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https://mathoverflow.net/questions/416497 | 6 | I was recently trying to find a numerical solution to a thermodynamics problem and the expression $x\ln x$ appeared in one of the computations. I did not have to find its value very near $0$, so the computer managed fine, but it got me thinking - can one make a stable numerical algorithm to compute $x\ln x$ for values near 0?
It is easy to prove that $\lim\limits\_{x\to 0^+} x\ln x=0$. However, simply multiplying $x$ by $\ln x$ is not a good solution for small $x$. The problem seems to be that we are multiplying a small number ($x$) by a large number ($\ln x$).
My first thought would be to approximate it somehow. But I quickly saw that Taylor series wouldn't work, as the derivative is $\ln x + 1$, which blows up (or rather down :-)) to $-\infty$. Some kind of iterative method like Newton's method does not seem to be the solution either, because the operations needed seem to be even more messy than what we are trying to compute.
So my question is - is there some numerically stable method to compute $x\ln x$ for small values of $x$? And preferably one that is more general, so that it could be used on functions like $x^n \ln x$ - but these at least have a finite first derivative at $0$ for $n>1$.
| https://mathoverflow.net/users/114143 | How to numerically compute $x \ln x$ and related functions near $0$? | Modern mathematics libraries should be able to find $\log x$ precisely for all floating-point numbers, as the algorithms for doing that have long been known and adopted. My experiments on a fairly recent Intel chip with gnu mathematics library and gcc 10 compiler confirm that.
Multiplication is even more definite: correct rounding of the last bit is guaranteed (though there can be different options available for what "correct rounding" means).
It might appear from the above that precise computation of $x\log x$ is guaranteed for any small $x>0$. However there is a reason why that doesn't happen for really tiny $x$ and there is no way to avoid it.
Floating-point numbers are usually stored with the mantissa normalized (leading bit 1, sometimes implicit). However, when the number is too small it may be impossible to normalize the number without the exponent underflowing. In this case (usually) the number is kept in unnormalized form and the number of bits of precision is reduced. This situation is called *partial underflow* and such numbers are *subnormal* or *denormalized*.
So, if you try to compute $x\log x$ when $x\log x$ is in the partial underflow range, $\log x$ will be computed precisely but the multiplication by $x$ cannot produce more bits of precision than numbers of the size of $x\log x$ have. Short of using different floating-point numbers, there is no solution.
If $x\log x$ is in the partial underflow range, then $x$ will be too, or maybe it will be so small as to underflow to 0. In practice $x$ will come from some earlier computation and the partial underflow means it may not be so precise as thought, which is another source of error. It isn't the fault of the function $x\log x$.
Incidentally I tested this explanation empirically and behaviour was exactly as predicted.
| 8 | https://mathoverflow.net/users/9025 | 416518 | 169,729 |
https://mathoverflow.net/questions/416464 | 2 | In all introductory texts I'm aware of, convex polytopes are dealt with strictly within $\mathbb R^n$. Being used to quite larger levels of generality elsewhere, I'm wondering if the same can be done here. Specifically, I'm looking for a generalization of convex polytopes in other spaces such that their face lattices still satisfy most of the properties they do in $\mathbb R^n$, like being graded lattices and satisfying the diamond property.
An obvious setting in which convex polytopes work are topological vector spaces of finite dimension. Unfortunately all of these are isomorphic to $\mathbb R^n$, so nothing new.
Interestingly, it seems like most if not all of the basic definitions of the theory can be made in terms only of a betweenness relation. Convex sets/hulls, affine subspaces/hulls, and closed or open half-spaces can be defined without even a single axiom attached to this relation. Of course, very little can actually be proved about these notions, and I don't know what axioms would be needed to get "expected behavior".
I believe hyperbolic space might satisfy my constraints. However I see no obvious way to generalize this example.
| https://mathoverflow.net/users/147705 | Convex polytopes in other spaces | Some time ago I was having the same question and I found these two:
* Robert Morelli *"[A theory of polyhedra](https://doi.org/10.1006/aima.1993.1001)"* (Adv. Math. 97, No. 1, 1-73 (1993)). He works in modules over ordered rings (and vector spaces over ordered fields). This was very helpful personally to understand valuations on polyhedral algebras in a more general context. zbMATH review [here](https://zbmath.org/?q=an%3A0779.52016).
* Walter Prenowitz and James Jantosciak *"[Join geometries. A theory of convex sets and linear geometry](https://doi.org/10.1007/978-1-4613-9438-9)"* (Springer UTM). zbMATH review [here](https://zbmath.org/?q=an%3A0421.52001). I haven't looked at this book in detail, but seems interesting (and perhaps related to your "betweenness relation"). The second paragraph from its introduction says:
>
> The familiar figures of classical geometry $-$points, segments, lines,
> planes, triangles, circles, and so on$-$ stem from problems in the physical
> world and seem to be conceptually unrelated. However, a natural setting
> for their study is provided by the concept of *convex set,* which is comparatively
> new in the history of geometrical ideas. The familiar figures can then
> appear as convex sets, boundaries of convex sets, or finite unions of
> convex sets. Moreover, two basic types of figure in linear geometry are
> special cases of convex set: *linear space* (point, line, and plane) and
> *halfspace* (ray, halfplane, and halfspace). Therefore we choose convex set
> to be the central type of figure in our treatment of geometry. How can the
> wealth of geometric knowledge be organized around this idea? By definition,
> a set is convex if it contains the segment joining each pair of its
> points; that is, if it is closed under the operation of joining two points to
> form a segment. But this is precisely the basic operation in Euclid. Our
> point of departure is to take the operation of joining two points to form a
> segment as fundamental, and to throw the burden of unifying the material
> on the consistent and relentless exploitation of this operation.
>
>
>
| 1 | https://mathoverflow.net/users/109085 | 416519 | 169,730 |
https://mathoverflow.net/questions/416517 | 2 | Let $M\_n$ denote the set of all $n\times n$ complex matrices. Let $1\leq p<\infty.$ For $A\in M\_n$ define $\|A\|\_p:=(Tr(A^\*A)^{p/2})^{1/p}$ where $Tr$ denotes the usual trace of a matrix. Then $\|.\|\_p$ is indeed a norm. We write $S\_p^n$ to be the Banach space whose vector space is $M\_n$ and norm is given by $\|.\|\_p.$ Define the Banach-Mazur distance as $d(S\_p^n,S\_q^n):=\inf\{\|T\|\_{S\_p^n\to S\_q^n}\|T^{-1}\|\_{S\_q^n\to S\_p^n}:T:S\_p^n\to S\_q^n\ \text{is an invertible linear map}\}.$ Can anyone tell me how does the constant $d(S\_p^n,S\_q^n)$ behaves as $n\to\infty$?
| https://mathoverflow.net/users/136860 | Banach-Mazur distance between Schatten-$p$ classes | Same order as $d(\ell\_p^n, \ell\_q^n)$. It is $n^{1/p-1/q}$ when they are on the same side of 2, and $n^{(1/2-1/q)\vee (1/p-1/2)}$ for $1\le p<2<q\le\infty$.
See Section 45 of [Tomczak-Jaegermann's book Banach Mazur distances and Finite dimensional Operator ideals](https://sites.ualberta.ca/%7Entj/bm_book/index.html) which she made freely available at the link.
| 4 | https://mathoverflow.net/users/3675 | 416525 | 169,731 |
https://mathoverflow.net/questions/416540 | 1 | Let $V = (v\_1,...,v\_n)$ be a set of vertices for a directed graph. We denote an edge between the two vertices $v\_i$ and $v\_j$ by $(i,j)$. The edge set $E$ for a graph $(V,E)$ is then a set of tuples $E \in P(\{(i,j) \mid i,j \in \{1,...,n\}\})$.
My Question is if for fixed $n \in \mathbb{N}$ we can efficiently define a map
$$
F\_n : P(\{(i,j) \mid i,j \in \{1,...,n\}\}) \rightarrow \mathbb{R}
$$
such that $F\_n(E)=F\_n(E')$ if and only if the two Graphs are the same modulo permutation of vertices, i.e. if there exists a permutation $\sigma \in S\_n$ such that $E = \{(\sigma(i),\sigma(j)) \mid (i,j) \in E'\}$.
Such a map should be relatively simple to construct with sums over all permutations. I am interested in simpler known invariants, which would ideally require less than $\mathcal{O}(n!)$ amount of work to calculate.
I'm new to graph-theory and thought someone must have already thought about this, since it is fundamental to classifying graphs. Any references or pointers in the right direction are much appreciated.
| https://mathoverflow.net/users/409412 | Is there a permutation invariant for graphs? | You are essentially asking about the graph isomorphism problem. Or, more precisely, you are asking about canonical representation of graphs. There is a huge literature about this. See <https://en.wikipedia.org/wiki/Graph_canonization>.
| 4 | https://mathoverflow.net/users/20598 | 416541 | 169,734 |
https://mathoverflow.net/questions/416547 | 2 | Does there exist any C\*-algebra $(A,\|\cdot\|)$ enjoying the following property?
$\bullet$ There exists a norm $|\cdot|$ on $A$ with $\|\cdot\|\leq|\cdot|$ such that $(A,|\cdot|)$ is a pre C\*-algebra (necessarily non-complete).
| https://mathoverflow.net/users/84390 | A C*-algebra enjoying some different C*-norms | No that's not possible (except the trivial case). Any $\*$-homomorphism between $C^\*$-algebras is automatically contractive, and if it is injective then it is isometric. You can apply this to the identity map of $A$ seens as a map from $(A,\Vert \cdot \Vert)$ to the completion of $(A,\vert \cdot \vert)$ and conclude that two norm are equal.
More concretely it follows from the fact that (for self adjoint elements) the norm in a $C^\*$-algebra coincide with the spectral radius, which force relations between the two norm. Here it forces $\vert \cdot \vert \leqslant \Vert \cdot \Vert$ hence the equality of the two norms.
| 10 | https://mathoverflow.net/users/22131 | 416550 | 169,736 |
https://mathoverflow.net/questions/416548 | 3 | Let $ f \colon E \to B $ be a morphism between (weak) homotopy types such that each (homotopy) fiber has a homotopy type of some finite CW complex.
Then does $ \Sigma f \colon \Sigma E \to \Sigma B $ have the same property?
| https://mathoverflow.net/users/119076 | Is the suspension of a finite fibration again finite? | Assuming you work with unpointed spaces (but the example can easily be adapted to the pointed case) the map $1 \to 2$ gives a counterexample : its fiber are $1$ and $\varnothing$ so they are both finite, but on applying the suspension you get $1 \to S^1$ whose homotopy fibers identitifes with $\mathbb{Z}$.
| 8 | https://mathoverflow.net/users/22131 | 416551 | 169,737 |
https://mathoverflow.net/questions/414575 | 1 | Intro
-----
I'm referring to the original paper [Fast unfolding of communities in large
networks](https://perso.uclouvain.be/vincent.blondel/publications/08BG.pdf) by Blondel et al. in this question and adopt their notation. Explanation for the used symbols is on page 4 below equation (2).
I'm wondering why Blondel et al. are dropping the coefficient $\frac{1}{m}$. First, let's simplify their equation (2) to see what I mean:
$$ \Delta Q= \Biggl[ \frac{\Sigma\_{in}+2k\_{i,in}}{2m} - \Bigl( \frac{\Sigma\_{tot}+k\_i}{2m} \Bigr)^2 \Biggr] - \Biggl[ \frac{\Sigma\_{in}}{2m} - \Bigl( \frac{\Sigma\_{tot}}{2m} \Bigr)^2 - \Bigl( \frac{k\_i}{2m} \Bigr)^2 \Biggr] $$
$$ \Delta Q = \frac{\Sigma\_{in}}{2m} + \frac{2k\_{i,in}}{2m} - \Bigl( \frac{\Sigma\_{tot}}{2m} + \frac{k\_i}{2m}\Bigr)^2 - \frac{\Sigma\_{in}}{2m} + \Bigl( \frac{\Sigma\_{tot}}{2m} \Bigr)^2 + \Bigl( \frac{k\_i}{2m} \Bigr)^2 $$
$$ \Delta Q = \frac{2k\_{i,in}}{2m} - \frac{\Sigma\_{tot}\cdot 2k\_i}{(2m)^2} $$
However, this is not the version they used in their code. On their [official code page](https://sites.google.com/site/findcommunities/), they've linked to [sourceforge](https://sourceforge.net/projects/louvain/) to download the C++ code. In the file `Modularity.h`, you'll find the following code:
```
inline long double
Modularity::gain(int node, int comm, long double dnc, long double degc) {
assert(node>=0 && node<size);
long double totc = tot[comm];
long double m2 = g.total_weight;
return (dnc - totc*degc/m2);
}
```
`dnc` is passed over in line 257 in `louvain.cpp` and calculated in the method `void Louvain::neigh_comm(int node)`. If you follow the calculations, you will see that this is actually our $k\_{i,in}$ in the formula. I'm now wondering where this line in the method `Modularity::gain(...)` comes from:
```
return (dnc - totc*degc/m2);
```
My guess
--------
Let's further "simplify" our formula to a point where it resembles the code used:
$$\Delta Q = \frac{k\_{i,in}}{m} - \frac{\Sigma\_{tot}\cdot 2k\_i \cdot \frac{1}{2m}}{(2m)^2 \cdot \frac{1}{2m}}$$
$$ \Delta Q = \frac{k\_{i,in}}{m} - \frac{\Sigma\_{tot}\cdot \frac{k\_i}{m}}{2m}$$
This is very similar to the mentioned codeline if we let `dnc` denote $\frac{k\_{i,in}}{m}$ and `degc` $\frac{k\_i}{m}$.
However, in the method where they calculate `dnc` and `degc` (`void Louvain::neigh_comm(int comm)`), Blondel et al. just use the numerator, dropping the coefficient $\frac{1}{m}$ here:
$$ \Delta Q = \frac{1}{m} \cdot \Bigl( k\_{i,in} - \frac{\Sigma\_{tot}\cdot k\_i}{2m} \Bigr) $$
Of course, this could be done to save a division. However, they do not get the correct absolute value for $\Delta Q$ then. Maybe that's why they also have to recalculate the quality again multiple times in every pass in line 272 in `Louvain.cpp`:
```
new_qual = qual->quality();
```
and can't simply do something like $Q\_{new}= Q\_{old} + \Delta Q$, which would save a lot of multiplications and divisions.
| https://mathoverflow.net/users/475789 | Louvain method: Why do they drop coefficient 1/m in the official implementation? | The function is used when they decide whether the node should stay in the community or not. Therefore, they only need to compare quality value differences and have no need for absolute numbers. The division factor is constant during the computation and therefore can be omitted since it does not affect the result of the comparison:
$$
k \cdot \Delta Q\_1 \lt k \cdot \Delta Q\_2 \iff \Delta Q\_1 \lt \Delta Q\_2, k \in \Bbb R^+
$$
That way, they safe a division in the inner loops which is called for every edge in the graph. Divisions in general are computationally quite expensive even on modern processors. I do not have actual measurements, but if you put both version in [llvm-mca](https://releases.llvm.org/13.0.0/docs/CommandGuide/llvm-mca.html), the block throughput is twice as fast without the extra division.
As you mentioned, they recalculate the overall quality value for the graph in each pass. However, this loop is called for each node whose count is often much less than the number of edges.
Another reason might be accuracy. In the original formula, they divide by $4m^2$, which might become too large to be stored as floating point value and the denominator becomes infinite. If you divide a finite number by an infinite value, the final result becomes zero. In their quality gain formula, they used $2m$ as denominator and the risk of overflow is reduced significantly.
| 2 | https://mathoverflow.net/users/477406 | 416566 | 169,744 |
https://mathoverflow.net/questions/416565 | 3 | ### Situation
Suppose that we have
* a commutative ring (or an $E\_{\infty}$-ring) $R$ and
* a homotopy type $X$.
Then we get a canonical morphism
$$
f \colon K(R ^ {\Sigma ^ \infty X\_+}) \to K(R) ^ {\Sigma ^ \infty X\_+}
$$
where $K$ denotes the **nonconnective** algebraic $K$-theory.
### Question
Is $f$ an equivalence for example when $X$ is finite?
| https://mathoverflow.net/users/119076 | (Nonconnective algebraic) K-theory cohomology = K-theory of cohomology? | The answer is typically going to be no.
Suppose for example that $R$ is discrete. Then $R^{\Sigma^\infty\_+X}= C^\*(X;R)$ is the usual cochain algebra, and perfect modules over that algebra are equivalent (via the homotopy fixed points functor) to the thick full stable subcategory of $Fun(X, Perf(R))$ generated by $R$ as a trivial local coefficient system.
Let's further assume that $R$ is regular noetherian so that $Perf(R)$ has the (bounded) Postnikov $t$-structure, and $Fun(X,Perf(R))$ has the pointwise $t$-structure. This pointwise $t$-structure is inherited by the full subcategory described above. Indeed, I claim that under these assumptions, a local system $L$ is in this thick subcategory if and only if each homotopy group $\pi\_kL$, as a $\pi\_1(X)$-module, has a filtration whose associated graded's are trivial $\pi\_1(X)$-modules - this condition is obviously preserved under truncations.
It follows that the heart of this $t$-structure on this full subcategory is exactly finitely generated $R$-modules, so by the theorem of the heart, $K(C^\*(X;R)) = K(R)$, where of course $K(R)^{\Sigma^\infty\_+X}$ is usually not $K(R)$ (for instance if $X = S^n$, then $\pi\_0$ of the former has a summand isomorphic to $\pi\_n$ of the latter)
| 4 | https://mathoverflow.net/users/102343 | 416573 | 169,746 |
https://mathoverflow.net/questions/416561 | 3 | This question was originally at Math Stackexchange, but had no answers:
<https://math.stackexchange.com/questions/4348707/is-projection-of-locally-connected-compact-subset-locally-connected>
### Problem
Let $(X, \mathcal{T}\_X)$ and $(Y, \mathcal{T}\_Y)$ be topological spaces, $Z = X \times Y$, $\mathcal{T}\_Z$ be the product topology on $Z$, $f : Z \to X$ be defined by $f(x, y) = x$, and $C \subset Z$ be compact and locally connected.
Is $f[C]$ locally connected?
### Background
A space $(Z, \mathcal{T}\_Z)$, where $\mathcal{T}\_Z$ is a topology on $Z$, is *locally connected*, if for each $z \in U \in \mathcal{T}\_Z$ there exists a connected $V \in \mathcal{T}\_Z$ such that $z \in V \subseteq U$.
A space is *compact* if every open cover has a finite subcover.
### Locally connected subset whose image is not locally connected
The following shows that some restrictions are necessary for the subset $C$. Let $X = Y = \mathbb{R}$, and $Z' = \{(0, 1)\} \cup \{(1/n, 0) : n \in \mathbb{N}^{> 0}\}$. Then $Z'$ is locally connected, but not compact, and $f[Z'] = \{0\} \cup \{1/n : n \in \mathbb{N}^{> 0}\}$ is not locally connected.
### Holds when $f\restriction C$ is a quotient map
Suppose $f\restriction C$ is a quotient map. [Quotient maps preserve local connectedness](https://math.stackexchange.com/questions/2426794/a-quotient-of-a-locally-connected-space-is-locally-connected). Therefore $f[C]$ is locally connected.
[This question](https://math.stackexchange.com/questions/4350112/is-restriction-of-a-projection-to-a-compact-subset-a-quotient-map) provides conditions for $f\restriction C$ being a quotient map. However, as shown there, $f\restriction C$ is not always a quotient map.
Non-quotient strategy
---------------------
There [exist maps](https://math.stackexchange.com/questions/259001/a-continuous-surjection-that-is-not-a-quotient-map) which are continuous, surjective, and preserve local connectedness, but are not quotient; in the linked example $X$ and $Y$ are both locally connected. If the claim does hold, then a general solution to this problem may need a stronger theorem for preservation of locally connectivity which includes these maps.
| https://mathoverflow.net/users/32487 | Is projection of locally connected compact subset locally connected? | A counterexample to this question can be constructed as follows.
Let $X=[0,1]$ be the closed interval with the standard Euclidean topology.
Let $Y=\omega$ and $\mathcal T\_Y$ be the topology on $Y$ consisting of the sets $W\subseteq Y$ satisfying two conditions:
$\bullet$ if $0\in W$, then $W=\omega$;
$\bullet$ if $1\in W$, then $\omega\setminus W$ is finite.
The definition of the topology $\mathcal T\_Y$ implies that $\omega\setminus\{0,1\}$ is an open discrete subspace of $(Y,\mathcal T\_Y)$.
It is easy to see that the space $(Y,\mathcal T\_Y)$ is not locally connected at $1$.
Choose any sequence $(U\_n)\_{n=1}^\infty$ of pairwise disjoint nonempty open sets in $X=[0,1]$ and let $g:X\to Y$ be the function defined by
$$g(x)=\begin{cases}n&\mbox{if $x\in U\_n$ for some $n\ge 1$};\\
0&\mbox{otherwise}.
\end{cases}
$$
The definition of the topology $\mathcal T\_Y$ ensures that the function $g:X\to Y$ is continuous.
Let $C=\{(x,g(x)):x\in X\}\subseteq X\times Y$ be the graph of the function $g$. It is clear that $C$ is homeomorphic to $[0,1]$ and hence is compact, connected and locally connected. But the projection of $C$ onto $Y$ is not locally connected.
| 3 | https://mathoverflow.net/users/61536 | 416575 | 169,747 |
https://mathoverflow.net/questions/416574 | 0 | I am looking for the best way to approximate $\exp(x)$ on a finite domain $[0,M]$ with a piecewise-linear function. My initial approach is to take $K$ evenly-spaced segments from $0$ to $M$. For each segment $[s\_k, s\_{k+1}]$, I add the line from point $(s\_k, \exp(s\_k))$ to point $(s\_{k+1}, \exp(s\_{k+1}))$.
But what I've noticed is that the approximation gets worse for segments closer to $M$. Intuitively, this suggests that the segments need to start large and then reduce in size as we approach $M$. But how would one approach this formally? I am also interested in heuristics, or approximations.
| https://mathoverflow.net/users/476491 | Approximating exp(x) with a piecewise-linear function accurately | For $b$ not much larger than $a$, the maximum difference between $e^x$ and the line from $(a,e^a)$ to $(b,e^b)$, for $x\in[a,b]$, is approximately $\frac 18 (b-a)^2 e^a$.
So you should choose your points so that $(b-a)^2 e^a$ is approximately constant for each interval. After selecting the first interval, use this rule to select further intervals. They will get gradually smaller.
| 2 | https://mathoverflow.net/users/9025 | 416582 | 169,750 |
https://mathoverflow.net/questions/354899 | 7 | For this question a manifold is a locally upper-Euclidean Hausdorff space; paracompactness or second-countability is not assumed, and boundary may be present.
Let $M$ be a manifold. What are sufficient conditions for there to exist open $U \subset M$ such that $\partial M \subset U$ and $U$ (strongly?) deformation retracts onto $\partial M$? (Could this be true for every manifold?)
Clearly $\partial M$ being collared in $M$ is a fairly general sufficient condition (this includes paracompact manifolds). However, I'd be interested in whether there is a more general sufficient condition than this.
| https://mathoverflow.net/users/32487 | When is a manifold boundary a deformation retract of its open neighborhood? | The following theorem and example give (partial) answers. By an *$n$-manifold* we understand a Hausdorff space whose any point has a neighborhood, homeomorphic to an open subspace of $\mathbb R^n\_+=\{(x\_1,\dots,x\_n)\in\mathbb R^n:x\_n\ge 0\}$.
**Theorem.** For each 1-manifold $M$ the boundary $\partial M$ is a deformation retract of some open set $U\subseteq M$ that contains $\partial M$.
*Proof.* By the definition of a 1-manifold, for every $x\in\partial M$ there exists a neighborhood $U\_x\subseteq M$, homeomorphic to $[0,1)$. Fix a homeomorphism $h\_x:U\_x\to[0,1)$ such that $h\_x(x)=0$. It can be shown that for distinct $x,y\in\partial M$ the sets $U\_x,U\_y$ are disjoint. Then $U=\bigcup\_{x\in \partial M}U\_x$ is an open neighborhood of $\partial M$ and the function $H:U\times [0,1]\to U$, $H:(u,t)=h\_x^{-1}(t\cdot h\_x(u))$ where $u\in U\_x$, is a deformation retraction of $U$ onto $\partial M$.
**Example.** There exists a separable 2-manifold $M$ whose boundary $\partial M$ is homeomorphic to $\mathbb R\times\mathfrak c$. Being non-separable, the boundary $\partial M$ cannot be a retract of a (necessarily separable) neighborhood $U$ of $\partial M$ in $M$.
This example is described by Peter Nyikos as Example 3.6 of his survey paper "[The theory of nonmetrizable manifolds](https://www.sciencedirect.com/science/article/pii/B9780444865809500173?via%3Dihub)" in the Handbook of Set-Theoretic Topology.
| 3 | https://mathoverflow.net/users/61536 | 416586 | 169,752 |
https://mathoverflow.net/questions/416474 | 0 | In the case of a square lattice, the exact number of points within a circle of radius r centered in the center is (see: [http://mathworld.wolfram.com/GausssCircleProblem.html](https://mathworld.wolfram.com/GausssCircleProblem.html):
$$N(r)=1+4Floor(r)+4 \sum\_{x=1}^{Floor(r)}{Floor(\sqrt{r^2−x^2)}}$$
And in the case of an hexagonal lattice, I found in this post [An exact counting solution for the number of points within a circle of radius $r$ centered on a lattice point in a $A\_2$ hexagonal lattice](https://mathoverflow.net/questions/110186/an-exact-counting-solution-for-the-number-of-points-within-a-circle-of-radius-r) that the number of points within a circle of radius r centered in the center is:
$$ N(r)= \sum\_{x = -Floor(\frac{r}{\sqrt{3}})}^{Floor(\frac{r}{\sqrt{3}})}( 1 + 2 Floor(\sqrt{r^2 - 3 x^2}) + \sum\_{x = -Floor(\frac{r}{\sqrt{3}} + \frac{1}{2}) + \frac{1}{2}}^{Floor(\frac{r}{\sqrt{3}} + \frac{1}{2}) - \frac{1}{2}}( 2 Floor(\sqrt{r^2 - 3 x^2} + \frac{1}{2}). $$
And I checked this expression with the values in <http://oeis.org/A053416> and I don't obtain the same values for an r.
Can you guide me where I am wrong?
In my research I want to obtain the number of the lattice points for square and hexagonal lattices in function of lattice constant and region size.
[This is the structure that I obtain numerrically](https://i.stack.imgur.com/tRtk5.png)
I am new to this subject and I appreciate all the suggestions.
| https://mathoverflow.net/users/97558 | The exact number of points within a circle of radius r centered on a lattice point in a hexagonal lattice? Review expression Gauss circle problem | By identifying the lattice points with numbers of the form $x - y\omega$, $\omega = e^{2\pi i / 3}$, $x, y \in \mathbb{Z}$, we find that we want to count Diophantine solutions to $x^2 + xy + y^2 \le r^2$. From $$\sum\_{m, n \in \mathbb{Z}} q^{m^2 + mn + n^2} = 1 + 6 \sum\_{n \ge 0} \left(\frac{q^{3n+1}}{1 - q^{3n+1}} - \frac{q^{3n+2}}{1-q^{3n+2}}\right)$$ (see e.g. Michael D. Hirschhorn (1999) [Three Classical Results on Representations of a Number](http://www.mat.univie.ac.at/%7Eslc/wpapers/s42hirsch.pdf), Séminaire Lotharingien de Combinatoire, B42f) we get $$N(r) = 1 + 6 \sum\_{k=0}^\infty \left\lfloor \frac{r^2}{3k+1} \right\rfloor - \left\lfloor \frac{r^2}{3k+2} \right\rfloor$$
| 5 | https://mathoverflow.net/users/46140 | 416587 | 169,753 |
https://mathoverflow.net/questions/416585 | 2 | Consider the circle $\mathbb{T}^1= \frac{\mathbb{R}}{\mathbb{Z}}$. We represent it as a union of disjoint subsegments $M\_j=[t\_j,t\_{j+1})$, $j = 0, \cdots, n$, $t\_0=t\_n$ and define the map $S$ by the formula
\begin{align}
S(t)=t+c\_j \quad \text{mod 1}, \quad t\in M\_j
\end{align}
Here $c\_j$ are real values. Such map is called interval translation map.
Consider sequences $\{t\_k^m\}$ and $\{c\_k^m\}$ of rational numbers that converge to $t\_k$ and $c\_k$ respectively. Let $S\_m$ be corresponding mappings.
Lemma: The approximating sequences $\{t\_k^m\}$ and $\{c\_k^m\}$ may be selected so that the following statement is true. Assume that numbers $i,j \in \{1,\cdots,n\}$: $i\neq j$, and $l\_1, \cdots ,l\_n \in \mathbb{Z}^+$ are such that
\begin{align}
t\_j-t\_i=\sum\limits\_{k=1}^nl\_kc\_k. \quad \*
\end{align}
Then for any $m \in \mathbb{N}$
\begin{align}
t\_j^m-t\_i^m = \sum\limits\_{k=1}^n l\_k c\_k^m
\end{align}
Proof: Let $L$ be the space of all values $t\_j$ and $c\_j$ satisfying all equations $\*$ that are true for the considered values of parameters. Since all coefficients $l\_k$ are integers, points with rational coordinates are dense in $L$.
---
My questions about the proof :
At first I emphasize that this lemma and its proof is in "Invariant measures for interval translations" written by S.kryzhevich.
I'm a little confused in this proof of the lemma.
If I understood right $L:=\{t\_j, c\_j ;\quad t\_j-t\_i=\sum\limits\_{k=1}^nl\_k c\_k\}$ which $t\_j \in M\_j$ and $c\_j$ could be any real numbers. Now since $l\_k$ are integers so $l\_k c\_k^m$ are still rational numbers even since $t\_j^m , t\_i^m$ are rational and since rational numbers are dense in $\mathbb{R}$ so points with rational coordinates are dense in $L$. Why being dense in $L$ guarantees the validation of the proof?
| https://mathoverflow.net/users/172647 | A lemma in approximating sequences | Let me answer your question.
The cxdimension of the space $L$ is finite (this is a subspace of ${\mathbb R}^N$ for some $N$.
Let us assume for simplicity that $codim L=1$ i.e. there is only one equation (\*). Then one of coordinates (let it be $c\_{n}$) can be expressed
as
$$c\_n=a\_1 t\_1+\ldots+ a\_n t\_n+b\_1 c\_1+\ldots b\_{n-1} c\_{n-1}$$
with all coefficients $a\_j$ and $b\_j$ being rational.
Then the converging sequence may be 'projected' to the space $L$ as follows: we replace the value $c\_n^{m}$ with
$$a\_1 t\_1^m+\ldots+ a\_n t\_n^m+b\_1 c\_1^m+\ldots b\_{n-1} c\_{n-1}^m.$$
The case $codim L >1$ can be considered similarly.
Sincerely yours,
Sergey
| 2 | https://mathoverflow.net/users/477424 | 416591 | 169,756 |
https://mathoverflow.net/questions/416042 | 2 | Could somebody please direct me to textbooks / literature that (perhaps lay the foundations for and) detail the method for determining the fusion rules for categories such as
* $\operatorname{Rep}(G)$ i.e. the category of representations for a finite group $G$; I've also seen references to things like $\operatorname{Rep}(G)/H$ for a group $H$ in the literature — I'm not sure how they're defined, but this is also of interest.
* $G\_{k}$ i.e. a Lie group $G$ at level $k$ e.g. $(G\_{2})\_2$.
* $\mathfrak{g}\_{k}$ i.e. a Lie algebra $\mathfrak{g}$ at level $k$ e.g. $\mathfrak{su}(n)\_{k}$
I'm happy to assume the base field is $\mathbb{C}$ where possible. Full disclosure, I'm not even sure what ‘at level $k$’ means in the last two points. Roadmaps for learning and elementary/pedagogical references are greatly appreciated. If it helps, I have a pretty basic understanding of the representation theory of finite groups and Lie algebras. I have a physics background (so I'm familiar with fusion categories in a sense, but not via the routes/perspectives above — a situation that I think would be useful to rectify!).
| https://mathoverflow.net/users/135817 | Calculating fusion rules for $\operatorname{Rep}(G)$ and $G_{k}$ [reference request] | In general you should not expect to have explicit formulas for the decomposition of tensor products. The situation is a bit better for Lie algebras than for finite groups. As Ehud Meir pointed out, even calculating the Kronecker coefficients for symmetric groups
<https://en.wikipedia.org/wiki/Kronecker_coefficient>
is very hard. If you take other finite groups (except for very small ones), you won't find any general rules. For small groups you can certainly do this via character computations. Good introductory texts are the books by Serre and Fulton-Harris on representation theory.
I don't know such a thing as a Lie group or Lie algebra at level k. What you probably mean are level $k$ integrable highest weight modules over some affine Lie algebra. But you need to have a good understanding of the classical theory first.
For semisimple Lie algebras there are several ways to compute the tensor product decomposition. One approach is using the Weyl character formula and $ch(V \otimes W) = ch(V) \cdot ch(W)$. In this way you can get some complicated formulas for the fusion coefficients (see the book of Humphreys on Lie algebras (Chapter 24, "Steinberg's formula") for an example).
For $\mathfrak{sl}(n,\mathbb{C})$ the character of an irreducible representation $L(\lambda)$ is given by the Schur function $s\_{\lambda}$. The structure coefficients $c\_{\lambda,\mu}^{\nu}$ in $s\_{\lambda} \cdot s\_{\mu} = \sum\_{\nu} c\_{\lambda,\mu}^{\nu} s\_{\nu}$ are known as Littlewood-Richardson coefficients, and the combinatorial algorithm to compute them is the Littlewood-Richardson rule
<https://en.wikipedia.org/wiki/Littlewood%E2%80%93Richardson_rule>
Similar combinatorial descriptions exist for $\mathfrak{so}(n,\mathbb{C})$ and $\mathfrak{sp}(2n,\mathbb{C})$.
A very general rule is the Littelmann path model
<https://en.wikipedia.org/wiki/Littelmann_path_model>
which can be used to get fusion rules for representations of symmetrizable Kac-Moody algebras.
If you move away from semisimple Lie algebras and go to non-semisimple representation categories such as representations of a quantum group at a root of unity, or representations of a supergroup, or representation theory in positive characteristic, very little general formulas are known.
Since you mention fusion categories and level $k$: What you are ultimately interested in might be the theory of fusion categories arising from a quantum group and the Kazhdan-Lusztig equivalence. But this is quite advanced and if you are not familiar with enough with the classical theory, the books of Humphreys and Fulton-Harris might be better in the beginning. An introduction to representations of quantum groups can be found in the book on Quantum Groups by Chari-Pressley (Chapter 10, 11). A good introduction to fusion rings is Sawin's article on Quantum groups at roots of unity and modularity (which includes the Racah formula for tensor products)
<https://arxiv.org/abs/math/0308281>
As for the Kazhdan-Lusztig equivalence you can have a look at these mathoverflow answers to see what it is about:
[Why should affine lie algebras and quantum groups have equivalent representation theories?](https://mathoverflow.net/questions/177519/why-should-affine-lie-algebras-and-quantum-groups-have-equivalent-representation?noredirect=1&lq=1)
[What's the state of affairs concerning the identification between quantum group reps at root of unity, and positive energy affine Lie algebra reps?](https://mathoverflow.net/questions/178113/whats-the-state-of-affairs-concerning-the-identification-between-quantum-group?noredirect=1&lq=1)
| 3 | https://mathoverflow.net/users/129060 | 416600 | 169,758 |
https://mathoverflow.net/questions/416596 | 0 | *[Repost from MSE.](https://math.stackexchange.com/questions/4374009/a-series-expansion-of-a-function-at-a-non-differentiable-point)*
The function $x\mapsto (\frac{x}{1+x})^{x+x^2}$ is not differentiable in $0$, but nevertheless I'm interested in a series expansion at $0$. [Wolfram Alpha](https://www.wolframalpha.com/input?i=series%20of%20%28x%2F%281%2Bx%29%29%5E%28x%2Bx%5E2%29%20in%200) tells me it has the *Puiseux series* expansion
$$1 + x \log x + \mathcal O(x^2).$$
How do you compute this series expansion (or any "non-Taylor" Puiseux series expansion) over the reals?
---
It may be that the series expansion is easy to compute by making an educated guess first. If that is the case I'd prefer an answer without this "hindsight".
The [Wikipedia article](https://en.wikipedia.org/wiki/Puiseux_series) on Puiseux series has a strong algebraic focus related to polynomials, so I can't easily tell whether that information is really all the relevant to the case of real functions.
The [Mathworld article](https://mathworld.wolfram.com/PuiseuxSeries.html) on this topic is also very short. I wonder what the reason is that according to their definition (iterated) logarithms (as in the above example) can also appear, alongside fractional powers.
| https://mathoverflow.net/users/78650 | A series expansion of a function at a non-differentiable point | Let $$f(x)=\left(\frac{x}{1+x}\right)^{x+x^2}.$$
Expand first
$$g(x)=\log f(x)=(x+x^2)(\log x-\log(1+x))=x\log x+x^2\log x-x^2-\frac{1}{2}x^3-\ldots.$$
This has only two terms with $\log$, the rest are powers.
So powers of $g$ have the same property: they contain powers of $\log$ and ordinary powers. And then use
$$f=e^g=1+g+\frac{g^2}{2}+\frac{g^3}{6}+\ldots,$$
and substitute the series of $g$ to this series.
| 5 | https://mathoverflow.net/users/25510 | 416609 | 169,759 |
https://mathoverflow.net/questions/416608 | 8 | $\newcommand{\KL}{\operatorname{KL}}$Let $X$ be a Polish metric space and $P(X)$ the space of probability measures on $X$. Given $\mu, \nu\in P(X)$, recall that
$$\KL(\mu\parallel\nu) = \begin{cases}\mathbb E\_\mu[\log\tfrac{d\mu}{d\nu}]&\text{if $\mu\ll\nu$;}\\+\infty&\text{otherwise.}\end{cases}$$
I know that both $\mu\mapsto \KL(\mu\parallel\nu)$ and $\nu\mapsto \KL(\mu\parallel\nu)$ are convex maps $P(X)\to\mathbb [0,+\infty]$. Are
$$\mu\mapsto \sqrt{\KL(\mu\parallel\nu)}$$
and
$$\nu\mapsto \sqrt{\KL(\mu\parallel\nu)}$$
convex as well?
If this is not true in general, does it exists a measure $\mu\in P(X)$ such that $\nu\mapsto \sqrt{\KL(\mu\parallel\nu)}$ is a convex map $P(X)\to \mathbb [0,+\infty]$? Or a measure $\nu$ such that $\mu\mapsto \sqrt{\KL(\mu\parallel\nu)}$ is convex?
| https://mathoverflow.net/users/475450 | Is the square root of the Kullback-Leibler divergence a convex map? | $\newcommand\de\delta\newcommand{\KL}{\operatorname{KL}}\newcommand{\p}{\,\|\,}$The maps
$$\mu\mapsto\sqrt{\KL(\mu\p\nu)}$$
and
$$\nu\mapsto\sqrt{\KL(\mu\p\nu)}$$
are not convex in general.
Indeed, let $\mu\_p:=p\de\_0+(1-p)\de\_1$, where $p\in(0,1)$ and $\de\_a$ is the Dirac measure supported on $\{a\}$.
Then the second partial derivative with respect to $p$ of $\sqrt{\mathrm{KL}(\mu\_p\p\mu\_r)}$ at $(p,r)=(1/10,1/11)$ is $-7.17\ldots<0$. So, $\sqrt{\mathrm{KL}(\mu\p\mu\_r)}$ is not convex in $\mu$.
Also, the second partial derivative with respect to $r$ of $\sqrt{\mathrm{KL}(\mu\_p\p\mu\_r)}$ at $(p,r)=(1/10,1/9)$ is $-11.50\ldots<0$. So, $\sqrt{\mathrm{KL}(\mu\_p\p\nu)}$ is not convex in $\nu$.
---
You also asked: "If this is not true in general, does it exists a measure $\mu\in P(X)$ such that $\nu\mapsto \sqrt{\mathrm{KL}(\mu\p\nu)}$ is a convex map $P(X)\to \mathbb [0,+\infty]$? Or a measure $\nu$ such that $\mu\mapsto \sqrt{\mathrm{KL}(\mu\p\nu)}$ is convex?"
The answer to each of these two questions is yes, at least when $X=\{0,1\}$, say.
For $p\in(0,1)$, let
\begin{equation}
F(p):=\sqrt{\mathrm{KL}(\mu\_p\p\mu\_{1/2})},
\end{equation}
\begin{equation}
f(p):=F''(p)4 \mathrm{KL}(\mu\_p\p\mu\_{1/2})^{3/2},
\end{equation}
\begin{equation}
f\_1(p):=f'(p)(1-p)^2 p^2.
\end{equation}
Then $f\_1(1/2)=f'\_1(1/2)=f''\_1(1/2)=0$ and
\begin{equation}
f'''\_1(p)=\frac{2+4 p(1-p)}{(1-p)^2 p^2}>0.
\end{equation}
It follows that $F''(p)\ge0$, so that $\sqrt{\mathrm{KL}(\mu\p\mu\_{1/2})}$ is convex in $\mu$.
For $r\in(0,1)$, let
\begin{equation}
G(r):=\sqrt{\mathrm{KL}(\mu\_{1/2}\p\mu\_r)},
\end{equation}
and
\begin{equation}
g(r):=G''(r)4\mathrm{KL}(\mu\_{1/2}\p\mu\_r)^{3/2}.
\end{equation}
Then $g(1/2)=g'(1/2)=g''(1/2)=g'''(1/2)=0$ and
\begin{equation}
g''''(1/2+h)\frac{(1 - 4 h^2)^4}{16}=9- 16 h^4 + 156 h^2 + 64 h^6 \\
>9- 1 + 156 h^2 + 64 h^6>0
\end{equation}
if $|h|<1/2$.
It follows that $G''(r)\ge0$, so that $\sqrt{\KL(\mu\_{1/2}\p\nu)}$ is convex in $\nu$.
---
*Remark 1:* The existence of a probability measure $\nu\in P(X)$ such that $\sqrt{\KL(\mu\p\nu)}$ is convex in $\nu$ holds for any Polish space $X$. Indeed, the case when $X$ is a singleton set is trivial. Otherwise, take any distinct points $x$ and $y$ in $X$ and let $\nu:=\frac12\,\de\_x+\frac12\,\de\_y$. Then the condition $\mu\ll\nu$ implies $\mu=p\,\de\_x+(1-p)\,\de\_y$ for some $p\in[0,1]$, so that here $\sqrt{\KL(\mu\p\nu)}=F(p)$.
*Remark 2:* Let us say that a probability measure $\mu\in P(X)$ is good if $\sqrt{\KL(\mu\p\nu)}$ is convex in $\nu$. Then it is easy to see that the Dirac measure $\de\_x$ is good for any $x\in X$. It also follows from above that $\mu:=\frac12\,\de\_x+\frac12\,\de\_y$ is good if $X=\{x,y\}$ for some $x,y$.
Finally, using reasoning similar to that above, one can show that, in the case when the cardinality of $X$ is $\ge3$, the only good probability measures $\mu\in P(X)$ are the Dirac measures. Since this answer has already grown rather long, I will leave the latter assertion as an exercise to interested readers.
Now the answer is quite complete.
| 10 | https://mathoverflow.net/users/36721 | 416612 | 169,760 |
https://mathoverflow.net/questions/416611 | 6 | I asked this question on MSE 9 days ago and it got a very helpful comment from Eric Towers providing the Palais Stewart reference, but no answers. So I'm crossposting it here.
Let
$$
T^n \to M \to T^m
$$
be a principal torus bundle over a torus. Then $ M $ is a solvmanifold, even a nilmanifold (in fact $ M $ is the total space of a principal torus bundle over a torus if and only if it is a compact nilmanifold for a 2 step nilpotent Lie group, this is theorem 3 of [Palais, Stewart, TORUS BUNDLES OVER A TORUS]).
What if the bundle is not necessarily principal? Is every torus bundle over a torus a solvmanifold? In other words, if we have a fiber bundle
$$
T^n \to M \to T^m
$$
then can we conclude that the total space $ M $ is a solvmanifold?
EDIT:
The answer, comments and references from Igor Belegradek prove that something much stronger is true: A manifold $ M $ is the total space of a bundle
$$
N \to M \to T^n
$$
where $ N $ is a compact nilmanifold and $ T^n $ is a torus if and only if $ M $ is homeomorphic to a compact solvmanifold.
The smooth case is also addressed. Corollary 2.3 of <https://arxiv.org/pdf/1307.3223.pdf> states that "If $ h: T^n → T^n $ , $ n ≥ 6 $, is a diffeomorphism given by Proposition 2.1
then the mapping torus $ M\_h $ is a fake torus." By fake torus the authors mean $ M\_h $ is homeomorphic to a torus but not PL homeomorphic. Since $ M\_h $ is not PL homeomorphic to a torus it is also not diffeomorphic to a torus. Thus already in dimension 7 we have an exotic homotopy torus $ M\_h $ which is the total space of a torus bundle over the torus
$$
T^6 \to M\_h \to S^1
$$
but which is not diffeomorphic to any solvmanifold. So in every dimension $ d \geq 7 $ there are smooth torus bundles over a torus (both base and fiber with the standard smooth structure) that are not diffeomorphic to any solvmanifold.
For $ d \leq 3 $ we have by Moise's theorem that there are no exotic smooth structures. So all torus bundles over a torus with total space of dimension $ d \leq 3 $ are diffeomorphic to a solvmanifold.
For $ d=4 $ the existence of exotic 4 tori (or any exotic closed aspherical manifold) seems to be an open problem see
[Do there exist exotic 4-tori?](https://mathoverflow.net/questions/133797/do-there-exist-exotic-4-tori)
In that same question it is pointed out that there are exotic tori in $ d=5,6 $ but it is not clear that these can be constructed as smooth fiber bundles with base and fiber standard tori.
That said, for the $ d=4 $ case a torus bundle over a torus has base and fiber with dimension $ \leq 3 $ so there shouldn't be any exotic gluing maps and it seems like the total space of such a bundle should just always have the standard smooth structure.
For $ d=5,6 $ I have no idea what kind of smooth structures might or might not be possible on torus bundles over the torus.
| https://mathoverflow.net/users/387190 | Torus bundles and compact solvmanifolds | A group $G$ is isomorphic to the fundamental groups of a compact solvmanifold if and only if it fits into the short exact sequence $1\to N\to G\to\mathbb Z^n\to 1$ where $N$ is a finitely generated torsion-free nilpotent group.
This is stated on p.253 and explained is chapter III of Auslander's [An exposition of the structure of solvmanifolds. Part I: Algebraic theory](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-79/issue-2/An-exposition-of-the-structure-of-solvmanifolds-Part-I/bams/1183534430.full). In particular, **every torus bundle over a torus is homotopy equivalent to a solvmanifold**.
You may also be interested in Theorem 3 of Wilking's paper [Rigidity of group actions on solvable Lie groups](https://www.uni-muenster.de/imperia/md/content/theoretische_mathematik/diffgeo/mr1764235.pdf) which gives an analogous result for infrasolvmanifolds.
| 7 | https://mathoverflow.net/users/1573 | 416616 | 169,763 |
https://mathoverflow.net/questions/416598 | 25 | What is known about the existence of other pairs of spheres (such as $S^2$ and $S^3$) whose homotopy groups coincide starting from some index.
A sufficient condition for this is the existence of a fiber bundle $S^m \to S^n$ with fiber having a finite number of nonzero homotopy groups (as in the case of the Hopf fibration)
P.S. I don't know if my question has a research level. If it is not, then feel free to close it.
| https://mathoverflow.net/users/148161 | Spheres with the same homotopy groups | It is a result from Serre's thesis that for $n\geq 3$ and a prime $p$, the first $p$-torsion in $\pi\_\*S^n$ occurs precisely for $\* = n+2p-3$. This shows that $(m,n) = (2,3)$ is the only pair of (edit: positive integers) $m<n$ with $\pi\_\*S^m \cong \pi\_\*S^n$ for $\*$ large enough.
| 51 | https://mathoverflow.net/users/2039 | 416618 | 169,764 |
https://mathoverflow.net/questions/416624 | 5 | Let $(M, g\_M)$ where $M= B \times\_f F$ and $g\_M=g\_B + f^2g\_F$, an Einstein warped product manifold (i.e., $Ric\_M= \lambda g\_M$), with Ricci flat fiber-manifold $F$, i.e., $Ric\_F=0$.
Then $M$ can admit only constant negative Ricci curvature or zero Ricci curvature (i e., $\lambda \le 0$) or $M$ could also have positive constant Ricci curvature (i.e., $\lambda >0$)?
In other words, $Ric\_F = 0$ necessarily implies $\lambda \le 0$, or can solutions be obtained with $\lambda > 0$?
| https://mathoverflow.net/users/111304 | Possible sign of scalar curvature for Einstein warped product manifold with Ricci-flat | It can not have constant positive Ricci curvature. By Bonnet-Myers constant positive Ricci curvature implies that $M$ is compact.
If $V$ is a vertical vector then by the formula for Ricci curvature of warped product (page 266 in Besse's book)
$$
Ric(V,V)=Ric\_F(V,V) -|V|^2(\frac{\Delta f}{f}+(p-1)\frac{|\nabla f|^2}{f^2})
$$
where $p=\dim F$.
If the fiber is Ricci flat then
at the point on the base where $f$ achieves minimum (which exists by compactness) it holds that $Ric(V,V)\le 0$
| 4 | https://mathoverflow.net/users/18050 | 416629 | 169,767 |
https://mathoverflow.net/questions/416627 | 9 | For $S \subset \mathbb{N}$ define $S-S=\{x-y:x \in S, y \in S\}$.
As noted below there is a simple example showing that a set $S \subset \mathbb{N}$ with positive upper density has a sumset $S+S=\{x+y:x \in S, y \in S\}$ with $S+S$ containing only finite length arithmetic progressions. However the case for the difference set seems not so obvious to me hence the question:
>
> What is an example of a set S with positive upper density in $\mathbb{N}$ such that $S-S$ does not contain an infinite arithmetic progression?
>
>
>
Here is the example for the sumset $S+S$, in fact for any $hS=S+\dots+S$, taken from Erdos, Nathason and Sarkozy's paper "Sumsets Containing Infinite Arithmetic Progressions":
*"Let $(t\_n)$ be a strictly increasing sequence of positive integers such that $t\_{n+1}/t\_n$ tends to infinity, and let the set $A$ be the union of the intervals $[t\_{2n}+1, t\_{2n+1}]$. Then $A$ has upper asymptotic density $d\_U(A) = 1$ and lower asymptotic density $d\_L(A)=0$. For fixed $h$ and all sufficiently large $n$, the sumset $hA$ is disjoint from the interval
$[h t\_{2n-1} + 1, t\_{2n}]$. Thus, $hA$ contains arbitrarily long gaps, and so cannot
contain an infinite arithmetic progression."*
| https://mathoverflow.net/users/7113 | A set with positive upper density whose difference set does not contain an infinite arithmetic progression | Let $\langle x\rangle$ denote the fractional part of a real number $x$ (i.e. $\langle x \rangle := x- \lfloor x\rfloor $, where $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$).
Let $\alpha \in \mathbb R$ be irrational and let $S:=\{n\in \mathbb Z: \langle n\alpha \rangle \in (0,1/4)\}$. The upper (and lower) density of $S$ is $1/4$; this is a consequence of [Weyl's theorem](https://en.wikipedia.org/wiki/Equidistribution_theorem) on uniform distribution. Also, $S-S\subseteq \{n\in \mathbb Z: \langle n\alpha\rangle \in (3/4,1)\cup [0,1/4)\}$.
To see that $S-S$ does not contain an infinite arithmetic progression $\{a+bn:n\in \mathbb N\}$, note that $b\alpha$ is irrational if $b\in \mathbb Z\setminus \{0\}$, so the values $\langle (a+bn)\alpha \rangle$ are dense in $[0,1]$. So if $S-S$ contained an infinite AP, the values $\{\langle n\alpha \rangle:n\in S-S\}$ would be dense in $[0,1]$, but $\langle n\alpha\rangle \in (3/4,1)\cup [0,1/4)$ for $n\in S-S$.
This example $S$ is a Bohr neighborhood in $\mathbb Z$. Generally, if you want an example or counterexample of some structure in $S-S$, where $S$ has positive upper density, it's natural to look among Bohr neighborhoods: Følner [ZBL0058.02302](https://zbmath.org/?q=an:0058.02302) proved that if $S$ has positive upper Banach density, then $S-S$ contains (up to upper Banach density 0) a Bohr neighborhood of $0$. Since every Bohr neighborhood $B$ of $0$ contains a set of the form $B'-B'$, where $B'$ is a Bohr neighborhood, $S-S$ itself is not too far from containing a difference set of a Bohr neighborhood.
Ruzsa's section *Sumsets and structure* in [ZBL1221.11026](https://zbmath.org/?q=an:1221.11026) and Hegyvári and Ruzsa's article [ZBL1333.05042](https://zbmath.org/?q=an:1333.05042) are both good references on the relationship between Bohr sets and difference sets.
| 13 | https://mathoverflow.net/users/10457 | 416632 | 169,769 |
https://mathoverflow.net/questions/416630 | 1 | Can I calculate the volume of a frustum if all I know is the volume of the pyramid the height of the pyramid and the height of the frustum?
| https://mathoverflow.net/users/477453 | Volume of a frustum knowing the volume and height of the pyramid and the height of the frustum | Let the pyramid $P$ have volume $V\_P$ and height $H$. Let the frustum $F$ that is a sliced part of this pyramid have height $h$, so $0 \leq h \leq H$. You want to know the volume of the frustum. Call it $V\_F$.
By calculus, $V\_P = (1/3)AH$, where $A$ is the area of the base of the pyramid. The part of the pyramid above the frustum is a scaled version of the original pyramid. It has height $H-h$, so the smaller pyramid is a scaled-down version of the original pyramid by the factor $(H-h)/H = 1 - h/H$. In particular, the base of the smaller pyramid has area $(1-h/H)^2A$, so the volume of the smaller pyramid is $(1/3)(1-h/H)^2A(H-h)$. Therefore
$$
V\_F = V\_P - \frac{1}{3}\left(1 - \frac{h}{H}\right)^2A(H-h).
$$
Replacing $A$ by $3V\_P/H$,
$$
V\_F = V\_P - \frac{1}{3}\left(1 - \frac{h}{H}\right)^2\frac{3V\_P}{H}(H-h) = \left(1 - \left(1 - \frac{h}{H}\right)^3\right)V\_P.
$$
So there is your frustum volume formula in terms of $V\_P$, $H$, and $h$. As a reality check, when $h = H$ we get $V\_F = V\_P$ (and the frustum $F$ in this case is the full pyramid $P$) and when $h = 0$ we get $0$ (and the frustum $F$ in this case is the base of the pyramid, so it is planar).
PS: Until less than two weeks ago, I always thought the term in English was *frustrum*. To learn so late in life that it is actually frustum was quite… frustating.
| 3 | https://mathoverflow.net/users/3272 | 416633 | 169,770 |
https://mathoverflow.net/questions/416628 | 3 | Let $X$ be a scheme and $\mathcal{F}$ a presheaf on $X\_{ét}$.
For each $x\_{i}\in X$, pick a geometric point $\bar{x}\_{i}$ over $x$ and denote by $i\_{\bar{x}\_{i}}:\text{Spec}(k\_{i})\_{\text{ét}}\rightarrow X\_{ét}$ the morphism of sites induced by the geometric point $\bar{x}\_{i}$ where $k\_{i}$ is algebraically closed. Then we can define a sheaf on $X\_{ét}$
$$\text{Esp}(\mathcal{F}):=\prod\_{\bar{x}\_{i}}(i\_{\bar{x}})\_{\*}\mathcal{F}\_{\bar{x}}.$$
What is the map $\mathcal{F}\rightarrow \text{Esp}(\mathcal{F})$? I assume you would send sections to their germs, however if $U\rightarrow X$ is étale then $$
(i\_{\bar{x}})\_{\*}\mathcal{F}\_{\bar{x}}(U)=\prod\_{\text{Hom}\_{X}(\bar{x},U)}\mathcal{F}\_{\bar{x}}.
$$ So would you have to send a section to the germ then compose with some kind diagonal embedding?
Cross post:[https://math.stackexchange.com/questions/4386913/sheaves-on-étale-site-and-sheafification?noredirect=1#comment9171600\_4386913](https://math.stackexchange.com/questions/4386913/sheaves-on-%C3%A9tale-site-and-sheafification?noredirect=1#comment9171600_4386913)
| https://mathoverflow.net/users/211978 | Sheafifcation for the étale site | It's not diagonal. A section of $\mathcal F$ on $U$ has a germ at each element of $\text{Hom}\_{X}(\bar{x},U)$, and you take the product of all those germs, which may be different.
It's not so hard to see that if you tried to do it diagonally, it wouldn't be functorial for automorphisms of $U$, say in the case when $U$ is a finite étale cover.
| 4 | https://mathoverflow.net/users/18060 | 416637 | 169,773 |
https://mathoverflow.net/questions/415718 | 3 | Is there any standard name for semigroups $S$ in which $xS=Sx$ for all $x\in S$?
Examples of such semigroups are commutative semigroups and Clifford inverse semigroups.
| https://mathoverflow.net/users/61536 | A name for semigroups in which left and right principal ideals coincide | People in factorization theory call a monoid $H$ *normalizing* if $aH = Ha$ for every $a \in H$; see, e.g.,
>
> A. Geroldinger, *Non-commutative Krull monoids: A divisor-theoretic approach and their arithmetic*, Osaka J. Math. 50 (2013), 503-539.
>
>
>
However, it makes a lot of sense to refer to the same objects as *duo monoids* (as also suggested by Benjamin Steinberg in a [comment to the OP](https://mathoverflow.net/questions/415718#comment-1066373)), in such a way that a unital ring is duo if and only if its multiplicative monoid is duo (a ring, with or without unity, is said to be *duo* if every left or right ideal is in fact a two-sided ideal).
Duo rings have been studied (under this name) at least since the late 1950s. To the best of my knowledge, they were first considered by E.H. Feller in
>
> *Properties of primary noncommutative rings*, Trans. Amer. Math. Soc. 89 (1958), 79-91.
>
>
>
| 2 | https://mathoverflow.net/users/16537 | 416640 | 169,774 |
https://mathoverflow.net/questions/416473 | 5 | Let $X$ be a smooth variety over a characteristic zero field $k$. It seems to be well-known that
"A coherent $\mathcal{D}\_X$-module is holonomic if and only if 'every element is annihilated by a nonzero differential operator'."
I know that, when $X=\mathbb{A}^1$, this is true in the sense that "a finitely generated $k[x]\langle\partial\rangle$-module $M$ is holonomic if and only if, for every $m\in M$, there exists $P\in k[x]\langle\partial\rangle\setminus 0$ such that $Pm=0$."
**Is a similar statement true in more generality? (For all $X$? For affine $X$?) If so, why?**
| https://mathoverflow.net/users/131975 | Holonomic = annihilated by some differential operator | "A coherent $D\_X$-module is holonomic if and only if 'every element is annihilated by a nonzero differential operator'."
This is very much not true in dim>1.
For example, let $D=k[x\_1,x\_2]\langle \partial\_1,\partial\_2\rangle$ be the ring of differential operators on $\mathbb A^2$, and consider the left $D$-module
$$M= D/D\partial\_1.$$
Then every element is annihilated by a suitably high power of $\partial\_1$, but $M$ is not holonomic (its characteristic variety is the hyperplane in $T^\ast\mathbb A^2= \operatorname{Spec} k[x\_1,x\_2,\xi\_1,\xi\_2]$ cut out by the equation $\xi\_1=0$).
EDIT: In fact, we can make a stronger statement. Let $M$ be a $D$-module, and suppose we can find $m\in M$ which is not annihilated by any non-zero differential operator. Then the homomorphism $D\to M$ given by $P \mapsto Pm$ is injective. It follows that $\mathit{SS}(D) \subseteq \mathit{SS}(M)$ (the characteristic variety operation takes short exact sequences to unions). Thus $\mathit{SS}(M)$ is the entire cotangent bundle in this case.
Put another way, any $D$-module whose characteristic variety is a proper subset of the cotangent bundle will satisfy the property that each element is annihilated by a non-zero differential operator.
In dimension 1, every proper closed coisotropic subset of the cotangent bundle is lagrangian, so every such $D$-module is holonomic. But this is definitely not the case in higher dimensions.
| 8 | https://mathoverflow.net/users/7762 | 416643 | 169,777 |
https://mathoverflow.net/questions/416644 | 4 | Let $X \subseteq \mathbb{C}^n$ be an irreducible algebraic set that forms a cone, and let $I=I(X) \subseteq \mathbb{C}[x\_1,...,x\_n]$. Let $m < n$ and $k\leq m$ be positive integers. Is it true that for a generic collection of elements $v\_1,..., v\_k \in X$ and $v\_{k+1},..., v\_m \in \mathbb{C}^n$, it holds that $I + I(\text{span}\{v\_1,...,v\_m\})$ is radical?
| https://mathoverflow.net/users/150898 | Is the sum of a radical ideal and the ideal of a generic linear space intersecting that ideal radical? | In general no. I'm only familiar with this question with $k=0$. Here are two basic counter-examples (both with $k=0$)
(1) Take $X$ singular and dimension $n-m$, so we are intersecting $X$ with a generic codimension $n-m$-plane. Then the intersection is a point of multiplicity equal to the multiplicity of $0$ as a singular point of $X$.
(2) Let $X$ be non-Cohen Macaulay with depth $\delta < \dim X$. Then intersecting $X$ with a codimension $\delta$ plane gives a variety with an embedded point.
If we impose that $X$ is Cohen-Macaulay, that $k=0$ and that $n-m < \dim X$, then the answer is "yes". Recall Serre's Theorem that "reduced" is equivalent to R0 and S1. $X$ Cohen-Macaulay means that $X$ is $S\_{\dim X}$, and $X$ reduced means that $X$ is $R\_0$. Taking $q$ generic hyperplane sections (through the singularity of $X$) turns this into $S\_{\dim X - q}$ and $R\_0$, so still reduced. I can't quickly find good references for this; I'm willing to try harder if someone else doesn't beat me too it.
The problem with $k>0$ seems more interesting to me, and I haven't seen it before.
| 5 | https://mathoverflow.net/users/297 | 416660 | 169,779 |
https://mathoverflow.net/questions/416668 | -3 | We have two particles $A$ and $B$ in a maximally entangled state $|\Psi\rangle \in \cal{H}\_A \times \cal{H}\_B$
$$
\left|\Psi\right\rangle = \frac{1}{\sqrt{2}} ( \left| 0
\right\rangle\_A\otimes \left| 0 \right\rangle\_B + \left| 1
\right\rangle\_A\otimes \left| 1 \right\rangle\_B ),
$$
where $\_A\left\langle i | j \right\rangle\_A = {\_B\left\langle i | j \right\rangle\_B} =
\delta\_{ij}$ and $U^T$ is the transpose of the matrix $U$.
It is claimed that if $U \in SU(\cal{H}\_A)$, then
$$
U \otimes I \left|\Psi\right\rangle = I \otimes U^T \left|\Psi\right\rangle
$$
How can we show this to be true? I am trying to make sense of a "quantum game theory" paper, but it assumes a high comfort level with the mathematical apparatus of quantum physics.
| https://mathoverflow.net/users/477478 | SU(2) and entangled particles | This is presumably simply an issue of notation -
\begin{eqnarray}
U\otimes I
\left|\Psi\right\rangle = \frac{1}{\sqrt{2}} \left( (U\_{11} \left| 0 \right\rangle\_A + U\_{21} \left| 1 \right\rangle\_A ) \otimes \left| 0 \right\rangle\_B \\
\hspace{3cm} + ( U\_{12} \left| 0 \right\rangle\_A + U\_{22} \left| 1
\right\rangle\_A ) \otimes \left| 1 \right\rangle\_B \right)
\end{eqnarray}
and
\begin{eqnarray}
I\otimes U^T \left|\Psi\right\rangle = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle\_A \otimes (U\_{11} \left| 0 \right\rangle\_B + U\_{12} \left| 1 \right\rangle\_B ) \\
\hspace{3cm} + \ \left| 1 \right\rangle\_A \otimes ( U\_{21} \left| 0 \right\rangle\_B + U\_{22} \left| 1 \right\rangle\_B ) \right)
\end{eqnarray}
They're indeed the same.
| 0 | https://mathoverflow.net/users/134299 | 416670 | 169,783 |
https://mathoverflow.net/questions/416653 | 1 | For a bounded smooth domain $\Omega$, let $H\_2^{s}(\Omega)$ be the usual Sobolev space on $\Omega$.
Define $A:=\{f\in H\_2^{s}(\Omega)| \lVert f\rVert\_{L^p(\Omega)}=1\}$ where $2<p<2\_{s}^\*$.
Can we show $A$ is a $C^2$-Hilbert manifold since I need to use the Morse lemma?
| https://mathoverflow.net/users/166368 | Is an $L^p$-sphere in Sobolev space $H_2^{s}(\Omega)$ a Hilbert manifold? | The norm map
$$
f\mapsto \int\_{\Omega} f^p dvol
$$
is a $C^k$-map for $k<p$ and in fact smooth if $p$ is an even integer as is shown in [this paper](http://www.jstor.org/stable/24901438.) of Bonic and Frampton. The implicit function theorem for maps of Banach spaces then applies to give your desired result.
| 3 | https://mathoverflow.net/users/12605 | 416676 | 169,785 |
https://mathoverflow.net/questions/416659 | 16 | Is it possible to find the determinant of an $n\times n$- matrix, only given the determinant of all $p\times p$ sub-matrices in it? Here $p\leq n$ is fixed. This is obviously true if $p=1,n$. But what happens in other cases?
| https://mathoverflow.net/users/44215 | Find the determinant of a matrix given the determinant of all $p\times p$ sub-matrices? | As [mentioned](https://mathoverflow.net/questions/416659/find-the-determinant-of-a-matrix-given-the-determinant-of-all-p-times-p-sub-ma#comment1069136_416659) by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e\_1,\dotsc,e\_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae\_{i\_1})\wedge\cdots\wedge(Ae\_{i\_p})\in\Lambda^p(K^n),$$
where $K$ is the field of scalars (e.g. $\mathbb C$).
Splitting
$$(Ae\_1)\wedge\cdots\wedge (Ae\_n) = [(Ae\_1)\wedge\cdots\wedge (Ae\_p)]\wedge\cdots\wedge[(Ae\_{n-p+1})\wedge\cdots\wedge (Ae\_n)],
$$
we see that $(Ae\_1)\wedge\cdots\wedge (Ae\_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae\_1)\wedge\dotsb\wedge (Ae\_n)=(\det A)e\_1\wedge\dotsb\wedge e\_n$, we deduce the value of $\det A$.
Let me describe how it works when $n=4$ and $p=2$. The minors are denoted
$$A\binom{i\alpha}{j\beta}=a\_{i\alpha}a\_{j\beta}-a\_{i\beta}a\_{j\alpha}.$$
Then
$$(Ae\_\alpha)\wedge(Ae\_\beta)=\sum\_{i<j}A\binom{i\alpha}{j\beta}e\_i\wedge e\_j.$$
We thus obtain
$$\det A=\sum\_{\substack{\rho\in{\frak S}\_4 \\ \rho(1)<\rho(2),\rho(3)<\rho(4)}}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$
| 20 | https://mathoverflow.net/users/8799 | 416677 | 169,786 |
https://mathoverflow.net/questions/408166 | 7 | Federer and Fleming's notion of "currents" is well established so far, and starting from the seminal work of Ambrosio and Kirchheim, the notion of metric currents is well studied also. The main underlying idea of Federer and Flaming's notion of Euclidean current generalizes to metric space thanks to the fact that one finds Bilipschitz maps that allow this sort of generalization.
My question is: what is the main difficulty of building currents in the sub-Riemannian settings? Is there some work on the development of an analogous concept of currents in sub-Riemannian manifolds like Carnot groups or in general stratified Lie groups?
I may guess that, on the negative side, we find important examples of metric spaces where the theory of rectifiable currents does not apply simply because the class of rectifiable currents is very poor. The simplest of these examples is the Heisenberg group, due to the following theorem:
>
> If $(X, d)$ is the Heisenberg group $H\_1$ endowed with any left
> invariant homogeneous metric, then any $k-$dimensional rectifiable
> current is identically $0$ for $k=2, 3 , 4.$
>
>
>
Do there exist more general notions of rectifiable currents in sub-Riemannian geometries?
| https://mathoverflow.net/users/170982 | Currents in sub-Riemannian geometry | I want to try to give you an answer, although it will be very sketchy. Take into account that the topic is still unclear to the experts themselves.
The key word is “Rumin currents”: they are defined on all Heisenberg groups (and contact manifolds) starting with a pre-selection of differential forms. So, instead of taking the dual of all k-forms, currents are defined taking the dual of Rumin k-forms.
I think the original paper where “Rumin forms” were introduced is
Rumin M., Formes differentielles sur les varietes de contact, J. Diff. Geom. 39, 281-330, 1994.
For example, in the first Heisenberg group, we roughly have:
* 0-currents: (generalisation of) points.
* 1-currents: (generalisation of) horizontal curves.
* 2-currents: (generalisation of) surfaces with finite sub-Riemannian area (Hausdorff measure of dimension 3).
* 3-currents: (generalisation of) volume.
Currents should take into account that:
* in dimension 1 we have less objects than in the Euclidean case (not every curve is horizontal).
* in dimension 2 we have more objects than in the Euclidean case (there are surfaces whose sub-Riemannian Hausdorff dimension is 2, but whose Euclidean Hausdorff dimension is >2).
Rumin currents are able to take these differences into account.
However, I think one should ask: what for?
A theory of rectifiability in Carnot groups is still so under-developed that it is not clear what a generalisation to rectifiable currents should look like.
As far as I know, the only application of Rumin currents is in D. Vittone’s paper (which contains a description of Rumin currents):
“Lipschitz graphs and currents in Heisenberg groups”,
<https://www.cambridge.org/core/journals/forum-of-mathematics-sigma/article/lipschitz-graphs-and-currents-in-heisenberg-groups/0524E2DEE0F4878D20FF2B0641A48CB4>
I hope this can help, at least as a start.
| 5 | https://mathoverflow.net/users/97130 | 416680 | 169,787 |
https://mathoverflow.net/questions/416678 | 1 | This question is motivated by [Linear Feedback Shift Registers](https://en.wikipedia.org/wiki/Linear-feedback_shift_register), which cycle through $\{0,1\}^n \setminus \{(0,\ldots,0)\}$ by shifting and applying a small set of XOR operations.
Let $n>1$ be an integer. To any map $f:\{0,1\}^{n} \to \{0,1\}$ we associate a "shifting function" $\sigma\_f: \{0,1\}^n\to \{0,1\}^n$ by $$(b\_0, b\_1, \ldots, b\_{n-1}) \in \{0,1\}^n \; \mapsto \; \big(b\_1, b\_2, \ldots, b\_{n-1}, f(b\_0,b\_1,\ldots,b\_{n-1})\big).$$
**Question.** For what positive integers $n > 1$ is there $f:\{0,1\}^n\to \{0,1\}$ such that $\sigma\_f:\{0,1\}^n \to \{0,1\}^n$ is a [cyclic permutation](https://en.wikipedia.org/wiki/Cyclic_permutation) of $\{0,1\}^n$ having length $2^n = |\{0,1\}^n|$?
| https://mathoverflow.net/users/8628 | Cycling through $\{0,1\}^n$ by shifting and applying a $n$-ary function | Such $f$ exist for all $n.$
The resulting sequences are de Bruijn sequences. Let $a=(a₀, a₁,\ldots )$ be a periodic sequence of period
$T$ with symbols taken from a finite alphabet $A.$
**Definition** The sequence $a$ is a de Bruijn sequence of span $n$ if every block of length $n$
occurs exactly once in (each period of) $a.$
It is straightforward to show that:
**Proposition** Suppose $a$ is a de Bruijn sequence of span $n.$ Then
1. The period of $a$ is $T = | A|^n$.
2. For any $t \leq n$ and for any block $b$ of length $t$ , the number of occurrences of $b$
within a single period of $a$ is $| A|^{n−t}.$
Letting $n=t,$ each block (subsequence) of length $n$ occurs exactly one so a function $f$ that you want can always be defined. for any alphabet size $|A|.$
One can also construct a de Bruijn sequence of span $n+1$ from one of span $n$ recursively. There are many de Bruijn sequences but an algebraic classification of functions yielding them in the general case is lacking, though some function families exists. Books by Golomb (Shift Register Sequences) and Goresky/Klapper (Algebraic Shift Register Sequences) cover some of this topic.
A linear $n-$variable function (LFSR) stated algebraically won't give full period. For an introduction to NLFSRs (which is what you ask) focusing on the important binary alphabet, you can start with Helleseth's talk in the slides linked below and follow the references. Some keywords for sequence families are GMW sequences, WG sequences.
[Helleseth survey talk](https://www.ricam.oeaw.ac.at/specsem/specsem2013/workshop4/slides/helleseth.pdf)
| 4 | https://mathoverflow.net/users/17773 | 416683 | 169,788 |
https://mathoverflow.net/questions/416684 | 7 | Let $(M^n,g)$ be a smooth Riemannian manifold. It is well known that if $sec(M)\geq \kappa$ then $Ric(M)\geq (n-1)\kappa$.
If I understand correctly in dimensions $n\geq 4$ a lower bound on $Ric(M)$ does not imply a lower bound on $sec(M)$. However in $n=2$ this implication is true for trivial reason.
**Is it true that in $n=3$ a lower bound on $Ric(M)$ implies a lower bound on $sec(M)$?**
| https://mathoverflow.net/users/16183 | Relation between Ricci curvature and sectional curvature for 3-manifolds | This is definitely false. In dimension 3 if $\lambda\_1,\lambda\_2,\lambda\_3$ are eigenvalues of the curvature operator then Ricci curvatures of eigenvectors are $\lambda\_1+\lambda\_2, \lambda\_1+\lambda\_3, \lambda\_2+\lambda\_3$. If one of $\lambda\_i$'s is very negative but the other two are very positive then all these sums can be bounded below. Every algebraic curvature tensor can be realized at a point so locally this can definitely happen. Complete examples exist too. I can't seem to find a reference to an explicit example but it's well known that there are examples of metrics on $\mathbb R^3$ with $Ric\ge 0$ which have some negative sectional curvature. If you rescale such metric by a small number Ricci curvature remains nonnegative but sectional curvature can become arbitrary negative in some 2-planes.
| 11 | https://mathoverflow.net/users/18050 | 416687 | 169,789 |
https://mathoverflow.net/questions/156257 | 10 | Let $K$ be a $p$-adic field and $\Omega^{(n)}\_K$ the $n$-dimensional Drinfeld upper half space over $K$ (which is a rigid analytic space over $K$).
>
> Is the Picard group of $\Omega^{(n)}\_K$ known? More generally, I would like to know
> the Picard group of $\Omega^{(n)}\_K$ base changed to any finite extension $L$ of $K$ (which is not the same as $\Omega^{(n)}\_L$).
>
>
>
I'd appreciate any references where this question, or similar questions, are considered.
| https://mathoverflow.net/users/519 | Picard group of Drinfeld upper half space | In [this](https://arxiv.org/abs/2012.12729) paper of Junger, all these Picard groups are shown to be zero.
I think this was known for a long time for $n=1$, see for example the book *Rigid Analytic Geometry and its Applications* by Fresnel and van der Put.
| 2 | https://mathoverflow.net/users/144285 | 416688 | 169,790 |
https://mathoverflow.net/questions/416686 | 0 | Update: Thanks to GJC20's answer on the existence and uniqueness. Let me reformulate my questions 3/4 as follows: There exists a unique non-increasing and continuously differentiable function $f:\mathbb R \to [0,1]$ s.t.
$$X\_t=X\_0+W\_t-1+f(t) \mbox{ and } f(t)= \mathbb E \left[ \exp\left(-\frac 1 \epsilon \int\_0^t X\_s^-ds\right)\right],\quad \forall t\ge 0.$$
What is a characterization for $f$?
Consider
$$X\_t=X\_0+ W\_t-1 + \mathbb E \left[ \exp\left(-\frac 1 \epsilon \int\_0^t X\_s^-ds\right)\right],\quad \forall t\ge 0,\quad\quad\quad(\ast)$$
where $X\_0>0$ is a random variable as nice as possible, $(W\_t)\_{t\ge 0}$ is an independent Brownian motion, $\epsilon>0$ is fixed and can be as small as possible.
1. Has this equation been studied (I don't even know how to call such equation)?
2. Do we have the existence/uniqueness result?
3. If $X\_t$ has a density function, denoted by $f(t,x)$, is there any analytic equation for $f$?
4. Let $\tau:=\inf\{t\ge 0: X\_t \le 0\}$. If $X\_t{\bf 1}\_{\{\tau>t\}}$ has a density function, denoted by $g(t,x)$, is there any analytic equation for $g$?
Many thanks for the answer, comments and references.
PS : Here I adopt the notation $a^-:=\max(-a, 0)$ for all $a\in \mathbb R$.
| https://mathoverflow.net/users/nan | Has this "stochastic differential equation" been studied? | Updates : This is an answer to the simplest question above, existence and uniqueness.
If the existence holds, then any solution must be a strong solution. Let $X, Y$ be two arbitrary solution, it appears that $|X\_t-Y\_t|$ is deterministic and satisfies further
$$\mathbb E|X\_t-Y\_t| = |X\_t-Y\_t| \le \mathbb E\left[\frac{1}{\epsilon}\int\_0^t |X^-\_s-Y^-\_s|ds\right] \le \frac{1}{\epsilon}\int\_0^t \mathbb E[|X\_s-Y\_s|]ds,\quad \forall t\ge 0,$$
which yields the uniqueness by Gronwall's inequality.
In the following, let us show the existence. To do so, we adopt the argument of fixed point. Let $C$ be the space of continous functions on $\mathbb R\_+$ and $C\_{\preceq}\subset C$ be the subset of non-increasing functions $f$ s.t. $f(0)=1$ and $\inf\_{t\ge 0} f(t)\ge 0$. Define further the operator $\Gamma: C\_{\preceq}\to C\_{\preceq}$ by
$$\Gamma[f](t):=\mathbb E\left[\exp\left(-\frac{1}{\epsilon}\int\_0^t\big(X^f\_s\big)^-ds\right)\right],\quad \forall t\ge 0,$$
where $X^f\_t:=X\_0+W\_t-1+f(t)$. Clearly $\Gamma[f]\in C\_{\preceq}$ for every $f\in C\_{\preceq}$. Then it suffices to show $\Gamma$ has a fixed point in $C\_{\preceq}$. First, we observe that $\Gamma$ is monotone in the following sense: if $f\preceq g$, then $\Gamma[f]\preceq \Gamma[g]$. Here we say $f\preceq g$ iff $f(t)\le g(t)$ for all $t\ge 0$. Next, set $f\_0\equiv 1\in C\_{\preceq}$ and define $f\_{n+1}:=\Gamma[f\_n]$ for every $n\ge 0$. By induction, one has that $n\mapsto f\_n(t)$ is non-increasing for all $t\ge 0$. Thus the pointwise $f(t):=\lim\_{n\to\infty}f\_n(t)$ exists. Finally, using the equality
$$f\_{n+1}(t)=\mathbb E\left[\exp\left(-\frac{1}{\epsilon}\int\_0^t\big(X\_0+W\_s-1+f\_n(s)\big)^-ds\right)\right]$$
and the dominated convergence theorem, we may conclude the desired existence.
| 1 | https://mathoverflow.net/users/261243 | 416698 | 169,794 |
https://mathoverflow.net/questions/416691 | 7 | Bases of a topological space in point set topology will in general form a coverage on its category of inclusion on open subsets and on its category of inclusion on basic opens, but it takes a bit more work to check whether either forms a Grothendieck pretopology. Is there a useful or natural criterion for when a (point-set) basis does give a (Grothendieck) basis?
The criteria may apply either to the bases themselves, or to any particularly nice property of a topological space that forces some class of bases to have that property.
| https://mathoverflow.net/users/174368 | When is a basis of a topological space a Grothendieck pretopology? | This is a matter of expanding the definition, in this case Definition II.1.3 in SGA 4, which defines pretopologies.
By a “base” in this answer I mean what appears to be the most common definition: a collection of subsets of a fixed set $A$ such that any finite intersection of elements in the base is a union of elements in the base.
There are also multiple constructions of a site (i.e., a category with a coverage) from a base of a topological space $A$.
One can either (A) take the category of all open subsets of $A$,
or (B) the category whose objects are open subsets of $A$ that belong to the base.
For covering families of some object $V$, one can either take (a) those open covers of $V$ whose elements belong to the base, or (b) open covers of $V$ whose elements are given by the intersection of $V$ and some element of the base.
Altogether, there are three different options: A-a, A-b, B-a,
and only option A-b produces a pretopology in the sense of SGA 4.
Axiom PT0 for pretopologies says that any morphism in a covering family admits base changes.
Such base changes are always given by the corresponding intersection, provided that the intersection belongs to the category.
Thus, PT0 is satisfied for options A-a, A-b and not satisfied for option B-a.
If the base is closed under intersections, then PT0 is also satisfied for option B-a.
Axiom PT1 says that base changes of covering families are covering families.
In our case, the base change is given by the intersection with some open subset $V$.
Thus, PT1 is satisfied for option A-b and not satisfied for options A-a, B-a.
If the base is closed under intersections, then PT1 is also satisfied for options A-a, B-a.
Axiom PT2 says that covering families can be composed.
This is trivially true for the case under consideration.
Axiom PT3 says that the singleton family consisting of the identity map is a covering family, which is tautologically true in our case.
Thus, option A-b always gives a pretopology,
whereas options A-a, B-a give a pretopology
if and only if the base is closed under intersections of pairs.
In particular, open balls in a metric space as a coverage on the category of open subsets trivially form a pretopology using option A-b, and do not for a pretopology in options A-a, B-a.
| 11 | https://mathoverflow.net/users/402 | 416706 | 169,796 |
https://mathoverflow.net/questions/416709 | 1 | This is a [cross-post from MSE.](https://math.stackexchange.com/questions/4376145/affine-semigroup-generating-a-lattice)
Everything is assumed to be finite-dimensional. Let $S$ be a finitely generated affine semigroup (i.e. a subsemigroup of a lattice $N$ of a Euclidean space). Assume that $S$ generates $N$ as a group. Is it true that it contains all but a finite number of points in $\mathbb{R}\_{\ge 0} S\cap N?$
I believe the result should be known, however, I am not a specialist in convex geometry so book reference will be much appreciated.
| https://mathoverflow.net/users/143549 | Affine semigroup generating a lattice | (Write $\mathbf{N}=\{0,1,2,\dots\}$.)
What about the submonoid of $\mathbf{Z}^2$ generated by $\{(2,0),(3,0),(0,1)\}$? It equals $\mathbf{N}^2\smallsetminus (\{1\}\times\mathbf{N}$). So here the complement is infinite.
| 3 | https://mathoverflow.net/users/14094 | 416712 | 169,797 |
https://mathoverflow.net/questions/416704 | 3 | $\newcommand{\R}{\mathbb{R}}$I have a map $f\in C^0(X,\mathbb{R})$, where $X$ is a compact and Hausdorff topological space, which is a manifold outside of a compact subset $K\subset X$.
I would like to modify $f$, to obtain $\tilde f\in C^0(X,\mathbb{R})$ such that
1. $\tilde f\rvert\_K = f\rvert\_K$ and
2. $\tilde f\rvert\_{X\setminus K} \in C\_b^k(X\setminus K,\mathbb{R})$ where $C\_b^k$ denotes *bounded* functions continuously differentiable $k$-times (also the derivatives are bounded so to make this a Banach space).
>
> Is this possible or are the hypothesis on $X$ too weak?
>
>
>
**EDIT: as shown by Jochen Wengenroth the answer is negative, however as shown in Joseph Van Name's answer the answer is positive if we drop the requirement that the derivatives are bounded (i.e. instead of requiring $\tilde f\rvert\_{X\setminus K} \in C\_b^k(X\setminus K,\mathbb{R})$ we ask for $\tilde f\rvert\_{X\setminus K} \in C^k(X\setminus K,\mathbb{R})$). I accepted Joseph's answer without noticing this subtlety (beforee Jochen posting his answer) but, a posteriori, I regard both answers as relevant and on point as they give a complete picture.**
| https://mathoverflow.net/users/99042 | Smoothing a map $f:X\to \mathbb{R}$ while fixing it over a closed $C\subset X$ | I claim that it is always possible (as long as one does not require the derivatives of $\tilde{f}\_{X\setminus K}$ to be bounded). This is actually an easy consequence of the approximation theorem of manifolds.
To state the approximation theorem of manifolds, suppose that
$0\leq s<\infty$, and $M,N$ are manifolds. Let $C^{s}(M,N)$ be the collection of $C^{s}$ mappings from $M$ to $N$. Then we shall let $C^{s}\_{S}(M,N)$ denote the topological space with underlying set $C^{s}(M,N)$ and where the basic open sets are the sets of the following form.
Suppose that $\iota\_{n}:U\_{n}\rightarrow M$ is a chart for each $n\geq 0$. Suppose that $(\iota\_{n}[U\_{n}])\_{n\geq 0}$ is locally finite. Suppose furthermore that $C\_{n}$ is a compact subset of $\iota\_{n}[U\_{n}]$ and $\epsilon\_{n}>0$ for all natural numbers $n$. Let $f\in C^{s}(M,N)$. Suppose now that $j\_{n}:V\_{n}\rightarrow N$ is a chart for each $n\geq 0$ and that $f[C\_{n}]\subseteq j\_{n}[V\_{n}]$. Let $O$ be the collection of all functions $g\in C^{s}(M,N)$ such that
$g[C\_{n}]\subseteq j\_{n}[V\_{n}]$ and where if $\alpha$ is a multi-index with $|\alpha|\leq s$, then
$$|D^{\alpha}(j\_{n}^{-1}f\iota\_{n})(x)-D^{\alpha}(j\_{n}^{-1}f\iota\_{n})(x)|<\epsilon\_{i}$$ for each $x\in \iota\_{n}^{-1}[C\_{n}]$. Then $O$ is a basic open set, and all basic open sets are of this form.
Theorem (approximation theorem): Let $M,N$ be $C^{q}$ manifolds where $1\leq q\leq\infty$. Suppose that $0\leq p<q$. Then the set $C^{q}(M,N)$ is dense in the space $C\_{S}^{p}(M,N)$.
A good textbook reference for the above result is the text Differential Topology by Morris Hirsch.
Suppose now that $X$ is a compact space, $K$ is a compact subset of $X$, and $X\setminus K$ is a $C^{\infty}$-manifold.
Then there exists a continuous function $h:X\rightarrow[0,1]$ where
$K=h^{-1}[\{0\}]$.
Let $O$ be the collection of all continuous functions $g:X\setminus K\rightarrow\mathbb{R}$ with $|(g-f)(x)|<h(x)$ for all $x\in X$. Then $O$ is open in $C\_{S}^{0}(X\setminus K,\mathbb{R})$, so by the approximation theorem, there exists a $C^{\infty}$-mapping
$g:X\setminus K\rightarrow\mathbb{R}$ with $g\in O$, and in this case, $|(g-f)(x)|<h(x)$ for each $x\in X$.
Observe that if $k\in\partial K$, then $\lim\_{x\in X\setminus K,x\rightarrow k}(g-f)(x)=0$, so $\lim\_{x\in X\setminus K,x\rightarrow k}f(x)=\lim\_{x\in X\setminus K,x\rightarrow k}g(x)$. Therefore, if we set $\overline{g}=g\cup f|\_{K}$, then $\overline{g}$ is a continuous function that extends $g$.
Pierre PC observed that we do not need the full strength of the approximation theorem and that the weaker result that is easier to state and prove will suffice:
Proposition: Whenever $M$ is a $C^{k}-$manifold for $1\leq k\leq\infty$, $f:M\rightarrow\mathbb{R}$ is continuous, and $h:M\rightarrow(0,\infty)$ is a continuous function, there exists some $C^{k}$-function $g:M\rightarrow\mathbb{R}$ with $|f-g|<h$.
To prove the above proposition, first observe that if $U\subseteq\mathbb{R}^{n}$, $U$ is open, and $f:U\rightarrow\mathbb{R}$ is continuous, then for all $\delta>0$, there is a $C^{\infty}$-function $g\_{\delta}:\mathbb{R}^{n}\rightarrow[0,\infty]$ where $g\_{\delta}(x)=0$ whenever $\|x\|>\delta$ and where $\int\_{\mathbb{R}^{n}}g\_{\delta}(\mathbf{x})d\mathbf{x}=1$. In this case, each $f\*g\_{\delta}$ is $C^{\infty}$ on $\{\mathbf{x}\mid B\_{\delta}(\mathbf{x})\subseteq U\}$ and $f\*g\_{\delta}\rightarrow f$ uniformly on compact sets. One can extend this approximation result to obtain a proof of our proposition using a $C^{k}$-partition of unity.
| 5 | https://mathoverflow.net/users/22277 | 416718 | 169,799 |
https://mathoverflow.net/questions/416707 | 4 | Let $n>1$ be an integer. For $a,b\in \{0,1\}^n$ let $d\_h(a, b)$ denote the [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) of $a$ and $b$. For $k\in \{1,\ldots,n-1\}$ let $H(n,k)$ be the graph on $\{0,1\}^n$ given by the edge set $$E(n,k) = \{(a, b)\in (\{0,1\}^n)^2 : d\_h(a, b) = k\}.$$
**Question.** For what values of $k\in \{2,\ldots n-1\}$ does $H(n,k)$ have a Hamilton cycle?
**Note 1.** Hamilton cycles in $H(n,1)$ are called *Gray codes*.
**Note 2.** A necessary (but maybe not sufficient) condition for $H(n,k)$ to have a Hamilton cycle is that $\text{gcd}(k,2^n) = 1$, otherwise $H(n,k)$ is not a connected graph.
| https://mathoverflow.net/users/8628 | Hamilton cycles in $\{0,1\}^n$ with fixed Hamming distance | I believe these are exactly odd $k$.
Indeed, it can be easily seen that if $k$ is even, then a cycle starting at $0^n$ can visit only those vectors that have even number of $1$'s, and so it cannot be Hamiltonian.
On the other hand, there exist [long-run Gray codes](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v10i1r27) where any consecutive $\leq n-3\log\_2 n$ bit changes happening at distinct positions. So, for odd $k\leq n-3\log\_2 n$ traversing such a code with step $k$ produces a Hamiltonian cycle in $H(n,k)$.
Also, for an even $n$, inverting every second element of a cycle in $H(n,k)$ produces a cycle in $H(n,n-k)$.
It remains to address the case of odd $n$ and odd $k>n-3\log\_2 n$.
| 4 | https://mathoverflow.net/users/7076 | 416720 | 169,800 |
https://mathoverflow.net/questions/416681 | 3 | Let $X$ be a smooth projective variety of dimension $2n+1$, let $i\colon Y\subset X$ be an ample hypersurface, by Lefschetz hyperplane theorem, the pullback $i^\*\colon H^{2n}(X,\mathbb{Z})\to H^{2n}(Y,\mathbb{Z})$ is an injection.
When $X$ is a product of two projective spaces or more general (e.g., projective bundle over projective space), is it necessarily true that $i^\*\colon H^{2n}(X,\mathbb{Z})\to (H^{2n}(Y,\mathbb{Z})/H^{2n}(Y,\mathbb{Z})\_{\mathrm{tors}})$ is always a primitive embedding of lattices (i.e., the cokernel of $i^\*$ is torsion free)?
(The question rose when trying to calculate lattice of primitive middle cohomology of $Y$, and hopefully it is primitive embedding. For $X$ being projective space, there is a reference Thm 2.3 [The primitive cohomology lattice of a complete intersection](https://arxiv.org/abs/0909.5319). The primitivity is obtained by Thm 2.1 in "Libgober, J. Wood:
[On the topological structure of even-dimensional
complete intersections](https://doi.org/10.2307/1998676)".)
| https://mathoverflow.net/users/nan | Primitivity of the Lefschetz embedding | This is actually a general consequence of the Lefschetz hyperplane theorem and the universal coefficient theorem.
For simplicity, write $A\_{\operatorname{tf}} = A/A\_{\operatorname{tors}}$. Note that this association is functorial, additive, and preserves injections and surjections, but is not exact in the middle! It follows from the universal coefficient theorem that $H^k(X,\mathbf Z)\_{\operatorname{tf}} = \operatorname{Hom}(H\_k(X,\mathbf Z)\_{\operatorname{tf}},\mathbf Z)$.
**Lemma.** *Let $X$ be a smooth projective complex manifold of (complex) dimension $n+1$, and let $Y \subseteq X$ be a smooth hypersurface. Then the image of $H^n(X,\mathbf Z)\_{\operatorname{tf}} \hookrightarrow H^n(Y,\mathbf Z)\_{\operatorname{tf}}$ is saturated.*
*Proof.* By the Lefschetz hyperplane theorem, the map $H\_n(Y,\mathbf Z) \to H\_n(X,\mathbf Z)$ is surjective, hence the same goes for $H\_n(Y,\mathbf Z)\_{\operatorname{tf}} \to H\_n(X,\mathbf Z)\_{\operatorname{tf}}$. Since $H\_n(X,\mathbf Z)\_{\operatorname{tf}}$ is finite free, we may choose a splitting, realising $H\_n(X,\mathbf Z)\_{\operatorname{tf}}$ as a summand of $H\_n(Y,\mathbf Z)\_{\operatorname{tf}}$, and hence $H^n(X,\mathbf Z)\_{\operatorname{tf}}$ as a summand of the finite free module $H^n(Y,\mathbf Z)\_{\operatorname{tf}}$. In particular, it is saturated. $\square$
Alternatively, if you don't like making these choices, use the pairings between $H\_n$ and $H^n$ together with the push-pull formula
$$(i^\* \psi)(\beta) = \psi(i\_\* \beta)$$
for $\psi \in H^n(X,\mathbf Z)$ and $\beta \in H\_n(Y,\mathbf Z)$. If $n\phi = i^\* \psi$ for some $n > 1$ and some $\phi \in H^n(Y,\mathbf Z)\_{\operatorname{tf}}$, then surjectivity of $i\_\*$ shows that $\psi(\alpha)$ is divisible by $n$ for all $\alpha \in H\_n(X,\mathbf Z)$, thus $\psi$ is divisible by $n$ in $H^n(X,\mathbf Z)\_{\operatorname{tf}} = \operatorname{Hom}(H\_n(X,\mathbf Z),\mathbf Z)$.
| 1 | https://mathoverflow.net/users/82179 | 416721 | 169,801 |
https://mathoverflow.net/questions/416708 | 7 | I am cross-posting this question from math stack exchange (link below) as it has not received any comments or answers over the past month. A *regular epimorphism* is a morphism $f: X \to Y$ that is the coequalizer of some parallel pair $a,b: Z \to X$. I believe regular epimorphisms in the category of topological groups are precisely the surjective open maps. Is this also true for the category of topological rings? If not, is there some other characterization of the regular epimorphisms in the category $\mathbf{TopRing}$? A reference would be very much appreciated.
<https://math.stackexchange.com/questions/4357004/regular-epimorphisms-in-the-category-of-topological-rings>
| https://mathoverflow.net/users/46484 | Regular epimorphisms in the category of topological rings | It is true and hods more generlaly for models of any algebraic theory with a Mal'tsev operation, i.e. a trinary operation $f$ such that $f(x,x,y)=y=f(y,z,z)$. For groups, you can take $f(x,y,z)=xy^{-1}z$. This is Theorem 10 in A.I. Mal'tsev, On the general theory of algebraic systems, Mat. Sb. (N.S.), 1954, Volume (35)(77), Number 1, 3-20.
A proof in English can be found on page 396 of
V.V. Uspesnki$\check{\text i}$. The Ma'tsev operation on countably compact spaces. Comment. mat. Univ. Carolinae30 (1989) 395-402. available at <https://dml.cz/bitstream/handle/10338.dmlcz/106758/CommentatMathUnivCarol_030-1989-2_22.pdf>
The idea is to first simplify the situation of a regular epimorphism $X\to Y$ by using the fact it is the coequalizer of its kernel pair, hence the quotient of an equivalence relation on $X$, whence we want to show that the set of equivalence classes intersecting an open subset $U$ of $X$ is open, i.e. that any $y\in X$ equivalent to $x\in U$ has an open neighborhood of points $z$ equivalent to points in $U$.
But if this relation is kernel pair in the category of models of a theory,
then whatever operations the theory has, the equivalence relation has to respect. In particular, $x\sim y$ implies $f(x,y,z)\sim f(x,y,z)\sim f(x,x,z)=z$. Then because $f$ is continuous, there is an open that contains $z$ if and only if $f(x,y,z)\in U$, in which case $z\sim f(x,y,z)\in U$ for any $z$ in that open. Moreover, $x=f(x,y,y)$ implies the open contains $y$ and so is the desired neighborhood.
| 9 | https://mathoverflow.net/users/75650 | 416730 | 169,803 |
https://mathoverflow.net/questions/416699 | 1 | Here's the problem: start with $2^n$, then take away $\frac{1}{2}a^2+\frac{3}{2}a$ starting with $a=1$, and running up to $a = 2^{n+1}-2$, evaluating modulo $2^n$. Does the resulting sequence contain representatives for all the congruence classes module $2^n$?
Some examples of these sequences:
\begin{align\*}
n=2\quad & 2,3,3,2,0,1\\
n=3\quad & 6,3,7,2,4,5,5,4,2,7,3,6,0,1\\
n=4\quad & 14,11,7,2,12,5,13,4,10,15,3,6,8,9,9,8,6,3,15,10,4,13,5,12,2,7,11,14,0,1
\end{align\*}
It seems that the answer is yes but I want to show this holds for all natural numbers $n$. How does one do this?
| https://mathoverflow.net/users/265714 | Modular arithmetic problem | This is correct. Denote $f(x)=x(x+3)/2 \pmod {2^n}$. Then $f(x)=f(y)$ if and only if $x(x+3)-y(y+3)=(x-y)(x+y+3)$ is divisible by $2^{n+1}$. Since $x-y$ and $x+y+3$ have different parity, this in turn happens if and only if one of them is divisible by $2^{n+1}$. So, if $x=1,2,\ldots,2^{n+1}$, then $f(x)$ takes each value (i.e., each residue modulo $2^n$) exactly twice, as these numbers are distinct modulo $2^{n+1}$ and are partitioned onto pairs with sum $-3$ modulo $2^{n+1}$. Two of these pairs are $(2^{n+1}-2,2^{n+1})$ and $(2^{n+1}-1,2^{n+1}-2)$, thus if you remove a number from each pair (correspondingly, $2^{n+1}$ and $2^{n+1}-1$, so that the numbers $1,\ldots,2^{n+1}-2$ remain), the remaining numbers still cover all possible values of $f$.
| 4 | https://mathoverflow.net/users/4312 | 416758 | 169,812 |
https://mathoverflow.net/questions/416757 | 13 | There are a number of criteria for determining whether a polynomial $\in \mathbb{Z} [X]$ is irreducible over $\mathbb{Q}$ (the traditional ones being Eisenstein criterion and irreducibility over a prime finite field).
I was wondering if the decision problem of "Given an arbitrary polynomial $\in \mathbb{Z} [X]$, is it irreducible over $\mathbb{Q}$ or not?" is decidable or undecidable.
| https://mathoverflow.net/users/111528 | Is irreducibility of polynomials $\in \mathbb{Z} [X]$ over $\mathbb{Q}$ an undecidable problem? | There is a polynomial-time algorithm that decomposes any non-zero polynomial in $\mathbb{Q}[X]$ into irreducible factors. The algorithm is due to Lenstra–Lenstra–Lovász ([Factoring Polynomials with Rational Coefficients](http://www.digizeitschriften.de/dms/img/?PID=PPN235181684_0261%7Clog91)).
| 22 | https://mathoverflow.net/users/11919 | 416759 | 169,813 |
https://mathoverflow.net/questions/416765 | 1 | Suppose $X\sim \mathcal{N}(0,\sigma^2)$, find the expectation $\mathbb{E}\left[\frac{1}{(1+X^2)^a}\right]$ where $a$ is a fixed positive real number.
Is there an explicit formula for the above expectation?
| https://mathoverflow.net/users/111204 | Expected value of a function of normal random variable | According to Mathematica, the expectation is
$$\frac{\Gamma \left(a-\frac{1}{2}\right) \, \_1F\_1\left(\frac{1}{2};\frac{3}{2}-a;\frac{1}{2 \sigma^2}\right)}{\sqrt{2} \sigma
\Gamma (a)}+\frac{2^{-a} \sigma^{-2 a} \Gamma \left(\frac{1}{2}-a\right) \, \_1F\_1\left(a;a+\frac{1}{2};\frac{1}{2
\sigma^2}\right)}{\sqrt{\pi }}.$$
| 1 | https://mathoverflow.net/users/36721 | 416772 | 169,817 |
https://mathoverflow.net/questions/416754 | 24 | Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. A standard fact is that $G$ is generated by $\exp(\mathfrak{g})$, i.e. every $g \in G$ can be written as $g=\exp(x\_1)\cdots\exp(x\_n)$ for some $x\_i\in\mathfrak{g}$. A natural follow-up question is whether there is a bound on the number $n$ of Lie algebra elements. I suppose this depends on the group, so let me formulate my question as follows.
**Question.** *What are natural conditions on a Lie group $G$ which imply that there is a finite number $n$ such that
$$G = \underbrace{\exp(\mathfrak{g}) \cdots \exp(\mathfrak{g})}\_{n \text{ times}}?$$
For instance, is this true for complex semisimple groups?*
Of course, there are many cases in which $\exp$ is surjective (e.g. $G$ compact or nilpotent) so that $n = 1$.
I'm looking for a larger class of groups, preferably in the linear algebraic category, or just interesting specific examples where $n > 1$.
| https://mathoverflow.net/users/385475 | Lie groups generated by finitely many Lie algebra elements | $\def\fg{\mathfrak{g}}$The following answer is a paraphrase of material from
*Philips, Christopher N.*, [**How many exponentials?**](http://dx.doi.org/10.2307/2375057), Am. J. Math. 116, No. 6, 1513-1543 (1994). [ZBL0839.46054](https://zbmath.org/?q=an:0839.46054).
We'll say that a Lie group $G$ has **exponential rank** $\leq k$ if every element of $G$ is a product of $\leq k$ exponentials.
We'll say that $G$ has **exponential rank** $\leq k+\epsilon$ if the set of products of elements $\leq k$ exponents is dense in $G$. In other words, in a group of exponential rank $\leq k+\epsilon$, for any open neighborhood $\Omega$ of the identity, we have $G = \exp(g\_1) \exp(g\_2) \cdots \exp(g\_k) u$ for $g\_1$, $g\_2$, ..., $g\_k\in\fg$ and $u \in \Omega$. Since it is known that $\exp(\fg)$ contains an open neighborhood of the identity, any group of exponential rank $\leq k+\epsilon$ also has exponential rank $\leq k+1$. We define the exponential rank of $G$ to be the minimum $r$ in $\{ 1 < 1+\epsilon < 2 < 2+\epsilon < 3 < \cdots \}$ such that $G$ has exponential rank $\leq r$.
Note that, if $G = AB$ (for Lie subgroups $A$ and $B$) and the exponential ranks of $A$ and $B$ are $(a,b)$, $(a+\epsilon, b)$, $(a, b+\epsilon)$ or $(a+\epsilon, b+\epsilon)$, then the exponential rank of $G$ is bounded above by $a+b$, $a+b+\epsilon$, $a+b+\epsilon$, $a+b+\epsilon$ respectively.
In this vocabulary, Phillips proves (Section 5) that
* Connected compact groups have exponential rank $\leq 1$.
* Unipotent groups have exponential rank $\leq 1$.
* Connected complex groups have exponential rank $\leq 1+\epsilon$. This bound cannot be improved, because $SL\_2(\mathbb{C})$ has exponential rank $1+\epsilon$ (the matrix $\left[ \begin{smallmatrix} -1&1 \\ 0&-1 \end{smallmatrix} \right]$ is not an exponential)).
* Connected solvable groups have exponential rank $\leq 1+\epsilon$. This bound cannot be improved: Consider the Borel $\left[ \begin{smallmatrix} z & u \\ 0 & z^{-1} \\ \end{smallmatrix} \right]$ inside $SL\_2(\mathbb{C})$ and the same matrix as before.
* All connected Lie groups have rank $\leq 2+\epsilon$.
* $SL\_2(\mathbb{R})$ has exponential rank $2$.
The proof that all connected Lie groups have rank $\leq 2+\epsilon$ is credited to Djokovic: The point is that any connected $G$ has an Iwasawa decomposition $K(ANR)$ where $K$ is compact connected (so exponential rank $\leq 1$) and $ANR$ is connected solvable (so exponential rank $\leq 1+\epsilon$).
Phillips states it as an open problem whether there is a connected Lie group of exponential rank $2+\epsilon$, or whether, in fact, exponential rank $2$ is always enough.
| 26 | https://mathoverflow.net/users/297 | 416776 | 169,819 |
https://mathoverflow.net/questions/396064 | 17 | In the 1970s Ol'shanskii constructed a non-cyclic finitely generated group $G$ with the following properties:
1. Every proper, non-trivial subgroup of $G$ is infinite cyclic.
2. If $X^m=Y^n$ for $X, Y\in G$ with $m,n\neq0$, then $\langle X, Y\rangle$ is cyclic i.e., any two maximal subgroups of $G$ have trivial intersection.
Ol'shanskii gave an easy proof that such a group is simple, which roughly goes: Suppose $N$ is a proper, non-trivial normal subgroup of $G$. If $N$ is maximal then $G/N$ is cyclic of prime order, so $G$ is virtually-$N$, so $G$ is a torsion-free virtually-$\mathbb{Z}$ group, so must itself be cyclic. If $N$ is not maximal then $N$ is contained in a maximal subgroup $M$ such that $M^g\cap M\neq1$ for all $g\in G$, so as $M^g$ is also maximal and as maximal subgroups intersect trivially (by (2)) we have that $M^g=M$ for all $g\in G$, i.e. $M$ is normal in $G$, which is impossible by the previous case.
Property (2) was used here. I was wondering if this can be dropped. So:
**Question.** Suppose $G$ is a non-cyclic finitely generated group with every proper, non-trivial subgroup of $G$ infinite cyclic. Is $G$ simple?
---
If $G$ instead satisfies that it is infinite and every proper, non-trivial subgroup has order $p$ for a fixed prime $p$ then $G$ is a "Tarski monster" group and is indeed simple: If $N$ is a proper, non-trivial normal subgroup of $G$ and $g\not\in N$ then $N\cap\langle g\rangle=1$, as both subgroups have prime order, so $N\langle g\rangle=N\rtimes\langle g\rangle$ has order $p^2$, a contradiction. However, this proof uses primality so does not extend to the setting here.
| https://mathoverflow.net/users/6503 | Are groups with every proper, non-trivial subgroup infinite cyclic simple? | The answer is no, there exist non-simple torsion-free Tarski monsters. Theorem 31.4 of Ol’shanskii's book "Geometry of defining relations in groups" [ZBL0676.20014](https://zbmath.org/?q=an:0676.20014) [MR1191619](https://mathscinet.ams.org/mathscinet-getitem?mr=1191619) is:
**Theorem.** There is a non-abelian group all of whose proper subgroups are infinite cyclic and the intersection of any two of them is non-trivial.
Such a group necessarily has infinite cyclic centre: take any non-commuting $x$ and $y$, then the centralizer of $\langle x \rangle \cap \langle y \rangle \cong \mathbb{Z}$ contains $\langle x, y \rangle$, which is the whole group. Thanks to Ashot Minasyan for pointing out Ol’shanskii's theorem.
| 9 | https://mathoverflow.net/users/24447 | 416786 | 169,822 |
https://mathoverflow.net/questions/416762 | 5 | For which $(a,b,n) \in \mathbb{Z}^3$ satisfying $a+b=n$ does $\frac{\Gamma(z+1)}{\Gamma(x+1)\Gamma(y+1)}$ approach a limit as $(x,y,z) \rightarrow (a,b,n)$ in $\mathbb{C}^3$, and what is that limit? (The case where $a$, $b$, and $n$ are all negative is the problematic one.) What if we restrict to $\{(x,y,z) \in \mathbb{C}^3 \ | \ x+y=z\}$?
I'm guessing that where the limit exists it agrees with the "windmill" shown in <https://i.stack.imgur.com/osFsj.png> (taken from page 197 of Hilton, Holton, and Pedersen's "Mathematical Reflections: In a Room with Many Mirrors").
| https://mathoverflow.net/users/3621 | Extended binomial coefficients and the gamma function | There's nothing special about the gamma function; the failure of the limit to exist when $a$, $b$, and $n$ are negative is exactly the same as the failure of
$$\lim\_{(x,y,z)\to(0,0,0)} \frac{xy}{z}$$ to exist.
I will renormalize in a way that I think makes clearer what's going on. First let's look at
$$\lim\_{(x,y,z)\to(0,0,0)} \frac{\Gamma(z)}{\Gamma(x)\Gamma(y)}.$$
We have
$$\frac{\Gamma(z)}{\Gamma(x)\Gamma(y)} = \frac{xy}{z}\frac{\Gamma(z+1)}{\Gamma(x+1)\Gamma(y+1)}.$$
Since $\Gamma(z+1)/\Gamma(x+1)\Gamma(y+1)$ is analytic and nonzero near $(0,0,0)$, $\Gamma(z)/\Gamma(x)\Gamma(y)$ behaves just like $xy/z$ near $(0,0,0)$. In particular, if you restrict to $x+y=z$, this is $xy/(x+y)$, for which the limit as $(x,y)\to(0,0)$ does not exist.
Now let's consider
$$\lim\_{(x,y,z)\to(0,0,0)} \frac{\Gamma(z-a)}{\Gamma(x-b)\Gamma(y-c)}$$ where $a$, $b$, and $c$ are nonnegative integers.
We have
$$\frac{\Gamma(z-a)}{\Gamma(x-b)\Gamma(y-c)} = \frac{\Gamma(z)}{\Gamma(x)\Gamma(y)}\cdot
\frac{(x-b)\cdots (x-1)(y-c)\cdots (y-1)}{(z-a)\cdots(z-1)}.$$
The second factor on the right is analytic and nonzero near $(0,0,0)$, so $\Gamma(z-a)/\Gamma(x-b)\Gamma(y-c)$ behaves just like $\Gamma(z)/\Gamma(x)\Gamma(y)$ (and thus like $xy/z$) near $(0,0,0)$.
| 6 | https://mathoverflow.net/users/10744 | 416792 | 169,824 |
https://mathoverflow.net/questions/416793 | 3 | A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H =(V, E)$ consists of a set $V$ and a set $E \subseteq {\cal P}(V)$ of subsets of $V$. A *hypergraph coloring* is a map $c: V\to \kappa$, where $\kappa \neq \emptyset$ is a cardinal and the restriction $c\restriction\_e: e \to \kappa$ is non-constant whenever $e$ has more than $1$ element. The *chromatic number* $\chi(H)$ is the least cardinal $\kappa \neq \emptyset$ such that there is a coloring $c: V \to \kappa$.
The map $c:V\to \kappa$ is said to be a *conflict-free* if every (non-empty) edge $e\in E$ contains at least one vertex $v$ of a color unique in $e$, or more formally, if there is $v\in e$ such that $$e \;\cap\; \bigl(c^{-1}(\{c(v)\})\bigr) = \{v\}.$$ The *conflict-free chromatic number* $\chi\_{\text{cf}}(H)$ is the least cardinal $\kappa \neq \emptyset$ such that there is a conflict-free coloring $c: V \to \kappa$.
(Conflict-free colorings were motivated by a frequency assignment problem in cellular networks, see the introduction of [Pach and Tardos - Conflict-free colorings of graphs and hypergraphs](https://www.math.nyu.edu/%7Epach/publications/ConflictFreeGraph052909.pdf).)
Of course, every conflict-free coloring is a coloring in the "traditional" sense, so $\chi(H) \leq \chi\_{\text{cf}}(H)$.
Let $\tau$ be the Euclidean topology on $\mathbb{R}$. The chromatic number $\chi(\mathbb{R}, \tau)$ equals $2$ (see [Chromatic number of a connected Hausdorff space](https://mathoverflow.net/questions/322042/chromatic-number-of-a-connected-hausdorff-space)).
**Question.** What is $\chi\_{\text{cf}}(\mathbb{R},\tau)$?
| https://mathoverflow.net/users/8628 | Conflict-free coloring of $\mathbb{R}$ with the Euclidean topology | By colouring every element of $\Bbb Q$ with a unique colour and every element of $\Bbb R\setminus\Bbb Q$ with the same colour (different from all the colours used so far), we see that $\chi\_\mathrm{cf}(\Bbb R)\leq\aleph\_0$.
But in fact $\aleph\_0$ is a lower bound in any reasonable space.
**Lemma 1:** Let $X$ be a $T\_1$ space with at least three points. Then $\chi\_{\mathrm{cf}}(X)>2$.
**Proof:** Suppose for a contradiction $\chi\_{\mathrm{cf}}(X)=2$ as witnessed by $c\colon X\to 2$. Let $x\in X$ be such that $c(x)\neq c(x')$ for all $x'\in X\setminus\{x\}$. But now we have that $X\setminus\{x\}$ is a monochromatic open set in $X$ with at least two points, a contradiction.
**Lemma 2:** Let $X$ be an infinite $T\_1$ space. Then $\chi\_{\mathrm{cf}}(X)\geq\aleph\_0$.
**Proof:** Suppose for a contradiction $\chi\_{\mathrm{cf}}(X)=n$ as witnessed by $c\colon X\to n$. Let $x\_1\in X$ be such that $c(x)\neq c(x')$ for all $x'\in X\setminus\{x\_1\}$. Now $c\upharpoonright X\setminus\{x\_1\}$ witnesses that $\chi\_{\mathrm{cf}}(X\setminus\{x\_1\})\leq n-1$. Proceed inductively to find $x\_1,\ldots,x\_{n-2}$ so that $c\upharpoonright X\setminus\{x\_1,\ldots,x\_{n-2}\}$ witnesses that the latter space has conflict-free chromatic number at most $2$, which contradicts Lemma 1.
In conclusion $\chi\_{\mathrm{cf}}(\Bbb R)=\aleph\_0$.
| 6 | https://mathoverflow.net/users/49381 | 416798 | 169,825 |
https://mathoverflow.net/questions/361757 | 1 | Let $ (\Lambda\_n) $ be a family of lattices, $ \Lambda\_n \subset \mathbb{Z}^n $, with $ \det\Lambda\_n \sim n $ as $ n \to \infty $ (meaning $ \lim\_{n\to\infty} n^{-1} \det\Lambda\_n = 1$). I am interested in the asymptotics of the number of points of $ \Lambda\_n $ in the hypercube $ [0,2)^n $. In particular, is it true that: $$ |\Lambda\_n \cap [0,2)^n| = |\Lambda\_n \cap \{0,1\}^n| \sim \frac{2^n}{n} $$ as $ n \to \infty $? Does someone know how to prove this type of statement? Would an additional condition on $ \Lambda\_n $, such as a lower bound on its minimum, be helpful in establishing such an asymptotic result?
| https://mathoverflow.net/users/83189 | Lattice points in hypercubes | The following answer is from [A Reverse Minkowski Theorem](https://arxiv.org/pdf/1611.05979.pdf). It deals with a sphere, rather than a hypercube. I am unaware of extensions to hypercubes, but the lower bound I will quote from it at least seems non-trivial, and of interest.
Minkowski's first theorem has a "point-counting" version, due to Blichfeldt and van der Corput.
>
> For any lattice $L \subseteq \mathbb{R}^n$ with $\det(L) \leq 1$ and $r > 0$,
> $$|L \cap rB\_2^n| \geq 2^{-n}\mathsf{vol}(rB\_2^n) = \frac{1}{\sqrt{\pi n}}\left(\frac{\pi e r^2}{2n}\right)^{n/2}(1 + o(1)).$$
>
>
>
This result is apparently given for arbitrary norms, so look at the linked paper for the proper citation/statement for $\ell\_\infty$.
Of course, to get an asymptotic, one needs an upper bound as well.
As the comments mention, one needs to rule out "skew" lattices (say of the form $t\mathbb{Z}\oplus \frac{1}{t}\mathbb{Z}$) somehow. The authors of the linked paper do this via the notion of *stable lattices*.
>
> A lattice $L\subseteq \mathbb{R}^n$ is *stable* if $\det L = 1$, and $\det(L') \geq 1$ for all sublattices $L'\subseteq L$.
>
>
>
They then give upper bounds for a number of point-counting contexts (assuming the underlying lattice is stable) in corollary 1.4. I will not copy them all here (there are 3 cases depending on what the particular value of $r$ one has). None of them appear to obviously (asymptotically) match the above upper bound (and in [later work](https://arxiv.org/pdf/2201.05201.pdf) on a similar problem, but without an obvious point-counting application, the authors mention the work I describe above is non-tight).
This is answer is mostly to say that there is recent work towards establishing these kinds of bounds, but I do not believe the hypercube has been examined, nor do I believe tight point-counting asymptotics have been established.
The assumption authors have found useful to get rid of the "skew" lattices described is known as *stability*, and has been used in other contexts going back a few decades.
| 1 | https://mathoverflow.net/users/101207 | 416810 | 169,830 |
https://mathoverflow.net/questions/416818 | 4 | Let $X$ denote the sequence [A200246](https://oeis.org/A200246): an $i$-th element of $X$ is equal to $w(p\_i) \bmod 2$, where $w(p\_i)$ is the number of ones in the base-$2$ representation of an $i$-th prime.
The first $564163$ bits of $X$ (corresponding to all primes less than $2^{23}$) contain $274615$ zeros and $289548$ ones.
The first $1077871$ bits of $X$ (corresponding to all primes less than $2^{24}$) contain $520574$ zeros and $557297$ ones.
It seems that the prevalence of ones is too visible to be ignored. How to explain this? What is the value of $$\lim\_{n \to \infty} \frac{f\_0(n)}{f\_1(n)},$$ where $f\_b(n)$ denotes the number of elements with value $b$ in the initial bits of $X$ corresponding to all primes less than $2^{n}$ for a positive integer $n > 1$ (if this value converges)?
| https://mathoverflow.net/users/122796 | Parities of binary weights of primes | [Mauduit and Rivat](https://annals.math.princeton.edu/2010/171-3/p04) showed that the sum of the binary digits of primes are equally distributed between the possibilities $0\bmod 2$ and $1\bmod 2$. So the limit you want is $1$. More generally they consider the distribution $\bmod q$ of the sums of digits of primes written in any base $b$.
| 8 | https://mathoverflow.net/users/38624 | 416824 | 169,834 |
https://mathoverflow.net/questions/416807 | 10 | In modern treatments of statistical mechanics, the natural base is conventionally used for the Gibbs and Boltzmann entropy without careful justification. While I am aware that the properties of the Shannon entropy are invariant to the choice of base of the logarithm, I suspect that physicists might have careful theoretical justifications for this particular choice.
>
> Question: Might there be a sound theory behind this convention?
>
>
>
So far a couple reasons occurred to me:
1. Stirling's log-factorial approximation:
\begin{equation}
\ln N! \sim N \cdot \ln N - N \tag{1}
\end{equation}
which is an essential tool in statistical mechanics.
2. The exponential function diagonalises the derivative operator:
\begin{equation}
\forall \lambda \in (0,1), \frac{\partial}{\partial H} \exp(\lambda \cdot H) = \lambda \cdot \exp(\lambda \cdot H) \tag{2}
\end{equation}
which may be useful whenever one wants to analyse variations in the exponential of entropy. The advantage of the exponential of entropy is that it is parameterisation invariant [as pointed out by Tom Leinster on a related question](https://mathoverflow.net/questions/398217/geometric-interpretations-of-the-exponential-of-entropy).
As there might be subtle reasons I have ignored, any useful references on this question are more than welcome.
References:
-----------
1. von Neumann, John (1932). Mathematische Grundlagen der Quantenmechanik (Mathematical foundations of quantum mechanics) Princeton University Press., . ISBN 978-0-691-02893-4.
2. Landau and Lifshitz. Statistical Physics. Butterworth-Heinemann. 1980.
3. David J.C. MacKay. Information Theory, Inference and Learning Algorithms. Cambridge University Press 2003.
| https://mathoverflow.net/users/56328 | The origin of the natural base in statistical mechanics | As Matt F. points out, we could just absorb a change of base of the logarithm into the coefficient. The reason that is not convenient in physics is that we would like the same coefficient $k$ to appear in the [ideal gas law](https://en.wikipedia.org/wiki/Boltzmann_constant#Roles_of_the_Boltzmann_constant) $pV=NkT$.
Note that in physics entropy changes can be defined in thermodynamic rather than statistical terms, as the absorbed heat in a reversible transition ($\Delta S=\int dQ/T$). So any change in the base of the logarithm has to be compensated by a change in the coefficient $k\mapsto k'$. And if we would define the entropy $S=k'\log\_2 W$ using the base-2 logarithm, the ideal gas law would read $pV\ln 2=Nk'T$.
| 13 | https://mathoverflow.net/users/11260 | 416840 | 169,838 |
https://mathoverflow.net/questions/416764 | 1 | Let $X\_k$ be a sequence of *iid* Bernoulli random variables of parameter $p$ and let $\hat{X}\_n=\frac1n\sum\_{k=1}^nX\_k$. Hoeffding's inequality states that for any $n$:
$$\mathbb{P}(\hat{X}\_n - p \ge \epsilon) \le e^{-2n\epsilon^2},$$
or said otherwise:
$$\mathbb{P}(\hat{X}\_n - p \ge \frac{\epsilon}{\sqrt{n}}) \le e^{-2\epsilon^2}.$$
The above holds if $n$ is deterministic. My question is: does it still hold if $n$ is a bounded stopping time? By bounded, I mean that there exists $T$ such that $n\le T$ a.s.
| https://mathoverflow.net/users/90045 | Hoeffing inequality is not true for stopping time | In fact, the answer is indeed contained in the question:
Let $N$ be the stopping time:
$$ N = \inf\{n : \hat{X}\_n - p \ge \frac{\epsilon}{\sqrt{n}}\}.$$
$N$ is finite almost surely (for instance, this is a consequence of the law of iterated logarithm). This shows that $$P( \hat{X}\_N - p \ge \frac{\epsilon}{\sqrt{N}} ) = 1,$$
which violates Hoeffding’s inequality for all $\epsilon$ such that $e^{-2\varepsilon^2}<1$.
Simulation indicates that the result is similar for $\tilde{N} = \min(N, T)$, when $T$ is large enough (e.g. when $p=0.5$, $\epsilon=1$ and $T=100$).
| 1 | https://mathoverflow.net/users/90045 | 416851 | 169,841 |
https://mathoverflow.net/questions/416855 | 2 | I am looking for an upper bound $R=R\_{n,\varepsilon}$ such that for given $\varepsilon>0$ and real numbers $\alpha\_1, \dotsc, \alpha\_n$ in, say, $[1,2]$, there is $x\in [1,R]$ such that
$$
\frac 1n\sum\_{k-1}^n\cos(\alpha\_k x)\geq 1-\varepsilon
$$
Dirichlet's principle states that $x$ can be (an integer) smaller than $\varepsilon^{-n}$.
We are actually interested in the case where the $\alpha\_i$ are iid, say, uniform over $[1,2]$, so pathological cases of badly approximable $n$-tuples can be ignored, and a result holding with high probability would be great.
The main problem is that $n$ goes to infinity, our feeling is that we could get something in $\varepsilon^{-n/2}$, using Fubini's theorem, but it is not clear.
| https://mathoverflow.net/users/16934 | When does a random trigonometric sum approximate $1$? | $\newcommand\ep\varepsilon\newcommand\al\alpha$Let $X\_k:=\cos(\alpha\_kx)$. Then $X\_1,X\_2,\dots$ are iid random variables such that $P(X\_1<1)=1$ and hence $\mu:=EX\_1<1$, so that $\mu<1-\varepsilon$ if $\varepsilon\in(0,1-\mu)$. So, by the law of large numbers, with high probability we get the inequality
$$
\frac 1n\sum\_{k=1}^n\cos(\alpha\_k x)<1-\varepsilon,
$$
opposite to your desired one, for all large enough $n$.
So, your idea of randomizing the $\alpha\_k$'s to exclude "pathological cases" is not working.
---
On the other hand, we can get the upper bound $2\pi\lceil(2\pi^2/\ep)^{n/2}\rceil$ on $x$ even without randomization. Indeed, by the [simultaneous version of Dirichlet's approximation theorem](https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem#Simultaneous_version), for any real $\al\_1,\dots,\al\_n$ and any natural $N$ there exist integers $p\_1,\dots,p\_n,q$ such that $1\le q\le N$ and $|2p\_i\pi-2q\pi\al\_i|\le2\pi/N^{1/n}$ for all $i\in[n]:=\{1,\dots,n\}$. So, letting $x:=2q\pi$ and using the inequality $\cos(2p\pi)-\cos(2p\pi+t)\le t^2/2$ for integral $p$ and real $t$, we have
$$1-\cos(\al\_k x)=\cos(2p\_k\pi)-\cos(2q\pi\al\_i)
\le2\pi^2/N^{2/n}\le\ep$$
for all $k\in[n]$ if $N=\lceil(2\pi^2/\ep)^{n/2}\rceil$.
Therefore and because $1\le x=2q\pi\le2\pi N$, we get the upper bound $2\pi\lceil(2\pi^2/\ep)^{n/2}\rceil$, of the desired form, on the smallest number $x\ge1$ such that
$$
\frac 1n\sum\_{k-1}^n\cos(\al\_k x)\ge1-\ep.
$$
| 3 | https://mathoverflow.net/users/36721 | 416856 | 169,843 |
https://mathoverflow.net/questions/416823 | 2 | Suppose $\Omega\subset\mathbb{R}^3$ is a bounded domain with smooth boundary. We note by $(-\Delta)^{-1}$ the inverse Laplacian i.e. $f\mapsto u$ where $u$ is the unique solution to
$$-\Delta u=f,\quad u\_{|\partial\Omega}=0$$
Now consider the elliptic operator $\Delta + b\cdot\nabla(-\Delta)^{-1}$ associated with, say, homogeneous Dirichlet boundary conditions, where $b$ is a smooth vector field (you can even take $b$ to be a constant vector field if that somehow makes this question answerable).
Is spectral analysis for such an operator possible? (i.e. existence of eigenvalues, properties of eigenfunctions, etc) I ask because such an operator (or rather a slightly more complicated version of it) appears as the linearization of a nonlinear operator that I'm studying. Any help/references would be greatly appreciated.
| https://mathoverflow.net/users/166785 | Spectral analysis for nonlocal elliptic operator | So that's
$$
-\Delta u -b\cdot\nabla v = \lambda u
$$
$$
-\Delta v + u =0
$$
which you can write, with $U=(u,v)$
$$
-\Delta U + B \nabla U + C U = \lambda Q U
$$
where $\Delta$ is applied line wise, and so is the gradient, whereas
$$
B=\begin{pmatrix} 0& b\cdot \\ 0 &0\end{pmatrix}, C=\begin{pmatrix} 0& 0 \\ 1 &0\end{pmatrix}\text{ and }Q=\begin{pmatrix} 1& 0 \\ 0 &0\end{pmatrix}.
$$
In the form you wrote it, you see immediately that there is an associated compact operator (hence properties of the spectrum), and that the problem is not self adjoint. Whether it is a cooperative or a non cooperative system will let you prove properties on the eigensolutions.
In any case, it is relatively nice, because it is weakly coupled, that is, the second order operator applies to each line separately. [Mitidieri and Sweers](https://onlinelibrary.wiley.com/doi/10.1002/mana.19951730115) have looked at these type of problems a while ago. More recent work probably quotes them.
| 2 | https://mathoverflow.net/users/40120 | 416861 | 169,845 |
https://mathoverflow.net/questions/416868 | 2 | Let $X$ be a compact metric space and $Y\subset X$ be a compact set. Assume that $f\_1, f\_2: Y \to \mathbb{P}\mathbb{R}^2$ are continuous functions. Let $N \subset \mathbb{P}\mathbb{R}^2$ be a countable set.
### Question
Is there a Borel measurable function $T:f\_1(Y) \to f\_2(Y)$ such that if $x \in Y$ and $f\_1(x) \notin N$, then $T(f\_1(x))=f\_2(x)$?
| https://mathoverflow.net/users/168351 | Existence of a Borel measurable function | Let $Y=[-1,1]$. We can also view $Y$ by homeomorphic embedding and abuse of notation as a subset of $\mathbb{P}\mathbb{R}^2$. For all $x\in Y$, let $f\_1(x)=|x|$ and $f\_2(x)=x$. For all but the countably many $x$ such that $f\_1(x)=|x|\in N$, one would need $$x=f\_2(x)=T(f\_1(x))=T(|x|)=T(|-x|)=T(f\_1(-x))=f\_2(-x)=-x,$$
which is impossible.
| 2 | https://mathoverflow.net/users/35357 | 416871 | 169,847 |
https://mathoverflow.net/questions/416490 | 3 | I define a subset $M$ of $\mathbb R^n$ to be a "homogeneous Euclidean manifold" if:
* it is a closed connected smooth submanifold of $\mathbb R^n$,
* for every $p, q$ in $M$, there is a Euclidean isometry $f$ of $\mathbb R^n$ sending $p$ to $q$ fixing $M$ (i.e., $f(M)=M$ and $f(p)=q$).
The problem is to classify the "homogeneous Euclidean manifolds".
I know there are several definitions of symmetric and homogeneous and that these spaces have been more or less classified so it may well be that my question is just a particular case.
| https://mathoverflow.net/users/477343 | Classification of "homogeneous" submanifolds of ℝⁿ | Such $M$ are called extrinsically homogeneous submanifolds of euclidean space.
Reference: for example the book by Berndt-Console-Olmos, Submanifolds and holonomy.
The set of all euclidean isometries that fix $M$ is a closed subgroup $G$ of the group of euclidean motions $E(n)$, and $M$ is an orbit of $G$. So the question is to classify all (compact connected) orbits of all closed subgroups of $E(n)$.
Since $M$ is assumed to be bounded, it has a euclidean center of mass, which one can assume to be the origin. That center of mass is a common fixed point of $G$. So $G$ is a closed subgroup of the orthogonal group, and the orbit $M$ is contained in a sphere around the origin. These are examples of isoparametric submanifolds of the sphere (Edit: This is not correct, see the comment by Claudio Gorodsky).
Classification is probably asking too much. For example, all compact homogeneous spaces occur as such orbits in the sphere (Mostow).
| 1 | https://mathoverflow.net/users/127309 | 416872 | 169,848 |
https://mathoverflow.net/questions/416556 | 4 | For an elliptic curve $E/\mathbb{Q}$, let us denote by $\Delta\_{\min}(E)$ the minimal discriminant of $E$ and $N(E)$ the conductor of $E$. Then it is well-known that $N(E) | \Delta\_\min(E)$.
The *Szpiro ratio* of $E$ is defined as the ratio
$$\displaystyle \beta\_E = \frac{\log |\Delta\_\min(E)|}{\log N(E)}.$$
This definition is motivated by *Szpiro's conjecture*, which asserts that for all $\epsilon > 0$ there exists only finitely many curves $E/\mathbb{Q}$ satisfying
$$\displaystyle \beta\_E > 6 + \epsilon.$$
Of course, it is known that Szpiro's conjecture is equivalent to the $abc$-conjecture.
Note that the value of $6$ in the conjecture is sharp, because one can construct explicit families with Szpiro ratio converging to $6$. For example, if $E$ is given by a short Weierstrass model
$$\displaystyle E\_{A,B} : y^2 = x^3 + Ax + B, A, B \in \mathbb{Z}$$
with the property that for all primes $p$ we have $p^4 | A \Rightarrow p^6 \nmid B$, then the discriminant is given by $\Delta(E) = 16(4A^3 - 27B^2)$. If $\Delta(E)$ is $12$-power-free, then $\Delta(E) = \Delta\_\min(E)$. We can then find an integer $c\_1$ such that the equation
$$\displaystyle 16(4x^3 - 27y^2) = c\_1 z^6$$
defines an elliptic generalized Fermat equation with a solution, which then gives at least one family of parametrized solutions. In this family, the Szpiro ratio will approach $6$.
Similar constructions can be given for the following families: elliptic curves with a rational 2-torsion point, with 3 rational 2-torsion points, and those with an abelian 2-torsion field.
My question is: is the Szpiro ratio $\beta\_E$, as $E$ varies over all $E/\mathbb{Q}$, dense in the interval $[1,6]$?
| https://mathoverflow.net/users/10898 | Szpiro ratios of elliptic curves over $\mathbb{Q}$ | The answer is "yes". Use one of the various families with Szpiro ratio exceeding $6$ and large conductor $N$, and then twist by a prime $p$ of appropriate size, coprime to $N$ (which increases the conductor by $p^2$ and the minimal discriminant by $p^6$). With a certain amount of care, this gives density in the interval $[3,6]$. A similar argument, starting with a curve of Szpiro ratio arbitrarily close to $1$ (there are various ways to construct these) and large conductor then gives density in $[1,3]$.
| 4 | https://mathoverflow.net/users/7302 | 416876 | 169,850 |
https://mathoverflow.net/questions/416883 | 5 | The Kodaira($-$Iitaka) dimension of a line bundle $L$ on a complex manifold $X$ can be defined either in three ways:
1. The maximal dimension of the image of the rational maps $φ\_{|mL|} : X \dashrightarrow \mathbb{P}(H^0(X, m L)^\*)$
2. The unique integer $k$ such that $h^0(m L) = O(m^k)$.
3. $\operatorname{trdeg} \operatorname{Frac} \left( \bigoplus\_{m ≥ 0} H^0(X, mL) \right) - 1$
I know why the first two definitions are equivalent but I'm struggling to find a reference treating the third. Several people cite Ueno's book *Classification Theory of Algebraic Varieties and Compact Complex Spaces* for this, but as far as I can see he only proves that first two are equivalent. Can anyone provide a proof of $(1) \iff (3)$ or $(2) \iff (3)$ or point me to a reference?
| https://mathoverflow.net/users/123448 | Equivalent definitions of Kodaira dimension | For $(1) \iff (3)$, we have the following chain of identities:
1. $\operatorname{trdeg} \operatorname{Frac} \left( \bigoplus\_{m ≥ 0} H^0(X, mL) \right)$ is equal to
2. $\max \operatorname{trdeg}(F)$ where $F$ is a finitely generated subfield of $ \operatorname{Frac} \left( \bigoplus\_{m ≥ 0} H^0(X, mL) \right)$, which is equal to
3. $\max \operatorname{trdeg}( \operatorname{Frac}(R))$ where $R$ is a finitely generated subring of $\bigoplus\_{m ≥ 0} H^0(X, mL) $ (take $R$ generated by the numerators and denominators of $F$ ), which is equal to
4. $\max \operatorname{trdeg}( \operatorname{Frac}(R))$, where $R$ is the subring generated by $\bigoplus\_{m \in S} H^0(X, mL) $ for $S$ a finite set of natural numbers (take the supports of the generators), which is equal to
5. $\max \operatorname{trdeg}( \operatorname{Frac}(R))$, where $R$ is the subring generated by $ H^0(X, nL) $ (take the least common multiple of $S$, and note the old generators are finite over this ring), which is equal to
6. the max of the dimension of the spectrum of $R$, where $R$ is the subring generated by $ H^0(X, nL) $, which is equal to
7. the max of the dimension of the affine cone on the closure of the image of the rational map $φ\_{|nL|} : X \dashrightarrow \mathbb{P}(H^0(X, n L)^\*)$ (since the spectrum is equal to that cone)
8. the max of the dimension of the image of the rational map $φ\_{|nL|} : X \dashrightarrow \mathbb{P}(H^0(X, n L)^\*)$, plus 1 (since the affine cone has dimension one higher).
| 10 | https://mathoverflow.net/users/18060 | 416885 | 169,851 |
https://mathoverflow.net/questions/416884 | 1 | Suppose a first order set theory $S$ has a countable model. Does it follow that there is a countable ordinal $\alpha$ so that $L\_\alpha$ in the constructible hierarchy is a model of $S$?
| https://mathoverflow.net/users/37385 | Are countable models constructible? | For the question in the text, the answer is no. Let $S$ consist of (1) a sufficient finite number of ZF axioms to provide an absolute (i.e., $\Delta\_1$ in the Lévy hierarchy) definition of the constructible hierarchy and (2) the sentence formalizing $V\neq L$.
For the question in the title, asking only for a constructible model and not necessarily one of the form $L\_\alpha$, the answer is yes if the theory itself is constructible, but not in general.
| 8 | https://mathoverflow.net/users/6794 | 416887 | 169,852 |
https://mathoverflow.net/questions/416888 | 1 | $\DeclareMathOperator\SO{SO}$I posted this on [MSE](https://math.stackexchange.com/questions/4381194/s-4-subgroups-and-so-3-mathbbr) 10 days ago and it got 3 upvotes but no answers or comments, so I'm cross-posting to MO.
Background: The group of rotations $ \SO\_3(\mathbb{R}) $ has a 24 element subgroup of integer points $ \SO\_3(\mathbb{Z}) $ which is isomorphic to the symmetric group $ S\_4 $.
Original Question: (answered almost immediately by LSpice with a great subgroup of $ \SO\_5(\mathbb{R}) $)
**Let $ G $ be a connected Lie group. For every injective group homomorphism $ \phi:\SO\_3(\mathbb{Z}) \to G $ does there exist a unique extension $ \tilde{\phi} : \SO\_3(\mathbb{R}) \to G $?**
New Question:
**Let $ G $ be a connected Lie group. Let $ \phi:\SO\_3(\mathbb{Z}) \to G $ be a group homomorphism whose image is a maximal closed subgroup of $ G $. Does there exist a unique extension $ \tilde{\phi} : \SO\_3(\mathbb{R}) \to G $?**
Note that the only quotients of $ S\_4 $ are $ S\_4 $, cyclic 2 group trivial group and the six element dihedral group $ D\_6 $. Of these only $ S\_4 $ itself is a maximal closed subgroup of $ \SO\_3(\mathbb{R}) $.
For $ G= \SO\_3(\mathbb{R}) $ the answer is yes to both questions. Here is a proof:
First we show existence of an extension. Define $ \Gamma:=\SO\_3(\mathbb{Z}) $ and $ \Gamma':=\phi(\SO\_3(\mathbb{Z})) $. Isomorphic finite subgroups of $ \SO\_3(\mathbb{R}) $ are conjugate. So there always exists some $ g \in G $ such that $ g \Gamma g^{-1}=\Gamma' $. So the map conjugation by $ g $ agrees with $ \phi $ up to some automorphism. From there we can conjugate by some $ x \in \Gamma $ to get an exact match (recall every automorphism of the symmetric group $ S\_4 $ is inner). So conjugation by $ xg $ is an automorphism of $ G $ extending $ \phi $.
Now we show uniqueness of extensions. Any two extensions $ \phi\_1 $ and $ \phi\_2 $ of $ \phi $ are automorphisms of $ \SO\_3(\mathbb{R}) $. Every automorphism of $ G $ is inner, so the extensions $ \phi\_1$, $\phi\_2 $ correspond to conjugation by $ g\_1$, $g\_2 $. Thus conjugation by $ g\_2^{-1}g\_1 $ fixes $ \Gamma $ pointwise. In other words $ g\_2^{-1}g\_1 $ centralizes $ \Gamma $. So in particular $ g\_2^{-1}g\_1 \in N\_G(\Gamma) $. But $ \Gamma $ is self-normalizing $ N\_G(\Gamma)=\Gamma $. So really $ g\_2^{-1}g\_1 \in \Gamma $. However $ \Gamma $ also has trivial center $ Z(\Gamma)=1 $. So we must have $ g\_1=g\_2 $ and thus uniqueness is proven.
EDIT: As Lspice points out the answer to the first question is no there does not always exist an extension and for the second question Lspice notes that if such an extension exists then $ G $ must be $ SO\_3(\mathbb{R}) $.
First we show uniqueness. If $ \phi $ is trivial there is nothing to prove. If $ \phi $ is trivial and there exists an extension $ \tilde{\phi} $ then $ \tilde{\phi} $ must be injective since $ \SO\_3(\mathbb{R}) $ is simple so since $ \Gamma':=\phi(\SO\_3(\mathbb{Z})) $ is maximal closed and contained in $ \tilde{\phi}(\SO\_3(\mathbb{R}))\cong \SO\_3(\mathbb{R})$ we can conclude that $ G=\tilde{\phi}(\SO\_3(\mathbb{R})) $.
Now we show existence. Suppose that $ \Gamma' $ is a maximal closed subgroup. Then $ G $ must be compact since $ \Gamma' $ is contained in a maximal compact. Also $ \Gamma' $ must contain the whole center otherwise it is not maximal closed (and $ G $ cannot be abelian because then it contains no maximal closed subgroups). So $ G $ must be semisimple since $ \Gamma' $ is finite. In fact I claim $ G $ must be simple. Otherwise we can pass to the universal cover and say $ G\_1\times G\_2 $ are two different semi simple factors with the lift of $ \Gamma' $ intersecting $ G\_1 $ non trivially. Then the direct product of $ G\_2 $ with the projection onto $ G\_1 $ of lift of $ \Gamma' $ is a closed subgroup. So descending back to $ G $ we have that $ \Gamma' $ is not maximal closed, a contradiction. This is a general proof that a Lie group with a finite maximal closed subgroup must be simple and compact. For here we specialize to the case of $ S\_4 $ subgroup. The only quotients are $ 1, C\_2,D\_6, S\_4 $. It is fairly clear that of these only $ S\_4 $ can be maximal closed in a Lie group since only an embedded $ S\_4 $ can act irreducibly with respect to the adjoint representation of a compact simple Lie group. Moreover, the only compact simple Lie group for which this is true. Thus any connected Lie group containing a quotient of $ S\_4 $ as a maximal closed subgroup must just be $ SO\_3(\mathbb{R}) $.
| https://mathoverflow.net/users/387190 | $ S_4 $ subgroups and $ \operatorname{SO}_3(\mathbb{R}) $ | Consider the group of matrices of the form $w \oplus \det(w)^{-1}$, with $w$ a $4\times4$ permutation matrix. This is an $\operatorname S\_4$ inside $\operatorname{SL}\_5(\mathbb R) \subseteq \operatorname{SL}\_5(\mathbb C)$, and $\mathbb C^5$ decomposes as a sum of the trivial, the sign, and the reflection representation of $\operatorname S\_4$. In particular, the space of fixed vectors for $\operatorname S\_4$ in $\mathbb C^5$ is $1$-dimensional.
If this $\operatorname S\_4$ were contained in an $\operatorname{SO}\_3(\mathbb R)$ in $\operatorname{SL}\_5(\mathbb R)$, or even in $\operatorname{SL}\_5(\mathbb C)$, then the resulting (complex) $5$-dimensional representation of $\operatorname{SO}\_3(\mathbb R)$ would be $0^5$, $2 + 0^2$, or $4$.
Since $\sum\_{g \in \operatorname{SO}\_3(\mathbb Z)} \chi\_4(g) = 1(5) + 6(1) + 3(1) + 8(-1) + 6(-1)$ equals $0$, there are no fixed vectors for $\operatorname{SO}\_3(\mathbb Z)$ in $4$. Clearly, there is at least a $2$-dimensional space of fixed vectors for $\operatorname{SO}\_3(\mathbb Z)$ in $2 + 0^2$ and in $0^5$.
| 4 | https://mathoverflow.net/users/2383 | 416889 | 169,853 |
https://mathoverflow.net/questions/416836 | 9 | Everyone knows the Heine-Borel theorem characterizing compact subsets of Euclidean space. For any $n \in \mathbb N$ a set $A \subseteq \mathbb R^n$ is compact just in case it is closed and bounded (in the sense that it is contained in some large enough finite diameter ball).
There is a similar theorem for the irrationals, due to Rothberger. The irrationals are homeomorphic to the space $\mathbb N^\mathbb N$ of functions $f:\mathbb N \to \mathbb N$ topologized by the product topology of the discrete topology on $\mathbb N$. With this presentation a set $A \subseteq\mathbb N^\mathbb N$ is compact just in case it is closed and bounded where bounded means there is an $f:\mathbb N \to \mathbb N$ so that for all $g \in A$ we have $\forall n \, g(n) \leq f(n)$.
My question is whether there is a similar (or at least relatively neat) characterization of compact sets of rationals (with the subspace topology). Does some relatively simple (topological, geometric, combinatorial) property characterize when a closed $A \subseteq \mathbb Q$ is compact?
| https://mathoverflow.net/users/114946 | Structure theorems for compact sets of rationals | It is easy to describe these up to homeomorphism:
* Every countable compact Hausdorff space is homeomorphic to a countable successor ordinal, see [Milliet - A remark on Cantor derivative](https://arxiv.org/abs/1104.0287v1).
* Conversely, it's easy to embed all countable limit ordinals in the rationals by transfinite induction, and successor ordinals will give you compact sets.
| 16 | https://mathoverflow.net/users/123634 | 416900 | 169,857 |
https://mathoverflow.net/questions/416902 | 2 | Let $\mu$ be a Borel probability measure on $\mathbb{R}^n$ such that there are $c,C,d,D>0$ satisfying: for every $x \in \mathbb{R}^n$ and every $r>0$
$$
c r^d \leq \mu(B(x,r)) \leq Cr^D.
$$
Let's call such a measure **almost Ahlors Regular.**
***Note:** when $d=D$ then we say that the measure is [*Ahlfors regular*](https://mathoverflow.net/questions/319955/origin-of-term-ahlfors-david-regular) and, if $c$ were allowed to be $=0$ then the measure would be called upper [*Ahlfors regular*](https://mathoverflow.net/questions/415903/terminology-upper-ahlfors-regular-measure).*
What are examples of Ahlors regular measures?
| https://mathoverflow.net/users/36886 | Examples of "almost" Ahlfors regular measures | $\newcommand\R{\mathbb R}\newcommand\la{\lambda}$A measure $\mu$ is Ahlfors regular (according to your definition) iff it has a density (with respect to the Lebesgue measure $\la$) bounded away from $0$ and $\infty$ (if $B(x,r)$ denotes the ball centered at $x$ and of radius $r$).
Indeed, by inscribing and circumscribing, we can replace $B(x,r)$ in your definition by the (left-open) $n$-cube $b(x,r):=\prod\_1^n(x\_i,x\_i+r]$ (say of size $r$) for $x=(x\_1,\dots,x\_n)\in\R^n$ (at the same time also appropriately changing the values of $c$ and $C$), so that
$$
c r^d\le\mu(b(x,r))\le Cr^d \tag{1}\label{1}
$$
for all $x\in\R^n$ and $r>0$.
Partitioning, in the natural manner, the unit $n$-cube $b(0,1)$ into $N^n$ $\ n$-cubes each of size $1/N$, from \eqref{1} we get
$$
c N^{n-d}\le\mu(b(0,1))\le CN^{n-d} \tag{2}\label{2}
$$
for all $N$, which is possible only when $d=n$ -- otherwise, $N^{n-d}$ will go either to $0$ or to $\infty$ as $N\to\infty$, which would contradict one of the two inequalities in \eqref{2}, since $\mu(b(0,1))\in(0,\infty)$.
Therefore and by partitioning and approximation, we get
$$
c \prod\_1^n r\_i\le\mu\Big(\prod\_1^n(x\_i,x\_i+r\_i]\Big)\le C \prod\_1^n r\_i
$$
for any real $x\_1,\dots,x\_n$ and any positive real $r\_1,\dots,r\_n$.
That is, $c\la\le\mu\le C\la$ on the semiring of all left-open $n$-boxes.
We conclude that $c\la\le\mu\le C\la$ on the Lebesgue $\sigma$-algebra, as claimed.
| 1 | https://mathoverflow.net/users/36721 | 416911 | 169,861 |
https://mathoverflow.net/questions/406120 | 10 | A monohedron is a convex polyhedron with all faces mutually congruent but with no other symmetry necessarily needed. So obviously, this is a wide class of polyhedrons that includes the Platonic solids and isohedra (<https://mathworld.wolfram.com/Isohedron.html>). An earlier post is <https://math.stackexchange.com/questions/3888486/what-are-the-known-convex-polyhedra-with-congruent-faces>
**Questions:** Are there monohedrons with odd number of faces (it is known that isohedrons necessarily have even number of faces - as stated in <https://mathworld.wolfram.com/Isohedron.html>)? What are the values for the number of edges on a face for which monohedrons are possible? Will relaxing convexity (of the body, not of the faces) have an impact on the answers to these questions?
| https://mathoverflow.net/users/142600 | Are there Monohedra with odd number of faces? | The answer to the question in the title is negative in dimension 3: it was shown by Grunbaum that every 3-polytope with congruent facets has an even number of facets. See p. 414 of his book "Convex Polytopes", 2nd edition. The original reference is:
Grunbaum, B. On polyhedra in $\mathbb{E}^3$ having all faces congruent. *Bull. Res. Council Israel*, **8**F (1960), 215-218.
| 5 | https://mathoverflow.net/users/110306 | 416912 | 169,862 |
https://mathoverflow.net/questions/416703 | 4 | I'm currently reading the article ["A Generalization of a Poincaré–Bendixson Theorem to Closed Two-Dimensional Manifolds" by Arthur Shwartz](https://www.jstor.org/stable/2373135). The paper first establishes a result for minimal sets, which are closed, non-empty subsets of $M$ which are invariant under the flow of some $C^2$ vector field, and contain no such proper subsets. In particular, the result can be summarized as follows:
>
> Suppose that $M$ is a compact, connected, two-dimensional manifold and $\Sigma \subset M$ is a minimal set. Then $\Sigma$ is either:
>
>
> 1. a fixed point,
> 2. a periodic orbit (homeomorphic to $S^1$), or
> 3. $M$ itself (which must be homeomorphic to $T^2$).
>
>
>
I was able to follow the proof of this theorem, but fail to understand the subsequent corollary, which may be stated as:
>
> Suppose that $M$ is a compact, connected, orientable, two-dimensional manifold such that $M$ is not a minimal set and for some $x \in M$, the $\omega$-limit set of $x$—$\omega(x)$—contains no fixed points. Then $\omega(x)$ is homeomorphic to $S^1$ and the flow of $x$—$\varphi\_{t}(x)$—winds towards $\omega(x)$ as $t \to \infty$.
>
>
>
The argument presented is that $\omega(x)$ must contain a minimal set (say $\Sigma$) which is homeomorphic to $S^1$ by the previous theorem. Now since $M$ is orientable, there exists a neighborhood $N$ of $\Sigma$ which is homeomorphic to $(-1, 1)\times S^1$ such that $\Sigma$ is mapped onto $\{0\}\times S^1$ and the line segments $(-1, 1)\times \{y\}$ are transversal arcs for $y \in S^1$. The setup is clear to me, but I don't follow the remaining steps of the proof: "It follows from the continuity of the flow, and the fact that $\Sigma \subset \omega(x)$, that there exist $0 < t\_1 < t\_2$, $q \in \omega(x)$, and a transversal $I\_q$ through $q$ such that $\varphi\_{t\_2}(x)$ is between $\varphi\_{t\_1}(x)$ and $q$ on $I\_q$. Thenceforth $\varphi\_{t}(x)$ is "trapped" between the closed curve formed by $\varphi\_{[t\_1, t\_2]}(x)$ and the portion of $I\_q$ between $\varphi\_{t\_1}(x)$ and $\varphi\_{t\_2}(x)$ and $\Sigma$. Thus we find the next intersection of $\varphi\_{t}(x)$ with $I\_q$ between $\varphi\_{t\_2}(x)$ and $q$. Repetition of this argument yields the conclusion of the corollary."
I believe that we're aiming to show that $\omega(x) = \Sigma$ (in particular by a monotonicity-type argument on the surface of the cylinder analogous to that used in the proof of the Poincaré–Bendixson theorem in $\mathbb{R}^2$). But how can we be sure that $q \in N$, or that the flow remains within $N$ (and hence its image under the homeomorphism remains on the surface of the cylinder, which I understand to be what "traps" it)? Even if so, how would this ultimately show that $q \in \Sigma$?
| https://mathoverflow.net/users/111199 | Poincaré–Bendixson Theorem on a compact, connected, orientable, two-dimensional manifold | To address your two questions:
1. Why is $q \in N$?
If I understand correctly, they could have (and should have) just started with $q \in \Sigma$, since for any $q \in \Sigma$ the orbit $\phi\_t(x)$ approaches $q$ arbitrarily closely.
To be precise, because $\Sigma \subset \omega(x)$, there's a $t\_1 > 0$ such that $\phi\_{t\_1}(x) \in I\_q \subset N$, and there's a $t\_2 > t\_1$ such that $\phi\_{t\_2}(x) \in I\_q$ and is even closer to $q$ than $\phi\_{t\_1}(x)$. By orientability of $M$, $\phi\_{t\_2}(x)$ is between $q$ and $\phi\_{t\_1}(x)$.
2. Why does the flow stay in $N$?
The proof doesn't need the flow to stay within $N$. Instead, it requires something weaker. Consider the subsegment $T$ of the transversal $I\_q$ between $q$ and $\phi\_{t\_1}(x)$. By continuity, if we choose $t\_1$ so that $\phi\_{t\_1}(x)$ is sufficiently close to $q$, then the orbit of any point $T$ will stay within $N$ up until it hits $I\_q$ again.
That is enough to rule out any "weird manifold topology" and conclude there is a "trapped" region bounded by $\phi\_{[t\_1, t\_2]}$, $\Sigma$, and a piece of $I\_q$, as in the Poincare-Bendixson proof.
| 1 | https://mathoverflow.net/users/1227 | 416917 | 169,865 |
https://mathoverflow.net/questions/416901 | 5 | This is a dublicate from stackexchange:
Consider two families of hyperplanes $F\_1$ and $F\_2$ in $\mathbb{R}^d$ both containing $n$ hyperplanes. We have that for all $f \in F\_1$ and $g \in F\_2$ that $f$ and $g$ intersect. Further we know that for each point $p \in \mathbb{R}^d$ we have that at most $c \cdot n$ hyperplanes from each family contain this point, $ c < 1$.
The second family induces $d-2$ dimensional planes on the hyperplanes of the first family. I want to show that there are at least $\epsilon \cdot n$ hyperplanes in the first family on which at least $\tau \cdot n$ planes are induced.
What I am able to show is: If there is a plane in which $k \cdot n$ hyperplanes form the second family intersect then, there this plane ist contained in at most $c \cdot n$ hyperplanes from the first family. Consider the $(1-c)n$ hyperplanes of the first family which do not contain the plane, than on each of them we get $k \cdot n$ induced planes.
On the other hand it is clear if at most a constant number, $k$, of hyperplanes intersect in one plane we get $\frac{n}{k}$ induced planes on each of the hyperplanes in the first family.
So somehow I can show the extremal cases but I'm stuck with the ones in between. Any help or suggestions?
**Edit:**
Let $n\_H$ denote the number of $d-2$ dimensional planes induced on the plane $H \in F\_1$. Further let $f\_{H,1},...,f\_{H,n\_H}$ denote this induced planes and $S(f\_{H,i}):=\{K \in F\_2 \mid f\_{H,i} \subset K\}$. Then for $H, H' \in F\_1$ we have:
\begin{equation}
\frac{n}{d} = \sum\_{i=1}^{n\_H} \sum\_{j=1}^{n\_K} \vert S(f\_{H,i}) \cap S(f\_{H',j})\vert \leq c \cdot \frac{n}{d}+ n\_{H} \cdot n\_{H'}
\end{equation}
Since $\vert S(f\_{H,i}) \cap S(f\_{H',j})\vert \leq 1$ for all $f\_{H\_i} \neq H \cap H'$ and $\vert S(f\_{H,i}) \cap S(f\_{H',j})\vert \leq c \cdot \frac{n}{d}$ if $f\_{H,i} = f\_{H',j}=H \cap H'$. Therefore $(1-c)\frac{n}{d} \leq n\_{H'} \cdot n\_H$. This holds for all $H, H' \in F\_1$ so $n\_H \gg \sqrt{n}$.
| https://mathoverflow.net/users/216305 | Counting intersections of hyperplanes | As by my comment above, it is enough to solve the problem for $d=2$.
Using point-line duality, we get the following:
Given $n$ red and $n$ blue points in the plane, such that no line contains more than $cn$ points, prove that through at least $\varepsilon n$ of the red points there are at least $\tau n$ lines that each go through some blue point.
This can be proved similarly to Beck's theorem that can be [found here](https://en.wikipedia.org/wiki/Beck%27s_theorem_(geometry)#Proof_2).
More precisely, there are $n^2$ red-blue point pairs.
At most $O(cn^2)$ of the lines of these pairs contains at least $1/c$ points (see the linked proof).
As there are no lines with $cn$ points, most pairs will contain at most $1/c$ points, which implies that there are at least $\varepsilon n$ red points with $\tau n$ lines through each with some blue point.
| 5 | https://mathoverflow.net/users/955 | 416920 | 169,867 |
https://mathoverflow.net/questions/416921 | 3 | Denote by $ \binom{n}{k}\_q = \prod\_{i=0}^{k-1} \frac{ q^{n-i} - 1 }{ q^{k-i} - 1 } $, $ k = 0, 1, \ldots, n $, the $ q $-binomial (Gaussian) coefficients. These numbers are symmetric, in the sense that $ \binom{n}{k}\_q = \binom{n}{n-k}\_q $, unimodal, and hence maximized for $ k = \lfloor n/2 \rfloor $ (or $ k = \lceil n/2 \rceil $). I am interested in the sum
$$
\sum^{n}\_{\substack{ k=0 \\ k \, \equiv \, h \pmod \ell }} \binom{n}{k}\_q
$$
where $ h $ is a parameter over which the sum is to be maximized. **Question:** Is it true that, for any fixed $ q $, $ n $, $ \ell $, the choice $ h = \lfloor n/2 \rfloor $ (or, equivalently, $ h = \lceil n/2 \rceil $) maximizes the sum?
*Special cases:*
* This is known to be true for binomial coefficients ($ q \to 1 $). It follows from the result of Katona (1972) stating that
$$
\sum^{n}\_{\substack{ k=0 \\ k \, \equiv \, \lfloor n/2 \rfloor \pmod \ell }} \binom{n}{k}
$$
is the maximum cardinality of a family of subsets $ \mathcal{F} \subseteq 2^{\{1, \ldots, n\}} $ satisfying the condition that if $ A, B \in \mathcal{F} $ and $ A \subsetneq B $, then $ |B| - |A| \geqslant \ell $. The proof relies on the theory of partial orders. A simpler and direct proof for this case would also be appreciated.
* The case $ \ell = 2 $. I saw the following identity in one of the answers to [this question](https://mathoverflow.net/questions/89934/sum-of-gaussian-binomial-coefficients/191803#191803):
$$
\sum\_{i=0}^n(-1)^i\binom ni\_q=\begin{cases}0,&n\, \text{odd}\\
\prod\_{j=1}^{n/2}(1-q^{2j-1}),&n\, \text{even}\end{cases}
$$
If this is true, it implies the statement I am interested in when $ \ell = 2 $. Does someone know how to prove this identity?
*Background:* The sum represents the number of subspaces of $ \mathbb{F}\_q^n $ whose dimension is congruent to $ h $ modulo $ \ell $. The set of all subspaces of $ \mathbb{F}\_q^n $ partially ordered by inclusion satisfies the so-called LYM condition. It follows from this fact and a result of Kleitman (1975) that
$$
\max\_h \sum^{n}\_{\substack{ k=0 \\ k \, \equiv \, h \pmod \ell }} \binom{n}{k}\_q
$$
is the maximum cardinality of a family of subspaces satisfying the condition that if $ U \subsetneq V $, then $ \dim V - \dim U \geqslant \ell $. I was wondering if the maximizer can be found explicitly.
| https://mathoverflow.net/users/83189 | Sum of $q$-binomial coefficients | Let's look at the ratio of two adjacent $q$-binomials as we move away from the center, for simplicity I'll do the even case.
$\binom{2n}{n-a}\_q / \binom{2n}{n-a-1}\_q = \frac{[n+a+1]\_q}{[n-a]\_q} > q^{2a+1}$
In particular these are decaying faster than geometrically. With some possible exceptions for say $q =2$ or $q=3$ with $n$ small, it looks like the middle binomial coefficient will be bigger than the sum of all the others.
| 4 | https://mathoverflow.net/users/39120 | 416926 | 169,870 |
https://mathoverflow.net/questions/416895 | 2 | I have been studying hyperplane arrangements from R. Stanley's notes and so far, I have read until lecture 5. But, Lecture 6 and end of Lecture 5 seems very combinatorial. I'm more interested in the "algebraic" direction. More specifically, in representation-theoretic perspective (if there exists any).
So, I was wondering if someone can suggest me some reference for more "algebraic" perspective of hyperplane arrangements. (It's OK if it's not representation-theoretic)
| https://mathoverflow.net/users/338456 | Possible "algebraic" direction in hyperplane arrangements | If I recall correctly, Stanley does not discuss the so-called "Tits monoid" associated to a hyperplane arrangement. The relationship between the Tits monoid and the lattice of flats of a hyperplane arrangement is the main focus of the book "Topics in hyperplane arrangements" by Aguiar and Mahajan (it's free online at [http://pi.math.cornell.edu/~maguiar/surv-226.pdf](http://pi.math.cornell.edu/%7Emaguiar/surv-226.pdf)). Actually, they have another more recent book on hyperplane arrangements as well - I'm not quite sure the difference between the two. Anyways, this is definitely an "algebraic" direction in hyperplane arrangement theory.
The study of free arrangements/the module of logarithmic derivations -- as championed by Terao -- is another completely different algebraic (or more specifically, algebro-geometric) direction in hypreplane arrangement theory. This one is arguably more connected (at least "morally") to the stuff in Stanley's notes (e.g. I believe Stanley mentions free arrangements, at least in an exercise). This is discussed e.g. in the book by Orlik and Terao mentioned in the comment above.
| 2 | https://mathoverflow.net/users/25028 | 416928 | 169,871 |
https://mathoverflow.net/questions/416905 | 1 | This question was previously posted [on MSE.](https://math.stackexchange.com/questions/4389798/eigenvectors-of-the-dual-of-positive-irreducible-operators)
Let $E$ be a Banach lattice such that $E$ is an $M$-space. Assume that $T\colon E\to E$ is a positive bounded non-compact irreducible linear operator with positive spectral radius. And define
\begin{align\*}
T’\colon E’ &\to E'\\
\phi&\mapsto [v\mapsto \phi(Tv)],
\end{align\*}
where $E’$ is the dual of $E$.
Under the above assumptions, I would like to know: Is it possible to guarantee that $T’$ has a positive eigenvector? If not, is there any “mild assumption” that can be added to this setting to ensure the existence of a positive eigenvector of $T’$?
| https://mathoverflow.net/users/125982 | Eigenvectors of the dual of positive irreducible operators | No, the dual operator $T'$ does have a positive eigenvector, in general.
As a counterexample, consider the space $E = c\_0(\mathbb{Z})$ if scalar-valued sequences indexed over the integers, endowed with the sup norm and the pointwise order. This is a Banach lattice and an M-space. Its dual space $E'$ can be identified with $\ell^1(\mathbb{Z})$.
Now let $L$ and $R$ denote the left and right shift on $E$, respectively, and define $T := (L+R)/2$.
Then $T$ is positive and irreducible and has norm $1$. The spectral radius of $T$ is $1$ (to see this, note that the bi-dual operator $T''$ on $\ell^\infty(\mathbb{Z})$ has the constant sequence with value $1$ as a fixed point).
However, $T' = (L' + R')/2$ does not have positive eigenvector. Indeed, assume to the contrary that $v \in E'$ is an eigenvector of $T'$ with entries $\ge 0$ and associated eigenvalue $\lambda$. Then $\lambda$ is a real number and $\lambda \ge 0$.
We distinguish between two cases:
1. case: $\lambda < 1$. In this case it follows that $T^nv \to 0$ as $n \to \infty$. Thus, the ideal
$$
I = \{x \in E': \; T^n |x| \to 0 \text{ as } n \to \infty\}
$$
in $E'$ contains the non-zero vector $v$. It is also closed (since $T'$ is power-bounded), and since the operator $T'$ can easily be checked to be irreducible, too, it follows that $I = E'$. But this cannot be true since $T'$ is norm preserving on the positive cone $E'\_+$.
2. case: $\lambda = 1$.
In this case, we can use that $T'$ commutes with the right shift $L'$ and the left shift $R'$ on $E'$, and hence every shifted version of $v$ is also an eigenvector for the eigenvalue $1$. Therefore, the eigenspace of $T'$ for the eigenvalue $1$ is infinite-dimensional. However, it follows from infinite-dimensional Perron-Frobenius theory that the fixed space of a power-bounded irreducible operator is either $\{0\}$ or one-dimensional, so was also have a contradiction in this case.
| 1 | https://mathoverflow.net/users/102946 | 416932 | 169,873 |
https://mathoverflow.net/questions/414577 | 2 | I'm interested in Sobolev space estimates for commutators involving a pseudodifferential operator and a Fourier multiplier. More specifically, suppose $p = p(x,\xi) \in S\_{1,0}^{m\_1}$ and let $q = q(\xi) \in S\_{1,0}^{m\_2}$, where $S\_{\rho,\delta}^m$ denotes the standard Hörmander symbol class. Does anyone know of a reference for a commutator estimate of the form
$$\| [\mathrm{Op}(p),\mathrm{Op}(q)](u) \|\_{H^\sigma} \lesssim \|u\|\_{H^s}$$
for some range of $\sigma, s$?
Or even something simpler like
$$\| [\mathrm{Op}(p),\partial\_j](u) \|\_{H^\sigma} \lesssim \|u\|\_{H^s}?$$
Any ideas would be appreciated! Thanks!
| https://mathoverflow.net/users/114164 | Reference for commutator estimate | Given your assumptions on $p$ and $q$, $[Op(p),Op(q)]\in S^{m\_1+m\_2-1}\_{1,0}.$ Given any $A\in OPS^m\_{\rho,\delta}$ with $0\leq\delta<\rho\leq 1$, one has that $A:H^s\rightarrow H^{s-m}$ boundedly (follows from a combination of Calderon-Valliancourt and the Fourier multiplier $\Lambda\_s$ with symbol $\langle \xi\rangle^s$). In particular, given $u\in H^s,$ we have that $$\|[Op(p),Op(q)]u\|\_{H^\sigma}\lesssim \|u\|\_{H^s}$$ for any $\sigma\leq s+m\_1+m\_2-1.$
| 2 | https://mathoverflow.net/users/477714 | 416946 | 169,879 |
https://mathoverflow.net/questions/416892 | 5 | We define a norm on $\mathbb C^2$ as $\|(\alpha,\beta)\|:=\max\left\{|\alpha|,|\beta|,\big|\frac{\alpha+\beta}{\sqrt{2}}\big|\right\}.$ Can the dual norm be calculated explicitly?
| https://mathoverflow.net/users/136860 | Dual norm of a subspace of $\ell_\infty^3$ | $\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$This is to detail and correct the answer by Onur Oktay, which is based on a nice idea, leading to simplified and speedier calculations of the dual norm. (That Onur Oktay's answer contains at least two mistakes was pointed out in comments by Nathaniel Johnston and me.)
As in Onur Oktay's answer, let $X:=\ell^3\_\infty$ and $V:=\{av\_1+bv\_2:a,b\in\C\}\subset X$, where
\begin{equation\*}
v\_1:=(1,0,1/\sqrt2),\quad v\_2:=(0,1,1/\sqrt2),\quad v\_3:=(1,1,-\sqrt2).
\end{equation\*}
Note that $v\_3\cdot v=0$ for all $v\in V$, where $\cdot$ denotes the dot product.
Consider
\begin{equation\*}
V^\perp:=\{x^\*\in X^\*\colon x^\*(v)=0\ \forall v\in V\},
\end{equation\*}
the so-called annihilator of $V$.
Note that, if, as usual, $X^\*=(\ell^3\_\infty)^\*$ is identified with $\ell^3\_1$, then $V^\perp$ will be identified with the span $\C v\_3=\{tv\_3\colon t\in\C\}$ of $\{v\_3\}$.
It is then said in Onur Oktay's answer that
>
> $V^\*=X^\*/V^\perp$, which is isometrically isomorphic to $\C^2$ equipped with the norm
> \begin{equation\*}
> \|(a,b)\|=\inf\{\|av\_1+bv\_2-tu\|\_{\ell^3\_1}: t\in\C\},
> \end{equation\*}
>
>
>
where $u$ is our $v\_3$.
This statement is incorrect. First here is the minor point that $V^\*$ is, not equal, but isometrically isomorphic to $X^\*/V^\perp$ ([Theorem 4.9](https://www.powells.com/book/functional-analysis-2nd-edition-9780070542365)), with the isometric isomorphism $\si$ given by the formulas
\begin{equation\*}
V^\*\ni v^\*\mapsto\si(v^\*):=x^\*+V^\perp
\end{equation\*}
and
\begin{equation\*}
\|x^\*+V^\perp\|:=\inf\{\|x^\*+y^\*\|\colon y^\*\in V^\perp\}, \tag{0}\label{0}
\end{equation\*}
where $x^\*\in X^\*$ is any extension of the continuous linear functional $v^\*$.
More importantly, the expression for $\|(a,b)\|$ in Onur Oktay's answer is incorrect.
Indeed, consider the isometric isomorphism
\begin{equation\*}
\C^2\ni(\al,\be)\mapsto\iota((\al,\be)):=\al v\_1+\be v\_2\in V\subset X=\ell^3\_\infty
\end{equation\*}
from $\C^2$ onto $V$, with the norm on $\C^2$ induced by $\iota$:
\begin{equation\*}
\|(\al,\be)\|=\|\al v\_1+\be v\_2\|\_\infty=\max(|\al|,|\be|,|\al+\be|/\sqrt2). \tag{1}\label{1}
\end{equation\*}
For any $a$ and $b$ in $\C$, define linear functionals $l\_{a,b}\in(\C^2)^\*$ and $L\_{a,b}\in V^\*$ by the conditions
\begin{equation\*}
l\_{a,b}((1,0))=a=L\_{a,b}(v\_1)\quad\text{and}\quad l\_{a,b}((0,1))=b=L\_{a,b}(v\_2);
\end{equation\*}
here, of course, the norm $\|\cdot\|\_{(\C^2)^\*}$ on $(\C^2)^\*$ is dual to the norm on $\C^2$ given by \eqref{1}.
Also for $a$ and $b$ in $\C$, define the vector $x\_{a,b}\in\R^3$ by the conditions
\begin{equation\*}
x\_{a,b}\cdot v\_1=a,\quad x\_{a,b}\cdot v\_2=b,\quad x\_{a,b}\cdot v\_3=0,
\end{equation\*}
so that
\begin{equation\*}
x\_{a,b}=\tfrac14\,(3 a-b,3 b-a,\sqrt{2}(a+b)).
\end{equation\*}
So, the vector $x\_{a,b}\in\R^3$ is the canonical representation of the extension -- say $\tilde L\_{a,b}$ -- of the linear functional $L\_{a,b}\in V^\*$ such that $\tilde L\_{a,b}\in X^\*=(\ell^3\_\infty)^\*\simeq\ell^3\_1$ and $\tilde L\_{a,b}(v\_3)=0$.
The canonical representation of $\tilde L\_{a,b}$ by $x\_{a,b}$ here is of course the one induced by the canonical isometric isomorphism $(\ell^3\_\infty)^\*\simeq\ell^3\_1$, so that $\tilde L\_{a,b}(x)=x\cdot x\_{a,b}$ for all $x\in\ell^3\_\infty$.
Then $l\_{a,b}=L\_{a,b}\circ\iota$ and, in view of \eqref{0},
\begin{equation\*}
\|l\_{a,b}\|\_{(\C^2)^\*}=\|L\_{a,b}\|\_{V^\*}
=\inf\_{t\in\C}\|x\_{a,b}+tv\_3/4\|\_1.
\end{equation\*}
Thus,
\begin{equation\*}
\|l\_{a,b}\|\_{(\C^2)^\*}
=\tfrac14\,\min\_{t\in\C}(|3 a - b + t|+|3 b-a + t|+\sqrt2\, |a + b - t|). \tag{2}\label{2}
\end{equation\*}
---
Finding the minimum in \eqref{2} is a problem of real algebraic geometry. So, in principle, the dual norm can be computed purely algorithmically; however, this calculation can take too much time -- as it actually is the case with the minimum in \eqref{2}.
However, finding the minimum in \eqref{2} numerically for any particular $(a,b)\in\C^2$ is not a problem. The numerical results obtained according to \eqref{2} completely agree with the numerical results obtained according to my other answer on this page, but the execution time using \eqref{2} is about 6 times as small.
---
As Matthew Daws noted in a comment, one could use the vector $y\_{a,b}:=(a,b,0)$ in place of $x\_{a,b}=\tfrac14\,(3 a-b,3 b-a,\sqrt{2}(a+b))$ -- in the sense that the linear functional $X\ni x\mapsto x\cdot y\_{a,b}$ is, just as the linear functional $\tilde L\_{a,b}$, an extension of the linear functional $L\_{a,b}$. This results in an expression for $\|l\_{a,b}\|\_{(\C^2)^\*}$ looking slightly simpler than the one in \eqref{2}:
\begin{equation\*}
\|l\_{a,b}\|\_{(\C^2)^\*}
=\inf\_{s\in\C}\|y\_{a,b}+sv\_3\|\_1
=\min\_{s\in\C}(|a + s|+|b + s|+\sqrt2\, |s|), \tag{2a}\label{2a}
\end{equation\*}
which seems to match the final expression in Nathaniel Johnston's answer. In fact, the ultimate expression in \eqref{2a} can be obtained from that in \eqref{2} by substitution $t=a+b+4s$.
| 4 | https://mathoverflow.net/users/36721 | 416948 | 169,881 |
https://mathoverflow.net/questions/416844 | 5 | Let $S$ and $\rm ku$ denote the sphere spectrum and the connective K-theory spectrum, respectively.
Then is the morphism ${\rm ku} \to F(F({\rm ku},S),S)$ an equivalence?
Here $F$ denotes the mapping spectrum.
| https://mathoverflow.net/users/119076 | Is ku reflexive as a spectrum? | Writing $D(-) = F(-, S)$, I believe that there is an equivalence $$D ku \simeq \prod\_{n=0}^\infty \Sigma^{-2n-1} H(\widehat{\mathbb Z} / \mathbb Z),$$ where $\widehat{\mathbb Z}$ is the profinite completion of the integers. This would split as a similar product after dualizing a second time, so can't be equivalent to $ku$.
The claimed equivalence is a combination of the following facts:
1. $D H \mathbb Q \simeq \Sigma^{-1} H \widehat{\mathbb Z} / \mathbb Z$: Writing $H \mathbb Q = \operatorname{colim}( \cdots \to S \xrightarrow n S \to \cdots)$, we find $D H \mathbb Q = \lim( \cdots \leftarrow S \xleftarrow n S \leftarrow \cdots)$, and the conclusion follows from the Milnor sequence and the finite torsion exponents of the higher homotopy groups of spheres.
2. $D H A \simeq 0$ for $A$ torsion abelian: Since $S$ is harmonic, $D H \mathbb F\_p \simeq 0$, and $A$ torsion abelian has a filtration whose quotients are prime torsion.
3. $D H \mathbb Q \simeq D H \mathbb Z$: There's a fiber sequence $$D H \mathbb Z \leftarrow D H \mathbb Q \leftarrow D H(\mathbb Q / \mathbb Z),$$ and the last spectrum is null by the above.
4. For $X$ coconnective, $D X \simeq D (\mathbb Q \otimes X)$: $X$ can be written as the colimit of a filtration whose quotients are Eilenberg–Mac Lanes, to which one applies the previous facts.
5. $L\_{H \mathbb Z} KU \simeq \mathbb Q \otimes KU$: Using Snaith's theorem $KU \simeq \Sigma^\infty\_+ \mathbb CP^\infty[\beta^{-1}]$, one calculates the integral homology of $KU$ to be rational.
6. $D KU \simeq \prod\_{n=-\infty}^\infty \Sigma^{-2n-1} H(\widehat{\mathbb Z} / \mathbb Z)$: Since $S$ is $H\mathbb Z$–local, $DKU \simeq D(L\_{H\mathbb Z} KU) \simeq D(\mathbb Q \otimes KU)$. Rational spectra are Eilenberg–Mac Lane, hence determined by their homotopy groups, so one combines Bott periodicity and the first fact.
The main claim then follows from the fiber sequence $$D ku \leftarrow D KU \leftarrow D(\tau\_{< 0} KU).$$ We've analyzed the middle term, and the last term starts off coconnective so (after applying the above facts) turns out to be the connective part of $D KU$, which finally determines the first term to be the remaining coconnective part.
| 8 | https://mathoverflow.net/users/1094 | 416949 | 169,882 |
https://mathoverflow.net/questions/416939 | 5 | Let $\mu$ be a finite non negative singular measure on $\mathbf{R}^d$. I would like to know if there exists some result on the infimum of the absolute value of its Fourier Transform $$\hat{\mu}(t)=\displaystyle\int \mathrm{e}^{2i\pi t\cdot x} ~\mu(dx).$$
It is known that we don't necessarily have $\hat{\mu}(t)\rightarrow0$ for singular measure. But does one have like Lebesgue measure $$\mathrm{inf}~|\hat{\mu}(t)|=0~?$$
Edit : To be more precise, $\mu$ is continuous singular, so does not have a discret part (<https://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem>)
| https://mathoverflow.net/users/477699 | Infimum of Fourier transform of singular measure | Yes, that's true. By Wiener's lemma we have $\sum\_{x\in \mathbb{R}^d} |\mu(\{x\})|^2 = \lim\_{R\to \infty} \frac{1}{(2R)^d} \int\_{[-R,R]^{d}} |\widehat{\mu}(t)|^2 dt$. If $\mu$ is continuous, then the left-hand side is $0$, so the infimum of $|\widehat{\mu}|$ has to be zero as well, since the right-hand side is bounded below by this infimum (squared).
| 4 | https://mathoverflow.net/users/24953 | 416958 | 169,885 |
https://mathoverflow.net/questions/416837 | 2 | It is a classical result of Wakimoto that the sheaf of chiral differential operators $D\_{ch}$ on $\mathbf{P}^1$ has global sections
$$D\_{ch}(\mathbf{P}^1)\ \simeq\ L\_{-2}(\mathfrak{sl}\_2)$$
the simple quotient affine vertex algebra at critical level $-2$. Apparently the map
$$L\_{-2}(\mathfrak{sl}\_2)\ \to\ D\_{ch}(\mathbf{A}^1)\ =\ \mathbf{C}[x\_0,x\_{-1},...,\partial\_{-1},\partial\_{-2},...\ ]$$
sends the fields $e(z),h(z),f(z)$ to $\partial(z)$, to $-2:x(z)\partial(z):-2$ and to $:x(z)^2\partial(z):+x'(z)$, where we have the fields
$$x(z)\ =\ \sum\_{n\in\mathbf{Z}} x\_nz^{-n},\ \ \ \partial(z)\ =\ \sum\_{n\in\mathbf{Z}} \partial\_nz^{-n-1}.$$
**Question:** I am having trouble showing that $-2:x(z)\partial(z):-2$, or equivalently just $\alpha=:x(z)\partial(z):$, glues to a global section of $D\_{ch}$.
Indeed, the transition function $\tau: D\_{ch}(\mathbf{C}^\times)\to D\_{ch}(\mathbf{C}^\times)$, where $D\_{ch}(\mathbf{C}^\times)=D\_{ch}(\mathbf{A}^1)[x\_0^{-1}]$, sends
$$x(z)\ \mapsto\ x(z)^{-1}, \ \ \ \partial(z)\ \mapsto \ :x(z)^2\partial(z):+2x'(z)$$
and so it sends
$$\alpha\ =\ :x(z)\partial(z):\ \mapsto\ :x(z)\partial(z):+2x'(z)x(z)^{-1}.$$
Thus $\alpha$ glues to a global section iff $\tau \alpha$ has no pole at $x\_0=0$.
But my **problem** is that $x'(z)x(z)^{-1}$ has a pole! For instance,
$$x'(z)x(z)^{-1}|0\rangle\ =\ -x\_{-1}x\_0^{-1}|0\rangle \text{ mod }z.$$
| https://mathoverflow.net/users/119012 | Confusion about Wakimoto's chiral differential operators on $\mathbf{P}^1$ | There are a couple of errors in the statement of the question. First in the transformation formula that I suppose you took from MSV "Chiral de Rham complex". Let us denote $\tilde{x} = x^{-1}$ and $\tilde{\partial} = x^2 \partial + 2 x'$. We assume the standard OPE $\partial(z) \cdot x(w) \sim \frac{1}{z-w}$ as the only non-zero OPE among generators.
In general for a function $f = f(x)$, denoting by $f(z)$ the corresponding field, we have $\partial(z) \cdot f(w) \sim \frac{f'(w)}{z-w}$
Notice that $x(z) \cdot \partial(w) \sim - \frac{1}{z-w}$ and in general $f(z) \cdot \partial(w) \sim - \frac{f'(w)}{z-w}$.
Now we compute $x^{-1}(z) \cdot \partial(w) \sim \frac{x^{-2}(w)}{z-w}$, from where we obtain
$$
\tilde{x}(z) \cdot \tilde{\partial}(w) = x^{-1}(z) \cdot x^2(w)\partial(w) \sim \frac{1}{z-w} \neq - \frac{1}{z-w} \sim x(z) \cdot \partial(w)
$$
So these don't glue to give a sheaf. The second error is that even assuming your transformation formula, the transformation for $\alpha$ is not correctly computed. I assume here that you missed quasi-associativity. Namely
$$
:x^{-1}(z) :x^2(z) \partial(z): :\ \neq\ ::x^{-1}(z) x^2(z): \partial(z):\ =\ :x(z) \partial(z):
$$
When you fix both issues everything will be fine.
| 3 | https://mathoverflow.net/users/17980 | 416972 | 169,888 |
https://mathoverflow.net/questions/416987 | 4 | Let $\zeta$ be the Riemann zeta-function, and let $t> 0$. I'm interested in the growth rate of
$$
\left|\frac{\zeta'}{\zeta}\left(-\frac{1}{2}+it\right)\right|
$$
as $t\to\infty$. It is easy to find references in the literature to estimations like $O(\log t)$. What is the actual growth rate?
| https://mathoverflow.net/users/3199 | Correct growth rate of logarithmic derivative of zeta, outside critical strip | It is $\log t + O(1)$. Without the absolute value, $-(\zeta'/\zeta)(-1/2 + it) = \log |t| + O(1)$ as $|t| \to \infty$.
First, by taking the logarithmic derivative of the functional equation
$$
\zeta(1-s) = 2(2\pi)^{-s}\Gamma(s)\cos\left(\frac{\pi}{2}s\right)\zeta(s)
$$
we have
$$
-\frac{\zeta'(1-s)}{\zeta(1-s)} = -\log(2\pi) + \frac{\Gamma'(s)}{\Gamma(s)} - \frac{\sin((\pi/2)s)}{\cos((\pi/2)s)}\frac{\pi}{2} + \frac{\zeta'(s)}{\zeta(s)}.
$$
Let $s = 3/2 - it$:
$$
-\frac{\zeta'(-1/2 + it)}{\zeta(-1/2+it)} = -\log(2\pi) + \frac{\Gamma'(3/2-it)}{\Gamma(3/2-it)} - \frac{\sin((\pi/2)(3/2-it))}{\cos((\pi/2)(3/2-it))}\frac{\pi}{2} + \frac{\zeta'(3/2-it)}{\zeta(3/2-it)}.
$$
All terms on the right side, as functions of $t$, are bounded except for the $\Gamma'/\Gamma$ term. The last term on the right side bounded since
$|\zeta'(s)/\zeta(s)| \leq |\zeta'(\sigma)/\zeta(\sigma)|$ for $\sigma = {\rm Re}(s) > 1$ by the Dirichlet series for $\zeta'(s)/\zeta(s)$ when ${\rm Re}(s) > 1$. The trigonometric term is bounded since on each vertical line in $\mathbf C$ we have $|\sin(\sigma + it)|, |\cos(\sigma + it)| \sim (1/2)e^{|t|}$ as $|t| \to \infty$, so $|\tan(\sigma + it)| \sim 1$ as $|t| \to \infty$. (That growth estimate on $|\sin(\sigma + it)|$ and $|\cos(\sigma + it)|$ as $|t| \to \infty$ is true not just on each vertical line in $\mathbf C$, but also in each vertical strip $a \leq \sigma \leq b$ in $\mathbf C$ and is uniform in $\sigma$, so $|\tan(\sigma + it)| \to 1$ as $|t| \to \infty$ in each vertical strip. Thus $\tan(\sigma + it)$ is bounded as $|t| \to \infty$ in each vertical strip.)
It remains to see what $\Gamma'(s)/\Gamma(s)$ looks like on the vertical line ${\rm Re}(s) = 3/2$. I'll show for $s = 3/2 - it$ that it is $\log|t| + O(1)$. The proof of the complex Stirling's formula for $\log \Gamma(s)$ in Lang's *Complex Analysis* (3rd edtion, Chapter XV, Sect. 2) gives an exact error term (see equation $\Gamma$ 13 on the top of p. 405) and differentiation of that formula gives a formula for $\Gamma'(s)/\Gamma(s)$ with an exact error term (see equation $\Gamma$ 15 on the top of p. 410) that implies
$$
\frac{\Gamma'(s)}{\Gamma(s)} = \log s + O\left(\frac{1}{|s|}\right)
$$
on the right half-plane ${\rm Re}(s) > 0$ (more precisely, on the complex plane outside of an angular sector symmetric with respect to the negative real axis, exactly the kind of region where the complex Stirling's formula is proved). Here $\log s$ means the usual logarithm defined on the complement of the real axis: if $s = re^{i\theta}$ where $-\pi < \theta < \pi$, then $\log s := \log r + i\theta$. So $\log s = \log r + O(1) = \log |s| + O(1)$.
Taking $s = 3/2 - it$,
$$
\frac{\Gamma'(3/2-it)}{\Gamma(3/2-it)} = \log\left|\frac{3}{2}-it\right| + O(1)
$$
as $|t| \to \infty$. Since $|3/2 - it| = \sqrt{9/4 + t^2} \sim |t|$ as $|t| \to \infty$, $\log|3/2 - it| = \log |t| + o(1)$, so
$$
\frac{\Gamma'(3/2-it)}{\Gamma(3/2-it)} = \log|t| + O(1)
$$
as $|t| \to \infty$. Thus $-(\zeta'/\zeta)(-1/2 + it) = \log|t| + O(1)$ as $|t| \to \infty$.
The same reasoning shows in each vertical strip $a \leq \sigma \leq b < 0$ to the left of the imaginary axis,
$-(\zeta'/\zeta)(\sigma + it) = \log |t| + O(1)$ as $|t| \to \infty$, uniformly in $\sigma$.
More generally, suppose $L(s)$ is an Euler product of degree $d$ with an absolutely convergent Dirichlet series for ${\rm Re}(s) > 1$ and $\Lambda(s) := cA^{s/2}\prod\_{j=1}^d \Gamma((\mu\_j/2)s + \nu\_j/2)L(s)$ has a meromorphic continuation to $\mathbf C$ that satisfies a functional equation $\Lambda(s) = w\overline{\Lambda(1-\overline{s})}$, where $c \not= 0$, $A > 0$, $\mu\_j > 0$, and
$w \not= 0$. Then for each $\sigma < 0$ (or in fact uniformly in each vertical strip to the left of the imaginary axis), we have $-(L'/L)(\sigma + it) = (\sum\_{j=1}^d \mu\_j)\log |t| + O(1)$ as $|t| \to \infty$. (In the terminology of the Selberg class, $\sum\_{j=1}^d \mu\_j$ is the "degree" of $L(s)$. In practice each $\mu\_j$ is $1$, so that sum is $d$.) The $\Gamma$-factor in the definition of $\Lambda(s)$ can be written in many ways by identities for the $\Gamma$-function, so no individual $\mu\_j$ is uniquely determined, but the estimate on $-(L'/L)(s)$ on vertical lines in the left half-plane, coming from a functional equation, shows $\sum\_{j=1}^d \mu\_j$ is well-defined.
| 10 | https://mathoverflow.net/users/3272 | 416990 | 169,891 |
https://mathoverflow.net/questions/416985 | 1 | I am wondering if there is a way to formulate or generate a matrix $X \in R^{n\times n}$ whose column vectors $\{x\_1,\dotsc,x\_i,\dotsc,x\_n\}$ are such that $x\_i$ and $x\_j$ are orthogonal iff $\lvert i-j\rvert>m$. The matrix $X^T X$ would be a band matrix with zeros off-band. I do not have a particular requirement for $x\_i \cdot x\_j$ when $\lvert i-j\rvert\le m$ other than being non-zero. I do require, however, that all vectors are distinct.
One example of $X$ would be a band matrix with bandwidth $2m+1$ and zero off-band. However, I am looking for a more general formula. I would like to somehow link it to SVD, but I am not sure how to. Any ideas would be appreciated!
| https://mathoverflow.net/users/173974 | Generate an ordered set of mostly orthogonal vectors $\{x_i\}$ where $x_i \cdot x_j =0$ iff $\lvert i-j\rvert >m$ | It seems that the band matrix example is an essentially sufficient characterization, up to a choice of basis.
The orthogonality constraint forces zeroes in the QR decomposition $X=QR$: $R$ is an upper triangular $m$-band matrix (and $Q$ is our usual orthogonal matrix.)
Conversely, any such pair $Q$, $R$ generates an $X$ where any two column vectors further than $m$ apart are constructed from different column vectors in $Q$ and are thus orthogonal.
| 2 | https://mathoverflow.net/users/69359 | 416992 | 169,892 |
https://mathoverflow.net/questions/315115 | 2 | For the integral kernel of the Laplacian $\Delta$ on $\mathbb{R}^n$, consider the resolvent $R(\lambda) := (\lambda - \Delta)^{-1}$ and let $R(\lambda; x, y)$ be its kernel, which is a smooth function away from the diagonal $\{x=y\}$. If I calculated correctly, we have estimates of the form
$$|R(\lambda;x, y)| \leq C e^{-\delta(\lambda)|x-y|},$$
whenever $|x-y| \geq 1$, where $\delta(\lambda)$ is a certain positive constant, depending (quite explicitly) on the distance of $\lambda$ to the spectrum of $\Delta$.
**Q:** On general Riemannian manifolds, do we still have estimates like this on the resolvent kernel of the Laplacian (or more general operators)?
More specifically, if $M$ is a complete Riemannian manifold, say with bounded geometry, and $\Delta$ is the Laplace-Beltrami operator, is it true that for any $\varepsilon>0$, there exist constants $C, \delta>0$ such that we have
$$|R(\lambda;x, y)| \leq C e^{-\delta\,d(x, y)}$$
whenever $d(x, y) \geq 1$ and $\lambda$ has distance at least $\varepsilon$ from the spectrum of $\Delta$? Here $d(x, y)$ is the Riemannian distance function.
| https://mathoverflow.net/users/16702 | Exponential decay of resolvent kernel | The local statement is correct for any positive self-adjoint extension of the Laplace operator on a Riemannian manifold (not necessarily complete). It is an old result by Hoermander. See the proof of Prop. 4.8 in
"L. HOERMANDER, On the Riesz means of spectral functions and eigenfunction expansions
for elliptic differential operators, in “Belfer Graduate School of Science Annual Science
Conference Proceedings” (A. Gelbart, Ed.), pp. 155-202, Vol. 2, 1969," where this is proved even much more generally for any differential operator. If you want uniform constants you need to assume bounded geometry and adapt his proof, using the appropriate calculus.
| 1 | https://mathoverflow.net/users/471159 | 417008 | 169,897 |
https://mathoverflow.net/questions/416989 | 2 | Let $M=(E,I)$ be a paving matroid with rank $n$. Let $A\subset E$ be an $n-1$ subset. How many bases of $M$ containing $A$ exist? (Note that every $n-1$ subset of $E$ is independent.)
| https://mathoverflow.net/users/165074 | Counting certain bases of a paving matroids | In an $n$-dimensionsl space, take a hyperplane $H$. Take a set of vectors in $H$ in a general position, and add one vector $x$ outside $H$. This is a paving matroid of rank $n$.
Now, if $|A|=n-1$ and $x\notin A$, then the only vector which can be added without breaking independency is $x$, so the minimal number of sought bases is $1$.
If $x\in A$, then any vector can be added in order to form a base So the maximal number is $|E|-n+1$.
| 2 | https://mathoverflow.net/users/17581 | 417016 | 169,899 |
https://mathoverflow.net/questions/416965 | 9 | This question is inspired by [xkcd #2585 (*Rounding*)](https://xkcd.com/2585/):
Let $u\_0,\ldots,u\_n$ be positive real numbers (we can assume w.l.o.g. that $u\_0=1$) or “units”.
Consider the following directed graph: its vertices are pairs $(i,k)$ where $0\leq i\leq n$ designates a unit, and $k\in\mathbb{Z}$ is a “measurement” performed in that unit. We place a directed edge from $(i,k)$ to $(j,\ell)$ whenever
$$
\ell = \lfloor k \cdot u\_i/u\_j \rceil
$$
where $\lfloor—\rceil$ means “closest integer to” (i.e., $\lfloor x\rceil = \lfloor x+\frac{1}{2}\rfloor$). Maybe assume that all $u\_i/u\_j$ are irrational so there is never any ambiguity as to what the closest integer means. In other words, we can get from $(i,k)$ to $(j,\ell)$ by converting unit $u\_i$ into unit $u\_j$ and rounding the result to the nearest integer.
**Question:** does there exist $u\_0,\ldots,u\_n$ such that the graph just defined has an infinite strongly connected component?
(In other words, can we find units such that infinitely many different values are reachable from one another by converting between these units and rounding to the nearest integer?)
| https://mathoverflow.net/users/17064 | How far away can we get by multiple rounding and unit change? | For a vertex $(i,k)$, it is reasonable to define its *value* as $ku\_i$. We show that the value cannot grow too large, which establishes a negative answer to the question.
Let $v$ be the current value, and let $V>v$ be a real number such that $\{V/u\_i\}<1/2$ for all $i$. Such a $V$ exists: we can start by choosing a very small $w>0$ such that $0<\{w/u\_i\}<\frac12$, and then approximate those values by $\{V/u\_i\}$ due to Kronecker.
We claim that the value cannot exceed $V$. Indeed, if the current value $ku\_i$ does not exceed $V$, then $ku\_i/u\_j$ rounds to at most $\lfloor V/u\_j\rfloor$ (check the brackets’ corners!), so the new value does not exceed $V$.
| 7 | https://mathoverflow.net/users/17581 | 417018 | 169,900 |
https://mathoverflow.net/questions/416976 | 4 | Let $\tau\_X$ denote the collection of open subsets of a topological space $X$ and let $\mathsf{RO}(X)$ be the subset of $\tau\_X$ made up of regular open subsets. With this terminology, the inequality $|\mathsf{RO}(X)|\leq |\tau\_X|$ is obvious. Furthermore, if $X$ is a semiregular space, it is easy to see that $|\tau\_X|\leq 2^{|\mathsf{RO}(X)|}$.
It is well-known (see for example section 10 of Hodel's chapter in the Handbook) that every perfectly normal Hausdorff space satisfies the equality $|\mathsf{RO}(X)|=|\tau\_X|$. My question goes in two directions:
* What are some examples of Tychonoff spaces $X$ such that $|\mathsf{RO}(X)|< |\tau\_X|$?
* Can the "perfect normality" condition be relaxed in such a way that the equality remains true? E.g., is it true that every perfect (closed sets are $G\_\delta$) Tychonoff space verifies the equality $|\mathsf{RO}(X)|=|\tau\_X|$?
| https://mathoverflow.net/users/146942 | $|\mathsf{RO}(X)|$ vs. $|\tau_X|$ for Tychonoff spaces | The compact Hausdorff space $X = \beta\mathbb{N}$ is another example. Every regular open subset is the closure of a subset of $\mathbb{N}$ and there are only $\frak{c}$ such subsets but $X$ has $2^{\frak{c}}$ points, and for each such point $p$, the set $X \setminus \{p\}$ is an open set.
For the second question, a scattered separable space of cardinality $\frak{c}$ and scattered height $2$ (such as a Cantor Tree, or a $\Psi$-space of size $\frak{c}$) gives a space in which every closed subset is a G$\_\delta$ but in which the equality fails for the same reason that it fails in the tangent-disk space.
| 4 | https://mathoverflow.net/users/89233 | 417024 | 169,902 |
https://mathoverflow.net/questions/417001 | 0 | Let $G$ be a finitely generated group and $\varphi:G\to \operatorname{Aut}(\mathbb C)$ a homomorphism, where $\operatorname{Aut}(\mathbb C)$ is the group of complex affine transfromations $a z+b$.
Can we find a torsion free-subgroup $H$ of $G$ with finite index? And can we find a normal subgroup $H$ which is torsion-free with finite index?
| https://mathoverflow.net/users/356774 | Torsion-free subgroup of affine group | Yes. More generally, for any field $K$ we have an embedding of $\operatorname{Aff}(K^n)$ in $\operatorname{GL}\_{n+1}(K)$, and so if $K$ has characteristic zero we can apply Selberg's lemma to conclude that a finitely generated group of affine transformations of $K^n$ is virtually torsion-free. The normal core of any finite index torsion-free subgroup will be a normal finite index torsion-free subgroup.
| 6 | https://mathoverflow.net/users/24447 | 417027 | 169,903 |
https://mathoverflow.net/questions/416998 | 0 | The game allows you to bet only \$1 at a time, and if you win, you end up \$1 richer, otherwise you end up \$1 poorer. Probability of winning is $p=0.45$. If you start with \$1 - what is expected number of games before bust (i.e., \$0 dollars)?
My initial approach was essentially just:
$\sum\_{n=1}^{\infty} (2n-1)(0.45^{n-1})(0.55^n) {\approx} 1.212$ where $2n-1$ is the number of games played (as it must be odd because you start with \$1 and therefore must lose one more game than you win.
| X = No. of Games | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | 19 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
| P(X) | 0.55 | 0.136125 | 0.0336909375 | 0.008338507031 | 0.00206378049 | 0.0005107856713 | 0.0001264194537 | 0.00003128881478 | 0.000007743981658 | 0.00000191663546 |
(i.e., a sum product of the table above)
This is apparently wrong and I'm unsure where my logic / thinking is wrong. Would love any insight!
| https://mathoverflow.net/users/477770 | Expected number of games until bust | Let $x$ be the expected time to get to \$0, starting from \$1.
What's the expected time to get to \$0, starting from \$2? It's $2x$. That's because to get from \$2 to \$0, we first need to get from \$2 to \$1, and then we need to get from \$1 to \$0. And getting from \$2 to \$1 behaves exactly the same as getting from \$1 to \$0.
Now we "condition on the first step". Starting from \$1, we make one step, and then we're either finished (with probability $0.55$), or we're at \$2 (with probability $0.45$). So we get
$$
x = 1 + 0.45 \times 2x.
$$
Solving this gives $\boxed{x=10}\,$.
(More generally, if $p<1/2$ is the probability of winning a dollar at each step, then $\boxed{x=1/(1-2p)}\,$.)
(As others observed, your original method doesn't work because your sum only counts one way of going bust after $2n-1$ games, for each $n$. But for any $n>2$, there are more. e.g. for $n=3$, you have two, WWLLL and WLWLL.)
| 12 | https://mathoverflow.net/users/5784 | 417035 | 169,906 |
https://mathoverflow.net/questions/417012 | 1 | Let $u\leq v$ be two locally bounded subharmonic functions in a domain in $\mathbb{R}^n$. Assume that $u=v$ on a dense subset.
**Is it true that $u=v$ everywhere?**
| https://mathoverflow.net/users/16183 | Equality of two subharmonic functions | This is not the case. Here is a counterexample in the case where $n=2$.
Suppose that $A$ is a countable dense subset of $\mathbb{R}^{2}=\mathbb{C}$. Let $r\_{a}>0$ for each $a\in A$, and suppose that $\sum\_{a\in A}r\_{a}\cdot(1+\max(0,\log(a)))<\infty$.
Let $p(z)=\sum\_{a\in A}r\_{a}\cdot\log(|z-a|)$. Let $s:\mathbb{C}\rightarrow(0,\infty)$ be any subharmonic function. Let $q(z)=p(z)+s(z)$. Then observe that $p(z)=q(z)=-\infty$ on a dense $G\_{\delta}$-set. In particular, $p=q$ on a dense set. The functions $p,q$ are not locally bounded, but we can obtain locally bounded functions from $p,q$.
The function $p$ cannot be $-\infty$ everywhere since
$$\frac{1}{2\pi}\int\_{0}^{2\pi}p(re^{i\theta}+z\_{0})d\theta=(\sum\_{a\in A,|z\_{0}-a|<r}r\_{a})\cdot\log(r)+\sum\_{a\in A,|z\_{0}-a|>r}r\_{a}\cdot\log(|z\_{0}-a|)>-\infty.$$
In fact, the set $p^{-1}[\{-\infty\}]$ is a polar set and polar subsets of $\mathbb{C}$ have Hausdorff dimension $0$ and therefore Lebesgue measure zero as well.
Let $u(z)=e^{p(z)},v(z)=e^{q(z)}$. As a consequence of Jensen's inequality, the functions $u(z),v(z)$ are subharmonic, locally bounded, and $u\leq v$. However, $u(z)=v(z)=0$ on a dense $G\_{\delta}$-set, and if $p(z)>-\infty$, then $p(z)<q(z)$, so $u(z)<v(z)$.
The fact that $u(z)=v(z)=0$ on a dense $G\_{\delta}$-set should be contrasted with the fact that if $f,g$ are subharmonic functions that are equal almost everywhere, then $f=g$.
| 2 | https://mathoverflow.net/users/22277 | 417044 | 169,909 |
https://mathoverflow.net/questions/416993 | 1 | Is there a standard phrase for the largest odd factor of a positive integer $n$, or more generally for $n$ divided by the largest power of $p$ that divides it (with $p$ some fixed prime)? Five minutes of online search failed to yield an answer.
(I’m hoping for an answer to my terminology question, but general advice about how to find this sort of information on my own would be appreciated as well. I hate to clutter up MathOverflow with questions that I could answer for myself if I knew how to search the internet more intelligently.)
| https://mathoverflow.net/users/3621 | Nomenclature for largest odd factor | Call it the $2$-free part of $n$ or the odd part of $n$. For a general prime $p$, the $p$-free part of $n$ is what you have after you divide $n$ by the highest power of $p$ dividing it.
| 5 | https://mathoverflow.net/users/3272 | 417047 | 169,912 |
https://mathoverflow.net/questions/416638 | 0 | Is it possible to get a good upper bound for $$\sum\_{1\leq |h|\leq q}\frac{c\_{q}(a-h)}{h}$$ with $(a,q)=1$ and $1\leq a\leq q$.
| https://mathoverflow.net/users/155294 | Ramanujan's type sum | If I'm not mistaken, the $h>0$ part of the sum is
\[ \sum \_{d|q}d\mu (q/d)\sum \_{h\leq q\atop {h\equiv a(d)}}\frac {1}{h}\]
and here the inner sum is
\[ \sum \_{0\leq h\leq (q-a)/d}\frac {1}{hd+a}\leq \frac {\log q}{d}+\frac {1}{a}\]
so that the whole sum is
\[ \leq \sum \_{d|q}d\left (\frac {\log q}{d}+\frac {1}{a}\right )\ll q^\epsilon +\frac {q}{a},\]
whilst for prime $q$ the ($h>0$ part of the) sum is literally
\[ -\sum \_{h\leq q\atop {h\not =a}}\frac {1}{h}+\frac {p-1}{a}\ll \log q+\frac {p}{a}.\]
What kind of bound were you expecting/hoping for?
| 2 | https://mathoverflow.net/users/110603 | 417059 | 169,916 |
https://mathoverflow.net/questions/417077 | 1 | I've been trying to read through the following paper:
<https://arxiv.org/pdf/0704.1751.pdf>
And I've been stuck on the proposition in the middle of page 8 which says that if a r.v. $X$ has independent entries, then:
$$\textbf{J}(X) = \text{diag}(J(X\_i))\_i$$
Where $\textbf{J}(X) := \text{Cov}(S(X))$ and $J(X):=\text{tr}(\textbf{J}(X))$.
What I have so far: Assume $X$ has independent entries, and p.d.f $f$. Then
$$S(X) = \frac{\nabla f(x)}{f(x)} = \frac{1}{\prod\_if\_i(x\_i)}\begin{bmatrix}\frac{\partial}{\partial x\_1}\prod\_if\_i(x\_i)\\ \vdots
\\ \frac{\partial}{\partial x\_n}\prod\_if\_i(x\_i)\end{bmatrix} = \begin{bmatrix}f'(x\_1)/f(x\_1)\\ \vdots
\\ f'(x\_n)/f(x\_n)\end{bmatrix} = \begin{bmatrix}S(X\_1)\\ \vdots
\\ S(X\_n)\end{bmatrix}$$
From this we get $\textbf{J}(X)\_{ii} = E[S(X\_i)^2] = J(X\_i).$ Where I'm having trouble is showing that the off-diagonal entries are 0, namely that if $i\neq j$ then
$$E[S(X\_i)S(X\_j)] = 0$$
Thanks
| https://mathoverflow.net/users/477840 | Covariance matrix of score of random vector with independent entries is diagonal | The equality $ES(X\_i)S(X\_j)=0$ for $i\ne j$ follows because $X\_i$ and $X\_j$ are independent and $ES(X\_i)=0$, so that
$$ES(X\_i)S(X\_j)=ES(X\_i)\,ES(X\_j)=0.$$
(The equality $ES(X\_i)=0$ holds because for the density $p\_i$ of $X\_i$ we have
$$ES(X\_i)=E\frac{p\_i'(X)}{p\_i(X)}=\int\_{-\infty}^\infty\frac{p\_i'(x)}{p\_i(x)}\,p\_i(x)\,dx
=\int\_{-\infty}^\infty p\_i'(x)\,dx=0$$
under standard assumptions on $p\_i$, such as $\int\_{-\infty}^\infty |p\_i'(x)|\,dx<\infty$.)
| 0 | https://mathoverflow.net/users/36721 | 417078 | 169,918 |
https://mathoverflow.net/questions/417068 | 2 | A simple, undirected graph is vertex-transitive if for any pair of vertices $x,y$, there exists an automorphism (adjacency-preserving self-bijection) $\phi$ such that $\phi(x)=y$.
What if, instead of **taking** $x$ to $y$ as above, we require the automorphism $\phi$ to **exchange** $x$ and $y$, i.e. $\phi(x)=y$ and $\phi(y)=x$?
1. Is there a name for this natural refinement of the notion of vertex-transitivity?
2. What is a simple example of a vertex-transitive graph which does not satisfy this?
Note that any Cayley graph whose generating set is conjugacy-invariant does satisfy this exchange property (take $\phi(u)=xu^{-1}y$).
| https://mathoverflow.net/users/477827 | Imposing reciprocity in the definition of vertex-transitivity | A permutation group $G$ acting on a set $X$ is called *generously transitive* if for any two elements $x$, $y \in X$ there is a permutation $g \in G$ such that $x^g = y$ and $y^g = x$.
It is fairly easy to find examples of vertex-transitive graphs whose automorphism group is not generously transitive, by searching MathOverflow or math.stackexchange.
For example: <https://math.stackexchange.com/questions/953856/automorphism-groups-of-vertex-transitive-graphs>
| 2 | https://mathoverflow.net/users/1492 | 417095 | 169,927 |
https://mathoverflow.net/questions/415789 | 0 | By the [Tennenbaum's theorem](https://en.wikipedia.org/wiki/Tennenbaum%27s_theorem),
there are no non-standard countable models of
[Peano Arithmetic](https://en.wikipedia.org/wiki/Peano_axioms#Peano_arithmetic_as_first-order_theory) that are computable using Turing machines. What about models of infinitary computation like [infinite time Turing machines](https://arxiv.org/abs/math/9808093),
or [ordinal Turing machines](https://www.researchgate.net/publication/2117812_Turing_Computations_On_Ordinals)?
**Question**: Is there a nonstandard model of PA (countable or not) where both,
or at least one of addition and multiplication are computable by an ITTM or an OTM?
| https://mathoverflow.net/users/116605 | Is there a non-standard model of PA computable with infinitary computation? | To move this off the unanswered queue, the answer is **yes** in a very strong way.
The relevant notion is that of a **PA degree**. There are many equivalent definitions of PA-ness, as well as many interesting results about them, but for the current purposes the relevant ones are the following I think:
* **Definition**: **d** is PA iff every infinite *computable* subtree of $2^\mathbb{N}$ has a **d**-computable infinite path.
* **Theorem** (Shoenfield?): The following are equivalent for each degree **d**:
1. **d** is PA.
2. There is a **d**-computable nonstandard model of PA (this justifies the name, and explains the relevance to the current question).
+ Every computable consistent first-order theory has a **d**-computable model (this shows that the model-theoretic side of PA-ness is extremely robust).
* **Theorem**: there are surprisingly simple PA-degrees - namely ones $\le\_T\emptyset'$ (Shoenfield), $<\_T\emptyset'$ (Kreisel), and even low (Jockusch/Soare). And there are other basis theorems too, but these three in order tell the simplest story in my opinion.
Consequently, any modification of the Turing machine model in which the usual halting problem is newly computable will be able to construct nonstandard models of PA. ITTMs, OTMs, and every other variation on the basic theme of infinitary computation goes well beyond this; the closest variant I can think of is *limit computability*, which [Shoenfield showed corresponds](https://en.wikipedia.org/wiki/Computation_in_the_limit) to computability relative to the halting problem.
See e.g. [Diamondstone/Dzhafarov/Soare](https://damir.math.uconn.edu/papers/pi01classes.pdf) for more information on PA degrees and the attendant notions.
| 4 | https://mathoverflow.net/users/8133 | 417098 | 169,930 |
https://mathoverflow.net/questions/417022 | 3 | I believe I found a complicated proof by bounding the spectral norm $||uu^T-vv^T||^2\_2:=\max\_{||x||=1}|(u^Tx)^2-(v^Tx)^2|$.
Using the fact that $dist(x,y):=\sin|x-y|$ is a distance function over unit vectors, we may prove
$$(v^Tx)^2-(u^Tx)^2=\sin^2(u,x)-\sin^2(v,x)=dist^2(u,x)-dist^2(v,x)
=(dist(u,x)-dist(v,x))(dist(u,x)+dist(v,x)) \leq 2dist(u,v),$$
but the approximation factor is too large.
Can you improve the bound using small changes to the above, or find another simple proof using trigo or Taylor's approx?
Note: This is a follow-up on a previous question that was stated wrong [here](https://mathoverflow.net/questions/414965/prove-that-vtx2-utx2-1-utv2-for-any-unit-vectors-u-v-x). I hope it is correct this time.
| https://mathoverflow.net/users/476110 | Prove that $(v^Tx)^2−(u^Tx)^2\leq \sqrt{1−(u^Tv)^2}$ for any unit vectors $u, v, x$ | This is a variation of Fedor Petrov's proof (found independently). First, it suffices to prove the statement when $x$ lies in $V:=\mathbb{R}u+\mathbb{R}v$. Indeed, we see this reduction readily by decomposing $x$ as $cy+z$, where $c\in[0,1]$ is a scalar, $y\in V$ is a unit vector, and $z\in V^\perp$. By this reduction, we can assume that $V=\mathbb{R}^2$, and
$$\qquad x=(1,0),\qquad u=(\cos\alpha,\sin\alpha),\qquad v=(\cos\beta,\sin\beta).$$
So we need to prove that
$$\cos^2\beta-\cos^2\alpha\leq|\sin(\alpha-\beta)|.$$
However, this is clear, because the left-hand side equals $\sin(\alpha+\beta)\sin(\alpha-\beta)$.
**P.S.** I upvoted the question and the other three answers, for fun and world peace.
| 7 | https://mathoverflow.net/users/11919 | 417107 | 169,933 |
https://mathoverflow.net/questions/417102 | 9 | Ellis Lemma on idempotent elements asserts that:
>
> **Lemma (Ellis).** Every compact semigroup has an idempotent.
>
>
>
The proof below is excerpted from Todorcevic's *Introduction to Ramsey Spaces*, [Lemma 2.1](https://books.google.com/books?id=Yhyi8vRfPNAC&pg=PA27).
>
> Let $S$ be a compact semigroup. Pick by Zorn’s Lemma a minimal compact subsemigroup $R \subseteq S$ and an arbitrary $s \in R$. Then $Rs$ is also a compact subsemigroup and $Rs \subseteq R$. Hence $Rs = R$. Let $P = \{x \in R : xs = s\}$. Then $P \neq \emptyset$, since $s \in Rs$. Note that $P$ is also a compact subsemigroup of $S$. Hence $P = R$ and therefore
> $s^2 = s$.
>
>
>
It appears that Zorn's Lemma is required to obtain a minimal compact subsemigroup $R \subseteq S$. I do not understand why, as it appears to me that we can simply let:
$$
R := \bigcap \{Q \subseteq S : Q \text{ is a compact subsemigroup of $S$}\}
$$
What is the error in reasoning here?
| https://mathoverflow.net/users/146831 | Why is choice needed in Ellis' Lemma? | The issue is that $R$ could be empty. When you apply Zorn's lemma, a nested intersection of non-empty compact sets is non-empty, guaranteeing minimal non-empty elements.
| 9 | https://mathoverflow.net/users/11552 | 417109 | 169,934 |
https://mathoverflow.net/questions/417085 | 1 | Let $X$ be a topological space and consider a continuous function $f:X\to [0,+\infty)$. For $c\geq 0$ set $X\_c := f^{-1} ([0,c])$.
Furthermore, suppose that $X\_0 \neq \emptyset$ and $f$ is proper.
**Motivating example**
If $X$ is a closed manifold and $f$ is smooth and $0$ is a regular value, then $X\_0$ is a smooth submanifold and $X\_c\cong X\_0\times [0,1]$ when $c$ is small enough. Consequently, the cohomology of the pair $H^\*(X\_c, X\_0)\simeq (0)$.
If $X$ is a closed manifold and $f$ is smooth (without the assumption of $0$ being a regular value)the same still holds even if $X\_0$ now can have singularities, indeed we can consider the flow of the normalized gradient of $f$, $\Phi:X\times \mathbb{R}\to X$ which gives a retraction of $X\_c$ onto $X\_0$.
What can be said in the general case?
>
> Is it still true that there for all $c$ sufficiently small $H^\*(X\_c, X\_0) \simeq (0)$? Does it hold at least when $X$ is a topological manifold?
>
>
>
| https://mathoverflow.net/users/99042 | Vanishing of $H^*(f^{-1}[0,c], f^{-1}(0))$ for small $c$, and $f\in C^0(X, [0,+\infty))$ | With this degree of generality I don't think you can say anything. Here are two pathological examples:
* If X is the [Hawaiian earring](https://en.wikipedia.org/wiki/Hawaiian_earring) and $f(x,y) = x$, then for any $c>0$, $f^{-1}([0,c])$ retracts on a copy of the Hawaiian earring (hence has very large $H^1$).
* If X is $\mathbb R^2$ and
$f(x,y) = e^{\frac{-1}{x^2+y^2}}h(x^2+y^2)$ where $h$ is a smooth positive bounded function on $\mathbb R\_{>0}$ that oscillates very fast near $0$, then there are arbitrary small values $c$ such that $f^{-1}([0,c])$ has small will have a connected component which is an annulus (hence has non-zero $H^1$). Choosing $h$ carefully enough, you can even arrange so that this is true for all $c$.
| 1 | https://mathoverflow.net/users/173096 | 417115 | 169,935 |
https://mathoverflow.net/questions/417108 | 6 | [Solèr’s theorem](https://arxiv.org/abs/math/9504224) says that for every star division ring $R$ and every $R$-module $H$ with an orthomodular Hermitian form $\langle (-),(-) \rangle:H \times H \to R$ such that there exists an infinite orthonormal sequence $e:\mathbb{N} \to H$, $R$ is either the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, or the quaternions $\mathbb{H}$, and $H$ is a Hilbert space over $R$. Assuming that the star division rings used are Heyting division rings (or else Solèr’s theorem is most likely false), is Solèr’s theorem true in constructive mathematics?
| https://mathoverflow.net/users/nan | Is Solèr’s theorem true in constructive mathematics? | Suppose you have a classical classification theorem saying
>
> Each structure (of a certain kind) is either an $A$ or a $B$.
>
>
>
Then you cannot exhibit constructively a $C$ which is neither $A$ nor $B$ because every constructive proof is also classical, and so that would contradict the classical classification.
What happens instead is that the classical meaning of “$p$ or $q$” corresponds to the constructive “not ($\neg p$ and $\neg q$)“ so the constructive reading of the classical classification theorem is
>
> Apart from $A$ and $B$, there is no other structure (of a certain kind).
>
>
>
To give you an idea on how to play tricks with excluded middle in constructive mathematics, consider any proposition $p$ and define
$$K\_p = \{ z \in \mathbb{C} \mid p \Rightarrow z \in \mathbb{R} \}.$$
It is easy to check, regardless of what $p$ is, that $K\_p$ is a subfield of $\mathbb{C}$.
Moreover, if $p$ holds then $K\_p = \mathbb{R}$, and if $\neg p$ holds then $K\_p = \mathbb{C}$. But stating that $K\_p$ is either $\mathbb{R}$ or $\mathbb{C}$ implies $\neg \neg p \lor \neg p$, which lets us decide $\neg p$, which is not generally possible in constructive mathematics. It is still the case that $K\_p$ cannot be different from both $\mathbb{R}$ and $\mathbb{C}$, because that amounts to $\neg (\neg p \land \neg\neg p)$, which is constructively true (obviously).
Consequently, if Solèr’s theorem were true constructively (in the version that says that every structure is either this or that), we could decide $\neg p$: just ask the theorem to classify $K\_p$ for you.
The usual solution to the conundrum is to strengthen the assumptions to something that does not matter classically. For example, you might want to assume that the vector space $E$ featuring in the definition of a Hermitian form has a given basis, and that the basis has size which is *either* a natural number *or* is countably infinite (this cannot be shown to hold for $K\_p$ above seen as a real vector space). But that is only the first step, there will be further complications, and one would have to dig into the details of the classification theorem. Unfortunately, I do not know whether anyone has done so.
| 8 | https://mathoverflow.net/users/1176 | 417128 | 169,941 |
https://mathoverflow.net/questions/417123 | 2 | Two $n\times n$ positive definite symetric matrices $A,B$ define two normed spaces $E\_A=({\mathbb R}^n;\|\cdot\|\_A)$ and $E\_B=({\mathbb R}^n;\|\cdot\|\_B)$, where
$$\|x\|\_A=\sqrt{x^TAx},\qquad \|x\|\_B=\sqrt{x^TBx}.$$
>
> Is it true that the interpolation space $[E\_A,E\_B]\_{1/2}$ equals $E\_{A\sharp B}$, where $A\sharp B$ is the geometrical mean of $A$ and $B$ ?
>
>
>
Let me recall that $A\sharp B$ is the 'middle' of the geodesic from $A$ to $B$, with respect to the Riemannian metric for which ${\bf SPD}\_n$ is a symmetric space:
$$D(A,B)=\left({\rm Tr}(\log(BA^{-1}))^2\right)^{\frac12}.$$
A pedestrian way to ask the question is
>
> Let $f$ be a linear form such that $|f(x)|^4\le(x^TAx)(x^TA^{-1}x)$ for every $x\in{\mathbb R}^n$. Is it true that $|f(x)|\le\|x\|$ for the standard euclidian norm ?
>
>
>
| https://mathoverflow.net/users/8799 | Interpolation of normed spaces *vs* geometrical mean of positive matrices | Yes. The proof of Theorem 1.1 from John E. McCarthy's "Geometric interpolation between Hilbert spaces," Ark. Mat. 30, 321-330 (1991) works for this case. Let $A\_i$, $B\_i$, be SPD matrices, $i = 1, 2$, and let $$X\_i = A\_i^{1/2}(A\_i^{-1/2} B\_i A\_i^{-1/2})^{1/2}A\_i^{1/2}$$ be the geometric means of each pair $(A\_i,B\_i)$. Given any matrix $L$ with compatible dimensions, the norm $\lVert L \rVert\_{X\_2,X\_1} = \sup\_{\lVert u \rVert\_{X\_1} \le 1} \lVert L u \rVert\_{X\_2}$ has square bounded by
$$ \begin{aligned} \lVert L \rVert\_{X\_2,X\_1}^2
&= \rho(X\_1^{-1} L^\* X\_2 L) \le
\lVert X\_1^{-1} L^\* X\_2 L \rVert\_{A\_1} \\
&\le \lVert X\_1^{-1} L^\* X\_2 \rVert\_{A\_1,A\_2}
\lVert L \rVert\_{A\_2, A\_1} \\
&= \lVert X\_2 L X\_1^{-1} \rVert\_{A\_2^{-1},A\_1^{-1}}
\lVert L \rVert\_{A\_2, A\_1} \\
&= \lVert L \rVert\_{B\_2,B\_1}
\lVert L \rVert\_{A\_2, A\_1}
\end{aligned} $$
so that $(E\_{X\_1}, E\_{X\_2})$ is an exact interpolation pair of exponent $1/2$ for the couples $(E\_{A\_1}, E\_{B\_1})$ and $(E\_{A\_2}, E\_{B\_2})$. Note the last step above used that, e.g., $\lVert X\_1 u \rVert\_{A\_1^{-1}} = \lVert u \rVert\_{B\_1}$, which follows since $X\_1^\*A\_1^{-1} X\_1 = B\_1$. The prior step used the identity $\lVert T^\* \rVert = \lVert T \rVert$, accounting for the fact that $\lVert f \rVert\_{Y^{-1}}$ is the norm dual to $\lVert u \rVert\_Y$ for SPD $Y$.
| 3 | https://mathoverflow.net/users/70005 | 417152 | 169,945 |
https://mathoverflow.net/questions/416969 | 3 | Let $(M, \bar g)$, where $M=\mathbb{R}^n$, with coordinates $(x\_1, x\_2, .. x\_n)$, $f:\mathbb{R}^n \rightarrow \mathbb{R}$ and $\phi:\mathbb{R}^n \rightarrow \mathbb{R}$ with $n \geq 3$ and $f$ a positive function.
Let $g=\epsilon\_i \delta\_{ij}$ and $\bar g=\frac{1}{\phi^2}g$.
I denote $\phi\_{, x\_i x\_j}$, $\phi\_{, x\_i}$, $f\_{, x\_i x\_j}$ and $f\_{, x\_i}$ the second and the first order derivatives of $\phi$ and $f$ with respect to $x\_ix\_j$.
I consider smooth functions $\phi(\xi)$ and $f(\xi)$, where $\xi=\sum\_{i=1}^n\alpha\_ix\_i$, with $\alpha\_i \in \mathbb{R}$ and whenever $\sum\_{i=1}^n\alpha\_i^2\epsilon\_i \neq 0$, whitout loss of generality we may consider
$\sum\_{i=1}^n\alpha\_i^2\epsilon\_i=-1$.
So I mean:
$\phi\_{, x\_i x\_j}=\phi''\alpha\_i \alpha\_j$,
$\phi\_{, x\_i}= \phi' \alpha\_i$,
$f\_{, x\_i x\_j}=f'' \alpha\_i \alpha\_j$,
$f\_{, x\_i}=f'\alpha\_i$
I am trying to solve this system with these three PDE for *positive constant $\lambda$*:
**(I)** $(n-2)(f+1)\phi''-n\phi f''-2n\phi'f'=0$ for $i \neq j$.
**(II)** $-(f+1) \phi \phi'' + (n-1)(f+1) \phi'^2 - n\phi\phi' f'=\lambda (f+1)$ for $i=j$.
**(III)** $(f+1)\phi^2 f''-(n-2)(f+1)\phi \phi' f' + (n-1)\phi^2 f'^2=\lambda (f+1)^2$ for $i=j$.
| https://mathoverflow.net/users/111304 | PDE system solution on manifold with conformal metric | I don't know what the OP means by **canonical answer**. I assume that what's desired is to understand the solutions of the system of *three* second order ordinary differeential equations for *two* unknown positive functions $\phi(\xi)$ and $f(\xi)$.
At first glance, there's no reason to believe that any nontrivial solutions exist, since the system is overdetermined. One must first check compatibility conditions. Fortunately, this is easy to do, but to save writing (and typing), I'm going to simplify the notation: Instead of $\bigl(\xi,\phi(\xi),f(\xi)\bigr)$, I'll write $\bigl(t, u(t), v(t){-}1\bigr)$, and I'll also write $\lambda = nm^2/2>0$. The system of equations then becomes
$$\begin{aligned}
0 &= (n{-}2)\,v\,\ddot u-n\,u\,\ddot v - 2n\,\dot u\,\dot v\\
0&=-u\,v\,\ddot u - (n{-}1)\,v\,{\dot u}^2-n\,\dot u\,\dot v -\tfrac12 nm^2\,v\\
0&=v\,u^2\,\ddot v - (n{-}2)\,u\,v\,\dot u\,\dot v + (n{-}1)\,u^2{\dot v}^2-\tfrac12 nm^2\,v^2
\end{aligned}\tag1
$$
Solving the second and third equations for $\ddot u$ and $\ddot v$ and substituting them into the first equation, we find that the first equation can be replaced by a *first* order equation, namely
$$
0 = (n{-}2)\,v^2\,{\dot u}^2 - 2n\,u\,v\,\dot u\,\dot v + n\,u^2\,{\dot v}^2 - nm^2\,v^2 =: F(u,v,\dot u,\dot v).\tag2
$$
One then finds that, upon differentiating $F$ with respect to $t$ and then eliminating $\ddot u$ and $\ddot v$ using the second and third equations of (1), the resulting expression in $(u,v,\dot u,\dot v)$ is a multiple of $F(u,v,\dot u,\dot v)$. This shows that the combined system of equations (1) and (2) satisfies the compatibility conditions, so that the system has solutions, in fact, a 3-parameter family of them.
To describe these solutions more explicitly, note that the equations are $t$-autonomous and have a 2-parameter family of scaling symmetries. In particular, the equations are invariant under the $3$-parameter group of transformations of the form
$$
\Phi\_{a,b,c}(t,u,v) = (at{+}c,\,au,\,bv)
$$
where $a$ and $b$ are nonzero constants and $c$ is any constant.
In fact, the equation (2) implies that there is a function $p(t)$ such that
$$\begin{aligned}
\dot u &= \frac{2mn\,p\,(p{-}1)}
{\bigl((n{-}2)\,p^2-2n\,p + n\bigr)}\\
\dot v &= \frac{m\,v\,\bigl((n{-}2)p^2-n\bigr)}
{u\bigl((n{-}2)\,p^2-2n\,p + n\bigr)}
\end{aligned}\tag 3
$$
and then the second and third equations of (1) imply that $p$ must satisfy
$$
\dot p = \frac{m\bigl(n+2n\,p - (3n{-}2)\,p^2\bigr)}{u}\tag 4
$$
Conversely, the combined system of (3) and (4) gives the general solution of the original system.
This latter system is easily integrated by the usual separation of variables method: Eliminating $t$ yields a system of the form
$$
\frac{du}{u} = R(p)\,dp\qquad\text{and}\qquad
\frac{dv}{v} = S(p)\,dp
$$
where $R(p)$ and $S(p)$ are rational functions of $p$. Once $u$ and $v$ are written as elementary functions of $p$, then we can write
$$
dt = u\,T(p)\,dp,
$$
where $T$ is a rational function of $p$, so that $t$ can be written as a function of $p$ by quadrature. Thus, we have the integral curves in $(t,u,v,p)$-space in terms of explicit functions.
| 5 | https://mathoverflow.net/users/13972 | 417154 | 169,946 |
https://mathoverflow.net/questions/417100 | 5 | For ordinals $\alpha<\beta$, we say $\alpha<\_{el}\beta$, if there is an elementary embedding with domain $L\_\beta$ and critical point $\alpha$.
Is $<\_{el}$ transitive?
| https://mathoverflow.net/users/170286 | Is this relation about elementary embedding transitive? | EDIT: If the codomain $N$ is allowed to be illfounded, then the answer is yes.
(Here if $\kappa=\mathrm{crit}(j)$ then this will mean that $\kappa\subseteq N$, but it might be that $N$ is illfounded and $\kappa$ is exactly the wellfounded part of the codomain, in which case $\kappa\notin N$.)
For let $\kappa\_0<\beta\_0=\kappa\_1<\beta\_1$ be ordinals and $j\_i:L\_{\beta\_i}\to N\_i$ be elementary (for $i=0,1$) with $\mathrm{crit}(j\_i)=\kappa\_i$. Let $U\_0$ be some non-principal ultrafilter derived from $j\_0$ (let $x\in N\_0$ such that $N\_0\models$"$\alpha\in x\in j\_0(\kappa\_0)$" for each $\alpha<\kappa\_0$, and let $U\_0$ be the filter over $\kappa\_0$ derived from $x$). Then $U\_0$ is an $L\_{\beta\_0}$-ultrafilter over $\kappa\_0$ which is $L\_{\beta\_0}$-$\kappa\_0$-complete (i.e. closed under length ${<\kappa\_0}$-intersections of sequences $\left<X\_\alpha\right>\_{\alpha<\gamma}\in L\_{\beta\_0}$). And because $\kappa\_1=\mathrm{crit}(j\_1)$, $\kappa\_1$ is a (regular) cardinal in $L\_{\beta\_1}$. Therefore $\mathcal{P}(\kappa\_0)\cap L\_{\beta\_1}\subseteq L\_{\beta\_0}=L\_{\kappa\_1}$. But then $U\_0$ is also an $L\_{\beta\_1}$-ultrafilter over $\kappa\_0$ which is $L\_{\beta\_1}$-$\kappa\_0$-complete. So letting $N=\mathrm{Ult}(L\_{\beta\_1},U\_0)$ and $j:L\_{\beta\_1}\to N$ the ultrapower map, then $j$ is elementary and $\mathrm{crit}(j)=\kappa\_0$.
---
EDIT 2: On the other hand, if the codomain is required to be wellfounded, and $0^\sharp$ exists, the answer is no. For let $\kappa\_0$ be the least $L$-indiscernible and let $\pi:L\_{\omega\_1}\to L\_{\omega\_1}$ be elementary with $\mathrm{crit}(\pi)=\kappa\_0$. (Here $\omega\_1$ is as computed in $V$.)
There is an elementary $\sigma:L\_{\kappa\_0}\to L\_{\kappa\_0}$. For there is an elementary $\sigma':L\_{\kappa\_\omega}\to L\_{\kappa\_\omega}$ where $\kappa\_\omega$ is the $\omega$th $L$-indiscernible, for example the one with $\sigma'(\kappa\_n)=\kappa\_{n+1}$. But the existence of some such $\sigma'$ is forced over $L$ by $\mathrm{Coll}(\omega,\kappa\_\omega)$, and so by indiscernibility, the existence of a $\sigma$ as stated is forced over $L$ by $\mathrm{Coll}(\omega,\kappa\_0)$, and therefore there is actually some such embedding $\sigma$ in $V$.
So letting $\kappa=\mathrm{crit}(\sigma)$, then $\kappa<\_{\mathrm{el}}\kappa\_0<\_{\mathrm{el}}\omega\_1$. But I claim $\kappa\not<\_{\mathrm{el}}\omega\_1$.
For since $\kappa<\kappa\_0$, $\kappa$ is not an $L$-indiscernible. But if $j:L\_{\omega\_1}\to L\_{\lambda}$ is elementary (where $\lambda$ is some ordinal) then $\mathrm{crit}(j)$ is an $L$-indiscernible. (This is a standard fact. The key is that if $U$ is the normal ultrafilter derived from $j$, then $\mathrm{Ult}(L,U)$ is wellfounded.)
---
These observations leave open the version where the codomain is required to be wellfounded, but $0^\sharp$ does not exist.
| 11 | https://mathoverflow.net/users/160347 | 417157 | 169,948 |
https://mathoverflow.net/questions/416774 | 5 | $\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$Let $S\_\infty$ be the group of all permutations of a countable infinite set (enriched with the product Polish topology).
Does there exist a (continuous) group embedding $\phi: S\_\infty\to S\_\infty$ such that $\phi(G)$ is conjugate to $\phi(H)$ for every pair of (closed) isomorphic groups $G, H\leq S\_\infty$?
Note that two subgroups $A, B$ of a group $G$ are conjugate in $G$ if there is $g\in G$ for which $A=gBg^{-1}$.
**Additional note**: It would be stil of interest for me, if there is an embedding $\phi: S\_\infty \to T$ for some Polish group $T$, such that $\phi(G)$ is conjugate to $\phi(H)$ in $T$ for every pair of (closed) isomorphic groups $G, H\leq S\_\infty$?
My first guess was to consider $T$ to be the semidirect product of $S\_\infty$ with $\Aut(S\_\infty)=\Inn(S\_\infty)=S\_\infty$.
| https://mathoverflow.net/users/128723 | Are isomorphic subgroups of the symmetric group conjugate in some larger group? | No, there's no such endomorphism. More generally:
>
> **Proposition.** for every infinite set $X$, there exists no set $Y$ and nontrivial continuous homomorphism $f:S\_X\to S\_Y$ for which the following assertion holds: for any two isomorphic discrete infinite cyclic subgroups of $S\_X$, the images $f(S\_X)$ and $f(S\_Y)$ are conjugate.
>
>
>
Note: every group endomorphism of $S\_\omega$ is continuous (automatic continuity, here essentially due to Dixon-Neumann-Thomas: see Prop 4.3 and Thm 4.4 in [Rosendal's survey](http://homepages.math.uic.edu/%7Erosendal/PapersWebsite/AutomaticContinuity06.pdf)). [The notation $S\_\infty$, which implicitly assumes that every infinite set is countable (Cantor disproved this) is inconvenient.]
Let $c$ be any nontrivial permutation of $X$ consisting only of infinite cycles (at least one) and fixed points. Let $c'$ be any permutation of $X$ with both at least one infinite cycle and finite cycles of all sizes.
Since both $c$ and $c'$ have an infinite cycle, the cyclic subgroups $\langle c\rangle$ and $\langle c'\rangle$ are infinite and discrete (hence closed).
>
> **Proposition.** For every set $Y$ and homomorphism $f:S\_X\to S\_Y$, the subgroups $\langle f(c)\rangle$ and $\langle f(c')\rangle$ are not conjugate.
>
>
> More precisely, $f(c)$ has no finite cycle of size $\ge 2$, while (unless $f$ is the trivial homomorphism) $f(c')$ has finite cycles of unbounded size.
>
>
>
(This immediately implies the first proposition, in a strong sense since the embedding property fails for a given pair of isomorphic closed subgroups.)
**Lemma** Let $H$ be an open subgroup of $S\_X$. Then there exists a finite subset $F$ of $X$ such that $H$ is trapped between the pointwise and global stabilizer of $F$. $\Box$
[This is classical. The main case is when $H$ is transitive, in which case the conclusion is $H=S\_X$. The general case easily follows.]
Through $f$, we can view $Y$ as an $S\_X$-set. For both statements it enough to assume that $Y$ is a transitive $S\_X$-set (through $f$). Hence $Y=S\_X/H$ for some open subgroup $H$.
For the statement about $c$ (no nonsingleton finite cycle), we can use the lemma and find an open finite index subgroup $H'$ of $H$, such that $H'$ is the pointwise stabilizer of a finite subset. We can identify $S\_X/H'$ with the set of pairwise distinct $n$-tuples in $X^n$ for the product action, for some $n\ge 0$. Indeed we see that $c$ has no nonsingleton finite cycle on $\omega^n$. Since the equivariant surjection $S\_X/H'\to S\_X/H$ is finite-to-one, we deduce that $c$ has also no nonsingleton finite cycle on $S\_X/H$.
For the statement about $c'$ (existence of unbounded finite cycles), we can use the lemma again, now defining $H'$ as finite index overgroup of $H$, such that $H'$ is the global stabilizer of a finite subset $F$. If $F$ is empty, we have $H=H'=S\_X$ and the statement is void. Assume now $F$ nonempty. We can identify $S\_X/H'$ with the set of $n$-elements subsets of $X$ for some $n\ge 1$. Since $c'$ admits a $kn$-cycle for every $k\ge 2$, we see that $c'$ has a $k$-cycle for its action on $S\_X/H'$. Hence $c'$ has a $kk'$-cycle for some $k'\ge 1$ on $S\_X/H$.
Using automatic continuity, let me mention, out of curiosity:
>
> **Corollary.** For $X=Y=\omega$ and $c,c'$ as above, the HNN extension of $S\_\omega$ over the isomorphism $\langle c\rangle \to \langle c'\rangle$ mapping $c$ to $c'$, has no faithful action on any countable set.
>
>
>
(Indeed, finding groups of at most continuum cardinal with no faithful action on any countable set has been an open problem long ago, and while various examples are now known, such a construction seems to be a new one. Note that no group topology is involved in the above statement.)
---
This argument doesn't address general Polish groups as targets. However it suggests that it already makes sense to attempt to make discrete subgroups conjugate, even infinite cyclic ones.
| 5 | https://mathoverflow.net/users/14094 | 417163 | 169,950 |
https://mathoverflow.net/questions/417105 | 4 | Given a finite field $\mathbb{F}$ with $|\mathbb{F}|=q=p^m\geq4$ where $p=\text{char}(\mathbb{F})$, I'm wondering is there a characterization of the kernel of the map $f:H\_3(\text{GL}\_3(\mathbb{F}))\to H\_3(\text{GL}\_4(\mathbb{F}))$?
Is it an isomorphism?(Here $H\_n(G)$ means the $n$-th integral homology of the group $G$)
Here is some material I have known. [Sprehn&Wahl](https://arxiv.org/abs/1812.08742v2) says $f$ must be a surjection. In the paper [1], it is shown that $f$ would be an isomorphism if $\mathbb{F}$ is an infinite field. But in the talk [[2]](https://people.math.harvard.edu/%7Ekupers/notes/nrwtalk.pdf), it is asserted that the techniques in [1] doesn't apply for finite fields. And with the result of [Galatius-Kupers-Randal-Williams](https://arxiv.org/abs/1810.11931v1), it is deduced that $f$ induces an isomorphism on $p$-primary part. I'm wondering what happens with the $\text{mod } l$ homology where $l\neq p$.
The reason I raise this question is that in the Chapter.VI in Weibel's [K-book, remark 5.12.1](https://sites.math.rutgers.edu/%7Eweibel/Kbook/Kbook.VI.pdf), he asserts the map $\varphi$ could extend to a map $H\_3(\text{GL}(\mathbb{F}))\to B(F)$. There one works on a general field $F$ with $|F|>3$. But all I can get now is $ H\_3(\text{GL}(\mathbb{F}))=H\_3(\text{GL}\_4(\mathbb{F})) $ and if we want to extend $\varphi$ to $H\_3(\text{GL}(\mathbb{F}))$ we have to work on $\ker(f)$.
reference:
[1]*Yu. P. Nesterenko and A. A. Suslin, Homology of the general linear group over a
local ring, and Milnor’s K-theory, Izv. Akad. Nauk SSSR Ser. Mat. 53 (1989), no. 1,
121–146. MR 992981*
[2]*Alexander Kupers, [NRW TALK: $E\_\infty$-CELLS AND THE HOMOLOGY OF GENERAL
LINEAR GROUPS OVER FINITE FIELDS](https://people.math.harvard.edu/%7Ekupers/notes/nrwtalk.pdf)*
| https://mathoverflow.net/users/471160 | The third homology stability of general linear groups over finite fields | The original paper by Suslin is available [here](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=tm&paperid=1914&option_lang=eng) and with some effort you should be able to read it.
If $|\mathbb{F}| = p^r$ then the homology groups $H\_\*(GL\_n(\mathbb{F});\mathbb{F}\_\ell)$ with $\mathbb{F}\_\ell$-coefficients for $\ell \neq p$ were computed completely by Quillen, in Theorem 3 of [On the Cohomology and K-Theory of the General Linear Groups Over a Finite Field](https://www.jstor.org/stable/1970825?seq=1#metadata_info_tab_contents). You'll see that the stabilisation map
$$H\_\*(GL\_{n-1}(\mathbb{F});\mathbb{F}\_\ell) \to H\_\*(GL\_n(\mathbb{F});\mathbb{F}\_\ell)$$
is always injective and surjective in a range $\* < 2n-1$. In particular, the map
$$H\_3(GL\_3(\mathbb{F});\mathbb{F}\_\ell) \to H\_3(GL\_4(\mathbb{F});\mathbb{F}\_\ell)$$
is an isomorphism. Combined with our improvement or Sprehn-Wahl's this gives what you want as long as $p^r \geq 4$.
| 7 | https://mathoverflow.net/users/798 | 417165 | 169,951 |
https://mathoverflow.net/questions/417184 | 2 | Could any one give a proof for this inequility here? I just know its some kind of Gagliardo-Nirenberg inequility, but where does the second term come from? Thx~
$$
\int\_{B\_r}|u|^q\le C\left(\int\_{B\_r}|\nabla u|^2\right)^a \left(\int\_{B\_r}|u|^2\right)^{\frac{q}{2}-a}+\frac{C}{r^{2a}} \left(\int\_{B\_r}|u|^2\right)^{\frac{q}{2}}
$$
$$
2\le q\le 6, a=\frac{3}{4}(q-2)
$$
Here $B\_r$ is a ball of radius $r$ and $C$ is a constant independent of $r$. If complete detailed proof could be provided, that will help me a lot!!
| https://mathoverflow.net/users/295572 | A kind of Gagliardo-Nirenberg inequality proof | By context, I infer that you are working in 3 dimensions.
**Step 1: reduce to case $r = 1$**
Consider the mappings $S\_r$ that map
$$ S\_r u(x) := r^{-3/q} u(r^{-1} x) $$
We have that $S\_r: L^q(B\_1) \to L^q(B\_r)$ is an isometric bijection.
Note that
$$ \|\nabla S\_r u \|\_{L^2(B\_r)} = r^{-3/q - 1 + 3/2} \|\nabla u\|\_{L^2(B\_1)} $$
and
$$ \| S\_r u\|\_{L^2(B\_r)} = r^{3/2 - 3/q} \|u\|\_{L^2(B\_1)} $$
A direct computation shows that if we know
$$ \|u\|\_{L^q(B\_1)}^q \leq C \|\nabla u\|\_{L^2(B\_1)}^{2a} \|u\|\_{L^2(B\_1)}^{q-2a} + C \|u\|\_{L^2(B\_1)}^{q} $$
Then for $u\in H^1(B\_r)$, applying the inequality to $S\_r^{-1} u\in H^1(B\_1)$ and using the scaling identity above we see that the desired inequality holds for all $r$.
**Step 2: Extension and GNS**
The unit ball has $C^1$ boundary and is a Sobolev extension domain. In particular, there exists a bounded linear operator $E: H^1(B\_1) \to H^1\_0(B\_2)$.
Gagliardo-Nirenberg-Sobolev inequality says that
$$ \|Eu\|\_{L^6(B\_2)} \leq C \|\nabla Eu\|\_{L^2(B\_2)} \leq C \|Eu\|\_{H^1(B\_2)} $$
So (for a potentially different $C$)
$$ \|u\|\_{L^6(B\_1)} \leq \|Eu\|\_{L^6(B\_2)} \leq C\|u\|\_{H^1(B\_1)}$$
**Step 3: interpolate**
Apply the standard interpolation inequality
$$ \|u\|\_{L^q} \leq \|u\|\_{L^2}^{1-\theta} \|u\|\_{L^6}^{\theta} $$
for an appropriate $\theta\in [0,1]$ and $q\in [2,6]$, you get
$$ \|u\|\_{L^q(B\_1)}^q \leq C \|u\|\_{L^2(B\_1)}^{1-a} \|u\|\_{H^1(B\_1)}^{a} $$
Finally expand
$$ \|u\|\_{H^1}^a \leq C'' \|u\|\_{L^2}^a + C'' \|\nabla u\|\_{H^1}^{a} $$
and you are done.
| 2 | https://mathoverflow.net/users/3948 | 417190 | 169,958 |
https://mathoverflow.net/questions/417191 | 5 | "A Primer on Mapping Class Groups" wrote
>
> Let $\mathrm{Homeo}\_+(S, \partial S)$ denote the group of orientation-preserving
> homeomorphisms of $S$ that restrict to the identity on $\partial S$. $\mathrm{Mod}(S)$ is the
> group of isotopy classes of elements of $\mathrm{Homeo}\_+(S, \partial S)$, where isotopies
> are required to fix the boundary pointwise.
>
>
>
and
>
> Let $\mathrm{PMod}(S\_{g,n})$ denote the pure mapping class group of $S\_{g,n}$, which is
> defined to be the subgroup of $\mathrm{Mod}(S\_{g,n})$ consisting of elements that
> fix each puncture individually.
>
>
>
What is the difference between the two definitions? Doesn't "restrict to the identity on $\partial S$" mean for for all $x\in \partial S$ we have $\phi(x)=\mathrm{id}(x)=x$, which implies $\phi$ fixes each puncture individually?
| https://mathoverflow.net/users/nan | Mapping class group and pure mapping class group | I do not recall the conventions adopted in the Primer, but there is a wide difference between boundary components (which will be embedded circles or lines) and punctures (which are “missing points” and are typically given by specifying a “punctured disk”). Under most conventions punctures are not “part” of the boundary (whatever that might mean…)
I greatly prefer, when possible, to avoid discussions of punctures and instead use the formalism of “marked points”. So, suppose that $S$ is a connected compact surface. Suppose that $Z \subset S$ is a finite set disjoint from the boundary. Then a mapping class of $(S, Z)$ is any homeomorphism of $S$ that permutes the points of $Z$. A pure mapping class of the pair $(S,Z)$ is required to fix each component of $\partial S$ setwise and also fix $Z$ pointwise.
There are further variations (if $Z$ is allowed to meet the boundary, if we fix the boundary pointwise, and so on). It really depends on what you want to do.
| 10 | https://mathoverflow.net/users/1650 | 417196 | 169,961 |
https://mathoverflow.net/questions/417167 | 1 | Before I begin, I apologize for the bad wording. Consider the following "sorting algorithm":
Suppose there are $n$ books on the bookshelf labeled $1$-$n$, and ordered from left to right in a random way. Your goal is to end up with a sorted bookshelf from left to right with $1$ at the left and $n$ at the right. The algorithm is as follows:
1. Pick a book at random that's not already in its correct position, say book $r$.
2. Remove the book from its current position and place it at its correct position (position $r$)
3. In the process, slide the books in between the original incorrect position of book $r$ and position $r$ towards the gap left open.
4. Repeat if it's not sorted already, and if it is sorted, the algorithm terminates.
For example, suppose $n = 5$ and our original permutation is $5, 4, 3, 2, 1$ and suppose $r = 2$. Then after one step of the algorithm, the permutation becomes $5, 2, 4, 3, 1$.
Prove that regardless of the choice of the random book at each step of the algorithm that the algorithm will eventually terminate.
What is the worst case runtime of this algorithm? (I believe it occurs with $2, \dots, n, 1$ with $r$ being the left most book at each step and runtime roughly $2^{n - 2}$).
| https://mathoverflow.net/users/142201 | Runtime for Terrible "Sorting Algorithm"? | Here is a quote of [my 2009 proof @ AoPS](https://artofproblemsolving.com/community/c6h256829p1419072) that the algorithm will eventually terminate:
For a permutation $ p$, consider a function $ f(p) := \sum\_{j=1}^n |p\_j - j|$, i.e., the sum of absolute differences of elements and their positions in $ p$.
In the sorting process this function never increases but occasionally can decrease. It is clear that $ f(p)\geq 0$ and $ f(p) = 0$ only if the permutation $ p$ is sorted.
In an infinite sorting this function must take a constant non-zero value after some step.
Suppose that there exists an infinite sorting process and let $ n$ be the smallest possible integer that admits such an infinite sorting.
Let $ x$ be the smallest elements that jumps infinite number of times. Then after some step in the sorting process no element smaller than $ x$ jumps at all and the function $ f(p)$ does not decrease. Without loss of generality assume that this happens straight from the beginning of the sorting process (simply by ignoring everything what happened before).
Since $ x$ jumps infinite number of times, it takes its correct place in the permutation an infinite number of times. However, it can leave its correct place only if there exists an element larger than $ x$ somewhere before $ x$ (whose jump would shift $ x$ from its correct place). Therefore, there exists an element $ y$ smaller than $ x$ somewhere after $ x$.
Since $ y < x$, it never jumps itself and can be shifted by jumps of other elements only towards the end of the permutation (as otherwise $ f(p)$ would decrease). But this way it cannot be shifted infinitely many times, so after some point $ y$ takes a fixed place in the permutation.
As soon as that happened, the sorting can be considered as two independent sortings -- at the left hand side of $ y$ and at the right hand side of $ y$ since no element jumps across $ y$ from one "half" to the other. Moreover, sorting of the left "half" must be infinite as it contains $ x$. Renaming the elements at the left hand side of $ y$ will give a permutation on smaller number of elements that admits an infinite sorting, a contradiction to the minimality of $ n$.
This contradiction proves that no infinite sorting is possible.
---
Another approach:
Alternatively, we can take $ y = 1$ right away (without considering any $x$) and notice that if it ever takes its correct place then we have an infinite sortings of the remaining $n-1$ elements, a contradiction to the minimality of $ n$. Otherwise, $ y$ will be shifted towards the end of the permutation and sooner or later will take a fixed position, again splitting the permutation into two smaller ones, at least one of which will be subject to infinite sorting in contradiction to the minimality of $n$.
---
PS. It is conjectured that $2^{n-1}-1$ is the tight upper bound for the number of steps (attained at permutation $(n,1,2,\dots,n-1)$, for example), but to best of my knowledge it was not proved.
| 4 | https://mathoverflow.net/users/7076 | 417201 | 169,962 |
https://mathoverflow.net/questions/417199 | 0 | Consider the equation
$$ -\Delta f+mf+\lambda f^p=0$$
on $\mathbb{R}^d$, where $d>2$,$m>0$, $p>1$ is integer, and $\lambda \in \mathbb{R}$. Are there any known results regarding the non-existence of non-zero solutions $f$ of this equation in $\mathcal{S}(\mathbb{R}^d)$ (the Schwartz space of rapidly decreasing smooth functions), at least for some values of $d$, $p$ and $\lambda$?
| https://mathoverflow.net/users/17971 | Non-existence of rapidly decaying solutions of certain elliptic semilinear equations | (To substantiate my comment above)
If $\lambda \geq 0$ and $p$ is odd, assume $f\in \mathcal{S}(\mathbb{R}^d)$ is a solution, then you can multiply the equation by $f$ and integrate by parts (everything converges appropriately) to find
$$ \int |\nabla f|^2 + m f^2 + \lambda f^{p+1} = 0 $$
Noting that $p+1$ is even and so $f^{p+1}\geq 0$, you find this implies $\|f\|\_{L^2} = 0$.
(Alternatively, you can also use the maximum principle: write the equation as $-\triangle f + Vf = 0$ where $V = m + \lambda f^{p-1}$. When $p$ is odd this function $V$ is manifestly positive, and by maximum principle any solution cannot have a positive local max nor a negative local min. Combined with the Schwartz assumption this forces $f\equiv 0$.)
| 5 | https://mathoverflow.net/users/3948 | 417210 | 169,965 |
https://mathoverflow.net/questions/417204 | -1 | I have the following question related to multivariable moment generating functions. Given two vectors $u,X\in\mathbb{R}^d$, I want to characterize the quantity $e^{u^T X}$ in terms of "powers" of the vector $X$. To see what I mean, note that by the power series expansion of the exponential:
$$e^{u^T X} = \sum\_{n=0}^\infty \frac{1}{n!}\left(u^T X\right)^n$$
which is "sort of" in terms of powers of $X$, but they're all mixed up with "powers of $u$". However, for $n\leq 2$ we can break up the dependence and make the dependence on $X$ very explicit. For $n=0$ there's nothing special (everything is constant), and for $n=1$ we are left with the inner product $u^T X = \langle u, X \rangle$, so $X$ itself characterizes the "first power of $X$". For $n=2$, we may write:
$$\left(u^T X\right)^2 = \left(u^T X\right)\left(u^T X\right) = \left(u^T X\right)\left(X^T u\right) = u^T \Sigma u$$
where $\Sigma = XX^T$ characterizes the "second power of $X$" in the matrix product $u^T \Sigma u$.
If we try to do the same for $n=3$ however, there is no way to characterize an "$n$th power of $X$" as a matrix - seemingly, we need an object of dimension $d\times d\times d$. I did some research into this in the setting of multidimensional Taylor series expansions, and it would seem that the appropriate object for the "$n$th moment of $X$" is a tensor of dimension $d\times d \times \overset{n}{...} \times d$. My question is, in this setting (the power series for $e^{u^T X}$), how can I write the product $(u^T X)^n$ as a general "tensor product", and how can I generally write the "power of $X$" that determines the product, given $n$ copies of the vector $u$?
My motivation for this question is in studying the moment generating function when $X$ is a random vector. In the one-dimensional case ($d=1$), assuming the moment generating function exists in a small neighbourhood around zero, the moments $(1, E[X], E[X^2], ...)$ uniquely determine the distribution. I'm interested in the multidimensional analogue, where given a positive radius of convergence around zero for the multidimensional moment generating function, the countable sequence $(1, E[X], E[XX^T], \overset{?}{...})$ uniquely determines the distribution of $X$.
Thank you!
Edit: I don't know much about the tensor product, but I found one resource that said the tensor product of two vectors $v, w$ is defined as $v\otimes w = vw^T$. I'd blindly guess then, since this is the form of the second-order term $E[XX^T]$, that the "$n$th moment of $X$" should be defined as
$$M\_n = E\left[\bigotimes\_{k=1}^{n} X\right]$$
where supposedly the empty product convention yields the scalar $1$. Is this the right idea? If so, does anyone have any resources that would be good for a beginner here? I'm not even sure how I would write the product of this object with $n$ copies of $u$ to get back to the scalar $\left(u^T X\right)^n$.
| https://mathoverflow.net/users/477962 | Multi-variable expansion of $e^{u^T X}$ as a power series in terms of tensors | Define the tensor $v^{\otimes k}$ with elements $[v^{\otimes k}]\_{i\_1,i\_2,\ldots i\_k}=v\_{i\_1}v\_{i\_2}\cdots v\_{i\_k}$ and denote the full contraction of two tensors by $A\odot B=\sum\_{i\_1,i\_2,\ldots i\_k}A\_{i\_1,i\_2,\ldots i\_k}B\_{i\_k,\ldots i\_2,i\_1}$, then you can write
$$e^{u^T X} = \sum\_{n=0}^\infty \frac{1}{n!}u^{\otimes n}\odot X^{\otimes n}.$$
| 1 | https://mathoverflow.net/users/11260 | 417213 | 169,967 |
https://mathoverflow.net/questions/414896 | 6 | I have been reading the famous [paper of Alves, Bonatti, and Viana](https://link.springer.com/article/10.1007/s002220000057) where they proved that there is an SRB measure for partially hyperbolic systems. Since I am new to this field, I have some basic questions.
1)All results were proved for $C^2$ diff, but I saw at some talks that they mentioned their results for $C^{1+\varepsilon}.$ Is it true that their results work for $C^{1+\varepsilon}.$ I am aware that the limsup condition of their result can be replaced by the liminf condition (Alves, Luzzatto. etc), but I want to know whether all their results work for $C^{1+\varepsilon}$ or not.
2)Can one give an example that a Gibbs-u state is not a hyperbolic SRB measure?
| https://mathoverflow.net/users/127839 | SRB measure and Gibbs u-state | The results are true for $C^{1+}$. [This paper](https://arxiv.org/pdf/1403.2937.pdf) that you mentioned uses $C^{1+}$ regularity. Actually, in this business the main thing that usually comes from $C^{1+}$ condition is control of distortion along an unstable disc.
For you second question, consider the following. Let $f$ be a $C^{1+}$ Anosov diffeomorphism of $\mathbb{T}^2$. We know that such system has only one $u$-Gibbs measure that is the SRB measure $\mu\_f$.
Now, let $g$ be the north/South Pole on $S^1$ and suppose that the hyperbolicity of $g$ is weaker then the hyperbolicity of $f$, so that $f\times g$ is a partially hyperbolic diffeomorphism of $\mathbb{T}^3$. Let $p\_1$ be the North Pole (repeller for $g$) and $p\_2$ be the South Pole (attractor). The measure $\mu\_f \times \delta\_{p\_1}$ is $u$-Gibbs but it is not SRB. However the measure $\mu\_f \times \delta\_{p\_2}$ is $u$-Gibbs and SRB.
| 1 | https://mathoverflow.net/users/117630 | 417216 | 169,968 |
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