. I would guess that in order to construct such an isomorphism maybe one has to use the fact that $Index(F\_{g,h}F\_{h,k})=Index(F\_{g,h})+Index(F\_{h,k})$.
Once we have this identification one can show that in fact $F\_{g,h}F\_{h,k}-F\_{g,k}$ is a trace-class operator.
>
> How to prove that if $S,T$ are two Fredholm operators with $S-T$ being trace-class then $detS \cong detT$ are *canonically* isomorphic?
>
>
>
I would be very grateful for any help.
| https://mathoverflow.net/users/24078 | Determinant line of Fredholm operators and composition of morphisms | As mentioned in the comments the first part of the question is in Abbonandolo and Majers "Infinite dimensional Grassmannians". My Hilbert spaces are real and separable and infinite dimensional.
I want to state that I basically know next to nothing about trace class operators, so take the rest with a grain of salt.
To every Fredholm operator one can associate its determinant line. An important fact is that this defines an actual line *bundle* over the space of Fredholm operators $\mathrm{det}\rightarrow \Phi(\mathbb{H})$. This is the unique (up to isomorphism) line bundle over $\Phi(\mathbb{H})$ that is non-trivial over each component of $\Phi(\mathbb{H})$. (This uses the fact that all components are homotopy equivalent and that $\pi\_1(\Phi(\mathbb H))\cong\mathbb{Z}/2)$.
A continuous family of Fredholm operators is a continuous map $f:X\rightarrow \Phi(\mathbb H)$. One can use this map to pull back the determinant line bundle over $X$. Let's assume that $X$ is connected. Then the bundle $f^\*\det$ is a trivial bundle over $X$ if and only if $f\_\*:\pi\_1(X)\rightarrow \pi\_1(\Phi(\mathbb H))$ is the trivial homomorphism.
Let $S$ be a Fredholm operator. Let $X$ be the space of Fredholm operators T such that $S-T$ is trace class, and $f:X\rightarrow \Phi(\mathbb H)$ the inclusion. I think that $X$ is an affine space, hence contractible, from which it follows that $f\_\*$ is trivial. This means that the determinant line bundle is trivial over $X$. You can use the triviality of the determinant line bundle to give a canonical isomorphism between two different operators of trace class.
| 1 | https://mathoverflow.net/users/12156 | 447916 | 180,357 |
https://mathoverflow.net/questions/447920 | 8 | In Lurie's treatment of the cobordism hypothesis, the domain is $\mathsf{Bord}^{fr}\_n$, the symmetric monoidal $(\infty,n)$-category of $k$-bordisms with $n$-framing for $0\leq k\leq n$.
If I understand his definitions correctly, for $n\neq 0,1,3,7$, $\mathsf{Bord}^{fr}\_n$ does not contain $S^n$ since $S^n$ does not have a framing.
In particular, among all closed 2-manifolds, only $T^2$ (with a framing) is contained in $\mathsf{Bord}^{fr}\_2$.
I want to have spherical partition functions but do not want to loosen the tangential structure too much.
So, can we cook up a symmetric monoidal $(\infty,n)$-category of stably-framed manifolds?
What's its relation to $\mathsf{Bord}^{fr}\_n$?
What's the cobordism hypothesis on this domain?
| https://mathoverflow.net/users/472749 | Stably-framed cobordism $(\infty,n)$-category | Yes, there is a stably-framed bordism category. Recall that tangential structures on smooth $n$-manifolds can be parameterized by maps $X \to BO(n)$; an $X$-structure on $M$ is then by definition a lift of the classifying map for the tangent bundle $T\_M : M \to BO(n)$ through $X$. In the stably-framed case, you take $X$ to be the fibre of $BO(n) \to BO = BO(\infty)$, equivalently $X = O/O(n)$. The cobordism hypothesis says what it always says: to give a fully-extended TQFT on manifolds with $X$-structure is to give a fully-dualizable object together with a trivialization of the $\Omega X$-action thereon. Remark: although people regularly say things like "oriented TQFT" and "framed TQFT" and the like, it would be more accurate to call the TQFT "cooriented" or "coframed", since it is the manifold which is oriented or framed — that way you can remember that a framing, say, is more data on the manifold, but a coframing is less data on the TQFT (since it needs to be defined on fewer manifolds).
The fibre of $BO(n) \to BO(n+1)$ is an $S^n$, and so this $X$ is filtered as $X = \dots.S^{n+2}.S^{n+1}.S^n$, i.e. it has one cell in each dimension $\geq n$. Thus to give a stable coframing to an $n$-dualizable object is to give one $n$-morphism, one $(n+1)$-morphism, one $(n+2)$-morphism, and so on. If your target $(\infty,n)$-category is just an $n$-category, then the $n$-morphism is data, and the $(n+1)$-morphism is an equation your data must solve, and all the higher cells are vacuous. The upshot is that many TQFTs are naturally co-framed.
| 8 | https://mathoverflow.net/users/78 | 447923 | 180,360 |
https://mathoverflow.net/questions/447890 | 1 | Let $X$ be a smooth connected proper scheme over field $k$. It is known that
correspondences $\alpha \subset X \times X$ regarded as
objects in Chow groups $\text{CH}^\*(X \times X)$
act on cohomology $H^\*(X, \mathbb{Z})$ via "push and pull",
namely we have a cycle class map
$$ \text{cl}: \text{CH}^\*(X) \to H^\*(X, \mathbb{Z} ) $$
which is compatible with pull-backs & push-forwards. If
$p\_1, p\_2: X \times X \to X $ are the projections and $\alpha $
a correspondence in $X \times X$, then it acts as
$$ \alpha^\*: H^\*(X \mathbb{Z}) \to H^\*(X \mathbb{Z}), \ \ \
\alpha^\*(s) := p\_{2\*}(p^\*\_1(s) \cap \text{cl}(\alpha)) $$
Clearly this action extends linearly to action by *rational*
Chow groups $\text{CH}\_r^\*(-)= \text{CH}^\* \otimes \mathbb{Q} $
on rational cohomology.
Let now specialize: Assume $X$ is a smooth proper curve and $D \subset X$
an effective divisor of degree $d$, ie a "multisection".
By dividing the degree $\beta:= \frac{1}{d} \cdot D $ becomes a
$\mathbb{Q}$-divisor of degree $1$, so a "virtual" section.
So far I understand it correctly in the [discussion here](https://mathoverflow.net/q/443274)
Dan Petersen uses that the induced action by
such $\mathbb{Q}$-divisor of degree $1$ in terms from above
as associated Chow correspondence
induces a splitting of the cohomology. (Dan Petersen used this more generally in relative setting
& associated action on derived object $Rf\_\* \mathbb{Q}$ but for
sake of simplicity I would like to understand it in absolute case for a
single curve as baby version of this mechanism.)
Question: I not see why this action gives a splitting of the cohomology?
In other words does such $\mathbb{Q}$-divisor "normed" to degree $1$
induce an idempotent endomorphism corresponding
to the associated action described above? It seems that the degree $1$ assumption is crucial, but I not see the connection. Then it would mean that every effective divisor turned into degree $1$ $\mathbb{Q}$-divisor by dividing it's degree would acting idempotently on the cohomology, ie give a splitting of the cohomology? That seems strange. Maybe I misunderstood somewhere Dan Petersen's argument.
| https://mathoverflow.net/users/501436 | Correspondences acting on cohomology groups $H^*(X)$ & splittings | Composition of a correspondence in $X \times Y$ and a correspondence in $Y \times Z$ is given by pulling both back to $X\times Y\times Z$, intersecting them, and pushing forward to $X \times Z$.
To check that $ D\times X \subset X\times X$ is idempotent under composition for $D$ of degree $1$, we pull back along two projections $X\times X \times X \to X\times X$, obtaining $D \times X \times X$ and $X\times D \times X$, then intersect $D \times X \times X$ and $X\times D \times X$, obtaining $D \times D \times X$, then finally pushforward $D \times D \times X$ along the projection to $X \times X$, which gives $D\times X$.
Here we use that for a cycle on $X \times Y \times Z$ obtained as a product of $A$ on $X\times Z$ with $B$ on $Y$, the pushforward to $X\times Z$ is $A$ times the degree of $B$. So this is where the degree enters into the picture.
| 3 | https://mathoverflow.net/users/18060 | 447932 | 180,365 |
https://mathoverflow.net/questions/447925 | 12 | This was previously posted to Math StackExchange. I was originally unsure whether it is suitable for posting here, but I've yet to get an answer there, so I'm just trying to see if people here can help.
As in the question title, let $A, B$ be a partition of the unit circle $S^1$, equipped with the Haar measure. Here, we *do not* require $A, B$ to be measurable. Also, assume neither $A$ nor $B$ is of measure zero, so they are either both non-measurable or both of positive measure. (Equivalently, they both have positive outer measure.) Is it possible, then, for $R\_\theta(A) \cap B$ to be measurable and have Haar measure zero for all $\theta$, where $R\_\theta$ is the rotation by degree $\theta$?
| https://mathoverflow.net/users/504602 | If $A, B$ is a non-trivial partition of $S^1$, is it possible that $R_\theta(A) \cap B$ has measure zero for all rotations $R_\theta$? | Assuming the Continuum Hypothesis, the answer is yes. On CH, the $\mathbf Q$-dimension of $\mathbf R$ has the cardinality of the first uncountable ordinal $\omega\_1$, so we can find a $\mathbf Q$-linear basis $e\_\alpha, \alpha \in \omega\_1$ of ${\bf R}$, and we can normalize $e\_0=1$. Thus every irrational real number $x$ has a unique representation of the form
$$ x = \sum\_{\alpha \leq \beta} q\_\alpha e\_\alpha\tag{1}\label{1}$$
where $0 < \beta < \omega\_1$ and $q\_\alpha, \alpha \leq \beta$ are rational numbers with all but finitely many $q\_\alpha$ non-zero, and $q\_\beta$ also non-zero. In particular, $q\_\beta$ is the final non-zero coefficient of $x$. We then assign $e^{2\pi i x}$ to lie in $A$ if $x$ is rational, or $x$ is irrational with representation \eqref{1} with $\nu\_2(q\_\beta)$ even, where $\nu\_2(q\_\beta) \in \mathbf Z$ is the number of times two divides the final non-zero coefficient $q\_\beta$, and $e^{2\pi i x}$ to lie in $B$ if $x$ is irrational with representation \eqref{1} with $\nu\_2(q\_\beta)$ odd. Since the final non-zero coefficient $q\_\beta$ is unaffected by shifts by integers (or even rationals) on the irrationals, this is a well-defined partition of $S^1$. By construction, if $z$ is not a root of unity, then $z$ lies in $A$ if and only if $z^2$ lies in $B$. Because of this, neither $A$ nor $B$ can have measure zero (otherwise $S^1 = A \cup B$ would also have measure zero).
On the other hand, if $\theta = e^{2\pi i s}$ is an angle, then $s$ is a finite $\mathbf Q$-linear combination of the $e\_\alpha$. Such rotations do not affect the final non-zero coefficient $q\_\beta$ in \eqref{1} unless $\beta$ is less than or equal to one of these $\alpha$ (or $x$ is rational), which only occurs countably often. Thus $R\_\theta(A) \cap B$ is countable (hence null) for all $\theta$.
| 13 | https://mathoverflow.net/users/766 | 447940 | 180,367 |
https://mathoverflow.net/questions/313516 | 16 | In his famous paper ["On a problem of Kurosh, Jonsson groups, and applications"](https://www.sciencedirect.com/science/article/pii/S0049237X08713466) of 1980, Shelah constructed a CH-example of an uncountable group $G$ equal to $A^{6640}$ for any uncountable subset $A\subset G$.
Let us call a group $G$
$\bullet$ *$n$-Shelah* if $G=A^n$ for each subset $A\subset G$ of cardinality $|A|=|G|$;
$\bullet$ *Shelah* if $G$ is $n$-Shelah for some $n\in\mathbb N$;
$\bullet$ *almost Shelah* if for each subset $A\subset G$ of cardinality $|A|=|G|$ there exists $n\in\mathbb N$ such that $A^n=G$;
$\bullet$ *Jonsson* if each subsemigroup $A\subset G$ of cardinality $|A|=|G|$ coincides with $G$.
$\bullet$ *Kurosh* if each subgroup $A\subset G$ of cardinality $|A|=|G|$ coincides with $G$.
It is clear that for any group $G$ the following implications hold:
>
> finite $\Leftrightarrow$ 1-Shelah $\Rightarrow$ $n$-Shelah $\Rightarrow$ Shelah $\Rightarrow$ almost Shelah $\Rightarrow$ Jonsson $\Rightarrow$ Kurosh.
>
>
>
In the mentioned paper, Shelah constructed a ZFC-example of an uncountable Jonsson group and also a CH-example of an uncountable 6641-Shelah group.
>
> **Problem 1.** Can an infinite (almost) Shelah group be constructed in ZFC?
>
>
> **Problem 2.** Find the largest possible $n$ (which will be smaller than [6640](https://mathoverflow.net/questions/314007/on-the-number-n-0-in-shelahs-construction-of-a-jonsson-group)) such that each $n$-Shelah group is finite.
>
>
>
[This result](https://link.springer.com/article/10.1007/s00012-010-0032-0) of Protasov implies
>
> **Theorem (Protasov).** Each countable Shelah group is finite.
>
>
>
It is easy to show that each 2-Shelah group is finite.
>
> **Problem 3.** Is each 3-Shelah group finite?
>
>
>
| https://mathoverflow.net/users/61536 | A Shelah group in ZFC? | In [this preprint](https://arxiv.org/pdf/2305.11155.pdf), Mark Poor and Assaf Rinot construct a ZFC-example of a $10120$-Shelah group $G$ of cardinality $|G|=\lambda^+$, where $\lambda$ is an arbitrary regular сardinal.
| 4 | https://mathoverflow.net/users/61536 | 447941 | 180,368 |
https://mathoverflow.net/questions/115982 | 9 | I am searching for R. Bott's lectures on characteristic classes and Gel'fand Fuks cohomology (New Mexico State Univ. 1973), apparently there are notes of these lectures taken by Mostow and Perchik which were published in some proceedings. But I was not able to find these proceedings.
Thank you in advance for your help.
| https://mathoverflow.net/users/27816 | R. Bott's lectures on characteristic classes | It turns out that what you are looking for is [here](https://math.ucr.edu/%7Eres/bott-nmsu/Bott%20-%20Lectures%20on%20Gelfand-Fuks%20Cohomology%20and%20Foliations.pdf)
which was noted by Mostow and Perchik.
| 4 | https://mathoverflow.net/users/136661 | 447947 | 180,370 |
https://mathoverflow.net/questions/447942 | 7 | Is the following statement (†) consistent with ZFC?
* If $E \subseteq [0,1]^2$ is such that $E\_x := \{y\in[0,1] : (x,y)\in E\}$ has measure zero for almost all $x$, then $E^y := \{x\in[0,1] : (x,y)\in E\}$ has measure zero for almost all $y$.
Of course I'm not assuming $E$ to be measurable, because that case follows trivially from Fubini's theorem. I know that (†) is not *implied* by ZFC, because it contradicts CH: namely, if $\prec$ is a well-ordering of $[0,1]$ with order type $\omega\_1$ then $E := \{(x,y) \in [0,1]^2 : y\prec x\}$ is such that $E\_x$ is countable for all $x$, and in particular has measure zero, whereas $E^y$ is co-countable for all $y$, and in particular has measure $1$. The consistency of (†) does not seem to follow from strong Fubini theorems such as [this one by Harvey Friedman](https://doi.org/10.1215/ijm/1256047607) because they assume that both iterated integrals exist in order to conclude that they are equal.
(The consistency of (†) would imply a partial reply to [this other question](https://mathoverflow.net/questions/447925/if-a-b-is-a-non-trivial-partition-of-s1-is-it-possible-that-r-thetaa/447929) in that (†) implies a negative answer to the latter.)
| https://mathoverflow.net/users/17064 | Consistency of a strong Fubini type theorem for measure zero sets | ZFC refutes this principle. Let $\kappa=\text{non}(\mathcal{L}),$ i.e. the least cardinality of a set of reals of positive outer measure. Let $X \subset [0,1]$ be such that $|X|=\kappa$ and $\lambda^\*(X)>0,$ and let $\prec$ be a well-ordering of $X$ of order type $\kappa.$ Then $E:=\{(x,y) \in X^2: y \prec x\}$ is a counterexample.
| 12 | https://mathoverflow.net/users/109573 | 447952 | 180,372 |
https://mathoverflow.net/questions/447950 | 6 | Let $N$ be a finite subset of $\mathbb{N}$, where $|N|>1$. For $i\in\mathbb{N}$, let $a\_i$ and $b\_i$ be chosen uniformly at random from $N$. Is it true that $\mathbb{P}[\sup\_{n\in\mathbb{N}}\{\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}\}=\infty]=1$? Note that given the symmetry involved, I believe this should be the same as asking if $\mathbb{P}[\inf\_{n\in\mathbb{N}}\{\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}\}=0]=1$.
This question came up while looking at the coarse geometry of trees, and if it is true then it has a nice corollary in that area. Unfortunately I do not have the background in probability to tell if this is an easy question or not, let alone answer it. I suspect it is true based only on the intuition of the people I have already asked, and doing some tests in python.
| https://mathoverflow.net/users/505965 | Probability that the ratio of products of randomly chosen natural numbers is unbounded | Yes, this is true. Indeed, let
$$S\_n:=\sum\_{i=1}^n Y\_i=\frac1s\,\ln\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n},$$
where $Y\_i:=X\_i/s$, $X\_i:=\ln\frac{a\_i}{b\_i}=\ln a\_i-\ln b\_i$, and $s:=\sqrt{Var\, X\_i}\in(0,\infty)$. Note that the $Y\_i$'s are i.i.d. zero-mean unit-variance random variables. So,
$$P\Big(\sup\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=\infty\Big)
=P(\sup\_{n\ge1}S\_n=\infty)\ge P\Big(\limsup\_n \frac{S\_n}{\sqrt{2n\ln\ln n}}>0\Big)=1,$$
by the [law of the iterated logarithm](https://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm).
So,
$$P\Big(\sup\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=\infty\Big)=1.$$
---
Similarly,
$$P\Big(\inf\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=0\Big)=1;$$
alternatively, write
$$P\Big(\inf\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=0\Big)=
P\Big(\sup\_{n\ge1}\frac{b\_1\cdots b\_n}{a\_1\cdots a\_n}=\infty\Big)=1.$$
---
Also, in view of Strassen's [converse to the law of the iterated logarithm](https://link.springer.com/article/10.1007/BF00539114) (and [Kolmogorov's zero–one law](https://en.wikipedia.org/wiki/Kolmogorov%27s_zero%E2%80%93one_law)), we do not need to assume that the $a\_i$'s and $b\_i$'s are chosen uniformly from a finite subset of $\mathbb N$ -- they can be sampled, independently, from any non-degenerate distribution whatsoever on $(0,\infty)$. Another proof of Strassen's result was given by [Heyde](https://www.cambridge.org/core/journals/journal-of-applied-probability/article/abs/on-the-converse-to-the-iterated-logarithm-law/B475B8821B2A214C43A6BA16FFC4E086).
| 8 | https://mathoverflow.net/users/36721 | 447953 | 180,373 |
https://mathoverflow.net/questions/447848 | 7 | We say that $\alpha$ is $\Sigma\_n$-extendable (to $\beta$), if there is $\beta>\alpha$ such that $L\_\alpha$ is a $\Sigma\_n$ elementary submodel of $L\_\beta$.
First ordinal: the least $\alpha\_0$ such that there are ordinals $\alpha\_0<\alpha\_1<...<\alpha\_{\omega^2}$, such that each $\alpha\_i$ is $\Sigma\_2$ extendable, and for any $i<j$, $\alpha\_i$ is $\Sigma\_1$ extendable to $\alpha\_j$.
Second ordinal: the least $\alpha\_0$ such that there are ordinals $\alpha\_0<\alpha\_1<...<\alpha\_{\omega}$, such that each $\alpha\_i$ is $\Sigma\_2$ extendable, and for any $i<j$, $\alpha\_i$ is $\Sigma\_1$ extendable to $\alpha\_j$, and $\alpha\_\omega$ is the limit of $\alpha\_i, i<\omega$.
Which one of these two ordinals is larger?
I'm able to prove that if we change $\omega^2$ to any smaller ordinal, then the second one is larger, but I don't know how to go further.
| https://mathoverflow.net/users/170286 | Which one of the following two ordinals is larger? | The second one is larger.
Because of the conflict of terminology with "extendible", I will say that an ordinal $\alpha$ is
*somewhere $\Sigma\_n$-stable*
if there is $\beta>\alpha$ such that $L\_\alpha\preccurlyeq\_n L\_\beta$.
Given a somewhere $\Sigma\_2$-stable $\alpha$,
let $\beta\_2(\alpha)$ be the least $\beta>\alpha$ such that $L\_\alpha\preccurlyeq\_2L\_\beta$.
Let $\left<\alpha\_n\right>\_{n\leq\omega}$
be as in the second situation.
So each $\alpha\_n$ (for $n\leq\omega$)
is somewhere $\Sigma\_2$-stable,
and for all $m\leq n\leq\omega$,
$L\_{\alpha\_m}\preccurlyeq\_{1}L\_{\alpha\_n}$,
and $\alpha\_m<\alpha\_{m+1}$ and $\alpha\_\omega=\sup\_{m<\omega}\alpha\_m$.
Now note that for $m<n\leq\omega$,
we have $\beta\_2(\alpha\_m)\neq\beta\_2(\alpha\_n)$,
because otherwise letting $\beta=\beta\_2(\alpha\_m)=\beta\_2(\alpha\_n)$,
we would have $L\_{\alpha\_m}\preccurlyeq\_2 L\_{\beta}$ and $L\_{\alpha\_n}\preccurlyeq\_2 L\_\beta$,
and $\alpha\_m<\alpha\_n<\beta$.
But then actually $L\_{\alpha\_m}\preccurlyeq\_2 L\_{\alpha\_n}$,
so in fact $\beta\_2(\alpha\_m)\leq\alpha\_n$, contradiction.
Now by refining the sequence $\left<\alpha\_n\right>\_{n<\omega}$
if necessary, we may assume that $\beta\_2(\alpha\_m)<\alpha\_{m+1}$ for all $m<\omega$ (though we might lose the minimality of $\alpha\_0$). For let $\gamma\_0=\inf\_{n<\omega}\beta\_2(\alpha\_n)$ and let $n\_0<\omega$ be the unique $n$
such that $\beta\_2(\alpha\_n)=\gamma\_0$.
Now let $\gamma\_1=\inf\_{n\in(n\_0,\omega)}\beta\_2(\alpha\_n)$ and let $n\_1\in(n\_0,\omega)$ be such that $\beta\_2(\alpha\_n)=\gamma\_1$. Etc,
producing $\left<\gamma\_i,n\_i\right>\_{i<\omega}$
with $n\_i<n\_{i+1}<\omega$ and $\gamma\_i<\gamma\_{i+1}$ and $\gamma\_i=\beta\_2(\alpha\_{n\_i})$.
By reindexing, we may assume that $n\_i=i$ for all $i<\omega$, so in other words,
$\beta\_2(\alpha\_n)<\beta\_2(\alpha\_{n+1})$ for all $n<\omega$. So
$$ L\_{\alpha\_n}\preccurlyeq\_2 L\_{\beta\_2(\alpha\_n)}\in L\_{\beta\_2(\alpha\_{n+1})}, $$
and
$$L\_{\alpha\_{n+1}}\preccurlyeq\_2 L\_{\beta\_2(\alpha\_{n+1})},$$
and the existence of $\beta\_2(\alpha\_{n})$
is then a $\Sigma\_1$ statement about $\alpha\_n$ which is true in $L\_{\beta\_2(\alpha\_{n+1})}$,
and therefore true in $L\_{\alpha\_{n+1}}$,
and so $\beta\_2(\alpha\_n)<\alpha\_{n+1}$.
Let $X$ be the set of all $\alpha<\alpha\_\omega$ such that $L\_\alpha\preccurlyeq\_1 L\_{\alpha\_\omega}$
and $L\_{\alpha\_\omega}\models$"$\alpha$ is somewhere $\Sigma\_2$-stable"; note the latter condition is equivalent to saying that $\alpha$ is somewhere $\Sigma\_2$-stable
(in $V$) and $\beta\_2(\alpha)<\alpha\_\omega$).
So we have (after thinning out as above)
$\alpha\_n\in X$ for all $n<\omega$,
and in particular, $X$ is cofinal in $\alpha\_\omega$.
I claim that $X$ has ordertype $\alpha\_\omega$.
For suppose not and let $\eta<\alpha\_\omega$
be the ordertype of $X$.
Let $\beta=\beta\_2(\alpha\_\omega)$.
Then $L\_\beta\models$"There is $\widetilde{\alpha}>\eta$ such that $L\_{\widetilde{\alpha}}\preccurlyeq\_1 L$ and $\widetilde{\alpha}$
is a limit of ordinals $\alpha'<\widetilde{\alpha}$
such that $L\_{\alpha'}\preccurlyeq\_1 L\_{\widetilde{\alpha}}$ and $L\_{\widetilde{\alpha}}\models$"$\alpha'$ is somewhere $\Sigma\_2$-stable", and the ordertype of the set $\widetilde{X}$ of those ordinals is $\eta$''
(in fact this is witnessed by $\widetilde{\alpha}=\alpha\_\omega$).
This statement satisfied by $L\_\beta$
is a $\Sigma\_2$ assertion about $\eta$.
(It is not a $\Sigma\_1$ assertion about $\eta$, because of the clause "$L\_{\widetilde{\alpha}}\preccurlyeq\_1 L$".)
Therefore $L\_{\alpha\_\omega}$ satisfies
the same statement about $\eta$.
Let $\widetilde{\alpha}<\alpha\_\omega$ and $\widetilde{X}$
be a witness. In particular, $L\_{\widetilde{\alpha}}\preccurlyeq\_1 L\_{\alpha\_\omega}$.
Note then that $\widetilde{X}\subseteq X$
(in particular since if $\alpha'\in\widetilde{X}$
then $L\_{\alpha'}\preccurlyeq\_1 L\_{\widetilde{\alpha}}\preccurlyeq\_1 L\_{\alpha\_\omega}$, so $L\_{\alpha'}\preccurlyeq\_1 L\_{\alpha\_\omega}$).
But $\widetilde{X}\subseteq\widetilde{\alpha}<\alpha\_\omega$,
and it follows that $X$ has ordertype $>\eta$, a contradiction.
So $X$ has ordertype $\alpha\_\omega$.
Now let $\alpha^{(1)}\_0$ be the first ordinal you defined and $\alpha^{(2)}\_0$ the second. Maybe $\alpha^{(2)}\_0<\alpha\_0$ (because of the thinning out we did), but anyway $L\_{\alpha^{(2)}\_0}\preccurlyeq\_1 L\_{\alpha\_\omega}$. From this and the fact that $X$ has ordertype $\alpha\_\omega$, it follows that there are chains of the form required for the definition of $\alpha^{(1)}\_0$ which appear in $L\_{\alpha^{(2)}\_0}$. Therefore $\alpha^{(1)}\_0<\alpha^{(2)}\_0$.
| 6 | https://mathoverflow.net/users/160347 | 447969 | 180,379 |
https://mathoverflow.net/questions/447980 | -2 | Show that the series
$$\sum\_{n=2}^{\infty}\frac{1}{[\frac{(1+\epsilon)\log n}{\log\log n}]!}$$converges for $\epsilon>0$.
Stirling's approximation gives that the convergence for the series is equivalent to the series
$$\sum\_{n=2}^{\infty}\frac{1}{\sqrt{2\pi [\frac{(1+\epsilon)\log n}{\log\log n}]}}\left(\frac{e}{[\frac{(1+\epsilon)\log n}{\log\log n}]}\right)^{[\frac{(1+\epsilon)\log n}{\log\log n} ]}.$$
But it is still hard to see the convergence.
| https://mathoverflow.net/users/484728 | convergence for a series | This is not a research level question, but I feel like answering it. Simply, use the elementary estimate
$$\log(m!)\sim m\log m.$$
The symbol $\sim$ means that the ratio of the two sides tends to $1$. For
$$m=\left[\frac{(1+\epsilon)\log n}{\log\log n}\right],$$
this becomes
$$\log(m!)\sim\frac{(1+\epsilon)\log n}{\log\log n}\log m\sim(1+\epsilon)\log n.$$
In particular, if $n$ is sufficiently large in terms of $\epsilon,$
then
$$m!>n^{1+\epsilon/2}.$$
Convergence is immediate from here.
| 2 | https://mathoverflow.net/users/11919 | 447982 | 180,382 |
https://mathoverflow.net/questions/447987 | 2 | Is the following embedding possible?
$\mathrm{Sp}\_{2m}(p)\leqslant S\_{p^m-1}$ where $S\_{p^m-1}$ is a symmetric group and $p$ is prime. I see that when $p=3$ and $m=3$, the order of the former does divide the order of the latter. I was also thinking along the lines of $\mathrm{PSp}$ being simple except for finite cases but haven't gotten an answer. I feel that it is not possible, though.
Thank you.
| https://mathoverflow.net/users/488802 | is the embedding $\mathrm{Sp}_{2m}(p)\leqslant S_{p^m-1}$ possible? | No. The smallest degrees of the faithful permutation representations of the finite simple groups are listed in Table 4.5 of [On the maximum orders of elements of finite almost simple groups and primitive permutation groups](https://arxiv.org/abs/1301.5166) by Guest, Morris, Praeger, and Spiga.
$\DeclareMathOperator\PSp{PSp}\DeclareMathOperator\Sp{Sp}$With a couple of small exceptions, for $\PSp(2m,q)$ with $m \ge 2$, the minimum degree is $(q^{2m}-1)/(q-1)$ for $q \ne 2$ and $2^{m-1}(2^m-1)$ when $q=2$.
So $\Sp(2m,2)$ has no proper subgroup of index less than this, and so it cannot have a faithful permutation representation of smaller degree.
I think the minimum degree for $\Sp(2m,q)$ is probably $q^{2m}-1$ for $q$ odd.
| 9 | https://mathoverflow.net/users/35840 | 447991 | 180,385 |
https://mathoverflow.net/questions/447951 | 3 | $\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\Proj{\mathbf{Proj}}$Let $S$ be a Noetherian scheme and $X$ finite $S$-scheme. The finite morphism $X \to S$ is projective in the sense of the definition in the Stacks Project (tag [01W8](https://stacks.math.columbia.edu/tag/01W8), part (1)).
This is proved in tag [0B3I](https://stacks.math.columbia.edu/tag/0B3I) of the Stacks Project, where a map $\sigma : X\to \textbf{Proj}\_S(f\_\*\mathcal{O}\_X)$ is constructed, associated to the surjective map
$$f^\*f\_\*\mathcal{O}\_X\to\mathcal{O}\_X.$$
However, when proving that $\sigma$ is a closed immersion, the Stacks Project invokes tag [01KT](https://stacks.math.columbia.edu/tag/01KT), saying that $\sigma$ is a section of a separated morphism.
Say $f : X\to S$ is also a locally free map for simplicity.
>
> What morphism is $\sigma$ a section of?
>
>
>
The proof seems to suggest there should be a morphism of $S$-schemes $\Proj\_S(f\_\*\mathcal{O}\_X)\to X$ and $\sigma$ should be a section to it, however, I don't see such a morphism.
Another question is this.
>
> Suppose $g : X\to X$ is an $S$-endomorphism of $X$. Is there an $S$-endomorphism of the "proj" construction, $g' : \Proj\_S(f\_\*\mathcal{O}\_X)\to \Proj\_S(f\_\*\mathcal{O}\_X)$ that is compatible with $g$? That is, such that $g'\circ \sigma = \sigma\circ g$.
>
>
>
To give some context, the questions came up while reviewing different definitions of projectivity for morphisms (in the Stacks Project, in R. Vakil's notes [The rising sea: Foundations of algebraic geometry](http://math.stanford.edu/%7Evakil/216blog/FOAGmay2523public.pdf), and in Hartshorne's book). The first question was about a slight inaccuracy in [0B3I](https://stacks.math.columbia.edu/tag/0B3I), now fixed. The second came up when R.Vakil's notes say, when $X=\text{Spec}(B)$ and $S=\text{Spec}(A)$, that there is a closed immersion of $X$ into $P:=$"$\text{Proj}(A\oplus B\oplus B\cdots)$" and I was confused as to how to extend endomorphisms of $X$ to $P$. Now it all makes sense.
| https://mathoverflow.net/users/nan | Finite subschemes of projective bundles | Welcome new contributor. There is no such morphism from the relative Proj to $X$: it would contradict Stein factorization. However, there is a simple proof that $\sigma$ is a closed immersion that does not need this.
For every quasi-coherent $\mathcal{O}\_S$-module $\mathcal{F}$, denote by $$(p\_\mathcal{F}:\mathbb{P}\_S(\mathcal{F})\to S,\ \ \alpha\_{\mathcal{F}}:p\_{\mathcal{F}}^\*\mathcal{F} \twoheadrightarrow \mathcal{O}(1)),$$ a universal pair of an $S$-scheme and an invertible quotient of the pullback of $\mathcal{F}$ to that $S$-scheme. Associated to every invertible $\mathcal{O}\_S$-module $\mathcal{L}$ and every $\mathcal{O}\_S$-module homomorphism, $$\beta:\mathcal{L}\to \mathcal{F},$$ there is an induced morphism of invertible sheaves on $\mathbb{P}\_S(\mathcal{F})$; namely the composition, $$p\_{\mathcal{F}}^\*\mathcal{L} \xrightarrow{p\_{\mathcal{F}}^\*\beta} p\_{\mathcal{F}}^\*\mathcal{F} \xrightarrow{\alpha\_{\mathcal{F}}}\mathcal{O}(1). $$ The maximal open $U\_\beta$ of $\mathbb{P}\_S(\mathcal{F})$ on which this composition is an isomorphism is the complement of a "pseudo divisor", i.e., the ideal sheaf of the zero scheme of the composition is everywhere locally principal (perhaps locally equal to the zero ideal sheaf). Of course if $\beta$ is locally a direct summand of $\mathcal{F}$, then this closed subscheme is normally flat. If also $\mathcal{F}$ is a flat $\mathcal{O}\_S$\_module, then this closed subscheme is an $S$-flat relative hyperplane section.
Now let $\mathcal{F}$ be the locally free $\mathcal{O}\_S$-module $f\_\*\mathcal{O}\_X$, and let $\beta$ be the locally direct summand, $$f^\#:\mathcal{O}\_S\to f\_\*\mathcal{O}\_X.$$ The corresponding open subscheme $U\_{f^\#}$ is a relative affine space bundle whose sheaf of relative differentials is canonically isomorphic to the pullback of the cokernel of $f^\#$. By adjointness of pushforward and pullback, the following map of $\mathcal{O}\_X$-modules is an isomorphism, $$\mathcal{O}\_X \xrightarrow{f^\*f^\#} f^\*f\_\*\mathcal{O}\_X \xrightarrow{\text{nat}\_f} \mathcal{O}\_X.$$ Therefore the morphism $\sigma$ factors through $U\_{f^\#}$.
The induced morphism from $X$ to $U$ is a morphism of affine $S$-schemes. This morphism is a closed immersion if and only if the induced morphism of pushforward structure sheaves is a surjective morphism of quasi-coherent $\mathcal{O}\_S$-modules. Since both of these quasi-coherent sheaves is the target of compatible morphisms from $\mathcal{O}\_S$, the original morphism of quasi-coherent sheaves is surjective if and only if the induced map of quotients modulo $\mathcal{O}\_S$ is surjective. For an affine space bundle, up to the choice of a zero section (which exists locally), this quotient sheaf is the direct sum of all positive symmetric powers of the relative sheaf of differentials, i.e., all positive symmetric powers of $f\_\*\mathcal{O}\_X/\mathcal{O}\_S$. So already the restriction of the $\mathcal{O}\_S$-module homomorphism to the first symmetric power $f\_\*\mathcal{O}\_X/\mathcal{O}\_S$ is an isomorphism onto the quotient $f\_\*\mathcal{O}\_X/\mathcal{O}\_S$, and thus the full $\mathcal{O}\_S$-module homomorphism is surjective.
For your second question, since pushforward is functorial, the pushforward along $f$ of the pushforward along $g$ of $\mathcal{O}\_X$ is canonically isomorphic to the pushforward along $f$ of $\mathcal{O}\_X$. This canonical isomorphism defines an automorphism of the pushforward along $f$ of $\mathcal{O}\_X$ functorially associated to $g$. You can use this to prove that the induced automorphism of relative Proj is compatible with $g$ and $\sigma$.
| 1 | https://mathoverflow.net/users/13265 | 447997 | 180,386 |
https://mathoverflow.net/questions/447949 | 18 | The Robertson–Seymour theorem concerns downwardly closed classes of isomorphism classes of finite undirected graphs. (Am I committing some sin by referring to a class of classes? An isomorphism class is a proper class, and a set, as opposed to a proper class, is a class that is a member of some other class. But ignore this present comment, unless you decide not to ignore it.)
Downwardly closed means that if any (isomorphism class of a) graph is a member, then so are (the isomorphism classes of) all of its minors, a minor being defined as a graph obtained by deleting vertices (and all of their incident edges) or contracting edges (so that two vertices become one vertex with each contraction).
The theorem says that the set of minimal non-members of every such class is finite.
For example, the set of minimal non-members of the class of planar graphs has just two members: $K\_5$ and $K\_{3,3}.$
Other such classes include outer-planar graphs (embeddable in a plane with vertices on a circle and edges as non-crossing chords of the circle), where the forbidden minors are $K\_4$ and $K\_{3,2},$ and the graphs linklessly embeddable in $\mathbb R^3$, or the knotlessly embeddable graphs, or the graphs embeddable in a torus, or some other classes defined in ways not relying on geometry (I think?). With some such classes, more than just two minimal non-members are needed, but, according to the theorem, always only finitely many.
So a set of forbidden minors can always be reduced to a finite subset.
Is this equivalent to the compactness of some topological space?
(And I'm wondering if someone's going to tell me the answer is *obviously* "yes.")
| https://mathoverflow.net/users/6316 | Is the Robertson–Seymour theorem equivalent to the compactness of some topological space? | The Robertson–Seymour graph minor theorem states that the set of all (isomorphism classes of) finite undirected graphs under the graph minor relation is a [well-quasi-ordering](https://en.wikipedia.org/wiki/Well-quasi-ordering) (or wqo for short). It is a theorem that a quasi-ordering $Q$ is a wqo if and only if the Alexandrov topology on $Q$ is Noetherian, meaning that every subspace of $Q$ is compact. See for example [Reverse mathematics, well-quasi-orders, and Noetherian spaces](https://arxiv.org/abs/1504.07452), by Emanuele Frittaion, Matt Hendtlass, Alberto Marcone, Paul Shafer, and Jeroen van der Meeren.
| 25 | https://mathoverflow.net/users/3106 | 448004 | 180,389 |
https://mathoverflow.net/questions/447979 | 0 | Is there a characterization of all continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying:
* $f(0)=0$
* $f$ is monotonically increasing
* $f$ is concave
My intuition is that $f$ should admit an integral representation:
$$
f(t) = \int\_0^t\, g(s)\,ds
$$
for some positive continuous function to satisfy 1 and 2, but I don't know what would be needed for 3?
| https://mathoverflow.net/users/36886 | Representation of continuous, monotone, concave functions | $\newcommand\R{\mathbb R}$Since $f$ is nondecreasing, concave, and continuous on $[0,\infty)$, it has a nonnegative nonincreasing right derivative $g=f'\_+$ on $[0,\infty)$, which is of course right continuous. So, there is a limit $a\_g:=\lim\_{u\to\infty}g(u)\in[0,b\_g]$, where $b\_g:=g(0)<\infty$.
Let $\mu\_g$ be the finite Lebesgue--Stieltjes measure on $(0,\infty)$ defined by the condition $\mu\_g((u,\infty))=g(u)-a\_g$ for $u\in(0,\infty)$; such a measure exists because $g$ is right continuous. Then for all $x\in[0,\infty)$
$$
\begin{aligned}
f(x)&=\int\_0^x du\,g(u) \\
&=\int\_0^x du\,\Big(a\_g+\int\_{(u,\infty)}\mu\_g(dv)\Big) \\
&=a\_g x+\int\_{(0,\infty)}\mu\_g(dv)\int\_0^{\min(x,v)} du \\
&=a\_g x+\int\_{(0,\infty)}\mu\_g(dv) f\_v(x),
\end{aligned}
$$
where
$$f\_v(x):=\min(x,v).$$
Vice versa, since for any $v\in(0,\infty)$ the function $f\_v$ on $(0,\infty)$ is nondecreasing, concave, and continuous, and $f\_v(0)=0$, we have the following: if
$$f(x)=a x+\int\_{(0,\infty)}\mu(dv) f\_v(x) \tag{1}\label{1}$$
for some real $a\ge0$, some finite measure $\mu$ on $(0,\infty)$, and all real $x\ge0$, then the function $f$ on $(0,\infty)$ is nondecreasing, concave, and continuous, and $f(0)=0$.
We conclude that the functions $f$ on $[0,\infty)$ that are nondecreasing, concave, and continuous, and with $f(0)=0$ are precisely the functions of the form \eqref{1}.
| 2 | https://mathoverflow.net/users/36721 | 448012 | 180,393 |
https://mathoverflow.net/questions/448006 | 4 | Let $p > 3$ be prime. Is is true that there exists $x \in \mathbb{Z}\_p$ such that
$$
(1+x^2)^3-1
$$
is not a square in $\mathbb{Z}\_p$? In particular, when $-1$ is not a square in $\mathbb{Z}\_p$, can we show that the equation
$$
-y^2 = (1+x^2)^3-1
$$
always has non-trivial solutions $(x,y) \ne (0,0)$ in $\mathbb{Z}\_p$?
| https://mathoverflow.net/users/506018 | Polynomial that is not always a square over $\mathbb{Z}_p$ | Yes, this is true.
We have $(1+x^2)^3-1 = x^2 (x^4 + 3x^2 + 3)$, so we're asking for
nonzero x such that $x^4 + 3x^2 + 3$ is not a square. If none exist then
the genus-1 curve $y^2 = x^4 + 3x^2 + 3$ has at least $2p-4$ rational points
over the $p$-element field (including the two points at infinity, and
subtracting at most 4 for the zeros of $x^4 + 3x^2 + 3 \bmod p$).
We can now use the Hasse bound, which says that the number of points is
at most $p + 2\sqrt{p} + 1$, to give an upper bound on $p$; explicitly,
if $2p-4 \leq p + 2\sqrt{p} + 1$ then $p < 7 + \sqrt{24} < 12$.
So we need only only exhibit solutions for $p=5,7,11$,
and it turns out that $x = \pm 4$ works for each of them. **QED**
| 10 | https://mathoverflow.net/users/14830 | 448017 | 180,396 |
https://mathoverflow.net/questions/448013 | 0 | Let $f : \mathbb{R} \to \mathbb{R}$ be a smooth function which has a smooth inverse and satisfies the estimate
\begin{equation}
\lvert f(x) \rvert \leq \lvert x \rvert.
\end{equation}
Also, let $d\mu$ be the standard normal Gaussian measure on $\mathbb{R}$.
Assume that
\begin{equation}
\int\_{\mathbb{R}} \lvert f(x) -f(y) \rvert^2 d\mu(x) d\mu(y) \leq \int\_{\mathbb{R}} \lvert x-y \rvert^2 d\mu(x) d\mu(y)
\end{equation}
as well.
Now, if we taylor-expand $f(x)$ around some fixed $w \in \mathbb{R}$, we have
\begin{equation}
f(x)-f(w)=f'(w)(x-w)+F(x,w)
\end{equation}
Here, the remainder term $F(x,w)$ satisfies the decay $o(\lvert x-w \rvert)$ when $\lvert x-y \rvert \to 0^+$. However, I wonder if there is any bound over
\begin{equation}
\frac{\int\_{\mathbb{R}}\int\_{\mathbb{R}} \lvert F(x,w)-F(y,w) \rvert^2 d\mu(x)d\mu(y)}{\int\_{\mathbb{R}}\int\_{\mathbb{R}} \lvert x-y \rvert^2 d\mu(x)d\mu(y)}
\end{equation}
with respect to quantities regarding $f$ at $w$
This seems to be related to the behavior of $F(x,w)$ at $x$ far away from $w$, but I cannot figure out such behaviors..
Could anyone please help me?
| https://mathoverflow.net/users/56524 | Estimating the bound of the integral over whole $\mathbb{R}$ of the Taylor remainder term? | $\newcommand\R{\mathbb R}$We want to bound the ratio
\begin{equation\*}
R:=\frac ND,
\end{equation\*}
where
\begin{equation\*}
N:=\iint\_{\R^2}(F(x,w)-F(y,w))^2 \mu(dx)\mu(dy),\quad
D:=\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy).
\end{equation\*}
We have $D=2$. Next,
\begin{equation\*}
A:=(F(x,w)-F(y,w))^2=(f(x)-f(y)-f'(w)(x-y))^2 \\
\le2(f(x)-f(y))^2+2f'(w)^2(x-y)^2
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
R=\frac N2&\le\iint\_{\R^2}(f(x)-f(y))^2 \mu(dx)\mu(dy) \\
&+f'(w)^2\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy) \\
&\le(1+f'(w)^2)\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy)
=2(1+f'(w)^2).
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
On the other hand,
\begin{equation\*}
A\ge\frac12\,f'(w)^2(x-y)^2-(f(x)-f(y))^2
\end{equation\*}
and hence
\begin{equation\*}
R=\frac N2\ge\frac14\,f'(w)^2\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy) \\
-\frac12\,\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy)
= \frac12\,f'(w)^2-1. \tag{2}\label{2}
\end{equation\*}
Thus, we have the upper bound on $R$ in \eqref{1} in terms of $|f'(w)|$. However, \eqref{2} shows that, if we cannot control $|f'(w)|$, then there is no finite bound on $R$.
| 3 | https://mathoverflow.net/users/36721 | 448023 | 180,400 |
https://mathoverflow.net/questions/448015 | 0 | Let $k$ be an algebraically closed field of characteristic zero. Let $\Sigma$ be the fan in $\mathbb{R}^2$ consisting of three cones, cone generated by $e\_1,e\_2$,cone generated by $e\_2,-e\_1-2e\_2$ and the cone generated by $-e\_1-2e\_2,e\_1$.Let $U$ be the toric variety associated to the cone in $\mathbb{R}^2$ which is generated by $e\_1,e\_2$. Then it is easy to show that $U$ is $\mathbb{A}^2\_k$. I am trying to compute all the $G$-theory groups of the toric variety associated to the fan $\Sigma$, which is the weighted projective space $\mathbb{P}(1,1,2)$. Here,$G$-theory means the algebraic $K$-theory of the abelian category of coherent sheaves.I am thinking of using the $G$-theory localization sequence induced by the homotopy fibration $G(Z)\rightarrow G(X)\rightarrow G(U)$, where $X$ is $\mathbb{P}(1,1,2)$, $Z$ is the complement of $U$ in $X$.
Since $G\_n(U)\cong G\_n(k)$ for all $n$, after passing to the reduced $G$-theory localization sequence, we get that $\tilde{G}\_{n}(Z)\rightarrow \tilde{G}\_n(X)$ are isomorphisms for all $n$. Here,I defined $\tilde{G}\_n(U)$ to be the cokernel of the homomorphism $G\_n(k)\rightarrow G\_n(U)$ induced by the structure morphism of the $k$-variety $U$. $\tilde{G}\_n(X)$ is defined similarly.
So now I need to understand the closed subvariety $Z$ of $X$, in order to compute the $G$-theory groups of $X$.
Also, are there known methods of calculating the $G$-theory groups of weighted projective spaces?
| https://mathoverflow.net/users/477848 | How to compute the $G$-theory of the weighted projective space $\mathbb{P}(1,1,2)$? | If $X$ is a projective rational normal surface then
$$
G\_0(X) \cong \mathbb{Z} \oplus \mathrm{Cl}(X) \oplus \mathbb{Z},
$$
see, e.g., Lemma 4.2 in [Karmazyn, Joseph; Kuznetsov, Alexander; Shinder, Evgeny. Derived categories of singular surfaces. J. Eur. Math. Soc. (JEMS) 24 (2022), no. 2, 461--526]. Since
$$
\mathrm{Cl}(\mathbb{P}(1,1,2)) \cong \mathbb{Z},
$$
it follows that $G\_0(\mathbb{P}(1,1,2)) \cong \mathbb{Z}^3$.
| 3 | https://mathoverflow.net/users/4428 | 448031 | 180,405 |
https://mathoverflow.net/questions/446397 | 4 | Adapted from [math stack exchange](https://math.stackexchange.com/questions/4694934/just-how-regular-are-the-sample-paths-of-1d-white-noise-smoothed-with-a-gaussian).
**Background**: the prototypical example of---and way to generate---smooth noise is by convolving a one-dimensional white noise process with a Gaussian kernel.
**My question**: beyond smoothness, does one-dimensional white noise convolved with a Gaussian kernel have sample paths that are analytic functions with probability 1? Since there is a lot of existing literature on the sample path regularity of Gaussian processes I am hopeful that this already has been investigated somewhere, however, I haven't found an answer so far. Any help, pointers, or references are appreciated!
**Definitions:**
* White noise: White noise $\omega\_t$ is a mean-zero stationary Gaussian process on $t\in \mathbb R$ with autocovariance function given by: $\mathbb E[\omega\_t \omega\_s]= \delta(t-s)$, where $\delta$ is the Dirac delta.
* Gaussian kernel: is the function $\Phi\_t = e^{-\beta t^2}$ for all $t \in \mathbb R$ and some parameter $\beta >0$.
* White noise convolved with a Gaussian kernel: the process $w\_t=\int\_{\mathbb R} \omega\_s \Phi\_{t-s} d s$
**Known so far:**
* Properties of $ w\_t$: this is a mean-zero stationary Gaussian process with smooth sample paths and autocovariance given by
$$\kappa(h):=\mathbb E[w\_t w\_{t+h}] =e^{-\frac\beta 2h^2}\sqrt{\frac{\pi}{2 \beta}}.$$
* For any $n \in\mathbb N$, the derivative process $w^{(n)}\_t:=\frac{d^n}{dt^n}w\_t$ is a mean zero stationary Gaussian process with autocovariance
$\mathbb E[w^{(n)}\_t w^{(n)}\_{t+h}]= (-1)^n\kappa^{(2n)}(h)$. The latter comes from the **relation**:
\begin{equation}
\label{eq: autocov n m}
\mathbb E[w^{(n)}\_t w^{(m)}\_{t+h}]= (-1)^n\kappa^{(n+m)}(h).
\end{equation}
To prove the relation we first **claim**:
$
\mathbb E[w\_t w^{(m)}\_{t+h}]= \kappa^{(m)}(h).
$
We **prove the claim** by induction on $m$. The case $m=0$ is the definition. We show the statement for $m\geq 1$ assuming it holds for $m-1$:
\begin{align\*}
\mathbb E[w\_t w^{(m)}\_{t+h}] &= \lim\_{s \to 0}\frac 1 s\mathbb E[w\_{t}(w^{(m-1)}\_{t+h+s}-w^{(m-1)}\_{t+h})]\\
&= \lim\_{s \to 0}\frac 1 s \left( \kappa^{(m-1)}(h+s)-\kappa^{(m-1)}(h)\right) \\
&= \kappa^{(m)}(h).
\end{align\*}
We now **prove the relation**.
Fix $m\geq 0$. We prove the relation by induction on $n$. The case $n=0$ is the claim. We show the statement for $n\geq 1$ assuming it holds for $n-1$:
\begin{align\*}
\mathbb E[w^{(n)}\_t w^{(m)}\_{t+h}]&= \lim\_{s \to 0}\frac 1 s\mathbb E[(w^{(n-1)}\_{t+s}-w^{(n-1)}\_t)w^{(m)}\_{t+h}]\\
&= \lim\_{s \to 0}\frac 1 s \left( \mathbb E[w^{(n-1)}\_{t+s} w^{(m)}\_{t+h}]- \mathbb E[w^{(n-1)}\_{t} w^{(m)}\_{t+h}]\right)\\ %intermediate step
&= (-1)^{n-1}\lim\_{s \to 0}\frac 1 s \left( \kappa^{(n-1+m)}(h-s)-\kappa^{(n-1+m)}(h)\right) \\
&=(-1)^{n}\lim\_{s \to 0}\frac 1 s \left( \kappa^{(n-1+m)}(h+s)-\kappa^{(n-1+m)}(h)\right) \\ %intermediate step
&=(-1)^{n}\kappa^{(n+m)}(h).
\end{align\*}
| https://mathoverflow.net/users/121501 | Just how regular are the sample paths of 1D white noise smoothed with a Gaussian kernel? | Yes, $t \mapsto w\_t$ extends to an entire function $\mathbb{C} \to \mathbb{C}$ with probability $1$.
Here is an explicit construction for $(w\_t)\_{t \in \mathbb{R}}$.
Let $(X\_n)\_{n \ge 0}$ be independent real standard normal random variables, and let
$$w\_t := (\frac{\pi}{2 \beta})^\frac{1}{4} e^{-\frac{\beta}{2} t^2} \sum\_{n \ge 0} \frac{X\_n}{\sqrt{n!}} (\sqrt{\beta} t)^n$$
for $t \in \mathbb{R}$.
A basic fact about normal distributions:
If $d \ge 1$, $v\_n \in \mathbb{R}^d$ for $n \ge 0$, and $\sum\_n \|v\_n\|^2 < \infty$, then $V = \sum\_n X\_n v\_n$ is a real $d$-dimensional normal random vector, with $E[V] = 0$ and $E[V V^T] = (\sum\_n (v\_n)\_p (v\_n)\_q)\_{p,q \in \{1,\dotsc, d\}}$.
So $(w\_t)\_{t \in \mathbb{R}}$ is joint normal, $E(w\_t) = 0$, and $E[w\_t w\_{t+h}] = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} (t^2 + (t+h)^2)} \sum\_{n \ge 0} \frac{1}{n!} (\sqrt{\beta} t)^n (\sqrt{\beta} (t+h))^n = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} h^2}$.
We can extend $w\_t$ to $t \in \mathbb{C}$, then $E[\overline{w\_t} w\_{t+h}] = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} (\overline{t}^2 + (t+h)^2) + \beta \overline{t} (t+h)} = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} (h + 2 \operatorname{Im}(t) i)^2}$.
Since $\lim\_{n \to \infty} \frac{C^n}{\sqrt{n!}} = 0$ for every $C>0$, to show that $t \mapsto w\_t$ is analytic with probability $1$, it is enough to show that if $q \in (0,1)$, then $\lim\_{n \to \infty} q^n X\_n = 0$ with probability $1$ (this is easy to check).
One could even use this method to simulate $(w\_t)\_{t \in \mathbb{R}}$ or $(w\_t)\_{t \in \mathbb{C}}$ for not too big $t$'s, by choosing a big enough $N$ and using $w\_t \approx (\frac{\pi}{2 \beta})^\frac{1}{4} e^{-\frac{\beta}{2} t^2} \sum\_{n = 0}^{N-1} \frac{X\_n}{\sqrt{n!}} (\sqrt{\beta} t)^n$ (for bigger $t$'s the convergence is slower, so one needs a bigger $N$).
I am sure there are faster algorithms, especially for big $t$'s, but at least this is pretty simple.
| 1 | https://mathoverflow.net/users/42355 | 448039 | 180,407 |
https://mathoverflow.net/questions/448035 | 5 | Let $f\_1(x)\in \mathbb{Z}[x]$ be a fixed irreducible degree 4 polynomial such that its splitting field $F\_1$ is an $S\_4$-Galois extension over $\mathbb{Q}$ and the discriminant of $F\_1$ is of the form $-k^2$ for some integer $k$. It is possible to show that $F\_1$ contains $\mathbb{Q}(\sqrt{-1})$.
Does there always exist another irreducible degree 4 polynomial $g(x)\in \mathbb{Z}[x]$ whose splitting field $F$ is an $S\_4$-Galois extension over $\mathbb{Q}$ and such that $F\cap F\_1 = \mathbb{Q}(i)$?
In fact, I suspect that there should be infinitely many such extensions $F/\mathbb{Q}$ which are $S\_4$-Galois over $\mathbb{Q}$ satisfying $F\cap F\_1 = \mathbb{Q}(i)$? If my guess is actually correct, is there a generic way to construct a family of such polynomials which give rise to these distinct $F$?
| https://mathoverflow.net/users/116598 | Splitting fields of degree 4 irreducible polynomials containing a fixed quadratic extension | Yes, this can be done. Let $K$ be a $S\_4$-quartic field, $C$ its cubic resolvent field, and $L$ the Galois closure of $K$. By Galois correspondence, $L$ contains a unique quadratic subfield $Q$ which corresponds to the alternating group $A\_4$; by our hypothesis, this quadratic subfield is equal to $\mathbb{Q}(\sqrt{-1})$. Note that the Galois closure $C^\prime$ of $C$ is also contained in $L$, and by Galois correspondence again $C^\prime$ has a unique quadratic subfield, which is then necessarily equal to $Q = \mathbb{Q}(\sqrt{-1})$. Therefore, to construct the required quartic polynomials (fields) we first look at the possible cubic resolvent fields.
It is known that the discriminant of a cubic field $C$ can be expressed uniquely in the form $\Delta(C) = df^2$, where $d$ is the discriminant of the quadratic resolvent field of $C$. By our assumption, we must have $d = -4$. We are thus looking for cubic fields whose discriminants are equal to $-(2k)^2$ for some $k \geq 1$. By the Delone-Faddeev correspondence, this is the same as looking for ($\text{GL}\_2(\mathbb{Z})$-equivalence classes of) binary cubic forms with integer coefficients and discriminant $-4k^2$.
Restricting to monic binary cubic forms of the shape $F(x,y) = x^3 - Axy^2 + By^3$ whose discriminant is $4A^3 - 27B^2$, this gives the equation
$$\displaystyle 4A^3 = 27B^2 - 4k^2 \text{ giving } A^3 = 27b^2 - k^2, b = B/2.$$
This equation defines a genus 0 curve and can be explicitly parametrized.
Now to answer the question: given an initial $F\_1$ with discriminant $-k^2$, construct infinitely many cubic fields $C\_\ell$ with discriminant equal to $-4\ell^2$ and $\gcd(k, \ell)$ equal a power of 2 as above. Then, using the results of this paper ([Dirichlet series associated to quartic fields with given cubic resolvent](https://link.springer.com/article/10.1007/s40993-015-0001-y)) by Cohen and Thorne, one can give infinitely many quartic fields with $K$ with cubic resolvent field equal to $C\_\ell$. The resulting Galois closures of the fields $K$ generated this way will satisfy your constraints.
| 6 | https://mathoverflow.net/users/10898 | 448054 | 180,409 |
https://mathoverflow.net/questions/448051 | 1 | ### Question
I am making a tree using the following two functions:
$$f(x)=\frac{x}{r},\quad g(x)=\frac{x+b}{r}$$
where $1<r<2$ and $0<b$ are rationals. Everything is a real number here.
The starting point is $x=0$. From here, we will compute $f(0)$ and $g(0)$ which will be the branches of $x=0$. We keep on repeating this process by composing the functions in every possible combination, e.g. $f(f(0))$ and $g(f(0))$ will be the branches of $f(0)$, and $f(g(0))$ and $g(g(0))$ will be the branches of $g(0)$, and so on. If we repeat this process $n$ times, we have $2^n$ branches.
My question is, will these tree branches eventually cover all rational numbers in a certain range?
### Some headway
Intuitively, when $0<x<\frac{b}{r-1}$, $f(x)$ decreases $x$, whereas $g(x)$ increases $x$, i.e. $f(x)<x$ and $g(x)>x$.
Also, let's not consider all the branches off of $f(0)$, since $f(0)=0$ (it's the same as the starting point $x=0$.) Therefore the real branching starts at the point $g(0)=b/r$.
Now let's consider the branch path where we keep on applying $g(\cdot)$. The $n$ number of compositions would be
$$ b\sum\_{i=1}^n\frac{1}{r^i}$$
which converges to $\frac{b}{r-1}$ as $n\rightarrow \infty$. Therefore, the extremities of the tree are $0$ and $\frac{b}{r-1}$, which sets the range.
I guess a related or sub-question is: can there even be branches that have the same value?
| https://mathoverflow.net/users/173974 | Will this "tree" cover all rational numbers in a range? | The main Question is answered in the negative by @Saúl RM in the comments.
We answer the sub-question
"can there even be branches that have the same value?":
not for any rational $r$ other than $1$ and $-1$
(neither of which is in the allowed range $1 < r < 2$).
Indeed the "branches" are precisely the sums of finite subsets of
$\{ b/r, b/r^2, b/r^3, \ldots, b/r^n, \ldots, \}$.
(Proof: by induction on $n$, after $n$ steps we have the $2^n$ subsets of
$\{ b/r, b/r^2, b/r^3, \ldots, b/r^n \}$.) So, some two branches coincide
if and only if $P(r) = 0$ for some nonzero polynomial $P$
each of whose coefficients is $0$ or $\pm 1$.
But such $P$ cannot have any rational roots other than
$0$ (which is not allowed because $r$ is the denominator of $f$ and $g$)
and $\pm 1$. **QED**
| 4 | https://mathoverflow.net/users/14830 | 448055 | 180,410 |
https://mathoverflow.net/questions/448057 | 4 | I want to calculate the centralizer of a subgroup of $\mathrm{GL}(n, \mathbb{Z})$ when its generators $G\_i$ (whose number is finite) are known.
For any matrix $S$ that commutes with the group: $G\_iS$ = $SG\_i$, and I get a system of linear equations. Any commuting element then can be written as $S = \sum\_k x\_k S\_k, x\_k \in \mathbb{Z} $, where $S\_k$ are matrices with integer entries, and the commuting matrices form a ring with a finite basis.
But how do I calculate from these equations the group of invertible matrices? Particularly since the centralizer can have infinite elements.
Some special cases for $n=4$ are addressed here:
<https://www.ams.org/mcom/1973-27-121/S0025-5718-1973-0333025-7/S0025-5718-1973-0333025-7.pdf>
but is there a general method of solving the problem?
| https://mathoverflow.net/users/173855 | Calculating the centralizer of a subgroup of $\mathrm{GL}(n, \mathbb{Z})$ | There is an algorithm to do this (and also to test two matrices in ${\rm GL}(n,{\mathbb Z})$ for conjugacy) described in [the paper](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms.12246):
The conjugacy problem in ${\rm GL}(n,{\mathbb Z})$
Bettina Eick, Tommy Hofmann, E. A. O'Brien, Journal of the London Mathematical Society,
Volume 100, Issue 3,
December 2019
Pages 731-756.
It is implemented in Magma as the function $\mathtt {GLCentraliser}$.
*Later edit*: In view of the comments, I should have mentioned that the algorithm returns a finite generating set of the centralizer.
It is also worth pointing out that, given a finitely generated subgroup $G$ of ${\rm GL}(n,{\mathbb Z})$, there is a surprisingly fast algorithm to decide whether $G$ is finite and, if so, to find an isomorphic image of $G$ in ${\rm GL}(n,{\mathbb F}\_q)$ for some finite field $q$, which facilitates further investigation of the structure of $G$.
If $G$ is infinite, then you can decide which of the two Tits' alternatives it satisfies. There is not much more you can do if it has a free subgroup (although you can often find a free subgroup). If it is virtually solvable, then you can do further structural computations.
| 8 | https://mathoverflow.net/users/35840 | 448061 | 180,413 |
https://mathoverflow.net/questions/448046 | 2 | I'm trying to structure a proof where there are several algebras instantiated over sets, where the properties that you get from the algebraic theories are important, but the properties of the sets themselves are also important.
This is why I want to ask whether extending a proof system with models of first order theories and first order theories is consistent.
Here is what I am saying:
Deductive rules include all of the deductive rules of system LK, as well as
from nothing infer $\vdash A$ where $A$ is an axiom of ZFC
Now, suppose you define first order formulas over a signature. Then for every such signature gives the set of formulas $\Phi$ with meta-variables $\phi \in \Phi$.
Then we unroll the satisfaction relation into the deduction rules as follows:
from $\vdash (\mathfrak{M}, \mathfrak{s}) \vDash \forall x, \phi$ infer $\vdash \forall y \in \mathfrak{M}, (\mathfrak{M}, \mathfrak{s}[x := y]) \vDash \phi$
and conversely, from $\vdash \forall y \in \mathfrak{M}, (\mathfrak{M}, \mathfrak{s}[x := y]) \vDash \phi$ infer $\vdash (\mathfrak{M}, \mathfrak{s}) \vDash \forall x, \phi$
Is this a standard way of "using multiple algebraic theories in a proof"? Would this cause the system to become inconsistent? Thanks.
| https://mathoverflow.net/users/324769 | Extending a first-order deductive system with satisfaction relation | Given that your proof system includes ZFC, the standard way to handle algebraic structure like this in a set-theoretic context is simply to interpret those concepts in set theory. In the language of set theory we can interpret the notions of signature, model, language, and it is simply a ZFC theorem that every model $\mathcal{M}$ in a given signature admits a satisfaction relation $\models$, as defined and proved by Tarski. And part of what that means is that $\mathcal{M}\models\forall x\,\phi$ if and only if for every $y$ in the domain of $\mathcal{M}$ we have $\mathcal{M}\models\phi[x:=y]$. That is, your biconditionals are just part of the Tarski recursion, defining what it means to have a satisfaction relation $\models$.
From this perspective, we don't need to add any new inference rules to the proof system—we are just proving things in ZFC.
However, a slightly revised and more formally correct perspective is that whenever we introduce defined terms to a theory, such as defining the concepts of ordinal, functions, sequences, models, signature, satisfaction $\models$ and so on in ZFC, then we have introduced a definitional expansion of the theory to include these defined terms. We can treat this formally by augmenting the formal system with the definitions of those defined terms. If one does this, then the right treatment of $\models$ would not be just what you wrote, but rather the full Tarski recursive definition of satisfaction.
| 2 | https://mathoverflow.net/users/1946 | 448063 | 180,415 |
https://mathoverflow.net/questions/448008 | 5 | In Libgober's paper [Alexander polynomial of plane algebraic curves and cyclic multiple planes](http://homepages.math.uic.edu/%7Elibgober/otherpapers/export/1982alexanderduke.pdf), Example 2 (p.850), Libgober claims that the complement to this curve (i.e. $x^2u=y^3$ relative to the line in infinity $u=0$) is a retract of the complement of the trefoil knot in $S^3$. I wonder how to see this. Is in general the complement of a plane curve (relative to a line in infinity) the retract of a knot in $S^3$? If not, when is that true?
I heard that the complement to an affine curve retracts on a complement to a knot (for any weighted homogenous curve). Is that true and is there a reference?
| https://mathoverflow.net/users/100624 | Complement of plane curve and knot | If you have a weighted homogeneous polynomial $f(z\_1,z\_2)$ then that
means there's a $\mathbb{C}^\times$ action which preserves both
the curve $C=\{f=0\}$ and its complement. Take a small sphere
$S$ centred at the origin which is transverse to the orbits of
this group action. Its intersection with the curve is a link $L$
(not necessarily a knot); these knots and links are examples of
algebraic links. They are well-studied and two of the classic
references are Milnor "Singular points of complex hypersurfaces" (Chapter 10)
and Brieskorn--Knoerrer "Plane algebraic curves" (Chapter 8.5). The subset
$S\setminus L\subset\mathbb{C}^2\setminus C$ is a deformation
retract: you can construct the deformation retraction using the
scaling action of $\exp(\mathbb{R})\subset\mathbb{C}^\times$ to
move points into $S$.
The simplest example is the curve $\{z\_1=0\}$, which gives the
unknot in $S^3$. If you take $\{z\_1z\_2=0\}$ then you get the
2-component Hopf link. You can see this by stereographically
projecting the 3-sphere $|z\_1|^2+|z\_2|^2=1$ from the point
$(0,i)$. The intersection of the curve with the sphere consists
of the unit circles in the $z\_1$ and $z\_2$ planes, which project
respectively to the unit circle in the $xy$-plane and the
$z$-axis. When you project instead from a nearby point that
isn't on the $z\_2$-axis, this vertical line becomes a very big
circle that links with the first one.
It's a bit harder to see why you get the trefoil from
$z\_1^2=z\_2^3$. More generally $z\_1^p=z\_2^q$ gives you the
$(p,q)$-torus link. Brieskorn and Knoerrer give a good
explanation of how to figure out what you get. More generally,
an algebraic link is obtained by intersecting a small sphere
with a curve, but the curve doesn't have to be weighted
homogeneous. In that case, you get an "iterated torus knot": you
take a torus knot and thicken it, then stick another torus knot
on the boundary of the thickening, etc. Which torus knots you
use depends on the Newton-Puiseux expansion of $f$. Edit: Of course if you don't have a weighted homogeneous polynomial then you won't be able to say anything global about the complement of the curve; the comment about iterated torus knots only applies to how the curve intersects a small sphere centred at a point on the curve. If you're interested in general curve complements then there are also methods for studying them with a knotty flavour (e.g. braid monodromy) but there's no reason to expect them to retract onto knot complements.
| 4 | https://mathoverflow.net/users/10839 | 448065 | 180,416 |
https://mathoverflow.net/questions/448044 | 2 | Is [Scholl, *Motives for modular forms*, Theorem 1.2.4 (ii)] proven for any $p$ independent of the weight?
Concretely, let $f$ be a normalized eigenform of weight $w$. Let $p$ be a prime not dividing the level of $f$. Then the $\pi$-adic ($\pi \mid p$) realization is crystalline at $p$ and the characteristic polynomial equal to the Hecke polynomial of $f$ for $p$.
[Scholl] says that forthcoming work of Faltings should remove his condition $p \geq w$, but I read that there might be a mistake in Faltings' article. Is this true?
If it is true and known (which is what several posts like [Galois representations attached to newforms](https://mathoverflow.net/questions/14508/galois-representations-attached-to-newforms) here suggest), what is a citable reference (article and theorem therein)?
| https://mathoverflow.net/users/471019 | $\pi$-adic Galois representations of attached to newforms at $p \nmid N$ are crystalline | Blasius and Rogawski's paper "Motives for Hilbert modular forms" (1993) proves a more general result for Hilbert modular forms over any totally-real field, which includes this as a special case.
| 4 | https://mathoverflow.net/users/2481 | 448067 | 180,417 |
https://mathoverflow.net/questions/448056 | -2 | Consider a simple branching process $Z\_0,Z\_1,Z\_2...$ such that at every discrete step, a particle splits into $k\geq1$ particles where $k$ follows a discrete distribution with probability mass $p(k)$.
From the theory of branching processes, a basic result is that the average number of particles at the $n$th step is given by:
$$\mu:=\mathbb{E}[Z\_n|Z\_0=1]=m^n,$$
where $m=\sum\_{k=1}^\infty k p(k)$ is the mean of the offspring distribution.
### Question
What happens to $\mu$ when the offspring distribution depends on the number of steps? That is, $p(k)\to p\_n(k)$ where $n$ is a parameter that represents the number of steps.
What is the typical vocabulary to describe this generalization? I could not find this with some naive searches on google scholar.
Finally, to illustrate it with an example: Let
$$p\_n(k)= p(k) e^{-\frac{k}{n}}c\_n,$$
where $c\_n$ is to normalise $p\_n(k)$ and $p\_0(k):=p(k)$. Is it still possible to express $\mu$ with a close formula?
(This question is motivated by population dynamics, where as a population grows, the number of new-borns decreases due to limited space, competition etc.)
| https://mathoverflow.net/users/481145 | Branching process with varying offspring distribution at each step | As far as terminology goes, I have seen this called "Branching process in a varying environment" and also "inhomogeneous Galton-Watson process".
Starting from one individual, the mean of $Z\_n$ is simply $\prod\_{i=1}^n m\_i$ where $m\_i$ is the mean of the offspring distribution for the $i$th generation. For example you can get this by induction using the conditional expectation $E(Z\_n|Z\_{n-1})=m\_n Z\_{n-1}$.
| 1 | https://mathoverflow.net/users/5784 | 448069 | 180,418 |
https://mathoverflow.net/questions/448049 | 5 | Let $A$ be a finite set of free generators and their inverses and $F$ the free group generated by elements in $A$ (some call $A$ the **alphabet** of $F$). For each $g\in F$, use $\vert\,g\,\vert$ to denote the length of $g$. The [**Gromov boundary**](https://en.wikipedia.org/wiki/Gromov_boundary) of $F$, denoted by $\partial F$, can be viewed as the set of words that consists of letters from $A$ and has infinite length starting from the left. For instance, $a\_1 a\_2 \cdots $ is the general form of elements in $\partial F$ where $a\_i\in A$ and $a\_i a\_{i+1}\neq e$ for each $i\in\mathbb{N}$.
Set $\overline{F} = F\cup\partial F$. Given $x, y\in\overline{F}$, if $x\neq y$, let $x\wedge y$ denote the common part between $x$ and $y$ starting from the left. For instance: $ (a\_1 \cdots a\_n )\wedge (a\_1 \cdots a\_n b\_{n+1} b\_{n+2} \cdots) = a\_1 \cdots a\_n$. Define a function $d: \overline{F}\times\overline{F}\rightarrow(0, 1)$ as follows: $d(x, y)=0$ if $x=y$ and $d(x, y)= \operatorname{exp}(-\vert\,x\wedge\,y\,\vert)$. One can check $(\overline{F}, d)$ is a compact metric space and $d$ is an ultrametric. From now on, we assume $\overline{F}$ is equipped with the $d$-metric topology. An action of $F$ on $\partial F$ can be defined as follows: given $g\in F$ and $x\in\partial F$, $g\cdot x = gx$ (after cancellation).
Given $\gamma\in\partial F$, for each $i\in\mathbb{N}$, let $\gamma\_i$ denote the $i$-th letter of $\gamma$ and define $[\gamma]\_{\leq i} = \gamma\_1 \cdots \gamma\_i$. Obviously, for each $\gamma\in\partial F$, we have $\gamma = \lim\_i [\gamma]\_{\leq i}$. Fix $\gamma\in\partial F$ such that $\gamma \neq \lim\_n gh^n$ for any $g, h\in F$. Then define:
$$
C = \overline{ \big\{ [\gamma]\_n^{-1} \big\}\_{n\in\mathbb{N}} }
$$
$C\cap\partial F$ is the set of clustered points of the sequence $\big\{ [\gamma]\_n^{-1} \big\}$ and, by compactness of $\overline{F}$, is non-empty. My question is: is it true that for each $x\in C$, $\lim\_n [\gamma]\_n x = \gamma$? If not, is the set of exceptions countable or uncountable?
**Update**: Thanks for Sam's answer, the answer to the question above is negative. Now I wonder, if the set of exceptions is always countable regardless of $\gamma$ that does not have the form $\lim\_n gh^n$. One of my attempts shows the following statement is true:
$$
\forall k\in\mathbb{N}\, \forall N\in\mathbb{N}\, \exists\,m>N \hspace{0.3cm} \text{ suh that } \vert\, [\gamma]\_N \wedge [\gamma]\_N[\gamma]\_m^{-1}\,\vert \geq k
$$
This tells for a fixed $k\in\mathbb{N}$ and for any subsequence of $\{[\gamma]\_n\}$, there exists another subsequence of $\{ [\gamma]\_m^{-1} \}$ such that the latter subsequence will not cancel all and leave at least the first $k$ letters of $\gamma$. I believeed applying the diagonalization method could give me something but then failed.
| https://mathoverflow.net/users/151332 | Cancellation of elements in the Gromov boundary of a free group | It is not true. Consider the following point at infinity:
$
\gamma = a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbbbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdots
$
That is, $\gamma$ is the limit of the sequence $(\gamma\_n)$ defined by $\gamma\_0 = a$ and $\gamma\_{n+1} = \gamma\_n \cdot b^{n+1} \cdot \gamma\_n$. It is an exercise to show that $C$ contains $x = \lim \gamma\_n^{-1}$. It is also an exercise to show that the sequence $([\gamma]\_k x)$ does not converge.
---
I will further guess that (for carefully choosen $\gamma$) there will be uncountably many $x \in C$ where the sequence $([\gamma]\_k x)$ does not converge.
| 2 | https://mathoverflow.net/users/1650 | 448079 | 180,420 |
https://mathoverflow.net/questions/448087 | 1 | I am starting to learn the $K$-theory of triangulated categories and is stuck with the following.
Let $\mathcal{T}$ be a triangulated category having a semi-orthogonal decomposition $\langle \mathcal{A\_1},\mathcal{A\_2},…,\mathcal{A\_n}\rangle$.
I saw in the paper of Orlov titled Smooth and Proper Noncommutative Schemes and Gluing of DG Categories mentioning an isomorphism $K\_i(\mathcal{T})\cong K\_i(\mathcal{A\_1})\oplus K\_i(\mathcal{A\_2})\oplus…\oplus K\_i(\mathcal{A\_n})$ for all $i$.How do we prove this result?
Are there any detailed references for this result?
I am thinking of applying this result to the case where $\mathcal{T}$ is the bounded derived category of coherent sheaves on an algebraic variety.
| https://mathoverflow.net/users/477848 | How to compute the higher $K$-theory of a triangulated category having a semi-orthogonal decomposition? | Assume $\mathcal{T} = D^b(X)$ and $X$ is smooth. Then $K\_0(X \times X)$ has an algebra structure (with the convolution product) and $K\_\bullet(X)$ is a graded module over this algebra. So, the idea is to use a decomposition of $1 \in K\_0(X \times X)$ (the class of the diagonal $\Delta\_X$) into a sum of idempotents, then the action of these idempotents will give the required direct sum decomposition.
To construct the required decomposition of the diagonal, note that there are objects
$$
R\_i \in D^b(X \times X)
$$
such that the corresponding Fourrier--Mukai functors are the projection functors for the semiorthogonal decomposition, see Theorem 7.1 in [Kuznetsov, Alexander. Base change for semiorthogonal decompositions. Compos. Math. 147 (2011), no. 3, 852--876]. Moreover,
$$
[\Delta\_X] = \sum [R\_i]
$$
and $[R\_i] \in K\_0(X \times X)$ are the required idempotents.
If $X$ is singular, the argument is essentially the same, but you need to replace $K\_0(X \times X)$ by the subgroup of $G\_0(X \times X)$ formed by classes that are perfect over each factor.
If $\mathcal{T}$ does not have a geometric origin, but is enhanced, say, by a DG-algebra $A$, the argument is still the same, but you need to replace $K\_0(X \times X)$ by the Grothendieck group of appropriate category of $A$-bimodules.
| 6 | https://mathoverflow.net/users/4428 | 448090 | 180,425 |
https://mathoverflow.net/questions/431926 | 14 | Wikipedia seems to have an answer
"The concept of 2-category was first introduced by Charles Ehresmann in his work on enriched categories in 1965. The more general concept of bicategory (or weak 2-category), where composition of morphisms is associative only up to a 2-isomorphism, was discovered in 1968 by Jean Bénabou."
<https://en.wikipedia.org/wiki/Strict_2-category>
But this answer seems to me a little bit odd.
SGA 4 of Grothendieck was held between 1963-1964. There we find the following remarks
"C’est le fait que les -topos (éléments d’un univers ) forment une 2-catégorie, et non plus seulement une catégorie ordinaire comme les espaces topologiques ordinaires, qui constitue du point de vue technique la différence la plus importante entre la théorie des
topos et celle des espaces topologiques. Ce fait est la source de certaines complications techniques auxquelles on a déjà fait allusion, mais aussi de faits essentiellement nouveaux par rapport à la topologie traditionnelle."
and yes, I know that SGA 4 was edited and published then in 1972 when the notion was already known. But there is another remark
"Le fait que les -topos (éléments d’un univers ) forment une 2-catégorie permet en particulier de définir la notion d’équivalence de deux -topos , ′."
And it is very likely that a notion of equivalence of topoi was already in the years 63, 64 in the original seminars (this is a speculative remark).
SGA4 cites the thesis of Monike Hakim as a former reference but that thesis was published later.
>
> Do we have some conclusive information to add to the hypothesis that was Grothendieck who introduced the notion of 2-categories?
>
>
>
We can also say a few words about the use of Cat, already in use in SGA 1 and suggested even in the Tohoku paper. Grothendieck wrote a paper related to Quillen in 1968 where it is already present the theory of n-categories.
If you think Wikipedia is right, can you please let me know how Grothendieck enters in this picture?
| https://mathoverflow.net/users/429204 | Who introduced the notion of 2-categories? | It appears that the definition of 2-category was introduced independently by two authors, both of whom independently introduced the modern notion of enriched category, for which 2-categories appeared as an example.
* Jean Bénabou gives 2-categories as example (5) in the 1965 paper [Catégories relatives](https://gallica.bnf.fr/ark:/12148/bpt6k4019v/f37.item), under the name "2-catégories".
* Jean-Marie Maranda gives 2-categories as an example on the second page (p. 759) in the 1965 paper [Formal categories](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/formal-categories/A7C463460EB8CAC64C2CA340F870CF80), under the name "categories of the second type". The definition is spelled out in §2.
Both authors mention the connection to Ehresmann's double categories.
---
Bénabou cites his own forthcoming thesis for the concept, which was eventually published under [Structures algébriques dans les catégories](http://www.numdam.org/item/?id=CTGDC_1968__10_1_1_0) (which is not precisely the name mentioned in the paper). He also points the reader to Ehresmann's 1963 [Catégories structurées](https://eudml.org/doc/urn:eudml:doc:81794), but the definition of 2-category does not appear here; the reference is most likely due to the similar notion of double category, of which the notion of 2-category is a special case.
Other authors, such as Eilenberg and Kelly in the 1965 paper [Closed categories](https://link.springer.com/chapter/10.1007/978-3-642-99902-4_22), also cite Ehresmann's 1963 paper for the notion of 2-category. However, as it does not appear there, it is more likely they intended to cite Ehresmann's similarly named 1965 book *Catégories et structures* (which is the reference currently [listed on Wikipedia](https://en.wikipedia.org/wiki/Strict_2-category)), which does contain a discussion of 2-categories on page 324 in a historical note.
To settle the matter, I emailed Andrée Ehresmann, who wrote:
>
> [Charles Ehresmann] did not introduced himself 2-catégories, which have been introduced soon after both by his student Jean Bénabou as special double categories, and by different other authors as categories 'enriched' in Cat
>
>
>
---
The appearance in SGA 4 of 2-categories appears to have either been introduced during editing, or was based on word-of-mouth ideas. SGA 4, for instance, cites Monique Hakim's 1972 thesis, who introduced 2-categories by writing:
>
> La notion de 2-catégorie est due a J. Benabou
>
>
>
| 9 | https://mathoverflow.net/users/152679 | 448093 | 180,427 |
https://mathoverflow.net/questions/448095 | 6 | Let $\epsilon\_1,\ldots,\epsilon\_n$ be a sequence of signs and $M(t)$ be the tridiagonal matrix whose diagonal entries are $\epsilon\_1 t,\ldots, \epsilon\_n t$ and off-diagonal entries equal to $1$. Is it true that the number of real zeroes of $P(t)=\det M(t)$ is equal to the absolute value of $\epsilon\_1+\dots+\epsilon\_n$?
I checked this numerically on various cases, tried to apply the Sturm algorithm but couldn't prove it. Thanks for any help!
| https://mathoverflow.net/users/14547 | Real zeroes of the determinant of a tridiagonal matrix | For $\epsilon\_1=\epsilon\_2=-1$ and $\epsilon\_3=\epsilon\_4=\epsilon\_5=1$ you get the counterexample $\operatorname{det}M(t)=t(t - 1)^2(t + 1)^2$.
Another example, with simple real roots, is $\epsilon\_1=\epsilon\_5=-1$ and $\epsilon\_2=\epsilon\_3=\epsilon\_4=1$ with $\operatorname{det}M(t)=(t - 1)t(t + 1)(t^2 + 1)$.
| 6 | https://mathoverflow.net/users/18739 | 448098 | 180,429 |
https://mathoverflow.net/questions/448101 | 0 | Consider a random variable $X\in \mathbb{R}^d$. Let ${\theta\_m}$ be a sequence of real numbers that converge to $\theta$. Let $f(x,y)$ be a function that is not continuous. To be specific, fix, $x=a$, $f\_a(y):= f(a,y)$ is not a continuous function in $y$. However, $E(f(X,y))$ is a continuous function in $y$. Example: take $f(x,y) = I(x\leq y)$, with the given fact $P(X\leq y)$ is continuous in $y$. Now, my question is: is it possible to show,
$$E|f(X,\theta\_m)-f(X,\theta)|\xrightarrow{}0?$$
It feels like the above should be true, but unable to show it rigorously. Any hint or help will be greatly appreciated.
Initially, I thought it would be straight-forward by interchanging limit and expectation, by virtue of Fubini's theorem,
$$\text{lim}\_{m\xrightarrow{}\infty} E|f(X, \theta\_m)-f(X,\theta)| = E|\text{lim}\_{m\xrightarrow{}\infty}f(X, \theta\_m)-f(X,\theta)|$$
But, can't proceed from here since $f$ is discontinuous.
| https://mathoverflow.net/users/506211 | Convergence in expectation of a discontinuous function | Without more assumptions, it's not true.
Take $d=1$ and let
$$f(x,y) = \begin{cases} x, & \text{if } y=0 \\
0, & \text{otherwise}
\end{cases}$$
Let $X$ be any nontrivial random variable with mean zero, such as $X= \pm 1$ with probability $1/2$ each. Then $E[f(X, y)]=0$ for all $y$, which is certainly continuous in $y$. But for any $y \ne 0$ we have
$E|f(X,0) - f(X,y)| = E|X| > 0$.
| 1 | https://mathoverflow.net/users/4832 | 448104 | 180,431 |
https://mathoverflow.net/questions/448103 | 4 | Suppose that $G=(G\_{ij})$ is a positive-semidefinite symmetric matrix with the diagonal entries all equal $1$ and all off-diagonal entries $\le0$. Does it then necessarily follow that
$$\sum\_{i,j}(G^5)\_{ij}\le\sum\_{i,j}(G^3)\_{ij}\,?$$
This is true if it is additionally assumed that all off-diagonal entries of $G$ are the same.
| https://mathoverflow.net/users/36721 | An inequality for certain positive-semidefinite matrices | If I have not committed any mistake, please, find below a counter-example.
>
> **Counter-example.** Let $G\in \mathbb{S}^3\_{+}$ be defined by
> $$
> G = \begin{pmatrix}
> 1 & -\frac{2}{5} & 0 \\
> -\frac{2}{5} & 1 & -\frac{2}{5}\\
> 0 & -\frac{2}{5} & 1
> \end{pmatrix}.
> $$
>
>
>
| 9 | https://mathoverflow.net/users/138242 | 448112 | 180,433 |
https://mathoverflow.net/questions/448110 | 2 | Can a computable partial order have a maximal chain of order-type $\omega\_1^{ck}$? My instinct is to say no, of course not, but I can't actually make the argument. If the p.o. also has chains of Harrison type, there seems to be no violation of $\Sigma^1\_1$-bounding.
*Edit:* The answer is yes. Let $T$ be the tree of descending sequences in a Harrison order, so $T$ has nodes of every computable rank. Let $P$ consist of all finite antichains of $T$, and define an ordering on $P$ by $F \le G$ if for every $x \in F$ there is a $y \in G$ such that $x$ extends $y$ (in the tree order).
One shows that for $F \in P$, if every element of $F$ is ranked, then the partial order below $F$ is well-founded (König's Lemma or just an argument on ranks), and further that if $\alpha = \max\_{x \in F} \text{rank}(x)$, then $F$ bounds a chain of order-type $\alpha$ (induction on $\alpha$). Then let $x\_0, x\_1, x\_2, \dots$ be the ranked children of the root; one shows that $\{x\_0\}, \{x\_0, x\_1\}, \dots$ can be extended to a maximal chain of type $\omega\_1^{ck}$.
*Second edit:* Here's how to show that $\mathcal{O}$ has continuum many paths of length $\omega\_1^{ck}$, so $\mathcal{O}^\*$ works as an example.
Fix $H$ a computable Harrison ordering for which the successor function and the set of limit points are both computable, and nonuniformly fix the least element. Fix $(a\_n)\_{n \in \omega}$ an increasing sequence of limit points which is cofinal in the well-founded part (said sequence will be noncomputable, but that's okay). The plan is to build notations corresponding to the elements of $H$ using effective transfinite recursion, but at each $a\_n$ we'll use padding to cause a bifurcation, giving us a perfect tree of notations. Since the sequence is noncomputable, our function giving the notations will have to be fed the $a\_n$.
To the details. Let $A$ be the set of (canonically given) finite partial functions from $H$ to $2$. Let $A' \subset A$ be those functions for which all the elements of the domain are from the sequence $(a\_n)\_{n \in \omega}$. We build a partial computable function $f: \omega \times H \times A \to \mathcal{O}^\*$. We define $f(e, x, \sigma)$ as follows:
If $x$ is the least element of $H$, $f(e, x, \sigma)$ is the notation for $0$.
If $x$ is the successor of $y$ in $H$, we compute $\phi\_e(x, \sigma)$. Assuming this converges to a notation $b$, we output the notation for the successor of $b$. ($2^b$ in Kleene's system.)
If $x$ is a limit point of $H$, then compute an increasing sequence $(b\_n)\_{n \in \omega}$ cofinal below $x$, and let $i$ be such that $\phi\_i(n) = \phi\_e(b\_n, \sigma)$. If $\sigma(x) = 1$, then we use padding to find an equivalent $i' > i$; otherwise, we keep this $i$. We then output the notation corresponding to this $i$ or $i'$. ($3\cdot 5^i$, in Kleene's system.)
Now apply the $s$-$m$-$n$ theorem to get a total computable $s$ with $\phi\_{s(e)}(x, \sigma) = f(e, x, \sigma)$, then the recursion theorem to get an $\hat{e}$ with $\phi\_{s(\hat{e})} = \phi\_{\hat{e}}$.
Now we consider $\phi\_{\hat{e}}(x, \sigma)$ for $x$ in the well-founded part of $H$ and $\sigma \in A'$. By a bunch of inductions: 1) $\phi\_{\hat{e}}(x, \sigma)\downarrow$ and is a notation for the order-type of $x$; 2) $\phi\_{\hat{e}}(x,\sigma) = \phi\_{\hat{e}}(x,\sigma\upharpoonright\_{\text{support}(\sigma)}) = \phi\_{\hat{e}}(x,\sigma\upharpoonright\_{\le x})$; 3) $\phi\_{\hat{e}}(x,\sigma) \le\_{\mathcal{O}} \phi\_{\hat{e}}(y,\sigma)$ iff $x \le y$; and 4) if $\sigma(a\_n) = 1-\tau(a\_n)$, then $\phi\_{\hat{e}}(a\_n, \sigma) \neq \phi\_{\hat{e}}(a\_n,\tau)$. So this gives us our perfect tree of notations.
| https://mathoverflow.net/users/32178 | Maximal chains of order type $\omega_1^{ck}$ in computable partial orders? | This is perhaps a simplification of the example you give: let $H$ be a Harrison order and let $P$ be the poset of pairs $(a,b)\in H$ with $a<b$, ordered by $$(a,b)\le (a',b') \quad\iff\quad a\le a'\mbox{ and }b\ge b'.$$ Note that in a chain in $P$, we do *not* require both coordinates to simultaneously change.
Let $(\alpha\_i)\_{i\in\omega}$ be cofinal in $\omega\_1^{CK}$. Fix a descending sequence $(h\_i)\_{i\in\omega}$ in the Harrison order such that the ill-founded part of $H$ is exactly the union of the final segments given by the $h\_i$s. Then - identifying elements of the well-founded part of $H$ with computable ordinals - the sequence $$(\alpha, h\_{\min\{k: \alpha\_k>\alpha\}})$$ is a maximal chain in $P$ with ordertype $\omega\_1^{CK}$.
---
Here's another example:
Let $\mathcal{O}^\*$ be the "computable analogue" of $\mathcal{O}$ (so $\mathcal{O}^\*$ is a computable poset which is an end-extension of $\mathcal{O}$; see Sacks' book for details). Fix an increasing sequence $(\alpha\_i)\_{i\in\omega}$ of computable ordinals cofinal in $\omega\_1^{CK}$ and an enumeration $(p\_i)\_{i\in\omega}$ of $\mathcal{O}^\*\setminus\mathcal{O}$. We can iteratively build a sequence of notations $n\_i$ for $\alpha\_i$ such that $n\_i<\_\mathcal{O}n\_{i+1}$ and $n\_i\not<\_{\mathcal{O}^\*}p\_i$; the downwards closure of $\{n\_i:i\in\omega\}$ then is a maximal chain in $\mathcal{O}^\*$ and clearly has ordertype $\omega\_1^{CK}$. The key point is that by picking a cofinal $\omega$-sequence in $\omega\_1^{CK}$ we avoid having to deal with limit stages in this construction, which would otherwise be a problem (we might become maximal too early).
(This is closely related to the construction above, which is unsurprising.)
| 2 | https://mathoverflow.net/users/8133 | 448116 | 180,436 |
https://mathoverflow.net/questions/447986 | 1 | Given a partial order $P$ on a set $S$ does the set of ordered pairs $(x,y)$ in $S\times S\setminus P$ such that $P\cup\{(x,y)\}$ is a partial order have a name? (If so then it would apply to all sorts of orders not just partial orders.)
The answer was "no" in this crosspost: <https://cs.stackexchange.com/questions/155928/relation-based-on-a-given-partial-order-does-it-have-a-name>
If MathOverflow concurs, then it must not have a name yet. (I named it the *envelope* of $P.)$
============== edit added 2 June 2023=================
[Some parrot-like ChatGPT answers were added then removed when I learned that ChatGPT content is banned.]
| https://mathoverflow.net/users/5090 | does this relation associated with a poset have a name? | It appears that there is no established name for this concept, but if you are looking for a suggestion, **"potential covers"** might be a reasonable name, since these are precisely the pairs $(x,y)$ which are not in the partial order $P$ but would be a cover in $P\cup \{(x,y)\}$.
| 5 | https://mathoverflow.net/users/25028 | 448121 | 180,438 |
https://mathoverflow.net/questions/447993 | 2 | A Banach space $E$ is called *Grothendieck* if every weak\* convergent sequence in the dual space $E^\*$ is weakly convergent. A typical example of a Grothendieck space is $\ell\_\infty$. Diestel proved the following characterization of Grothendieck Banach spaces:
>
> For every Banach space $E$, TFAE:
>
>
> 1. $E$ is Grothendieck,
> 2. for each separable Banach space $F$, every bounded operator $T\colon E\to F$ is weakly compact,
> 3. every bounded operator $T\colon E\to c\_0$ is weakly compact.
>
>
>
**My question:** Can we replace the space $c\_0$ in item 3 by some/any other fixed separable space? In particular, can we use here $\ell\_1$?
(By "other" I of course mean "non-isomorphic to $c\_0$".)
| https://mathoverflow.net/users/15860 | Weakly compact operators into $c_0$ and other separable spaces | In the light of Professor Johnson's comment, we cannot replace $c\_0$ with another $X$ unless $X$ contains a copy of $c\_0$.
Indeed, let $X$ be a Banach space that contains no isomorphic copy of $c\_0$. Consider a $C(K)$ space that is not a Grothendieck space (e.g., $C([0,1])$ ). Any bounded linear $T:C(K)\to X$ is unconditionally converging since it does not fix a copy of $c\_0$, thus weakly compact since $C(K)$ has property $(V)$.
Generally, if $E$ (may not be a Grothendieck space) has property (V) & $X$ contains no copy of $c\_0$, then every $T\in B(E,X)$ is weakly compact.
| 3 | https://mathoverflow.net/users/164350 | 448127 | 180,440 |
https://mathoverflow.net/questions/447894 | 2 | Dirichlet's theorem says that all numbers $x\in [0,1]$ can be approximated by infinitely many fractions $p/q \in \mathbb{Q}$ with error $|x - p/q| \le 1/q^2$.
I am interested in the following question:
Fix $m>0$. Let $S(m)$ denote the set of $x\in [0,1]$ for which no fraction $p/q \in \mathbb{Q}$ with $q<100\sqrt{m}$ approximates $x$ with error smaller than $1/m$, i.e., $|x-p/q| > 1/m$ for any such choice of $p,q$.
What bounds are there for the measure of $S(m)$?
EDIT:
Let $\Phi(q)$ denote the set of numbers smaller than $q$ coprime to $q$.
And let
$$I(q) = \bigcup\_{p\in \Phi(q)} [p/q, p/q+1/m]$$
For $q\_1,q\_2<\sqrt{m}$ the sets $I(q\_1),I(q\_2)$ will be disjoint.
This can give a simple lower bound (by analyzing $\sum\_{k<\sqrt{m}} \phi(k)$, which is a well known problem). I'm interested in if anything stronger than this bound can be given.
| https://mathoverflow.net/users/155604 | Simultaneously approximating all $x \in [0,1]$ with fractions of bounded denominator | Fix $T$ ($T=100$ in your example) and denote by $S\_T(n)$ the set of numbers $x\in [0,1]$ such that $|x-p/q|>1/n^2$ whenever $p,q$ are integers and $1\leqslant q\leqslant Tn$ (so, my $n$ is your $\sqrt{m}$).
Let $0=r\_0<r\_1<\ldots<r\_K=1$ be Farey sequence of all irreducible fractions with denominator at most $Tm$. Denote $\delta\_i=r\_{i+1}-r\_i$ for $i=0,1,\ldots,K-1$ the gap between corresponding Farey fractions. Then $S\_T(n)\cap [r\_{i},r\_{i+1}]$ has length $\max(\delta\_i-2/n^2,0)$, so, the measure of $S\_T(n)$ equals
$$
f(T,n):=\sum\_{i:\delta\_i>2/n^2} (\delta\_i-2/n^2).
$$
Recall that if $r\_i=a/b$, $r\_{i+i}=c/d$ are irreducible fractions, then $\delta\_i=1/(bd)$ and $b+d>Tn$. Denote $M=\max(b,d)$, $m=\min(b,d)$. Then $M> Tn/2$, and the inequality $1/(Mm)=\delta\_i>2/n^2$ yields $1/m>2M/n^2>T/n$, so, $m<n/T$. And $\delta\_i=1/(mM)\leqslant 2/(Tnm)$. Thus, for fixed $m$, the fractions with denominator $m$ give a contribution at most
$$
2\varphi(m)\left(\frac{2}{Tnm}-\frac2{n^2}\right)\leqslant \frac4{Tn}.
$$
Summing up over $m<n/T$ gives you at most $4/T^2$.
On the other hand, if $m<n/(3T)$, say, then $mM\leqslant mTn\leqslant n^2/3$ and thus $1/bd\geqslant 3/n^2$. Therefore, each of these intervals (with $m<n/(3T)$) give contribution at least $1/n^2$, and there are about $$2\sum\_{m<n/(3T)} \varphi(m)\sim \zeta(2)\frac{n^2}{9T^2}$$
such intervals that gives lower bound also of order $1/T^2$.
So, $$c\_1/T^2<f(T,n)<c\_2/T^2$$
for some universal explicit constants $c\_1,c\_2$. If you care on sharp constant, please let me know, this requires more accurate analysis but looks doable.
| 3 | https://mathoverflow.net/users/4312 | 448129 | 180,442 |
https://mathoverflow.net/questions/448130 | 3 | Lets take the intersection of the theory of $L\_{\omega\_1^{CK}}$ and $\sf ZF + [V=L]$, this is [equivalent](https://mathoverflow.net/a/316109/95347) to the theory of [constructability from below + limit stages](https://mathoverflow.net/questions/316099/what-is-the-strength-of-this-strict-constructible-iterative-hierarchy).
>
> Can this theory prove the following definability $\omega$-rule?
>
>
>
$\textbf{Definability: }$ if $\phi\_1,\phi\_2, \phi\_3,...$ are all formulas in which only symbol "$y$" occurs free, and "$y$" never occur bound, and that doesn't use the symbol "$x$", and $\psi$ is a formula in which only symbol "$x$" occurs free, and "$x$" never occur bound; then:
$\underline {[i=1,2,3,...; \\ \forall x \, (x=\{y \mid \phi\_i\} \to \psi)]} \\ \forall x: \psi$
In English: if a parameter free formula holds for all parameter free definable sets, then it holds for all sets.
In other words, are all models of this theory [pointwise definable models](https://arxiv.org/abs/1105.4597)?
The context is generally related to having an extremely concrete kind of sets, and so it is in continuation to the context of this [posting](https://math.stackexchange.com/questions/4710877/is-sf-v-hod-satisfied-in-l); and also to the [hyperarithmetical hierarchy](https://en.wikipedia.org/wiki/Hyperarithmetical_theory#Hyperarithmetical_sets_and_iterated_Turing_jumps:_the_hyperarithmetical_hierarchy)
| https://mathoverflow.net/users/95347 | Are all constructible from below sets parameter free definable? | There are two issues with your question.
First, your statement "in other words" is not correct, since there are theories whose models have the property that whenever a statement holds of every parameter-free definable element, then it holds of every element, but not all models are pointwise definable.
**Theorem.** In any model of ZFC+V=HOD, if a property $\psi$ holds of every parameter-free definable element, then it holds of every element.
**Proof.** This theory has a parameter-free definable well ordering of the universe. So if there is a counterexample to any statement $\psi$, then there is a definable counterexample, since there will be one that is least with respect to the well ordering.
$\Box$
But not every model of ZFC+V=HOD is pointwise definable, since this is a first-order theory which, if consistent, has uncountable models, which cannot be pointwise definable.
What your inference rule is expressing is not that the models should be pointwise definable, but rather, that the definable elements form an elementary substructure of the universe. This is true whenever there is a definable global well-ordering or more generally whenever there are definable Skolem functions, since the class of definable elements will be closed under the Skolem functions and therefore constitute an elementary substructure.
What is more, I claim that your deduction rule is simply equivalent in models of ZFC (and also weaker theories) to the axiom V=HOD. The reason is that if a model has V=HOD, then it has a parameter-free definable well-ordering and hence definable Skolem functions, which makes the definable elements an elementary substructure. Conversely, if your rule is valid for a particular theory extending ZFC, then the models of that theory will have their definable elements forming an elementary substructure (by the Tarski-Vaught test). So the definitions will still work inside the substructure, where everything is definable, and so that will be a model of V=HOD, since you don't even need the ordinal parameters. So the original model also has V=HOD. In conclusion, your deduction rule is equivalent over ZFC to the set-theoretic axiom V=HOD.
This brings us to the second point. There can be no first-order theory with infinite models but only pointwise definable models. The upward Löwenheim-Skolem theorem shows that every consistent theory with infinite models has models of arbitrarily large cardinality. Once the model is larger than the number of definitions, it cannot be pointwise definable.
| 7 | https://mathoverflow.net/users/1946 | 448132 | 180,443 |
https://mathoverflow.net/questions/448142 | 2 | I am trying to understand paragraph 1.6 of Lusztig's paper "Character Sheaves I". The basic setup is that $X$ is a smooth irreducible variety over a field $k=\overline{k}$, $D\_i, i=1,...,r$ are smooth divisors with normal crossings and $\mathcal{L}$ is $\overline{\mathbb{Q}}\_l$-local system of rank one such that the corresponding monodromy action factors through a finite quotient of order invertible in $k$. Now the IC sheaf $IC(X,\mathcal{L})$ is represented by a single constructible $\overline{\mathbb{Q}}\_l$-sheaf $\overline{\mathcal{L}}$.
Then he claims that if the local monodromy of $\mathcal{L}$ around one $D\_i$ is trivial, then $\overline{\mathcal{L}}$ restricted to $U\bigcup D\_i$ is a local system (this is a special case of his claim).
What is the local monodromy of $\mathcal{L}$ around a $D\_i$ and how to show his claim?
| https://mathoverflow.net/users/173314 | Extending IC sheaves across smooth divisors with normal crossings | The local monodromy around $D\_i$ can be obtained by taking a $\eta$ a geometric generic point of $D\_i$, $R$ the etale local ring of $X$ at $\eta$ with uniformizer $\pi$, then pulling $\mathcal L$ back to $\operatorname{Spec} R[ \pi^{-1} ]$. We then obtain a representation of the étale fundamental group of $\operatorname{Spec} R[ \pi^{-1} ]$. The image of this representation is called the local monodromy.
Rather than $U \cup D\_i$, it would be more precise to say $X \setminus \cup\_{j\neq i} D\_j$. To check that $\overline{\mathcal L}$ restricted to this space is a local system, we first check that for $j \colon U \to X \setminus \cup\_{j\neq i} D\_j$ the open immersion, $j\_\* \mathcal L$ is a local system, and then that $j\_\* \mathcal L$ is the restriction of $\overline{\mathcal L}$.
The pullback of $j\_\*\mathcal L$ to $\operatorname{Spec} R$ is the pushforward from $\operatorname{Spec} R[ \pi^{-1} ]$ to $\operatorname{Spec} R$ of the restriction of $\mathcal L$ to $\operatorname{Spec} R[ \pi^{-1} ]$. So it is the pushforward of a constant sheaf from the generic point to the whole spectrum and thus is a constant sheaf. Thus $j\_\* \mathcal L$ is locally constant at the generic point of $D\_i$. Thus it is locally constant on some neighborhood of the generic point. The complement of the largest open set on which $j\_\* \mathcal L$ is locally constant is a closed set contained in $D\_i$ but not containing the generic point of $D\_i$ and thus has codimension $\geq 2$.
If that complement is nonempty, consider a generic point $\eta'$ of that complement and a Henselian local ring $R'$ at $\eta$. The lisse sheaf gives a representation of the fundamental group of the punctured spectrum of $R'$, which by purity (since the codimension is $\geq 2$) is trivial. So again the sheaf is the constant sheaf and the pushforward is just the constant sheaf, contradicting the assumption that the sheaf is not locally constant at $\eta$, so the complement is indeed empty and $j\_\* \mathcal L$ is lisse.
Since the intermediate extension is the unique perverse extension with no irreducible components supported outside $U$, and $j\_\* \mathcal L$, being lisse, is perverse and has no irreducible components supported outside $U$, $j\_\* \mathcal L$ must give the intermediate extension.
| 7 | https://mathoverflow.net/users/18060 | 448147 | 180,446 |
https://mathoverflow.net/questions/448133 | 5 | Recently I'm reading the paper [Ramsey–Milman phenomenon, Urysohn metric spaces, and extremely amenable groups](https://arxiv.org/pdf/math/0004010.pdf) by Pestov. When it comes to the definition of an extremely amenable topological group, it claims (without proof) that
>
> (the extreme amenability) is equivalent to the existence of a left invariant **multiplicative** mean on the space of all bounded right uniformly continuous functions on the group.
>
>
>
I'm not aware of any references on this result.
**Question:** Is there any reference on extreme amenability of topological groups and invariant means?
Thanks for any comments or answers!
| https://mathoverflow.net/users/140578 | Extreme amenability of topological groups and invariant means | The action $G\curvearrowright\beta G$ is continuous iff $G$ is discrete, so for nondiscrete groups it is not true that $G$ is extremely amenable iff this action has a fixed point.
What one should look at instead, is the biggest compactification on which $G$ acts continuously, this is known as the Samuel compactification, $S(G)$, and can be constructed as the spectrum of the algebra $RUCB(G)$ of right uniformly continuous bounded functions on $G$. This is where the uniform continuity condition comes from. It is then true that $G$ is extremely amenable iff $G\curvearrowright S(G)$ has a fixed point. (Of course on a discrete group all continuous functions are uniformly continuous so $\beta G$ and $S(G)$ are isomorphic)
Note also that if you write explicitly what it means, for a fixed $f\in C(G)$, that the orbit map $G\to C(G)$, $g\mapsto gf$, is $\|\cdot\|\_\infty$-continuous, you'll end up with a uniform continuity condition on $f$!
Pestov wrote a whole book on the subject, called *Dynamics of Infinite-dimensional Groups: The Ramsey-Dvoretzky-Milman Phenomenon*, it is a great introduction to extreme amenability which contains all the details of the above and much more
| 7 | https://mathoverflow.net/users/49381 | 448150 | 180,447 |
https://mathoverflow.net/questions/448107 | 2 | Let $R$ be a normal complete local domain of dimension $n \geq 4$. Does there exist a prime ideal $\mathfrak{p}$ of height $\dim(R) - 1$ such that $R\_{\mathfrak{p}}$ is a regular local ring? In general, can the dimension of the smooth locus be as low as possible or can the codimension of the singular locus be exactly $2$ in excellent schemes?
| https://mathoverflow.net/users/130925 | Krull dimension of the smooth locus | $\DeclareMathOperator\Spec{Spec}$
Maybe I'm misreading this, but I don't see why you need dimension $\geq 4$, normal, domain, etc.
**EDIT:** I originally wrote this requiring R1 but I don't think we need R1 (regular in codimension 1). Instead, we should just require R0, regular at the minimal primes (ie, $R\_Q$ is a field if $Q$ is a minimal prime). The proof is modified to reflect this.
**Theorem.** *If $(R,m)$ is excellent and regular in codimension 0 and $\dim R > 0$ then there exists a prime $P$ such that $\dim R\_P = \dim R - 1$ and such that $R\_P$ is regular.*
Let's prove it.
Let $J$ denote a radical ideal such that $\dim V(J) \leq \dim R - 1$ and such that the singular locus of $R$ is contained in $V(J)$ (we can do this since $R$ is excellent). Finally set $U = \Spec R \setminus V(J)$ (note $U$ is nonsingular).
If $J = m$, then the statement is obvious as any height one prime will do the job. In particular, we may suppose $\dim R/J > 0$.
We proceed by induction on dimension of $R$. If $R$ is dimension 1, then since $R$ is local and R0, we can take $J = m$ and we are done with the base case.
Note $V(J)$ has codimension $\geq 1$ and say minimal associated primes $Q\_1, ... Q\_t$ of $V(J)$ of heights $\geq 1$. By Flenner-Trivedi local Bertini, there exists $x \in m$ such that $x$ is not in any of the $Q\_i$, and not in any minimal prime of $R$, and such that $x \notin P^{(2)}$ for any $P \in U$.
We set $R' = R/xR$, $J' = (J + xR)/xR$, and $U' = U \cap V(x)$.
The first statement of the local Bertini theorem guarantees that $\dim( R/J ) = \dim (R' / J') + 1$ (we also have $\dim(R) = \dim(R') + 1$). Next, for any $P \in U$, $R\_{P}$ is a regular ring, the second statement in the local Bertini means that $(R/x)\_P$ is also regular. Hence $R/x$ is R0 too (its regular locus contains $U \cap V(x)$).
Thus, by the induction hypothesis, there exists $P' = (P + xR)/(xR) \in U' \subseteq \Spec R/x$ of height $\dim(R/x) - 1 = \dim R - 2$ such that $(R/x)\_{P'}$ is regular. Hence $R\_P$ is also regular by the above ($P \in U$). But then $\dim R\_P = \dim R'\_{P'} + 1$ and so $P$ has height $\dim R - 1$ as desired.
| 4 | https://mathoverflow.net/users/3521 | 448155 | 180,449 |
https://mathoverflow.net/questions/447595 | 4 | Let $R$ be a ring. An $R$-module $M$ is called FL (FP) if it has a finite resolution consisiting of finitely generated free (projective) modules.
Given an exact sequence of $R$-modules, $0\to M\_1\to M\_2\to M\_3\to 0$, if two of the modules $M\_1$, $M\_2$ or $M\_3$ are FP, then the third is FP as well. This follows, for example, from Proposition 1.4 and Proposition 4.1b of
Bieri, Robert Homological dimension of discrete groups. Second edition. Queen Mary College Mathematics Notes. Queen Mary College, Department of Pure Mathematics, London, 1981.
My question is
>
> If two of the modules $M\_1$, $M\_2$ or $M\_3$ are FL, is it true that
> the third one is also FL?
>
>
>
| https://mathoverflow.net/users/10482 | Exact sequences with two FL-modules | Yes, this is true. See Bourbaki, N. Éléments de mathématique. Algèbre. Chapitre 10. Algèbre homologique. (French) [Elements of mathematics. Algebra. Chapter 10. Homological algebra], Theorem 3.9.1
Let $K\_0(R)$ be the Grothendieck group of $R$ and $\widetilde K\_0(R)=K\_0(R)/\langle [R]\rangle$.
Then $[P]=0$ in $\widetilde K\_0(R)$ if and only if $P$ is stably free.
For any $FP$ $R$-module $M$ having the following resolution consisting of finitely generated projective $R$-modules
$$0\to P\_k\to \ldots \to P\_1\to P\_0\to M\to 0,$$ we denote
$$\chi\_u^R(M)=\sum\_{i=0}^k (-1)^i [P\_i]\in \widetilde K\_0(R).$$
Observe that $\chi\_u^R(M)=0$ if and only if $M$ is $FL$.
Thus, in order to prove the claim it is enough to show that if $0\to M\_1\to M\_2\to M\_3\to 0$ is an exact sequence of $FP$ $R$-modules, then $\chi\_u(M\_2)=\chi\_u(M\_1)+\chi\_u(M\_3)$.
This can easily be done by induction on
$pd\_R(M\_1)+pd\_R(M\_2)+ pd\_R(M\_3)$, where $pd\_R(M)$ is the projective dimension of $M$.
| 1 | https://mathoverflow.net/users/10482 | 448156 | 180,450 |
https://mathoverflow.net/questions/448153 | 2 | The question is as in the title.
I know that a traceless matrix can be written as a commutator of two matrices.
Then, let $v : \mathbb{R}^3 \to \mathbb{R}^3$ be a divergence-free smooth vector field. That is, $\nabla \cdot v=0$.
Then, the matrix $A(x)=\bigl[\partial\_i v\_j(x) \bigr]$ is smooth as a mapping from $\mathbb{R}^3$ into $M\_{3 \times 3}(\mathbb{R})$ and tracelss for each $x$.
Then, what does any two matrices $B(x)$ and $C(x)$ that satisfy $A(x)=B(x)C(x)-C(x)B(x)$ look like?
This is quite new and original for me..Could anyone please help me?
| https://mathoverflow.net/users/56524 | For a divergence free smooth vector field $v : \mathbb{R}^3 \to \mathbb{R}^3$, how to find the commutator form of the matrix $A=(\partial_i v_j)$? | You might want to first carry out a unitary transformation $A(x)\mapsto U(x)A(x)U^\top(x)$, such that all diagonal elements are zero.$^\ast$
Then $A(x)=BC(x)-C(x)B$ with
$$B=\begin{pmatrix}
1&0&0\\
0&2&0\\
0&0&3
\end{pmatrix},\;\;
C\_{ij}(x)=\begin{cases}
0&\text{if}\;i=j\\
\frac{A\_{ij}(x)}{B\_{ii}-B\_{jj}}&\text{if}\;\;i\neq j.
\end{cases}
$$
---
$^\ast$ This is always possible for a traceless $A$ ([proof](https://math.stackexchange.com/q/267192/87355)).
For a $3\times 3$ matrix the unitary has the two-parameter form
$$U(x)=\begin{pmatrix}
\cos\alpha(x)&\sin\alpha(x)&0\\
-\sin\alpha(x)&\cos\alpha(x)&0\\
0&0&1
\end{pmatrix}
\begin{pmatrix}
1&0&0\\
0&\cos\beta(x)&\sin\beta(x)\\
0&-\sin\beta(x)&\cos\beta(x)
\end{pmatrix}
.$$
You can solve first for $\alpha(x)$,
$$(\partial\_1 v\_1) \cos ^2\alpha+(\partial\_2v\_2) \sin ^2\alpha+ (\partial\_1v\_2+\partial\_2 v\_1)\sin\alpha\cos\alpha=0,$$
and then for $\beta(x)$.
| 2 | https://mathoverflow.net/users/11260 | 448163 | 180,454 |
https://mathoverflow.net/questions/448152 | 4 | Let $A$ be a densely-defined, positive, self-adjoint operator with compact resolvent on a Hilbert space $H$. Then, $\text{Range}(1+A)=H$ and there is a basis for $H$ consisting of eigenvectors of $1+A$. Assume also that $D\subset H$ is a core domain for $A$; that is, $D$ is dense in $\text{Dom(A)}$ with respect to the graph norm.
Is it true that we can pick an eigenvector basis for $H$ to consist of elements in $D$ instead of $\text{Dom(A)}$?
Thank you in advance.
| https://mathoverflow.net/users/161393 | Diagonalizing selfadjoint operator on core domain | This sounds suspicious right away since the eigenvectors are what they are (nothing to choose here, unless you have degeneracies), but there is much choice for $D$ and we should be able to avoid eigenvectors.
For a concrete example, you can take $H=\ell^2$, $Ae\_n=ne\_n$, which is self-adjoint on its natural domain $D(A)=\{ x: \sum n^2 |x\_n|^2<\infty \}$. Then consider
$$
D= L( e\_1+e\_k/k^2: 1<k)
$$
($L$ = linear span).
This is a core of $A$: Clearly $e\_1+e\_k/k^2$ is close to $e\_1$ in graph norm for large $k$, and then $e\_1+e\_2/4-e\_1-e\_k/k^2$ is close to $e\_2/4$ etc. So all $e\_j$ are in the domain of the operator closure.
On the other hand, $D$ doesn't contain any of the eigenvectors $e\_j$.
| 9 | https://mathoverflow.net/users/48839 | 448172 | 180,460 |
https://mathoverflow.net/questions/448185 | -1 | Let $\mathcal{B}([0, 1])$ be the Boolean algebra of measurable subsets of $[0, 1]$ modulo almost everywhere equivalence, i.e., two measurable sets which differ only by a Lebesgue null set are identified. For each $t \in [0, 1]$, let $\mathcal{D}\_t$ be the filter on $\mathcal{B}([0, 1])$ generated by open neighborhoods of $t$. Does these exists a filter $\mathcal{F}$ on $\mathcal{B}([0, 1])$ s.t. for each $t \in [0, 1]$, the filter generated by $\mathcal{D}\_t$ and $\mathcal{F}$ is an ultrafilter?
| https://mathoverflow.net/users/504602 | The existence of a maximal “cross-sectional” filter on the Boolean algebra of measurable subsets of [0, 1] modulo almost everywhere equivalence | No. Suppose $\mathcal{F}$ is such a filter. Clearly each $X \in \mathcal{F}$ has positive measure intersection with every positive-length interval. Then neither $[0,1/2]$ nor $[1/2, 1]$ are in the filter generated by $\mathcal{D}\_{1/2}$ and $\mathcal{F},$ contradiction.
| 1 | https://mathoverflow.net/users/109573 | 448188 | 180,464 |
https://mathoverflow.net/questions/447937 | 3 | Suppose that $X$ is a standard Borel space (meaning it is endowed with a $\sigma$-algebra coming from some Polish topology on $X$) and $G$ is a Polish group acting in a Borel way on $X$. Denote by $E\_G$ the resulting orbit equivalence relation on $X$.
There are (at least) three competing definitions, that I know of, for this action or equivalence relation being *smooth*:
1. There is a standard Borel space $Y$ and a Borel measurable function $f:X\to Y$ such that for all $x,y\in X$, $xE\_G y$ if and only if $f(x)=f(y)$.
2. The quotient Borel structure on $X/E\_G$ is standard.
3. The quotient Borel structure on $X/E\_G$ is countably separated (meaning there is a sequence of Borel sets which separates points).
The first is what I'd call "the descriptive set theorists definition" (and can be found in any number of papers on Borel equivalence relations), while the third is "the ergodic theorists definition" (see, e.g., Zimmer, *Ergodic Theory and Semisimple Groups*). Clearly, the second implies the other two. And in the case when $G$ is countable, all three are equivalent (see Propositions 2.8-2.9 in <https://filippoc.people.uic.edu/2020/Spring/CBER.pdf>).
**My question:** Are these three definitions equivalent for general Polish groups $G$? If not, what about locally compact $G$? And is there a standard reference for these facts?
| https://mathoverflow.net/users/16107 | Competing definitions of smooth orbit equivalence relation | Let me turn my comment into an answer now that I have time to write some details. The short answer is that yes, those three definitions are equivalent, the long answer is below.
Let's forget about orbit equivalence relations and let's look at an arbitrary Borel equivalence relations $E$ on a Polish space $X$ for a second. Then we have (using a mix of the question's and Hjorth's terminology) the following three properties:
1. $E\leq\_B\mathrm{Id}\_{\Bbb R}$, that is there is a Borel $f\colon X\to\Bbb R$ such that for all $x,y\in X$, $xEy\iff f(x)=f(y)$. Here $\Bbb R$ can be replaced by any uncountable Polish space as they are all Borel isomorphic, or by a standard Borel space as in the question.
2. There exists a sequence $(A\_n)\_{n\in\Bbb N}$ of Borel $E$-invariant sets, such that for all $x,y\in X$, $xEy\iff\forall n(x\in A\_n\iff y\in A\_n)$, which is the same as the third condition in the question.
3. The quotient structure on $X/E$ generated by the Borel $E$-invariant sets is standard.
Then we have $(1)\iff (2)$ and $(3)\implies (2)$. If $E$ is the orbit equivalence relation induced by a continuous action of a Polish group, then all three conditions are equivalent. The first part is exercise 2.27 in Hjorth's *Classification and orbit equivalence relations*, let me give a sketch.
$(3)\implies (2)$ is obvious.
$(1)\implies (2)$ Suppose $f$ is as in $(1)$ and let $(U\_n)\_{n\in\Bbb N}$ enumerate a countable basis for $\Bbb R$. Let $A\_n=f^{-1}(U\_n)$.
$(2)\implies (1)$ Let $(A\_n)\_{n\in\Bbb N}$ be a sequence as in $(2)$. We reduce $E$ to $\mathrm{Id}\_{2^\Bbb N}$ with the map $\theta\colon X\to\Bbb 2^\Bbb N$, defined by $\theta(x)(n)=1\iff x\in A\_n$.
If $E=E^X\_G$ is the orbit equivalence relation induced by a continuous action $G\curvearrowright X$ of a Polish group $G$ we have that all three conditions are equivalent. This was proved by Burgess in *A selection theorem for group actions* (Pacific Journal of Mathematics, 80(2), pp.333-336.). To be precise Burgess shows that under the hypothesis above $E^X\_G$ admits a Borel transversal, that is a Borel set $A\subseteq X$ such that $|A\cap[x]\_{E^X\_G}|=1$ for all $x\in X$. Once such a transveral is available it is easy to check smoothness of $E^X\_G$ by reducing it to $\mathrm{Id}\_X$ with the function $f\colon X\to X$, $f(x)=$the unique element of $A$ which is $E^X\_G$-related to $x$.
| 1 | https://mathoverflow.net/users/49381 | 448194 | 180,468 |
https://mathoverflow.net/questions/448193 | 3 | I have been working on minimal surfaces, only recently learnt about maximal surfaces and "maxfaces" in Lorentz spaces.
I can clearly see the mathematical motivations. But I wonder if zero-mean-curvature hypersurfaces in Lorentz spaces ($\mathbb{R}^2\_1$ or $\mathbb{R}^3\_1$), like minimal surfaces in Euclidean spaces, also find applications in nature sciences or other branches of mathematics.
| https://mathoverflow.net/users/20595 | Applications of maximal surfaces in Lorentz spaces | Maybe you already encountered such maximal surfaces in the context of General Relativity. Still, the one application of spacelike maximal surfaces that I am aware of is as special kinds of initial data (Cauchy) surfaces for the Einstein equations (a popular broader class are constant mean curvature (CMC) surfaces, of which the maximal condition (vanishing mean curvature) is a special case). The vanishing (or constant) mean curvature condition simplifies the algebraic structure of the Einstein equations when they are split into constraint and evolution subsystems, hence also simplifying their analysis. The existence of maximal Cauchy surfaces in asymptotically flat Lorentzian geometries (spacetimes) was studied for instance in
>
> *Bartnik, Robert*, [**Existence of maximal surfaces in asymptotically flat spacetimes**](https://doi.org/10.1007/BF01209300), Commun. Math. Phys. 94, 155-175 (1984). [ZBL0548.53054](https://zbmath.org/?q=an:0548.53054). (cf. Thm.5.4)
>
>
>
That work has quite a few citations.
| 5 | https://mathoverflow.net/users/2622 | 448196 | 180,469 |
https://mathoverflow.net/questions/195267 | 7 | $\DeclareMathOperator\SL{SL}\DeclareMathOperator\Aut{Aut}$EDITED Let $G=\SL\_{n,{\mathbb{C}}}$, the special linear group over ${\mathbb{C}}$.
Let $H\subset G$ be a finite subgroup.
Set $X=G/H$ be the corresponding homogeneous space, it is a complex variety.
Let $\tau\in \Aut({\mathbb{C}})$ be an automorphism of the field of complex numbers,
not necessarily continuous.
Consider the conjugate variety $\tau X$.
We have $\tau X=\tau G/\tau H$.
We take the standard $\mathbb{Q}$-form $\SL\_{n,\mathbb{Q}}$ of $G$, then we identify $\tau G$ with $G$,
and we obtain $\tau X= G/\tau(H)$.
**Question 1.** Is it always true that $\tau X\simeq X$ as homogeneous spaces of $G=\SL\_{n,\mathbb{C}}$?
**Question 2.** Is it always true that $\tau X\simeq X$ as algebraic varieties over ${\mathbb{C}}$?
**Question 3.** Is it always true that that the $(\tau X)({\mathbb{C}})\simeq X({\mathbb{C}})$ as $C^\infty$-manifolds?
I expect the answer NO to Question 1.
Let us try to construct a counter-example.
We wish to construct a finite subgroup $H\subset G$ such that $G/\tau(H)\not\simeq G/H$,
i.e., $\tau(H)$ is not conjugate to $H$ in $\SL(n,\mathbb{C})$.
We come to the following question:
**Question 4.** Let $\rho\colon H\to \SL(n,\mathbb{C})$ be an $n$-dimensional complex representation of a finite group $H$.
Is it possible that $\tau(\rho(H))$ is not conjugate to $\rho(H)$ in $\SL(n,\mathbb{C})$ for some $\tau\in \Aut({\mathbb{C}})$?
In the last question by $\tau(\rho(H))$ we mean $\{ \tau(\rho(h))\, |\ h\in H\}$,
where $\tau(\rho(h))$ is the matrix with matrix elements $\tau(\rho(h)\_{i,j})$.
| https://mathoverflow.net/users/4149 | Conjugation of the quotient of $\mathrm{SL}(n,\mathbb{C})$ by a finite subgroup | $\DeclareMathOperator\SL{SL}
\DeclareMathOperator\GL{GL}
\DeclareMathOperator\PSL{PSL}
\DeclareMathOperator\Aut{Aut}
\DeclareMathOperator\Out{Out}
\DeclareMathOperator\Ad{Ad}
$I can only address Question 1/Question 4. Let me replace $\SL\_n$ with $\GL\_n$; our $H$ will be simple, so it will provide an example for $\SL\_n$ also.
If $\rho, \psi: H \to \GL\_n(\mathbb C)$ are faithful representations with the same image, then $\psi^{-1} \circ \rho$ is an automorphism of $H$. Hence asking whether $(\tau\rho)(H)$ is conjugate to $\rho(H)$ as a set for faithful $\rho$ is asking whether there is an automorphism $\varphi \in \Aut(H)$ such that $\tau \rho$ is isomorphic to $\rho \varphi$ as a complex representation of $H$. This reduces the problem to character theory. Note also that the action of $\Aut(H)$ on characters factors through $\Out(H)$.
An example is given by $H = \PSL\_2(\mathbb F\_p)$ for $p > 13$.
For any nonsquare $a \in \mathbb F\_p^\times$,
$$\Out(H) = \left\langle \Ad \begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}\right\rangle \cong \mathbb Z/2\mathbb Z.$$
See [1] or [2] for this.
Let
$$T = \left\{ \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mid x \in \mathbb F\_p^\times \right\} / \{\pm 1\}$$
be the diagonal torus in $H$.
If $\zeta$ is a $(p-1)/2$th root of unity, let $\alpha: \mathbb F\_p^\times \to \mathbb C^\times$ be a multiplicative character sending a primitive element of $\mathbb F\_p^\times$ to $\zeta$. Note that $\alpha(-1) = 1$.
There is an irreducible representation of degree $p+1$ given by parabolically inducing the character $\begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mapsto \alpha(x)$ of $T$ up to $H$. Its character $\chi$ satisfies
$$ \chi\begin{pmatrix} x & 0 \\ 0 & x^{-1}\end{pmatrix} = \alpha(x) + \alpha(x^{-1}).$$
See e.g. [3] for a description of the characters of $\mathrm{PSL}\_2(\mathbb F\_p)$. If $\varphi$ is the automorphism of conjugation by $\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$, then $\varphi$ fixes $T$ elementwise, so $\chi\varphi$ has the same values on $T$ as $\chi$.
>
> Lemma: If $\zeta$ is a primitive $m$th root of unity, then $\zeta^{-1} + \zeta$ is irrational for $m \notin \{1,2,3,4,6\}$.
>
>
>
This lemma also appears in the theory of root systems.
If $\zeta$ is chosen to be a primitive $(p-1)/2$th root of unity, then $\zeta + \zeta^{-1}$ is irrational for $(p-1)/2 \notin \{1,2,3,4,6\}$, i.e. $p = 11$ or $p > 13$. This provides an example of a character $\chi$ and $\tau \in \Aut(\mathbb C)$ such that $\tau \chi \neq \chi \varphi$ for all $\varphi \in \Out(H)$.
---
Proof of the lemma: We may assume $m > 2$.
The degree $[\mathbb Q(\zeta):\mathbb Q]$ is $\phi(m)$.
As $\zeta + \zeta^{-1}$ is real, $[\mathbb Q(\zeta): \mathbb Q(\zeta + \zeta^{-1})] > 1$, but $\zeta$ satisfies the quadratic equation
$$\zeta^2+1=(\zeta+\zeta^{-1})\zeta $$
over $\mathbb Q(\zeta + \zeta^{-1})$, so $[\mathbb Q(\zeta):\mathbb Q(\zeta + \zeta^{-1})] = 2$. Hence the degree of $\zeta^{-1} +\zeta$ is $\phi(m)/2$. If $m = p\_1^{k\_1}\cdots p\_\ell^{k\_\ell}$, then $\phi(m) = p\_1^{k\_1 - 1}(p\_1 - 1)\cdots(p\_\ell^{k\_\ell})(p\_\ell - 1)$. If $\phi(m) = 2$, then primes $p\_i$ appearing must have $p\_i - 1$ divides $2$, whence $p\_i$ can be only $2$ or $3$. Further, only $m=3$, $m=4$, and $m=6$ are possible.
---
**References**
[1] Steinberg, R. (1960). Automorphisms of Finite Linear Groups. Canadian Journal of Mathematics, 12, 606-615. doi:10.4153/CJM-1960-054-6
[2] [What is the outer automorphism group of $\operatorname{SL}(2,\mathbb{F}\_q)$?](https://mathoverflow.net/questions/348440/what-is-the-outer-automorphism-group-of-operatornamesl2-mathbbf-q/348441#348441)
[3] Bonnafé, *Representations of $SL\_2(\mathbb F\_q)$*, §5.3, or Fulton and Harris, *Representation Theory*, §5.2
| 3 | https://mathoverflow.net/users/125523 | 448202 | 180,470 |
https://mathoverflow.net/questions/448198 | 4 | What is the complexity (e.g. is it $\Sigma^0\_1$, arithmetic, fully $\Pi^1\_1$) of the relation $|a| < |b|$ given two notations $a, b \in \mathscr{O}$ (Kleene's O)?
What about the case where only one of the notations must be in $\mathscr{O}$ (where $b \not\in \mathscr{O} \implies |b| = \infty$)?
For instance, what is the complexity of a predicate $P(a,b)$ which is true whenever $a \in \mathscr{O} \land b \not\in \mathscr{O}$ and false whenever $a \not\in \mathscr{O} \land b \in \mathscr{O}$ (no restriction on the other cases)?
Note that P(a,b) is sorta the analog of a PA degree for well-foundedness.
--
For backgound, the Selection and Reduction chapter in Higher Recursion Theory tells us that there is a *computable* function t(a,b) such that t(a,b) is in $\mathscr{O}$ iff either a or b is in $ \mathscr{O}$ and $|t(a,b)| \geq \min(|a|,|b|)$ (but this doesn't obviously guarantee that $t(a,b) >\_{\mathscr{O}} a$ or $t(a,b) >\_{\mathscr{O}} b$).
However, I'm guessing that despite this being able to guess which of two relations is well-founded is pretty powerful. So I'm guessing that P(a,b) can be either $\Pi^1\_1$ or $\Sigma^1\_1$ depending on how you define it on the unrestrained cases but can't be $\Delta^1\_1$.
| https://mathoverflow.net/users/23648 | Complexity of |a| < |b| for ordinal notations? | For your first question, Spector showed that the relation $|a| < |b|$ is uniformly computable from $\emptyset^{|b|+1}$. It's not $\Delta^1\_1$, but it is both the restriction of a $\Sigma^1\_1$ relation and of a $\Pi^1\_1$ relation to $\mathcal{O}$. The $\Sigma^1\_1$ relation is "there exists an embedding of $|a|+1$ into $|b|$", while the $\Pi^1\_1$ relation is "there is no embedding of $|b|$ into $|a|$".
For your second question, it again cannot be $\Delta^1\_1$, but can be either $\Sigma^1\_1$ or $\Pi^1\_1$. For $\Sigma^1\_1$, just take "$b$ is ill-founded". For $\Pi^1\_1$, take "$a$ is well-founded".
Less trivially, there is a collection of such $P$s which form a $\Sigma^1\_1$ class: take the collection of $P$ which are true when $a$ enters $\mathcal{O}$ before $b$, and are false when $b$ enters before $a$. So there is such a $P$ which is low for $\omega\_1^{ck}$.
| 7 | https://mathoverflow.net/users/32178 | 448205 | 180,471 |
https://mathoverflow.net/questions/448197 | 1 | Let $ K $ be finite degree extension of $ \mathbb{Q} $ such that $ -1 $ is not a square in $ K $. Let $ L = \frac{K[x]}{\langle x^2 +1\rangle}$. Thus every element of $ L $ is of the form $ a + ib $ where $ i^2 = -1 $ and $ a,b \in K $. Let $ \eta\_{2^i} $ is the primitive $ 2^i $-th root of unity contained in $ L $ where $ i \geq 2 $ . Can $ N\_{L/K}(\eta\_{2^i})$ be $ -1 $? Where $ N\_{L/K} $ is a norm map from $L$ to $K$ and defined by $ N\_{L/K}(b) = b \sigma(b) $ for all $ b \in L $ where $ \sigma $ is the automorphism of the $ \operatorname{Gal}(L/K)$.
If $ L $ is real subfield then it is clear.
| https://mathoverflow.net/users/215016 | Norm of $2^{i}$-th primitive root | Let $K=\mathbb{Q}[\sqrt{-2}]$. Then $\frac{1}{2}\sqrt{-2}(1-i)$ is an eighth root of unity in $L$ whose norm is $\frac{1}{2}\sqrt{-2}(1-i)\frac{1}{2}\sqrt{-2}(1+i)=\frac{1}{4}(-2)(1-i^2)=-1\in K$. [I'm assuming that when you write $2^i$ it's a different $i$]
| 4 | https://mathoverflow.net/users/460592 | 448208 | 180,473 |
https://mathoverflow.net/questions/448186 | 2 | I'm trying to understand Bourgain's paper "Besicovitch type maximal operators and applications to Fourier analysis". Let $\xi\in S^2\subset\mathbb{R}^3$ be a unit vector and $\delta>0$, by a $(\xi,\delta)$ tube in $\mathbb{R}^3$, we mean a cylinder $\tau$ of unit length in direction $\xi$ and of radius $\delta$. For $f$ a locally integrable function on $\mathbb{R}^3$, we define $$f\_\delta^\ast(\xi)=\sup\_\tau\frac{1}{|\tau|}\int\_\tau f(x)\,dx$$ where the supremum refers to all $(\xi,\delta)$ tubes $\tau$ and $|\tau|$ stands for the measure of $\tau$. Bourgain proves the following bound
$$||f\_\delta^\ast||\_{L^p(S^2)}\leq C\_\epsilon (1/\delta)^\epsilon ||f||\_{L^P(\mathbb{R}^3)}$$
On page 152 of the paper, Bourgain says we only need to check that for every $A\subset B(0,1)$ and $\sigma\in[0,1]$, we have $$|A|\geq c\delta^{2/3+\epsilon}\sigma^{7/3}|(\chi\_A)^\ast\_\delta>\sigma\}|$$
Bourgain says this is due to Stein's maximal principle, as in the paper "Limtis of sequences of operators" by Stein. Also found here <https://terrytao.wordpress.com/2011/05/12/steins-maximal-principle/> But why? Stein's maximal principle gives weak Lp type bound of maximal functions, Bourgain is saying weak Lp type estimates give strong Lp type estimates. What are we taking to be the sequence of operators in Stein's paper here?
| https://mathoverflow.net/users/147078 | Use of Stein's maximal principle in Bourgain's paper on Besicovitch sets | This may be a slight misattribution. It is the implication $(1.22) \implies (1.21)$ which is essentially in Stein's paper; the implication $(1.21) \implies (1.9)$ is much simpler, following from Marcinkiewicz interpolation. Roughly speaking, if one has a weak-type $L^p$ estimate on the Kakeya maximal operator, one can interpolate it with the $L^2$ estimate (1.20) to get a strong-type $L^{p-\varepsilon}$ estimate, and then interpolate again with the trivial $L^\infty$ estimate to get back a strong $L^p$ estimate, losing some factors of $\delta^{-O(\varepsilon)}$ in the process.
[A more general principle here is that, in situations where one is willing to concede epsilon losses in the exponents, strong type and weak type estimates are equivalent; also, singular integrals are all basically bounded, rendering a large part of Calderon-Zygmund theory trivial in this regime.]
| 7 | https://mathoverflow.net/users/766 | 448209 | 180,474 |
https://mathoverflow.net/questions/448157 | 3 | Let $Q$ be a probability measure on $\mathbb{R}$. Let $$Q\_h(dy) = e^{y \cdot h} Q(dy) / M(h) \quad \text{where} \quad M(h) = \int e^{y \cdot h} Q(dy)$$ defined for $h \in (-c,\infty)$ with some $c > 0$. Let $Y\_h \sim Q\_h$ and $V(h) := EY\_h^2 - (EY\_h)^2$. Then, does it follow that $$\exists c' > 0,\ h\_0 > -c\ \text{ s.t. }\ \forall h \geq h\_0,\ V(h) \geq e^{-c' h} \text{?}$$
| https://mathoverflow.net/users/146981 | Variance lower bound for natural exponential family | The desired result obviously fails to hold if the probability measure $Q$ is degenerate, that is, supported on a singleton set. Indeed, then $V(h)=0$ for all $h$.
It is much harder to construct a counterexample with a non-degenerate $Q$. The idea of such a counterexample, given below, is to make the distribution $Q$ highly lacunary, in order to make the tilted distribution $Q\_h$ very close to a degenerate one, for large $h$.
Indeed, let
\begin{equation\*}
Q:=\sum\_{i\ge1}p\_i\delta\_{2^i},
\end{equation\*}
where $\delta\_a$ is the Dirac distribution supported on the set $\{a\}$, $p\_i:=Ce^{-4^i}$, and $C$ is the normalizing constant. Then
\begin{equation\*}
V(h)=\frac IJ, \tag{10}\label{10}
\end{equation\*}
where
\begin{equation\*}
I:=I(h):=\frac12\sum\_{i,j\ge1}(2^j-2^i)^2z\_i(h)z\_j(h) \\
\le S:=S(h):=\sum\_{1\le i<j}4^j z\_i(h)z\_j(h), \tag{20}\label{20}
\end{equation\*}
\begin{equation\*}
J:=J(h):=M(h)^2=\Big(\sum\_{i\ge1}z\_i(h)\Big)^2, \tag{30}\label{30}
\end{equation\*}\begin{equation\*}
z\_i(h):=e^{h2^i}p\_i.
\end{equation\*}
For large natural $k$, let now
\begin{equation\*}
h=h\_k:=2^{k+1}, \tag{35}\label{35}
\end{equation\*}
so that
\begin{equation\*}
z\_i(h\_k)=y\_i:=y\_{k,i}:=C\exp(2^{k+1+i}-4^i).
\end{equation\*}
Then
\begin{equation\*}
\frac{y\_{i+1}}{y\_i}=\exp(2^{k+i}(2-3\times2^{i-k})) \\ \le\exp(-2^{k+i})\le\exp(-2^{i+1})
\text{ for natural $i\ge k$} \tag{40}\label{40}
\end{equation\*}
and
\begin{equation\*}
\frac{y\_{i+1}}{y\_i}\ge\exp(2^{k-1+i})\ge1 \\
\text{ for natural $i<k$.} \tag{50}\label{50}
\end{equation\*}
In particular, \eqref{30} and \eqref{40} imply
\begin{equation\*}
\frac{y\_{k+1}}{y\_k}\le\exp(-4^k)\quad\text{and}\quad \frac{y\_k}{y\_{k-1}}\ge\exp(4^{k-1}). \tag{60}\label{60}
\end{equation\*}
Now recall \eqref{20} and write
\begin{equation\*}
S=S\_1+S\_2+S\_3, \tag{70}\label{70}
\end{equation\*}
where
\begin{equation\*}
S\_1:=\sum\_{i,j\colon k\le i<j}4^jy\_iy\_j, \quad
S\_2:=\sum\_{i,j\colon 1\le i<j\le k}4^jy\_iy\_j, \quad
S\_3:=\sum\_{i,j\colon i<k<j}4^jy\_iy\_j.
\end{equation\*}
In view of \eqref{40}, \eqref{50}, and \eqref{60},
\begin{equation\*}
S\_1\le c\_1y\_ky\_{k+1}\le c\_1 e^{-4^k}y\_k^2,
\end{equation\*}
where $c\_1:=1+\sum\_{j\ge1}j4^j e^{-2^j}\in(0,\infty)$;
\begin{equation\*}
S\_2\le k^2 4^k y\_{k-1}y\_k\le k^2 4^k e^{-4^{k-1}}y\_k^2\le e^{-4^{k-2}}y\_k^2,
\end{equation\*}
since $k$ is large;
\begin{equation\*}
S\_3\le c\_2 k y\_{k-1}y\_k\le e^{-4^{k-2}}y\_k^2,
\end{equation\*}
where $c\_2:=\sum\_{j\ge1}4^j e^{-2^j}\in(0,\infty)$.
So, by \eqref{20} and \eqref{70},
\begin{equation\*}
I\le e^{-4^{k-3}}y\_k^2.
\end{equation\*}
On the other hand, clearly, $J\ge y\_k^2$. Thus, recalling \eqref{35},
\begin{equation\*}
V(h)=V(h\_k)\le e^{-4^{k-3}}<e^{-c'2^{k+1}}=e^{-c'h\_k}
\end{equation\*}
for any real $c'>0$ and all large enough $k$. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 448211 | 180,475 |
https://mathoverflow.net/questions/448182 | 6 | Suppose that $G=(G\_{ij})$ is an $n\times n$ positive-definite symmetric matrix with the diagonal entries all equal $1$ and all off-diagonal entries $\le0$. Let $a$ be a column $n\times1$ matrix with nonnegative entries such that $a^\top G^{-1}a=1$. Does it then necessarily follow that $a^\top Ga\le1$?
Certain numerical experiments suggest that this is true.
| https://mathoverflow.net/users/36721 | An inequality for certain positive-definite matrices | The answer seems to be **yes**.
Let $G$ be the Gram matrix of a base $(e\_1,\dots,e\_n)$ in some Euclidean space. Then $G^{-1}$ is the Gram matrix of the dual base $(f\_1,\dots,f\_n)$, i.e., the one satisfying $\langle e\_i,f\_j\rangle=\delta\_{ij}$.
Denote $u=\sum\_i a\_ie\_i$ and $v=\sum\_j a\_jf\_j$; then $\|u\|^2=a^TGa$ and $\|v\|^2=a^TG^{-1}a$. . The properties of $G$ immediately yield that
$$
\langle u,u\rangle=a^TGa=\sum\_ia\_i^2-(\text{nonnegative})\leq\sum\_ia\_i^2=\langle u,v\rangle.
$$
So $\|u\|\|v\|\geq \langle u,v\rangle\geq \|u\|^2,$ which yields $|v\|\geq \|u\|$, as desired.
| 5 | https://mathoverflow.net/users/17581 | 448225 | 180,479 |
https://mathoverflow.net/questions/248337 | 6 | Consider a Wang tiling (given a subset of $C^4$ for a finite set $C$ of colours, e.g.). It's well-known to be undecidable whether there exists a tiling, and also whether there exists a periodic tiling. It's also well-known that there exists a periodic tiling iff there exists a doubly periodic tiling.
On the other hand, for any given n, then it's decidable whether there exists a periodic tiling with period $(n,0)$: construct a directed graph with vertex set $C^n$ and an edge from $v$ to $w$ if there exists a tiling of the $n\times1$ rectangle with identical left and right labels and $v,w$ at the bottom and top respectively. Then there exists such a periodic tiling iff this graph has a loop.
Here is a question I can't answer: for any given $n$, is it decidable whether there exists a periodic tiling with period $(n,m)$ for some $m$?
| https://mathoverflow.net/users/10481 | Decidability of (restricted) periodicity of Wang tilings | This is undecidable. I refer to the proof of the periodic tiling problem which is Theorem 5.7 in the lecture notes of Jarkko Kari
<https://users.utu.fi/jkari/wp-content/uploads/sites/1251/2021/10/part2.pdf>
but I give a construction that works with any such proof where the period is suitably marked.
In the proof I cite, in the periodic configurations, there are explicit horizontal and vertical fault lines, which mark the period. (I'll need only the horizontal fault lines.) More precisely, this construction has the property that the set of tiles (constructed from some Turing machine $M$) can be (algorithmically) partitioned as $T = T\_1 \sqcup T\_2$ so that
* If the machine $M$ halts, there is a periodic configuration with period $(0,m)$ such that the tiles on rows $mk$ come from $T\_1$, and others from $T\_2$.
* In every periodic configuration (should one exist), tiles from $T\_1$ are used.
* If the machine $M$ does not halt, there is no periodic configuration.
On any such construction we can perform a post-processing step where we add a constant shear to the horizontal fault lines when we see tiles $T\_1$. Wang tiles are equivalent to subshifts of finite type, so it suffices to explain how to make a subshift of finite type. We first add the constraint that each row is explicitly enforced to have only tiles from $T\_1$, or only from $T\_2$. Then we make $T\_1$ tiles check their color constraint not with their top neighbor, but their top neighbor's right$^n$ neighbor.
Now I claim the resulting subshift of finite type has an $(n,m)$-periodic configuration if and only if $M$ halts. Proof:
* If $M$ halts, take a valid Wang tiling by $T$ coming from the first item. To get it in our SFT, for each line over $T\_1$, shear the half-plane above by $n$ steps to the right (and take a limit). Because the period was $(0,m)$ initially, and $T\_1$-rows appear on every $m$th row, this gives an $(n,m)$-periodic configuration.
* Suppose our new subshift of finite type has a periodic point. If it doesn't use tiles from $T\_1$, this is a periodic Wang tiling over $T\_2$, which is forbidden by item 2 above. It tiles from $T\_1$ are used, they are used on entire rows, and the sequence of rows where they are used is periodic as well. Clearly we get a periodic configuration of the original Wang tile set by shearing by $-n$ on top of the $T\_1$-rows. So $M$ halts.
So we have reduced the halting problem of $M$ algorithmically to the existence of an $(n,m)$-periodic point in a subshift of finite type (for our fixed $n$). As mentioned we can easily convert to Wang tiles if we wish.
(One can also phrase the proof as reducing a technical variant of the periodic tiling problem "with marked periods" directly to OP's problem, and noting that the proof in the lecture notes proves this technical variant.)
| 2 | https://mathoverflow.net/users/123634 | 448239 | 180,483 |
https://mathoverflow.net/questions/448204 | 2 | I am quite confused between Helmholtz decomposition and Laplacian vector fields in the periodic case.
Let $\mathbb{T}^3$ be the $3$-dimensional torus. Then, I thought any divergence-free smooth vector field $v : \mathbb{T}^3 \to \mathbb{R}^3$ can be expressed as a curl of another smooth periodic vector field $V : \mathbb{T}^3 \to \mathbb{R}^3$. That is. $v=\nabla \times V$ according to the following link:
<https://en.wikipedia.org/wiki/Helmholtz_decomposition>
However, I came across the notion of Laplacian vector fields, which have both zero curl and zero divergence. Since the Laplace equation has nontrivial solutions on $\mathbb{T}^3$ with the periodic boundary conditions, there must exist nontrivial Laplcian vector fields on $\mathbb{T}^3$, I believe.
For such vector fields, is it possible to apply the Helmholtz decomposition? I am quite confused...
Edit : by "apply the Helmholtz decomposition" I mean the possibility of the Laplacian vector fields on $\mathbb{T}^3$ expressed as a curl of another smooth periodic vector field.
| https://mathoverflow.net/users/56524 | Helmholtz decomposition vs Laplacian vector fields on $\mathbb{T}^3$? | The correct generalization to $\mathbb{T}^3 = (\mathbb{R}/\mathbb{Z})^3$ is the [Hodge decomposition](https://en.wikipedia.org/wiki/Hodge_theory). Every vector field on $\mathbb{T}^3$ is uniquely expressible as $u = \nabla \times v + \nabla w + h$, where $\nabla \cdot h = \nabla \times h = 0$ is a Laplace vector field in your terminology. It is the summands that are unique in the decomposition, obviously $v \mapsto v + \nabla \chi$ and $w \mapsto w + C$ with constant $C$ produce the same $u$. If $\nabla \cdot u = 0$ then the $\nabla w$ term can be dropped, same for $\nabla \times v$ when $\nabla \times u = 0$. There are topological obstructions to expressing the $h$ term as either a curl or a divergence, but $h$ belongs only to a finite dimensional space. In fact, the space of Laplace vector fields is isomorphic to the de Rham cohomology $H^1(\mathbb{T}^3; \mathbb{R}) \cong H^2(\mathbb{T}^3; \mathbb{R}) \cong \mathbb{R}^3$, which in this case consists explicitly of the span $\langle \partial\_x, \partial\_y, \partial\_z \rangle$, using the standard $(x,y,z)$ coordinates on $\mathbb{R}^3$ (before quotienting by $\mathbb{Z}^3$).
Above, I have simply translated the general statement of the Hodge decomposition to $\mathbb{T}^3$ and the language of the OP. The main observation is that, using appropriate [Hodge dualization](https://en.wikipedia.org/wiki/Hodge_star_operator), the sequence of operators [(gradient, curl, divergence)](https://en.wikipedia.org/wiki/Hodge_star_operator#Example:_Derivatives_in_three_dimensions) becomes the sequence of de Rham differentials $d$ acting on differential forms (dualizing once more, the same sequence becomes the sequence of de Rham codifferentials $\delta$ on forms). The harmonic forms (of degree 1 or 2) of Hodge theory then become the Laplace vector fields of the OP (harmonic forms of degree 0 or 3 become constant functions).
| 3 | https://mathoverflow.net/users/2622 | 448248 | 180,486 |
https://mathoverflow.net/questions/448245 | 5 | Let $k/\mathbb{Q}$ be a number field and $\mathbb{A}$ its ring of adèles. As usual $\mathbb{A} = \mathbb{A\_f} \times \mathbb{A\_{\infty}}$.
The standard definition of an *automorphic representation* $(\pi,V)$ for $\textrm{GL}\_n(\mathbb{A})$ is its realization as (irreducible) subquotient of the space of automorphic forms $\mathcal{A}(\textrm{GL}\_n(k) \backslash \textrm{GL}\_n(\mathbb{A})), \omega)$, where $\omega$ is some central character. This is for example in Bump's Automorphic Forms and Representations on p.300.
This implies that $(\pi,V)$ is a representation for the finite part $\textrm{GL}\_n(\mathbb{A\_f})$ and a $(\mathfrak{g}\_{\infty}, K\_{\infty})$-module for $\textrm{GL}\_n(\mathbb{A\_{\infty}})$ with the property that each two actions commute. By abuse of notation I write $\pi$ for 'all' actions.
Then there is the notion of *admissibility* of such a representation, which is fairly standard too.
However, in the Corvallis proceedings, Flath gives the notion of an (admissible) automorphic representation in a purely algebraic way, i.e. he abstracts the properties from above, but it is not clear that his definition is 'embeddable' into the space of automorphic forms, i.e. that the 2 definitions are equivalent.
What is specially not clear to my; if there is a notion of **cuspidality** in the pure algebraic description of Flath? (I am working over $\mathbb{C}$, so cuspidality = supercuspidality).
It is I think due to Jacquet that cuspidality at local non-archimedean places has equivalent meanings; by a vanishing integral and that one is *not* (properly) parabolically induced. I could therefore imagine that one can define cuspidality for $(\pi,V)$ as a representation of $\textrm{GL}\_n(\mathbb{A\_f})$ in an analogue way. But I do not know what should be the notion of cuspidality for a $(\mathfrak{g}, K)$-module.
I am concerned with construction of $L$-functions attached to cuspidal representations. I imagine that the best way to introduce cuspidal representations is how Bump does it. However, I wondered if I can skip all the 'analytic conditions' on automorphic forms.
| https://mathoverflow.net/users/484997 | On the notion of cuspidality | To every local admissible representation $\pi\_v$ of $\mathrm{GL}\_n(k\_v)$, there is a local $L$-function $L(s,\pi\_v)$. For a global admissible representation $\pi=\otimes\_v \pi\_v$ of $\mathrm{GL}\_n(\mathbb{A}\_k)$, the corresponding global $L$-function is defined as $L(s,\pi)=\prod\_v L(s,\pi\_v)$. However, unless $\pi$ is automorphic, this global $L$-function will not have the usual analytic properties (cf. [converse theorems](https://en.wikipedia.org/wiki/Converse_theorem)). So it is very special for a global admissible representation to be automorphic. Automorphicity of $\pi=\otimes\_v \pi\_v$ connects the various local factors $\pi\_v$. In particular, if you change finitely many local factors $\pi\_v$ in a given cuspidal automorphic representation $\pi=\otimes\_v \pi\_v$, the resulting global admissible representation will no longer be cuspidal automorphic (cf. [multiplicity one theorems](https://en.wikipedia.org/wiki/Multiplicity-one_theorem)).
BTW I recommend Goldfeld-Hundley's two-volume textbook on automorphic forms (in the same series as Bump's textbook). Also, there is this [related earlier post of mine](https://mathoverflow.net/questions/349122/are-the-l-functions-of-a-normalized-newform-and-the-corresponding-cuspidal-repre/349124#349124) that you might find useful.
| 8 | https://mathoverflow.net/users/11919 | 448251 | 180,489 |
https://mathoverflow.net/questions/447827 | 7 | $\DeclareMathOperator\SL{SL}$ Let $p>3$ and $G$ be an open subgroup of the special linear group $\SL\_2(\mathbf{Z}\_p)$ over the ring $\mathbf{Z}\_p$ of $p$-adic integers. Suppose that $G$ is topologically generated by a finite set $S=\{g\_1,\cdots,g\_r\}$ such that all $g\_i$ have finite order. Is it true that $G=\SL\_2(\mathbf{Z}\_p)$? If we denote the natural surjective morphism $ \SL\_{2}(\mathbf{Z}\_{p})\to \SL\_{2}(\mathbf{F}\_{p}) $ by $ \eta $, then it is equivalent to ask whether $\eta(S)$ generates $\SL\_2(\mathbf{F}\_p)$.
| https://mathoverflow.net/users/504912 | Topological generators for $\mathrm{SL}_2(\mathbf{Z}_p)$ | $\def\ZZ{\mathbb{Z}}\def\SL{\text{SL}}\def\Id{\text{Id}}$This seems false to me.
**Lemma:** $e:=\left[ \begin{smallmatrix} 1&p\\0&1 \\ \end{smallmatrix} \right]$ and $f:=\left[ \begin{smallmatrix} 1&0\\p&1 \\ \end{smallmatrix} \right]$ topologically generate an open subgroup of $\SL\_2(\ZZ\_p)$.
**Proof:** Let $G$ be the group topologically generated by $e$ and $f$. Let
$$\Gamma(p^k) = \{ g \in \SL\_2(\ZZ\_p) : g \equiv \Id\_2 \bmod p^k \}.$$
We will show that $\Gamma(p^2) \subset G$.
To this end, note that $G$ contains $e^{p^{k-1}} = \left[ \begin{smallmatrix} 1&p^k\\0&1 \\ \end{smallmatrix} \right]$, $f^{p^{k-1}} = \left[ \begin{smallmatrix} 1&0\\ p^k&1 \\ \end{smallmatrix} \right]$ and $[e, f^{p^{k-2}}] \equiv \left[ \begin{smallmatrix} 1+p^k&0\\ 0&1-p^k \\ \end{smallmatrix} \right] \bmod p^{k+1}$ for all $k \geq 2$. Thus, $G$ contains generators for $\Gamma(p^k)/\Gamma(p^{k+1})$ for all $k \geq 2$, so any element of $\Gamma(p^k)$ can be written as an infinite convergent product of elements of $G$. $\square$
Now, let $\zeta$ be a $(p-1)$-st root of unity in $\ZZ\_p$ other than $\pm 1$. (Here I use $p>3$.) Let $d = \left[ \begin{smallmatrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{smallmatrix} \right]$.
The matrices $d$, $de$ and $df$ all have order $p-1$. (They are all upper or lower triangular with diagonal entries $\zeta$, $\zeta^{-1}$ and, since $\zeta \neq \zeta^{-1}$, they are diagonalizable.) Clearly, the group topologically generated by $d$, $de$, $df$ contains the group topologically generated by $e$ and $f$, so it is open.
However, reducing $d$, $de$ and $df$ modulo $p$ gives diagonal matrices, so the map onto $\SL\_2(\mathbb{F}\_p)$ isn't surjective.
| 8 | https://mathoverflow.net/users/297 | 448256 | 180,491 |
https://mathoverflow.net/questions/448261 | 2 | We consider the heat kernel
$$
g :\mathbb R\_{>0} \times \mathbb R^d \to \mathbb R,\quad (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ).
$$
Let $0 < t\_1 < t\_2 <\infty$ and $1\le\lambda<\infty$ such that $\frac{t\_2}{t\_1} \le \lambda$. I would like to ask if there are constants $C\_1, C\_2>0$ (depending only on $\lambda$) such that
$$
\frac{g(t\_1, x)}{t\_1} \le C\_1 \frac{g(C\_2 t\_2, x)}{\sqrt{t\_2}}
\quad \forall x \in \mathbb R^d.
$$
Thank you so much for your elaboration!
| https://mathoverflow.net/users/477203 | Let $g$ be the heat kernel. Are there constants $C_1, C_2>0$ such that $\frac{g(t_1, \cdot)}{t_1} \le C_1 \frac{g(C_2 t_2, \cdot)}{\sqrt{t_2}}$? | The answer is no. E.g., let $t\_1\sim t\_2\downarrow 0$ and $|x|\sim\sqrt{t\_2}$.
| 6 | https://mathoverflow.net/users/36721 | 448262 | 180,492 |
https://mathoverflow.net/questions/448231 | 0 | During developing a new statistical estimator, I faced the following problem.
Let $\mathbf{x}\_i$ be a sequence of i.i.d. $d$-dimensional random vectors with
\begin{align\*}
\mathbf{x}\_i = \mathbf{O}\_i \mathbf{\mu} + \mathbf{\varepsilon}\_i,
\end{align\*}
where $\mathbf{\mu}$ is a mean vector, $\mathbf{O}\_i$ is an random orthogonal matrix (so $\mathbf{O}\_i^\top \mathbf{O}\_i = \mathbf{I}$), and $\mathbf{\varepsilon}\_i$ is an i.i.d. mean-zero noise vector. Then the question is
>
> **Question**
> Is there any way to test $H\_0: \{\mathbf{\mu} = 0\} $?
>
>
>
---
**Motivation**
This problem is important in estimating the risk premium of the factor model with time-varying factor loading. From the PCA, we can only estimate the factor loading up to rotation, so conducting Fama-MacBeth regression with the aggregated factor loadings would be problematic since they have different rotations for each component.
---
**Here is what I have done.**
To find the answer, I have tried the simplest case: $d=1$ (a univariate case). For $d=1$, the orthogonal matrix $\mathbf{O}$ becomes just $+1$ or $-1$. Then we have
\begin{align\*}
X\_i = o\_i \mu + \varepsilon\_i, \qquad o\_i = \begin{cases} +1 & \text{ with prob }p \\ -1 & \text{ with prob }1-p, \end{cases}
\end{align\*}
where the variance of $\varepsilon\_i$ is $\sigma^2$. Notice that $\bar{X} = n^{-1} \sum\_{i=1}^n X\_i$ is not a consistent estimator of $\mu$ (but $\bar{X} \xrightarrow{P} (2p - 1) \mu$), so we should consider another approach. Notice that $\mu \neq 0$ imples $\mu^2 > 0$, so it may be useful to use $n^{-1} \sum\_{i=1}^n X\_i^2$ for the test:
\begin{align\*}
\frac{1}{n} \sum\_{i=1}^n X\_i^2 &= \frac{1}{n} \sum\_{i=1}^n (o\_i \mu + \varepsilon\_i)^2 \\
&= \frac{1}{n} \sum\_{i=1}^n \big(\mu^2 + 2 o\_i \varepsilon\_i + \varepsilon\_i^2\big) \\
&\xrightarrow{P} \mu^2 + \sigma^2
\end{align\*}
by the law of large numbers. So we should debias $\sigma^2$. Notice also that the classical estimator of $\sigma^2$ is not consistent: $n^{-1} \sum\_{i=1}^n (X\_i - \bar{X})^2 \xrightarrow{P} \big( 1 - (2p-1)^2 \big) \mu^2 + \sigma^2$ from the fact $\bar{X} \xrightarrow{P} (2p - 1) \mu$, so it is not clear how to debias it. This problem is generically due to the degrees of freedom of sample first and second moment is $2$ while we have to infer $3$ parameters ($p, \mu, \sigma^2$).
(My previous answer was wrong.)
It would be really appreciated if you have any idea or comments.
Thank you for reading!
Seunghyeon
| https://mathoverflow.net/users/159685 | Estimation on rotationally-disturbed random vectors | $\newcommand\ep\varepsilon$This is impossible to do even for $d=1$, as your [model is not identifiable](https://en.wikipedia.org/wiki/Identifiability#Definition), that is, the parameters of the model are not identifiable even if the distribution of the $X\_i$'s is fully known.
Indeed, let $R\_1,\dots,R\_n$'s be independent Rademacher random variables, with $P(R\_i=\pm1)=1/2$. Then the following two scenarios result in the same distribution of the $X\_i$'s:
(i) $\mu=0$ and $\ep\_i=R\_i$ for all $i$;
(ii) $\mu=1$, and $O\_i=R\_i$ and $\ep\_i=0$ for all $i$.
We see that even the function $1(\mu=0)$ of the parameter $\mu$ is not identifiable, that is, the value of
$1(\mu=0)$ cannot be determined even if the distribution of the $X\_i$'s is fully known.
| 1 | https://mathoverflow.net/users/36721 | 448264 | 180,493 |
https://mathoverflow.net/questions/447494 | 2 | Let $K/\mathbb Q\_p$ be a finite extension, and $\mathcal O\_K$ the ring of integers of $K$. I am asking for a reference for a structure theorem
of finitely generated modules over the completed group algebra $\mathcal O\_K[[\mathbb Z\_p]]$.
I am familiar with structure theorems for $\Lambda = \mathbb Z\_p[[\mathbb Z\_p]]$, and
more generally for $\mathbb Z\_p[[G]]$ where $G$ is a $p$-adic Lie group. I believe that analogous structure theorems should hold in the $\mathcal O\_L$ coefficent case, but I was unable to find an explicit reference.
Motivation: rings of the form $\mathcal O\_K[[\mathbb Z\_p]]$ arise naturally in the
$p$-adic Langlands program.
| https://mathoverflow.net/users/161063 | Structure theorem for Iwasawa modules over $p$-adic rings of integers | Serre, Jean-Pierre, *Classes des corps cyclotomiques* (Séminaire Bourbaki décembre 1958) lemme 5 p. 90 contains a statement and proof of a structure theorem (up to pseudo-isomorphism) for finitely generated modules over a noetherian integrally closed domain $A$, of which $\mathcal O\_{K}[[\mathbb Z\_{p}]]$ is a very particular example. Not that the proof in that general setting is no harder than in the case of $A=\mathbb Z\_{p}[[X]]$.
Applied to a regular local ring of dimension 2, this lemma translates in what you want (provided you know that a reflexive module over such a ring is free, which is also proved in the same reference, lemme 6 p. 91).
| 3 | https://mathoverflow.net/users/2284 | 448276 | 180,497 |
https://mathoverflow.net/questions/448274 | 0 | Let $p>0$ and consider a metric space $(X,d)$. I have recently come across a problem where the space $(X,d^q)$ provides is natural; where $q>1$. However, the triangle inquality break (i.e. it is a ["semimetric space"](https://en.wikipedia.org/wiki/Metric_space#Semimetrics)).
I'm wondering, in this case, does $(X,d^p)$ satisfy some kind of "generalized triangle inequality"?
| https://mathoverflow.net/users/496781 | Generalized Triangle Inequality for Snowflakes | More generally, let $p>1$ and $a,b>0$, and set $\tilde{d}:X\times X\rightarrow [0,\infty)$ defined by
$$
\tilde{d}(x,z)\mapsto a \,d(x,z) + b\,d(x,z)^p \qquad (\text{ for all }x,z\in X)
$$
Then, for every $x,y,z\in X$ we have that
$$
\begin{aligned}
\tilde{d}(x,z) = &\, a \,d(x,z) + b\,d(x,z)^p \\
\le & a(d(x,y)+d(y,z)) + b(d(x,y)+d(y,z))^p\\
\le & a(d(x,y)+d(y,z)) + 2^{p-1}\,b(d(x,y)^p+d(y,z)^p)\\
\le & 2^{p-1}\,a(d(x,y)+d(y,z)) + 2^{p-1}\,b(d(x,y)^p+d(y,z)^p)\\
= & 2^{-1}\big(\tilde{d}(x,y)+\tilde{d}(x,z)\big).
\end{aligned}
$$
For more details, see [Qinglan Xia - The Geodesic Problem in Quasimetric Space, J. Geom. Anal. 2009](https://link.springer.com/article/10.1007/s12220-008-9065-4).
| 4 | https://mathoverflow.net/users/36886 | 448278 | 180,499 |
https://mathoverflow.net/questions/448235 | 11 | Say that a **long model** is an $\mathfrak{A}\models\mathsf{I\Sigma\_1}$ such that $\mathfrak{A}$ has strictly greater cardinality than each of its proper initial segments (in the case $\vert\mathfrak{A}\vert=\aleph\_1$ this is "$\omega\_1$-like"ness). If $\mathfrak{A}$ is a long model and $\vert\mathfrak{A}\vert$ is regular, then in fact $\mathfrak{A}\models\mathsf{PA}$ since the bounding scheme is trivially satisfied in $\mathfrak{A}$ (see Lemma 2.15 of [Hajek/Pudlak](https://projecteuclid.org/eBooks/perspectives-in-logic/Metamathematics-of-First-Order-Arithmetic/toc/pl/1235421926)). Moreover, by the MacDowell/Specker theorem (see [Kossak, *63 years of the MacDowell-Specker theorem*](https://arxiv.org/pdf/2208.03156.pdf)) the converse holds: if $\kappa$ is an uncountable cardinal and $\varphi$ is true in every long model of $\mathsf{PA}$ of size $\kappa$, then $\mathsf{PA}\models\varphi$ already. (Given $\mathsf{PA}\not\models\varphi$, let $\mathfrak{B}\models\mathsf{PA}+\neg\varphi$ be countable and iterate MD/S $\kappa$-many times.)
Interestingly, the first observation seems to need regularity:
>
> Is there a singular cardinal $\kappa$ such that the common theory of long models of size $\kappa$ is strictly weaker than $\mathsf{PA}$?
>
>
>
| https://mathoverflow.net/users/8133 | Can singular long models require less than PA? |
>
> The answer to the question is strongly in the positive, since the following theorem of Kaye implies that for *every* singular cardinal $\kappa$, and for any fixed natural number $n$, the common theory of $\kappa$-like models of $\mathrm{I}\Sigma\_{n}$ is strictly weaker than $\mathsf{PA}$.
>
>
>
**Theorem.** *Let $\kappa$ be a singular cardinal, $n \geq 1$, and let* $M\models \mathrm{B}\Sigma\_n + \mathsf{exp} + \lnot\mathrm{I}\Sigma\_n$. *Then there is a $\kappa$-like model $K$ that is elementarily equivalent to $M$.*
The above theorem appears as Theorem 2.4 of Kaye's paper *Constructing κ-like models of arithmetic*. **J. London Math. Soc.** (2) 55 (1997), no. 1, pp.1–10. According to Kaye, the paper arose out of a question of Slaman: Are there $\kappa$-like models of arithmetic for singular $\kappa$ that do not
satisfy $\mathsf{PA}$?
A key ingredient of Kaye's proof is the technology of "doubly indexed indiscernibles" invented by Keisler, who showed that for any consistent extension $T$ of Zermelo set theory, and any given singular cardinal $\kappa$, $T$ has a $\kappa$-like model. This result of Keisler appears in his paper *Models with orderings*, **Logic, methodology and philosophy of science III** (eds. B. van Rootselaar and J. F. Staal; North-Holland, Amsterdam, 1968) 35–62.
It is worth noting that Kaye has another paper related to this topic: *The theory of $\kappa$-like models of arithmetic*. (English summary)
**Special Issue: Models of arithmetic.
Notre Dame J. Formal Logic** 36 (1995), no. 4, pp.547–559.
| 11 | https://mathoverflow.net/users/9269 | 448280 | 180,500 |
https://mathoverflow.net/questions/448218 | 1 | Let $M$ be a simply connected Hadamard manifold. That is, $M$ is a complete Riemannian manifold with sectional curvature bounded above by 0. Let $f(x,y)=\frac{1}{2}d^2(x,y)$, where $x,y \in M$, and $d(.,.)$ is geodesic distance. I'm curious if $f:M^2 \to \mathbb{R}$ is a Morse-Bott function, and how to show it.
I know that $\nabla f(x,y)=(-\log\_x(y),-\log\_y(x))=0$ if and only if $x=y$. If I understand correctly, this is because $\exp\_x:T\_xM \to M$ and $\exp\_y:T\_yM \to M$ are both diffeomorphisms. Following from this, the critical points of $f$ are precisely $C^\* =\{(x,x):x \in M\}$, hence a closed submanifold. It follows then to show that the kernel of $\nabla^2 f(x,x)$ is precisely the tangent bundle of $C^\*$. This is something I find difficult as I do not really know how to compute the hessian of $f(x,y)=\frac{1}{2}d^2(x,y)$. Could I possibly get some help on this?
| https://mathoverflow.net/users/141449 | Is squared geodesic distance a Morse-Bott function on simply-connected Hadamard manifolds? | Choose $V,W \in T\_p M$ and note that
$$
D^2f((V,W),(V,W)) = \frac{d^2}{dt^2}\Big|\_{t=0} \frac 12 d(\exp\_p(tV),\exp\_p(tW))^2 .
$$
By the computation in [Section 1.3 here](https://www2.math.upenn.edu/%7Ewziller/math660/TopogonovTheorem-Myer.pdf)
$$
\frac 12 d(\exp\_p(tV),\exp\_p(tW))^2 = \frac{t^2}{2} |V-W|^2 + O(t^3).
$$
Thus we find
$$
D^2f((V,W),(V,W)) = |V-W|^2.
$$
Now, since both sides are quadratic bilinear forms the parallelogram identity implies
$$
D^2f((V\_1,W\_1),(V\_2,W\_2)) = \langle V\_1-W\_1,V\_2-W\_2\rangle.
$$
Now, recall that $(V\_1,V\_2)$ is in the kernel of the Hessian if and only if $D^2 f((V\_1,V\_2),(W\_1,W\_2))$ vanishes for ALL $(V\_1,V\_2)$ (this is the definition of the kernel of a symmetric bilinear form). In particular, we see that $(V\_1,W\_1)$ is in the kernel if and only if $V\_1-W\_1=0$ (by nondegeneracy of the form).
Thus, we see that $\ker D^2 f\_p =\{(V,V) : V\in T\_pM\} = T\_{(p,p)}\Delta$.
| 2 | https://mathoverflow.net/users/1540 | 448290 | 180,504 |
https://mathoverflow.net/questions/448286 | 2 | Let $a\_i (i \in\{1...k\})$ be $k$ IID standard Gaussian random variables, $P\_i$ are $d$-dimensional constant vectors. How to prove with a probability of at least $1/e$,
$$
\left\|\sum\_{i=1}^k a\_{i} P\_{i}\right\|\_2 \geq\left\|P\_{1}\right\|\_2 .
$$
holds.
| https://mathoverflow.net/users/501704 | prove with a probability of at least $1/e$: $\left\|\sum_{i=1}^k a_{i} P_{i}\right\|_2 \geq\left\|P_{1}\right\|_2$ holds | $\newcommand\si\sigma$Let
$$p:=P\Big(\Big\|\sum\_{i=1}^k a\_i P\_i\Big\| \ge\|P\_1\|\Big).$$
Here $\|\cdot\|:=\|\cdot\|\_2$.
The inequality $p\ge1/e=0.367\dots$ does not hold in general.
Actually, the best lower bound on the probability $p$
is
$$p\_\*:=P(|a\_1|\ge1)=0.317\dots,$$
which is $<1/e=0.367\dots$.
Indeed, on the one hand, if $P\_1\ne0=P\_2=\cdots=P\_k$, then clearly $p=p\_\*$. So, there is no lower bound on $p$ that is $>p\_\*$.
On the other hand, there is a unit $d$-dimensional vector $u$ such that $u\cdot P\_1=\|P\_1\|$, where $\cdot$ denotes the dot product. Then
$$\Big\|\sum\_{i=1}^k a\_i P\_i\Big\|\ge\Big|u\cdot\sum\_{i=1}^k a\_i P\_i\Big|
=\Big|\sum\_{i=1}^k c\_i a\_i\Big|=|X|, \tag{1}\label{1}$$
where $c\_i:=u\cdot P\_i$ and $X:=\sum\_{i=1}^k c\_i a\_i$, so that $c\_1=\|P\_1\|$. Without loss of generality (wlog), $\|P\_1\|\ne0$ -- otherwise, $p=1$. So, by rescaling, wlog $\|P\_1\|=1$ and hence $c\_1=1$. So, $X$ is a Gaussian random variable with mean $0$ and variance
$$\si^2:=\sum\_{i=1}^k c\_i^2=1+\sum\_{i=2}^k c\_i^2\ge1.$$
So, in view of \eqref{1},
$$p\ge P(|X|\ge1)=P(\sqrt{\si^2}\,|a\_1|\ge1)\ge P(|a\_1|\ge1)=p\_\*.$$
This concludes the proof that $p\_\*$ is the best lower bound on $p$. $\quad\Box$
| 8 | https://mathoverflow.net/users/36721 | 448295 | 180,506 |
https://mathoverflow.net/questions/448297 | 2 | For a non zero rational $r=p/q$ ($p,q\in\mathbb Z$ coprimes), define the height of $r$ by $\mathrm{ht}(r)=\max(|p|,|q|)$ (by convention $\mathrm{ht}(0)=0$). For a polynomial $P\in\mathbb Q[X]$, define the height of $P$ by the maximum of height of its coefficients. Let $A$ and $C$ be two non zero polynomials of $\mathbb Z[X]$ such that $A$ divides $C$ in $\mathbb Q[X]$. Denote by $B$ the quotient of the division of $C$ by $A$. My question: can one bound the height of $B$ in function of the height of $A$ and $C$?
Thanks in advance for any answer
| https://mathoverflow.net/users/33128 | Bounds of heights of coefficients of rational polynomials | In general, no. Take $A=(x-1)^2$, $C=(x^n-1)^2$ for large $n$. Then $B=(1+x+\ldots+x^{n-1})^2$ has height $n$.
| 5 | https://mathoverflow.net/users/4312 | 448300 | 180,507 |
https://mathoverflow.net/questions/448288 | 3 | Let $p$ be an odd prime, $g$ a primitive root of $p$, and
$f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}\_p$
is defined by
$$f(i,x) = g^{i+1} + g^{-i-1} - g^i x^2\mod p.$$
**Is it true that $f$ attains all values modulo $p$?**
It is not hard to see that $g^{i+1} + g^{-i-1}$ produces $(p+1)/2$ distinct values and $x^2$ produces $(p+1)/2$ distinct values. So, $g^{i+1} + g^{-i-1} - x^2$ does attain all values. Unfortunately, $g^i$ added to $x^2$ creates a twist.
I suspect that the statement is true -- I checked primes up to 3500 using a computer.
| https://mathoverflow.net/users/506363 | The range of values of $f(i,x) = g^{i+1} + g^{-i-1} - g^i x^2$ modulo a prime p | This is true.
Denote $g^{i+1}=y$, then $y$ takes any non-zero value mod $p$. Fix an element $a\in \mathbb{F}\_p$ and assume that the equation $y+1/y-g^{-1}yx^2=a$ does not have a solution. It yields that $y+1/y-a\ne g^{-1}yx^2$ for all $y\ne 0$ and all $x$. Thus, in particular, $y+1/y-a\ne 0$ for all $y\ne 0$, and for Legendre symbols we have $(\frac{y+1/y-a}p)=(\frac yp)$ (otherwise there exists $x$ such that $y+1/y-a=g^{-1}yx^2$). Therefore, $$(y+1/y-a)^{(p-1)/2}=y^{(p-1)/2}$$
for all non-zero $y$, and the polynomial
$$
f(y):=(y^2+1-ay)^{(p-1)/2}-y^{p-1}
$$
vanishes for all $y\ne 0$. But $\deg f<p-1$, so this may happen only if $f\equiv 0$. But $f(0)=1\ne 0$.
| 6 | https://mathoverflow.net/users/4312 | 448306 | 180,512 |
https://mathoverflow.net/questions/448279 | 5 | Let $S$ be a smooth closed connected hyperbolic surface. On $S$ we have the Laplace operator $\Delta$, whose eigenvalues form a discrete sequence
\begin{equation}
0=\lambda\_0<\lambda\_1\leq \lambda\_2\leq ... \leq \lambda\_k\nearrow\infty
\end{equation}
To my knowledge, a generic surface will not have eigenvalues of multiplicity $>1$ (this is a result of K.Uhlenbeck, see <https://www.jstor.org/stable/2374041>). On the other hand, any surface with degenerate eigenvalues (such as the Bolza surface or Klein quartic), which I am aware of, usually has a lot of symmetries (in the sense that the automorphism group of $S$ is large), which cause the high multiplicities.
Are there any known inverse results of the type "If a hyperbolic surface $S$ has an eigenvalue of multiplicity >1, then it has non-trivial automorphism group". I was unable to spot any related results in the literature, so I want to to ask whether there is anything known in this direction or if there are any examples of hyperbolic surfaces violating this idea.
| https://mathoverflow.net/users/155336 | Multiplicity of Laplace eigenvalues and symmetry | Let me extend, and correct, the argument expressed in the comment made by user378654.
Let us start with a surface $S$ for which $\Delta$ admits a double eigenvalue $\lambda$. For instance, you may choose an $S$ with a non-trivial symmetry group. Denote $V=\ker(\lambda-\Delta)$. Consider now a smooth $m$-parameter family $p\mapsto S\_p$ around $S$, with say $S\_0=S$. By a Lyapunov-Schmidt procedure, which essentially involves the fact that the spectrum of $\Delta$ is discrete, there exists a unique smooth map $p\mapsto V(p)$ over a neighbourhood $U$ of $0$, where $V(p)$ is a plane in $C^\infty(S\_p)$, such that $V(0)=V$ and $V(p)$ is stable under $\Delta$. Choosing a smoothly varying unitary basis ${\cal B}\_p$ of $V(p)$, the action of $\Delta$ over $V(p)$ is represented by a $2\times2$ symmetric real matrix $M(p)$.
Now if $m=3$ and the family is "generic", in the sense that the map $p\mapsto M(p)$ is a submersion at $p=0$, then the $3$-parameter family $S\_p$ contains a $1$-parameter family of surfaces that admit a double eigenvalue.
Notice the importance of the assumption that $m\ge3$. A one-parameter family $p\mapsto S\_p$ does not suffice. This phenomenon (for symmetric matrices in general) is described in a book of V. I. Arnold.
| 3 | https://mathoverflow.net/users/8799 | 448312 | 180,514 |
https://mathoverflow.net/questions/448314 | 2 | Let $\kappa:\lbrace 1,2,3,\ldots\rbrace\longrightarrow \mathbb Z$
be the convolutional inverse in the Dirichlet ring of $n\longmapsto {n+1\choose 2}$. It is defined by $\kappa(1)=1$ and
by the functional equation $\sum\_{d\vert n}{d+1\choose 2}\kappa(n/d)=0$ for $n\geq 2$.
It is not hard to show that $\kappa(3\cdot 5\cdot p^2)=0$
for every prime number $p\geq 7$.
There seem to be very few other values $n$ such that $\kappa(n)=0$.
My computer found only two : $2^4\cdot 13\cdot 23$ and $5^3\cdot 7\cdot 37$.
*Is the set of such exceptional zeros of $\kappa$ finite or not?*
| https://mathoverflow.net/users/4556 | Exceptional zeros of a convolutional inverse | There are many more examples. An infinite series is for $n=5^8\cdot7\cdot p^2$ for primes $p$ different from $5$ and $7$.
Another infinite series is $n=5\cdot11\cdot17^3\cdot p^2$ for primes $p$ different from $5,11,17$. And there are many more infinite series.
| 4 | https://mathoverflow.net/users/18739 | 448319 | 180,516 |
https://mathoverflow.net/questions/448266 | 4 | It seems to me from a quick glance at several sources describing the complex and modular irreducible representations of $\mathrm{GL}(2,p)$ that any field $K$ containing a primitive $(p-1)$-root of unity is a splitting field for $\mathrm{GL}(2,p)$ (but maybe I need to read the details more carefully). Is this correct? And if so, what is a good source to cite for this besides folklore?
I flipped through a number of books, but they never bother to talk about field of definition for the irreducibles. But the constructions seem to always involve characters of $\mathbb F\_p^\times$, which makes me suspect this is correct.
**Edit.** The original version of this question was incorrect, as pointed out in the comments. Here is the revised version.
1. Is any field of characteristic $0$ containing a primitive $(p^2-1)$-root of unity a splitting field for $GL(2,p)$? Note that $GL(2,p)$ has exponent $p(p^2-1)$, so I am trying to dump the factor $p$.
2. Is a any field of characteristic $p$ a splitting field for $\mathrm{GL}(2,p)$?
| https://mathoverflow.net/users/15934 | Splitting field for $\mathrm{GL}(2,p)$ - reference request | I realised that my previous version of this answer was irrelevant to the question (though what was said was accurate, as far as it went).
I think you can make the desired conclusion if you resort to modular representation theory, specifically, the theory of blocks with cyclic defect group (for this question we only need the cyclic defect theory with defect group of order $p$, which was done by R. Brauer himself).
In a block with cyclic defect group, all decomposition numbers are $0$ or $1$. This means that every irreducible character occurs with multiplicity $1$ in some projective indecomposable module (over an appropriate local ring of characteristic zero).
Let $\chi$ be an irreducible character of a finite group $G$ which has order divisible by $p$ and not $p^{2}.$ Let $\theta$ be the character of a suitable projective indecomposable module as above, with
$\langle \theta, \chi \rangle = 1.$
It is a theorem of W. F. Reynolds (a consequence of Brauer's characterization of characters), that $\theta$ is a $\mathbb{Z}$-linear combination of characters, each of which is induced from
a Brauer elementary $p^{\prime}$-subgroups of $G$.
This means that $\chi$ occurs with multiplicity $1$ in the difference
$\alpha - \beta$, where $\alpha$ and $\beta$ are characters of representations which are visibly respectively realizable over $\mathbb{K} = \mathbb{Q}[\omega]$, where $\omega$ is a primitive
$m$-th root of unity and $m$ is the $p^{\prime}$-part of the exponent of $G$ ( in this case, just $\frac{e}{p},$ where $e$ is the exponent of $G$).
The Schur index $m\_{\mathbb{K}}(\chi)$ divides both $\langle \alpha, \chi \rangle$ and $\langle \beta, \chi \rangle$ by the defining properties of the Schur index . Since $\langle \alpha - \beta, \chi \rangle = 1,$ we conclude that $m\_{\mathbb{K}}(\chi) = 1.$
In the case of $G = {\rm GL}(2,p),$ this gives a positive answer to part $1$ of the question (every element of order divisible by $p$ has central $p^{\prime}$-part, and every element of order $p$ is conjugate to all its powers. Hence all character values on $p$-singular elements take values in the field $\mathbb{K}$ as above. and clearly all character values at $p$-regular elements take values in $\mathbb{K}$
(recall that $p$-regular means of order prime to $p$ and $p$-singular means of order divisible by $p$)).
Well, I suppose that strictly speaking, the answer as it stands does not now cover the irreducible characters of degree (divisible by) $p$,it only deals with the irreducible charcters in $p$-blocks of defect $1$. But for any irreducible character $\chi$ in a $p$-block of defect zero of any finite group $G$, the irreducible character $\chi$ is afforded by a projective $RG$-module, where $R$ is a suitable local ring of characteristic zero with residue field of characteristic $p$. Using the result of Reynolds, and arguing as above, we obtain $m\_{\mathbb{K}}(\chi) = 1$, where $\mathbb{K}$ is the cyclotomic field extension of $\mathbb{Q}$ generated by a primitive $m$-th root of unity, with $m$ being the $p^{\prime}$-part of the exponent of $G$. In this case, $\chi$ vanishes on elements of order divisible by $p$, so that all character values taken by $\chi$ are in $\mathbb{K}.$
| 4 | https://mathoverflow.net/users/14450 | 448320 | 180,517 |
https://mathoverflow.net/questions/448281 | 2 | Baur-Monk quantifier elimination theorem asserts that any formula in the language of modules is modulo the theory a boolean combination of BG-Invariants and positive primitive formulas. However, in p.54 Model theory of Modules by Prest, just right after proving the theorem, it was immediately concluded as Cor 2.15 that every sentence is a boolean combination of BG-Invariant conditions. Why does it follow? Thanks.
| https://mathoverflow.net/users/506361 | Question on Baur-Monk quantifier elimination for modules | By Baur-Monk QE, every sentence is equivalent to a boolean combination of BG-invariants and positive primitive *sentences* (which have no free variables).
A positive primitive sentence $\psi$ has the form $\exists x\_1\dots\exists x\_n\,\varphi(x\_1,\dots,x\_n)$, where $\varphi$ is a finite conjunction of equations between linear combinations of the variables $x\_1,\dots,x\_n$. Since $\varphi(0,\dots,0)$ is true in every module, $\psi$ is true in every module.
It follows that the Boolean combination of BG-invariants and positive primitive sentences can be simplified to remove the positive primitive sentences.
| 3 | https://mathoverflow.net/users/2126 | 448337 | 180,521 |
https://mathoverflow.net/questions/448330 | 1 | This question is cross-posted from <https://math.stackexchange.com/questions/4711799/asymptotic-behavior-of-a-markov-process-on-the-set-of-0-1-polynomials>
I am trying to study the asymptotic behavior of a stochastic process defined on the space of single variable polynomials whose coefficients are either $0$ or $1$.
Letting $\mathbb{B}=\{0,1\}$, I will denote by $\mathbb{B}[x]$ this set of polynomials, and I will denote by $\mathbb{B}[x]\_{t}\subset\mathbb{B}[x]$ the subset of such polynomials whose degree is less or equal to $t$. Moreover, given a polynomial $b\_t(x)\in \mathbb{B}[x]\_{t}$ notice that $xb\_t(x)$ and $1+xb\_t(x)$ are elements of $\mathbb{B}[x]\_{t+1}$: these are the two possible "shifted" version of the polynomial.
Fix two parameters $\kappa,a\in (0,1)$. Define a process $(A\_t)\_{t=0}^\infty\subset\Delta(\mathbb{B}[x])$ as follows
* $A\_0=\delta\_0\in\Delta(\mathbb{B}\_{0}[x])$. In words $A\_0$ is a Dirac on the null polynomial.
* $A\_t\in\Delta(\mathbb{B}\_{t-1}[x])$ is defined as follows:
\begin{align\*}
A\_{t+1}(b\_{t})=\begin{cases}
A\_{t}(b\_{t-1})P(b\_{t-1})\quad &if\quad b\_{t}=1+xb\_{t-1}\\
A\_{t}(b\_{t-1})(1-P(b\_{t-1})) \quad &if\quad b\_{t}=xb\_{t-1}\\
0\quad & else
\end{cases}
\end{align\*}
where, letting $b\_{t-1}(\kappa)$ be the evaluation of $b\_{t-1}$ at $\kappa$, $P(b\_{t-1})=F(\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa))$ where $F:[0,1]\to[0,1]$ is a decreasing function such that $F(x)=1$ for $x$ close to 0, $F(1/2)=1/2$ and $F(x)=1$ for $x$ close to 1.
Hence, transition probabilities are as follows. $b\_{t-1}\in\mathbb{B}[x]\_{t-1}$ transitions to $1+xb\_{t-1}\in\mathbb{B}[x]\_{t}$ with probability $P(b\_{t-1})=F(\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa))$ or to $xb\_{t-1}\in\mathbb{B}[x]\_{t}$, with probability $1-P(b\_{t-1})$.
This should be a Markov process, where intuitively, the state $b\_{t-1}$ is split in the two "shifted" states $1+xb\_{t-1}$ and $xb\_{t-1}$. Since each distribution is supported in a new set, there is no hope that there is an ergodic distributions. In simulations,though, it appears that, asymptotically $A\_t$ concentrates on those polynomials in $\mathbb{B}[x]\_{t-1}$ such that the value $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ is closer to $1/2$.
This is intuitive, given the definition of $P$. While for polynomials such that the value $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ is very close to $0$, surely or with high probability the mass $A\_t(b\_{t-1})$ is transfered completely on $1+xb\_{t-1}$ (and viceversa for those who have $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ close to $1$) the polynomials with $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ close to $1/2$ are those where the mass is split equally. There is a sort of "balance" on this states.
But how can I show this formally?
Any help or reference would be immensely appreciated.
| https://mathoverflow.net/users/143913 | Asymptotic behavior of a Markov process on the set of $\{0,1\}$-polynomials | Just a comment really, but too long to fit in a comment box! I think you can simplify the formulation considerably. Let $X\_t$ be your $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$. Then you can describe the process $X\_t$ as a Markov chain in the following way: $X\_1=\kappa a$, and for $t\geq 1$,
\begin{equation\*}
X\_{t+1}=\begin{cases}
\kappa X\_t &\text{with probability $1-F(X\_t)$};\\
\kappa X\_t + (1-\kappa) &\text{with probability $F(X\_t)$}.
\end{cases}
\end{equation\*}
Now $(X\_t)\_{t\geq 1}$ is a nice Markov process taking values in $[0,1]$, and you can indeed look for a stationary distribution to which $X\_t$ converges in distribution as $t\to\infty$. Because of your assumption on $F$, such a stationary distribution will be supported on some sub-interval $[u,v]$ with $0<u<1/2<v<1$ (but it's not the case that it's concentrated on the point $1/2$).
| 2 | https://mathoverflow.net/users/5784 | 448343 | 180,524 |
https://mathoverflow.net/questions/448360 | 5 | Hall's (weak) conjecture is the statement that for all $\varepsilon > 0$ there exists a positive number $c(\varepsilon) > 0$ such that for all $x,y \in \mathbb{Z}$ with $y^2 \ne x^3$, that
$$\displaystyle \left \lvert y^2 - x^3 \right \rvert \geq c(\varepsilon) |x|^{\frac{1}{2} - \varepsilon}.$$
My question concerns a related question. What is the smallest $\theta \geq 0$ for which it is known that there exist infinitely many $x,y \in \mathbb{Z}$ with $y^2 \ne x^3$ satisfying
$$\displaystyle |y^2 - x^3| \ll |x|^{\frac{1}{2} + \theta}?$$
The choice $\theta = 1/2$ can be achieved as follows: write $x^3 = (x-1)^2(x+2) + 3x - 2$, and choose $x = t^2 - 2$ and $y = (t^2 - 3)t$. Then
$$y^2 - x^3 = (t^2 - 3)^2t^2 - (t^2 - 3)^2 t^2 + 3(t^2 - 2) - 2 = 3t^2 - 8,$$
which is manifestly $O(t^2) = O(|x|)$. Here we can freely choose $t \in \mathbb{Z}$, so this gives infinitely many integers satisfying the required inequality.
Can one do better than $\theta = 1/2$?
| https://mathoverflow.net/users/10898 | Upper bound for Hall's conjecture on separation of squares and cubes | The best $\theta$ is $0$. It is known that there are infinitely many
solutions of 0 < $|x^3 - y^2| \ll x^{1/2}$, parametrized by certain
"Pell equations"; indeed one such family attains $|x^3 - y^2| \sim C x^{1/2}$
with $C = 5^{-5/2} 54 \approx 0.966$. See [D].
The underlying polynomial identity
$$
(t^2 + 10t + 5)^3 - (t^2 + 22t + 125)(t^2 + 4t − 1)^2 = 1728t
$$
is connected with the degree-6 cover
${\rm X}\_0(5) \to {\rm X}(1)$ of modular curves [E].
**References**
[D] Danilov, L.V.: The Diophantine equation $x^3 - y^2 = k$ and Hall's conjecture, *Math. Notes Acad. Sci. USSR* **32** (1982), 617-618.
[E] Elkies, N.D.: Rational points near curves and small nonzero $|x^3-y^2|$ via lattice reduction, *Lecture Notes in Computer Science* **1838** (proceedings of ANTS-4, 2000; W.Bosma, ed.), 33-63 (arXiv: math.NT/0005139).
| 15 | https://mathoverflow.net/users/14830 | 448363 | 180,528 |
https://mathoverflow.net/questions/448258 | 9 | I have some basic question about real analytic sets which I've asked in MSE but couldn't get an answer there.
Let $M$ be a real analytic manifold, and let $X \subset M$ be an analytic set. We know by the definition of an analytic set that for every $x \in X$ there exists an open neighborhood $U$ of $x$ in $M$ and a real analytic function $f : U \to \mathbb{R}$ such that
$$ X \cap U = \left\{ p \in U \mid f(p) = 0 \right\}. $$
So, locally analytic sets are zero sets of real analytic functions. Can this be said globally? That is, can we find an open neighborhood $V$ of $X$ in $M$ and a real analytic $f : V \to \mathbb{R}$ such that $X = \left\{ p \in V \mid f(p) = 0 \right\}$? If that's not always possible, is it possible under the further assumption that $X$ can be expressed as a finite union of analytic sets, each of which is a smooth submanifold of $M$?
| https://mathoverflow.net/users/30048 | Can a real analytic set be expressed as the zero set of a single real analytic function? | Let me expand on my comment. In the terrific book "Topics on real analytic spaces" by Guaraldo, Macri and Tancredi one can find a discussion of such questions. See page 64 & 65 and in particular Theorem 2.1, Definition 2.2 and Remark 2.3. Let me quote Theorem 2.1 and make some remarks.
**Theorem 2.1.** Let $Y$ be a coherent real analytic variety of finite dimension [e.g. a connected real analytic manifold]. For a [real analytic] subvariety $X$ of $Y$ the following statements are equivalent.
$i)$ There exist $p$ analytic functions, $f\_1,....,f\_p\in \Gamma(Y,\mathcal{O}\_Y)$, such that $X=\left\{x\in Y \mid f\_1(x)= \dots = f\_p(x)=0\right\}$.
$ii)$ There exists a coherent sheaf of $\mathcal{O}\_Y$-modules $F$ such that $X=\mathrm{supp}F$.
$iii)$ There exists a [coherent] real analytic subspace of $(Y,\mathcal{O}\_Y)$ such that its reduction is $X$.
$iv)$ There exists a closed complex analytic subvariety $\tilde{X}$ of the complexification $\tilde{Y}$ of $Y$ such that $\tilde{X}\cap Y = X$.
Note that the inserted comments marked with [...] were added by me and note that $\mathcal{O}\_Y$ denotes the sheaf of real analytic functions on $Y$. Real analytic subvarieties that satisfy one of these equivalent conditions are often given the name $\mathbb{C}$-analytic or $C$-analytic. This name is in my opinion very misleading and I prefer just saying "$X$ has global defining functions" or "$X$ admits global equations". It should be noted that admitting global equations ***is*** dependent on the embedding and hence it is not an intrinsic aspect on $X$.
In general and as a word of caution, real analytic subsets/subvarieties are quite wild objects and there are many pathological examples. For example, in the mentioned book in example 2.4 on page 65, the authors construct a real analytic subset $X$ in $\mathbb{R}^3$ such that for every real analytic $f\colon U \to \mathbb{R}$ with $f\mid\_X=0$ and $U$ being an open neighbourhood of $X$ in $\mathbb{R}^3$, one has $f=0$. Similar examples were also already known to H. Cartan (see for example the [paper](http://www.numdam.org/item/BSMF_1957__85__77_0/) "Varietes analytiques reelles et varietes analytiques complexes" (1957)). For a "modern" example with some visualizations of the same phenomenon you can also look at this [paper](https://arxiv.org/abs/1412.4838) by Jiri Lebl.
To your second question: Suppose that $Y$ is a real analytic manifold and $X=\bigcup\_{i=1}^k M\_i$, where each $M\_i$ is a real analytic submanifold of $Y$. Real analytic submanifold are, in particular, coherent real analytic subspaces and by Theorem 2.1 have global defining functions. Hence, there exist finitely many $f\_{i,j}$ such that $M\_i$ is equal to the vanishing set of the $f\_{i,j}$, with fixed $i$. As was already mentioned in the comments: In the real analytic situation one can combine finitely many defining functions to one defining function by considering $g\_i:= \sum\_j f\_{i,j}^2$, as $g\_i$ is zero if and only if all the summands are zero. Therefore, all the $M\_i$ have a global defining function $g\_i$. Because $X$ is a union of the $M\_i$ it follows that $h:=\prod\_i g\_i$ is a global defining function of $X$.
| 11 | https://mathoverflow.net/users/109193 | 448372 | 180,530 |
https://mathoverflow.net/questions/448373 | 4 | Assume that $ (M,g) $ and $ (N,h) $ are two smooth closed manifold and $ N $ is embedded isometrically into $ \mathbb{R}^K $ for some $ K\in\mathbb{Z}\_+ $. Assume that $ u\in C^{\infty}(M\times\mathbb{R}\_+,N) $ satisfies the equation of harmonic heat flow
$$
\frac{\partial u}{\partial t}-\Delta\_g u=A(u)(\nabla u,\nabla u),
$$
where $ u=(u^1,u^2,...,u^K) $ and $ A(u)=(A\_{jk}^i(u))\_{1\leq i,j,k\leq K} $ is the second fundamental form of $ N $ at $ u(x) $. In the book "The analysis of harmonic maps and their heat flows" by F. Lin and C. Wang, page 115, there is a formula
$$
(\partial\_t-\Delta\_g)|\partial\_tu|^2=-|\nabla\partial\_tu|^2+R^N(\nabla u,\partial\_tu,\nabla u,\partial\_t u),
$$
where $ R^N $ denote the Riemannian curvature tensor. I have some difficulty in obtaining this formula since the I cannot simply use the arguments in the proof of Bochner formula. Can you give me some hints or references?
| https://mathoverflow.net/users/241460 | A formula in harmonic heat flow | Actually there is a typo in the formula, i.e.
$$
(\partial\_t-\Delta\_g)|\partial\_tu|^2=-2|\nabla\partial\_tu|^2+2R^N(\nabla u,\partial\_tu,\nabla u,\partial\_t u).
$$
Firstly for $ x\in M $, we can choose an orthorgonal basis $ \{e\_{\alpha}\} $. Then it follows from simple calculations that
$$
\begin{aligned}
\partial\_t|\partial\_tu|^2&=\langle\partial\_tu,\partial\_{tt}^2u\rangle\\
\nabla\_{e\_{\alpha}}\nabla\_{e\_{\alpha}}\langle\partial\_tu,\partial\_tu\rangle&=2\langle\partial\_tu\_{,\alpha},\partial\_tu\_{,\alpha}\rangle+2\langle\partial\_tu,\partial\_tu\_{,\alpha\alpha}\rangle.
\end{aligned}
$$
Therefore
$$
(\partial\_t-\Delta\_g)|\partial\_tu|^2=-2|\partial\_t u\_{,\alpha}|^2+2\langle\partial\_tu,(A(u)(\nabla u,\nabla u))\_t\rangle
$$
Define $ P(y):\mathbb{R}^K\to T\_uN $ as an orthogonal projectopn map. We can obtain that
$$
\begin{aligned}
(\partial\_t-\Delta\_g)|\partial\_tu|^2&=-2|\nabla \partial\_tu|^2-2|A(u)(\partial\_tu,\nabla u)|^2+2\langle P(u)\partial\_tu,(A(u)(\nabla u,\nabla u))\_t\rangle\\
&=-2|\nabla \partial\_tu|^2-2|A(u)(\partial\_tu,\nabla u)|^2-2\langle \partial\_tP(u)\partial\_tu,(A(u)(\nabla u,\nabla u))\_t\rangle,
\end{aligned}
$$
where we have used the fact that
$$
\langle P(u)\partial\_tu,(A(u)(\nabla u,\nabla u))\_t\rangle=0,\,\,\langle P(u)\partial\_{tt}^2u,(A(u)(\nabla u,\nabla u))\_t\rangle=0.
$$
Since $ \partial\_tP(u)\partial\_tu=P(u)\_{,\alpha}(\partial\_tu,\partial\_tu)=-A(u)(\partial\_tu,\partial\_tu) $, it can be got that
$$
(\partial\_t-\Delta\_g)|\partial\_tu|^2=-2|\nabla \partial\_tu|^2-2|A(u)(\partial\_tu,\nabla u)|^2+2\langle A(u)(\partial\_tu,\partial\_tu),(A(u)(\nabla u,\nabla u))\_t\rangle.
$$
Therefore the formula above is true by using Gauss-Codazi equation, i.e.
$$
R^N(X,Y,X,Y)=\langle A(X,X),A(Y,Y)\rangle-\langle A(X,Y),A(X,Y)\rangle.
$$
| 1 | https://mathoverflow.net/users/241460 | 448386 | 180,534 |
https://mathoverflow.net/questions/448371 | 6 | We set :
* $\phi\_1, \phi\_2 : \mathbb{R} \to \mathbb{R}^M$ with compact support
* $\theta \in \mathbb{R}^M$ non-zero, $\theta\_{p-1} = (\operatorname{sign}(\theta\_i)|\theta\_i|^{p-1})\_{1 \leq i \leq M} \in \mathbb{R}^M$ with $p \geq 2$.
We assume that the matrix
$$ \int\_\mathbb{R} \phi\_1 \phi\_1^T - \int\_\mathbb{R} \phi\_2 \phi\_2^T $$
is positive definite. I then want to show that (if it is true) :
$$ \int\_\mathbb{R} \theta\_{p-1}^T \phi\_1(t) \phi\_2(t)^T \theta < \int\_\mathbb{R} \theta\_{p-1}^T \phi\_1(t) \phi\_1(t)^T \theta $$
If $p=2$ the problem is easy with Cauchy-Schwarz on the second term + assumption. I struggle to prove the general form (which seems to be true according to some code but I am not sure).
***Edit***
**The proof of the case $p=2$** :
$$
\int\_\mathbb{R} \theta^T \phi\_1(t) \phi\_2(t)^T \theta \leq \sqrt{\int\_\mathbb{R} (\theta^T \phi\_1(t))^2 \int\_\mathbb{R} (\theta^T \phi\_2(t))^2 } $$
using Cauchy-Schwarz and
$$
\int\_\mathbb{R} (\theta^T \phi\_2(t))^2 < \int\_\mathbb{R} (\theta^T \phi\_1(t))^2
$$
Finally :
$$
\int\_\mathbb{R} \theta^T \phi\_1(t) \phi\_2(t)^T \theta < \sqrt{\int\_\mathbb{R} (\theta^T \phi\_1(t))^2 \int\_\mathbb{R} (\theta^T \phi\_1(t))^2 } = \int\_\mathbb{R} (\theta^T \phi\_1(t))^2
$$
**Attempt of a general proof (false !!!)**
We write :
$$ \theta\_{p-1} = D \theta $$
with
$$D = \operatorname{DiagonalMatrix}(|\theta\_i|^{p-2})$$
From now (third step is false):
$$\begin{aligned}
\int\_\mathbb{R} \theta\_{p-1}^T \phi\_1(t) \phi\_2(t)^T \theta
&= \int\_\mathbb{R} \phi\_1(t)^T \theta\_{p-1} \theta^T \phi\_2(t) \\
&= \int\_\mathbb{R} \phi\_1(t)^T D \theta \theta^T \phi\_2(t) \\
&= \int\_\mathbb{R} \phi\_1(t)^T \sqrt{D} \theta \theta^T \sqrt{D} \phi\_2(t) \\
&\leq \sqrt{ \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \,dt \cdot \int (\phi\_2(t)^T \sqrt{D} \theta)^2 dt } \\
&< \sqrt{ \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \,dt \cdot \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \,dt }
\end{aligned} $$
Then
$$ \begin{aligned}
\int\_\mathbb{R} \theta\_{p-1}^T \phi\_1(t) \phi\_2(t)^T \theta &< \sqrt{ \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \, dt \cdot \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \, dt } \\
&= \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \, dt \\
&= \int \theta^T \sqrt{D} \phi\_1(t) \phi\_1(t)^T \sqrt{D} \theta \, dt \\
&= \int \theta^T D \phi\_1(t) \phi\_1(t)^T \theta \, dt \\
&= \int \theta\_{p-1}^T \phi\_1(t) \phi\_1(t)^T \theta \, dt
\end{aligned} $$
| https://mathoverflow.net/users/402235 | Cauchy-Schwarz-like inequality with a power $p$ term | $\newcommand\th\theta\newcommand\R{\mathbb R}$This is false for any real $p>2$ (actually, this is false for any real $p>1$ such that $p\ne2$).
Indeed, if this were true, then, by continuity, we could replace "positive definite" with "positive semi-definite", at the same time replacing $<$ in the desired inequality by $\le$.
Next, assuming that each of the functions $\phi\_1$ and $\phi\_2$ takes only finitely many values, the positive answer to your question would imply the positive answer to the following question:
>
> Suppose $A$ and $B$ are $m\times n$ real matrices such that
>
> $$\|A\th\|\le\|B\th\|\quad\text{for all $\th=[\th\_1,\dots,\th\_n]^\top\in\R^n=\R^{n\times1}$ },\tag{1}\label{1}$$
> where $\|\cdot\|$ is the Euclidean norm. Let $p\in(1,\infty)$. Does it then necessarily follow that
> $$(B\th^{[p-1]})\cdot(A\th)\le(B\th^{[p-1]})\cdot(B\th)\tag{2}\label{2}$$
> for all $\th=[\th\_1,\dots,\th\_n]^\top\in\R^n$, where $\cdot$ is the dot product, $\th^{[p-1]}:=[\th\_1^{[p-1]},\dots,\th\_n^{[p-1]}]^\top$ and $u^{[p-1]}:=|u|^{p-1}\text{sign}\, u$ for real $u$?
>
>
>
(See the detail on this reformulation at the end of this answer.)
However, it is easy to find (say) $2\times2$ real matrices $A$ and $B$ and some $\mu\in\R^2$ such that \eqref{1} holds, $\|A\mu\|=\|B\mu\|$, $A\mu$ is in the same direction with $B\mu^{[p-1]}$, but $B\mu$ is not in the same direction with $B\mu^{[p-1]}$. Then \eqref{2} will fail to hold for $\th=\mu$.
E.g., suppose that $B=I\_2$ and $\mu=[2,1]^\top$, so that $B\mu=\mu$ is not in the same direction with $B\mu^{[p-1]}=\mu^{[p-1]}$. Let now $A$ be the rotation matrix such that $A\mu$ is in the same direction with $B\mu^{[p-1]}=\mu^{[p-1]}$. Then \eqref{1} will hold and, in particular, we will have $\|A\mu\|=\|B\mu\|$, whereas \eqref{2} will fail to hold for $\th=\mu$.
---
**Detail on the highlighted reformulation:** This follows because
$$\int\_\R\rho^\top\phi\psi^\top\th \\
=\sum\_{j,k}\rho\_j\th\_k\int\_\R\phi^{(j)}\psi^{(k)}
=\sum\_{j,k}\rho\_j\th\_k\sum\_i a\_{ij}b\_{jk}
=(A\rho)\cdot(B\th),$$
where $A:=[a\_{ij}]$, $B:=[b\_{ik}]$, $\rho=[\rho\_1,\dots,\rho\_M]^\top$, $\th=[\th\_1,\dots,\th\_M]^\top$, $\phi=[\phi^{(1)},\dots,\phi^{(M)}]^\top$, $\psi=[\psi^{(1)},\dots,\psi^{(M)}]^\top$, $\phi^{(j)}=\sum\_{i=1}^m a\_{ij}1\_{[i,i+1)}$, and $\psi^{(k)}=\sum\_{i=1}^m b\_{ik}1\_{[i,i+1)}$.
| 4 | https://mathoverflow.net/users/36721 | 448402 | 180,537 |
https://mathoverflow.net/questions/448370 | 2 | I am reading O. Fujino's book [*Iitaka Conjecture*](https://link.springer.com/book/10.1007/978-981-15-3347-1). In page 42, Lemma 3.1.19 he restated one result due to Viehweg to use the base change arguments.
There exists some details in the proof of **Step 2** in Lemma 3.1.19 that I can't figure out as follows:
>
> Let $X$ be a reduced Gorenstein analytic space which may be reducible, denote by
> $$\nu:\widetilde X\rightarrow X$$ the normalization of $X$. Then he claim that one can get an inclusion $$\iota:\omega\_{\widetilde X}\hookrightarrow{\nu^\*\omega\_X}.$$
>
>
>
$\color{red}{\bullet}$
Thanks to relative trace map (one reference is Kollar-Mori's book Prop. 5.77), we have a natural map $$\textrm{Tr}:\nu\_\*\omega\_{\widetilde X}\rightarrow \omega\_X.$$
$\color{blue}{\bullet}$
As $\nu$ is finite, we have the following exact sequence:
$$0\rightarrow\mathcal T\rightarrow \nu^\*\nu\_\*\omega\_{\widetilde X}\rightarrow \omega\_{\widetilde X}\rightarrow 0,$$
here the sheaf $\mathcal T$ denotes the torsion part of $\nu^\*\nu\_\*\omega\_{\widetilde X}.$
These two steps seem suitable for me.
However, he then said that $\iota$ holds thanks to $\color{red}{\bullet}$ and $\color{blue}{\bullet}$. This is somewhat too quick for me.
Now I will state some of my thoughts on how to get $\iota$.
Pulling back $\mathrm{Tr}$ by $\nu$, we get the morphism
$$\mu: \nu^\*\nu\_\*\omega\_{\widetilde X}\rightarrow \nu^\*\omega\_X.$$
Here appears my first question,
>
> **Q1:** Do $\mu$ must factor through $\omega\_\widetilde X$? If so, why?
>
>
>
If Q1 holds, we indeed have
$$\nu^\*\nu\_\*\omega\_{\widetilde X}\xrightarrow{\alpha} \omega\_{\widetilde X} \xrightarrow{\iota} \nu^\*\omega\_X$$
with $\mu=\iota\circ\alpha.$
Here appears my second question,
>
> **Q2:** do we have $\mathrm{Ker} {\mu} \subseteq \mathrm{Ker} \alpha=\mathcal T$? If so, why?
>
>
>
The last question is that
>
> **Q3:** how is the sheaf map $\mu$ defined explicltly?
>
>
>
Thanks in advance.
| https://mathoverflow.net/users/141609 | Inclusion of (pulling back of) dualizing sheaves under normalization | $X$ is Gorenstein, so $\omega\_X$ is a line bundle and hence so is $\nu^\*\omega\_X$. In particular, it is torsion-free and hence $\mu(\mathcal T)=0$, so $\mu$ indeed factors through $\omega\_{\widetilde X}$.
As $X$ is reduced, $\nu$ is generically an isomorphism, so ${\rm Ker}\ \iota$ is a torsion sheaf, and then because, as Leo pointed out, $\omega\_{\widetilde X}$ is a torsion-free sheaf, it has to be zero. So, $\iota$ is injective as claimed.
In case your second question is still a question: both morphisms map to a torsion-free sheaf and as $\nu$ is generically an isomorphism, so are $\mu$ and $\alpha$. This means that the kernel of each morphism is exactly the torsion subsheaf of $\nu^\*\nu\_\*\omega\_{\widetilde X}$.
I am not really sure what you are asking in Q3. $(\ )^\*$ is a functor, so it is defined on morphisms. If you look up its definition for sheaves anywhere, it should be clear how it is defined on morphisms.
| 4 | https://mathoverflow.net/users/10076 | 448425 | 180,540 |
https://mathoverflow.net/questions/448347 | 10 | Consider a set of $n$ elements $S=\lbrace 1,\dots,n\rbrace$ and $\mathcal{P}(S)$ to be the power set of $S$, which is a well-defined poset with respect to the inclusions. Now consider $\emptyset\neq T\varsubsetneq S$ and define $A=\lbrace \emptyset\neq U\in\mathcal{P}(S)\mid T\nsubseteq U\rbrace$. I want to compute the homotopy type of the geometric realization $\left|A\right|$.
My intuition tells me that $\left|A\right|\simeq\mathbb{S}^{\left|T\right|-2}$, however I don't know if this is right and how to prove it.
I know some facts such that if the poset has a maximal element or a minimal element then the geometric realization is contractible. Also, for some small cases I am able to do it by hand, however I don't know how to do it in general and whether there is some ''general'' procedure to study this problem.
Any help or references about this problem will be appreciated.
| https://mathoverflow.net/users/482329 | Homotopy type of the geometric realization of a poset | First recall that geometric realisation of posets preserves products: the projections $P\xleftarrow{p}P\times Q\xrightarrow{q}Q$ give a map $(|p|,|q|)\colon |P\times Q|\to|P|\times|Q|$, and it is a standard fact that this is a homeomorphism.
Next, if $f\_0,f\_1\colon P\to Q$ are morphisms of posets and $f\_0(x)\leq f\_1(x)$ for all $x$ then we can define a poset morphism $h\colon \{0,1\}\times P\to Q$ by $h(i,x)=f\_i(x)$, and this gives a map $|h|\colon [0,1]\times|P|\to|Q|$ which is a homotopy between $|f\_0|$ and $|f\_1|$.
Now put $A=\{U\subseteq S\;:\;U\neq\emptyset, U\not\supseteq T\}$ as in the question. We can define $f\_0,f\_1,f\_2\colon A\to A$ by $f\_0(U)=U$ and $f\_1(U)=U\cup T^c$ and $f\_2(U)=T^c$. Then $f\_0\leq f\_1\geq f\_2$, so the identity is homotopic to the constant map $|f\_2|$, so $|A|$ is contractible.
Now suppose instead we consider the poset $B=\{U\subseteq S\;:\;T\cap U\neq\emptyset,U\not\supseteq T\}\subset A$, and the poset $C=\{U\;:\;\emptyset\subset U\subset T\}$. We have an inclusion $i\colon C\to B$ and a retraction $r\colon B\to C$ given by $r(U)=T\cap U$. These have $ri=1$ and $ir\leq 1$ so $|i|$ and $|r|$ are mutually inverse homotopy equivalences.
Alternatively, we can say that $B\simeq C\times D$, where $D$ is the poset of all subsets of $T^c$. This gives $|B|\simeq|C|\times|D|$, where $|D|$ is homeomorphic to $[0,1]^{|T^c|}$ and in particular is contractible.
Now put $M=\{m:T\to\mathbb{R}\;:\;\sum\_{t\in T}m(t)=0\}$ (which is a vector space of dimension $|T|-1$ with an obvious inner product). For $U\in C$ define $f(U)=\chi\_U-|U||T|^{-1}\in M$. Extend this linearly over simplices to get a map $f\colon |C|\to M$. One can check that this is nowhere zero, so we can define $f\_1(x)=f(x)/\|f(x)\|$, and this gives a homeomorphism $|C|\to S(M)\simeq S^{|T|-2}$.
| 6 | https://mathoverflow.net/users/10366 | 448433 | 180,541 |
https://mathoverflow.net/questions/448369 | 2 | This posting is generally related to a prior posting titled ["Are all constructible from below sets parameter free definable?"](https://mathoverflow.net/questions/448130/are-all-constructible-from-below-sets-parameter-free-definable)
If we work in infinitary language $\mathcal L\_{\omega\_1, \omega}$, then we can define *parameter free definable*, denoted "$D$", as:
$Dx \iff \bigvee x= \{ y \mid \Phi \}$
where $\Phi$ range over all formulas in $\mathcal L\_{\omega, \omega}$ in which only the symbol "$y$" occurs free, and the symbol "$y$" never occurs bound.
**Axiom of definability:** $\forall x Dx$
Let $\sf ZF + Def$ be the theory that extends $\mathcal L\_{\omega\_1, \omega}$ with axioms of $\sf ZF$ (written in $\mathcal L\_{\omega, \omega}$) and the axiom of definability.
>
> Is $\sf ZF +Def$ consistent ?
>
>
>
>
> If so then does $\sf ZF + Def$ have all of its models being pointwise definable models?
>
>
>
| https://mathoverflow.net/users/95347 | Does adding definability axiom expressed in infinitary language to ZF, let all models be pointwise definable? | Yes, the theory is consistent, if ZF is consistent, because there are pointwise definable models of ZF. Any such model is a model of your theory, which is therefore satisfiable and hence consistent.
And yes, clearly every model of your theory is pointwise definable (in the first-order language), because that is precisely what your axiom of definability asserts. The only way to make this axiom true in a model of ZF is for every set to be definable.
The models of ZF+Def are exactly the pointwise definable models of ZF.
| 3 | https://mathoverflow.net/users/1946 | 448434 | 180,542 |
https://mathoverflow.net/questions/448426 | 14 | Let $R$ be a commutative unital ring. Let $n\in\mathbb{N}\_+$. Consider a $n\times n$-matrix $A=(a\_{ij})\_{i,j=1}^n$ with entries in $R$. A diagonal matrix is defined obviously. We can define several notions about diagonalizability (for some prime $\mathfrak{p}\subseteq R$):
1. $A$ is *diagonalizable in $R$*: if there exists an $n\times n$-matrix $P$ with entires in $R$ such that $\det(P)\in R^\times$ and $PAP^{-1}$ is diagonal;
2. $A$ is *diagonalizable in $R\_{\mathfrak{p}}$*: consider $A$ as a matrix over $R\_{\mathfrak{p}}$ and similarly defined;
3. $A$ is *diagonalizable in $\kappa(\mathfrak{p})$*: similarly defined;
4. $A$ is *diagonalizable in $\overline{\kappa(\mathfrak{p})}$*: similarly defined, where we consider an algebraic closure.
Obviously $(1)\implies(2)\implies(3)\implies(4)$. Moreover $(4)$ has the minimal polynomial criterion. I have two observations:
* (2) is an open condition on $\mathfrak{p}$: the primes $\mathfrak{p}$ such that $A$ is diagonalizable in $R\_{\mathfrak{p}}$ form an open subset of $\mathrm{Spec}(R)$;
* (4) is a *constructible condition* on $\mathfrak{p}$: the primes $\mathfrak{p}$ such that $(4)$ holds form a constructible subset. This is seen from the morphism, whose image is where $(4)$ holds
$$\mathrm{GL}\_n\times\mathbb{A}^n\to \mathbb{A}^{n\times n},\quad P,\lambda\_1,\cdots,\lambda\_n\mapsto P\begin{pmatrix}\lambda\_1\\&\ddots\\&&\lambda\_n\end{pmatrix}P^{-1}. $$
I do not expect either $(3)\implies(2)$ or $(4)\implies(3)$, though my intuition of local diagonalizability is $(4)$. My question is about $(2)\implies(1)$. For the title I mean the following.
>
> If $A$ is diagonalizable in $R\_{\mathfrak{p}}$ for all primes $\mathfrak{p}$, can we claim that $A$ is diagonalizable?
>
>
>
| https://mathoverflow.net/users/105537 | Is diagonalizability a local property? | Here is a simple counterexample (simplified and generalised after Mohan's comment): let $R$ be a domain with a non-free finite projective module $P$, and let $Q$ be a finite projective module with $P \oplus Q \cong R^n$. This gives an idempotent matrix $A \in M\_n(R)$ corresponding to $(p,q) \mapsto (p,0)$.
For each prime ideal $\mathfrak p \subseteq R$, the modules $P\_{\mathfrak p}$ and $Q\_{\mathfrak p}$ are free. If $S \colon P\_{\mathfrak p} \oplus Q\_{\mathfrak p} \stackrel\sim\to R\_{\mathfrak p}^n$ is the isomorphism above and $D \colon P\_{\mathfrak p} \oplus Q\_{\mathfrak p} \to P\_{\mathfrak p} \oplus Q\_{\mathfrak p}$ the diagonal matrix $(p,q) \mapsto (p,0)$, then $A\_{\mathfrak p} = SDS^{-1}$ is diagonalisable. (We can really think of $D$ as a matrix, at least after choosing bases for $P\_{\mathfrak p}$ and $Q\_{\mathfrak p}$.)
But $A$ is not globally diagonalisable since $P$ is not free: if $A = SDS^{-1}$ for some diagonal idempotent matrix $D$ and some invertible matrix $S$, then $S$ induces an isomorphism $S \colon \operatorname{im}(D) \stackrel\sim\to \operatorname{im}(A)$. But $\operatorname{im}(D)$ is free whereas $\operatorname{im}(A) \cong P$ is not free.
**Example.** Carrying this out for $R = \mathbf Z[\sqrt{-5}]$ and $P = (2,1+\sqrt{-5}) \subseteq R$ produces the matrix
$$A = \begin{pmatrix} -2 & -1-\sqrt{-5} \\ 1-\sqrt{-5} & 3 \end{pmatrix}.$$
The image contains the vectors $\left(\begin{smallmatrix} -2 \\ 1-\sqrt{-5} \end{smallmatrix}\right)$ and $\left(\begin{smallmatrix} -1-\sqrt{-5} \\ 3\end{smallmatrix}\right)$, which are linearly dependent but not common multiples of some other vector in $R^2$.
**Motivation.** If $A\_{\mathfrak p}$ is diagonalisable for all $\mathfrak p \subseteq R$, then there exists an open cover $U\_1 \cup \ldots \cup U\_m = \operatorname{Spec} R$ such that $A|\_{U\_i}$ is diagonalisable for all $i$. This gives matrices $S\_i \in \operatorname{GL}\_n(U\_i)$ such that $D\_i := S\_i^{-1}A|\_{U\_i}S\_i$ is diagonal.
Assuming for simplicity that $R$ is a domain, we conclude in particular that all eigenvalues $\lambda\_1,\ldots,\lambda\_r$ of $A$ lie in $R$, since they lie in $R\_{\mathfrak p}$ for all prime ideals $\mathfrak p$. Choose some diagonal matrix $D \in M\_n(R)$ with the same eigenvalues/multiplicities as $A$, and assume without loss of generality that $D\_i = D|\_{U\_i}$ for all $i$. Then $S\_j^{-1}S\_i \in \operatorname{GL}\_n(U\_i \cap U\_j)$ is a matrix commuting with $D$, giving a cocycle in $H^1(\operatorname{Spec} R,G)$, where $G = C\_{\operatorname{GL}\_n}(D)$ is the centraliser of $D$. This cocycle is a coboundary if and only if $A$ and $D$ are globally conjugate matrices.
Under the additional hypothesis that $\lambda\_i - \lambda\_j \in R^\times$ whenever $i \neq j$, a straightforward computation shows $G = \prod\_{i=1}^r \operatorname{GL}\_{m\_i}$, where $m\_i$ is the algebraic multiplicity of the eigenvalue $\lambda\_i$. Thus, in this case the only obstruction is in $\prod\_{i=1}^r H^1(\operatorname{Spec} R, \operatorname{GL}\_n)$, leading to the example above.
But in general, $G$ need not be a flat group scheme over $R$, so the situation is considerably more complicated. For instance, if $D$ is the diagonal matrix $\left(\begin{smallmatrix}2 & 0 \\ 0 & 5\end{smallmatrix}\right)$, then $G \times \operatorname{Spec} \mathbf F\_3$ is the full group $\operatorname{GL}\_{2,\mathbf F\_3}$, so $G$ has a vertical component above $p=3$. Even in the case of a PID, it is not so clear to me what is going on...
**Remark.** If $R$ is no longer a domain, it is probably more natural to look at the centraliser of $A$ instead of $D$ (which is some sort of 'Zariski inner form' of $C\_{\operatorname{GL}\_n}(D)$ in the domain case, so it should have the same $H^1$). I expect there might be other global obstructions coming from the non-injectivity of the maps $R \to R\_{\mathfrak p}$: perhaps there are too many global choices of $D$ to run the argument above.
You could also imagine an obstruction in $H^1(\operatorname{Spec} R, S\_n)$ for globally ordering the eigenvalues in the matrix $D$, although all examples I know where $H^1(\operatorname{Spec} R, S\_n) \neq 1$ also have $H^1(\operatorname{Spec} R, \operatorname{GL}\_m) \neq 1$ for $m \gg 0$. (An irreducible scheme has no higher cohomology in constant sheaves, even for $H^1$ with non-abelian coefficients, so we didn't see this problem in the case above.)
| 18 | https://mathoverflow.net/users/82179 | 448447 | 180,547 |
https://mathoverflow.net/questions/448441 | 2 | I am looking at showing that a complex symmetric invertible matrix always has a complex symmetric square root and I refer to [this Q&A](https://mathoverflow.net/a/438232/497608) for the answer to this question. I am little confused at the reasoning given in the answer however, specifically in two places.
1. Firstly, how can we show that the polynomial $Q(z) := P(z)^2 -z$ and its $m\_j-1$ derivatives are all $0$ at each $\lambda\_j$? I have done out a the first few examples which all work, but how do we know this in general? Note that $\lambda\_j$ are the eigenvalues of some $n \times n$ complex matrix $A$ and $m\_j$ are the indices of $\lambda\_j$ in the Jordan Canonical Form of $A$.
2. Once we have established the result in (1) above, the answer to the OP says that this implies that $Q(z)$ is a multiple of the characteristic polynomial of $A$, denoted here by $C\_A(z)$.
My understanding for (2) is this: For $Q(z)$ and $C\_A(z)$ to be a multiplicative constant, they both must have the same roots and these roots must have the same multiplicities. Both $C\_A(z)$ and $Q(z)$ have $\lambda\_j$ as their roots, this is obvious. Initially, I thought that the derivatives of $Q(z)$ up to order $m\_j-1$ being $0$ implied something about the multiplicities of each of the roots but now I don't think that is true because the multiplicity and the index aren't necessarily the same.
Since then I have hit a bit of a wall. Does anyone have any insight into these two problems?
| https://mathoverflow.net/users/497608 | Questions regarding answer to complex symmetric square root of a complex symmetric invertible matrix | $\newcommand\la\lambda$
1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la\_j)=g^{(k)}(\la\_j)$ for each $j$ and all $k=0,\dots,m\_j-1$, where the $\la\_j$'s are the distinct eigenvalues of $A$ and the $m\_j$'s are their multiplicities. Then, by the [Leibniz rule](https://en.wikipedia.org/wiki/General_Leibniz_rule), $Q^{(k)}(\la\_j)=0$ for each $j$ and all $k=0,\dots,m\_j-1$.
2. So, for some polynomial $R(z)$,
\begin{equation}
Q(z)=R(z)\prod\_j(z-\la\_j)^{m\_j}
=R(z)C(z), \tag{1}\label{1}
\end{equation}
where $C(z):=\prod\_j(z-\la\_j)^{m\_j}$ is the characteristic polynomial of $A$ (see details on the first equality in \eqref{1} below). So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.
(The [previous answer](https://mathoverflow.net/a/438232/36721), referred to in your question, does not even mention indices.)
---
**Details on the first equality in \eqref{1}:**
>
> **Lemma 1:** Let $q(z)$ be a polynomial such that $q^{(k)}(\la)=0$ for some $\la$ and all $k=0,\dots,m-1$, where $m$ is an integer $\ge1$. Then $(z-\la)^m$ is a divisor of the polynomial $q(z)$ -- that is,
> $q(z)=(z-\la)^m S(z)$ for some polynomial $S(z)$.
>
>
>
*Proof of Lemma 1:* By shifting, without loss of generality $\la=0$.
Divide $q(z)$ by $z^m$ with a remainder $r(z)$ of degree $\le m-1$, so that $q(z)=z^m s(z)+r(z)$ for some polynomial $s(z)$. Then for all $k=0,\dots,m-1$ we have $0=q^{(k)}(0)=r^{(k)}(0)$. Because $r(z)$ is of degree $\le m-1$, it follows that $r(z)$ is the zero polynomial, so that $q(z)=z^m s(z)$, which completes the proof of Lemma 1. $\quad\Box$
Now, by part 1 of the answer, $Q^{(k)}(\la\_j)=0$ for each $j$ and all $k=0,\dots,m\_j-1$. So, for each $j$, by Lemma 1, the polynomial $(z-\la\_j)^{m\_j}$ is a divisor of the polynomial $Q(z)$. Also, the polynomials $(z-\la\_j)^{m\_j}$ are coprime for different values of $j$, since the $\la\_j$'s are distinct. So, $\prod\_j(z-\la\_j)^{m\_j}$ is a divisor of $Q(z)$; that is, the first equality in \eqref{1} holds.
It remains to provide
| 3 | https://mathoverflow.net/users/36721 | 448449 | 180,549 |
https://mathoverflow.net/questions/448444 | 2 | Consider $\mathbb{R}\_t \times \mathbb{R}\_x ^n$ , let $b\_1(t,x)$ and $b\_2 (t,x)$ be two velocity fields with all the regularity you want and consider the flow of the point $(0,x\_0)$ for a time $T$. Now, I can view the trajectories of $(0,x\_0)$ as curves in $\mathbb{R}^{n+1}$ like $ \gamma\_1 =(t, \Phi\_1 (t,x\_0))$ and $\gamma \_2= (t, \Phi\_2 (t,x\_0))$ with $\Phi\_i$ the flux relative to $b\_i$. The question is: if I consider $\tilde{\gamma} =\gamma\_1 \cup \gamma\_2 \cup I\_T$ where $I\_T$ is a segment in $\mathbb{R}^{n+1}$ joining $(T,\Phi\_1 (T, x\_0) )$ and $(T, \Phi\_2 (T, x\_0))$ , I can consider the Plateau problem associated to the rectifiable set $\tilde{\gamma}$, and by the general theory I'll have a minimal surface whose boundary is $\tilde{\gamma}$. The question is: given how my $\gamma$ is constructed, is there a way to estimate the area of such surface in terms of $b\_1$ and $b\_2$?
EDIT:
Actually, what I want is to integrate such an estimate, starting from a certain mass distribution at time 0, so I would like something whixh takes that into account.
| https://mathoverflow.net/users/109382 | Plateau problem for fluxes of curves | Almgren (<https://mathscinet.ams.org/mathscinet-getitem?mr=855173>) proves that the Plateau solution will satisfy the 2-dimensional Euclidean isoperimetric inequality. So you can bound
$$
\textrm{Area(Plateau Solution)} \leq (4\pi)^{-1}(L(\gamma\_1)+L(\gamma\_2)+L(I\_T))^2.
$$
If you just want a coarse estimate you should be able to bound each term by the $L^\infty$ norm of $b\_1,b\_2$.
Of course, this inequality is unlikely to be close to optimal. (For example, if $b\_1=b\_2$ then the area of the Plateau solution is $=0$ while the RHS above is nonzero). If you want an estimate that works better in this case you might construct the explicit competitor
$$
\Gamma : = \bigcup\_{t=0}^T I\_{\Phi\_1(t,x\_0),\Phi\_2(t,x\_1)} \times \{t\}
$$
namely you union up all the lines between the points in the flow. You should be able to compute the area of this in terms of $b\_1,b\_2$ using the co-area formula.
EDIT: As requested here is a formula for the area of $\Gamma$. I did it via a parametrization instead of co-area but I think co-area would also work.
I parametrize $\Gamma$ by $F: [0,1]\times [0,T]\to\mathbb{R}^{n+1}$
$$
F(s,t) : = (s\Phi\_1(t)+(1-s)\Phi\_2(t),t)
$$
(dropping the $x\_0$ notation). Then we have
$$
\partial\_s F = \Phi\_1-\Phi\_2, \qquad \partial\_t F = s\partial\_t\Phi\_1 +(1-s)\partial\_t \Phi\_2 + \mathbf{e}\_{n+1}.
$$
In particular, the induced metric is
\begin{align\*}
g\_{ss} & = |\Phi\_1-\Phi\_2|^2 \\
g\_{st} & = \langle \Phi\_1-\Phi\_2,s\partial\_t\Phi\_1 +(1-s)\partial\_t \Phi\_2\rangle\\
g\_{tt} & = 1 + |s\partial\_t\Phi\_1 +(1-s)\partial\_t \Phi\_2|^2.
\end{align\*}
Hence
$$
\det g = (1 + |s\partial\_t\Phi\_1 +(1-s)\partial\_t \Phi\_2|^2)|\Phi\_1-\Phi\_2|^2 - \langle \Phi\_1-\Phi\_2,s\partial\_t\Phi\_1 +(1-s)\partial\_t \Phi\_2\rangle^2.
$$
This gives the explicit formula
$$
\textrm{area}(\Gamma) = \int\_{[0,1]\times [0,T]} \sqrt{(1 + |s\partial\_t\Phi\_1 +(1-s)\partial\_t \Phi\_2|^2)|\Phi\_1-\Phi\_2|^2 - \langle \Phi\_1-\Phi\_2,s\partial\_t\Phi\_1 +(1-s)\partial\_t \Phi\_2\rangle^2}.
$$
If you want to write this in terms of $b\_1$ and $b\_2$ you can do the following (perhaps one can be more precise here to get a better estimate). First of all
$$
|\partial\_t \Phi\_i| = |b\_i| \leq \Vert b\_i\Vert\_{C^0}
$$
Secondly, by the argument [here](https://mathoverflow.net/a/300284/1540) we can bound
$$
|\Phi\_1-\Phi\_2| \leq L^{-1}(e^{Lt}-1)|b\_1-b\_2|\_{L^\infty}.
$$
for $L$ the Lipschitz constant of one of the vector fields.
Using this above we should get
$$
\textrm{area}(\Gamma) \leq (1 + (|b\_1|\_{C^0}+|b\_2|\_{C^0}) |b\_1-b\_2|\_{C^0}^2 L^{-2}(e^{LT}-LT-1).
$$
Note that I discarded the $g\_{st}$ term, which could improve this estimate if you had some way of knowing it was big. But it does have the nice feature that it vanishes if $b\_1=b\_2$ and it's $O(T^2)$ for $T$ small.
I'm sure you can improve the estimate used above in the case of large $T$ (e.g. if $b\_i$ are bounded, then the flows should diverge at most linearly).
| 4 | https://mathoverflow.net/users/1540 | 448453 | 180,550 |
https://mathoverflow.net/questions/448452 | 0 | I'm wondering about the following:
1. Every continuous map between smooth manifolds is homotopic to a smooth map.
2. By density of polynomials in space of continuous functions on [0,1], continuous functions can be approximated by smooth ones with respect to the maximum norm.
Now, assume $G\_1$ and $G\_2$ are Lie groups; maybe asume $G\_1$ and $G\_2$ compact. Let $f:G\_1\rightarrow G\_2$ be a continuous map. Can $f$ approximated (in some sense) by a continuous group homomorphism?
The answer by Dmitri raises another question.
What classes of continuous functions between Lie groups can be approximated by continuous group homomorphisms?
| https://mathoverflow.net/users/506544 | Continuous map between Lie groups approximation | Let $G\_1=G\_2=U(1)$. Take $f$ such that its image is in an $1/2$-neighborhood of the identity. On the other hand any group homomorphism is either constant or surjective. So there is no chance of approximating continuous maps by homomorphisms.
| 4 | https://mathoverflow.net/users/13842 | 448455 | 180,551 |
https://mathoverflow.net/questions/448450 | 4 | This is closely related to the question [here](https://mathoverflow.net/questions/82613/riemann-mapping-theorem-and-smoothness-on-the-boundary). The setup is that $U\subset\mathbb{C}$ is an open bounded simply connected domain with $C^\infty$ boundary. If $\phi:U\rightarrow\mathbb{D}$ is a biholomorphic mapping from $U$ to the unit disk, [it is known](http://www.ams.org/journals/bull/1990-22-02/S0273-0979-1990-15879-3/S0273-0979-1990-15879-3.pdf) that $\phi$ extends to a smooth map $\overline{U}\rightarrow\overline{\mathbb{D}}$. However, the question that is not addressed is whether this extension is actually a diffeomorphism. Now a comment by the OP on the linked question states that Graeme Segal showed something like this in some unpublished notes, which I can't find anywhere. So let me now ask this precisely:
Does $\phi$ as above extend to a diffeomorphism? The answer seems to be yes, but I would appreciate any reference that confirms this.
Incidentally, I'm also interested in analogous $C^k$-boundary and boundary-with-corners statements, but those seem even more exotic.
| https://mathoverflow.net/users/480683 | Riemann mapping theorem with smooth boundary | The main reference on this topic is the book "Boundary behavior of conformal maps" by Pommerenke.
If the curve is $C^\infty$, then the biholomorphic mapping extends to a smooth map on the closure of the unit disk, and the derivative is non-vanishing. This is implicit in Theorem 3.2 of Pommerenke's book, see also Theorem 3.5.
For boundaries with corners, you might be interested in Section 3.4.
| 7 | https://mathoverflow.net/users/1162 | 448458 | 180,553 |
https://mathoverflow.net/questions/448309 | 2 | Let $X$ be a surface $P$ be a point on $X$. Let $Z$ be a subscheme supported on the single point $P$ ($Z$ not equal $P$), denote its local multiplicity by $\mu\_P(Z)$.
Now blow up $\pi: \tilde{X} \to X$ along $Z$, we get corresponding exceptional divisor $E$. I would like to know what is $E^2$. I made my guess that it might be $-\mu\_P(Z)$ but I am not sure.
I tried to do the same calculation as how monoidal transform being done in Hartshorne. The self intersection is defined to be (Hartshorne V 1.4.1) $E^2=\deg\_E \mathcal{N}\_{E/\tilde{X}}$, where $\mathcal{N}\_{E/\tilde{X}}$ is the normal sheaf. From Hartshorne II Theorem 8.24, which suits for any blow up along an ideal sheaf, the normal sheaf is isomorphic to $\mathcal{O}\_{E}(-1)$. So $E^2$ looks still like $-1$.
But clearly this blow up is not monoidal, by Castelnuovo contract criterion, $E^2$ should not be $-1$. I am confused here.
Any comment is appreciated! If there is any reference for this situation would be great!
| https://mathoverflow.net/users/80167 | Self-intersection of exceptional divisor from blowing up a subscheme | I'm afraid it does not work the way you are hoping. At least not right out of the box. First of all, why do you think that $E^2$ is even defined? Also, what exactly is $E$? Is it the pre-image of $Z$, or the reduced exceptional divisor? I assume you are thinking of the former, because the latter could be rather tricky...
Furthermore, the proof of Hartshorne II Theorem 8.24 **does not** suit blowing up an arbitrary ideal sheaf. At best it requires the ideal sheaf to be locally generated by a regular sequence. In other words, it would require $Z$ to be a local complete intersection in $X$. OK, so assume that. Then if you blow up $Z$, then the proof works, and gives you $\mathscr O(1)$, but then the question is: This is $\mathscr O(1)$ of **what**? If you look at the proof, it is $\mathscr O\_{\mathbb P(\mathscr I/\mathscr I^2)}(1)$. However, since you blew up a fat point, $\mathbb P(\mathscr I/\mathscr I^2)$ is not a "usual" $\mathbb P^1$. It is $\mathbb P^1\_Z$, i.e., a $\mathbb P^1$ over your fat point. In particular, it is non-reduced. So, in order to define $E^2$ you need to figure out a degree function for invertible sheaves on $\mathbb P^1\_Z$. I suppose one possibility is to multiply the degree function of $\mathbb P^1$ by the length of $Z$. If you do that, then you actually get what you want (assuming that $P$ was a non-singular point of $X$), but this $\widetilde X$ is not going to be even normal, so you can't expect a very good intersection theory. Perhaps you would be better off, trying to keep track of the sheaves associated to your codimension 1 subvarieties.
Finally, you could try looking at examples. The simplest is probably blowing up the ideal $(x^2,y^2)$ in $\mathbb A^2$. That gives you a pinch point and your $E$ is the singular locus (doubled). You can try out your expectations on this example.
| 0 | https://mathoverflow.net/users/10076 | 448461 | 180,555 |
https://mathoverflow.net/questions/444841 | 9 | Namba forcing is stationary-preserving and forces $cf(\omega\_2^{\mathbf{V}}) = \omega$. Ronald Jensen used $\mathcal{L}$-forcing to iterate Namba posets in order to solve the extended Namba problem: for any strongly inaccessible $\kappa$, he constructed a stationary-preserving forcing notion $\mathbb{P}\_{\kappa}$ that forces all ground model regular cardinals in the interval $(\omega\_1, \kappa)$ to have cofinality $\omega$, while preserving the cofinality of $\kappa$. His construction allows $\mathbb{P}\_{\kappa}$ to add no reals.
To formalise my question, let me use $Nb(\alpha, \beta)$ to denote the statement
$``$there is a stationary-preserving forcing notion $\mathbb{P}$ such that
* for all $\lambda \in \mathbf{V} \cap (\alpha, \beta)$, if $\lambda$ is regular, then
$\Vdash\_{\mathbb{P}} cf(\lambda) = \omega$, and
* $\Vdash\_{\mathbb{P}} cf(\beta) > \omega$.$"$
In solving the extended Namba problem in the affirmative, Jensen showed that $Nb(\omega\_1, \beta)$ is true when $\beta$ is strongly inaccessible. In another example of $\mathcal{L}$-forcing he gave, he used a modified construction to get $Nb(\omega\_1, \beta^+)$ for any cardinal $\beta$ of cofinality $\omega\_1$. The forcing notions he presented have the additional property of not adding reals, which I do not require.
My question is thus, is it known if $Nb(\omega\_1, \beta)$ is true for all regular cardinals $\beta$?
| https://mathoverflow.net/users/29231 | Extending Namba forcing to arbitrary lengths | An alternative to Andreas’ reply in the comments; using side condition methods one can construct forcings that perform similar jobs as the forcings by Jensen that were mentioned. The following is provable in ZFC by such rather elementary tools making use of countable models as side conditions:
**Theorem.**
For every regular cardinal $\lambda$ and every set $\mathcal{K} \subset \lambda$ that consists of uncountable regular cardinals different from $\omega\_1$, there exists a forcing $\mathbb{P}$ with the following properties
1. $\mathbb{P}$ is ssp,
2. for every $\kappa \in \mathcal{K}$, forcing with $\mathbb{P}$ changes the cofinality of $\kappa$ to $\omega$,
3. for every regular cardinal $\kappa \in (\omega\_1, \lambda] \setminus \mathcal{K}$, forcing with $\mathbb{P}$ changes the cofinality of $\kappa$ to $\omega\_1$.
Working with Boban Velickovic, I consider such a forcing in my thesis (forthcoming), in the context of constructing combinatorial analogues to $\mathcal{L}$-forcings. The conditions are basically finite configurations of suitable countable elementary substructures of some $H\_\theta$, together with working parts that grow into countable sequences which are cofinal in the cardinals $\kappa \in \mathcal{K}$. To infer that $\mathbb{P}$ is ssp, one needs to know that there exist projective stationary many of these suitable models, this can be proved using arguments involving games similar to those in Section 3 of
*Foreman, M., Magidor, M.* Mutually stationary sequences of sets and the non-saturation of the non-stationary ideal on $P\_\kappa(\lambda)$. *Acta Math.* **186**, 271–300 (2001).
This particular forcing $\mathbb{P}$ does always add reals.
| 4 | https://mathoverflow.net/users/506456 | 448463 | 180,556 |
https://mathoverflow.net/questions/448462 | 2 | Let $M, N$ be monads of rank $\lambda$, where $\lambda$ is a regular cardinal (I'm primarily interested in the case of finitary monads). Is there a known characterization of functors $\mathrm{Alg}~N \to \mathrm{Alg}~M$ (of course, I mean the Eilenberg-Moore category) coming from monad morphisms $M \to N$ (that is, a morphism of monoids in the monoidal category $\mathrm{End}(\mathrm{Set})$)?
| https://mathoverflow.net/users/148161 | What functors between categories of algebras are induced by morphisms of monads on $\mathrm{Set}$? | The functor $U\_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's [Some remarks on triples](https://link.springer.com/article/10.1007/BF01110118)). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.
Given such a functor $F : \mathrm{Alg}(N) \to \mathrm{Alg}(M)$, we define a monad morphism $M \to N$ as follows. For each set $x$, we have a free $N$-algebra $Nx$, which is sent by $F$ to an $M$-algebra on $Nx$. Concretely, such an algebra comprises a function $MN x \to N x$ satisfying two laws. Given this function, we precompose by the unit $M\eta\_x : Mx \to MNx$ of $N$, producing a function $Mx \to Nx$. This defines a natural transformation $M \Rightarrow N$, and it can be checked that this is a monad morphism.
In fact, the corresponding statement is true for monads on any category, not just $\mathrm{Set}$.
| 6 | https://mathoverflow.net/users/152679 | 448464 | 180,557 |
https://mathoverflow.net/questions/448226 | 9 | Let $r\leq n$ and $d$ be positive integers. A *probability vector* is a vector of non-negative entries that sum to 1. For each probability vector $\lambda$ of length $n$, let
$$s(\lambda)=(\dim[\pi] \cdot s\_{\pi}(\lambda))\_{\substack{\pi \vdash d}},$$
where $\pi \vdash d$ means that $\pi$ is an integer partition of $d$, $[\pi]$ denotes the irreducible representation of $\mathfrak{S}\_d$ indexed by $\pi$, and $s\_{\pi} (\lambda)$ is the Schur polynomial indexed by $\pi$ evaluated at $\lambda$. It can be checked that $s(\lambda)$ is a probability vector.
**Question:** Consider the probability vector
$$ t = \left(\frac{1}{\sum\_{ \pi \vdash d, \ell(\pi)\leq r } s\_{\pi}(1^n)^2}\right) \big(s\_{\pi}(1^n)^2\big)\_{ \pi \vdash d, \ell(\pi)\leq r },$$
where $\ell(\pi)$ is the number of integers appearing in the integer partition $\pi$. Is it true that $t$ is in the convex hull of the $s(\lambda)$, where $\lambda$ ranges over all probability vectors containing at most $r$ non-zero entries?
The answer is trivially yes when $r=1$.
**Notes:** The dimension of $[\pi]$ can be computed using the [Hook-Length formula](https://en.wikipedia.org/wiki/Hook_length_formula) and the Schur polynomials can be found [here](https://en.wikipedia.org/wiki/Schur_polynomial).
| https://mathoverflow.net/users/150898 | The convex hull of Schur polynomial evaluations | If I understand the question correctly then this is false. Since you want $t$ to live in the same vector space as the $s(\lambda)$'s, I am assuming that the vector $t$ has zero coordinates at each $\pi \vdash d, \ell(\pi)> r$.
Taking $d=3, r=2, n=6$ and indexing the vectors in $\mathbb C^3$ as $(x\_{(3)}, x\_{(2,1)}, x\_{(1,1,1)})$ gives us the vectors
$$s(\lambda)=(\lambda\_1^2+\lambda\_2^2, 2\lambda\_1\lambda\_2, 0)$$
and
$$t=(\frac{16}{41}, \frac{25}{41}, 0)$$
Now, the convex hull of the $s(\lambda)$'s is given by the family of vectors $(a, 1-a, 0)$ with $\frac{1}{2}\le a\le 1$, so $t$ is not contained in this convex hull.
---
All that being said, the answer is *positive* for the special case $r=n$, where the probability measure induced by $t$ is (a specialization of) a Schur measure on partitions as studied by Okounkov. This is known to induce a harmonic function on the Young graph and therefore is in the convex hull of the $s(\lambda)$'s (known as *central measures*) by the Edrei-Thoma machinery that was mentioned by Sam above in the comments.
| 2 | https://mathoverflow.net/users/2384 | 448470 | 180,558 |
https://mathoverflow.net/questions/448471 | 2 | Working in [$\mathcal L\_{\omega\_1, \omega\_1}$](https://plato.stanford.edu/entries/logic-infinitary/), add symbol $=$ with its axioms; add symbol $\in$ and axiomatize:
$\textbf{Extensionality: } \forall x \forall y : \forall z (z \in x \leftrightarrow z \in y) \to x=y$
$\textbf{Foundation: } (\forall v\_n)\_{n \in \omega} \, \exists x: \bigvee\_{n \in \omega} (x=v\_n) \land \bigwedge\_{n \in \omega} (v\_n \not \in x)$
$\textbf{Define: } x=\{y \mid \phi\} \equiv\_{def} \forall y \, (y \in x \leftrightarrow \phi)$
$\textbf{Construction: } \\(\forall v\_n)\_{n \in \omega} \, \exists x : x=\{y \mid \bigvee\_{n \in \omega} ( y=v\_n)\}$
$ \textbf{Countability: } \\ \forall x \, (\exists v\_n)\_{n \in \omega} : x=\{y \mid y \neq v\_0 \land \bigvee\_{n \in \omega} ( y=v\_n)\} $
$\textbf {Empty set: } \exists x \forall y: y \not \in x$.
>
> Is this theory Complete?
>
>
>
>
> Is this theory Categorical?
>
>
>
| https://mathoverflow.net/users/95347 | Is this theory of well founded countable sets formalized in infinitary logic, complete and categorical? | Yes, this is categorical and hence complete (with respect to any satisfactory notion of proof, that is). Specifically, I claim that any model $M$ of your theory is isomorphic to $\mathsf{HC}$, the set of hereditarily countable sets.
First, given $M$ we can construct recursively an embedding $i:\mathsf{HC}\rightarrow M$. (Note that this uses the "construction" axiom.) So we just need to show that $i$ is surjective. I think the cleanest way to do this is to note that the image of $i$ is $\mathcal{L}\_{\omega\_1,\omega\_1}$-definable in $M$, as "$a\in im(i)$ iff there is a well-founded countably-branching tree such that, when we recursively label the nodes of $T$ by the sets of labels of their children, the root gets labelled with $a$." (In fact, since $M$ correctly computes well-foundedness and countability this is a first-order definition in $M$, but that doesn't matter.) Now every element of $M\setminus im(i)$ must consist only of elements of $M\setminus im(i)$, but this contradicts well-foundedness.
Note that the ability to characterize well-foudnedness makes $\mathcal{L}\_{\omega\_1,\omega\_1}$ diverge wildly from the standard model-theoretic constraints we are used to for first-order logic or even $\mathcal{L}\_{\omega\_1,\omega}$. That said, the situation shouldn't be overstated: see [S. Friedman's *Model theory for $\mathcal{L}\_{\infty\omega\_1}$*](https://www.sciencedirect.com/science/article/pii/0168007284900125) for a discussion of what sort of "local" compactness phenomena (analogous to Barwise compactness) can still occur.
| 8 | https://mathoverflow.net/users/8133 | 448475 | 180,559 |
https://mathoverflow.net/questions/448405 | 5 | This question is an extension of my earlier question [here](https://mathoverflow.net/questions/448360/upper-bound-for-halls-conjecture-on-separation-of-squares-and-cubes), answered by Noam Elkies.
Let $A,B \in \mathbb{Z}$. Consider the inequality
$$\displaystyle |y^2 - x^3 - Ax - B| = O(|x|^{1/2 + \theta}).$$
For general $A,B$, what is the smallest $\theta \geq 0$ (independent of $A,B$) for which the above inequality has infinitely many solutions in integers $x,y$?
The construction I gave in the question linked above still works in the general case, which shows that $\theta = 1/2$ is admissible. The polynomial identity which works for the case $A = B = 0$ does not seem easy to generalize to the general case.
| https://mathoverflow.net/users/10898 | Integral points near elliptic curves | You can take $\theta = 0$, even $\theta = -1/6$ works.
Fix an integer $A \ne 0$ and an integer $B$. If $r$ is an integer, the elliptic curve $E : y^{2} = x^{3} + Ax + r^{2} A^{2}$ has the obvious point $P = (0,rA)$. Somewhat miraculously, the point $3P$ is integral. Setting $y = 512 A^{3} r^{9} + 96A^{2} r^{5} + 3Ar$ and $x = 64 A^{2} r^{6} + 8Ar^{2}$ gives
$$
|y^{2} - x^{3} - Ax - B| = |A^{2} r^{2} - B| = O(x^{1/3})
$$
if $r$ is large.
| 10 | https://mathoverflow.net/users/48142 | 448479 | 180,560 |
https://mathoverflow.net/questions/448465 | 8 | Let $M$ be a closed smooth manifold of dimension $n$ and $z\in H\_l(M,\mathbb{Z})$ a $k$-dimensional integral homology class. Theorem II.4 of Thom's classical 1954 paper states that for $l< n/2$ or $n-l$ odd, some multiple of $z$ can be represented by a smoothly embedded submanifold with the trivial normal bundle.
I wonder if this conclusion is sharp. In other words, do there exist some $M$ and a non-torsion integral homology class $z$ on $M$, so that no multiples of $z$ can admit a smoothly embedded representative with a trivial normal bundle? (Non-torsion examples are what I need. Torsion class examples are also welcome.) I'm sorry if this is a trivial question, as my field is not adjacent to algebraic topology. Many thanks.
| https://mathoverflow.net/users/482183 | Integral homology classes of which no multiples admit embedded representatives with trivial normal bundle | With the trivial normal bundle condition, it's fairly easy to produce non-realizable examples using Théorème II.2 of Thom's paper. Namely, a class $z\in H\_l(M^n;\mathbb{Z})$ is realizable by an embedding with trivial normal bundle if and only if the Poincaré dual class $y\in H^{n-l}(M;\mathbb{Z})$ is *spherical*, meaning that there is a map $f:M\to S^{n-l}$ such that $y=f^\*(s\_{n-l})$, where $s\_{n-l}\in H^{n-l}(S^{n-l};\mathbb{Z})$ is the generator.
For $y$ to be spherical, any cohomology operation which vanishes on $s\_{n-l}$ must vanish on $y$. In particular, $y^2$ must be zero.
To give an explicit torsion-free example, no nonzero element $z\in H\_2(\mathbb{C}P^2;\mathbb{Z})\cong\mathbb{Z}$ is realized by an embedding with trivial normal bundle.
| 9 | https://mathoverflow.net/users/8103 | 448487 | 180,563 |
https://mathoverflow.net/questions/448486 | 2 | Working in $\mathcal L\_{\omega\_1, \omega}$, can Foundation be captured?
My idea is to formalize a theory where all of its models are the well founded pointwise definable models of $\sf ZFC$. I attempt to formalize it as:
$\textbf{Foundation: } \\\forall x: \neg [ \bigwedge\_{n \in \omega} (\exists v\_0,..,\exists v\_n: \bigwedge\_{i \in n} (v\_{i+1} \in v\_i) \land v\_0 \in x)] $
In English: for every set $x$ it is not the case that for every finite size there is a descending membership set from $x$ of that size.
This way all sets would be truly well founded.
So we add all axioms of $\sf ZF$ written as usual in $\mathcal L\_{\omega,\omega}$, and finally add the definability axiom written in $\mathcal L\_{\omega\_1, \omega}$, which is:
$\textbf{ Axiom of Definability: } \forall x D(x)$,
Where "$D$" is defined as:
$$Dx \iff \bigvee x= \{ y \mid \Phi \}$$
where $\Phi$ range over all formulas in $\mathcal L\_{\omega, \omega}$ in which only the symbol "$y$" occurs free, and the symbol "$y$" never occurs bound.
>
> Are all models of this theory well founded!
>
>
>
If that works, then we have all models of this theory countable, all pointwise definable, and all are well founded. And I think that all ought to be arithmetically sound.
>
> But, do all models of this theory prove exactly the same $\mathcal L\_{\omega, \omega}$ set theoretic sentences.
>
>
>
My point is that every set theoretic sentence in $\mathcal L\_{\omega, \omega}$ can be written arithmetically, then if this theory captures standard arithmetic, then it cannot prove opposing arithmetical sentences.
| https://mathoverflow.net/users/95347 | Can Foundation be captured in $\mathcal L(\omega_1,\omega)$? | Your Foundation axiom does not assert that there is no infinite descending sequence, but rather merely rules out sets at infinite set-theoretic rank. For example, if $x=\omega$, then we can find for every $n$ a descending $\in$ sequence of length $n$.
You can formulate well-foundedness in $L\_{\omega\_1,\omega\_1}$ by saying directly that there is no infinite $\in$-descending sequence. But this is not possible in $L\_{\omega\_1,\omega}$, by a result due to Morley.
| 5 | https://mathoverflow.net/users/1946 | 448488 | 180,564 |
https://mathoverflow.net/questions/448438 | 6 | Let $f,g \in \mathbb{R}[x\_0,\dots,x\_k]$ be homogeneous polynomials and $X:=Z(f) \subset \mathbb{RP}^k$ be the projective variety defined by $f$.
Assume that $X$ is smooth and has codimension $1$.
Then it's a folklore statement that $X\_\epsilon:=Z(f+\epsilon g)$ is diffeomorphic to $X$ for small $\epsilon$.
>
> Question 1: How is this proved?
>
>
> Question 2: What is an upper bound for $\epsilon$ such that all $X\_\epsilon$ are diffeomorphic to $X$? (Possibly given in terms of $grad(f)$, obtained by the implicit function theorem in some way.)
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>
>
| https://mathoverflow.net/users/164084 | How small need a perturbation be to not change the diffeomorphism type of a variety? | Let me prove $(1)$.
First of all, I guess that $f, \, g$ are homogeneous polynomials *of the same degree* $d$, otherwise $Z(f+ \varepsilon g)$ is not well-defined as a subvariety of $\mathbb{RP}^k$.
That said, note that the locus $S\_d$ of singular hypersurfaces of degree $d$ in $\mathbb{RP}^k$ is closed in the locus of all hypersurfaces, because it is given by a finite number of polynomial equations (obtained imposing the vanishing of the $k+1$ partial derivatives).
Since, by assumption, $Z(f+\varepsilon g)$ is smooth for $\varepsilon =0$ and $S\_d$ is closed, it follows that $Z(f + \varepsilon g)$ is smooth for $|\varepsilon|<t$, where $t$ is sufficiently small. A similar continuity argument shows that $Z(f + \varepsilon g)$ has still codimension $1$ for $t$ sufficiently small (and I guess that this
"sufficiently small" can be made explicit in terms of grad$(f)$, also answering $(2)$).
Now, set $$\mathscr{X}=\{(x, \, t) \, | \, f(x)+tg(x)=0 \} \subset \mathbb{RP}^k \times (-t, \, t).$$
Then, projecting over the factor $(-t, \,t)$ we have a surjective smooth submersion $F \colon \mathscr{X} \to (-t, \, t)$, whose (compact) fibre $X\_{\varepsilon}=F^{-1}(\varepsilon)$ is precisely $Z(f + \varepsilon g)$.
By [Ehresmann Lemma](https://en.wikipedia.org/wiki/Ehresmann%27s_lemma), it follows that $F$ is a locally trivial fibration and so all its fibres are diffeomorphic.
| 6 | https://mathoverflow.net/users/7460 | 448490 | 180,565 |
https://mathoverflow.net/questions/448485 | 6 | Let $X$ and $Y$ be standard Borel measurable spaces. A **Markov kernel** $f : X \rightsquigarrow Y$ is a map $f(-|-) : \Sigma\_Y \times X \to [0,1]$ such that:
* $f(-|x)$ is a probability measure on $Y$ for every $x \in X$,
* $f(S|-)$ is a measurable function $X \to [0,1]$ for every $S \in \Sigma\_Y$.
The **set of atoms** of a Markov kernel $f$ is
$$A := \{y \in Y \mid \exists x \in X : \: f(\{y\}|x) > 0 \}.$$
>
> Question: Is $A$ necessarily a Borel subset of $Y$? If not, is it at least universally measurable?
>
>
>
Clearly if $X$ or $Y$ is countable, then $A$ is indeed Borel. The non-obvious case is $X \cong Y \cong \mathbb{R}$.
For a concrete example, let $f : X \rightsquigarrow X$ be the identity kernel defined by $f(S|x) = \delta\_x(S)$. Then the set of atoms is clearly $Y$ itself, hence trivially Borel.
| https://mathoverflow.net/users/27013 | Atoms for Markov kernels | Here probably a partial answer.
I claim that the set $A$ is an analytic subset of $Y$ and thus universally measurable.
If we assume that $X$ and $Y$ are uncountable standard Borel spaces, then wlog we can assume that $X=Y=[0,1]$.
There we can then consider the cumulative distribution functions:
\begin{align}
F(y|x) &:= f([0,y]|x), & G(y|x) &:= f([0,y)|x).
\end{align}
One can show by making use of the their one-sided continuity in the first argument that both $F$ and $G$ are jointly measurable maps w.r.t. the product $\sigma$-algebra:
\begin{align}
F,G: Y \times X &\to [0,1].
\end{align}
This shows that also the map:
\begin{align}
Y \times X &\to [0,1], & (y,x) & \mapsto f(\{y\}|x)=F(y|x) - G(y|x),
\end{align}
is measurable w.r.t. the product $\sigma$-algebra.
This shows that the following set:
\begin{align}
C&:= \{ (y,x) \in Y \times X \,|\, f(\{y\}|x) >0 \}
\end{align}
is a Borel set of $Y \times X$.
Since both $X$ and $Y$ are standard Borel spaces we see that the set:
\begin{align}
A &= \mathrm{pr}\_Y(C),
\end{align}
is an analytic subset of $Y$ and thus universally measurable.
| 5 | https://mathoverflow.net/users/506586 | 448496 | 180,569 |
https://mathoverflow.net/questions/448494 | 6 | Let $R$ be a ring , $n \gt 1$, such that for all $x \in R$: $x^n-x \in Z(R)$, the center of $R$. Does it follow that $R$ is commutative?
For $n=2,3$ this is pretty straightforward to prove. But what about higher powers? This is a generalization of Nathan Jacobson's theorem that if $x^n=x$ for all $x \in R$, then $R$ is commutative.
| https://mathoverflow.net/users/11629 | Ring in which $x^n-x$ is central for every $x$ | Herstein, "A generalization of a theorem of Jacobson" Amer. J. Math. 73 (1951), 756–762 proves this. Also part III, Amer. J. Math. 75 (1953), 105–111 proves something a bit more general. Namely, the hypothesis can be weakened to $\forall x\ \exists\,n>1\ \cdots$
| 10 | https://mathoverflow.net/users/460592 | 448498 | 180,570 |
https://mathoverflow.net/questions/448499 | 3 | Let $\mathrm{AlgTh}$ be the category of one-sorted algebraic theories (synonym: Lawvere theories; morphisms are functors that are identical on objects and strictly preserve products). It is known that it is locally representable, hence it is bicomplete.
I wonder how the limits and colimits are described in it.
I know that (see e.g. [The Category Theoretic Understanding of
Universal Algebra: Lawvere Theories and
Monads](https://www.dpmms.cam.ac.uk/%7Emartin/Research/Publications/2007/hp07.pdf))
1. The initial object is an empty theory (algebras over it are sets)
2. The terminal object is a trivial theory (the category of algebras is $1$)
3. The coproduct of two theories is a disjoint union of the operations of theories with the preservation of internal axioms, without adding new ones
How are products, equalizers, and coequalizers described?
| https://mathoverflow.net/users/148161 | Limits and colimits in the category of algebraic theories | Typically, limits of multisorted algebraic theories (by which I mean a pair of a set $S$ and an $S$-sorted algebraic theory $\mathbb F(S) \to L$) are most easily described in terms of their presentation as categories with finite products (i.e. a [cartesian (monoidal) category](https://ncatlab.org/nlab/show/cartesian+monoidal+category)), whereas colimits are most easily described in terms of their presentation as equational presentations, i.e. by sets of operations and equations.
* The limit of a diagram of multisorted algebraic theories is given by the limit of the underlying cartesian categories. This means, for instance, that the product of an $S$-sorted algebraic theory with an $S'$-sorted algebraic theory will be an $(S \times S')$-sorted algebraic theory. When $S = S' = 1$, this recovers the product of one-sorted algebraic theories, since $1 \times 1 \cong 1$. Limits of algebraic theories typically do not have nice descriptions in terms of the operations and equations (e.g. the product of two finitely presented algebraic theories will not usually be finitely presented).
* Take a morphism of multisorted algebraic theories, from an $S$-sorted algebraic theory $\mathbb F(S) \to L$ to an $S'$-sorted algebraic theory $\mathbb F(S') \to L'$, to comprise a pair $(s, f)$ of a function $s : S \to S'$ and a functor between the codomains $f : L \to L'$ forming an evident commutative square\*. Syntactically $f : L \to L'$ maps $S$-sorted terms of $L$ to $S'$-sorted terms of $L'$. The coequaliser of two such morphisms is given by the $S'$-sorted algebraic theory $L''$ obtained by imposing equations on $L'$ that equate any two terms whose preimage under $f$ is equal. When $S = S' = 1$, this recovers the coequaliser of one-sorted algebraic theories.
(\*More generally, we could take $s$ to be a function $S \to |\mathbb F(S')|$, but this definition is a little less intuitive syntactically.)
| 4 | https://mathoverflow.net/users/152679 | 448503 | 180,571 |
https://mathoverflow.net/questions/448491 | 0 | We consider the heat kernel
$$
g :\mathbb R\_{>0} \times \mathbb R^d \to \mathbb R,\quad (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ).
$$
Then
$$
\partial\_t g(t, x) = \Delta g(t, x) = \left(\frac{|x|^2-2 d t}{4 t^2}\right) g(t, x).
$$
*Corollary 1.3* and *Theorem 1.2* in the paper [Upper Bounds of Derivatives of the Heat Kernel on an Arbitrary Complete Manifold](https://www.sciencedirect.com/science/article/pii/S0022123685710166) by Alexander Grigor'yan imply there is a constant $C$ such that
$$
|\partial\_t g| (t, x) \le C\frac{g(2t, x)}{t}
\quad \forall t>0, \forall x \in \mathbb R^d.
$$
This upper bound is not good enough for my purpose because $\int\_0^t \frac{\mathrm d s}{s} = +\infty$ for any $t>0$. I would like to ask if the following improvement is possible, i.e.,
>
> There exist a constant $C \ge 1$ and a measurable function $f:(0, \infty) \to \mathbb R$ such that
>
>
> * $\int\_0^t f(s) \, \mathrm d s < +\infty$ for all $t>0$.
> * $|\partial\_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$.
>
>
>
Any reference is appreciated. Thank you so much for your help!
| https://mathoverflow.net/users/477203 | Is there $f$ such that $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$? | Such a function $f$ does not exist.
Indeed, if the inequality
$|\partial\_t g|(t,x)\le f(t)g(Ct,x)$ holds for all $t>0$ and $x\in\mathbb R^d$, then it holds for $x=0$, so that
$$f(t)\ge f\_\*(t):=\frac{|\partial\_t g|(t,0)}{g(Ct,0)}
=C^{d/2}\frac{d }{2 t}$$
for all real $t>0$.
So, $\int\_0^t f\ge\int\_0^t f\_\*=\infty$ for all $t>0$.
(Instead of choosing $x=0$, here we can more generally let $|x|=O(\sqrt t)$.)
| 4 | https://mathoverflow.net/users/36721 | 448522 | 180,578 |
https://mathoverflow.net/questions/448385 | 2 | $\DeclareMathOperator\Inv{Inv}$Baur-Monk quantifier elimination implies that a sentence in the language of modules is a combination of BG invariant statements.
A BG invariant sentence is a boolean combination of $ \Inv(A,\phi, \psi) \* k$ where $\*$ is one of $=,>,<$ where $\phi$ and $\psi$ are positive primitive formulae.
We define $\Inv(A,\phi, \psi)$ as the cardinality of the quotient group $\phi(A) / \psi(A)$ and $\Inv(A,\phi, \psi) > k$ is equivalent modulo the theory of modules $T$ to the following first-order statement:
\begin{equation\*}
\forall \overline{v\_1},...,\overline{v\_k} \exists \overline{v} (\phi(\overline{v}) \wedge {\bigwedge\_{i=1}^{k}} \neg \psi (\overline{v}-\overline{v\_i})).
\end{equation\*}
It is then concluded from quantifier elimination that $\Inv(A,\phi, \psi)=\Inv(B,\phi, \psi)$ iff $A
\equiv B$ (elementarily) where $\phi$ and $\psi$ are positive primitive in *one* free variable.
My question is why it follows that we can restrict to invariants with formulas in one free variable? (Instead of just invariants). I assume that it is because the positive primitive formulas are conjunction closed. Is this correct?
| https://mathoverflow.net/users/506361 | Baur-Monk quantifier elimination (BG-invariants in 1-free variable) | This doesn't follow directly from the statement of QE (and I don't think it has anything to do with the fact that pp formulas are conjunction-closed). To understand it, you really have to look at the proof of QE. For example, see the proof of Theorem 1.1 on p. 155 of Ziegler's paper *[Model Theory of Modules](https://doi.org/10.1016/0168-0072(84)90014-9)*.
The proof of QE works one quantifier at a time. For a fixed module $M$, you show that if $\varphi(x,\overline{y})$ is equivalent (in $M$) to a Boolean combination of pp formulas, then so is $\forall x\, \varphi(x,\overline{y})$. Here $x$ is a single variable. By induction, it follows that every formula is equivalent (in $M$) to a Boolean combination of pp formulas.
But now, examining the proof, you observe that the proof almost works uniformly in $M$: when you show that $\forall x\, \varphi(x,\overline{y})$ is equivalent to the boolean combination of pp formulas $\theta(\overline{y})$, the formula $\theta$ only depends on the values of finitely many $\mathrm{Inv}(M,\psi,\chi)$ where $\psi$ and $\chi$ are pp formulas in the single variable $x$. So if you know all the values of the BG invariants $\mathrm{Inv}(M,\psi,\chi)$ when $\psi$ and $\chi$ are pp formulas in a single variable, then you know which quantifier-free formula every formula is equivalent to. In particular, these values determine whether every sentence is true or false in $M$.
| 3 | https://mathoverflow.net/users/2126 | 448528 | 180,581 |
https://mathoverflow.net/questions/448536 | 0 | In a closed model category, is the identity $\textrm{id}: A \to A$ a cofibration? Does it only hold on some special cases? Or is it never true?
| https://mathoverflow.net/users/492610 | Is the identity a cofibration? | Yes: it is always a cofibration. Fix a trivial fibration $f : X \to Y$. It suffices to show that $\hom(A,X) \to \hom(A,Y) \times\_{\hom(A,Y)} \hom(A,X)$ is surjective (informally: $\mathsf{id}$ lifts against $f$). Inspecting the codomain of this map, we see that it's isomorphic to $\hom(A,X)$ so the map is not only surjective, it's an isomorphism.
The same argument shows that any isomorphism is a cofibration and even---as weak equivalences contain all isomorphisms---a trivial cofibration. This argument also does not require the model category to be closed or satisfy any additional properties. One consequence is that this fact then dualizes to show that isomorphisms are also (trivial) fibrations.
| 5 | https://mathoverflow.net/users/76636 | 448537 | 180,583 |
https://mathoverflow.net/questions/448398 | 5 | The Giry monad $G : \textbf{Meas} \to \textbf{Meas}$ maps a measurable space $(X, \mathcal{F})$ to its set of probability measures. The $\sigma$-algebra of $G(X, \mathcal{F})$ is the smallest algebra such that the evaluation map $p \mapsto p(E)$ is measurable for all $E \in \mathcal{F}$. I am aware that this condition is needed to define the integral in $\mu\_X : \mathcal{G}^2X \to \mathcal{G}X$, but what is the intuitive meaning of the $\sigma$-algebra? Is there an intuitive example of a subset $A \subseteq \mathcal{G}(X, \mathcal{F})$ that is **not** in the $\sigma$-algebra?
| https://mathoverflow.net/users/477321 | Intuitive meaning of Giry monad's $\sigma$-algebra | (Disclaimer: I’m not at all seriously experienced with the Giry monad; I’ve not read much beyond the original Giry paper and what I’ve picked up on the street.)
Here are two kinds of sets which typically aren’t in the Giry $\sigma$-algebra $G(X)$:
* $\{ p \mid p(E) \in A \}$, for some measurable $E \subseteq X$ and *non-measurable* $A \subseteq [0,1]$. (For specific spaces $X$, it’s not hard to check this is non-measurable; I guess that should be true in fairly wide generality, but I don’t quickly see how to prove that.)
* $\{ p \mid p(E\_i) \in A\_i,\, \text{for all}\ i \in I \}$, for some arbitrary *uncountable family* of measurable sets $E\_i \subseteq X$, $A\_i \subseteq [0,1]$. For instance, take $X$ to be an uncountable set with the $\sigma$-algebra of just its countable and co-countable subsets. Then it’s not hard to check that for every measurable set $S$ of $G(X)$, the set $\{ p(\{x\}) \mid p \in S \}$ is an interval $[0,a]$ for all but countably many $x \in X$.
My overall intuition is that **very crudely, the Giry $\sigma$-algebra is a bit like the (infinite-indexed) product topology:** on the one hand there’s a “componentwise” constraint, as exemplified by the former counterexample above; on the other hand theres an “across-components” constraint, as seen with the latter.
| 3 | https://mathoverflow.net/users/2273 | 448553 | 180,588 |
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