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https://mathoverflow.net/questions/421940 | 5 | The Gamma transform of a measure is defined as follows. If $\alpha$ is a $\mathbf{Z}\_p$-valued measure on $\mathbf{Z}\_p$, then the Gamma transform of $\alpha$ is:
$$\Gamma\_{\alpha}(s) = \int\_{\mathbf{Z}\_p^{\times}} \langle x \rangle^s \, d\alpha(x). $$
So the Gamma transform takes as input a measure $\alpha$, and returns an analytic function of the variable $s$, which we call $\Gamma\_{\alpha}(s)$. But we can also think of the Gamma transform in a different way: as taking as input a *power series* and returning as output a *power series*. Namely:
* as input, the Gamma transform takes in the power series $F\_{\alpha}(T)$ corresponding to the measure $\alpha$
* as output, the Gamma transform returns the power series $G$ such that $G((1+p)^s - 1) = \Gamma\_{\alpha}(s)$.
My question is: can one *explicitly* describe the Gamma transform as a map from power series to power series? That is, given a power series $f(T) = \sum a\_nT^n$, is there an *explicit* formula for the power series expansion of $\Gamma\_F(T)$ in terms of the power series expansion of $F(T)$?
Here is my motivation for asking this. Washington has a very nice article, "[On Sinnott's Proof of the Vanishing of the lwasawa Invariant $\mu\_p$](https://doi.org/10.2969/aspm/01710457)", where he gives a different proof of the Ferrero–Washington Theorem, inspired by a proof of Sinnott. On page 3, Washington does a few calculations with power series and says that "this is essentially the Gamma-transform". But I don't know what the Gamma transform looks like as a map from power series to power series, so I don't understand what that comment means. It seems to lie at the heart of the proof, so I want to ask this question to make sense of that step in the paper.
| https://mathoverflow.net/users/394740 | Describing the Gamma-transform explicitly in terms of power series | This is a hard problem (and one which is easily overlooked by the unwary)! Just to be clear, I'll summarize (how I think about) the problem: as a relation between *additive* and *multiplicative* Fourier transforms. We start with a measure $\mu$, i.e. a linear map
$\mu$: (continuous functions on $\mathbf{Z}\_p$) $\to \mathbf{Z}\_p$.
For any $t$ with $|t| < 1$, there is a unique character $\kappa\_t$ of $\mathbf{Z}\_p$ as an *additive* group which sends 1 to $1 + t$; and we have $\mu(\kappa\_T) = F\_\mu(T)$ for some power series $F\_\mu(T) \in \mathbf{Z}\_p[[T]]$: the additive Fourier transform of $\mu$.
On the other hand, for each $u$ with $|u| < 1$, there is also a character $\chi\_u$ of $\mathbf{Z}\_p^\times$ as a *multiplicative* group, which is trivial on roots of unity and sends $1 + p$ to $1 + u$. This is given by $\mu(\chi\_u) = G\_\mu(u)$ for some power series $G\_\mu \in \mathbf{Z}\_p[[U]]$: the multiplicative Fourier transform of $\mu$. (I am deliberately using a different name for the variable here!)
Your question is, then, how $F\_\mu$ and $G\_\mu$ are related. The answer is: "not in any straightforward way". For a given $t$, the functions $\chi\_u$ and $\kappa\_u$ are totally different as functions on $\mathbf{Z}\_p$, so the values of $F\_\mu$ and $G\_\mu$ at a specific $u$ do not determine each other. On the other hand, if $\mu$ is supported in $1 + p\mathbf{Z}\_p$ (so we lose nothing by restricting), then the two series $F\_\mu$ and $G\_\mu$ do determine each other: we get a bijection (the "Mellin transform")
$$\mathfrak{M}: (1 + T) \cdot \mathbf{Z}\_p[[(1+T)^p - 1]] \longleftrightarrow \mathbf{Z}\_p[[U]]$$
which is a bijection of additive groups mapping $F\_\mu(T)$ to $G\_\mu(U)$ (but not respecting multiplication on either side). As an example, if $\mu$ is the Dirac measure sending a function to its value at $a$, for some $a \in 1 + p\mathbf{Z}\_p$, then $F\_\mu(T) = (1 + T)^a$, and $G\_\mu(U) = (1 + U)^{log\_p(a) / log\_p(1 + p)}$. (This formula actually determines $\mathfrak{M}$ uniquely, since the Dirac measures are dense.)
One can pull various information about the power series through this isomorphism -- for instance, the Iwasawa $\lambda$ and $\mu$-invariants of the two series are the same, and the Newton polygons coincide beyond a certain point (Sarah Zerbes and I proved this in [an old paper of ours](https://dx.doi.org/10.1515/crelle.2012.012)). But there is no simple formula that would allow you to read off the coefficients of $F\_\mu$ from those of $G\_\mu$ or vice versa.
| 6 | https://mathoverflow.net/users/2481 | 421954 | 171,613 |
https://mathoverflow.net/questions/368736 | 3 | Is there any published, somewhat comprehensive, list of (almost?) all the many ways in which the Leibniz notation ($dx,$ $P(dx),$ $d\mu(x),$ $du\wedge dv,$ etc., etc.) gets used in the various areas of mathematics?
(I posted this question [here](https://math.stackexchange.com/questions/3783281/dictionary-of-the-leibniz-notation) on m.s.e. and it was closed although nobody verbally expressed any specific objections to it or even hinted at such.)
| https://mathoverflow.net/users/6316 | Reference request: Dictionary of the Leibniz notation | Not quite sure in which direction you are hoping for an answer, but to set a first data point I offer the [Pantheon of Derivatives](https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/)
Part 1 – Directional Derivatives
Part 2 – Manifolds
Part 3 – Vector Bundles
Part 4 – Lie Theory
Part 5 – Theorems
| 2 | https://mathoverflow.net/users/11260 | 421962 | 171,617 |
https://mathoverflow.net/questions/419518 | 2 | It is known that the limit of a sequence of non injective operators is not necessarily non-injective, for instance, the operator \begin{eqnarray\*}
T\_n &:&\ell ^{2}\rightarrow \ell ^{2} \\
x &\longmapsto &(x\_{1},\frac{x\_{2}}{2},...,\frac{x\_{n}}{n},0,0,...).
\end{eqnarray\*}
does the job.
My question is the following: what supplementary assumption can guarantee the non-injectivity of the limit operators of a sequence of non-injective operators acting from $E \rightarrow F$ ($E$ and $F$ are Hilbert spaces).
Thank you in advance.
| https://mathoverflow.net/users/106804 | Non-injectivity of the limit of non-injective sequence of operators | Let $(T\_n:E\to F)$ be a sequence of noninjective operators such that $\|T\_n-T\|\to 0$ as $n\to\infty$ for some bounded linear $T:E\to F$. Since $\ker(T\_n)\neq \{0\}$, for each $n\in\mathbb{N}$, we can pick $u\_n\in E$, $\|u\_n\|=1$ such that $T\_nu\_n=0$. Consequently,
$$\|Tu\_n\|\leq \|Tu\_n-T\_nu\_n\| \leq \|T-T\_n\|\|u\_n\| \to 0,$$
that is $0\in\sigma(T)$ and $(u\_n)$ is an approximate eigenvector for $T$.
Second, the unit ball of $E$ is weakly compact since $E$ is reflexive. An approximate eigenvector $(u\_n)$ of $T$ of the form above lies in the unit sphere of $E$, thus $(u\_n)$ has weak limit points. There are 2 disjoint cases:
1. Either $u\_n\to 0$ weakly for all approximate eigenvectors of the form above,
2. or there exists $(u\_n)$ that possess a nonzero weak limit point, say $u\in E$.
In the second case, there exists a subsequence $(u\_{n\_k})$ such that $u\_{n\_k}\to u\neq 0$ weakly in $E$ by Eberlein-Šmulian theorem. Next, for each $f\in F^\*$,
$$|f(Tu)|
\leq |f(Tu - Tu\_{n\_k})| + |f(Tu\_{n\_k} - T\_{n\_k}u\_{n\_k})|
\leq |foT(u-u\_{n\_k})| + \|f\|\|T-T\_{n\_k}\|
\to 0$$
as $k\to\infty$. Thus, $f(Tu) = 0$ for all $f\in F^\*$, so $Tu=0$. Consequently, $T$ is not injective.
**We can conlude:** if there exists $u\_n\in\ker(T\_n)$, $\|u\_n\|=1$ such that $(u\_n)$ has nonzero weak limit points, then $T$ is not injective.
| 1 | https://mathoverflow.net/users/164350 | 421965 | 171,618 |
https://mathoverflow.net/questions/421439 | 7 | I came across the following conjecture. If you have any thoughts on how to approach it, let me know.
**Conjecture.** For any integer $n > 3$ and any Gaussian integer $z$ that is not a unit, if $z^n - z$ or $z^n + z$ is a rational integer, then $z$ is a rational integer.
| https://mathoverflow.net/users/22733 | Given that $n > 3$ and $z$ is a Gaussian integer, when can $z^n \pm z$ be a rational integer? | The question can be rephrased: Find the integers $n > 3$ for which there exist Gaussian integers $z$ such that
$z^n \pm z = \overline{z}^n \pm \overline{z}$. Rewriting the equation as $(z^n -\overline{z}^n)/(z-\overline{z}) = \mp 1$, one recognizes an instance of the problem of finding the terms of a Lucas sequence with no primitive divisors. The complete answer to an even more general problem (Lehmer sequences instead of Lucas sequences) was given by Bilu, Hanrot, Voutier [J. Reine Angew. Math.
539, 75-122 (2001)](https://www.degruyter.com/document/doi/10.1515/crll.2001.080/html) and Abouzaid [J. Théor. Nombres Bordx.
18, No. 2, 299--313 (2006)](https://jtnb.centre-mersenne.org/articles/10.5802/jtnb.545/) . To settle the conjecture, it suffices to look up the tables in these papers .
| 2 | https://mathoverflow.net/users/315992 | 421972 | 171,621 |
https://mathoverflow.net/questions/421971 | 1 | By a Narayana-enumerated object I mean an object whose count is given by the Narayana number $N(n,k)=\frac{1}{n} {n \choose k} {n \choose k-1}$. Can you give me a reference to some good big list of Narayana-enumerated objects? I have found two Narayana-enumerated objects (the two objects being closely related one to another) and would like to see whether my two objects are in the list.
| https://mathoverflow.net/users/481690 | A big list of Narayana-enumerated objects | [Elements of the sets enumerated by super-Catalan numbers](http://math.haifa.ac.il/toufik/enumerative/supercat.pdf) contains many Narayana-enumerated objects. (The super-Catalan number $s\_n$ is related to the Narayana numbers by $s\_n=\sum\_{k=1}^n 2^{k-1}N\_{n,k}$.)
Some examples:
The following five parameters of Dyck paths are enumerated by Narayana numbers:
(i) number of high peaks;
(ii)number of valleys;
(iii)number of doublerises;
(iv) number of rises at an even level;
(v) number of nonfinal ascents and descents of length greater than 1.
132-avoiding permutations with given number excedances ($a\_i > i$) are counted by Narayana numbers, as well as those with given number descents ($a\_i<i$).
| 1 | https://mathoverflow.net/users/11260 | 421974 | 171,622 |
https://mathoverflow.net/questions/421978 | 15 | I asked this question on [M.SE](https://math.stackexchange.com/questions/4443148/grothendiecks-relative-point-of-view-and-yoneda-lemma), but didn't get any answers.
Occasionally I hear people saying that one of Grothendieck's big insights was that often when interested in an object $X$ it's better to study *morphisms* into that object, $-\to X$. Apparently that's called the [relative point of view](https://en.wikipedia.org/wiki/Grothendieck%27s_relative_point_of_view).
**First question.** How is that principle applied in practice? What are some concrete examples in mathematics where the relative point of view is useful?
Wikipedia mentions the Riemann–Roch theorem and a [similar MSE question](https://math.stackexchange.com/questions/1624553/grothendiecks-relative-point-of-view) mentions a theorem about coherent sheaves. Unfortunately, I don't know any algebraic geometry yet. Are there more down-to-earth applications of the relative point of view that an undergraduate can understand, say, in linear algebra, group theory, ring theory, Galois theory, or maybe even in basic category theory?
What are (some of) the most important theorems that feature the relative point of view?
---
I recently heard about the Yoneda lemma in category theory (I know the statement and can prove it). I know that it can be used to prove that two objects are isomorphic whenever they have the same universal property. In Awodey's category theory book, there's a concrete application of that: in categories with enough structure, $(A\times B)+(A\times C)\cong A\times (B+C)$. That proof is elegant, I agree. But it doesn't live up with the praise many people give to the Yoneda lemma, does it?
Maybe a more concrete application in non-category theory would help me to get convinced of the contrary. For instance, I read on Wikipedia (and elsewhere) that Grothendieck used the Yoneda lemma in his famous book EGA (which a lot of people seem to talk about). (In fact, it seems this was another [insight of him](https://people.math.harvard.edu/~mazur/papers/Thinking.about.Grothendieck(5).pdf): that Yoneda is useful.)
**Second question.** So what were Grothendieck's main applications of the Yoneda lemma in algebraic geometry? (In contrast to the first question, here it suffices for me to just know roughly what kind of statement he proved with the Yoneda lemma---rather than understanding it in detail, because I already know one application of the Yoneda lemma.)
---
**Third question.** Is the second question related to the first one, i.e., is there a connection between the relative point of view and the Yoneda lemma? (At least the Wikipedia page linked above mentions the Yoneda lemma.)
| https://mathoverflow.net/users/481861 | Grothendieck's relative point of view and Yoneda lemma | Let me answer your questions in reverse order.
For the last question, yes, Yoneda's lemma is absolutely crucial to the relative point of view, as it essentially postulates that passing from a scheme $X$ (or more generally an object of any category) to the Hom functor $Hom(-,X)$ does not lose information. More precisely, the functor $Hom(-,X)$ determines the object $X$ up to isomorphism, and moreover natural transformations between functors $Hom(-,X)$ and $Hom(-,Y)$ are in correspondence with morphisms from $X$ to $Y$ (so you recover not only objects, but also morphisms). In the case of schemes, it turns out that you don't need to consider $Hom(Z,X)$ for all schemes $Z$, it suffices to consider them for $Z=\operatorname{Spec}R$ for a ring $R$ - in this case, $Hom(Z,X)$ should be thought of as the set of points of $X$ with coefficients in $R$. This is not something that follows automatically from categorical considerations, but rather to the fact schemes are "locally affine".
As for the applications of Yoneda, one point is that in general describing a morphism of schemes directly can be rather annoying - working at the level of underlying topological spaces and structure sheaves would be quite tedious. Fortunately, using the remarks of my previous paragraph, to specify a morphism it is enough to specify maps at the level of $R$-valued points for all rings $R$ (in a functorial manner). A good example of this is given by group schemes - for instance, if you have an elliptic curve $E$, then for any ring you can define a group structure on the set $E(R)$ using the formulas which are not hard to derive from the geometric construction of the addition law. This gives a function from $E(R)\times E(R)\cong(E\times E)(R)$ to $E(R)$, and induces a morphism $E\times E\to E$.
Talking about any of the Grothendieck's results won't be easy if you are not familiar with algebraic geometry, but let me sketch the idea behind one such, namely the theory of the Picard functor. To any (smooth projective) algebraic curve $C$ over a field, one can construct another variety $J(C)$, known as its Jacobian, which parametrizes divisors (linear combinations of points) on $C$ modulo a certain equivalence relation. There are many ways to construct it, but relative point of view gives a nice explanation of what it "is". Specifically, for any other variety $V$, we can consider "families of divisors on $C$ indexed by $V$" - those are certain ("nice") divisors on $V\times C$. Jacobian $J(C)$ then has the following property: such families of divisors indexed by $V$ are completely determined by maps $V\to J(C)$. We say that $J(C)$ *represents* the functor taking a variety $V$ to the set of all such families of divisors. This is something that only makes sense once we consider a relative point of view, generalizing the curve $C$ to the "family of curves" $V\times C\to V$.
In general, Grothendieck has described an analogous functor for any "relative curve" over another scheme; in particular one can consider such over arbitrary ring (not just a field) and has shown that such functors are always representable, by a scheme called the Picard scheme. This is again something we couldn't even make sense of without considering the relative point of view.
| 19 | https://mathoverflow.net/users/30186 | 421986 | 171,623 |
https://mathoverflow.net/questions/421861 | 2 | Let $\{X\_i^n\}$ be a sequence of $n$-dimensional Alexandrov spaces with curvature uniformly bounded from below which converges in the Gromov-Hausdorff sense to a compact $n$-dimensional Alexandrov space (i.e. without collapse). Let $E\_i\subset X\_i$ be extremal subsets. Assume that $E\_i$ converge to a compact subset $E\subset X$ in the Hausdorff sense.
**Is it true that $E$ is an extremal subset of $X$?**
| https://mathoverflow.net/users/16183 | Convergence of extremal subsets in Alexandrov spaces | The limit of extremal subsets is an extremal subset, see Lemma 4.1.3 in Petrunin's [Semiconcave functions in Alexandrov geometry](https://arxiv.org/abs/1304.0292). The non-collapsing assumption is not needed.
| 4 | https://mathoverflow.net/users/1573 | 421987 | 171,624 |
https://mathoverflow.net/questions/421984 | 7 | Let a compact Lie group $G$ act smoothly on a compact smooth manifold $M$. For any compact subgroup $H\subset G$ denote by $E^H$ the image in $M/G$ of the fixed point set of $H$ in $M$.
**Is it true that the family of all such subsets $\{E^H\}$ is finite when $H$ runs over all compact subsgroups of $G$?**
| https://mathoverflow.net/users/16183 | On fixed point sets of actions of compact Lie groups | The quotient $M/G$ carries a stratification by orbit type (see e.g. [this MO question](https://mathoverflow.net/questions/191988/local-structure-of-the-quotient-of-a-lie-group-action) for references). More precisely, for any closed subgroup $F\subseteq G$ the stratum $(M/G)\_{(F)}$ is the set of all orbits which are isomorphic to $G/F$. The set $E^H$ is the union of all strata such that $H$ is conjugate to a subgroup of $F$. Since $M$ is compact, the stratification is finite. So also only finitely many subsets of the form $E^H$ are possible.
| 11 | https://mathoverflow.net/users/89948 | 421994 | 171,626 |
https://mathoverflow.net/questions/421999 | 2 | I heard "Feller made a famous mistake in 1954 and fixed by A.D. Wentell in 1959" from one lecture. There is no further explain what is that mistake? Is there someone know it? Is it possible to explain a little bit it?
| https://mathoverflow.net/users/147009 | What is famous mistake made by Feller? | For reference, the two papers are
[1] W. Feller. [Diffusion processes in one dimension,](https://doi.org/10.1090/S0002-9947-1954-0063607-6%20) Trans. Amer. Math. Soc. **97**, 1-31 (1954).
[2] A. D. Wentzell. [On boundary conditions for multidimensional diffusion processes,](https://doi.org/10.1137/1104014) Theor. Probability Appl. **4**, 164-177 (1959).
Differential operators with boundary conditions containing diffusion terms were introduced by Feller [1] for one-dimensional diffusion and by Wentzell [2] for higher dimensions.
My best guess to what is going on is that the the lecturer was referring to the limitation to one dimension in Feller's 1954 work, a limitation removed by Wentzell's 1959 paper. I am not aware of any error in [1].
| 1 | https://mathoverflow.net/users/11260 | 422005 | 171,629 |
https://mathoverflow.net/questions/421985 | 7 | Can we find a counterexample to the following assertion?
Assume that $f:[-1,1]\to [-1,1]$ an odd function of class $C^3$, and assume thaht $f$ is a concave increasing diffeomorphism of $[0,1]$ onto itself. Then my examples say that $$\frac{1-f(x)^2}{1-x^2}\le f'(x),\; x\in (0,1).$$
| https://mathoverflow.net/users/409893 | A counterexample to: $\frac{1-f(x)^2}{1-x^2}\le f'(x)$ — revisited | A counterexample is provided by any function that equals $f(x)=1+m(x-1)$ near $x=1$, with $0<m<1$. (Maybe this is in fact just restating Anthony's comment, with a typo corrected?)
What is actually true is the trivial observation that (by the mean value theorem)
$$
\frac{1-f^2(x)}{1-x^2}=f'(c) \frac{1+f(x)}{1+x} \le f'(x)\frac{1+f(x)}{1+x} ,
$$
and it seems this is as far as we can go in general. (In my example, we have equality here, and $(1+f)/(1+x)>1$, since $f>x$.)
| 3 | https://mathoverflow.net/users/48839 | 422011 | 171,633 |
https://mathoverflow.net/questions/257889 | 13 |
>
> Let $R$ be a commutative ring. Let $A$ be an $R$-algebra (i.e., an $R$-module
> equipped with an $R$-bilinear multiplication map that turns $A$ into a unital
> ring). We do **not** require $A$ to be commutative. Assume that $A$ is free as an $R$-module, with a finite basis. Let
> $\left( e\_{1},e\_{2},\ldots,e\_{n}\right) $ be a basis of the $A$-module $R$.
>
>
>
>
> We define a *trace map* $\operatorname{Tr}\_{A/R}:A\rightarrow R$ as
> follows: For every $b\in A$, we let $\operatorname{Tr}\_{A/R}b$ be
> the trace of the endomorphism $A\rightarrow A,\ a\mapsto ba$ of the $R$-module
> $A$. (We are using the fact that $A$ has a finite basis here.) Clearly, the
> map $\operatorname{Tr}\_{A/R}$ is $R$-linear.
>
>
>
>
> Let $\Delta=\det\left( \left( \operatorname{Tr}\_{A/R}\left(
> e\_{i}e\_{j}\right) \right) \_{1\leq i\leq n,\ 1\leq j\leq n}\right) \in R$.
>
>
>
>
> Is it true that $\Delta=u^{2}+4v$ for some elements $u$ and $v$ of $R$ ?
>
>
>
---
The above is an elaborate generalization of [Stickelberger's discriminant
theorem](https://math.stackexchange.com/questions/394785/proof-of-stickelberger-s-theorem). Indeed, if we assume $A$ to be commutative, then $\Delta
=u^{2}+4v$ is true; this is proven in Remark 5.4 of [Owen Biesel, Alberto
Gioia, *A new discriminant algebra construction*, arXiv:1503.05318v3](http://arxiv.org/abs/1503.05318v3) (published in [Documenta Mathematica 21 (2016), pp. 1051--1088](https://www.emis.de/journals/DMJDMV/vol-21/28.html)).
(More precisely, they
prove it in the case when $n\geq2$; but the remaining case is obvious.) If we
furthermore assume that $R=\mathbb{Z}$ and $A$ is the integer ring of a number
field $K$, then $\Delta$ becomes the discriminant $\Delta\_{K}$ of $K$, and the
$\Delta=u^{2}+4v$ claim becomes $\left( \Delta\_{K}\equiv0 \mod %
4\ \vee\ \Delta\_{K}\equiv1 \mod 4\right) $, which is the
classical claim of Stickelberger's discriminant theorem.
The claim does not significantly depend on the choice of basis $\left(
e\_{1},e\_{2},\ldots,e\_{n}\right) $, since any change of basis causes $\Delta$
to be multiplied by a square in $R$ (namely, by the square of the determinant
of the matrix responsible for the change of basis... or of its inverse,
depending on how you define that matrix).
I do not know whether to expect the conjecture to be true or not. All examples
I have tried (the complexity of computing $\Delta$ for high $n$ limits my
abilities here) satisfy $\Delta=u^{2}+4v$ for rather stupid reasons (often,
$\Delta$ will either be a square or be divisible by $4$ to begin with); but
this says more about the poverty of my examples than about the correctness of
the conjecture. For what it's worth, here are my examples:
* If $A$ is the group ring of a group $G$, and $\left(e\_1, e\_2, \ldots, e\_n\right)$ is the standard basis $G$ of $A$ (abusing notation as usual), then $\Delta = \left(-1\right)^{n\left(n-1\right)/2} n^n$. This is either of the form $4v$ or of the form $1+4v$; thus, the conjecture holds here.
* If $A$ is the matrix ring $R^{m\times m}$ (so that $n=m^2$), and $\left(e\_1, e\_2, \ldots, e\_n\right)$ is the basis consisting of the matrix units in their usual order, then $\Delta = \left(-1\right)^{m\left(m-1\right)/2} m^n$. This is either of the form $4v$ or of the form $1+4v$; thus, the conjecture holds here.
* If $A$ is the (standard) quaternion algebra, then $\Delta = -4^4$; thus, the conjecture holds here. Other quaternion algebras may have different $\Delta$, but the factor $4^4$ will still be present.
Notice that the trace map $\operatorname{Tr}\_{A/R}$ defined above
has a "right analogue": the map $\operatorname{Tr}\_{A/R}^{\prime
}:A\rightarrow R$ sending each $b\in A$ to the trace of the endomorphism
$A\rightarrow A,\ a\mapsto ab$ of the $R$-module $A$. If we do not require
commutativity of $A$, these maps will in general not be identical (for a
specific example, extend a left-trivial semigroup by a $1$ to obtain a
monoid, then take the monoid algebra of this monoid).
I do not know whether the $\Delta$ computed via the other map will be different; this is another interesting question.
| https://mathoverflow.net/users/2530 | Is the discriminant of a free (as a module) $R$-algebra always congruent to a square modulo 4? | I see this question is still being viewed. To avoid misleading anyone, let me put a placeholder answer here: The question has been **answered positively** (with a beautiful proof) by John Voight, Asher Auel and Owen Biesel:
[Asher Auel, Owen Biesel, John Voight, *Stickelberger's discriminant theorem for algebras*, submitted to AMM, arXiv:2208.06138v1](https://arxiv.org/abs/2208.06138v1).
(Theorem 3.28 proves my conjecture.)
| 5 | https://mathoverflow.net/users/2530 | 422030 | 171,640 |
https://mathoverflow.net/questions/422029 | 7 | Let $A = (a\_{i,j})$ be a [double stochastic matrix](https://en.wikipedia.org/wiki/Doubly_stochastic_matrix) with positive entries. That is, all entries are positive real numbers, and each row and column sums to one. A [permanent](https://en.wikipedia.org/wiki/Permanent_(mathematics)) of a matrix $A = (a\_{i,j})$ is defined as
$$
\text{perm}(A) = \sum\_{\pi \in S\_n}\prod\_{i=1}^{n}a\_{i,\pi(i)}.
$$
**Is it true that $\text{perm}(A^2) \leq \text{perm}(A)$ with the equality achieved only by $P\_n = (1/n)$?**
**More generally, is it true that $\text{per}(A^n) \leq \text{per}(A^m)$ for $n > m$ with the equality achieved only by $P\_n = (1/n)$?**
Van der Waerden's conjecture states that the minimum permanent among doubly stochastic matrices is achieved precisely by the matrix $P\_n = (1/n)$. Since $P\_n$ is the only doubly stochastic matrix with positive entries and $P\_n^2 = P\_n$, the conjecture above implies Van der Waerden's conjecture for matrices with positive entries, so I don't assume it's easy to prove it if it's true.
I wrote a simple code (well, I mostly got it from [here](https://mathematica.stackexchange.com/questions/119485/how-to-generate-a-markov-matrix-efficiently)) in Wolfram Mathematica to test the conjecture on random doubly stochastic matrices but couldn't find any counterexamples.
```
n := 5
While[Min[
dsm := FixedPoint[
Standardize[Transpose[Standardize[#, 0 &, Total]], 0 &, Total] &,
RandomReal[1, {n, n}],
SameTest -> (Norm[#1 - Transpose[#2], "Frobenius"] <
1.*^-12 &)]] < 0.1]
Do[
mat = dsm;
If[Permanent[mat] < Permanent[mat.mat],
Print[mat]];
, {i, 1, 10000}]
```
The intuition behind the conjecture is from the two-sided Markov chains. Since $A$ is doubly stochastic, one can run two Markov chains on both sides. The permanent can be interpreted as some statistics on collisions between particles that start at certain positions. Since the Markov chain, after many steps, "wash out" original biases, the number of collisions should get smaller and achieve the minimum on the limit, which is $P\_n = (1/n)$ for a doubly stochastic matrix.
**Update**: Relevant partial result: [Sinkhorn, Richard. "Doubly stochastic matrices whose squares decrease the permanent."](https://www.tandfonline.com/doi/abs/10.1080/03081087608817142) Linear and Multilinear Algebra 4, no. 2 (1976): 153-158.
**Update**: Joseph Van Name showed in the answer that the conjecture is wrong: the following doubly stochastic matrix is a counterexample.
\begin{pmatrix}
15/32 & 1/64 & 1/2 & 1/64 \\
1/64 & 1/64 & 15/32 & 1/2 \\
1/2 & 15/32 & 1/64 & 1/64 \\
1/64 & 1/2 & 1/64 & 15/32
\end{pmatrix}
| https://mathoverflow.net/users/nan | On permanent of a square of a doubly stochastic matrix | No. This is not true.
For example, set
$$A=\frac{1}{2}\begin{bmatrix}
1&0&1&0\\
0&0&1&1\\
1&1&0&0\\
0&1&0&1
\end{bmatrix}.$$
Then $\text{Per}(A)=1/8$ and $\text{Per}(A^2)=9/64.$
Since the collection of all matrices $B$ where $\text{per}(B^2)>\text{per}(B)$ is open, there is some $\epsilon$ where if $\|B-A\|<\epsilon$, then $\text{per}(B^2)>\text{per}(B)$. In this case, one can select a doubly stochastic $B$ with strictly positive entries with $\|B-A\|<\epsilon$ and therefore $\text{per}(B^2)>\text{per}(B)$.
It looks like the class of arithmetic means of two permutation matrices is the best source of examples of doubly stochastic matrices $A$ with $\text{Per}(A^2)>\text{Per}(A).$
More generally, if $P$ is a permutation matrix, and $A=\frac{P+P^{\*}}{2}$, then in the computer experiments that I have ran, we usually have $\text{Per}(A^2)>\text{Per}(A).$
**Testing a conjecture on convex sets**
We observe that the permutation matrices are the extreme points among all doubly stochastic matrices. Recall that the Krein-Milman theorem states that every compact convex subset of a Hausdorff locally convex topological vector space is in the convex closure of its extreme points. Furthermore, recall that Caratheodory's theorem states that if $C\subseteq\mathbb{R}^{n}$ is the convex closure of a set $A$, then every point in $C$ is a convex combination of at most $n+1$ many points of $A$. By combining these two theorems together, we conclude that for every compact convex subset $C$ of $\mathbb{R}^n$, every element in $C$ is in the convex closure of at most $n+1$ many extreme points in $C$.
In general, if $U$ is a totally bounded convex subset of a real affine space $V$ with $\dim(V)=n$, and one is trying to find an element in $U$ that satisfies a property using a computer search, one should test elements near the extreme points of $\overline{U}$. If this does not work, then whenever $1\leq d\leq n+1$, one should test points near convex combinations of $d+1$ many extreme points.
For this kind of problem, it also appears that it is a good idea to run all these tests to see if there is a symmetric counterexample. If this still did not work, then I would use a gradient descent to minimize $\text{per}(B)-\text{per}(B^2).$
| 5 | https://mathoverflow.net/users/22277 | 422032 | 171,641 |
https://mathoverflow.net/questions/422034 | 2 | I‘m currently reading [Arveson’s “A Short Course on Spectral Theory”](https://link.springer.com/book/10.1007/b97227), and I’m stuck at Exercise 3.1 (1). The question is:
Let $l^{\infty}(\mathbb{N})$ be the set of all bounded sequences of complex numbers. A Banach limit is a linear functional $\Lambda : l^{\infty}(\mathbb{N}) \to \mathbb{C}$ which satisfies $|| \Lambda ||=\Lambda((1,1,\ldots))=1$, and $\Lambda (Ta) = \Lambda(a)$ for any $a\in l^{\infty}(\mathbb{N})$, where $T$ denotes the left shift operator.
Prove that every Banach limit $\Lambda$ is a positive linear functional in the sense that
$$
\forall n \ a\_n\geq 0 \implies \Lambda(\{a\_n\}\_n) \geq 0
$$
I have no idea to prove it. Would you please give me some hints?
| https://mathoverflow.net/users/481886 | Positiveness of Banach limit | You do not need the shift invariance, what is important is that $\|\Lambda\|=\Lambda((1,1,\ldots))=1$.
Denote ${\bf 1}=(1,1,\ldots)$, ${\bf a}=(a\_1,a\_2,\ldots)$, $\Lambda(a)=\theta$. Assume at first that $\theta$ is not real (here we do not need that $a\_n$'s are non-negative, it is sufficient that they are real). Then for small real $t$ we have $\|{\bf 1}+it{\bf a}\|=1+o(t)$, while $|\Lambda({\bf 1}+it{\bf a})|=|1+it\theta|\geqslant \Re(1+it\theta)=1-t\Im \theta$, and if we choose sign of $t$ opposite to the sign of $\Im \theta$, we get $|\Lambda({\bf 1}+it{\bf a})|>\|\Lambda({\bf 1}+it{\bf a})\|$ for small $t$ of the chosen sign, a contradiction.
Now if additionally $a\_n\geqslant 0$ for all $n$, and $\theta<0$, do a similar trick comparing $\Lambda({\bf 1}-t{\bf a})=1-t\theta$ and $\|{\bf 1}-t{\bf a}\|\leqslant 1$ for small positive $t$.
| 11 | https://mathoverflow.net/users/4312 | 422036 | 171,643 |
https://mathoverflow.net/questions/420021 | 2 | Disclaimer: I'm just starting to read Sieve Methods by Halberstam and Richert, so my present knowledge of the subject is close to zero, but it made me wonder if some connection to physics could exist, so I dare share this question here.
As I said, I'm reading Halberstam and Richert's book on sieve theory, and as of now I just try to understand the very principle of sieving (sifting? Some help with English would be welcome too) a set of numbers.
Still, it seems one wants to minimize the error term in estimating the size of the considered set, so what we deal with is an optimization problem. That made me wonder if some variational principle like the maximal entropy principle or the least action principle which are widely used in physics could be used: namely can one define some "sieve action" from first principles depending on the specific problem to be solved, and whose extremum would lead to an optimal result?
What kind of mathematical tools would its implementation require? Have such ideas been suggested so far? If yes, could I get relevant references?
| https://mathoverflow.net/users/13625 | Sieve theory through variational principles | As suggested in the comment by Stanley Yao Xiao, Selberg's choice of $\lambda\_d$ in his clever upper bound sieve is an application of a "discrete" variational principle. This answer is going to present another concept arising from sieve theory that may possibly be connected with physics.
In physics, objects are studied via differential equations, and in analytic number theory theory, sieves can be studied via **differential-difference equations**.
In 1964, Ankeny and Onishi proved that if $\sigma(u)$ is the solution to the IVP that
$$
\sigma(u)={e^{-\gamma\kappa}\over\Gamma(\kappa+1)}\left(\frac u2\right)^\kappa\quad(0<u\le2),
$$
$$
[u^{-\kappa}\sigma(u)]'=-\kappa u^{-\kappa-1}\sigma(u-2)\quad(u>2),
$$
then we have
**Theorem (Ankeny & Onishi, 1964; Halberstam & Richert, 1974):** For $u>0$, if $\nu(d)$ denote the number of distinct prime divisors of $d$, $\omega(d)$ is a multiplicative function and $X,R\_d$ are numbers satisfying
$$
|\mathcal A\_d|={\omega(d)\over d}X+R\_d,
$$
$$
-L\le\sum\_{w\le p<z}{\omega(p)\log p\over p}-\kappa\log\frac zw\le O(1),
$$
then
$$
S(\mathcal A,\mathcal P,z)<X\prod\_{p<z}\left(1-{\omega(p)\over p}\right)\left\{{1\over\sigma(u)}+O\left(L\cdot{u^{2\kappa}+u^{-\kappa-1}\over\log z}\right)\right\}+\sum\_{\substack{p|d\Rightarrow p<z\\d<z^u}}3^{\nu(d)}|R\_d|
$$
| 3 | https://mathoverflow.net/users/449628 | 422039 | 171,645 |
https://mathoverflow.net/questions/414576 | 4 | Just to fix the environment, let's work in the Baire space $\omega^\omega$, the space of infinite sequences of natural numbers with the product of the discrete topology over $\omega$. We say that a subset $A\subseteq \omega^\omega$ satisfies Hurewicz dichotomy if either it's $F\_\sigma$ or there exists a Cantor set (a closed subset with no isolated points) $\mathcal{C}\subseteq \omega^\omega$ such that $\mathcal{C}\setminus A$ is countable dense in $\mathcal{C}$.
Now Hurewicz proved that
>
> Every analytic subset $A\subseteq\omega^\omega$ satisfies the Hurewicz dichotomy.
>
>
>
Now my questions are:
1. What is the consistency strength of the statement "Every subset of the Baire space satisfies the Hurewicz dichotomy"?
2. What is the relationship between the Hurewicz dichotomy property and the perfect set property (also consistency-wise)?
Thanks!
| https://mathoverflow.net/users/141146 | Consistency of the Hurewicz dichotomy property | The anwer is in: *Tall, Franklin D.; Todorcevic, Stevo; Tokgöz, Seçil*, [**The strength of Menger’s conjecture**](http://dx.doi.org/10.1016/j.topol.2020.107536).
In the paper they prove (among other things) that the Hurewicz dichotomy extended to all subsets of the reals is equiconsistent with an inaccessible cardinal.
| 1 | https://mathoverflow.net/users/141146 | 422059 | 171,650 |
https://mathoverflow.net/questions/422062 | 1 | Let $X$ a variety over an algebraically closed field $k$ (which we can assume to be actually $k=\mathbb{C}$) and $G$ a connected reductive algebraic group acting freely on $X$ (we can actually assume $G=Gl\_n$). Can we relate somehow the (compactly supported) etale cohomology $H\_c(X/G,\overline{\mathbb{Q}}\_{\ell})$ and $H\_c(X/B,\overline{\mathbb{Q}}\_{\ell})$? Better
We have the morphism $f:X/B \to X/G$ which should be a fibration with fibre $G/B$ so the better possibile should be something like $$H\_c(X/B,\overline{\mathbb{Q}}\_{\ell})=H\_c(X/G,\overline{\mathbb{Q}}\_{\ell}) \otimes H\_c(G/B,\overline{\mathbb{Q}}\_{\ell}) $$
To get this or something similar my idea would have been to study the local systems $R^qf\_\*\overline{\mathbb{Q}\_{\ell}}$ and use the spectral sequence: $$E^{p,q}\_2=H^p\_c(X/G,R^qf\_\*\overline{\mathbb{Q}\_{\ell}}) \Rightarrow H^{p+q}\_c(X/B,R^qf\_\*\overline{\mathbb{Q}\_{\ell}}) .$$ However it seems to me that there should be no reason a priori for these local systems to be trivial or the sequence to abrupt at the second page.
| https://mathoverflow.net/users/146464 | Comparing cohomology of quotient by algebraic group and Borel subgroup | The spectral sequence degenerates on the second page since $X/B \to X/G$ is a smooth projective morphism (as $G/B$ is smooth projective) by a result of [Deligne and Blanchard](https://en.wikipedia.org/wiki/Leray_spectral_sequence#Degeneration_theorem).
The local systems are trivial because, first, they can be trivialized on open sets $U$ where the bundle is trivial, and then on the intersection between two open sets $U \cap V$, the gluing data is given by the action of some function $U \cap V \to G$ on $H^\* ( G/B, \overline{\mathbb Q}\_\ell)$, but this action is trivial since $G$ is connected so the gluing data is trivial.
So there is indeed a tensor product isomorphism as you suggest (though not usually one compatible with cup product structure).
| 2 | https://mathoverflow.net/users/18060 | 422064 | 171,652 |
https://mathoverflow.net/questions/407758 | 3 | Let $f(n)$ be [A007814](https://oeis.org/A007814), the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Let $g(n)$ be [A025480](https://oeis.org/A025480), $g(2n) = n$, $g(2n+1) = g(n)$.
Then we have an integer sequences given by
\begin{align}
a\_1(0)=a\_1(1)&=1\\
a\_1(2n+1) &= a\_1(n)+a\_1(g(n-1))\\
a\_1(2n)& = a\_1(n-2^{f(n)})+a\_1(2n-2^{f(n)})+a\_1(g(n-1))
\end{align}
and
\begin{align}
a\_2(0)=a\_2(1)&=1\\
a\_2(2n+1) &= a\_2(n)+a\_2(g(n-1))\\
a\_2(2n)& = a\_2(n) + a\_2(n-2^{f(n)}) = a\_2(2n-2^{f(n)})+a\_2(g(n-1))
\end{align}
Let
$$s\_k(n)=\sum\limits\_{j=0}^{2^n-1}a\_k(j)$$
Then I conjecture that $s\_1(n)$ is [A011965](http://oeis.org/A011965), second differences of Bell numbers and $s\_2(n)$ is [A026012](http://oeis.org/A026012), second differences of Catalan numbers.
I also conjecture that $a\_1(\frac{4^n-1}{3})$ is [A141154](http://oeis.org/A141154) and $a\_2(\frac{4^n-1}{3})$ is [A000958](http://oeis.org/A000958).
Is there a way to prove it?
Similar questions:
* [Sum with Stirling numbers of the second kind](https://mathoverflow.net/questions/406738/sum-with-stirling-numbers-of-the-second-kind)
* [Recurrence for the sum](https://mathoverflow.net/questions/405174/recurrence-for-the-sum)
* [Pair of recurrence relations with $a(2n+1)=a(2f(n))$](https://mathoverflow.net/questions/406902/pair-of-recurrence-relations-with-a2n1-a2fn)
* [Sequence that sums up to INVERTi transform applied to the ordered Bell numbers](https://mathoverflow.net/questions/407290/sequence-that-sums-up-to-inverti-transform-applied-to-the-ordered-bell-numbers)
| https://mathoverflow.net/users/231922 | Sequences that sums up to second differences of Bell and Catalan numbers | Let me address the case of $s\_2(n)$.
First we notice that $g(n-1) = k$ whenever $n=(2k+1)2^t$. Then
for $n=2^{t\_1}(1+2^{1+t\_2}(1+\dots(1+2^{1+t\_\ell}))\dots)>1$ with $t\_j\geq 0$, we have
$$a\_2(n) = \begin{cases}
a\_2\bigg(2^{t\_2}(1+\dots(1+2^{1+t\_\ell}))\dots)\bigg) + a\_2\bigg(2^{t\_3}(1+\dots(1+2^{1+t\_\ell}))\dots)\bigg) & \text{if}\ t\_1=0; \\
a\_2\bigg(2^{t\_1-1}(1+2^{1+t\_2}(1+\dots(1+2^{1+t\_\ell}))\dots)\bigg) + a\_2\bigg(2^{t\_1+t\_2}(1+\dots(1+2^{1+t\_\ell}))\dots)\bigg) & \text{if}\ t\_1>0.
\end{cases}
$$
In contrast to previous questions by the OP, obtaining an explicit formula for $a\_2(n)$ here seems to be quite challenging. Still, thanks to the connection
$$s\_2(n) = \sum\_{\ell=0}^n \sum\_{t\_1+\dots+t\_\ell\leq n-\ell} a\_2\bigg( 2^{t\_1}(1+2^{1+t\_2}(1+\dots(1+2^{1+t\_\ell}))\dots) \bigg),$$
we can translate the above recurrences over to $s\_2(n)$.
---
Let's partition $s\_2(n)$ into smaller sums depending on the 2-adic valuation of the summands:
$$s\_2(n) = \sum\_{k\geq 0} s^{(k)}\_2(n),$$
where
$$s^{(k)}\_2(n) := \sum\_{j=0\atop \nu\_2(2^n+j)=k}^{2^n-1} a\_2(j).$$
From the recurrence above, it follows that for $n>1$,
$$s\_2^{(0)}(n) = \sum\_{t=0}^{n-1} s\_2(t) = s\_2(n-1) + s\_2^{(0)}(n-1) = 2s\_2^{(0)}(n-1) + \sum\_{k\geq 1} s^{(k)}\_2(n-1) $$
and for $k>0$
$$s\_2^{(k)}(n) = \sum\_{m\geq k-1} s\_2^{(m)}(n-1).$$
That is, in the matrix form we have:
$$
\begin{bmatrix} s\_2^{(0)}(n)\\ s\_2^{(1)}(n)\\ s\_2^{(2)}(n) \\ \vdots \end{bmatrix}
=
M\cdot
\begin{bmatrix} s\_2^{(0)}(n-1)\\ s\_2^{(1)}(n-1)\\ s\_2^{(2)}(n-1) \\ \vdots \end{bmatrix} = M^{n-1}\cdot \begin{bmatrix} 1\\ 1\\ 0\\ 0\\ \vdots \end{bmatrix}
$$
where
$$M: = \begin{bmatrix}
2 & 1 & 1 & 1 & \\
1 & 1 & 1 & 1 & \ddots \\
0 & 1 & 1 & 1 & \ddots \\
0 & 0 & 1 & 1 & \ddots \\
& \ddots & \ddots & \ddots & \ddots \\
\end{bmatrix}.$$
---
By induction on $n$ it can be verified that the generating functions for sequences $\big(s\_2^{(k)}(n)\big)\_{n\geq 0}$ are expressed in terms of the one for Catalan numbers $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ as follows:
$$\sum\_{n\geq 0} s\_2^{(k)}(n)\cdot x^n = (1-x) x^k C(x)^{k+2},$$
or in other words:
$$s\_2^{(k)}(n) = \frac{k+2}{n+2}\binom{2n-k+1}{n-k} - \frac{k+2}{n+1}\binom{2n-k-1}{n-k-1}.$$
Summing the above generating functions over $k\geq 0$ yields:
$$\sum\_{n\geq 0} s\_2(n)\cdot x^n = \frac{(1-x)C(x)^2}{1-xC(x)} = \frac{(1-x)^2C(x) - 1 + x}{x^2},$$
proving the conjecture for $s\_2(n)$.
| 5 | https://mathoverflow.net/users/7076 | 422067 | 171,653 |
https://mathoverflow.net/questions/422023 | 4 | $\newcommand\bHom{\mathbf{Hom}}\newcommand\bOb{\mathbf{Ob}}\newcommand\bRel{\mathbf{Rel}}$This question is probably stupid and definitely bureaucratic, but
>
> Is writing $f\circ g$ for the composition of morphisms in the ‘many hom-classes’ definition of a category unambiguous?
>
>
>
The ‘many hom-classes’ definition of a category (as given e.g. on the [nlab](https://ncatlab.org/nlab/show/category)) says that for each pair of arrows
$(f,g)\in\bHom\_\mathcal{C}(Y,Z)\times\bHom\_\mathcal{C}(X,Y)$ we ‘have an arrow’ $f\circ g\in\bHom\_\mathcal{C}(X,Z)$, but if the hom-classes may not be disjoint how do we know that the arrows we ‘have’ from identical composable arrow pairings in differing hom-class pairs match up?
Rephrased using the language of composition functions, the above definition is the same as a collection of functions $$\{\circ\_{XYZ}:\bHom\_\mathcal{C}(Y,Z)\times\bHom\_\mathcal{C}(X,Y)\to\bHom\_\mathcal{C}(X,Z)\}\_{X,Y,Z\in\bOb\_\mathcal{C}}.$$
The axioms then specify associativity and unitarity, but how do we know that we don't have objects $X,Y,Z,X',Y',Z'\in\bOb\_\mathcal{C}$ and arrows $$f\in\bHom\_\mathcal{C}(Y,Z)\cap\bHom\_\mathcal{C}(Y',Z'), \\ g\in\bHom\_\mathcal{C}(X,Y)\cap\bHom\_\mathcal{C}(X',Y'),$$ so $$(f,g)\in\bigl(\bHom\_\mathcal{C}(Y,Z)\times\bHom\_\mathcal{C}(X,Y)\bigr)\cap\bigl(\bHom\_\mathcal{C}(Y',Z')\times\bHom\_\mathcal{C}(X',Y')\bigr),$$ such that $$\circ\_{XYZ}(f,g)\neq\circ\_{X'Y'Z'}(f,g)?$$ Note that these composites live in $\bHom\_\mathcal{C}(X,Z)$ and $\bHom\_\mathcal{C}(X',Z')$, respectively, and as David Roberts points out in the [comments](https://mathoverflow.net/questions/422023/missing-axiom-in-the-typed-definition-of-a-category#comment1084413_422023) these classes can be disjoint even if the original hom-classes aren't.
If it is possible to have objects and arrows as above the notation $f\circ g$ is obviously ambiguous — also obvious is that we could fix this with an additional axiom (scheme?) if it were a problem.
>
> Is this situation already precluded by the other axioms/data present in the many hom-classes definition of a category?
>
>
>
---
For an example where we have identical arrow pairings in differing hom-classes but composition still matches up, consider any two composable relations $R,S\in\bOb\_{\bRel}$. By definition $R$ and $S$ are subsets of some Cartesian squares $Y\times Z$ and $X\times Y$ (respectively), but we can take $X'$, $Y'$, $Z'$ to be any strict superclasses of $X$, $Y$, $Z$ (respectively) and observe that $R$ and $S$ are also subsets of $Y'\times Z'$ and $X'\times Y'$ (respectively), so $$R\in\bHom\_{\bRel}(Y,Z)\cap\bHom\_{\bRel}(Y',Z'), \\ S\in\bHom\_{\bRel}(X,Y)\cap\bHom\_{\bRel}(X',Y').$$ Here we obviously still have that the composition functions coincide, but this is arguably due to the fact that $\bRel$ is most naturally presented as a one hom-class category. Again, I understand that this question is very pedantic at best — the patience involved in any clarification is greatly appreciated.
| https://mathoverflow.net/users/92164 | Missing axiom in the typed definition of a category? | **There is no missing axiom. The *notation* is potentially ambiguous, but rarely (if ever) so in practice.**
The situation is just the same as writing addition in arbitrary abelian groups as $x + y$. Formally, the operation “$+$” depends not just on $x,y$ but (before them) on a choice of group $G$ — explicitly, we could write it as $x +\_G y$. And so the notation $x+y$ can potentially be ambiguous — $x$ and $y$ could belong to two different Abelian groups, whose addition operations disagree on $x+y$. But it would be nonsense to suggest that algebraists need to assume different groups considered must always be disjoint, or that their addition must always agree on intersections. We simply agree that the notation $x + y$ implicitly depends on the intended group, and make sure that this is always clear from context. And in the very rare situations where ambiguity would arise, we distinguish it explicitly by writing $+\_G$ or some similar annotation. Similarly, composition $f \circ g$ is formally also dependent on the category and objects involved, $\mathrm{comp}(\mathcal{C},X,Y,Z,f,g)$ (so for a fixed category, it is a function of the objects and arrows) — but in the *notation*, we usually write just $f \circ g$, with the rest inferred from context.
**This notational point is formally addressed in computer proof assistants, where functions have *implicit arguments*.** This is essential for formalising many notations used in mathematical practice. In particular, the many-hom-sets definition of categories has been formalised by multiple authors many times (including myself), and shown equivalent to the class-of-arrows definition for instance [here](https://github.com/UniMath/TypeTheory/blob/master/TypeTheory/Categories/ess_and_gen_alg_cats.v), in the TypeTheory Coq library over UniMath. So we can be very confident that there is no missing axiom.
(Counter to [Duchamp Gérard’s answer](https://mathoverflow.net/a/422028/2273), I do not think all (or even most) category theorists assume hom-sets are always disjoint. What I think all agree is that if we start from the many-hom-sets definition, then “the class of all arrows of $C$” must mean the *disjoint* union of the hom-sets, not the pure set-theoretic union. But when working with the many-hom-sets definition, the “class of all arrows” is not taken as primary — it is fairly rarely used, and when it’s used, there’s no problem with the original hom-sets being sub-objects of it rather than literal set-theoretic subsets.)
(Also, as Andreas Blass and Simon Henry note in comments, if we’re working in a type-theoretic or structural set-theoretic foundation, then asking if abstract sets are disjoint is not even well-defined. Intersection is defined between *subsets* of any given set/type; but in structural foundations, abstract sets not automatically subsets of an ambient universal class.)
| 19 | https://mathoverflow.net/users/2273 | 422068 | 171,654 |
https://mathoverflow.net/questions/422076 | 7 | For a manifold $M$ a **vector field** is a derivation of the algebra $C^{\infty}(M)$ of smooth functions on $M$. What happens if look instead as derivations on the continuous functions of a manifold. I guess we get fewer derivations . . . but I'm not sure how one might prove this.
| https://mathoverflow.net/users/128876 | Derivations on the continuous functions of a manifold | More is true: if $X$ is a topological manifold, then in fact $\operatorname{Der}(C(X)) = 0$, where $C(X)$ denotes the $\mathbb{R}$-algebra of $\mathbb{R}$-valued continuous functions on $X$. In particular, this is so for smooth manifolds $M$.
Here is one proof: <https://ncatlab.org/nlab/show/derivation#DerOfContFuncts>
| 11 | https://mathoverflow.net/users/1849 | 422078 | 171,655 |
https://mathoverflow.net/questions/349712 | 9 | The Wikipedia article on [Hahn Series](https://en.wikipedia.org/wiki/Hahn_series) ~~mentions~~ mentioned that these were studied by Hahn "in his approach to Hilbert's seventeenth problem".
>
> Is this correct? If so, what was this approach, and where can I read about it?
>
>
>
I have read most of Hahn's paper [über die nichtarchimedischen Größensysteme](https://zbmath.org/?q=an%3A38.0501.01), where Hahn series were introduced, but I have not seen Hilbert's 17th problem on positive polynomials being mentioned there. I have also skimmed his list of publications and not found anything else that looks relevant.
| https://mathoverflow.net/users/27013 | Hahn's approach to Hilbert's 17th problem? | As you suspect, Hahn does not discuss Hilbert's 17th problem in the paper you cite, and, like you, I am not aware of any of Hahn's publications that applies his work on non-Archimedean ordered systems to Hilbert's 17th problem.
However, in the paper you mention Hahn does discuss Hilbert's (arithmetic) completeness condition, which is discussed by Hilbert in the second problem of his famous list of problems. There, however, Hahn is not concerned with the consistency of the real number system, which is the subject of the second problem, but rather with generalizing Hilbert's completeness condition so as to be applicable to non-Archimedean as well as Archimedean ordered Abelian groups and ordered fields.
| 4 | https://mathoverflow.net/users/18939 | 422083 | 171,657 |
https://mathoverflow.net/questions/422085 | 5 | Im reading a paper by Matomaki [here](https://link.springer.com/content/pdf/10.1007/s10474-009-8163-5.pdf), and the following is stated (I'm paraphrasing):
"By the Cauchy-Schwarz inequality and the large sieve, we have
$$\sum\_{q \leq Q}\frac{q}{\phi(q)}\sum\_{\substack{\chi\text{(mod $q$)}\\\text{primtiive}}} \big{|}\sum\_{a \in \mathcal{A}}\chi(a)\sum\_{b \in \mathcal{B}}\chi(b) \big{|} \leq (Q^2 + N)(AB)^{1/2}$$
where $Q,N$ are positive integers and $\mathcal{A}, \mathcal{B}\subseteq \{1,...,N\}$ and $|\mathcal{A}| = A$ and $|\mathcal{B}| = B$."
Now I am not so concerned with the application of the large sieve, but I am a little confused about how she applied Cauchy-Schwarz. Of course the large sieve she is referring to states
$$\sum\_{q \leq Q}\frac{q}{\phi(q)}\sum\_{\substack{\chi\text{(mod $q$)}\\\text{primtiive}}} \big{|}\sum\_{n \leq N }a\_n\chi(n)\big{|}^2 \leq (Q^2 + N)\sum\_{n \leq N}|a\_n|^2.$$
But I am unsure of how she used Cauchy-Schwarz, especially with multiplicative characters. Does anyone have any thoughts?
| https://mathoverflow.net/users/423821 | Specific application of Cauchy-Schwarz and Large Sieve | As she writes, first apply Cauchy-Schwarz, and only then apply the large sieve (twice).
The relevant instance of Cauchy-Schwarz is
$$|x\_1 x\_2| \le \frac{|x\_1|^2+|x\_2|^2}{2},$$
which, by replacing $x\_1$ and $x\_2$ by $x\_1\sqrt{C}$ and $x\_2/\sqrt{C}$ ($C>0$) becomes
$$|x\_1 x\_2| \le \frac{C |x\_1|^2 + C^{-1} |x\_2|^2}{2}.$$
We apply it with $x\_1=\sum\_{a \in \mathcal{A}} \chi(a)$ and $x\_2 = \sum\_{b \in \mathcal{B}}\chi(b)$ and with $C$ to be determined later (but independent of $\chi$). We obtain that the relevant sum is
$$\le \frac{1}{2}\left( C\sum\_{q \le Q} \frac{q}{\phi(q)}\sum\_{\substack{\chi \bmod q\\ \text{primitive}}}\left|\sum\_{a \in \mathcal{A}} \chi(a)\right|^2 + C^{-1}\sum\_{q \le Q} \frac{q}{\phi(q)}\sum\_{\substack{\chi \bmod q\\ \text{primitive}}} \left|\sum\_{b \in \mathcal{B}} \chi(b)\right|^2 \right),$$
which, by two applications of the large sieve, is
$$\le \frac{1}{2}(Q^2 + N)\left( CA+ C^{-1} B\right).$$
Now take $C=\sqrt{B/A}$.
| 8 | https://mathoverflow.net/users/31469 | 422088 | 171,659 |
https://mathoverflow.net/questions/390402 | 9 |
>
> I've already asked this question [on Math StackExchange](https://math.stackexchange.com/questions/3981370/is-mathbbzi-varphi-a-euclidean-domain) but having gotten no responses this may be more obscure than I had initially believed.
>
>
>
---
Here $\varphi=\frac{1+\sqrt{5}}{2}$. It's true that $\mathbb{Z}[\varphi]=\mathcal{O}\_{\mathbb{Q}(\sqrt{5})}$ is Euclidean since $\mathbb{Q}(\sqrt{5})$ is norm Euclidean, and I've read that $A=\mathbb{Z}\left[\frac{1+i}{\sqrt{2}}\right]$ is Euclidean as well, though I'm not certain what the Euclidean function is there (the reasonable candidate being $x\mapsto N\_{K/\mathbb{Q}}(x)$, $K$ being $A$'s fraction field). So, my questions are if anything is known about:
* Is $R=\mathbb{Z}[i,\varphi]$ Euclidean?
* + All I know is that $R=\mathcal{O}\_L$ ($L$ being $R$'s fraction field), by computing its discriminant and comparing with $D\_L$ given by [LMFDB](https://www.lmfdb.org/NumberField/4.0.400.1), so $R$ is a PID.
* If so, what is the Euclidean function?
* + The reasonable candidate is $x\mapsto N\_{L/\mathbb{Q}}(x)$, but this has proved difficult to work with.
* If not, is $R$ a (finite-index) subring of a (nice) Euclidean domain?
* My main goal is to compute $\gcd$s in $R$, so if all of the above don't have known/affirmative answers, a Euclidean algorithm in $R$ (sans a Euclidean function) would be just as great instead.
Thanks in advance for any answers.
| https://mathoverflow.net/users/159965 | Is $\mathbb{Z}[i,\varphi]$ a Euclidean domain? | It turns out that $K=\mathbb{Q}[i,\varphi]$ **is** norm-Euclidean. The proof of this fact appears as Appendix A in <https://arxiv.org/abs/2205.03007>. I'll explain the heft of the argument here.
Let $R=\mathbb{Z}[i,\varphi]$ and $N=N\_{K/\mathbb{Q}}$. We use this formulation of norm-Euclidean: for all $\alpha\in K$ there exists $\beta\in R$ with $N(\alpha+\beta)<1$.
Model $K\cong\mathbb{Q}^4$ via $w+x\varphi+yi+zi\varphi\mapsto(w,x,y,z)$ whence $R$ identifies with $\mathbb{Z}^4$. If $K$ were "extra nice" we could simply verify that for all $\alpha\in\left[-\frac{1}{2},\frac{1}{2}\right)^4\cap K=C$, we have $N(\alpha)<1$ whence for $\alpha\in K$ we could simply let $\beta=-[\alpha]$ where $[\alpha]$ is $\alpha$ with each coordinate rounded to the nearest integer. (This is how the standard proof goes for the Gaussian number field being norm-Euclidean.) Unfortunately, this approach fails, particularly near the corners where exactly three components have the same sign.
Notice, however, that $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ has very small norm, so that if we shift only *some* points in $C$ then we may be okay, since this would resolve e.g. $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)$, and all points nearby, since $N$ is a polynomial in the coordinates, so it is definitely continuous. The trick then is to exactly quantify continuity. We will consider $C\_n=\left[-\frac{1}{2},\frac{1}{2}\right)^4\cap(\frac{1}{n}\mathbb{Z}^4)$ with $n$ to be decided later. If we can show for each $\alpha\in C\_n$ that there exists $\delta\in\mathbb{Z}^4$ such that $N(\alpha+\delta+B\_n)<1$ where $B\_n$ is the radius-$\frac{1}{n}$ $L^\infty$ ball in $\mathbb{Q}^4$, then we are done.
We consider now finding an explicit bound for $N(\alpha+\beta)$ where $\alpha=w+x\varphi+yi+zi\varphi\in K$ is any point and $\beta=d\_1+d\_2\varphi+d\_3i+d\_4i\varphi$ has $\|\beta\|\_\infty<\varepsilon$, i.e. $\lvert d\_1\rvert<\varepsilon$. $N(\alpha+\beta)$ is quartic in the eight variables, so write it out and group terms by "$d$-type," that is, $N(\alpha+\beta)=\sum\_{e\in\mathbb{N}^4}d^eP\_e(\alpha)$ for polynomials $P\_e\in\mathbb{Q}[w,x,y,z]$. (e.g., $d\_1d\_2^2x+d\_2d\_3d\_4z+2d\_1d\_2^2y+8d\_4wyz$ would have $P\_{1200}=x+2y$, $P\_{0111}=z$, $P\_{0001}=8wyz$, and $P\_e=0$ otherwise.) Then, apply the triangle inequality (and positivity of $N$): $N(\alpha+\beta)=\left\lvert\sum\_{e\in\mathbb{N}^4}d^eP\_e(\alpha)\right\rvert\le\sum\_{e\in\mathbb{N}^4}\lvert d\rvert^e\lvert P\_e(\alpha)\rvert\le\sum\_{e\in\mathbb{N}^4}\varepsilon^{\sum e}\lvert P\_e(\alpha)\rvert$. (For the same example, the bound becomes $8\varepsilon\lvert wyz\rvert+\varepsilon^3(\lvert x+2y\rvert+\lvert z\rvert)$.) This now depends entirely on $\alpha$ and $\varepsilon=\frac{1}{2n}$. Let that bound be $f(\alpha,n)$ (also a polynomial in $\alpha$'s coordinates).
The strategy is then to check with a computer all $n^4$ elements $\alpha\in C\_n$, and for each evaluate $f(\alpha,n)$. If $f(\alpha,n)<1$ then we move on, otherwise check $f(\alpha\pm e\_j,n)$ for $e\_j$ one of the four Kronecker basis vectors. It turns out that when $n=6$ we only ever need to check $\alpha$ and $\alpha\pm e\_j$ (i.e. 0 or 1 steps) for this number field.
---
Curiously, it would seem that this approach could work to show that other biquadratic number fields $\mathbb{Q}\left[i,\sqrt{m}\right]$ of relatively small discriminant are norm-Euclidean, but this approach does not yield any results for $5<m,n<100$.
| 5 | https://mathoverflow.net/users/159965 | 422095 | 171,663 |
https://mathoverflow.net/questions/421801 | 4 | I am reading the one lecture note [Dynamics for Spherical Models of Spin-Glass and Aging](https://doi.org/10.1007/978-3-540-40908-3_5).
On page 126. In the Sherrington-Kirkpatrick (SK) model, we suppose that there are $N$ people labeled as $[N]:=\{1,2,\ldots,N\}$, $\sigma \in \{+1,-1\}^N$. Let $\mathbf{J}=(J\_{ij})\_{1\le i, j\le N}$ be symmetric and let $J\_{ij}$ be a centered independent random variable such that
$$
E[J\_{ij}^2]=1/n, \, E[J\_{ii}^2]=2/n,
$$
often assumed to be Gaussian for simplicity. Assume that $x\_0$ is independent of $\mathbf{J}$.
If $\{\lambda\_i\}\_{1\le i\le N}$ denotes the eigenvalue of $\mathbf{J}=\{J\_{ij}\}$, then for a vector $\mathbf{x}\_0$,
$$
\frac{1}{N}\langle \mathbf{x}\_0, e^{\mathbf{J}}\mathbf{x}\_0\rangle=^d\frac{1}{N}\langle \mathbf{x}\_0, \mathbf{x}\_0\rangle\sum\_{i=1}^N e^{\lambda\_i}u\_i^2
$$
where $u$ is independent of $\lambda\_i$ and $\mathbf{x}\_0$, and follows the law of the sphere $S\_{\sqrt{N}}^{N-1}$ with radius $\sqrt{N}$.
***(I think $u$ follows the law of the unit sphere?)***
---
From the Weak law of large numbers, the first term
$$
\frac1N\langle x\_0, x\_0\rangle\to \int x^2d\mu(x)
$$
as $n\to \infty$.
The main question is as follows.
>
> Question: Why do we have
> $$
> \lim\_{N\to \infty} \sum\_{i=1}^N e^{\lambda\_i}u\_i^2=\int e^{\lambda}d\sigma(\lambda)
> $$
>
>
>
From semi-circle law, we have the empirical measure $\frac{1}{N}\sum \delta\_{\lambda\_i}\to \sigma(dx)=C\sqrt{4-x^2}dx$. But how to get the above limit? I think there is missing the scaling $1/N$ on the left-hand side.
---
I try to show that
$$
law(\frac{1}{N}\langle \mathbf{x}\_0, e^{\mathbf{J}}\mathbf{x}\_0\rangle)=law(\frac{1}{N}\langle \mathbf{x}\_0, \mathbf{x}\_0\rangle\sum\_{i=1}^N e^{\lambda\_i}u\_i^2)
$$
Let $\mathbf{J}=G^\*DG$ where $D$ is the diagonal matrix with eigenvalues $\lambda\_i$. Then $e^\mathbf{J}$ can be expressed via spectral decomposition as
$$
e^\mathbf{J}=G^Te^{D}G=\sum e^{\lambda\_i} s\_is\_i^T
$$
(the second one is wrong?)
where $s\_i\in R^N$ are eigenvectors of $\mathbf{J}$ and $G=[s\_1,\dots, s\_N]$.
I am confused about where $u\_i^2$ comes from? If we plug in the spectral decomposition, it becomes that
$$
\frac{1}{N}\langle \mathbf{x}\_0, e^{\mathbf{J}}\mathbf{x}\_0\rangle=\frac{1}{N}\langle \mathbf{x}\_0, \sum e^{\lambda\_i} s\_is\_i^T\mathbf{x}\_0\rangle
$$
Here
$$
E[\sum e^{\lambda\_i} s\_is\_i^T]=E[\frac{1}{N}I]
$$
| https://mathoverflow.net/users/168083 | How to get $\lim_{N\to \infty} \sum_{i=1}^N e^{\lambda_i}u_i^2=\int e^{\lambda}d\sigma(\lambda)$? | The eigenvector components $u\_i$ have zero mean and variance $1/N$ (since $\sum\_i u\_i^2=1$); they are independent of the eigenvalues $\lambda\_i$. We therefore have the expectation value
$$\lim\_{N\to \infty}\mathbb{E}\left[ \sum\_{i=1}^N e^{\lambda\_i}u\_i^2\right]=\lim\_{N\to \infty}\mathbb{E}\left[\frac{1}{N} \sum\_{i=1}^N e^{\lambda\_i}\right]=\int e^{\lambda}d\sigma(\lambda)=I\_2(2),$$
with $p(\lambda)=d\sigma(\lambda)/d\lambda=(2\pi)^{-1}\sqrt{4-\lambda^2}\,\mathbb{1}\_{[-2,2]}(\lambda)$ the semi-circular probability distribution of the eigenvalues (normalized to unity).
Fluctuations arond the expectation value are smaller by a factor $1/\sqrt N$ and may be neglected in the large-$N$ limit.
| 1 | https://mathoverflow.net/users/11260 | 422097 | 171,664 |
https://mathoverflow.net/questions/422091 | 7 | Suppose $g$ is a total computable injective function and $f$ is a total computable function satisfying $$g(x)<f(x)$$ for all sufficiently large $x$. Then we have $ran(f)\le\_Tran(g)$; basically, elements enter $ran(f)$ only when "permitted" to do so by $g$, and the injectivity of $g$ prevents this from happening unexpectedly.
It just occurred to me that I don't know if an appropriately-formulated *converse* is true. Suppose $g$ is a total computable injective function with noncomputable range, and $X$ is a noncomputable c.e. set with $X\le\_Tran(g)$. Is there a total computable $f$ outputting *canonical codes for finite sets* with $\bigcup\_{n\in\omega}f(n)=X$ and $g(x)<\min(f(x))$ for all sufficiently large $x$? What if we simply require $\bigcup\_{n\in\omega}f(n)\equiv\_TX$? After looking for a positive proof for a while, I'm starting to suspect that the answer is in fact negative; however, the construction of a counterexample $g, X$ seems a bit difficult.
(I suspect the answer to this question is in Soare's old book somewhere, but I don't have my copy at hand right now.)
| https://mathoverflow.net/users/8133 | Does permission always work? | No, there is a counterexample. The idea is that the use of the computation $X \le\_T ran(g)$ can be much worse than identity, and since we only care about the reduction in one direction, we can drive that use up very large. In contrast, a function $f$ trying to maintain $X \equiv\_T ran(f)$ can't freely increase the use on one side without making changes on the other.
We are building $g$ and $X$, and a basic module must diagonalize against a tuple $(f, \Phi, \Psi, k)$, ensuring that if $f$ is total and $g(y) < \min(f(y))$ for all $y > k$, then it's not the case that $\Phi(X) = \bigcup\_n f(n)$ and $\Psi(\bigcup\_n f(n)) = X$. For ease of notation, let $Z = \bigcup\_n f(n)$.
For this module, we choose large $m\_1 > m\_0$ and enumerate the axiom $m\_0 \not \in X$ with use $m\_1 \not \in ran(g)$. We require that all future definitions of $g$ must use values greater than $m\_1$, and we wait until we see $f$ converge on all $y \le k$ and all $y$ where we have already defined $g(y) < m\_0$. We wait further until we see an expansionary stage: some $s$, $r\_0$ and $r\_1$ with $\Phi\_s(X\_s\upharpoonright r\_0) = Z\_s\upharpoonright r\_1$ and $\Psi(Z\_s\upharpoonright r\_1) = X\_s\upharpoonright m\_0+1$. We pick an $m\_2 > r\_1$, we use our next definition of $g$ to enumerate $m\_1$ into $ran(g)$, and we enumerate the axiom $m\_0 \not \in X$ with use $m\_2 \not \in ran(g)$. We then require that there be no enumerations into $X$ below $r\_0$ and no further definitions of $g$ with values below $r\_1$.
Now, we wait until $f$ converges on all $y$ where we have already defined $g(y) < r\_1$. If $f$ uses any of these values to enumerate a new element into $Z$ below $r\_1$, we then have $\Phi\_s(X\_s\upharpoonright r\_0)$ incompatible with $Z$, so we win by maintaining the restraint on $X$ below $r\_0$.
If $f$ does not take this opportunity to enumerate new elements into $Z$ below $r\_1$, then $f$ will have lost the opportunity to ever again enumerate further elements into $Z$ below $r\_1$ (assuming that our choice of $k$ was correct and we maintain the restraint on $g$). So $Z\upharpoonright r\_1 = Z\_s\upharpoonright r\_1$. Now we enumerate $m\_0$ into $X$ and $m\_2$ into $ran(g)$, giving us $\Phi(Z)(m\_0) = \Phi\_s(Z\_s\upharpoonright r\_1)(m\_0) = 0 \neq X(m\_0)$, so we have won.
Now just arrange these modules into a finite injury priority argument.
| 10 | https://mathoverflow.net/users/32178 | 422100 | 171,666 |
https://mathoverflow.net/questions/422114 | 3 | In the end of the *[Voevodsky’s lectures on cross functors](https://www.math.ias.edu/vladimir/sites/math.ias.edu.vladimir/files/2015_transfer_from_ps_delnotes01.pdf)*, P. Deligne considers a couple of axioms which define (using the vocabulary of Ayoub's thesis) a stable homotopical 2-functor. Among them, we have that
1. **(Homotopy invariance)** If $p$ is the projection $\mathbb{A}^1\_X\to X$, the adjunction morphism $\operatorname{id}\to p\_\*p^\*$ is an isomorphism;
2. **(Stability)** If $s$ is the zero-section of $p$, then $p\_\#\circ s\_+$ is an equivalence of categories. (Where $p\_\#$ is the left adjoint of $p^\*$, which exists since $p$ is smooth.)
He then affirms that the two axioms above are well known in the $\ell$-adic setting. My first question then is: how are they proven? (I think a description of the proof would be nice for the MO community, but I would also be happy with a reference.)
The axiom of homotopy invariance surely axiomatises what its name describes: in Sheaves and Manifolds, M. Kashiwara and P. Shapira deduce the homotopy invariance of sheaf cohomology from the fact that the projection $X\times [0,1]\to X$ satisfies the axiom above. The axiom 1 then refers to this.
Now, I don't really understand whats the role of the axiom 2 (of "stability"). (Perhaps because I don't really have much of an intuition for $p\_\#$.) So my second question is: how should one think about this axiom? Perhaps it is more intuitive in the $\ell$-adic context?
| https://mathoverflow.net/users/131975 | Homotopy invariance of $\ell$-adic cohomology | Proof of homotopy invariance: This follows from a base change/Kunneth type statement and the calculation of the cohomology of $\mathbb A^1$.
Specifically, Lemma 7.6.7 of Lei Fu's etale cohomology theory, specialized to $S$ a point, $f$ the map from $X$ to a point, $g$ the map from $\mathbb A^1$ to a point, so that $g' = p$, $K$ an arbitrary complex on $X$, and $L$ the constant sheaf, implies that $p\_\* p^\* K = K \otimes f^\* g\_\* \mathbb Z\_\ell$ as soon as $K$ is strongly locally acyclic relative to $f$, which it is by Lemma 9.3.4.
Then $g\_\* \mathbb Z\_\ell$ is the cohomology of the affine line, which is simply $\mathbb Z\_\ell$, so $K \otimes f^\* g\_\* \mathbb Z\_\ell = K \otimes f^\* \mathbb Z\_\ell = K \otimes \mathbb Z\_\ell= K$.
Proof of stability: As Deligne notes in the very last paragraph, in the etale setting $p\_\#$ is $p\_!$ up to a shift and twist: For a smooth morphism, $p^\*$ and $p^!$ agree up to a shift and twist, and $p^!$ has a left adjoint $p\_!$, which thus agrees with $p\_\#$ up to the dual shift and twist.
Since shifting and twisting are equivalences of categories, it suffices to check that $p\_! s\_\*$ is an equivalence of categories. Now $s\_\*$ of any sheaf is compactly supported over the base $X$ (since a section is compact), which means $p\_! s\_\* = p\_\* s\_\*$, and $p\_\* s\_\*$ is the identity, and thus an equivalence, by the Leray spectral sequence.
Not sure on the motivation.
| 8 | https://mathoverflow.net/users/18060 | 422119 | 171,671 |
https://mathoverflow.net/questions/422107 | 5 | I am wondering if the following statement holds.
>
> If $u\in \mathscr{S}'$ satisfies $\left< u,\Phi\right>=0$ for all $\Phi \in \mathscr{S}$ with $\mathrm{div}\, \Phi=0$, then there exists $p\in \mathscr{S}'$ such that $u=\nabla p$ in $\mathscr{S}'$. Here <,> denotes the dual pairing.
>
>
>
It is well known that if we replace $\mathscr{S}$ with $\mathscr{D}$, then this is well-known theorem due to de Rham.
It is hard for me to find the analog version of tempered distribution.
Thank you very much for your time.
| https://mathoverflow.net/users/88462 | de Rham theorem for tempered distributions | This works. As explained in my comment, we need only show that if $p\in\mathcal D'(\mathbb R^n)$ and $\nabla p\in\mathcal S'$ (vector valued), then $p\in\mathcal S'$.
The condition for a function $\varphi$ to be a divergence of a vector field is $\int\varphi=0$; see [here.](https://mathoverflow.net/questions/112180/which-test-functions-are-the-divergence-of-a-vector-field%3F) So if we fix a $\varphi\_0\in\mathcal D$ with $\int\varphi\_0=1$ and $\varphi\in\mathcal S$ is arbitrary, then we can write $\varphi=\textrm{div }\Phi + c\varphi\_0$, for some $\Phi\in\mathcal S$, and with $c=\int\varphi$.
Let $a=(p,\varphi\_0)$. We then obtain
$$
(p,\varphi)=(p,\textrm{div }\Phi) + a \int\varphi = -(\nabla p, \Phi) + a\int\varphi .
$$
This shows that $p$ is well defined and continuous on $\mathcal S$, so is a tempered distribution. (Strictly speaking, I was too lazy here to actually show the continuity explicitly and I more relied on the meta-principle that anything that has a natural definition as a functional on $\mathcal S$ is continuous. To do it properly, we'd have to establish that $\Phi$ can be made to depend continuously on $\varphi$ in the topology of $\mathcal S$.)
| 4 | https://mathoverflow.net/users/48839 | 422133 | 171,676 |
https://mathoverflow.net/questions/422144 | 10 | For convex domains $\Omega \subset \mathbb{R}^n$ with Dirichlet boundary conditions, it's known that any first Laplacian eigenfunction is log-concave. In particular, it has a unique maximum.
These are functions $f$ satisfying $-\Delta f = \lambda\_1 f$ for the smallest possible $\lambda\_1 > 0$ and $f|\_{\partial \Omega} = 0$.
**Q:** Is there such a concavity result for homogeneous spaces, ie. Riemannian manifolds with a transitive action of the isometry group? Does $f$ at least have a unique maximum?
I imagine that this might even be apparent from an explicit formula for these eigenfunctions, but I haven't been able to find one in general.
I can at least check that it holds for spherical harmonics. Actually for spheres I think it can be derived from the domain result by separation of variables (take $\Omega = \mathbb{R}^n$). Perhaps such a method can work for any homogeneous space with a nice enough embedding, such as as the boundary of a convex domain?
I'd appreciate any thoughts or relevant references!
| https://mathoverflow.net/users/81773 | Does the first Laplacian eigenfunction on a homogeneous space have a unique maximum? | The flat torus $\mathbb{T} = \mathbb{R}^2/\Lambda$ gives a counterexample: The first nontrivial eigenvalue is of the form $\lambda\_1 = \xi\_1^2+\xi\_2^2$, where $\xi = (\xi\_1,\xi\_2)$ is a nonzero element of the dual lattice $\Lambda^\*$ of smallest norm, and the correspnding eigenfunctions are of the form $f(x\_1,x\_2) = a\cos(\xi\_1 x\_1 + \xi\_2 x\_2 + b)$ for some constants $(a,b)$. This function has a whole circle of maxima.
In general, on a compact manifold, an eigenfunction with nonzero eigenvalue must change sign because its average value on the manifold must be zero. (Integration by parts.)
Oh, another example occurred to me that you might find more interesting: Let $M = \mathrm{SO}(3)$ with its standard biïnvariant Riemannian metric. The first nontrivial eigenvalue has multiplicity 9, and the corresponding eigenfunctions are the 9 entries $a\_{ij}$ of the standard matrix embedding of $\mathrm{SO}(3)$ into the space of 3-by-3 matrices. Each $a\_{ij}$ has its maximum and minimum values equal to $\pm 1$, but it attains each on a circle embedded in $\mathrm{SO}(3)$. Meanwhile $f = -a\_{11}-a\_{22}-a\_{33}$ has a maximum value of $1$, attained on a copy of $\mathbb{RP}^2$ embedded in $\mathrm{SO}(3)$, and a minimum value of $-3$, attained only at $I\_3\in\mathrm{SO}(3)$.
| 12 | https://mathoverflow.net/users/13972 | 422153 | 171,680 |
https://mathoverflow.net/questions/422146 | 4 | I have a question about the discrete spectrum of the Laplace operator on hyperbolic manifolds with infinite volume. I understand the case of infinite area surfaces: see Chapter 7 (Sections 1 and 2) of Borthwick's [book](https://mathscinet.ams.org/mathscinet/search/publdoc.html?pg1=INDI&s1=328585&sort=Newest&vfpref=html&r=6&mx-pid=3497464), where one finds the statement that (roughly speaking) infinite area hyperbolic surfaces with finitely many cusps and funnels have finitely many eigenvalues less than $1/4$, and the continuous spectrum is $[1/4, \infty)$ with no embedded eigenvalues.
My question is, what is the corresponding statement for higher dimensional infinite volume (let's say geometrically finite) hyperbolic manifolds? Borthwick mentions that it is contained in a collection of papers of Lax and Phillips, but I am not at all familiar with this work, and I was hoping for a more precise reference. Thanks!
| https://mathoverflow.net/users/65799 | Spectral theory of infinite volume hyperbolic manifolds | In dimension $n$, there are at most finitely many eigenvalues in $[ 0, (n-1)^2/4 )$ and that the continuous spectrum is $[ (n-1)^2/4 , \infty )$ with no embedded eigenvalues. The following survey article has a good discussion and an extensive bibliography (including the papers of Lax and Phillips that you asked for):
* Perry, Peter: [The spectral geometry of geometrically finite
hyperbolic manifolds](http://dx.doi.org/10.1090/pspum/076.1/2310208). (English summary) Spectral theory and
mathematical physics: a Festschrift in honor of Barry Simon's 60th
birthday, 289–327, Proc. Sympos. Pure Math., 76, Part 1, Amer. Math.
Soc., Providence, RI, 2007. [MR2310208](https://mathscinet.ams.org/mathscinet-getitem?mr=2310208)
| 6 | https://mathoverflow.net/users/121820 | 422155 | 171,682 |
https://mathoverflow.net/questions/422150 | 2 | **The background**: We recall/define the following:
* $\Omega\_n=\{1,\dots,n\}$.
* $M\_n$ is the Mathieu group of degree $n$. We follow the Wikipedia article "Mathieu group" and define these groups for each of the 10 $n$ values 8-12 and 20-24. In each of these two series, $M\_{k-1}$ is the point stabilizer in $M\_k$ of $k\in\Omega\_k$.
* Given an action of a group $G$ on a set $\Omega$ with a subset $\Delta\subseteq\Omega$, ${\mathbf C}\_G(\Delta)$ denotes the pointwise stabilizer of $\Delta$ in $G$ (with set braces omitted when $\Delta$ is a singleton $\{\delta\}$).
* An action (of $G$ on $\Omega$) is *half-transitive* when either (a) $|G|>1$ and every orbit has the same nontrivial cardinality or (b) [the degenerate case] $|G|=|\Omega|=1$.
**The question**: Given the Mathieu group $M\_{20}$ as a permutation group on $\Omega\_{20}$, is the induced action of ${\mathbf C}\_{M\_{20}}(20)$ on $\Omega\_{19}$ half-transitive?
**The motivation**: Just like transitivity and primitivity, half-transitivity has higher-fold gradations $(n+\frac 12)$-transitivity for each integer $n\geq 1$. Indeed, like the way $n$-transitivity and $n$-primitivity interleave, it is true that $(n+\frac 12)$-transitive implies $n$-transitive, which then implies $(n-\frac12)$-transitive. It is also known that an $n$-transitive action is $(n+\frac 12)$-transitive if and only if for each $\Delta\subseteq\Omega$ with $|\Delta|=n$, the induced action of ${\mathbf C}\_G(\Delta)$ on $\Omega\setminus\Delta$ is half-transitive. The parallel statement involving a reduction by one on $n$ holds when points $\delta\in\Omega$ replace the cardinality $n$ subsets $\Delta\subseteq\Omega$.
In particular, these lead to the following statements being equivalent; either they are all true or they are all false.
* The action of $M\_{24}$ on $\Omega\_{24}$ is $\frac{11}2$-transitive.
* The action of $M\_{23}$ on $\Omega\_{23}$ is $\frac 92$-transitive.
* The action of $M\_{22}$ on $\Omega\_{22}$ is $\frac 72$-transitive.
* The action of $M\_{21}$ on $\Omega\_{21}$ is $\frac 52$-transitive.
* The action of $M\_{20}$ on $\Omega\_{20}$ is $\frac 32$-transitive.
* The action of ${\mathbf C}\_{M\_{20}}(20)$ on $\Omega\_{19}$ is $\frac 12$-transitive.
My personal curiosity revolves around the $4$- and $5$-transitive actions interacting with higher-fold half-transitivity at the upper end of this list.
The Wikipedia article goes on to say that $M\_{20}$ has the form $2^4:A\_5$, but I have no idea what action this entails which leaves as unknown the form of the point stabilizer in this group.
As an aside, the corresponding question applied to the lower 5 Mathieu groups $M\_8$ through $M\_{12}$ similarly reduces to whether or not ${\mathbf C}\_{M\_8}(8)$ is half-transitive on $\Omega\_7$. In that case, $M\_8$ acts regularly on $\Omega\_8$ (indeed, the action is the regular action of the quaternion group on itself), guaranteeing ${\mathbf C}\_{M\_8}(8)$ is trivial. As $|\Omega\_7|=7>1$ is not trivial, this action is not half-transitive.
**EDIT** (9 Sep 2022) I previously commented on not knowing what the action of $A\_5$ on $2^4$ was. I now understand it to be $SL\_2(\mathbb F\_4)\cong A\_5$ in its natural action on $\mathbb F\_4{\!}^2$.
| https://mathoverflow.net/users/128140 | Is a point stabilizer in the Mathieu group $M_{20}$ half-transitive? | No it is not: half-transitive means that all orbits have equal size (and the groups acts non-trivially), but this is not the case here. One can verify this e.g. using GAP:
```
gap> M21:=MathieuGroup(21);
Group([ (1,4,5,9,3)(2,8,10,7,6)(12,15,16,20,14)(13,19,21,18,17),
(1,21,5,12,20)(2,16,3,4,17)(6,18,7,19,15)(8,13,9,14,11) ])
gap> M20:=Stabilizer(M21,21);
Group([ (1,5,18,16,17)(2,20,14,7,8)(3,12,11,19,10)(4,15,6,9,13),
(2,18,10)(3,4,5)(6,16,14)(7,15,11)(8,20,17)(12,19,13) ])
gap> M19:=Stabilizer(M20,20);
Group([ (1,18,19)(2,8,7)(3,13,16)(4,12,17)(5,15,9)(6,11,10),
(1,9,8)(2,5,11)(3,13,16)(4,14,12)(6,15,18)(7,10,19), (1,19,14)
(2,10,9)(3,13,16)(4,7,5)(6,17,15)(8,12,11) ])
gap> Orbits(M19);
[ [ 1, 18, 9, 19, 6, 5, 8, 2, 7, 14, 11, 15, 17, 4, 12, 10 ],
[ 3, 13, 16 ] ]
gap> OrbitLengths(M19);
[ 16, 3 ]
```
One can also try to see this geometrically: note that $M\_{21}$ is in fact $PSL\_3(\mathbb{F}\_4)$, which I will denote as $PSL(3,4)$, acting on the projective space $\mathbb{P}\_3(\mathbb{F}\_4)$. This space contains $\frac{4^3-1}{4-1}=21$ projective points; any two points span a unique projective line with 5 points on it; there are 21 such lines, inducing a block system on the 21 points. Stabilizing two points then stabilizes the unique line through them. The stabilizer of this projective line can be thought of as the stabilizer in $SL(3,4)$ of a 2-dimensional subspace, which induces the action of $PSL(2,4)$ on that subspace, which acts 3-transitively on the 5 points. But we did not just fix the line, we also fixed two points on it; so this stabilizer is 1-transitive on the remaining three points. This is exactly the orbit `[3, 13, 16]` in the GAP computation above.
I am too rusty in this stuff, so I don't immediately see a simple argument why it acts transitively on the remaining 21-5 = 16 points (perhaps someone can fill that in). However, that's not even necessary: clearly, it is impossible to divide these remaining points into orbits of size 3. Thus the action can't be half-transitive.
| 4 | https://mathoverflow.net/users/8338 | 422160 | 171,684 |
https://mathoverflow.net/questions/422145 | 4 | Let $k$ be a finite field of characteristic $p$, and $R$ a complete local noetherian algebra with residue field $k$. It is well known that $R$ has a natural structure of an algebra over the ring Witt vectors $W(k)$. Define $n:= dim\_{k}(\mathfrak{m}/(p,\mathfrak{m}^{2}))$ to be the dimension of the $mod-p$ Zariski tangent space. In theorem 2.4 in [A local-to-global principle for deformations of Galois representations](https://typo.iwr.uni-heidelberg.de/fileadmin/groups/arithgeo/templates/data/Gebhard_Boeckle/LocalGlobalRev3.pdf) it is stated that there is a surjection $W(k)[[t\_{1},\cdots,t\_{n}]]\rightarrow R$. I would like to understand this surjection. In the context of this paper, $R$ would be a universal deformation ring that pro-represents the functor of deformations of a pro-finite group acting continuously on a finite-dimensional vector space over $k$.
| https://mathoverflow.net/users/476832 | On presentations of universal rings of deformations | Doesn't this kind of prove itself? Pick some elements $\alpha\_1, \dots, \alpha\_n \in \mathfrak{m}$ which represent $\mathfrak{m} / (p, \mathfrak{m}^2)$. Clearly sending $t\_i$ to $\alpha\_i$ defines a map $W(k)[[[t\_1, \dots, t\_n]] \to R$ and it suffices to show that it is surjective modulo $\mathfrak{m}^r$ for every $r$. This is obvious for $r = 0$, and the assumption on the $\alpha\_i$'s shows precisely that $(p, \alpha\_1, \dots, \alpha\_n)$ generate $\mathfrak{m} / \mathfrak{m}^2$ so they also generate $\mathfrak{m}^r / \mathfrak{m}^{r+1}$ for every $r$, and you are done by induction on $r$.
| 2 | https://mathoverflow.net/users/2481 | 422168 | 171,689 |
https://mathoverflow.net/questions/422171 | 8 | See [Grushko decomposition theorem](https://en.wikipedia.org/wiki/Grushko_theorem#Grushko_decomposition_theorem).
Are the non-free factors of Grushko decomposition of a finitely generated convex–cocompact (but not cocompact) subgroup of $\operatorname{PSL}(2,\mathbb{R})$ finite?
In the cocompact case it is not true, since the group is not a free group and cannot be split into a non-trivial free product.
For convex–cocompact but not cocompact I know of particular examples with affirmative answer. Is it always the case?
| https://mathoverflow.net/users/153319 | Are the non-free factors of Grushko decomposition of a finitely generated convex-cocompact (but not cocompact) subgroup of PSL$(2,\mathbb{R})$ finite? | For a discrete, noncocompact subgroup $\Gamma < \text{PSL}(2,\mathbb R)$, the quotient $\mathbb H^2 / \Gamma$ is a noncompact, 2-dimensional oriented orbifold, i.e. a noncompact surface with an orbifold locus consisting of cone singularities forming a closed, discrete subset.
Assuming in addition the finite type hypothesis, the underlying surface of the quotient orbifold is obtained from some closed oriented surface by removing a finite subset, and the orbifold locus is a finite set.
Such an orbifold has a spine (an orbifold deformation retract, thus having the same fundamental group) which is a finite graph of groups, whose vertices include all of the cone singularities and perhaps some other points with trivial vertex group, and whose edges all have trivial group.
The Grushko decomposition of the graph-of-groups fundamental group therefore has the form $A\_1 \* ... \* A\_K \* F\_n$ for some finite cyclic groups $A\_1,...,A\_K$ and some finite rank free group $F\_n$.
So yes, the non-free factors of the Grushko decomposition are all finite.
| 9 | https://mathoverflow.net/users/20787 | 422175 | 171,692 |
https://mathoverflow.net/questions/422120 | 0 | $\DeclareMathOperator\Syl{Syl}$Let $G$ be a finite group, $\Phi(G)$ is the Frattini subgroup of $G$. And $G/\Phi(G)$ is a simple group. Let $P\in \Syl\_{p}(\Phi(G))$, where $p\in \pi(\Phi(G))$ and $\pi(\Phi(G))$ is the set of prime divisors of $|\Phi(G)|$.
Suppose that every subgroup $H$ of $P$ of order $p$ is normal in $G$.
Let $P\_1\in \Syl\_{p}(G)$. Then for any subgroup $H$ of $P$ with $|H|=p$, $H\leq Z(P\_1)$. So $P\_1\leq C\_G(H)\unlhd G$.
Under the above conditions, can we use the simplicity of $G/\Phi(G)$ to conclude that $H\leq Z(G)$?
| https://mathoverflow.net/users/478670 | The center of Sylow subgroups | Yes we can.
If $P\_1=P$ then, by the Schur-Zassenhaus Theorem,$P$ has a complement in $G$, contradicting $P \le \Phi(G)$.
Otherwise $P\_1\Phi(G)/\Phi(G)$ is a nontrivial Sylow $p$-subgroup of the simple group $G/\Phi(G)$, and its conjugates generate $G/\Phi(G)$, so $C\_G(H)\Phi(G) = G$, and hence $C\_G(H)=G$.
| 3 | https://mathoverflow.net/users/35840 | 422183 | 171,694 |
https://mathoverflow.net/questions/422103 | 1 | Denote by $R = \mathbb{C}\{x\_1, \dots, x\_n\}$ the ring of germs of analytics maps at the origin in $n$ variables and let $f \in R$ such that $Sing(V(f))=V(x\_1, \dots, x\_{n-1})$ as sets. In addition, assume that $V(\partial\_{1}(f), \dots, \partial\_{n-1}(f))=V(x\_1, \dots,x\_{n-1})$ as sets as well. (By "as sets" we mean without considering multiplicity, only as reduced schemes).
We know that $R/\langle \partial\_{1}(f), \dots, \partial\_{n-1}(f) \rangle$ is a Cohen-Macaulay ring, as it is a quotient of a Noetherian local Cohen-Macaulay ring by a regular sequence (see "Pellikaan, R. (1989). Series of isolated singularities" for proof why this sequence is regular.)
But can we conclude that $R/\langle \partial\_{1}(f), \dots, \partial\_{n}(f) \rangle$ is Cohen-Macaulay as well?
| https://mathoverflow.net/users/113200 | Cohen-Macaulyness of Milnor algebra | I am just posting my comment as one answer. Using the local flatness criterion, the ring $R/\langle \partial\_1(f),\dots,\partial\_n(f)\rangle$ is Cohen-Macaulay if and only if it is flat as a module over $\mathbb{C}\{x\_n\}$, i.e., if and only if multiplication by $x\_n$ is injective on the ring. In the answer to a previous question, I wrote an example where multiplication by $x\_n$ is not injective on the ring, i.e., I wrote an example where the image of $x\_n$ is a zero divisor in the ring. Here is the link to the previous question: [Deformation of isolated singularities and non zero divisors](https://mathoverflow.net/questions/421209/deformation-of-isolated-singularities-and-non-zero-divisors)
| 0 | https://mathoverflow.net/users/13265 | 422193 | 171,698 |
https://mathoverflow.net/questions/422003 | 0 | Let $S\_1 = S\_2 = S^d$ be two copies of the $d$-dimensional sphere. Let $p\_i : S\_1 \times S\_2 \to S\_i$ be the projection, $j : U \to S\_1 \times S\_2$ the inclusion of the complement of the diagonal and $q\_i : U \to S\_i$ the restriction of $p\_i$.
I want to know how to get the following formula for a constructible sheaf $\mathcal F \in D^b(Sh(S^d))$ : $${q\_2}\_! q\_1^\*(\mathcal F) \cong {p\_2}\_\* (p\_1^\* \mathcal F \otimes j\_! \mathbb Q\_U[1])$$ It looks like some kind of base change but I don't understand where the $[1]$-shift is coming from.
| https://mathoverflow.net/users/104742 | Fourier transform for constructible sheaves on spheres | We have $q\_1 = p\_1 \circ j$ and $q\_2 = p\_2 \circ j$ so
$$ {q\_2}\_! q\_1^\*(\mathcal F) \cong {p\_2}\_! j\_! j^\* p\_1^\*(\mathcal F) \cong {p\_2}\_! j\_! ( j^\* p\_1^\*(\mathcal F) \otimes \mathbb Q\_U) \ \cong {p\_2}\_! ( p\_1^\*(\mathcal F) \otimes j\_! \mathbb Q\_U) $$ where the last isomorphism is the projection formula (indeed a variant of proper base change).
I do not know where the $[1]$ shift is coming from either. It seems to me that ${p\_2}\_! ( p\_1^\*(\mathcal F) \otimes j\_! \mathbb Q\_U[1])$ is the correct formula for $SR$ so the earlier formula should be $T\_1(\mathcal F) = q\_{2!} (q\_1^\* \mathcal F [1])$.
Also the shift appearing in the formula for $T\_1'$ should be $[d-1]$, with the $[d]$ coming from the shift in $L$ (which I think should be $[d]$ instead of $[-d]$).
With these shifts, I can calculate that that they are quasi-inverses following the suggested approach. So one possible interpretation is that the claims are morally right but the shifts are all wrong. (Or maybe my claims are morally right but the shifts are all wrong.)
| 2 | https://mathoverflow.net/users/18060 | 422205 | 171,703 |
https://mathoverflow.net/questions/422182 | 0 | Let $K\in L^2([0,T]^2)$, and for each $t\in [0,T]$, let $\mathcal{T}\_t $ be such that for all $f\in L^2([0,T])$, $\mathcal{T}\_t f(s)=\int\_0^T K(s,t)K(u,t)f(u)\, d u$ for all $s\in [0,T]$. One can show by the integrability of $K$ that $\mathcal{T}\_t $ is a well-defined bounded linear operator acting on $L^2([0,T])$ into itself (i.e., $\mathcal{T}\_t\in \mathcal{L}(L^2([0,T]))$). Is it possible to prove that the map $[0,T]\ni t\mapsto \mathcal{T}\_t\in \mathcal{L}(L^2([0,T]))$ is [strongly measurable](https://en.wikipedia.org/wiki/Bochner_measurable_function)?
---
I observed that the map $t\mapsto \mathcal{T}\_t $ can be decomposed into
the map $t\in \mathcal{S}\_t\in \mathcal{L}(L^2([0,T]);\mathbb{R})$ such that $\mathcal{S}\_t f=\int\_0^T K(u,t)f(u)\, d u$ for all $f\in L^2([0,T])$, and the map $t\mapsto K(\cdot, t)\in L^2([0,T])$. But I am not sure whether this observation helps the proof.
| https://mathoverflow.net/users/91196 | Strong measurability of operator-valued map induced by a kernel | It helps to note that $\mathcal{T}\_tf(s) = K(s,t)\langle f(\cdot), \overline{K}(\cdot,t)\rangle$.
If $K\_n = \sum a\_i1\_{A\_i\times B\_i}$ is a finite linear combination of characteristic functions of rectangles then the map $\mathcal{T}\_t^n: f \mapsto K\_n(s,t)\langle f(\cdot),\overline{K}\_n(\cdot, t)\rangle$ is measurable in $t$ with finite range. Find a sequence of such functions $(K\_n)$ that converges in $L^2([0,T]^2)$ to $K$. Then for almost every $t$ we have $K\_n(\cdot, t)\to K(\cdot, t)$ in $L^2[0,T]$, and from here an easy estimate shows that $\mathcal{T}\_t^n f \to \mathcal{T}\_tf$ in $L^2[0,T]$ for those values of $t$. So yes, $\mathcal{T}\_t$ is Bochner measurable.
| 2 | https://mathoverflow.net/users/23141 | 422208 | 171,705 |
https://mathoverflow.net/questions/422209 | 4 | Let $X$ be a smooth closed subvariety of a complex abelian variety $A$. Assume $X$ is of general type and of codimension one with $\omega\_X$ ample.
Often, people speak about the stabilizer $\mathrm{Stab}\_A(X)$ of $X$ in $A$. This is the group of $a$ in $A$ such that $X+a = X$.
>
>
> >
> > What is the relation of $\mathrm{Stab}\_A(X)$ to $\mathrm{Aut}(X)$?
> >
> >
> >
>
>
>
They are both finite. Are they equal? If the stabilizer is trivial, does that imply $\mathrm{Aut}(X)$ is trivial? What about vice versa? Does $\mathrm{Stab}\_a(X)$ inject into $\mathrm{Aut}(X)$?
Crossposted from stackexchange, because I didn't get any replies there unfortunately: <https://math.stackexchange.com/questions/4446811/difference-between-stabilizer-and-automorphism-group>
| https://mathoverflow.net/users/200661 | Difference between stabilizer and automorphism group of subvariety of an abelian variety | They have absolutely no reason to be equal. Consider the case where $A$ is the Jacobian of a genus 2 curve $C$, and $X=C$ embedded in $A$ by $x\mapsto [x]-[p]$ for some fixed point $p\in C$. Then $X$ is a Theta divisor, so $\operatorname{Stab}\_A(X) $ is trivial. But $X$ has always a nontrivial automorphism, the hyperelliptic involution, and it may have more in some cases.
Of course $\operatorname{Stab}\_A(X) $ injects into $\operatorname{Aut}(X) $, since a nontrivial translation does not fix any point of $A$. But that is all you can say.
| 7 | https://mathoverflow.net/users/40297 | 422210 | 171,706 |
https://mathoverflow.net/questions/419465 | 7 | It is well-known that the category of commutative and cocommutative Hopf algebras is abelian (see <https://arxiv.org/abs/1502.04001v2> and its references). But does it have enough injectives? What about projectives?
| https://mathoverflow.net/users/36720 | Does the category of commutative and cocommutative Hopf algebras have enough injectives? | Over a field $k$, the answer is yes for injectives; I'm not sure about projectives. Over $\mathbb Z$ or other commutative rings, I really don't know -- the use of the fundamental theorem of coalgebra below seems pretty essential (and the fundamental theorem of coalgebra fails over $\mathbb Z$).
Over a field, in fact more is true:
**Claim:** Let $k$ be a field. Then the following categories are [locally finitely presentable](https://ncatlab.org/nlab/show/locally+presentable+category):
1. The category of coalgebras over $k$;
2. The category of cocommutative coalgebras over $k$;
3. The category of algebra objects in either (1) or (2);
4. The category of commutative algebra objects in (1) or (2);
5. As in (3) or (4), but with restricting to objects with antipodes.
Moreover, the natural tensor product on each of these categories is a symmetric monoidal structure preserving filtered colimits, with compact unit.
**Corollary:** The category of commutative, cocommutative Hopf algebras over a field $k$ is a Grothendieck abelian category (and in particular has enough injectives).
**Proof:** Every locally finitely presentable abelian category is Grothendieck.
**Proof of Claim:**
The conclusion for (1) and (2) follows from the [fundamental theorem of coalgebra](https://planetmath.org/fundamentaltheoremofcoalgebras). Then (3) and (4) follow: in general if you have a locally finitely-presentable category with a monoidal structure with compact unit and which preserves filtered colimits, its category of monoids will be locally finitely-presentable, and similarly for commutative monoids. Finally, (5) follows because Hopf algebras are closed among bialgebras under filtered colimits (when the monoidal product has compact unit and preserves filtered colimits).
| 4 | https://mathoverflow.net/users/2362 | 422248 | 171,717 |
https://mathoverflow.net/questions/422239 | 1 | Let $M$ be a smooth (embedded or immersed) surface in $\mathbb{R}^3$.
Let $Z\_1,Z\_2$ be two vector fields along $M$, thought of as $\mathbb{R}^3$-valued functions, satisfying the following differential equation:
$$\langle XZ\_i,Y \rangle + \langle X,YZ\_i \rangle =0$$
for all vector fields $X,Y$ on $M$. Here $XZ$ means a derivative with respect to the vector field $X$ of a $\mathbb{R}^3$-valued function $Z$. We will call such $Z\_1,Z\_2$ *infinitesimal bendings*, since the derivative of a one-parameter family of isometric immersions must satisfy this equation.
**Is it true, that the pointwise cross product $M \ni m\mapsto Z\_1(m)\times Z\_2(m)$ is also an infinitesimal bending?**
I don't have any particular reason to believe that it might be true, i couldn't prove it by algebraic manipulations, and not trained to produce non-flat surfaces admitting independent infinitesimal deformations to find a counterexample.
| https://mathoverflow.net/users/13842 | Cross product of two infinitesimal bendings | Let $M$ be a rotationally symmetric cylinder, $Z\_1$ be a vector field tangent to M generating the isometric translation along the cylinder and $Z\_2$ be a vector field also tangent to M generating the isometric rotation around the cylinder. Then $Z\_1\times Z\_2$ is a vector field which radially expanding or shrinking the cylinder, which is probably a counter example.
| 1 | https://mathoverflow.net/users/475031 | 422251 | 171,718 |
https://mathoverflow.net/questions/422075 | 10 | Let $t\_1,t\_2,\dots,t\_k$ be non-negative integers. Can the following sum
$$f\_k(t\_1,t\_2,\dots,t\_k):=\sum\_{j\_1=0}^{t\_1} \sum\_{j\_2=0}^{t\_2+j\_1} \sum\_{j\_3=0}^{t\_2+j\_2} \dots \sum\_{j\_k=0}^{t\_k+j\_{k-1}} 1$$
be explicitly expressed as a polynomial in $t\_1,t\_2,\dots,t\_k$ or via known combinatorial entities?
We surely have a recurrence formula:
$$f\_{k+1}(t\_1,t\_2,\dots,t\_{k+1}) = \sum\_{j=0}^{t\_1} f\_k(j+t\_2,\dots,t\_{k+1}),$$
which does not seem to easily unroll.
Just in case, first few terms are
\begin{split}
f\_0 &= 1,\\
f\_1(t\_1) &= 1+t\_1,\\
f\_2(t\_1,t\_2) &= (1+t\_1)(1+t\_2) + \frac{t\_1(1+t\_1)}2,\\
f\_3(t\_1,t\_2,t\_3) &= \left[ (1+t\_1)(1+t\_2) + \frac{t\_1(1+t\_1)}2 \right](1+t\_3) + \frac{t\_2(1+t\_2)}2 + \frac{3t\_2^2 + 6t\_2 + 2}6t\_1 + \frac{1+t\_2}2t\_1^2 + \frac16t\_1^3.
\end{split}
---
**UPDATED.** Billy Joe found that
$$f\_k(n,d,d,\dots, d) = \frac{n+1}k \binom{n+k(d+1)}{k-1}.$$
In particular, at $f\_k(1,1,\dots,1)$ gives $(k+1)$-st Catalan number.
| https://mathoverflow.net/users/7076 | Explicit expression for recursive sums | Claim: The iterated sum $f\_k(t\_1,\ldots,t\_k)$ counts the number of elements the interval $[\emptyset,\lambda]$ of Young's lattice, where $\lambda = (\lambda\_1,\lambda\_2,\ldots,\lambda\_k)$ is the partition determined by $\lambda\_{k-i+1} = t\_1 + \cdots + t\_i$. Equivalently, the function $f\_k$ counts the number of subdiagrams of $\lambda$.
For an arbitrary partition $\lambda$, we have
$$|[\emptyset,\lambda]| = \text{det} \left[\binom{\lambda\_i + 1}{i-j+1}\right]\_{1 \leq i,j \leq k}$$
which is a result due to P. A. MacMahon. The answer to Exercise 149 in Chapter 3 of [Stanley's Enumerative Combinatorics, volume 1, 2nd edition](https://math.mit.edu/%7Erstan/ec/ec1/) provides a good reference of references for this result, with various extensions and specializations, including some of the results mentioned in the comments. For a short visual proof using Lindström-Gessel-Viennot, see [Ciucu -
A short conceptual proof of Narayana's path-counting formula](https://arxiv.org/abs/1602.02085).
If the claim is true, MacMahon's result implies
$$\sum\_{j\_1=0}^{t\_1}\sum\_{j\_2=0}^{t\_2+j\_1}\cdots\sum\_{j\_k=0}^{t\_k+j\_{k-1}} = \text{det} \left[\binom{t\_1 + \cdots + t\_{k - i + 1} + 1}{i-j+1}\right]\_{1 \leq i,j \leq k}$$
which implies $f\_k(t\_1,\ldots,t\_k)$ is a polynomial in $t\_1,\ldots,t\_k$.
Note that $f\_k(t\_1,\ldots,t\_k)$ counts the number of $(j\_1,\ldots,j\_k)$ such that $0 \leq j\_1 \leq t\_1$ and $0 \leq j\_{i+1} \leq j\_i + t\_{i+1}$ for $i \geq 1$. To establish the claim, it suffices to find a bijection between the set of $\mu \subseteq \lambda$ and the set of tuples satisfying the above constraints.
Sketch: Map $\mu \subseteq \lambda$ to $(j\_1,\ldots,j\_k)$, where $j\_i = \lambda\_{k-i+1} - \mu\_{k-i+1}$. The visual interpretation is that each $j\_i$ measures the distance between the walls of the $i$-th row from the bottom of the Young diagrams (English convention) for $\mu$ and $\lambda$. The $t\_i$ specify how many boxes are added to the diagram for $\lambda$ in moving from the $(i-1)$-st row from the bottom to the $i$-th row. The constraints express the fact that in going from bottom to top in the diagram, the distance between walls increases by at most $t\_i$. For a more direct definition chase, note that $\lambda\_{k-i} - \lambda\_{k-i+1} = t\_{i+1}$. Since $\mu$ is a partition, we have $\mu\_{k-i+1} - \mu\_{k-i} \leq 0$. Combining the definitions and inequalities gives $j\_{i+1} \leq j\_i + t\_{i+1}$.
| 11 | https://mathoverflow.net/users/22379 | 422252 | 171,719 |
https://mathoverflow.net/questions/422228 | 2 | In [this](https://www.worldscientific.com/doi/10.1142/S0219498820501017) paper, the conception of the difference sequence and $\infty$-difference length of a subset of groups is introduced. As an important case, subsets of the additive group of integers are considered as follows:
Let $0\in A\subseteq \mathbb{Z}$ and put $A\_0:=A$, $A\_n:=A\_{n-1}-A\_{n-1}$, for all $n\in \mathbb{Z}^+$. Note that $A-A=\{a\_1-a\_2: a\_1,a\_2\in A\}$, $A\subseteq A\_1\subseteq A\_2\subseteq
\cdots \subseteq A\_n\subseteq \cdots \subseteq \bigcup\_{n=1}^\infty A\_n$, and $\bigcup\_{n=1}^\infty A\_n$ is a subgroup of $(\mathbb{Z},+)$.
Now, there are two special questions:
**(a)** if $A=\{2^m:m\in \mathbb{Z}^+\}\cup\{0\}$, then is there any positive integer $N$ such that $A\_N=A\_{N+1}$? (if yes, what is the least such $N$?)
**(b)** what about $A=\{m^k:m\in \mathbb{Z}\}$, where $k\geq 3$ is a fixed integer?
**Remark.**
We know that if the following properties hold, then the answer of **(a)** is negative:
(1) for every even integer $n$, there exists an integer $k=2^q$ and $a\_1,\cdots,a\_k\in A$ such that
$$
n=a\_1+\cdots+a\_{\frac{k}{2}}-(a\_{\frac{k}{2}+1}+\cdots+a\_k);
$$
(2) denoting by $k(n)$ the least $k$ obtained from (1), the set of all $k(n)$
where $n$ runs over all even integers, is unbounded above.
| https://mathoverflow.net/users/40520 | Difference sequences of sets of integers | Sketch of a proof for **(a)**: $A\_0$ has $m$ elements smaller than or equal to $2^m$. You can form $m^2$ pairs of them, so $A\_1$ has (at most) $m^2$ elements with an absolute value smaller than or equal to $2^m$ (the larger elements don't play a role, which can be seen by looking at the binary expansion). That means $A\_n$ has at most $m^{2^n}$ elements with an absolute value smaller than or equal to $2^m$.
Every positive even number $s$ can be written (uniquely) as the sum of powers of 2 (again, look at the binary expansion): $s = \sum\_{i=1}^{t}{2^{s\_i}}$ with $t$ and each $s\_i$ positive integers. Now $-(2^m) \in A\_1$ for every positive integer $m$, and we can write $s = 2^{s\_t} - (-(2^s\_{t-1})) - (-(2^s\_{t-2})) ... - (-(2^s\_1))$ so $s \in A\_t$. A similar sum/difference works for negative even numbers. So if an $N$ would exist as described in **(a)**, $A\_N$ should contain *all* even numbers.
Now take $m = N^N$, then $A\_N$ has $N^{N2^N}=2^{N \log\_2 N 2^N}$ elements with an absolute value smaller than or equal to $2^{N^N}$, while there are $2^{N^N}+1$ even numbers, which is (much) more.
Conclusion: such an $N$ does not exist.
| 2 | https://mathoverflow.net/users/70594 | 422267 | 171,725 |
https://mathoverflow.net/questions/422260 | 10 | Let $X$ be an algebraic variety defined over a field $k$ of characteristic zero. Suppose $X(K)$ is non-empty for some extension $K/k$. Is it true that $X(L)$ is non-empty where $L$ is the algebraic closure of $k$ in $K$? An equivalent formulation is: suppose $X(k)$ is empty; does that force $X(K)$ to be empty for every purely transcendental extension $K/k$?
In my particular application $k$ is a number field with a real place and $K=\mathbb{R}$ and then claim is true since the theory of real closed fields eliminates quantifiers, but I'm wondering about the general case. Relatedly Chevalley's Theorem (elimination of quantifiers for algebraically closed fields) does produce $\bar{k}$-points, but I don't see why they have to be $\bar{k}\cap K$-points.
| https://mathoverflow.net/users/327 | Can a variety acquire points in a purely transcendental extension? | The following answers your second question.
>
>
> >
> > Let $X$ be a finite type separated scheme over an infinite field $k$.
> > Let $K/k$ be a purely transcendental extension.
> > If $X(k)$ is empty, then $X(K)$ is empty.
> >
> >
> >
>
>
>
*Proof.* We may assume that $K$ has finite transcendence degree over $k$. Then, any element of $X(K)$ extends to a morphism $U\to X$, where $U$ is a dense open of some $\mathbb{P}^n$. Since $k$ is infinite, the set $U(k)$ is dense (hence non-empty). This contradicts $X(k)$ being empty.
Your first question is answered by Laurent Moret-Bailly.
>
>
> >
> > Let $X$ be a geometrically connected finite type scheme over $k$ (and thus non-empty). Let $K=K(X)$ be the function field of $X$. Then $X(K)$ is non-empty. However, there is no reason for $X(k)$ to be non-empty. Take $X$ to be a smooth proper conic over $\mathbb{Q}$ with no $\mathbb{Q}$-points, for example.
> >
> >
> >
>
>
>
| 11 | https://mathoverflow.net/users/4333 | 422276 | 171,726 |
https://mathoverflow.net/questions/422279 | 27 | I was looking at a [bio-movie of Ramanujan](https://www.imdb.com/title/tt0787524/) last night. Very poignant.
Also impressed by Jeremy Irons' portrayal of G.H. Hardy.
In [G.H. Hardy's wiki page](https://en.wikipedia.org/wiki/G._H._Hardy), we read:
*. . . "Hardy cited as his most important influence his independent study of Cours d'analyse de l'École Polytechnique by the French mathematician Camille Jordan, through which he became acquainted with the more precise mathematics tradition in continental Europe."*
and
*. . . "Hardy is credited with reforming British mathematics by bringing rigour into it, which was previously a characteristic of French, Swiss and German mathematics. British mathematicians had remained largely in the tradition of applied mathematics, in thrall to the reputation of Isaac Newton (see Cambridge Mathematical Tripos). Hardy was more in tune with the cours d'analyse methods dominant in France, and aggressively promoted his conception of pure mathematics, in particular against the hydrodynamics that was an important part of Cambridge mathematics."*
Are we to understand from this that up to the late 1800s, British mathematics used only partial or inductive proofs or what ?
On the face of it, this would have been quite a state of affairs.
What exactly - in general or by a specific example - did Hardy bring to mathematics by way of rigour that had previously been absent ?
If someone introduced a new and sketchily proven theorem in the days of Hardy's childhood - and we are talking about Victorian times here (...) - then surely all the mean old men of the profession would have been disapproving of it and would obstruct its publication ?
| https://mathoverflow.net/users/216345 | Unrigorous British mathematics prior to G.H. Hardy | [Rigor and Clarity: Foundations of Mathematics in France and England, 1800-1840](https://doi.org/10.1017/S0269889700000983) explains in some detail how British mathematicians in the early 19th century viewed the role of rigor in the formulation and proof of mathematical theorems.
>
> Rigor is now accepted as a universal good in mathematics. The
> differences between the French and the English at the turn of the
> century indicate that this was not always the case. [...] For Cauchy
> mathematical rigor was achieved when mathematical terms were defined
> unambiguously, so that they could be confidently used in subsequent
> proofs. The English did not agree that the essence of mathematics was
> captured in the abstract notion of rigor advocated by Cauchy and his
> school.
>
>
> For the nineteenth-century English, mathematical theorems, no matter
> how beautifully proved, did not stand alone. Their validity lay in the
> concepts they illuminated; these concepts existed independently of the
> systems describing them. In this view mathematics was not created, it
> was discovered, and the value of the discovery lay in the
> understandings it generated rather than in the mathematical structure
> itself.
>
>
> The English constructed for the subject a conceptual foundation that
> they found both strong and appropriate. Rigor as Cauchy and his
> followers understood it failed to capture the true spirit of
> legitimate mathematical development. The English would have agreed
> with the French that mathematics must be exact, but for them
> exactitude concerned the fit of mathematical definition to underlying
> concept, rather than precision in use. This way of seeing the issue
> supported an English style, just as Cauchy's notions of rigor came to
> support a French style, throughout the century.
>
>
>
| 32 | https://mathoverflow.net/users/11260 | 422280 | 171,727 |
https://mathoverflow.net/questions/422218 | 2 | We say a rational function $F(z)$ is positive if the coefficients of its Maclaurin expansion, in the variable $z$, are non-negative.
In this context, let
$$F\_r(z):=\frac{1 - 2z + z^r - (1 - z)^r}{(1 - z)^{r - 1}(1 - 2z)}.$$
Is the following true? Note: $F\_2(z)=0$ and $F\_3(z)$ is easier to manage.
>
> **QUESTION.** For $r\geq4$, each of the rational functions $F\_r(z)$ is positive.
>
>
>
**Example.** After simplifications, $F\_4(z)=\frac{2z}{(1-z)^2}$.
| https://mathoverflow.net/users/66131 | Prove positivity of rational functions | Notice that
$$F\_r(z) = \frac{1}{(1-z)^{r-1}} - \sum\_{k=0}^{r-1} \left(\frac{z}{1-z}\right)^k$$
and therefore for $r\geq 4$ and $n\geq 1$, we have
\begin{split}
[x^n]\ F\_r(z) &= \binom{n+r-2}{r-2} - \sum\_{k=1}^{r-1} \binom{n-1}{k-1} \\
& = \binom{n+r-2}{r-2} - \binom{n-1}{r-2} - \binom{n-1}{r-3} - \sum\_{k=1}^{r-3} \binom{n-1}{n-k} \\
&\geq \binom{n+r-2}{r-2} - \binom{n}{r-2} - \sum\_{k=1}^{r-3} \binom{n-1+r-3-k}{n-k} \\
&= \binom{n+r-2}{r-2} - \binom{n}{r-2} - \binom{n+r-4}{r-3} \\
&\geq \binom{n+r-2}{r-2} - \binom{n+r-4}{r-2} - \binom{n+r-4}{r-3} \\
&= \binom{n+r-3}{r-3}\\
&> 0.
\end{split}
---
**ADDED.** The above bound implies a stronger statement: for $r\geq 2$ the function
$$F\_r(z) + 1 - \frac{1}{(1-z)^{r-2}}$$
is non-negative.
| 9 | https://mathoverflow.net/users/7076 | 422282 | 171,729 |
https://mathoverflow.net/questions/422269 | 0 | Given a closed Riemannian manifold $(X,g)$ and let $p\colon TX\to X$ be the usual projection, the paper I'm reading asserts that the Levi-Civita connection induces a splitting $T(TX)= H(TX)\oplus V(TX)\cong p^\* TX\oplus p^\*TX$, so that
$$
\begin{aligned}\Omega^\bullet(TX) &= \{ \Gamma(H(TX))\to C^\infty(TX)\}\otimes\_{C^\infty(TX)} \{\Gamma(V(TX))\to C^\infty(TX) \} \\
&\cong p^\*\Omega^\bullet (X)\otimes\_{C^\infty(TX)} p^\*\Omega^\bullet(X).
\end{aligned}
$$
I want to investigate this isomorphism explicitly, in the sense that I can compute the images of the local coordinate functions on $X$ under the two maps
$$ \phi\_1\colon C^\infty(X)\xrightarrow{d}\Omega^1(X)\xrightarrow{p^\*} p^\*\Omega^\bullet(X) \cong \{\Gamma(V(TX))\to C^\infty(TX) \} \hookrightarrow \Omega^\bullet(TX),$$
$$\phi\_2\colon C^\infty(X)\xrightarrow{d}\Omega^1(X)\xrightarrow{p^\*} p^\*\Omega^\bullet(X) \cong \{\Gamma(H(TX))\to C^\infty(TX) \} \hookrightarrow \Omega^\bullet(TX). $$
But I failed finding any material that could tell me how the horizontal vector bundle is splitted out using the Levi-Civita connection, without which I cannot do any computation.
Also, I found that $V(TX) = \ker ( d p\colon T(TX)\to TX)$ on Wikipedia, but I'm not quite sure how the isomorphism $V(TX)\cong p^\*TX$ is given. The isomorphism that I came up with is the following:
>
> Let $x\_1,\cdots,x\_n$ be the coordinate functions of a coordinate
> neighborhood $U$ of $X$, then $x\_1,\cdots,x\_n,y\_1:=d x\_1,\cdots, y\_n:=dx\_n$
> are the coordinates of $p^{-1}(U)\cong U\times\mathbb{R}^n$
> of $TX$. Then the fibres of $V(TX)|\_{p^{-1}(U)}$ are exactly the spans of $\frac{\partial}{\partial y\_i}$'s. Since the fibres of $p^\*TX|\_{p^{-1}(U)}$ are the spans of $\frac{\partial}{\partial x\_i}$'s, the isomorphism is given by $\frac{\partial}{\partial x\_i}\mapsto \frac{\partial}{\partial y\_i} $ along each fibre.
>
>
>
So my questions are:
1. How the horizontal vector bundle is constructed using the Levi-Civita connection?
2. Is the above isomorphism the correct one?
3. What are the images of the local coordinate functions on $X$ under the maps $\phi\_1$ and $\phi\_2$ given above? (Probably I will be able to do this by myself with (1) and (2) solved, but I'm putting it here anyway, since it is the ultimate goal of mine)
To be honest, I'm not familiar to Riemannian geometry, so it would be the best if there exists a self-contained material that could provide a throughout treatment to this matter.
Thanks in advance for any help.
---
EDIT: I thought it over and a new question (4) arises: See, a choice of coordinates $x\_1,\cdots,x\_n$ of a neighborhood $U$ of $X$ induces coordinates $x\_1,\cdots,x\_n,y\_1:=d x\_1,\cdots, y\_n:=dx\_n$ of $p^{-1}(U)$ of $TX$, which gives a basis $\frac{\partial }{\partial x\_1},\cdots,\frac{\partial }{\partial x\_n},\frac{\partial }{\partial y\_1},\cdots,\frac{\partial }{\partial y\_n}$ of $T(TX)|\_{p^{-1}(U)}$. While the span of $\frac{\partial }{\partial y\_i}$’s gives $V(TX)| \_{p^{-1}(U)} $, why isn’t $H(TX)|\_{p^{-1}(U)}$ simply the span of $\frac{\partial}{\partial x\_i}$’s? I don’t see any trouble gluing these choices of $H(TX)|\_{p^{-1}(U)}$‘s up as $U$ varies, which seems to have no difference with $p^\*TX$. Am I missing something in this construction, or is there any other reason to define $H(TX)$ differently? This approach looks canonical — not even a Riemannian metric is required.
| https://mathoverflow.net/users/167862 | Explicit computation of the vertical and horizontal vector bundles | I am just answering to (1) and (2): The horizontal bundle is defined as follows. Consider a point $v\in T\_pX\subset TX,$ and $w\in T\_pm$. Consider a curve $\gamma$ through $p$ with $\gamma'(0)=w.$ Consider the parallel transport $V$ through $v$ along $\gamma,$ i.e. the curve $V\colon (-\epsilon, \epsilon)\to TX$ satisfying $$\nabla\_\gamma' V=0\quad\text{and}\quad V(0)=v.$$ Then, $V'(0)\in T\_vTX$ is per Definition horizontal, i.e.
$$V'(0)\in H\_v(TX).$$ Of course, one has to show that this gives a well-defined bundle, which is isomorphic to $p^\*TX.$ The later isomorphism is given by $$w\mapsto V'(0),$$ with inverse $$V'(0)\mapsto d\_vp(V'(0)).$$
(2) seems correct. There is the following invariant way describing the isomorphism: Take $v\in T\_pX\subset TX,$ and $w\in T\_pX.$ Then, $$t\in(-\epsilon,\epsilon)\mapsto v+t w\in T\_pX$$ is a vertical curve $\gamma\_w$. The isomorphism then is $$w\in T\_{p(v)} X\mapsto \gamma\_w'(0)\in V\_vV.$$
| 2 | https://mathoverflow.net/users/4572 | 422284 | 171,730 |
https://mathoverflow.net/questions/422285 | 3 | Let $A=k\mathbb \Pi$ be the group algebra of an abelian group $\Pi$ and let $B(A)=\bigoplus\_{k=0}^\infty\,B^k(A)$ be the unnormalized bar complex of $A$ with generators $[a\_0,\dots,a\_k] \in B^k(A)=A^{\otimes (k+1)}$. Geometrically, we can think of these generators as labelled $k$-simplices. That is, vertices are labelled by $a\_i$'s, edges are labelled by pairs $(a\_i,a\_j)$ for $i<j$ and so on.
On $B(A)$ we have a shuffle product $\ast\_{p,q}:B^p(A)\otimes B^q(A) \to B^{p+q}(A)$.
My question is, do you have a reference for a geometric interpretation of the shuffle $[a\_0,\dots,a\_p]\ast[b\_0,\dots,b\_q]$ as a simplicial complex? For simplicity, we assume that there are no relations between the $a\_i$'s and $b\_j$'s.
| https://mathoverflow.net/users/58211 | Geometric interpretation of shuffle product | Perhaps this is more naïve than you are looking for, but here is one interpretation: one representation of an $n$ simplex is the following
$$\Delta\_n=\{(t\_1,t\_2,\dotsc,t\_n)\mid 0\leq t\_1\leq t\_2\leq\dotsb\leq t\_n\leq 1\}.$$
The product of two such simplices has a canonical decomposition into a disjoint union of subsimplices
$$\Delta\_m\times\Delta\_n=\bigsqcup \Delta\_{m+n,\sigma}$$
where, if we use $s\_i$ ($t\_i$) for coordinates on $\Delta\_m$ ($\Delta\_n$), $\sigma$ represents a total order on the union $\{s\_1,\ldots,s\_m,t\_1,\ldots,t\_n\}$ compatible with the internal orders of each set of coordinates. In particular, $\sigma$ is a shuffle. So every term in a shuffle product corresponds to a subsimplex in this decomposition
Then, thinking of the elements of $A$ as functions on $[0,1]$ and of $[a\_1,\dotsc,a\_m]$ as an integral over $\Delta\_m$, and similarly for the $b$ term, the product $[a\_1,\dotsc,a\_m]\star [b\_1,\dotsc, b\_n]$ represents an integral on the product of simplices, which is equal to the sum of the integrals over the subsimplices.
A good example here is Chen's iterated integrals, and the iterated integral representation of multiple zeta values, where this is used to show that the product of two iterated integrals is equal to a sum over shuffles.
| 6 | https://mathoverflow.net/users/161009 | 422287 | 171,731 |
https://mathoverflow.net/questions/422291 | 7 | The following fact (slightly reworded here) is proven in [this answer](https://mathoverflow.net/questions/422260/can-a-variety-acquire-points-in-a-purely-transcendental-extension/422276#422276):
>
> If $K$ is a purely transcendental extension of an infinite¹ field $k$, then whenever a separated scheme $X$ of finite type over $k$ has points over $K$, it already has points over $k$ (viꝫ., $X(k) = \varnothing$ implies $X(K) = \varnothing$).
>
>
>
Let us say that the field extension $K$ of $k$ is “**astigmagenic**”² when the conclusion of this statement (the part that follows the “then”) holds. Thus, the above states that purely transcendental extensions of infinite fields are astigmagenic. But they are not the only ones: if $k$ is algebraically closed, then any extension of $k$ is astigmagenic (and it need not be purely transcendental: consider the field of functions of an algebraic variety that isn't rational).
**Question:** Can we characterize these “astigmagenic” field extensions in a simpler way? (E.g., do the finite type ones coincide with rational function fields of algebraic varieties for which the rational points are Zariski-dense?)
**Note:** I used “separated scheme of finite type” in the definition because this is what the linked answer does. But we can also try variations around this: e.g., maybe say that $K$ is **weakly astigmagenic** when $X(k) = \varnothing$ implies $X(K) = \varnothing$ for $X$ a *geometrically integral* separated scheme of finite type. It is very unclear to me how much this changes the condition, so I'm interested in the relations between the various variations: feel free to answer with whatever variation seems to make the question most natural or interesting.
1. As pointed out by Laurent Moret-Bailly in a comment, the word “infinite” was missing from the answer.
2. Pardon my Greek. This is supposed to mean “that does not create points”.
| https://mathoverflow.net/users/17064 | Field extensions over which algebraic varieties cannot acquire points | Let $K$ be a finite type field extension of $k$ which corresponds to a rational function field of an algebraic variety $V$ for which the rational points are *not* Zariski dense.
Let $U$ be the complement in $V$ of the Zariski closure of the rational points.
Then $U$ has no $k$-point, but has a $K$-point (the generic point), so $K/k$ is not astigmagenic.
Since every finite type extension is the rational function field of a variety, it follows that a finite type extension is astigmagenic if and only if it is the rational function field of an algebraic variety for which the rational points are Zariski-dense. (Well, that proves "only if", and "if" follows from the same proof as for purely transcendental - given a $K$-point, we obtain a rational map from the variety, hence from an open subset of the variety, and then look at the image of a rational point).
| 10 | https://mathoverflow.net/users/18060 | 422293 | 171,733 |
https://mathoverflow.net/questions/422278 | 2 | Let $T\_n$ be the set of all labelled trees with $n$ vertices. For any $T \in T\_n$ let $D(T)$ be the 'doubled tree', where each edge of $T$ is replaced by one directed edge in each direction. $D(T)$ is now an Eulerian directed graph with $2(n-1)$ edges and by the B.E.S.T theorem it has $\prod\limits\_{i=1}^n (\deg\_{D(T)}(v\_i)-1)!$ many Euler circuits.
Let $A\_n \subset T\_n$ be the set of all labelled trees with $n$ vertices, with the property that vertices $v\_1$ and $v\_2$ have a distance of $2$, i.e. there is no connection between $v\_1$ and $v\_2$, but there is a $t \in \{3,...,n\}$ such that there are edges $(v\_1,v\_t)$ and $(v\_2,v\_t)$.
**Question:** Is there a simpler formula for the expression
$$
\sum\limits\_{T \in A\_n} \prod\limits\_{i=1}^n (\deg\_{D(T)}(v\_i)-1)! \ ,
$$
which counts the number of Euler circuits through trees from $A\_n$.
If the distance is prescribed as $1$ instead of $2$, then by using the fact that there are ${n-2 \choose d\_1-1,...,d\_{n}-1}$ many $T \in T\_n$ with $d\_i = \deg\_{T}(v\_i) = \deg\_{D(T)}(v\_i)$ the formula is found to be $2 \frac{(2n-3)!}{n!}$.
Another way of formulating the question would be: For given $d\_1,...,d\_{n} \in \{1,...,n-1\}^n$ with $d\_1+...+d\_n = 2n-2$, how many of the ${n-2 \choose d\_1-1,...,d\_{n}-1}$ many $T \in T\_n$ with $d\_i = \deg\_{T}(v\_i) = \deg\_{D(T)}(v\_i)$ satisfy the condition that $v\_1$ and $v\_2$ have distance two?
Any help is much apprechiated.
| https://mathoverflow.net/users/409412 | Counting Euler circuits through labelled trees where $v_1$ and $v_2$ have distance two | Wlog we can use $v\_{n-2}$ and $v\_{n-1}$ instead of $v\_1$ and $v\_2$. Then if we let $B\_n \subset T\_n$ be the set of labelled trees with edges $(v\_{n-2}, v\_n)$ and $(v\_{n-1}, v\_n)$, the count for $A\_n$ is just $(n-2)$ times the count for $B\_n$ (there are $n-2$ choices for $v\_t$; swap $v\_t$ for $v\_n$ if they're different).
The advantage to numbering things this way is that when constructing the Prüfer codes for the trees in $B\_n$ the three labelled vertices are the last survivors. The final label in the Prüfer code will be $n$, and the penultimate one (if $n > 3$) will be in $\{n-2, n-1, n\}$. This appears empirically also be to sufficient, although a proof eludes me at present.
If so then the number of trees with given vertex degrees $d\_i$ (where each $d\_i \ge 1$ and $d\_n \ge 2$) having subtree $v\_{n-1} - v\_n - v\_{n-2}$ is a sum of three multinomial coefficients which simplifies to $$\binom{n-2}{d\_1-1, \ldots, d\_n-1} \frac{(d\_n - 1)(d\_{n-2} + d\_{n-1} + d\_n - 4)}{(n-3)(n-2)}$$
Then taking the alternative formulation
>
> For given $d\_1,...,d\_{n} \in \{1,...,n-1\}^n$ with $d\_1+...+d\_n = 2n-2$, how many of the ${n-2 \choose d\_1-1,...,d\_{n}-1}$ many $T \in T\_n$ with $d\_i = \deg\_{T}(v\_i) = \deg\_{D(T)}(v\_i)$ satisfy the condition that $v\_1$ and $v\_2$ have distance two?
>
>
>
the answer is given by relabelling to put the special vertices at $v\_1$, $v\_2$ and summing over all possible intermediate vertices. $$\sum\_{i=3}^n [d\_i \ge 2] \binom{n-2}{d\_1-1, \ldots, d\_n-1} \frac{(d\_i - 1)(d\_1 + d\_2 + d\_i - 4)}{(n-3)(n-2)}$$ simplifies to $$\frac{1}{(n-3)(n-2)} \binom{n-2}{d\_1-1, \ldots, d\_n-1} \left((n - d\_1 - d\_2)(d\_1 + d\_2 - 5) - (n-2) + \sum\_{i=3}^n d\_i^2 \right)$$
| 1 | https://mathoverflow.net/users/46140 | 422298 | 171,734 |
https://mathoverflow.net/questions/422262 | 2 | This is a repost from the [computer science stackexchange](https://cs.stackexchange.com/questions/151172/verify-if-array-is-orthogonal). The question has been offered a bounty, but received no answers. Therefore, I would like to ask this question here.
[Orthogonal arrays](https://en.wikipedia.org/wiki/Orthogonal_array) often appear in probabilistic algorithms. They can be efficiently constructed from, e.g., [BCH codes](https://en.wikipedia.org/wiki/BCH_code). But is there an efficient algorithm (better than brute force) that can verify if a binary array is orthogonal?
| https://mathoverflow.net/users/41145 | Verify if array is orthogonal | I don't see how there could be a deterministic algorithm to do this faster than brute force. Assume you have a $k\times N$ orthogonal array of strength $t.$ The statement of your question implies that no structural information other than the array itself is available.
This means any $t$ columns have all of the $2^{t}$ possible binary $t-$tuples occur the same number of times, say $\lambda,$ as one scans down the rows.
Now take a single entry of the array and switch it from $0$ to $1$ (complementing the OA gives another OA so this is fine). The problem will be that you cannot do a deterministic check without examining this entry.
If the strength $t$ is unknown, and you think an array is maybe not an OA, the most efficient method is checking that all columns are balanced. At worst, you might need to check $kN$ entries until finding the single entry that destroys the OA property. Then you'd start checking pairs of columns, etc.
If you know the strength is claimed to be $t$ and you want to check this you can have worst case complexity $\binom{k}{t}N$ until you examine all possible $t-$tuple balance conditions involving this entry.
As an aside, OAs are equivalent to what's called *resilient functions* in cryptography, see Stinson and Massey's paper "An infinite class of counterexamples to a conjecture concerning nonlinear resilient functions" in the Journal of Cryptology (vol. 8, pp. 167-172, 1995) [here].[1](https://link.springer.com/content/pdf/10.1007/BF00202271.pdf) In some sense these are functions that leak minimal information if some of their inputs (meant to be secret) are revealed. They have shown that there are (a large set of) nonlinear OAs for this application where linear OAs with the same parameters do not exist.
\**Edit:* For linear OAs a faster test is possible as in Theorem 3.29 of *Orthogonal Arrays: Theory and Applications* by Hedayat et al.
| 1 | https://mathoverflow.net/users/17773 | 422299 | 171,735 |
https://mathoverflow.net/questions/418804 | -1 | This question is going to be a bit rambling; for those who just want the essence of it, please see the definition in the middle of the post and the question at the end.
---
After some recent discussion in the MO comments on a separate question, it became apparent that the 'one hom-class' definition of a category gives the wrong intuition for 'what a category looks like in nature' if taken together with domain and codomain *functions*, since this immediately gives a category with disjoint hom-classes and consequently very canonical things like the 'EM-category of a monad' fail to actually be categories - they have arrows with multiple domains/codomains. It is always possible to consider arrows as triplets with specified domains and codomains, but this feels very 'clunky'.
For type theorists this issue [is easily resolved](https://mathoverflow.net/a/297693/92164) by pointing out that the 'hom-class for each pair of objects' typed definition of a category 'matches nature', so EM-categories are categories by their definition, and there is a $2$-equivalence (not an isomorphism) between the $2$-categories consisting of categories by either definition with appropriately defined notions of functor for each (the same notion of natural transformation works for both), so working with disjoint hom-classes is okay 'up to equivalence'. (This is distressing to me because disjoint hom-classes [dramatically simplify some arguments](https://mathoverflow.net/questions/417316/uniqueness-of-comparison-functors) in a way that feels like cheating, but that's a topic for another question.)
As someone who uses set theory for a foundation very comfortably and is uncomfortable with type theory foundationally, I wanted to find a 'non-typed' definition of a category that 'matched nature' and gave the correct intuition for how to reason about/inside them -- I found the notion of a [protocategory](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/protocategory) after a little web searching, but it seemed very unsatisfactory for the purposes of doing actual category theory.
Seeing as the only real 'problem' with the 'non-typed definition' is that the domain and codomain notions being formalized as *functions* (total functional relations) gives rise to disjoint hom-classes, I figured we could just drop the requirement that these relations be functional and see how the remaining structure matched nature.
>
> **Definition** A **category** $\mathcal{C}$ consists of
>
>
> 1. A class of **objects**, denoted ${\bf Ob}\_\mathcal{C}$, whose members are denoted by capital roman letters from the end of the alphabet $X,Y,Z,\dots$.
> 2. A class of **arrows**, denoted ${\bf Hom}\_\mathcal{C}$, whose members are denoted by lowercase roman letters from the middle of the alphabet $f,g,h,\dots$.
> 3. Entire relations $dom,cod:{\bf Hom}\_\mathcal{C}\leftrightarrow{\bf Ob}\_\mathcal{C}$, which specify a subset ${\bf Com}\_\mathcal{C}\subseteq{\bf Hom}\_\mathcal{C}\times{\bf Hom}\_\mathcal{C}$ given by $${\bf Com}\_\mathcal{C}=\{(f,g)\in{\bf Hom}\_\mathcal{C}\times{\bf Hom}\_\mathcal{C}:dom(f)\cap cod(g)\neq\emptyset\}.$$
> 4. A function $1:{\bf Ob}\_\mathcal{C}\to{\bf Hom}\_\mathcal{C}$.
> 5. A function $\circ:{\bf Com}\_\mathcal{C}\to{\bf Hom}\_\mathcal{C}$.
>
>
> We write $1(X)=1\_X$ and $\circ(f,g)=f\circ g$. These classes, relations and functions satisfy the following axioms for all $X\in{\bf Ob}\_\mathcal{C}$ and $(f,g)\in{\bf Com}\_\mathcal{C}$:
>
>
> 1. $X\in cod(1\_X)\cap dom(1\_X),$
> 2. $dom(f\circ g)=dom(g),$
> 3. $cod(f\circ g)=cod(f),$
> 4. $\forall Y\in cod(g)\big(1\_Y\circ g=g\big),$
> 5. $\forall Y\in dom(f)\big(f\circ 1\_Y=f\big),$
> 6. $(f\circ g)\circ h=f\circ(g\circ h).$
>
>
> For any two objects $X,Y\in{\bf Ob}\_\mathcal{C}$ we define the subclass ${\bf Hom}\_\mathcal{C}(X,Y)\subseteq{\bf Hom}\_\mathcal{C}$ as $${\bf Hom}\_\mathcal{C}(X,Y)=\{f\in{\bf Hom}\_\mathcal{C}:X\in dom(f)\wedge Y\in cod(f)\}.$$
>
>
>
The standard 'one hom-class' definition of a functor works for this definition of category, where we interpret the axioms $$F(dom(f))=dom(F(f))$$ $$F(cod(f))=cod(F(f))$$ as statements about classes instead of elements. This seems to suffice for the standard usage of those axioms, which serve essentially to guarantee compositionally of the image of composable arrows under a functor.
This definition seems to do the trick for matching what categories look like in nature, and we have an actual $2$-isomorphism between the $2$-category of these categories and the typed-definition categories: we pass back and forth simply by forming hom-classes as above to obtain the typed-definition from the one hom-class definition, or begin with the typed one and form the union of its hom-classes then define the domain and codomain relations by $$dom=\{(f,X):\exists Y(f\in{\bf Hom}\_\mathcal{C}(X,Y)\},$$ $$cod=\{(f,Y):\exists X(f\in{\bf Hom}\_\mathcal{C}(X,Y)\}.$$
This seems more natural than the $2$-equivalence described [on the nlab](https://ncatlab.org/nlab/show/category#equivalence_between_the_two_definitions) between the 'domain/codomain functions' definition of a category and the typed definition, where we form a disjoint union of the hom-classes from the typed definition to get to the 'domain/codomain functions' definition and we have to use the triplets trick to move in the other direction.
This also allows for more natural definition of categories like ${\bf Set}$; we can say that the morphisms are functions, the domain relation is actually a function sending a function to its domain, and the codomain relation assigns a function to all supersets of its image. There are other categories that are more naturally presented using the typed definition, but it seems like the 'domain/codomain functions' definition is ultimately less natural than the other two (made explicit by the fact that the typed definition and this one are isomorphic as $2$-categories, while the 'domain/codomain functions' definition is only equivalent).
In light of all of this, the disjoint hom-classes definition actually feels like a 'strictification' of the other definitions, allowing for easier arguments etc. which are 'guaranteed to be safe' because we know that the place we're working is equivalent to the place we really care about, similar to strictification for $2$/$3$-categories allowing us to work in a $2$-category/Gray category when we really care about a bicategory/tricategory. My question is:
>
> Has this notion appeared in the literature anywhere?
>
>
>
I'm sure the founders of the theory were aware of this business with disjoint hom-classes, but from what I can find it seems like 'protocategories' were what was viewed as the 'set theoretical fix' to the issue (immediately and appropriately discarded as unwieldy), whereas this fix requires almost no effort in tweaking other definitions and fixes the problem 'set theoretically'.
I'm primarily interested in reading more about it to make sure there aren't any pitfalls waiting down the line using this definition; in no sense do I think this is any kind of 'contribution', just bookkeeping. Any references are appreciated.
| https://mathoverflow.net/users/92164 | Category with domain/codomain relations | This **doesn't work**. In particular,
>
> we have an actual 2-isomorphism between the 2-category of these categories and the typed-definition categories: we pass back and forth simply by forming hom-classes as above to obtain the typed-definition from the one hom-class definition, or begin with the typed one and form the union of its hom-classes then define the domain and codomain relations by [...]
>
>
>
is false. Since you didn't give them a separate name, I'll call the things you defined using entire relations "categories" to distinguish them from the usual notion.
---
Example: Let $C$ be the "category" with three objects $a$, $b$, $c$, their identity morphisms, and two (nonidentity) morphisms $f : a \to c$, $g : b \to c$. Let $D$ be the "category" with the same objects but only a single nonidentity morphism $h$, which can be viewed as either $h : a \to c$ or $h : b \to c$. So $C$ and $D$ both have a single morphism from $a$ to $c$ and a single morphism from $b$ to $c$, but in $C$ these are different morphisms while in $D$ they are the same morphism.
I will write $U$ for your functor which turns a "category" into a standard (typed) category by forming the hom-sets as you describe. There is a unique isomorphism $G : UC \to UD$ such that $Ga = a$, $Gb = b$ and $Gc = c$. If $U$ is supposed to be an equivalence then there should also be a (unique) isomorphism $F : C \to D$ of "categories" such that $UF = G$. But $C$ and $D$ aren't isomorphic--one has a total of 5 morphisms and the other only 4.
---
The key point is that the "overlappiness" of different hom-sets in a typed category is **not part of the structure** of a category. (If you like, this is because the action of a functor on morphisms consists of separate functions $\mathrm{Hom}\_C(X, Y) \to \mathrm{Hom}\_D(FX, FY)$ which are (importantly!!) **not** required to agree in the situation that by accident $f \in \mathrm{Hom}\_C(X, Y)$ and also $f \in \mathrm{Hom}\_C(X', Y')$.) Consequently,
>
> begin with the typed one and form the [literal set-theoretic] union of its hom-classes
>
>
>
is not a functorial or isomorphism-invariant operation and so it cannot be inverse to the operation $U$.
---
You can still introduce a notion like this, of course, but to avoid confusion it would be better not to call or think of these objects as categories, but rather some related kind of objects.
| 5 | https://mathoverflow.net/users/126667 | 422307 | 171,736 |
https://mathoverflow.net/questions/421951 | 3 | I posted this question on math.stackexchange earlier, but didn't see any response. So, I am posting it here, in case someone else has an answer.
Original question: <https://math.stackexchange.com/questions/4443845/on-finding-an-upper-bound-on-the-error-of-a-sparse-approximation>
---
$x \in R^n$ is a non-negative vector such that $ \sum\_{i=1}^n x\_i = 1$ ($\forall i, 0 \le x\_i \le 1$).
The components are ordered: $x\_1 \ge x\_2 \ldots \ge x\_n$.
We are also given : $ \sum\_{i=1}^n x\_i^2 \ge t$ for some constant $t$ ($0 \le t \le 1$).
Clearly, the larger the constant $t$, the more concentrated the components $x\_i$ are going to become.
I want to make a claim on the approximate sparsity of $x$.
In other words, I want to place an upper bound on the total energy taken up by the smallest $(n-K)$ components.
Say, something like : there exists an integer $K(t)$, $1 \le K \le n$, such that
$$ \sum\_{i =K+1}^n x\_i^2 \le \phi(t) $$ where $\phi(t)$ is some decreasing function of $t$.
How do I get such a relation?
| https://mathoverflow.net/users/176364 | On finding an upper bound on the error of a sparse approximation | Take any integer $k\in[1,n]$ and any $t\in[1/2,1]$. Consider the problem of finding the exact upper bound on $\sum\_{i=k+1}^n x\_i^2$ given the conditions
\begin{equation\*}
x\_1\ge\cdots\ge x\_n\ge0,\quad\sum\_{i=1}^n x\_i=1,\quad\sum\_{i=1}^n x\_i^2=t. \tag{10}\label{10}
\end{equation\*}
By continuity and compactness, there is a maximizer $x=(x\_1,\dots,x\_n)$ of $\sum\_{i=k+1}^n x\_i^2$ given the conditions \eqref{10}. In what follows, let $x$ be such a maximizer (unless specified otherwise).
The cases when $n=k$ or $t=1$ are trivial: then conditions \eqref{10} imply $\sum\_{i=k+1}^n x\_i^2=0$. So, assume $n\ge k+1$ and $t\in[0,1)$. Then there is some integer $i\in[k+1,n]$ such that $x\_i>0$. Indeed, if $x\_1=\sqrt t$, $x\_2=\cdots=x\_{k+1}=\dfrac{1-\sqrt t}k$, and $x\_i=0$ for $i>k+1$, then conditions \eqref{10} hold and $\sum\_{i=k+1}^n x\_i^2>0$.
So,
\begin{equation\*}
\sum\_{i=k+1}^n x\_i^2>0 \tag{11}\label{11}
\end{equation\*}
for the maximizer $x=(x\_1,\dots,x\_n)$ of $\sum\_{i=k+1}^n x\_i^2$.
To obtain a contradiction, suppose that the maximizer $x=(x\_1,\dots,x\_n)$ is such that the $x\_i$'s take at least three distinct values $u\_1,u\_2,u\_3$ with respective multiplicities $m\_1,m\_2,m\_3$ such that $u\_1>u\_2>u\_3>0$. Without loss of generality, the "run" $R\_3:=\{i\colon x\_i=u\_3\}$ of the repeated value $u\_3$ is the rightmost run of a strictly positive value; that is, if $i\_3:=\max R\_3$, then $x\_i=0$ for all $i>i\_3$. So, in view of \eqref{11}, $i\_3\ge k+1$.
Let us apply respective infinitesimal changes $\dfrac{du\_1}{m\_1},\dfrac{du\_2}{m\_2},\dfrac{du\_3}{m\_3}$ to $u\_1,u\_2,u\_3$ such that
\begin{equation\*}
du\_1+du\_2+du\_3=0,\quad u\_1du\_1+u\_2du\_2+u\_3du\_3=0,\quad u\_3du\_3>0; \tag{12}\label{12}
\end{equation\*}
to satisfy conditions \eqref{12}, let
\begin{equation\*}
du\_3>0,\quad du\_2:=-\frac{u\_1-u\_3}{u\_1-u\_2}\,du\_3, \quad du\_1:=\frac{u\_2-u\_3}{u\_1-u\_2}\,du\_3.
\end{equation\*}
Then the conditions \eqref{10} will continue to hold, for $u\_1+\dfrac{du\_1}{m\_1},u\_2+\dfrac{du\_2}{m\_2},u\_3+\dfrac{du\_3}{m\_3}$ in place of $u\_1,u\_2,u\_3$, whereas the target sum $\sum\_{i=k+1}^n x\_i^2$ will strictly increase (since $i\_3\ge k+1$),
which will contradict the assumption that $x=(x\_1,\dots,x\_n)$ is a maximizer.
Thus, the $x\_i$'s take at most two distinct values $u,v$ such that $u\ge v>0$. So, for some integers $j\in[1,k]$ and $l\ge1$ we have
\begin{equation\*}
x\_1=\cdots=x\_j=u,\quad x\_{j+1}=\cdots=x\_{k+l}=v, \quad x\_i=0\text{ for }i>k+l.
\end{equation\*}
If $j\ge2$, then the condition $\sum\_{i=1}^n x\_i=1$ in \eqref{10} implies $u<\frac12$ and hence $x\_i<\frac12$ for all $i$, so that $\sum\_{i=1}^n x\_i^2<\frac12\,\sum\_{i=1}^n x\_i=\frac12\le t$, which contradicts the condition $\sum\_{i=1}^n x\_i^2=t$ in \eqref{10}. So, $j=1$ and we have
\begin{equation\*}
x\_1=u,\quad x\_2=\cdots=x\_{k+l}=v, \quad x\_i=0\text{ for }i>k+l.
\end{equation\*}
Now the equations $\sum\_{i=1}^n x\_i=1$ and $\sum\_{i=1}^n x\_i^2=t$ become $u+(k+l-1)v=1$ and $u^2+(k+l-1)v^2=t$. Solving the latter two equations for $u$ and $v$, we see that the maximum
of $\sum\_{i=k+1}^n x\_i^2$ given the conditions \eqref{10} is of the form
\begin{equation\*}
M\_k(l):=\frac l{(k+l)^2}\,\Big(1-\sqrt{1-\frac{k+l}{k+l-1}\,(1-t)}\Big)^2
\end{equation\*}
for some integer $l\ge1$. We shall assume that $n\ge k+l$, not to let $n$ play an essential role.
Now, to get the maximum of $\sum\_{i=k+1}^n x\_i^2$ given \eqref{10},
it remains to maximize $M\_k(l)$ in $l$. This maximization seems rather complicated to be done explicitly, even though the defining expression for $M\_k(l)$ is algebraic.
Instead of the exact maximization of $M\_k(l)$ in $l$, let us get a good upper bound on $M\_k(l)$. Since $\frac{k+l}{k+l-1}\le2$ and $\sqrt z\ge z$ for $z\in[0,1]$, we have
\begin{equation\*}
M\_k(l)\le\frac l{(k+l)^2}\,2^2(1-t)^2\le\frac{(1-t)^2}k.
\end{equation\*}
---
In my other answer on this page, it was shown (see (3) there, with $s=t$ when $t\in[1/2,1]$) that no upper bound on $\sum\_{i=k+1}^n x\_i^2$ given \eqref{10} can be better than $\asymp \frac{(1-t)^2}k$ for $t\in[1/2,1]$. It was also shown in that answer that the the optimal upper bound on $\sum\_{i=k+1}^n x\_i^2$ is $\asymp\dfrac1k$ for $t\in[0,1/2]$.
>
> Thus, the optimal upper bound on $\sum\_{i=k+1}^n x\_i^2$ given \eqref{10} is $\asymp \frac{(1-t)^2}k$ for all $t\in[0,1]$.
>
>
>
| 4 | https://mathoverflow.net/users/36721 | 422309 | 171,737 |
https://mathoverflow.net/questions/422319 | 4 | Let $G$ be an algebraic group defined over $\mathbb{Q}$ with maximal unipotent radical $N$. Let $\pi$ be an admissible representation of $G(\mathbb{Q}\_p)$, we say that this representation is supercuspidal if $\pi/\langle\pi(n)v-v\rangle = 0$. This condition is equivalent to the fact that their matrix coefficients have compact support modulo $Z(\mathbb{Q}\_p)$, the centre of $G(\mathbb{Q}\_p)$.
Let $K$ be a maximal compact subgroup of $G(\mathbb{Q}\_p)$, we say that $\pi$ is spherical or unramified if $\pi^K$, the space of $K$-fixed vectors of $\pi$ has dimension bigger than $0$.
We say that $\pi$ is a discrete series representation if the matrix coefficients of $\pi$ are $2$-integrable modulo $Z(\mathbb{Q}\_p)$.
Usually I have found that people divide the admissible representations into three disjoint sets: Supercuspidals, non supercuspidals and discrete series and spherical. Is this decomposition true? Can a supercuspidal representation be spherical? (If not, why not?) Can a discrete series be spherical? (If not, why not?) Why the classical principal series representation (induction of characters of the Torus) are not discrete series?
| https://mathoverflow.net/users/173538 | Supercuspidal, spherical and discrete series representation | As it happens, (admissible) supercuspidals cannot be spherical, because (by Borel–Casselman–Matsumoto) admissible repns with Iwahori-fixed vectors have non-trivial maps to and from unramified principal series. The Jacquet-module vanishing condition for supercuspidals is exactly that (via Frobenius Reciprocity, etc.) they admit no (non-zero) homomorphisms to principal series.
Many people would count supercuspidals as discrete series.
For $p$-adic groups, I myself do not know much about discrete series that are not supercuspidals.
And, again, if a discrete series repn were spherical, it would be a sub and quotient of principal series, which is not possible (EDIT … (thanks @Amitay for [comments](https://mathoverflow.net/questions/422319/supercuspidal-spherical-and-discrete-series-representation#comment1085257_422323)) for hyperspecial maximal compacts. For $\operatorname{GL}\_n$, all maximal compacts are conjugate, and are hyperspecial. For $\operatorname{SL}\_n$, they are all hyperspecial, but there are $n$ conjugacy classes. At least every (EDIT: thanks @LSpice) reductive group that splits over an unramified extension *has* at least one (conjugacy class of) hyperspecial maximal compact, but/and some (EDIT: split groups!) do have non-hyperspecials: for example, the affine apartments of $\operatorname{Sp}\_4$ have two different types of vertices, one with fewer edges touching it. The latter is *not* hyperspecial.
Principal series are generically irreducible. For classical groups, we can explicitly compute the $L^2$ norm (mod center) of the (essentially unique) spherical vector, (EDIT for hyperspecial maximal compact) and it's not finite….
Yes, there are some square-integrable subrepns of principal series, at some special values of the parameters. The archimedean case has the well-known examples of holomorphic discrete series subreps of principal series far away from the "unitary range".
| 10 | https://mathoverflow.net/users/15629 | 422323 | 171,742 |
https://mathoverflow.net/questions/422300 | 1 | I am stuck trying to understand certain claims made in [this](https://arxiv.org/abs/1805.01226) paper, and for completeness I will reproduce some definitions from it.
A Lorenz map $f$ on $I = [0,1]$ is a monotone increasing function that is continuous except at a critical point $c\in (0,1)$, where it has a jump discontinuity, and $f(I \setminus c)\subset I.$ The branches $f\_0:[0,c]\to I, f\_1:[c,1]\to I$ are assumed to satisfy $$f\_0(c) = 1, f\_1(c) = 0$$ $$f\_k(x) = \varphi\_k(|c-x|^\alpha)$$ for some exponent $\alpha>0$, $C^2$-diffeomorphisms $\varphi\_k, k=0,1.$
A Lorenz map $f$ is said to be *renormalizable* if there exist $p,q \geq 1$ such that $J = [f^p(0), f^q(1)]$ is contained in $I$ and has $c$ in its interior, and such that the first-return map restricted to $J$ is also a Lorenz map after rescaling. This rescaled map is known as the renormalization $Rf$ of $f$.
The symbolic coding of the branches defines the *combinatorics* $w = (w\_0, w\_1)$ of the renormalization, i.e . if $J\_k = J\cap[k,c)$ define $w\_k$ as the finite word on $\{0,1\}$ such that $f^j(J\_k)\subset[w\_k(j),c)$. A Lorenz map is said to be of $(a,b)$-type if its combinatorics are of the form $(011\cdots1, 10\cdots0)$ with $a$ 1's and $b$ 0's in the first and second positions respectively.
Now suppose one has a once $(2,1)$-renormalizable map with combinatorics $(011,10).$ The paper suggests (pg. 13) that a twice $(2,1)$-renormalizable map will have combinatorics $(0111010, 10011)$ and this can be obtained by substituting 011 for 0 and 10 for 1 in the original combinatorics.
My question: why does this rule work? And what would the combinatorial type of (say) a thrice $(2,1)$-renormalization be? I can't find any resources that handle these computations.
| https://mathoverflow.net/users/482093 | Computing kneading sequences for renormalizations of Lorenz maps | Perhaps this will not be a good answer, but, firstly you might have a look at [this paper](http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.zamlynska-bd295345-6842-4dfe-a29a-4ec72d7ba65d/c/rm38201.pdf) In particular, have a look at Chapter 4. Also, you might want to look at the work of Glendinning and Hall [here](https://iopscience.iop.org/article/10.1088/0951-7715/9/4/010)
Now, the way I understand it is the following (maybe is not the best because I only understand the symbolic setting). Recall that the kneading invariant of a Lorenz map, say $(\alpha, \beta)$ is *renormalisable* if there exist two finite words $\omega = w\_1 \ldots w\_m \quad \text{and} \quad \nu = v\_1 \ldots v\_n$
and two sequences $\{M\_i\}\_{i=1}^\infty, \, \{N\_j\}\_{j=1}^\infty \subset \mathbb{N}\cup \{\infty\}$ such that:
1. $n+m > 3$;
2. $(\omega^\infty, \nu^\infty)$ is also a kneading invariant;
3. $\alpha = \nu \omega^{M\_1}\nu^{M\_2}\omega^{M\_3}\nu^{M\_4}\ldots;\\
\beta = \omega\nu^{N\_1}\omega^{N\_2}\nu^{N\_3}\omega^{N\_4}\ldots,
$
If $(\alpha,\beta)$ is not renormalisable then we say that $(\alpha, \beta)$ is *prime*. Also, if $n+m = 3$ then $(\alpha,\beta)$ is called *trivially renormalisable.*
In the references I gave above (you might check also [this](https://www.sciencedirect.com/science/article/abs/pii/S0001870818304729)), it is mentioned/proved that if a Lorenz map is renormalisable if the kneading invariant is renormalisable. So, you only need to worry about the combinatorics of the kneading invariant to understand the Lorenz map (of course, if you can define a kneading invariant). I think this might answer your first question.
Regarding the second question: having a (2,1) renormalisable map, then the only words that will help to renormalise are $(011, 10)$ (if it was 1,2 renormalisable, then you would have $(01, 101)$). In particular, there are certain conditions needed in order to have a pair $\omega, \nu$ to be candidates to renormalise a kneading invariant as I mentioned in the definition; namely, $(\omega^\infty, \nu^\infty)$ must be a kneading invariant of another Lorenz map, and in fact it has to be prime or trivially renormalisable (this is because when you renormalise the kneading invariant you need that the words $\omega$ and $\nu$ to have minimal length). You can check some extra conditions for the linear case in the works of Glendinning ([here](https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/abs/topological-conjugation-of-lorenz-maps-by-transformations/D0557245F7C4C4EF7A04159A2D87292D)) and Barnsley et.al. ([here](https://people.clas.ufl.edu/avince/files/CriticalMPC.pdf)).
Now, if you apply the substitution you are mentioning above, you will get:
$011 \to 0111010 \to 01110101001110011 \quad \text{and} \quad 10 \to 10011 \to 100110111010$. Observe that the final words can be written concatenating 011 and 10. Now, the words $011$ and $10$ have minimal length, in the sense that $((011)^\infty, (10)^\infty)$ is trivially renormalisable. Take the final word and apply the inverse substitution you mentioned. In the first case, you will get the pair $(011)^\infty,(10)^\infty$ and in the second case you need to perform the inverse substitution twice (it is straightforward from the way I wrote it). Now, you can get a different (2,1) renormalisation with the same combinatorics. For example, if you take the sequences $(01101)^\infty, (100100)^\infty$ (which are kneading invariants) you can get a new $(2,1)$ renormalisable pair. Namely, apply the substitutions again, i.e.
$01101 \to 011101001110 \to 01110101001110011011101010011 = \omega'$
and
$100100 \to 1001101110011011 \to 10011011101001110101001101110100111010 = \nu'.$
It is not difficult to check that, if you have two sequences $\alpha$ and $\beta$ that satisfy 3. using the words $\omega'$ and $\nu'$ you can apply twice the inverse substitution you mentioned above. This will tell you that your kneading invariant is $(2,1)$ twice renormalisable, although it could be renormalisable using different combinatorics. In conclusion, if you take **any** kneading invariant of a Lorenz map, if you apply $K$ times a substitution of combinatorial type $(m,n)$ the resulting pair must be a $K$ times renormalisable kneading invariant with $(m,n)$ combinatorics.
Hope I helped.
| 0 | https://mathoverflow.net/users/10518 | 422332 | 171,745 |
https://mathoverflow.net/questions/421676 | 11 | Let $M$ be an open, simply connected, 3-manifold. Suppose $M$ admits a properly discontinuous, co-compact topological action by a finitely generated group.
**Question 1:** If $M$ is 1-ended, must it be homeomorphic with $\mathbb{R}^3$?
More generally:
**Question 2:** Is $M$ determined up to homeomorphism by its number of ends?
(By a classical result of Hopf, this number is 1, 2, or uncountably infinite, I think.)
More generally, I'm interested in results of the form: if a 3-manifold $M$ covers a compact manifold/orbifold, then it cannot be as wild as a generic open 3-manifold such as e.g. the Whitehead manifold.
---
Edit: Having discussed this with an expert, I believe that my questions above boil down to the following conjecture:
**Conjecture:** Let $G$ be the fundamental group of a closed 3-manifold $M$. Then $G$ is simply connected at infinity. (Equivalently, the universal cover of $M$ is simply connected at infinity.)
This conjecture is implicit in this paper by Funar & Otera:
[https://www-fourier.ujf-grenoble.fr/~funar/funote.pdf](https://www-fourier.ujf-grenoble.fr/%7Efunar/funote.pdf)
| https://mathoverflow.net/users/69681 | How wild can an open topological 3-manifold be if it has a compact quotient? | The possible universal covers of closed 3-manifolds are $S^3-C$, where $|C|=0, 1, 2$ or $C$ is a tame Cantor set, corresponding to the space of ends of the fundamental group as you suspect. This follows from the [geometrization theorem](https://en.wikipedia.org/wiki/Geometrization_conjecture), known to experts but might not be written down. I gave a survey talk on this once, you can find the [notes here.](https://www.dropbox.com/sh/cxdm5tu5hsnho0i/AACG_sRIgiHR9E79vhNOJiR3a/public_html/cover?dl=0&subfolder_nav_tracking=1)
If $M$ is a closed orientable connected 3-manifold, and $\pi\_1 M$ is finite, then its universal cover is $S^3$ by the [Poincaré conjecture](https://en.wikipedia.org/wiki/Poincar%C3%A9_conjecture).
If $\pi\_1M$ is infinite and $\pi\_2 M=0$, then the universal cover $\tilde{M} \cong \mathbb{R}^3$. In this case, $M$ has a geometric decomposition. If the decomposition is trivial, then $M$ is modeled on one of the six geometries homeomorphic to $\mathbb{R}^3$, and hence the universal cover is $\mathbb{R}^3$. Otherwise, $M$ has an essential torus, hence is a Haken manifold. Waldhausen proved that Haken manifolds have universal cover $\mathbb{R}^3$ - see [Theorem 8.1](https://doi.org/10.2307/1970594).
If $\pi\_1 M$ is infinite but $\pi\_2 M\neq 0$, then either $M$ is modeled on the $S^2\times \mathbb{R}$ geometry and $M$ is homeomorphic to $\mathbb{RP}^3\#\mathbb{RP}^3$ or $S^2\times S^1$. In this case the number of ends of $\pi\_1 M$ is 2. Otherwise, $M$ is a non-trivial connect sum by the [sphere theorem](https://en.wikipedia.org/wiki/Sphere_theorem_(3-manifolds)) and its universal cover is $S^3-C$ where $C$ is a tame Cantor set. The connect summands have universal cover either $S^3$, $S^2\times \mathbb{R}$ or $\mathbb{R}^3$. When you take connect sums, you remove open balls from each manifold and glue the sphere boundaries together. The universal cover is obtained by gluing the universal covers of each summand punctured along balls, either finitely many in $S^3$ or infinitely many in $S^2\times\mathbb{R}$ or $\mathbb{R}^3$.
One can see that such manifolds are built out of thrice punctured spheres, and hence the universal cover can be decomposed into thrice punctures spheres. Such a manifold is homeomorphic to $S^3-C$.
| 7 | https://mathoverflow.net/users/1345 | 422343 | 171,750 |
https://mathoverflow.net/questions/422288 | 1 | In the comments and answer to another recent [question](https://mathoverflow.net/q/422023/92164), it became apparent that category theorists who work with the ‘many hom-class’ definition of a category implicitly view composition as a function of five variables, three of them the objects that define the hom-classes the composable arrows live in.
This is, to put it mildly, very counterintuitive to me. I can’t think of a single time in my years working with (fibered) categories where composition depended on the objects involved, but maybe I just didn’t notice due to not looking for it. To this end,
>
> What are some examples of ‘naturally occurring’ categories $\mathcal{C}$ where we have objects $X,Y,Z,X’,Y’,Z’$ and arrows $$f\in{\bf Hom}\_\mathcal{C}(Y,Z)\cap {\bf Hom}\_\mathcal{C}(Y’,Z’),$$ $$g\in{\bf Hom}\_\mathcal{C}(X,Y)\cap{\bf Hom}\_\mathcal{C}(X’,Y’)$$ such that $$f\circ\_{\_{XYZ}}g\neq f\circ\_{\_{X’Y’Z’}}g?$$
>
>
>
Any relevant input is appreciated.
| https://mathoverflow.net/users/92164 | Naturally occurring examples of categories where composition depends on objects | Here is a way to see that [Brian Shin’s example](https://mathoverflow.net/a/422314/2273) can be obtained from (standard many-hom-sets presentations of) more general constructions — so this shows clearly that constructions from the standard literature can lead to cases where composition genuinely depends on the objects.
$\newcommand{\C}{\mathcal{C}}\newcommand{\op}{\mathrm{op}}\newcommand{\id}{\mathrm{id}}$First, take the [**category of elements**](https://en.wikipedia.org/wiki/Category_of_elements) $\int\_{\C^\op} F$ of a presheaf $F : \C^\op \to \mathrm{Set}$ to have:
* objects are pairs $(c,x)$, where $c \in \C$, $x \in F(c)$;
* maps $(c',x') \to (c,x)$ are maps $f : c' \to c$ such that $x|\_f = x'$.
In particular, applying this over the terminal category $\C = \{\*\}$ yields a construction of *discrete* categories $DX = \int\_1 X$ in which all identities are equal: $DX((\*,x),(\*,x)) = \{ \id\_\*\}$.
Now, take the [**Grothenedieck construction**](https://en.wikipedia.org/wiki/Grothendieck_construction) of a functor $\newcommand{\D}{\mathcal{D}}\D : \C \to \mathrm{Cat}$ to have:
* objects are pairs $(c,d)$, where $c \in \C$, $d \in \D(c)$;
* maps $(c,d) \to (c',d')$ are pairs $(f,g)$, where $f : c' \to c$, and $g : \D(f)(d) \to d'$.
Now take $G$ and $G'$ to be two different groups on the same underlying set, viewed as one-object categories; and consider the functor $\mathcal{G} : D2 \to \mathrm{Cat}$ picking out $G$ and $G'$. Then the Grothendieck construction of $\mathcal{G}$ recovers Brian Shin’s example: its hom-sets are $\{\id\_\*\} \times G$ and $\{\id\_\*\} \times G'$, i.e. the same set with different composition operations.
---
That said, let me address where you write:
>
> it became apparent that category theorists who work with the ‘many hom-class’ definition of a category implicitly view composition as a function of five variables […] This is, to put it mildly, very counterintuitive to me.
>
>
>
I really want to convince you it *shouldn’t* be counterintuitive: it’s just being slightly more explicit about something you’re doing all the time anyway, in many situations.
As in my [previous answer](https://mathoverflow.net/a/422068/2273), think about group operations (I’ll write multiplicatively). We think of multiplication as a function of two variables — and for a fixed group, of course, it is. But with varying groups, it also depends on the group. And this really matters — for instance, when we take the product of a family of groups $\prod\_i G\_i$, we define its pointwise multiplication as $(x \cdot y)\_i = x\_i \cdot y\_i$. Formally of course this must mean $x\_i \cdot\_i y\_i$ — the multiplication used depends on $i$. But this doesn’t disrupt our intuition of it as a binary operation; it just *also* depends on another parameter *i*.
Of course, you could avoid that dependence by assuming that the groups in the family must be disjoint, or compatible where they intersect, in order to form the product. But I think most modern algebraists would find such a restriction highly artificial, and unnecessary anyway — the dependence is nothing to be worried by. Indeed, once you look for it, this sort of thing is *everywhere* in mathematical practice. A large part of Martin-Löf’s original motivation for dependent type theory was analysing how mathematicians use this sort of dependency in practice. For me, meeting dependent type theory was like [Molière’s character](https://en.wikipedia.org/wiki/Le_Bourgeois_gentilhomme) learning he’d been speaking prose all his life.
Coming back to the typed-hom-sets definition of category: The point isn’t to *focus* on the extra dependence on objects. The point is to let us *ignore* the question intersection/disjointness of objects, and view it as just as irrelevant as the question of whether different abstract groups are disjoint; and we achieve it by exactly the same mechanism.
| 6 | https://mathoverflow.net/users/2273 | 422358 | 171,753 |
https://mathoverflow.net/questions/422361 | 2 | Let $X$ be a normal integral variety over $\mathbb{C}$ and $D \subset X$ be a Cartier divisor in $X$. Is the associated reduced scheme $D\_{\mathrm{red}}$ also necessarily a Cartier divisor in $X$?
| https://mathoverflow.net/users/45397 | Is the reduced scheme associated to a Cartier divisor always Cartier? | No. Consider, for instance, the quadratic cone
$$
X = \{xz - y^2 = 0\} \subset \mathbb{A}^3
$$
and the double line
$$
D = X \cap \{x = 0\} = \{x = y^2 = 0\}
$$
on $X$. Then $D$ is a Cartier divisor, but
$$
D\_{\mathrm{red}} = \{x = y = 0\}
$$
is not Cartier (this is the simplest example of a Weil divisor which is not Cartier).
| 9 | https://mathoverflow.net/users/4428 | 422362 | 171,754 |
https://mathoverflow.net/questions/422356 | 11 | Unless I made a mistake, the expected value of the largest exponent in the prime factorization of random positive integer (defined in the appropriate way) is $$\eta := \sum\_{n=1}^\infty \Big(1-\zeta(n)^{-1}\Big)$$ with the convention $\zeta(1)^{-1} = 0$ (for aesthetics). I was just wondering whether this constant $\eta$ has a name, whether it's been studied, etc..
| https://mathoverflow.net/users/129185 | Has this number-theoretic constant been studied? | If you calculate it to a few decimals, you find
$$
1.705211140105\ldots
$$
which is enough to locate it in the OEIS.
It's Niven's constant: [MathWorld](https://mathworld.wolfram.com/NivensConstant.html), [Wikipedia](https://en.wikipedia.org/wiki/Niven%27s_constant), [OEIS](https://oeis.org/A033150).
As [mentioned](https://mathoverflow.net/questions/422356/has-this-number-theoretic-constant-been-studied#comment1085332_422366) by GH from MO in the comments, it was in fact proven by Niven in 1969 that the average largest exponent tends to $\eta$.
Since the question is essentially about finding literature related to a given numerical constant, I should probably mention [this answer](https://math.stackexchange.com/questions/4263890/finding-a-closed-form-for-constant/4265349#4265349) by myself on the other site, exhibiting some other methods (Steven Finch's book *Mathematical Constants* & how to google the decimals effectively).
| 33 | https://mathoverflow.net/users/171662 | 422366 | 171,756 |
https://mathoverflow.net/questions/422365 | 16 | Let $M$ be a differential manifold and $\mathcal H^k$ the presheaf of real vector spaces associating to the open subset $U\subset M$ the $k$-th de Rham cohomology vector space: $\mathcal H^k(U)=H^k\_{DR}(U)$. Is this presheaf a sheaf?
Of course not! Indeed, given *any* non-zero cohomology class $0\neq[\omega]\in \mathcal H^k(U)$ represented by the closed $k$-form $\omega\in \Omega^k\_M(U)$ there exists (by Poincaré's Lemma) a covering $(U\_i)\_{i\in I}$ of $U$ by open subsets $U\_i\subset U$ such that $[\omega]\vert U\_i=[\omega\vert U\_i]=0\in \mathcal H^k(U\_i)$, and thus the first axiom for a presheaf to be a sheaf is violated.
But what about the second axiom?
**My question:**
Suppose we are given a differential manifold M, a covering $(U\_\lambda)\_{\lambda \in \Lambda}$of $M$ by open subsets $U\_\lambda \subset M$, closed differential $k-$forms $\omega\_\lambda \in \Omega^k\_M(U\_\lambda)$ satisfying $[\omega\_\lambda]\vert U\_\lambda \cap U\_\mu=[\omega\_\mu]\vert U\_\lambda \cap U\_\mu\in \mathcal H^k(U\_\lambda\cap U\mu)$ for all $\lambda,\mu \in \Lambda$.
Does there then exist a closed differential form $\omega\in \Omega^k(M)$ such that we have for the restrictions in cohomology: $[\omega]\vert U\_\lambda=[\omega \_\lambda]\in \mathcal H^k(U\_\lambda)$ for all $\lambda\in \Lambda$ ?
**Remarks**
1. This is an extremely naïve question which, to my embarrassment, I cannot solve.
I have extensively browsed the literature and consulted some of my friends, all brilliant geometers (albeit not differential topologists), but they didn't know the answer offhand.
For what it's worth, I would guess (but not conjecture!) that such glueing is impossible.
2. If the covering of $X$ has only two opens then we can glue.
This follows immediately from Mayer-Vietoris's long exact sequence
$$\cdots \to \mathcal H^k(M) \to \mathcal H^k(U\_1) \oplus \mathcal H^k(U\_2) \to \mathcal H^k(U\_1\cap U\_2)\to \cdots$$
**Update**
My brilliant friends didn't answer offhand but a few hours later, unsurprisingly, they came back to me with splendid counterexamples! See below.
| https://mathoverflow.net/users/450 | Can one glue De Rham cohomology classes on a differential manifolds? | No.
Make $M$ by gluing three strips to two discs to form a thrice-punctured sphere. Take three open sets $U\_\lambda$, each made by both discs and two of the strips. Then each $U\_\lambda$ is homeomorphic to annulus and thus has $1$-dimensional $H^1$.
The pairwise intersections, made from one strip connecting two discs, are contractible and so their $H^1$ vanishes. Thus, for $k=1$, the agreement condition on the pairwise intersections is vacuous.
If your claim held, then we could choose a $1$-form on $M$ that restricts to an arbitrary cohomology class on each of the three $U\_\lambda$, making the first de Rham cohomology of $M$ at least three-dimensional. But in fact it is only two-dimensional. Instead, there is a relation where the integrals around three clockwise loops around the three punctures sum to $0$, because these loops form the boundary of a particular subset of $M$.
It's true if the intersections $U\_\lambda \cap U\_\kappa \cap U\_\mu$ are empty for all distinct $\lambda,\kappa,\mu$, by iteratively applying the Mayer-Vietoris sequence or applying a single exact sequence in sheaf cohomology.
| 18 | https://mathoverflow.net/users/18060 | 422367 | 171,757 |
https://mathoverflow.net/questions/422352 | 3 | Let $M$ be a smooth Riemanniann manifold. For $\varepsilon \geq 0$ we call an $\varepsilon$-geodesic (I am not sure that this is a standard name) a smooth map
$$\gamma\colon [a,b]\to M$$
such that for any $t\in [a,b]$ there is a neighborhood $[u,v]$ of $t$ such that for any $x,y\in [u,v]$ one has
$$(1-\varepsilon)dist\_M(\gamma(x),\gamma(y))\leq length[\gamma(x),\gamma(y)]\leq (1+\varepsilon)dist\_M(\gamma(x),\gamma(y)),$$
where in the middle one has the length of the path $\gamma$ between the points $x$ and $y$, and $dist\_M$ denotes the distance in $M$.
Note that $0$-geodesics are the usual geodesics.
**Given a point $p\in M$, does there exist a neighborhood $U$ of $p$ and positive $\varepsilon >0$ such that there are no closed $\varepsilon$-geodesics contained in $U$?**
Remark. For $\varepsilon=0$ this result is well known (the injectivity radius is positive).
| https://mathoverflow.net/users/16183 | Closed almost geodesics in a Riemannian manifold | Any curve $\gamma:[a,b]\to M$ parametrized by arc length is an $\varepsilon$-geodesic for any $\varepsilon>0$.
The inequality $(1-\varepsilon)dist\_M(\gamma(x),\gamma(y))\leq length[\gamma(x),\gamma(y)]$ is obvious (for $\varepsilon=0$).
To prove the other inequality, for each $t\in [a,b]$ we will pick the geodesic $\alpha\_t$ based at $\gamma(t)$ and with direction $\gamma'(t)$.
As $\gamma$ and $\alpha\_t$ coincide up to first order in $t$, we have that $d(\gamma(t+\delta),\alpha\_t(t+\delta))$ is $o(\delta)$. By compactness we can find for every $\varepsilon$ some $\delta$ such that, $\forall t\in[a,b]$ and $\forall s$ with $|s|\leq\delta$, $d(\alpha\_t(t+s),\gamma(t+s))<\varepsilon |s|$.
This means that, if for some $t\_0$ we pick the neighborhood $[u,v]$ of $t\_0$ with $v-u<\delta$, then $\forall x,y\in [u,v]$ we have
$$dist\_M(\gamma(x),\gamma(y))\geq dist\_M(\gamma(x),\alpha\_x(y))-dist\_M(\alpha\_x(y),\gamma(y))\geq |y-x|-\varepsilon|y-x|=(1-\varepsilon)length[\gamma(x),\gamma(y)].$$
So $length[\gamma(x),\gamma(y)]\leq (1+2\varepsilon)dist\_M(\gamma(x),\gamma(y))$, as we wanted.
| 2 | https://mathoverflow.net/users/172802 | 422368 | 171,758 |
https://mathoverflow.net/questions/422379 | 2 | The moduli stack $\mathcal{M}\_{0,4}$ of 4 points on $\mathbb{P}^1$ is isomorphic to $\mathbb{P}^1 - \{0,1,\infty\}$, where $[a,b,c,d]$ is identified with the point $\lambda\_{a,b,c}(d)$, where $\lambda\_{a,b,c}\in\text{Aut}(\mathbb{P}^1)$ sends $a,b,c\mapsto 0,1,\infty$. We will use $\lambda$ to identify $\mathcal{M}\_{0,4}$ and $\mathbb{P}^1 - \{0,1,\infty\}$. Over $\mathcal{M}\_{0,4}$, it has a universal family $\mathcal{M}\_{0,4}\times\mathbb{P}^1$, with four canonical sections $s\_1 = 0,s\_2 = 1,s\_3 = \infty,s\_4 = \Delta$, where $\Delta$ denotes the diagonal. The complement of the four sections is naturally identified with $\mathcal{M}\_{0,5}$.
The Deligne-Mumford compactification of $\mathcal{M}\_{0,4}$ is obtained by adding in the boundary points, so we have $\overline{\mathcal{M}\_{0,4}} \cong\mathbb{P}^1$.
On the other hand [page 5 of these notes](https://anddil.github.io/_pages/CohFT/Lecture1.pdf) indicate that the compactification of $\mathcal{M}\_{0,5}$ should be obtained by blowing up $\overline{\mathcal{M}\_{0,4}}\times\mathbb{P}^1\cong\mathbb{P}^1\times\mathbb{P}^1$ at 3 points, and that this should now be the universal family over $\overline{\mathcal{M}\_{0,4}}$. Essentially, $\sigma\_4 = \Delta$ intersects $\sigma\_1,\sigma\_2,\sigma\_3$ each at 1 point, the intersection points lying over $\lambda = 0,1,\infty\in\overline{\mathcal{M}\_{0,4}}$ respectively.
This seems odd to me, since my feeling is that the sections $\sigma\_1,\sigma\_2,\sigma\_3,\sigma\_4$ should all be symmetric -- of course their definition is not, but their role in the moduli problem is. However, $\sigma\_1,\sigma\_2,\sigma\_3$ each meet the exceptional divisors at exactly 1 point, whereas $\sigma\_4$ meets them at 3 points, so if I haven't made a mistake we should have $\sigma\_1^2 = \sigma\_2^2 = \sigma\_3^2 = -1$, whereas $\sigma\_4^2 = -3$.
Is this correct? If so, can someone explain why the symmetry present in the moduli formulation is not present in the geometry of the sections?
| https://mathoverflow.net/users/88840 | Self intersections of the canonical sections on $\overline{\mathcal{M}_{0,5}}$ | Your calculation of the self-intersection of $\sigma\_4$ is wrong. You're correct that the self-intersection drops by $1$ for each time it intersects the exceptional divisor, but you've assumed the self-intersection before you blow up is $0$ (appropriate for a constant section) and not the correct value of $2$ (the self-intersection of the diagonal of $\mathbb P^1 \times \mathbb P^1$).
| 5 | https://mathoverflow.net/users/18060 | 422381 | 171,762 |
https://mathoverflow.net/questions/422351 | 1 | I am studying the work
*Ionescu, A. D.; Kenig, C. E.*, Local and global wellposedness of periodic KP-I equations, Bourgain, Jean (ed.) et al., Mathematical aspects of nonlinear dispersive equations. Lectures of the CMI/IAS workshop on mathematical aspects of nonlinear PDEs, Princeton, NJ, USA, 2004. Princeton, NJ: Princeton University Press (ISBN 978-0-691-12955-6/pbk; 978-0-691-12860-3/hbk). Annals of Mathematics Studies 163, 181-211 (2007). [ZBL1387.35528](https://zbmath.org/?q=an:1387.35528).
I have one particular question for which I wasn't able to find an answer.
>
> In equation (9.3.14), why did they assume that $l \in [j, 7j/4]$?
>
>
>
Thanks in advance.
| https://mathoverflow.net/users/471464 | Explanation of a step in a work by C. E. Kenig and A.D. Ionescu | $\newcommand{\ep}{\epsilon}$Formula (9.3.14) in the linked paper states that
\begin{equation}
L:=\Big|\sum\_{m\ge0}(m/2^j)^{1/2}\psi\_1^2(m/2^j)e^{i(mx'+m^3t)}\Big|
\le C\_\ep R, \tag{1}\label{1}
\end{equation}
where
\begin{equation}
R:=2^{(3/4+2\ep)j}2^{2(l-j)/5},
\end{equation}
$\ep>0$, $x'$ and $t$ are real numbers, $j$ and $l$ are integers such that $l\ge j$, and (by a description at the top of p. 187 in the paper) the function $\psi\_1$ is such that $0\le\psi\_1\le1$ and $\psi\_1(x)=0$ if $|x|\ge4$.
So,
\begin{equation}
L\le\sum\_{0\le m\le2^{j+2}}(m/2^j)^{1/2}. \tag{2}\label{2}
\end{equation}
It follows that, if $j\le-3$, then $L=0$ and hence \eqref{1} trivially holds. If $-3<j\le0$, then $L\ll1\ll e^{4\ep}R$ and hence \eqref{1} still holds. (We write $A\ll B$ if $A\le CB$ for some universal real constant $C>0$.)
Finally, consider the case $j>0$. Then, by \eqref{2}, $L\ll2^j$. On the other hand, if $l>7j/4$, then $R>2^{(3/4)j}2^{2\times3j/(4\times5)}=2^{21j/20}>2^j$. So, \eqref{1} holds in this case as well.
Thus, we may assume that $l\le7j/4$ and hence $l\in[j,7j/4]$.
| 2 | https://mathoverflow.net/users/36721 | 422383 | 171,763 |
https://mathoverflow.net/questions/422283 | 1 | Assume that $f:[0,1]\to [0,1]$ is an diffeomorphism so that $(f''(x)/f'(x))'<0$ and that $f''(0)=0$. It seems to me that $$\frac{1-f(x)^2}{1-x^2}\le f'(x)$$ on $[0,1]$. But no proof so far.
The answer posted below is correct. I however need $f$ to be increasing.
| https://mathoverflow.net/users/409893 | A condition on the inequality $f'(x)/(1-f(x)^2)-1/(1-x^2)\ge 0$ | For $x\in[0,1]$, let
\begin{equation\*}
f(x):=c\int\_x^1 e^{-t^2}\,dt,
\end{equation\*}
where $c:=1/\int\_0^1 e^{-t^2}\,dt$.
Then $f\colon[0,1]\to [0,1]$ is a diffeomorphism such that $(f''(x)/f'(x))'<0$ for all $x\in[0,1]$ and $f''(0)=0$.
However, the inequality in question,
\begin{equation\*}
L(x):=\frac{1-f(x)^2}{1-x^2}\le f'(x), \tag{1}\label{1}
\end{equation\*}
fails to hold for any $x\in[0,1]$, since $f'<0$ and $L>0$ on $[0,1)$, whereas $L(1)$ is undefined.
| 3 | https://mathoverflow.net/users/36721 | 422391 | 171,765 |
https://mathoverflow.net/questions/422344 | 0 | In page 16 of [these notes](https://da380198-735d-4021-b7f2-6247f0586806.filesusr.com/ugd/946d8a_60256058271940f7b9d2c14dc721bf91.pdf) on $p$-adic $L$-functions, it makes the following claim:
>
> Let $\alpha$ be a $p$-adic measure on $\mathbf{Z}\_p$ which corresponds to a power series $F\_{\alpha}(T) \in \mathbf{Z}\_p[[T]]$ under the Iwasawa isomorphism. Fix any $a \in \mathbf{Z}\_p^{\times}$. Then the measure $\alpha \circ (a)$ corresponds to the power series $F\_{\alpha}((1+T)^{a}-1) \in \mathbf{Z}\_p[[T]]$ under the Iwasawa isomorphism.
>
>
>
*My question is*: what exactly does $F\_{\alpha}((1+T)^{a}-1)$ mean? That is, for a general $a \in \mathbf{Z}\_p^{\times}$, the expression $(1+T)^{a}-1$ is a *power series* (not necessarily a polynomial). So when we write $F\_{\alpha}((1+T)^{a}-1)$, we are plugging in a power series into another power series.
Does it make sense to do this? Even once we expand out all the brackets and group all like terms, wouldn't we still have to add infinitely many coefficients for each $T^n$ term? So we'd have to worry about convergence issues for the coefficients. Is it known that the coefficients for each sum would always converge?
| https://mathoverflow.net/users/394740 | A confusion about power series and p-adic measures | You are asking about substitution of one power series into another. This can be interpreted in two ways, which ultimately amount to the same thing (like two different ways of thinking about *anything*) but one method is elementary and rather clunky while the other takes longer to set up but ultimately is more slick and better for actually proving properties in a non-tedious way.
**Method 1**: the combinatorial method (using finitely many terms at a time). If $f(x) = a\_0 + a\_1x + a\_2x^2 + \cdots$ is an element of $A[[x]]$, where $A$ is a commutative ring, and $g(x) = b\_1x + b\_2x^2 + \cdots$ is another element of $A[[x]]$ with constant term $0$, then $f(g(x))$ makes sense as
$$
a\_0 + a\_1g(x) + a\_2g(x)^2 + \cdots
$$
because $g(x)^n = b\_1^nx^n + \cdots$ has no terms in degree below $n$, so to find the coefficient of $x^d$ in $f(g(x))$ you only need to work with
the finite sum $a\_0 + a\_1g(x) + \cdots + a\_{d}g(x)^{d}$ and extract the coefficient of $x^d$ in that sum. It "doesn't make sense" to compose $f(g(x))$ if $g(x) = b\_0 + b\_1x + b\_2x^2 + \cdots$ has a nonzero constant term because the constant term of $f(g(x))$ would then be $f(g(0)) = f(b\_0) = a\_0 + a\_1b\_0 + a\_2b\_0^2 + \cdots$ and this is an infinite series that doesn't make *algebraic* sense unless $b\_0$ is nilpotent or some other reason causes all but finitely many terms to be $0$.
Example: if $f(x) \in \mathbf Z\_p[[x]]$ then
$f((1+x)^a - 1)$ makes sense since $(1+x)^a - 1 = \sum\_{n \geq 1} \binom{a}{n}x^n$ is a power series with constant term $0$.
I call this the combinatorial method because I have seen combinatorialists describe computations with formal power series in this way, which is purely algebraic and emphasizes the finitely many calculations needed to work out any particular coefficient. It seems like a clunky way of setting up formal power series if you want to prove properties of them beyond the simplest ones, but it has the advantage of being intuitive and getting to the point of things in a direct way. To prove things from this point of view you typically need to work them out degree-by-degree with induction. See, for example, how proofs look in the *Amer. Math. Monthly* article on formal power series by Ivan Niven [here](https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/IvanNiven.pdf).
Now let's try something "completely different" that is more in the $p$-adic spirit.
**Method 2**: the $x$-adic topology. For a commutative ring $A$, define the $x$-adic absolute value on $A[[x]]$ by $|f(x)|\_x = (1/2)^n$ when $f(x) = c\_nx^n + \cdots$ is nonzero with its first nonzero term occurring in degree $n$, and $|0|\_x = 0$. For example, all nonzero $a \in A$ have $|a|\_x = 1$ when we view $a$ in $A[[x]]$ as a constant power series.
For all $f$ and $g$ in $A[[x]]$, $|f+g|\_x \leq \max(|f|\_x,|g|\_x)$ and $|fg|\_x \leq |f|\_x|g|\_x$. If $A$ is an integral domain, such as $A = \mathbf Z\_p$ or $\mathbf Q\_p$, then $|fg|\_x = |f|\_x|g|\_x$. From $|\cdot|\_x$ we get a topology on $A[[x]]$ in which $x^n \to 0$ as $n \to \infty$, and $|f|\_x < 1$ if and only if $f$ has constant term $0$. The $x$-adic distance between $f$ and $g$ is defined to be $|f-g|\_x$, which is a non-Archimedean metric on $A[[x]]$ that makes it complete. Just as in the $p$-adic numbers, a sequence $\{f\_n\}$ in $A[[x]]$ is $x$-adically convergent if and and only if $|f\_{n+1} - f\_n|\_x \to 0$ as $n \to \infty$, so an infinite series of elements in $A[[x]]$ is $x$-adically convergent if and only if its general term tends to $0$.
Now let's talk about the composition $f(g(x))$ for $f(x) \in A[[x]]$ and $g(x) \in A[[x]]$. What does it mean and when does it make sense?
To take advantage of $x$-adic convergence, let's view $f(x)$ as a power series in a different indeterminate, say $f(y) \in R[[y]]$ where $R = A[[x]]$ is a non-Archimedean complete ring and $f(y)$ happens to have all "constant" coefficients (lying in $A$, not just in $R$). For example, if $f(x) = 1 + x + x^2 + x^3 + \cdots$ then we view it as $1 + y + y^2 + y^3 + \cdots$ in $R[[y]]$ whose coefficients are all $1 \in A$. Asking about the meaning of $f(g(x))$ is asking about where a power series in $R[[y]]$ converges on $R$. Just like a $p$-adic power series in $\mathbf Z\_p[[y]]$ has a $p$-adic disc of convergence, a power series $f(y)$ in $R[[y]]$ has a disc of convergence in $R$: if $f(y) = \sum\_{n \geq 0} c\_ny^n$, then $f(r)$ converges if and only if $|c\_nr^n|\_x \to 0$ as $n \to \infty$. Since $|c\_nr^n|\_x \leq |c\_n|\_x|r|\_x^n \leq |r|\_x^n$,
$f(r)$ converges *if* $|r|\_x < 1$. Since $r^n - s^n$ is divisible by $r-s$ for all $n \geq 0$, $|f(r) - f(s)|\_x \leq |r-s|\_x$ when $f(r)$ and $f(s)$ converge, so the power series $f(y)$ is $x$-adically uniformly continuous on its disc of convergence.
Writing $r \in R = A[[x]]$ as $g(x)$, saying $|r|\_x < 1$ means $g(x)$ has constant term $0$. This shows $f(g(x))$ converges $x$-adically in $A[[x]]$ when $g(x)$ has constant term $0$. It's left to you to check, using the $x$-adic uniform continuity estimate $|f(g(x)) - f(h(x))|\_x \leq |g(x) - h(x)|\_x$ when $g(x)$ and $h(x)$ have constant term $0$, that the calculation of the coefficients of $f(g(x))$ as an $x$-adic limit (finding each of its coefficients as an element of $A[[x]]$) can be done by the combinatorial method ("finitely many terms at a time") because $g(x) \equiv g\_d(x) \bmod x^{d+1}$ where $g\_d(x)$ is the truncation of $g(x)$ to the sum of its terms up through degree $d$.
Example: if $f(y) \in \mathbf Z\_p[[y]] \subset (\mathbf Z\_p[[x]])[[y]]$ then
$f((1+x)^a - 1)$ makes sense in $\mathbf Z\_p[[x]]$ since $|(1+x)^a - 1|\_x < 1$, and that's because
$(1+x)^a - 1 = \sum\_{n \geq 1} \binom{a}{n}x^n$ has constant term $0$.
If $A$ is an integral domain (like $\mathbf Z\_p$ or $\mathbf Q\_p$) and $f(y) = \sum\_{n \geq 0} c\_ny^n$ is a genuine power series in $A[[y]]$ -- it has infinitely many nonzero coefficients in $A$ -- then $|c\_nr^n|\_x = |c\_n|\_x|r|\_x^n = |r|\_x^n$ infinitely often (whenever $c\_n \not= 0$), and this tends to $0$ only if $|r|\_x < 1$, so $f(r)$ converges *if and only if* $|r|\_x < 1$. That means $f(g(x))$ converges $x$-adically in $A[[x]]$ if and only if $g(x) \in A[[x]]$ has constant term $0$.
**Remark 1**. Armed with the $x$-adic topology, we gain access to Hensel's lemma and topological arguments that can replace tedious degree-by-degree arguments in the combinatorial method. (It is not lost on me that the proof of Hensel's lemma itself is based on a degree-by-degree argument, so my point of comparison between Methods 1 and 2 is that Method 1 uses degree-by-degree arguments *very* often, while Method 2 can often avoid them with simple continuity arguments.) For example, the formal derivative on $A[[x]]$ is $x$-adically uniformly continuous (since
$|f'|\_x \leq (1/2)|f|\_x$), so to prove that the usual rules of formal derivatives are valid on $A[[x]]$, by its continuity and the denseness of $A[[x]]$ we are reduced to verifying the rules on polynomials, where the rules are "already known". There is no need to prove the identities on actual power series, but you may need to be more careful in the case of proving the chain rule to reduce to the polynomial case.
**Remark 2**. Above I am *not* saying $f(g(x))$ can never make sense in $A[[x]]$ if $f(x)$ is a genuine power series (not a polynomial) and $g(x)$ has a nonzero constant term, but only that it does not make $x$-adic sense (if $A$ is an integral domain). If $A = \mathbf Z\_p$, so $A$ itself has a nontrivial notion of convergence, then you may be able to evaluate $f(g(x))$ when $g(x)$ has a nonzero constant term $b\_0$ such that $f(b\_0)$ converges in $A$. For example, if $f(y) = 1 + y + y^2 + y^3 + \cdots$ has all coefficients equal to $1$ then $f(p+x)$ makes sense in $\mathbf Z\_p[[x]]$ but *not* with the $x$-adic topology since $|p+x|\_x = 1$. You'd have to use something like the $(p,x)$-adic topology instead, where elements of $\mathbf Z\_p[[x]]$ are small not just if they start off with a large power of $x$, but if they are in a high power of the ideal $(p,x)$, meaning below some high power of $x$ the coefficients are very divisible by $p$. Since $(x) \subset (p,x)$ we have $(x)^n \subset (p,x)^n$, so $x$-adically small power series in $\mathbf Z\_p[[x]]$ are also $(p,x)$-adically small, but not conversely (e.g., $p^n+x \to x$ in the $(p,x)$-adic topology but not in the $x$-adic topology).
| 6 | https://mathoverflow.net/users/3272 | 422393 | 171,766 |
https://mathoverflow.net/questions/422388 | 4 | Let $X$ be a variety over a field $k$ with separable closure $k\_s$. Let $A$, $B$ be étale sheaves on $X$. Consider now the Hochschild–Serre spectral sequences.
\begin{align\*}
E\_2^{pq}: H^p(k, H^q(X\_{k\_s}, A)) & {}\Rightarrow H^{p + q}(X, A) \\
E\_2^{pq}: H^p(k, H^q(X\_{k\_s}, B)) & {}\Rightarrow H^{p + q}(X, B) \\
E\_2^{pq}: H^p(k, H^q(X\_{k\_s}, A \otimes B)) & {}\Rightarrow H^{p + q}(X, A \otimes B).
\end{align\*}
The cup product for $H^{\bullet}(X\_{k\_s}, \cdot)$ gives a bilinear pairing $H^q(X\_{k\_s}, A) \times H^{q'}(X\_{k\_s}, B) \to H^{q + q'}(X\_{k\_s}, A \otimes B)$. This induces via the cup product on $H^{\bullet}(k, \cdot)$ a bilinear pairing $E^{pq}\_2 \times E^{p' q'}\_2 \to E^{(p + p')(q + q')}\_2$. This should be the same bilinear pairing (up to a sign) as the one coming from the usual cup product pairing $H^{p + q}(X, A) \times H^{p' + q'}(X, B) \to H^{p + p' + q +q'}(X, A \otimes C)$ (part of the statement being that the cup product is compatible with the filtration).
For the Serre spectral sequence in algebraic topology this is described in [Hatcher - Algebraic topology - Spectral sequences](https://pi.math.cornell.edu/%7Ehatcher/AT/ATch5.pdf) page 543 or, [Hutchings - Introduction to spectral sequences](https://math.berkeley.edu/%7Ehutching/teach/215b-2011/ss.pdf). Does anyone know of a good reference where this is explained for the Hochschild–Serre spectral sequence? The Leray spectral sequence would also be fine, since Hochschild–Serre is a special case. Especially the signs seems like they might be subtle.
| https://mathoverflow.net/users/479261 | The Hochschild–Serre spectral sequence and cup products | The original reference is
```
Hochschild, G.; Serre, J.-P.
Cohomology of group extensions.
Trans. Amer. Math. Soc. 74 (1953), 110–134.
```
On page 118 they introduce a filtration $\{A\_j\}\_j$ of the cochain complex $A = C^\*(G; M)$, which is compatible with cup products pairings and gives the desired pairing of Hochschild-Serre spectral sequences, as stated on page 119. The same kind of filtration was used by Serre in the context of cubical cohomology to define the multiplicative structure on the Serre spectral sequence of a fibration. Perhaps Chapters 5 and 6 of
<https://www.uio.no/studier/emner/matnat/math/MAT9580/v21/dokumenter/spseq.pdf>
are helpful.
| 3 | https://mathoverflow.net/users/9684 | 422394 | 171,767 |
https://mathoverflow.net/questions/422226 | 6 |
>
> Consider the affine space given by four $2\times 2$ matrices, i.e., $\mathbb{A}^{16}\cong M(\mathbb{C})\_{2\times 2}^4$. Now, consider the algebraic set $V$ given by the vanishing of the relation $AB-CD=0$, where the matrices are as follows: $A=(a\_{ij}), B=(b\_{ij}), C=(c\_{ij})$ and $D=(d\_{ij})$.
>
>
>
In other words, I have the following ring
$$R=\mathbb{C}[a\_{11},a\_{12},a\_{21},a\_{22}, b\_{11},\dotsc, d\_{21},d\_{22}]/I,$$ where $$I=(a\_{11}b\_{11}+a\_{12}b\_{21}−c\_{11}d\_{11}−c\_{12}d\_{21},\,a\_{11}b\_{12}+
a\_{12}b\_{22}−c\_{11}d\_{12}−c\_{12}d\_{22},\,a\_{21}b\_{11}+a\_{22}b\_{21}−c\_{21}d\_{11}−c\_{22}d\_{21},\,a\_{21}b\_{12}+a\_{22}b\_{22}−c\_{21}d\_{12}−c\_{22}d\_{22}).$$
The question is:
>
> Prove that $\overline{a}\_{11}$ is a prime element in $R$.
>
>
>
I don't know if there are any nice trick/strategy using commutative algebra (maybe using some change of coordinate or defining some norm of the ideal).
I'm thinking in terms of quiver representation. The above setup can be interpreted as:
$$\require{AMScd}\begin{CD}
\mathbb{C}^2 @>B>> \mathbb{C}^2\\
@VDVV @VVAV \\
\mathbb{C}^2 @>>C> \mathbb{C}^2.
\end{CD}$$
The ring $R$ is the coordinate ring of the representation space of the above quiver with relation, where the dimension vector is $(2,2,2,2).$
Any idea/suggestion will be apreciated.
| https://mathoverflow.net/users/338456 | Prove that $\overline{a}_{11}$ is a prime element in $R$ | Now that I have thought about this further, I realize that you need much less than the Samuel Conjecture to solve this problem. By a dimension count, the ring $R$ is a complete intersection ring, hence Cohen-Macaulay. Thus, by the Unmixedness Theorem, to prove that the ideal $I=\langle a\_{1,1} \rangle$ is prime, it suffices to prove that the corresponding closed subscheme of $\text{Spec}(R)$ is irreducible and generically reduced. This can be proved using Bertini's Theorem and a homogeneity argument.
| 1 | https://mathoverflow.net/users/13265 | 422397 | 171,770 |
https://mathoverflow.net/questions/422346 | 3 | I have recently read a paper by Talagrand: Embedding subspaces of $L\_{1}$ into $l^{N}\_{1}$, and part of the chapter: Finite dimensional subspaces of $L\_{p}$, written by Jonson and Schechtman. The theory of finite dimensional subspaces of $L\_{p}$ is beautiful and fruitful, which has been well studied.
It is quite natural to expect similar results for noncommutative Lp spaces (or, more specifically, the Schatten p-classes, denoted by $S\_{p}$). However, it seems that the theory of finite dimensional subspaces of Schatten p-classes is not well developed.
So, I am pretty interested in this topic. Are there any related references to embedding finite dimensional subspaces of $S\_{p}$ into $S^{N}\_{p}$? And what is the difficulty of such investigations in noncommutative settings?
PS: $S\_{p}$ (resp. $S^{N}\_{p}$) is the space of all infinite (resp. $N\times N$) matrices equipped with the Schatten-p norm $\|A\|\_{p}=(Tr(|A|^{p}))^{1/p}$ with $|A|=(A^{\*}A)^{1/2}$.
| https://mathoverflow.net/users/129593 | Embedding finite dimensional subspaces of Schatten p-classes | See
Schechtman, Gideon (IL-WEIZ)
Three observations regarding Schatten p classes. (English summary)
J. Operator Theory 75 (2016), no. 1, 139–149.
46B28 (47B10)
| 5 | https://mathoverflow.net/users/2554 | 422404 | 171,772 |
https://mathoverflow.net/questions/422399 | 4 | What is know about the homotopy groups of $S/3$ where $S/3 = \mathrm{hocofib}(S \xrightarrow{\cdot 3} S)$? Otherwise, is there some reference I can consult for the $BP$-ANSS for $S/3$?
| https://mathoverflow.net/users/114267 | $BP$-Adams Novikov Spectral Sequence or Homotopy groups of $S/3$ | For $3$-primary homotopy of $S$ there is early work by
```
Nakamura, Osamu
Some differentials in the mod 3 Adams spectral sequence.
Bull. Sci. Engrg. Div. Univ. Ryukyus Math. Natur. Sci. No. 19 (1975), 1–25.
```
and
```
Tangora, Martin (4-OX)
Some homotopy groups mod 3. Conference on homotopy theory (Evanston, Ill., 1974), 227–245,
Notas Mat. Simpos., 1, Soc. Mat. Mexicana, México, 1975.
```
through degree 103, extended to degree 108 by Ravenel in his green book, but beware of some issues in that extended range. There is ongoing work by Eva Belmont on this that uses modern machine computations.
From these 3-local results you can backtrack to extract quite a lot of information on $\pi\_\*(S/3)$. If you need to go further, maybe try
```
Arita, Yoshiko (J-HROSES); Shimomura, Katsumi (J-TOTED)
The chromatic E1-term H1M11 at the prime 3.
Hiroshima Math. J. 26 (1996), no. 2, 415–431.
Arita, Yoshiko (J-HROSES); Shimomura, Katsumi (J-TOTED)
On products of some β-elements in the homotopy of the mod 3 Moore spectrum.
Hiroshima Math. J. 27 (1997), no. 3, 477–486.
Shimomura, Katsumi (J-KOCHS)
The homotopy groups of the L2-localized mod 3 Moore spectrum.
J. Math. Soc. Japan 52 (2000), no. 1, 65–90.
```
and compare with Nassau's Adams spectral sequence charts at
[http://nullhomotopie.de/charts](http://nullhomotopie.de/charts/m3ext0.pdf)
<http://nullhomotopie.de/charts/m3ext0.pdf>
| 8 | https://mathoverflow.net/users/9684 | 422409 | 171,775 |
https://mathoverflow.net/questions/421182 | 5 | Joint with Qing-Hu Hou at Tianjin Univ., we seek for explicit criteria via coefficients for the solvability of an algebraic equation by radicals. In this direction, we formulate the following conjecture.
**Conjecture.** Let $p\ge5$ be a prime. Suppose that
$$f(x)=ax^p+bx^{p-1}+cx+d$$ is irreducible over $\mathbb Q$, where $a,b,c,d$ are pairwise coprime integers with $a\not=0$. Then the equation $f(x)=0$ over $\mathbb Q$ is not solvable by radicals.
We have checked this for $p=5,7$ and $|a|+|b|+|c|+|d|\le 108$. For example, the Galois group of the polynomial $$x^5+3x^4+5x+23$$ over $\mathbb Q$ is the alternating group $A\_5$ which is not solvable, and the Galois group of the polynomial $$x^7+x^6-49x-25$$ over $\mathbb Q$ is the alternating group $A\_7$.
**QUESTION.** Can one prove or disprove the above conjecture?
Your comments are welcome!
| https://mathoverflow.net/users/124654 | On the solvability of the equation $ax^p+bx^{p-1}+cx+d=0$ by radicals | This answer is experimentally driven. I tried to construct as many as possible solvable irreducible polynomials of the shape $f=x^5+bx^4+cx+d$ with integers $b,c,d$, then search for a message or a counterexample in the produced experimental list. The condition of having *pairwise coprime* integers turned out,
after seeing the examples,
to be very restricting for them, many examples have by the statistical "pseudo-law of small numbers" some common prime divisor of two coefficients.
(This condition leads formally immediately to $b,c,d\ne 0$, so it rules out something like $x^p-2$. The condition $a\ne 0$ was explicitly given for $f=ax^p+bx^{p-1}+cx+d$, and if i correctly get the message, it is superfluous.)
The counterexample found by chance was
$$
\color{blue}{
f = x^5 +3x^4+23x+1301\ ,
}
$$
the principal coefficient is one, the others are different prime numbers.
Computer check:
```
R.<x> = PolynomialRing(QQ)
f = x^5 + 3*x^4 + 23*x + 1301
G = f.galois_group()
print(f'f = {f}')
print(f'G = Gal(f) has order {G.order()}')
print(f'G has structure {G.structure_description()}')
print(f'Is G solvable? {G.is_solvable()}')
```
Results:
```
f = x^5 + 3*x^4 + 23*x + 1301
G = Gal(f) has order 10
G has structure D5
Is G solvable? True
```
The reader in hurry may please stop here.
---
---
Since i was trying to understand structurally as much as possible, and since the byproduct examples can be interesting also elsewhere, here are some details on the way the above $f$ was produced.
The search for solvable irreducible $f$ polynomials with $p=5$, $a=1$, was done along the usual path. While searching for a reference i could join, i found:
* [LJF2004] the master thesis of Lau Jing Feng [On Solvable Septics](https://core.ac.uk/download/pdf/48627385.pdf)
* [AS2022] a preprint of Abdeljalil Saghe, researchgate provides the link <https://www.researchgate.net/publication/358595406> for it.
* [SW1996] Spearman and Williams, [On Solvable Quintics...](https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-26/issue-2/On-Solvable-Quintics-X5aXb-and-X5aX2b/10.1216/rmjm/1181072083.full), Rocky Mountain Journal of Mathematics, Volume 26, Number 2, 753-772, 1996. (The link provides a [Download PDF] button.) See also the wiki note on this: [Quintics in Bring-Jerrard form](https://en.wikipedia.org/wiki/Quintic_function#Quintics_in_Bring%E2%80%93Jerrard_form).
It is a good idea to check first the **Proposition** from [SW1996] at page 754. This is cited with variations in the other sources. And [SW1996] further delegated the proof at an AMM article from 1994.
For the convenience of the reader i am reproducing here from [AS2022] a theorem, it also appears as a short mention in the same form in [LJF2004]. The latter points explicitly to [SW1996]. Although it is formulated for trinomial quintics (with a missing term in $x^4$), it may be useful to have an explicit example, and see the structure of the solution and the implications for the coefficients in the trinomial. Please skip if this feels not related to the question.
---
**Theorem 13.8** in [LJF2004]: Let $a$ and $b$ be rational numbers
such that the quintic trinomial
$
x^5 + ax + b
$ is irreducible.
Then, the equation
$$x^5 + ax + b = 0$$
is solvable
by radicals if and only if there exist rational numbers
$\epsilon$ among $\pm 1$, $c\ge 0$, and $e\ne 0$ such that
$$
\tag{$13.9$}
\begin{aligned}
a &= \frac 1D\cdot 5e^4(3 − 4\epsilon c)\ ,\\
b &= −\frac 1D\cdot 4e^5(11\epsilon + 2c)\ ,\text{ where $D$ is }\\
D &:= c^2 + 1\ ,
\end{aligned}
$$
in which case the roots of $x^5 + ax + b = 0$ are
$$
\tag{$13.10$}
x\_j = e\Big(\ \omega^j u\_1
+ \omega^{2j} u\_2
+ \omega^{3j} u\_3
+ \omega^{4j} u\_4\ \Big)\ ,\ j = 0, 1, 2, 3, 4,
$$
for $j$ among $0,1,2,3,4$. Here
$\omega$ is a primitive fifth root of the unit,
$\omega^5=1$, $\omega\ne 1$, and
$$
\tag{$13.11$}
$$
$$
u\_1 = \left(\frac{v\_1^2v\_3}{D^2}\right)^{1/5}\ ,\
u\_2 = \left(\frac{v\_3^2v\_4}{D^2}\right)^{1/5}\ ,\
u\_3 = \left(\frac{v\_2^2v\_1}{D^2}\right)^{1/5}\ ,\
u\_4 = \left(\frac{v\_4^2v\_2}{D^2}\right)^{1/5}\ ,
$$
and
$$
\tag{$13.12$}
$$
$$
\begin{aligned}
v\_1 &= +\sqrt D + \sqrt{D-\epsilon\sqrt D}\ ,\\
v\_2 &= -\sqrt D - \sqrt{D+\epsilon\sqrt D}\ ,\\
v\_3 &= -\sqrt D + \sqrt{D+\epsilon\sqrt D}\ ,\\
v\_4 &= +\sqrt D - \sqrt{D-\epsilon\sqrt D}\ .
\end{aligned}
$$
---
It is maybe best to have a concrete example for this situation,
my choice of the computational weapon is [sage](https://www.sagemath.org).
```
eps, c, e = 1, 3/5, 34
D = c^2 + 1
a, b = 5*e^4*(3 - 4*eps*c)/D, -4*e^5*(11*eps + 2*c)/D
R.<x> = PolynomialRing(QQ)
f = x^5 + a*x + b
G = f.galois_group()
print(f'f = {f}')
print(f'Is f irreducible? {f.is_irreducible()}.')
print(f'Let G be the Galois group associated to f. Its order is {G.order()}.')
print(f'Is G solvable? {G.is_solvable()}.')
print(f'Structurally, G is {G.structure_description()}.')
```
Results:
```
f = x^5 + 2947800*x - 1630329920
Is f irreducible? True.
Let G be the Galois group associated to f. Its order is 20.
Is G solvable? True.
Structurally, G is C5 : C4.
```
We can also ask for the field in between.
```
K.<v> = NumberField(f)
H.<w> = K.galois_closure()
subfields_data = H.subfields(degree=4)
subfield_data = subfields_data[0]
F, mor, nothing = subfield_data
F.inject_variables()
print('Let H be the Galois closure of f.')
print('Then H has a subfield F of degree 4:')
print(F)
```
This gives:
```
Defining w0
Let H be the Galois closure of f.
Then H has a subfield F of degree 4:
Number Field in w0 with defining polynomial
x^4 + 51114852000000*x^2 + 188270112726637200000000000
```
We can ask for the roots of the above polynomial, in a dialog with the sage interpreter:
```
sage: var('T');
sage: pol = F.defining_polynomial()
sage: for sol in solve(pol(T) == 0, T, solution_dict=True):
....: print(sol[T].simplify_full())
....:
-173400*sqrt(5)*sqrt(11*sqrt(170) - 170)
173400*sqrt(5)*sqrt(11*sqrt(170) - 170)
-173400*I*sqrt(5)*sqrt(11*sqrt(170) + 170)
173400*I*sqrt(5)*sqrt(11*sqrt(170) + 170)
```
Yes, we have an irreducible, solvable $f$, but the coefficients come with
many common primes. This can be traced back to the way they appear after fixing the parameters $\epsilon,c,e$.
---
The case of a more general quintic is similar. For a solvable, irreducible quintic
$$
f = x^5 +bx^4 + cx+d\ ,\qquad b,c,d\in\Bbb Q\ ,\ bcd\ne 0\ ,
$$
the Galois group is a subgroup of the Frobenius group
$F\_{5\cdot 4}=F\_{20}$.
Let $x\_1,x\_2,x\_3,x\_4,x\_5$ be the roots of $f$. Consider
$$
\theta\_1 :=\sum x\_2^2(x\_1x\_3+x\_4x\_5)\ ,
$$
the sum being cyclic. Consider $\theta\_1,\theta\_2,\dots,\theta\_6$ obtained from $\theta\_1$ by acting with permutations representing the cosets of $S\_5/F\_{20}$. Then the polynomial
$$
f\_{20}=f\_{20}(T)=\prod\_{1\le j\le6}(T-\theta\_j)
$$
is symmetric in the $f$-roots, and can be rewritten as a polynomial in $T;b,c,d$. (See also §12 in [AS2022].) Then $f$ is solvable iff the associated $f\_{20}$ has a rational root. Here is the explicit form of $f\_{20}$ computed in sage:
$$
\begin{aligned}
f\_{20}(T)
&=
T^6 + A\_1T^5 + A\_2T^4+ A\_3T^3+A\_4T^2 +A\_5T+A\_6\ ,\text{ where}\\
A\_1 &= 8c\ ,
\\
A\_2 &=-8b^3d + 40c^2\ ,
\\
A\_3 &=-2b^4c^2 - 44b^3cd - 50b^2d^2 + 160c^3
\\
A\_4 &=16b^6d^2 - 8b^4c^3 - 144b^3c^2d - 200b^2cd^2 + 400c^4
\\
A\_5 &=8b^7c^2d + 48b^6cd^2 + 3b^4c^4 - 56b^5d^3 - 364b^3c^3d - 550b^2c^2d^2 \\
&\qquad + 512c^5 + 2500bcd^3 - 3125d^4
\\
A\_6 &=b^8c^4 + 12b^7c^3d + 86b^6c^2d^2 + 17b^4c^5 - 468b^5cd^3\\
&\qquad - 276b^3c^4d + 625b^4d^4 - 1450b^2c^3d^2 + 256c^6 + 7500bc^2d^3 - 9375cd^4
\ .
\end{aligned}
$$
Then some random search for cases when $f\_{20}(T)$ has at least one rational root (this is the one and only thing we need) delivered the following tuples
$(a,b,c,d)$ with $a=1$, together with the corresponding $T$-root:
$$
\begin{array}{|c|c|c|c||c|}
\hline
a & b & c & d & T\\\hline
1 & 1 & -13 & 13 & 52\\\hline
1 & 1 & 30 & -10 & -80\\\hline
1 & 1 & 85 & -340 & -255\\\hline
1 & 1 & 173 & 1038 & -519\\\hline
1 & 1 & 197 & -1182 & 1182\\\hline
%
1 & 2 & -360 & 880 & 1120 \\\hline
1 & 2 & -208 & 416 & 832\\\hline
1 & 2 & 26 & 26 & -26\\\hline
1 & 2 & 208 & 104 & -416\\\hline
1 & 2 & 212 & -636 & - 636 \\\hline
1 & 2 & 320 & -240 & -880\\\hline
1 & 2 & 480 & -320 & -1280 \\\hline
1 & 2 & 728 & 624 & -1248 \\\hline
%
1 & 3 & -42 & -534 & 144\\\hline
1 & 3 & -13 & 13 & 52\\\hline
1 & 3 & 6 & -6 & 0\\\hline
1 & 3 & 13 & -39 & -26\\\hline
1 & 3 & 15 & 15 & 0\\\hline
\color{blue}{1} & \color{blue}{3} & \color{blue}{23} & \color{blue}{1301} & 1612\\\hline
1 & 3 & 54 & 90 & 0\\\hline
1 & 3 & 135 & -135 & 0\\\hline
%
1 & 4 & -40 & 160 & 160 \\\hline
1 & 4 & 116 & 174 & - 348 \\\hline
1 & 4 & 148 & 368 & - 408 \\\hline
1 & 4 & 208 &-208 & - 624 \\\hline
1 & 4 & 312 & 728 & - 104 \\\hline
1 & 4 & 416 & 832 & - 416 \\\hline
1 & 4 & 1010 & 909 & -2121 \\\hline
%
1 & 5 & -349 & 1396 & 2443\\\hline
1 & 5 & -65 & 143 & 520\\\hline
1 & 5 & -50 & 238 & 240\\\hline
1 & 5 & -25 & 275 & 700\\\hline
1 & 5 & -10 & 6 & 80\\\hline
1 & 5 & 5 & 3 & 30\\\hline
1 & 5 & 5 & 15 & 110\\\hline
1 & 5 & 30 & 46 & 160\\\hline
1 & 5 & 55 & 67 & 60\\\hline
1 & 5 & 61 & -61 & -122\\\hline
1 & 5 & 65 & 91 & 130\\\hline
1 & 5 & 80 & 48 & -240\\\hline
1 & 5 & 85 & -1717 & 1870\\\hline
1 & 5 & 85 & 34 & -255\\\hline
1 & 5 & 85 & 68 & -255\\\hline
1 & 5 & 95 & -705 & 840\\\hline
1 & 5 & 100 & 20 & -300\\\hline
1 & 5 & 100 & 100 & -300\\\hline
1 & 5 & 100 & 164 & 100 \\\hline
1 & 5 & 125 & 150 & -375 \\\hline
1 & 5 & 135 & 81 & -360 \\\hline
1 & 5 & 135 & 567 & 720 \\\hline
1 & 5 & 200 & 432 & -100 \\\hline
1 & 5 & 205 & -82 & -615 \\\hline
1 & 5 & 205 & 328 & -615 \\\hline
1 & 5 & 157 & 314 & -314 \\\hline
1 & 5 & 160 & -32 & -480 \\\hline
1 & 5 & 160 & 224 & -480 \\\hline
%
1 & 6 & -72 & -1008 & 864\\\hline
1 & 6 & 96 & -192 & 0\\\hline
1 & 6 & 120 & -420 & 360\\\hline
1 & 6 & 135 & 270 & 0\\\hline
%
1 & 7 & 37 & 37 & 222 \\\hline
%
1 & 10 & -160 & 192 & 1280\\\hline
1 & 10 & -40 & 176 & 1120\\\hline
1 & 10 & -10 & 126 & 530\\\hline
1 & 10 & 80 & 96 & 480\\\hline
1 & 10 & 135 & 162 & 720\\\hline
%
1 & 12 & -72 & 96 & 864 \\\hline
%
1 & 15 & -153 & 153 & 1836\\\hline
1 & 15 & -105 & 93 & 1440\\\hline
1 & 15 & -25 & 25 & 700 \\\hline
1 & 15 & -15 & 141& 1170\\\hline
1 & 15 & 15 & 69 & 1080\\\hline
1 & 15 & 85 & 204 & 1870\\\hline
\end{array}
$$
(The above list shows only some solutions, it is not systematic, i picked some entries for my taste and purposes.)
---
---
I was also looking for examples in degree $p=7$.
This is hard and becomes harder and harder when $p$ gets bigger.
Still, i'll try to say some words on this case.
In the *solvable* case, for $p=7$, the related Galois group associated to $f$ is and must be a subgroup of the Frobenius group $F\_{42}$, this is [LJF2004] Corollary **2.1.7**. The following post was targeting the solvable trinomials $f=x^7+ax+b$: [Is there an irreducible but solvable septic trinomial...?](https://mathoverflow.net/questions/146769/is-there-an-irreducible-but-solvable-septic-trinomial-x7axnb-0/146780) Given the answers in *loc. cit.*, to obtain
more, we have to use the one more term in $bx^6$ allowed in
$a^7+bx^6+cx+d$, i will eventually search next days for a solution.
The problem is that the group of Galois symmetries should "drop" from $5040=7!=|S\_7|$ to such a small divisor like $42$ (or even below it).
In order to obtain examples, instead of the polynomial $f\_{20}=\prod(T-\theta\_i)$ of degree $|S\_5|/|F\_{20}|=5!/20=3!=6$ for the quintic case we have a polynomial $f\_{42}=\prod(T-\theta\_i)$ of degree $|S\_7|/|F\_{42}|=7!/42=5!=120$ for the septic case, one can compute it, then search for cases where it has a rational root.
Examples with a slightly bigger divisor like $168$ and decent coefficients can be given. A simple trinomial example with this bigger, but **not solvable** Galois order $168=4\cdot 42$ is $g = x^7 - 154x + 99$, computer check for it:
```
sage: g = x^7 -154*x + 99
sage: G = g.galois_group()
sage: G.order()
168
sage: G.structure_description()
'PSL(3,2)'
sage: G.is_solvable()
False
```
| 8 | https://mathoverflow.net/users/122945 | 422412 | 171,778 |
https://mathoverflow.net/questions/422416 | 12 | In my recent research, I need to know if the symmetric spaces $\operatorname{SU}(n)/{\operatorname{SO}(n)}$ are always nontrivial in the unoriented and oriented bordism rings for $n>2$. (For the motivation, by Benny Cheng, the cones over them are special Lagrangians in $\mathbb{C}^{n^2+n}$. I want to know if the cones are topologically smoothable. The $n=3$ case comes from Haskins & Pacini.)
I know that for $n=3$ we are dealing with the Wu manifold, and there are many references here. For $n>3$, I tried to find references about the Stiefel-Whitney numbers of these, which by Thom, is enough for the unoriented case. However, I've only been able to locate the cohomology ring of these in Topology of Lie Groups I & II by Mimura and Toda. I don't know much algebraic topology, but I think they did not state what the Stiefel–Whitney classes are.
| https://mathoverflow.net/users/482183 | Are the symmetric spaces $\operatorname{SU}(n)/{\operatorname{SO}(n)}$ always nontrivial in the bordism rings for $n>2$? | There is a fibration $SU(n) \overset p\to SU(n)/SO(n) \overset j\to BSO(n)$, where the $j$ is the classifying map of $p$, viewed as (the projection of) a principal $SO(n)$-bundle. The Stiefel–Whitney classes for your Wu-esque manifolds are the $j^\*$-images of the universal Stiefel–Whitney classes; this is in Borel and Hirzebruch's "Characteristic classes and homogeneous spaces I."
To find a nonzero Stiefel–Whitney number is to find a product of these classes of total degree $\dim SU(n)/SO(n)$. The computation in Mimura–Toda shows that the cohomology over $\mathbb F\_2$ is an exterior algebra on one generator each of degrees $2$ through $n$, and that these are the images of the universal Stiefel–Whitney classes other than $w\_1$. The product of these generators thus does represent the fundamental class.
| 15 | https://mathoverflow.net/users/5792 | 422417 | 171,780 |
https://mathoverflow.net/questions/422345 | 1 | Let $|\cdot|$ denote the usual Euclidean norm on $\mathbb{C}^3$ and fix some arbitrary metric $\rho$ on $\mathbb{CP}^2$. For $\delta > 0$ and any set $\hat{P} \subset \mathbb{CP}^2$, define the $\delta$-neighborhood of $\hat{P}$ by
$$N\_{\delta}(\hat{P})= \{\hat{x} \in \mathbb{CP}^2: \rho(\hat{s},\hat{x})<\delta \text{ for some } \hat{s} \in \hat{P}\}.$$
Here, the hat notation refers to the obvious projection that sends nonzero vectors in $\mathbb{C}^3$ to their equivalence class in complex projective space via $v = \lambda w$ for some $\lambda \in \mathbb{C}-\{0\}$.
Now let's fix some nonzero $y\in \mathbb{C}^3$ with unit modulus and fix a complex 2-plane $\Sigma \subset \mathbb{C}^3$ going through the origin. We can write $y = v +w$ with $v\in \Sigma$ and $w\in \Sigma^{\perp}$. I'm reading a paper that uses the following fact without proof:
>
> If $\hat{y} \notin N\_{\delta}(\hat{\Sigma})$, then there is a constant $K>0$, depending only on the metric $\rho$, so that $|w| \geq K \delta$.
>
>
>
Essentially the fact is saying that if a vector lies far away from a plane at the projectivized level, then the component orthogonal to the plane can't be too small (up to a multiplicative constant that depends on the details of how you measure a vector being 'far away' from a plane).
I'd love to see some sketch of a proof for this.
I can sort of buy it for particular choices of metric like Fubini-Study, but the statement is made for arbitrary metric, so I imagine there is a slick proof using only general facts about how/whether metrics on $\mathbb{CP}^2$ "distort" if lifted to $\mathbb{C}^3-\{0\}$. I suppose there may be an equivalent formulation of this statement for $\hat{y} \in N\_{\delta}(\hat{\Sigma})$ (with the inequality switched and strict), but let me leave it as written in the paper.
| https://mathoverflow.net/users/34409 | Lower bound on a norm of $\mathbb{CP}^2$ inducing a lower bound on the Euclidean norm of $\mathbb{C}^3$ | Hmm, I think I've worked my way to exactly such a 'slick' proof:
Suppose that there is no such uniform constant. Then for each $L > 0$, we may find some $\hat{y} \notin N\_{\delta}(\hat{P})$ with $|w| < L \delta$, where $w$ denotes the perpendicular component of a representative $y\in \hat{y}$ having unit modulus. Indeed, for each positive integer $n$ there exists $\hat{y}\_n \notin N\_{\delta}(\hat{P})$ with $|w\_n| < \delta/n$. This gives us a sequence $\{y\_n\}$ on the unit sphere. Choose a convergent subsequence, again denoted $\{y\_n\}$ with $y\_n \to y\_\*$. Clearly the perpendicular component of $y\_\*$ must be zero. Hence $y\_\* \in P$, and thus $\hat{y}\_\* \in N\_{\delta}(P)$.
On the other hand, $\hat{y}\_n \to \hat{y}\_\* \notin N\_{\delta}(P)$ since the quotient map is continuous and the complement of $N\_{\delta}(P)$ is closed, hence containing all of its limit points. We have obtained our contradiction.
| 0 | https://mathoverflow.net/users/34409 | 422423 | 171,781 |
https://mathoverflow.net/questions/422407 | 5 | In the [folk model structure](https://arxiv.org/abs/0712.0617) on the category $sCat\_\omega$ of strict $\omega$-categories, the cofibrations are generated by the boundary inclusions $\{\partial \mathbb G\_n \to \mathbb G\_n \mid n \in \mathbb N\}$, where $\mathbb G\_n$ is the $n$-globe and $\partial \mathbb G\_n$ is its boundary.
**Question:** Is every folk cofibration a monomorphism in $sCat\_\omega$?
Folk cofibrations (or at least the cellular ones) are also called "relative computads / polygraphs".
I'm pretty sure the answer is *yes*, but I've been unable to find a reference.
Note that [as mentioned here](https://mathoverflow.net/questions/416279/are-monomorphisms-of-strict-omega-categories-stable-under-pushout-along-folk), monomorphisms in $sCat\_\omega$ are not stable under cobase change in general. So an affirmative answer doesn't follow immediately from knowing that the maps $\partial \mathbb G\_n \to \mathbb G\_n$ are monomorphisms.
| https://mathoverflow.net/users/2362 | Is every folk cofibration of strict $\omega$-categories a monomorphism? | I just thought (or maybe remember) a neat proof of this fact. It involve ideas I worked on a few years ago but never published - but that's short enough so that I can explain the key ideas on MO. Let me know if you need it : I might write up the details of this proof in a short paper if you want to have a reference for it...
Fix $T$ a topos (It will be globular set) and $M$ a [cartesian](https://ncatlab.org/nlab/show/cartesian+monad) monad on $T$ (it will be the free $\infty$-category monad - so that the category of $M$-algebra is the category of strict $\infty$-categories).
Let $\Omega$ be the sub-object classifier of $T$. Then I claim:
**Theorem 1 :** $\Omega$ has a $M$-algebra structure. Moreover as a $M$-algebra it classifies the "cartesian sub-objects" of $M$-algebras.
By a cartesian subobject we mean a subalgebra $X \subset A$ such that the inclusion morphism $A \to M$ is a cartesian morphism of $M$-algebra. That is, the square witnessing it is a morphism of algebras is a pullback square.
**Proof :** consider $E$ the category whose objects are monomorphisms in $T$ and whose morphisms are cartesian squares. $M$ acts on this category by sending $A \subset B$ to $M(A) \subset M(B)$, and because its unit and multiplication are cartesian transformation, they give a monad structure to $M$ acting on $E$.
Now the terminal object of $E$ is $1 \to \Omega$, it automatically inherits a $M$-algebra structure and is terminal amongst $M$-algebras in $E$.
One then easily check that a $M$-algebra in $E$ is exactly a $M$-algebra with a cartesian sub-algebra and hence the theorem follows.
*Remark : I had initially noticed this in the context of $\infty$-category a few years ago using the object classified instead of the subobject classifier. I used that to give a general proof that in an $\infty$-categorical context, if one defines "computads" for such a cartesian (or rather PRA) monad on a presheaf $\infty$-category then the category of computads is always a presheaf $\infty$-category. I gave a couple of talks about it in 2019 but never published the results because it seemed unclear what it was good for at the time. But it can be used to give a simple proof of the fact you mentioned.*
**Theorem 2 :** If one defines cofibration in the category of $M$-algebra to be the left class of the weak factorization system cofibrantly generated by the $M(A) \to M(B)$ for $A \subset B$ an inclusion in $T$. Then all cofibration are cartesian monomorphism in the category of $M$-algebra.
In particular, the cofibrations of the Folk model structure are cartesian monomorphisms.
**Proof :** Using that $\Omega$ classified cartesian monomorphism it is easy to check that cartesian monomorphism are stable under pushout and transfinite composition. For example if $U \to V$ is a cartesian mono, classified by $\chi : V \to \Omega$, and $U \to X$ is any map, then one can define a map $X \coprod\_U V \to \Omega$ that factor through to $1 \to \Omega$ on $X$ and is $\chi$ on $V$. One can then check\* that this map classifies $X$, and hence that $X \to X \coprod\_A B$ is a cartesian mono.
Finally, the $M(A) \to M(B)$ are cartesian monomorphism, so this conclude the proof.
(\*) : ok here I'm hiding a part of the proof. To do this, One need to use that if $T\_i$ is a diagram of $\infty$-category with a map $colim T\_i \to \Omega$ then the subobject it classifies is the colimit of the subobject classified by the $T\_i \to \Omega$. This is showed by generalizing theorem 1 to show that for every $\infty$-category $X$, the partial classifier $\tilde{X}$ is equiped with a $M$-algebra structure such that $M$-maps $B \to \tilde{X}$ correspodns to cartesian subobjects $A \subset B$ together with a $M$-map $A \to X$. This is proved exactly as Theorem 1 using the category $E$ whose object are inclusion $A \subset B$ together with a map $A \to X$ ($X$ being fixed), whose terminal object is $X \subset \tilde{X}$.
Once this is done, one can use it to show that the subobject classified by $colim T\_i \to \Omega$ as the correct universal property to be the colimit of the object classified by the $T\_i \to \Omega$ and really conclude the proof.
| 5 | https://mathoverflow.net/users/22131 | 422424 | 171,782 |
https://mathoverflow.net/questions/422420 | 2 | Let $p$ be a prime and $n$ a positive integer not divisible by $p$. When working on a fixed field in the cyclotomic field $Q(e^{2i\pi/n})$, I tumbled into the condition: $p$ does not divide $\frac{\phi(n)}{\mathrm{ord}(n)},$ where $\phi(n)$ is Euler's totient function, and $\mathrm{ord}(n)$ is the multiplicative order of $p$ mod $n$. That is, $\mathrm{ord}(n)$ is the smallest positive integer $k$ such that $p^k \equiv 1 (\bmod n)$. I would appreciate very much if you could provide some examples for this condition to be satisfied. A family (in either $p$ or $n$) of examples is even nicer.
| https://mathoverflow.net/users/64643 | Euler's totient phi and a prime | Below I write $\mathrm{ord}(x) $ for $\mathrm{ord}\_x(p)$. As usual, $\nu\_p(t)$ denotes the maximal $m$ for which $p^m$ divides $t$.
Let $n=\prod q\_i^{\alpha\_i}$ be a prime factorization of $n$. Then $\nu\_p(\phi(n))=\sum\_i \nu\_p(q\_i-1)$. On the other hand, $\mathrm{ord}(n)=\mathrm{lcm} \{\mathrm{ord}(q\_i^{\alpha\_i}):i=1,2,\ldots\}$. Since $\mathrm{ord}(q\_i^{\alpha\_i})$ divides $\phi(q\_i^{\alpha\_i})=(q\_i-1)q\_i^{\alpha\_i-1}$, we get $\nu\_p(\mathrm{ord}(n))\leqslant \max\_i \nu\_p(q\_i-1)$.
Therefore, if there exist at least two indices $i$ for which $p$ divides $q\_i-1$, we get $\nu\_p(\mathrm{ord}(n))<\nu\_p(\phi(n))$, and your condition does not hold.
If $p$ divides no $q\_i-1$, then your condition obviously holds as $p$ does not divide even $\phi(n)$ itself.
So, the interesting case to consider is when there exists unique $i$ for which $p$ divides $q\_i-1$. Say, $q\_i=1+p^Ab$, where $p$ does not divide $b$ (and $A>0$). Then your condition holds if and only if $p^A$ divides $\mathrm{ord}(q\_i^{\alpha\_i})$. Note that $\mathrm{ord}(q\_i)$ divides
$\mathrm{ord}(q\_i^{\alpha\_i})$, and the ratio is a certain power of $q\_i$ (indeed, $q\_i$ divides $p^r-1$, where $r=\mathrm{ord}(q\_i)$, hence $q\_i^2$ divides $p^{rq\_i}-1=(p^r-1)(1+p^r+\ldots+p^{r(q\_i-1)})$ and therefore $\mathrm{ord}(q\_i^2)$ divides $rq\_i$, and so on. This argument is an easy version of the so called Lifting the Exponent Lemma).
Thus, your condition holds if and only if $p^A$ divides $\mathrm{ord}(q\_i)$, in other words, if $q\_i$ does not divide $p^{(q\_i-1)/p}-1$. Or, in yet other words, if $p$ is not a $p$-th power modulo $q\_i$.
I doubt that there is a good characterization of such pairs $(p,q\_i)$. If $p=2$, this holds if and only if $2$ is not a quadratic residues modulo $q\_i$, that is, $q\_i$ is congruent to 3 or 5 modulo 8.
| 4 | https://mathoverflow.net/users/4312 | 422425 | 171,783 |
https://mathoverflow.net/questions/422325 | 7 | I am interested in the general theory of deformations locally ringed spaces in the same language of the deformation theory of schemes/varieties that is already widely available. I am aware for example of what is written down in the stacks project (for example [Tag 08UX](https://stacks.math.columbia.edu/tag/08UX)) but I'm failing to find a more through treatment specially one that would put in contrast the differences with the special case of schemes/varieties. In my literature review, however, I stumbled into the following quote in
*Lowen, Wendy; van den Bergh, Michel*, [**Deformation theory of abelian categories**](http://dx.doi.org/10.1090/S0002-9947-06-03871-2), Trans. Am. Math. Soc. 358, No. 12, 5441-5483 (2006). [ZBL1113.13009](https://zbmath.org/?q=an:1113.13009).
>
> These results confirm the fundamental insight of Gerstenhaber and Schack [6,
> 8] that one should define the deformations of a ringed space $(X, \mathcal{O}\_X)$ not as the deformations of $\mathcal{O}\_X$ as a sheaf of k-algebras, but rather as the deformations of the k-linear category $\mathfrak{u}$ (or of the “diagram” $(\mathcal{B}, \mathcal{O}\_{\mathcal{B}})$ in case $X\in \mathcal{B}$). These “virtual” deformations are nothing but the deformations of the abelian category $Mod(\mathcal{O}\_{X})$.
>
>
>
The cited category $\mathfrak{u}$ is an appropriately defined category which is used to show that the deformations of the category of presheaves of modules over an appropriate basis $\mathcal{B}$ are equivalent to deformations of sheaves of modules over $\mathcal{O}\_{X}$
I have skimmed through the literature including the cited papers and while I have found some indication of the exact meaning of this claim I am still confused. My interpretation is that the claim is either strictly noncommutative in nature ( so passing through a reconstruction theorem ), or deformations of this abelian category directly give the 'correct' deformation theory of the space in some sense ( for example the relationship with Hochschild cohomology in *Lowen, Wendy; van den Bergh, Michel*, [**Hochschild cohomology of Abelian categories and ringed spaces**](http://dx.doi.org/10.1016/j.aim.2004.11.010), Adv. Math. 198, No. 1, 172-221 (2005). [ZBL1095.13013](https://zbmath.org/?q=an:1095.13013).)
My first question would then be, could somebody clarify what exactly is the quote saying?
My second question is then, if the space is a (sufficiently nice) scheme then am I to understand that the deformation theory of the category of quasi-coherent sheaves controls the deformations of the scheme as I would find it written in classical texts?
Thanks in advance
The cited papers on the quote are, respectively:
*Gerstenhaber, M.; Schack, S. D.*, [**On the deformation of algebra morphisms and diagrams**](http://dx.doi.org/10.2307/1999369), Trans. Am. Math. Soc. 279, 1-50 (1983). [ZBL0544.18005](https://zbmath.org/?q=an:0544.18005).
and,
*Gerstenhaber, Murray; Schack, Samuel D.*, [**The cohomology of presheaves of algebras. I: Presheaves over a partially ordered set**](http://dx.doi.org/10.2307/2001114), Trans. Am. Math. Soc. 310, No. 1, 135-165 (1988). [ZBL0706.16021](https://zbmath.org/?q=an:0706.16021).
| https://mathoverflow.net/users/44499 | Deformation of (locally) ringed spaces and of their abelian categories of modules | As Jon Pridham notes in the comments, the quote should be understood noncommutatively. In fact, in the introduction Lowen and Van den Bergh write
>
> Deformation theory of abelian categories is important for non-commutative algebraic geometry. One of the possible goals of non-commutative algebraic geometry is to understand the abelian (or triangulated) categories which have properties close to those of the (derived) category of (quasi-)coherent sheaves on a scheme.
>
>
>
As you observe, this is a natural point of view when identifying a ringed space $(X, \mathcal O\_X)$ with its category of quasi-coherent sheaves by considering spectra of Abelian categories as in the [Gabriel–Rosenberg reconstruction theorem](https://ncatlab.org/nlab/show/Gabriel-Rosenberg+theorem).
To answer your second question, the deformations of the Abelian category $\mathrm{Qcoh} (X)$ of quasi-coherent sheaves contain classical deformations of $X$ as a special case, but in general admit *more* deformations than $X$.
The difference becomes quite clear when looking at the space of first-order deformations up to equivalence (i.e. the tangent space of the deformation functor).
For example, for **smooth** $X$ one gets
* $\mathrm H^1 (\mathcal T\_X)$ for deformations of $X$
* $\mathrm H\_{\mathrm{Ab}}^2 (\mathrm{Qcoh} (X)) \simeq \mathrm{HH}^2 (X)
\simeq \mathrm H^0 (\Lambda^2 \mathcal T\_X) \oplus \mathrm H^1 (\mathcal T\_X) \oplus \mathrm H^2 (\mathcal O\_X)$ for deformations of $\mathrm{Qcoh} (X)$, i.e. you get two new types of deformations: the direct summand $\mathrm H^0 (\Lambda^2 \mathcal T\_X)$ corresponds to quantizations of algebraic Poisson structures and $\mathrm H^2 (\mathcal O\_X)$ to twists of the structure sheaf.
For **affine** $X = \mathrm{Spec} (A)$ with $A$ commutative one gets
* $\mathrm{Har}^2 (A, A)$ for deformations of $X$ (Harrison / André–Quillen cohomology which controls *commutative* deformations of $A$ which are trivial if $A$ is smooth)
* $\mathrm{HH}^2 (A, A)$ for deformations of $\mathrm{Qcoh} (X)$ (Hochschild cohomology which controls *associative* deformations of $A$).
If you want to compare deformations of $\mathrm{Qcoh} (X)$ and $X$ on equal footing, you can look at deformations of $\mathcal O\_X \vert\_{\mathfrak U}$ for some affine open cover $\mathfrak U$ closed under $\cap$ either
* as a twisted presheaf of associative algebras — this is equivalent to deformations of $\mathrm{Qcoh} (X)$ as Abelian category by the results of Lowen and Van den Bergh
* as a presheaf of commutative algebras — this is equivalent to classical deformations of $X$ (see for example [1]).
The "insight" by Gerstenhaber and Schack was that the former type of deformations of the diagram $\mathcal O\_X \vert\_{\mathfrak U}$ give a useful generalization of "deformation" of a ringed space, and Lowen and Van den Bergh show that this corresponds to the deformation theory of $\mathrm{Qcoh} (X)$ as Abelian category.
[1] *Lepri, Emma; Manetti, Marco*, [**On deformations of diagrams of commutative algebras**](http://dx.doi.org/10.1007/978-3-030-37114-2_6), Colombo, Elisabetta (ed.) et al., Birational geometry and moduli spaces. Collected papers presented at the INdAM workshop, Rome, Italy, June 11–15, 2018. Cham: Springer. Springer INdAM Ser. 39, 77-107 (2020). [ZBL1440.13066](https://zbmath.org/?q=an:1440.13066).
| 4 | https://mathoverflow.net/users/163420 | 422432 | 171,784 |
https://mathoverflow.net/questions/422437 | 0 | I'm reading the proof of Kantorovich duality from Villani's book *Topics in Optimal Transportation*. In page 28, the author said that
>
> **Exercise 1.11.** Let us try to extend this proof to the non-compact case. Why would we like to replace $C\_{b}(X \times Y)$ by $C\_{0}(X \times Y) ?$ Show that if we do so in the definition of $\Xi$, then the latter turns out to be trivial: $\Xi \equiv 0$. This shows that the proof as such does not work in a non-compact setting. Still, a variation of it can be used, as we shall see later in Appendix 1.3.
>
>
>
Let $X$ and $Y$ be Polish spaces. Let $E:=C\_{0}(X \times Y)$ be the space of all continuous vanishing at infinity functions on $X \times Y$, equipped with its usual supremum norm $\|\cdot\|\_{\infty}$. Let
$$
\Xi: u \in E \longmapsto\left\{\begin{array}{l}
\int\_{X} \varphi d \mu+\int\_{Y} \psi d \nu &\begin{align\*}
&\text {if } u = \varphi \oplus \psi \text{ for some } (\varphi, \psi) \in C\_b(X) \times C\_b(Y)
\end{align\*}\\
+\infty &\text {else. }
\end{array}\right.
$$
I could not see how $\Xi \equiv 0$. For example, take some $(\varphi, \psi) \in C\_0(X) \times C\_0(Y)$. Then define $u$ by $u (x,y) := \varphi(x)+ \psi(y)$. Then $u \in E$. However, $\int\_{X} \varphi d \mu+\int\_{Y} \psi d \nu$ is possibly nonzero.
>
> Could you explain why $\Xi \equiv 0$ if $E=C\_{0}(X \times Y)$?
>
>
>
| https://mathoverflow.net/users/99469 | Why is $\Xi \equiv 0$ if $E=C_{0}(X \times Y)$? | A function $u(x,y) = \varphi(x) + \psi(y)$ with $\varphi\in C\_0(X)$ and $\psi\in C\_0(Y)$ will not be in $C\_0(X,Y)$ if $\varphi$ or $\psi$ is not constant zero: If $x$ "tends to infinity", we have $\varphi(x)\to 0$, but $u(x,y)\to\psi(y)$ which is not zero in general.
| 3 | https://mathoverflow.net/users/9652 | 422440 | 171,787 |
https://mathoverflow.net/questions/421763 | 4 | Consider an elliptic curve $E \subset \mathbb{P}^2$ with the zero point $\mathcal{O}$. There are classical articles about complete systems of addition laws on $E$ (see
[Lange and Ruppert - Complete systems of addition laws on abelian varieties](https://doi.org/10.1007/BF01388526),
[Bosma - Complete systems of two addition laws for elliptic curves](https://www.math.ru.nl/%7Ebosma/pubs/JNT1995.pdf)). It is known that the group law on $E$ can be represented by bi-homogenous polynomials of bidegree $(2, 2)$. More precisely, those articles show that for the divisor $D := 3(2V + 2H -\triangledown)$ the global section space $H^{0}(E^2, D)$ is non-zero, where $\triangledown$ is the anti-diagonal, $V := E \times \{\mathcal{O}\}$, and $H := \{\mathcal{O}\} \times E$. Moreover, $\lvert D\rvert$ is a base-point free linear system.
I am interested in the addition of three points $P := P\_0 + P\_1 + P\_2$ on $E$. Of course, $P = (P\_0 + P\_1) + P\_2$. However, expressing $P$ in this way
gives rise to tri-homogenous polynomials of tridegree $(4,4,2)$. Alternatively, I guess that there are tri-homogenous polynomials of tridegree $(3,3,3)$. Since $3\cdot 3 = 9 < 10 = 2\cdot 4+2$, it seems that the latter polynomials can be evaluated more efficiently. This may have applications in elliptic cryptography.
Now let $D := 3(3A\_0 + 3A\_1 + 3A\_2 - B)$, where
$$
A\_0 := \{\mathcal{O}\} \times E^2, \quad A\_1 := E \times \{\mathcal{O}\} \times E,
\quad A\_2 := E^2 \times \{\mathcal{O}\}, \quad B := \big\{(P\_0, P\_1, -P\_0-P\_1)\big\} \quad \subset \quad E^3.
$$
If I am not mistaken, to confirm my conjecture it is necessary to prove that the space $H^{0}(E^3, D)$ is non-zero. Help me please to do this. And what about base-point freeness of $\lvert D\rvert$?
| https://mathoverflow.net/users/69852 | Tri-homogenous polynomials of tridegree $(3,3,3)$ to add three points on an elliptic curve | The divisor $D$ is rationally equivalent to an effective divisor, hence $H^0(E^3,D) \neq 0$.
To see this, let $p\_0,p\_1,p\_2 \colon E^3 \to E$ be the canonical projections. For $i,j \in \{0,1,2\}$, write $p\_{i,j}=p\_i+p\_j$. By the theorem of the cube [1, Corollary 6.4], $B$ is rationally equivalent to
\begin{equation\*}
B \sim p\_{0,1}^\* (0) + p\_{0,2}^\* (0) + p\_{1,2}^\* (0) - A\_0 - A\_1 - A\_2.
\end{equation\*}
Moreover, on $E \times E$ we have the relation $\Delta + \nabla - 2H -2V \sim 0$, where $\Delta$ is the diagonal, $\nabla = \{(x,y) \in E \times E: x+y=0\}$ is the anti-diagonal, $H = E \times \{0\}$ and $V = \{0\} \times E$. Therefore
\begin{align\*}
D & \sim 3 (3A\_0+3A\_1+3A\_2 - p\_{0,1}^\*(0) - p\_{0,2}^\*(0) - p\_{1,2}^\*(0) +A\_0+A\_1+A\_2) \\
& \sim 3 \bigl(4A\_0+4A\_1+4A\_2 - (2A\_1+2A\_0-\Delta\_{0,1}) - (2A\_2+2A\_0-\Delta\_{0,2}) - (2A\_2+2A\_1-\Delta\_{1,2})\bigr) \\
& = 3 (\Delta\_{0,1} + \Delta\_{0,2} + \Delta\_{1,2}),
\end{align\*}
where $\Delta\_{i,j} = p\_{i,j}^\* \Delta$ is the pull-back of the diagonal under the projection $p\_{i,j} \colon E^3 \to E^2$ onto the factors with indices $i,j$. This proves the claim.
**[1]** Milne, J. S. Abelian varieties. Arithmetic geometry (Storrs, Conn., 1984), 103–150, Springer, New York, 1986.
| 2 | https://mathoverflow.net/users/6506 | 422454 | 171,792 |
https://mathoverflow.net/questions/422444 | 2 | Suppose that the sequence $(a\_{j})\_{j \in \mathbb{N}}$ is an increasing sequence of positive integers that satisfies $$(1)\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } d | a\_{d}$$ and $$ (2)\text{ }\text{ }\text{ }\text{ } |\{v| v\in (a\_{j})\_{j \in \mathbb{N}}, v \leq N \}| \geq C\log(N).$$
for a given constant $C > 0$ and every $N \in \mathbb{N}$. For a given $x \in \mathbb{R}$, do we have
$$(3)\text{ }\text{ }\text{ }\text{ }\text{ } \lim\_{N \rightarrow \infty}\min\_{1 \leq j \leq N}\|a\_{j}x\| = 0$$
(Here $\|y\|$ refers to the distance from $y$ to the nearest integer).
---
Note that if the sequence $(a\_{j})\_{j \in \mathbb{N}}$ satisfies the property $(3)$ then it is a Heilbronn sequence. <https://en.wikipedia.org/wiki/Heilbronn_set>
A Heilbronn sequence must contain multiples of every number, furthermore, this is not a sufficient condition. If the underlying sequence is sparse like $(j!)\_{j \in \mathbb{N}}$ then it is not a Heilbronn sequence.
This conjecture is related to Dirichlet's approximation theorem, the paper "Small values of $n^2 \alpha$ $\mod 1$" and "Analytischer Beweis des Satzes uber die Verteilung der Bruchteile eines ganzen Polynoms" where they consider the case where $a\_{j}$ is some fixed power of $j$.
| https://mathoverflow.net/users/134295 | A condition on $(a_{j})_{j\in \mathbb{N}}$ so that for all $x \in \mathbb{R}$ we have $\min_{1 \leq j \leq N}\|a_{j}x\|=o(1)$ | I will expand upon @mathworker21's comment.
Let $\theta = \frac{1 + \sqrt{5}}{2}$, and define the sequence $a\_d$ greedily to be the smallest positive integer greater than $a\_{d - 1}$ such that $d | a\_d$ and $\|a\_d \theta\| > \frac{1}{3}$.
Let $m$ be the smallest multiple of $d$ which is greater than $a\_{d - 1}$. Clearly, $m \leq a\_{d - 1} + d$. To find $a\_d$, we take the number $m \theta$, and repeatedly add $d \theta$ until we get a number whose fractional part is between $\frac{1}{3}$ and $\frac{2}{3}$.
By Dirichlet's approximation theorem, for some $1 \leq k \leq 3$ we have
$$\| kd \theta \| \leq \frac{1}{3}$$
and it is well known (for this specific value of $\theta$) that we have
$$\| kd \theta \| \geq \frac{1}{3 k^2 d^2} \geq \frac{1}{27 d^2}.$$
Therefore, if we take $m \theta$ and add $kd \theta$ repeatedly, after at most $27 d^2$ steps we will get a number with fractional part between $\frac{1}{3}$ and $\frac{2}{3}$, which shows that
$$a\_d \leq a\_{d - 1} + 30 d^3.$$
By induction, we have
$$a\_d \leq 30 d^4$$
which is sufficient for condition (2).
| 2 | https://mathoverflow.net/users/88679 | 422455 | 171,793 |
https://mathoverflow.net/questions/422447 | 1 | I am reading the paper "[A universal characterization of higher algebraic K-theory](https://arxiv.org/abs/1001.2282)" by Blumberg, Gepner, and Tabuada, and I am stuck on Corollary 4.25:
>
> …the fact that we have accessible localizations provides the following corollary about the
> structure of $\mathrm{Cat}\_\infty^\mathrm{ex}$ and $\mathrm{Cat}\_\infty^\mathrm{perf}$.
>
>
>
>
> Corollary 4.25. The ∞-categories $\mathrm{Cat}\_\infty^\mathrm{ex}$ and $\mathrm{Cat}\_\infty^\mathrm{perf}$ are compactly generated, complete, and cocomplete.
>
>
>
Here, $\mathrm{Cat}\_\infty^\mathrm{ex}$ (resp. $\mathrm{Cat}\_\infty^\mathrm{perf}$) is the ∞-category of small stable ∞-categories (resp. of small idempotent-complete stable ∞-categories).
The "accessible localizations" they are referring to are Theorem 4.22 and Theorem 4.23, which state that $\mathrm{Cat}\_\infty^\mathrm{ex}$ and $\mathrm{Cat}\_\infty^\mathrm{perf}$ are accessible localizations of the ∞-category of small spectral categories.
My thought is as follows: Dugger's theorem shows that the ∞-category of small spectral categories can be modelled by a combinatorial simplicial model category, so it is presentable. Therefore $\mathrm{Cat}\_\infty^\mathrm{ex}$ and $\mathrm{Cat}\_\infty^\mathrm{perf}$ are presentable. However, I cannot see how to prove that they are $\omega$-accessible, i.e. compactly generated. Can anyone help me fill in this gap?
| https://mathoverflow.net/users/328254 | Compact generation of the infinity category of stable infinity categories | Proposition 4.20 in the same paper shows that the localization is in fact $\omega$-accessible, i.e. the fully faithful right adjoint in question preserves filtered colimits. This implies that the localized category is $\omega$-accessible, in particular compactly generated (because the category you're localizing from is compactly generated)
Alternatively, you can observe that the forgetful functor $Cat^{ex}\_\infty\to Cat\_\infty$ preserves filtered colimits and arbitrary limits, and that therefore its left adjoint preserves compacts. Because this forgetful functor is furthermore conservative, the claim also follows.
| 2 | https://mathoverflow.net/users/102343 | 422456 | 171,794 |
https://mathoverflow.net/questions/422463 | 1 | Let's take the minimal transitive model of $\sf ZFC$ which, I came to know, is some minimal $L\_\kappa$ for a countable $\kappa$, that models $\sf ZFC$, and since its minimal so no subset of it can be a transitive model of $\sf ZFC$ and at the same time not isomorphic to it, call this stage $L\_{\sf ZFC}$.
What kind of cardinalities $Th(L\_{\sf ZFC})$ prove? I know it proves $\sf GCH$ because it's a model of $\sf ZFC + [V=L]$, but what theorems about cardinal existence it proves? Since this theory is *complete*, then it must decide on for example whether inaccessible cardinals exist or not?
>
> A which large cardinal, $Th(L\_{\sf ZFC})$ starts to prove its non-existence?
>
>
>
I tend to think that because it is minimal so it must prove the non-existence of inaccessibles whether strong or weak. Is that correct? And even if so, then what's the exact argument for that? If no, then at which large cardinal it starts proving its absence?
| https://mathoverflow.net/users/95347 | At which large cardinal, the theory of the minimal transitive model of ZFC starts proving its absence? | I don't like using the same notation to denote theories and ordinals, so I'll just use "$\alpha$" for the smallest ordinal such that $L\_\alpha\models\mathsf{ZFC}$ (and assume that there is one). We have $L\_\alpha\models$ "There is no transitive model of $\mathsf{ZFC}$." Consequently, $L\_\alpha\models$ "There are no weakly inaccessible cardinals," since $\mathsf{ZFC}$ (which $L\_\alpha$ satisfies!) proves "If $\kappa$ is weakly inaccessible then $L\_\kappa$ is a transitive model of $\mathsf{ZFC}$."
In general, nothing deserving to be called a large cardinal principle will be consistent with even the fragment $T\_0$ of (the "Platonic") $Th(L\_\alpha)$ consisting of sentences which $\mathsf{ZFC}$ proves are satisfied by the least transitive model of $\mathsf{ZFC}$ if such exists. (Each model of $\mathsf{ZFC}$ + "There is a transitive model of $\mathsf{ZFC}$" will have something it thinks is $Th(L\_\alpha)$, but different models may have disagreements about the details; however, they'll all agree about basic things like "$Th(L\_\alpha)\models$ "There are no weakly inaccessible cardinals,"" and this is what I'm trying to capture with $T\_0$ above.)
| 5 | https://mathoverflow.net/users/8133 | 422465 | 171,798 |
https://mathoverflow.net/questions/422436 | 0 | In the "Exposé XV" of the 1972-1973 of the Maurey-Schwartz Seminar of functional analysis ("Théorèmes de factorisation pour les opérateurs linéaires à
valeurs dans un espace $L^p(\Omega, \mu)$, $0<p \leq \infty$"), which can be found in French at <http://www.numdam.org/item/?id=SAF_1972-1973____A14_0> the following result appears
### Corollary
Let $0<p\leq q \leq s$ and $r^{-1}=p ^{-1}- q^{-1}$. Assume that $0<p \leq q \leq 2 \leq s < \infty$. Then, any bounded operator $u:L^s\_\nu \rightarrow L^p\_\mu$ factorizes as $u=T\_f \circ v$, with $v:L^s\_\nu \rightarrow L^q\_\mu$ and $T\_f$ the multiplication by a function $f\in L^r\_\mu$.
I would like to know whether there are any recent surveys or articles where this or similar results are discussed
| https://mathoverflow.net/users/164872 | References and updates on a $L^p$ Factorization theorem by Maurey | I mentioned in another of your posts the book of Diestel, Jarchow,and Tonge. Chapter 7 in the book of Albiac and Kalton, "Topics in Banach space theory", contains a nice exposition of Maurey's theorem and related topics.
Pisier proved and used non commutative factorization theorems, in case you are interested in non commutative $L\_p$ spaces.
| 1 | https://mathoverflow.net/users/2554 | 422468 | 171,799 |
https://mathoverflow.net/questions/422464 | 10 | Let $(X,d)$ be a compact and contractible metric space. Let $\operatorname{Isom}(X)=\{\phi\colon X\to X\}$ be its group of isometries.
>
> **Question:** Is there a point $x\in X$ fixed by all $\phi\in\operatorname{Isom}(X)$?
>
>
>
I am happy to assume some additional niceness conditions for $X$, enough to ensure that $X$ satisfies some fixed-point theorem, guaranteeing that every continuous map $\phi\colon X\to X$ has a fixed point (e.g. triangulable, locally contractible; see [A version of Brower's fixed point theorem for contractible sets?](https://math.stackexchange.com/questions/4449267/a-version-of-browers-fixed-point-theorem-for-contractible-sets) for details).
The emphasize is therefore on whether all isometries have a *common* fixed point.
This post is a refinement/generalization of [Symmetries of contractable subsets of $\Bbb R^n$](https://mathoverflow.net/questions/422384/symmetries-of-contractable-subsets-of-bbb-rn).
| https://mathoverflow.net/users/108884 | Does a compact contractible metric space have a point that is fixed by all isometries? | [There are](https://mathoverflow.net/questions/78501/do-finite-groups-acting-on-a-ball-have-a-fixed-point) finite groups that act smoothly on a disk without a global fixed point. You can arrange the metric to be isometric, e.g. via the [Mostow-Palais embedding theorem](https://en.wikipedia.org/wiki/Mostow%E2%80%93Palais_theorem%20), which equivariantly and smoothly embeds the disk into a Euclidean space where the group acts orthogonally, and taking the metric induced by Euclidean one. The full isometry group is potentially bigger, so it cannot fix a point.
| 15 | https://mathoverflow.net/users/1573 | 422474 | 171,802 |
https://mathoverflow.net/questions/422452 | 3 | Given a dynamical system $(X, \Omega, \mu)$ ($\Omega$ the $\sigma$-algebra and $\mu$ a measure), we assume there is a group $G$ acting on the system in the sense that, for each group element $g$, $g$ corresponds to an invertible measurable mapping $T\_g$ such that $g \cdot x = T\_g x$. We also define $T\_g\mu$ by, for each $A\in\Omega, T\_g\mu(A) = \mu[T\_g^{-1}(A)]$.
Then, we call the group action in $(X, \Omega, \mu, G)$ ergodic iff, for each $g\in G$, any $T\_g$ invariant set ($T\_g(A)=A$) will have $\mu$-measure zero; call the group action in $(X, \Omega, \mu, G)$ non-singular iff for each $g\in G$, $\mu$ is equivalent to $T\_g\mu$; call the group action conservative iff for each $A\in\Omega\_1$ with $\mu(A)>0$, we can find $h\in G$ such that $\mu[A\cap T\_h^{-1}(A)]>0$. Now, given a ergodic, non-singular and conservative system $(X, \Omega, G, \mu)$, the definition of a **ratio set** of $(X, \Omega, \mu, G)$, denoted by $r\_{\mu}(G)$, is defined below:
A positive real number $r$ is in $r\_{\mu}(G)$ iff, for each $A\in\Omega\_1$ with positive measure and $\epsilon>0$, we can find $g\in G$ such that $A\cap T\_g^{-1}(A)\cap \{x\in X\,\vert\,\frac{d\,T\_h\mu}{d\,\mu}(x)\in(r-\epsilon, r+\epsilon)\}$ has positive $\mu$ measure. Now we want to prove: given two equivalent measures $\mu\_1, \mu\_2$ defined on $\Omega$ such that both $(X, \Omega, \mu\_1, G)$ and $(X, \Omega, \mu\_2, G)$ are ergodic, non-singular and conservative, $r\_{\mu\_1}(G)=r\_{\mu\_2}(G)$ (i.e. the ratio set of $\mu\_1$ and $\mu\_2$ are the same).
Notice that, for each $A\in\Omega$ with positive measure, given $g\in G$, the following set will have positive measure (either $\mu\_1$ or $\mu\_2$).
$$
\{x\in A\,\vert\,\frac{d\,T\_g\mu\_1}{d\,\mu\_1}(x)\in(\frac{T\_g\mu\_1(A)}{\mu\_1(A)}-\epsilon, \frac{T\_g\mu\_1(A)}{\mu\_1(A)}+\epsilon)\}
$$
because, otherwise, the integral $\int\_A\frac{d\,T\_g\mu\_1}{d\,\mu\_1}d\,\mu\_1$ will not be $T\_g\mu\_1(A)$. The statement will also be true when we replace $\mu\_1$ by $\mu\_2$. Now it suffices to show that $r\_{\mu\_1}(G)\subseteq r\_{\mu\_2}(G)$. Given $r$ in the ratio set of $\mu\_1$, if we fix an arbitrary $\epsilon$ and set $Q\_A = A\cap T\_h^{-1}(A)\cap \{x\in X\,\vert\,\frac{d\,T\_h\mu}{d\,\mu}(x)\in(r-\epsilon, r+\epsilon)\}$ where $h$ is given by the definition of ratio set, then the following set will have positive measure:
$$
Q\_{A, 1} = \{x\in Q\_A\,\vert\,\frac{d\,\mu\_1}{d\,\mu\_2}(x)\in(\frac{\mu\_1(Q\_A)}{\mu\_2(Q\_A)}-\epsilon, \frac{\mu\_1(Q\_A)}{\mu\_2(Q\_A)}+\epsilon)\}
$$
I want to approach the problem by using $\frac{d\,T\_h\mu\_2}{d\,\mu\_2}(x)=\frac{d\,T\_h\mu\_2}{d\,T\_h\mu\_1}(x)\cdot\frac{d\,T\_h\mu\_1}{d\,\mu\_1}(x)\cdot\frac{d\,\mu\_2}{d\,\mu\_1}(x)$. From **p.153** in [Ergodic dynamics](https://link.springer.com/book/10.1007/978-3-030-59242-4) written by **Jane Hawkins**, we have that, for each $g\in G, \frac{d\,T\_g\mu\_1}{d\,T\_g\mu\_2}(x) = \frac{d,\mu\_1}{d\,\mu\_2}(T\_g x)$. Now $Q\_{A, 1}$ has positive measure and, for each element $x\in Q\_{A, 1}, \frac{d\,T\_g\mu\_1}{d\,\mu\_1}(x)\in (r-\epsilon, r+\epsilon)$. We want to find another set, say $B$ such that, for each $y\in B, \frac{d\,T\_g\mu\_2}{d\,T\_g\mu\_1}(y)$ is close to $\frac{\mu\_2(Q\_A)}{\mu\_1(Q\_A)}$ and this set need to have non-null intersection with $Q\_{A, 1}$. I thought about $T\_g^{-1}(Q\_{A, 1})$ but did not know if $Q\_{A, 1}$ and $T\_g^{-1}(Q\_{A, 1})$ will have non-null intersection. Any other hints will be appreciated
| https://mathoverflow.net/users/151332 | Questions about ratio set in a dynamical system | In what follows I'll write $g$ for $T\_g$, and i'll write $\alpha \_i (g,x)$ for $\tfrac{dg^{-1}\mu \_i}{d\mu \_i} (x)$. Here is one way to argue that the ratio sets are the same: given $r>0$ in the ratio set of $\mu \_2$ along with a positive measure set $A$ and an open neighborhood $U$ of $r$, find a neighborhood $V$ of $1$ with $V^2rV^{-2} \subseteq U$. The sets $Vs$, with $s>0$ rational, cover the positive reals, so we can find some $s>0$ such that the set $B$, of all $x\in A$ with $\tfrac{d\mu \_1}{d\mu \_2}(x)\in Vs$, has positive measure. Since $r$ is in the ratio set of $\mu \_2$ we can find some $g\in G$ and a positive measure $C\subseteq B$ with $gC\subseteq B$ and $\alpha \_2(g,x)\in VrV^{-1}$. Then each $x\in C$ satisfies $\alpha \_1(g,x) = \tfrac{d\mu \_1}{d\mu \_2}(gx)\alpha \_2(g,x)\tfrac{d\mu \_1}{d\mu \_2}(x)^{-1}\in V^2rV^{-2}\subseteq U$, so $r$ is in the ratio set of $\mu \_1$.
I think it is nice to understand this from the broader perspective of cocycles of dynamical systems. Without going into any real details, i'll just mention some relevant definitions/jargon can be looked into more if you are interested in this perspective. The function $\alpha \_i$ defined above is called the "Radon-Nikodym cocycle" of the system $(G,X,\mu \_i)$. It satisfies the cocycle equation $\alpha \_i (gh,x)=\alpha \_i (g,hx )\alpha \_i (h,x)$. The measures $\mu \_1$ and $\mu \_2$ being equivalent yields the equation $\tfrac{d\mu \_1}{d\mu \_2}(gx)\alpha \_2(g,x)\tfrac{d\mu \_1}{d\mu \_2}(x)^{-1}=\alpha \_1(g,x)$, so that $\alpha \_1$ and $\alpha \_2$ are cohomologous cocycles. The ratio set of the system $(G,X,\mu \_i )$ coincides with what is sometimes called the ``essential range'' of the cocycle $\alpha \_i$.
| 2 | https://mathoverflow.net/users/1243 | 422475 | 171,803 |
https://mathoverflow.net/questions/422430 | 4 | I am convinced I have seen results along the lines of: if $ u \ge 0$ is an $H\_0^1(\Omega)$ solution of
$$-\Delta u = u^{q-1}$$ in $\Omega$ with $ u=0$ on $ \partial \Omega$ (here $\Omega$ is a smooth bounded domain in $R^N$ and $ q=2^\*$, then $u$ is smooth. Any idea how one proves the regularity result? Since $ q=2^\*$ it appears the standard iteration method fails. Maybe one uses some $ \epsilon$ regularity approaches (which I don't know). Or maybe I am wrong and the result is false. thanks
| https://mathoverflow.net/users/66623 | Smoothness of critical elliptic problem | The result is true. As you suggest, the starting regularity assumption $u \in H^1\_0$ is *critical* in the sense that, if you additionally knew $u \in L^p$ with $p>2^\*$, then you could bootstrap to smoothness (at least, when $2^\*$ is an integer, as in dimension $N=3$, for which $2^\*=6)$. In the case $p=2^\*$, the key observation is that $H^1\_0$ solutions to $-\Delta u = Vu$ *do* satisfy higher integrability in the critical case $V \in L^{N/2}$. That is, such solutions belong to all $L^p$, $p < +\infty$. The idea is to split the potential $V$ into a bounded part and a part which is small in $L^{N/2}$ and run Moser's iteration. The argument is given in the book *Elliptic Partial Differential Equations* by Qing Han and Fanghua Lin, see Theorem 4.4, p. 76.
| 5 | https://mathoverflow.net/users/137457 | 422476 | 171,804 |
https://mathoverflow.net/questions/422445 | 2 | Suppose that the positive random variable $X$ is infinitely divisible and supported on $\mathbb R\_+$. Due to Lévy-Khintchine, its moment generating function then writes :
$$M(t) = \mathbb E\left(e^{tX}\right) = \exp\left\{\int\_{\mathbb R\_+} \left(e^{ty}-1\right)L(dy)\right\}.$$
**Question:** Is there a way, from a $n$-sample $X\_1,...X\_n$ of i.i.d. random variables, to estimate the (non-negative) Lévy measure $L$ ?
| https://mathoverflow.net/users/143783 | Estimation of Lévy measure of ID distribution | As is done in estimation of regression or pdf, you can parametrize the problem -- say by assuming that
$$\frac{L(dy)}{dy}=\sum\_{j=1}^k c\_j g\_j(y),$$
where the $g\_j$'s are known nonnegative functions and the $c\_j$'s are unknown nonnegative parameters. For instance, if you choose your $g\_j$'s to be the indicators of finite intervals or truncated normal densities (so that
$$g\_j(y)=\frac1{b\_j\sqrt{2\pi}}e^{-(y-a\_j)^2/(2b\_j^2)}$$
for some real $a\_j$'s and some positive real $b\_j$'s), then you can evaluate $\int\_{\mathbb R\_+} \left(e^{ty}-1\right)L(dy)$ explicitly. Then your moment generating function (mgf) will be explicitly determined by the $k$-tuple $c:=(c\_1,\dots,c\_k)$ of the parameters: $M=M\_c$.
Then you can fit the parametric mgf $M\_c(t)$ to the
empirical mgf
$$\hat M(t):=\frac1n\,\sum\_{i=1}^n e^{tX\_i}$$
to the true mgf $M(t)$ --- say, by minimizing
$$\int\_{\mathbb R}(M\_c(t)-\hat M(t))^2 w(t)\,dt$$
in $c=(c\_1,\dots,c\_k)$ for some nonnegative weight function $w$.
| 2 | https://mathoverflow.net/users/36721 | 422478 | 171,806 |
https://mathoverflow.net/questions/422496 | 2 | Let $I$ be a non trivial interval of $\mathbb R$, let $f : I \times \mathbb R^n \to \mathbb R^n$ and consider the following ordinary differential equation (ODE):
\begin{equation}\tag{$\mathscr E$}\label{ode}
y'(t) = f\big(t,y(t)\big)
\end{equation}
Suppose that:
1. all the maximal solutions of \eqref{ode} are global (defined on $I$) ;
2. and the set $S$ of such global solutions is a linear space.
Is it true that \eqref{ode} is linear ? *i.e.* that it exists $A : I \to \mathrm M\_n(\mathbb R)$ (the square matrices of size $n \times n$) such that every differentiable function $y: I \to \mathbb R^n$ is a solution of \eqref{ode} if and only if it is a solution of the linear ODE
\begin{equation}\tag{$\mathscr L$}\label{ode2}
y'(t) = A(t)\,y(t)
\end{equation}
*Remark.* I did not make any assumptions on $f$ and on the dimension of $S$ but if needed, we can assume for $f$ that the Picard–Lindelöf theorem holds ($f$ is continuous in $t$ and Lipschitz continuous in $y$, or even one can assume that $f$ is $\mathscr C^1$) and we can assume that $\dim S = n$, but maybe 1 and/or 2 implies these assumptions.
| https://mathoverflow.net/users/80602 | An ODE is linear if and only if the maximal solutions are a linear space? | This is a partial solution, assuming that $f$ is analytic on $I\times R$, and so
solutions are analytic as well, and that
the space $E$ of solutions is of the same dimension $n$ as
vectors $y$ and $f$.
We represent solutions $y$ and the function $f$ in the right hand side as column vectors, as usual.
Let $y\_1,\ldots,y\_n$ be a basis in $E$. Then every
vector $y\in E$ can be written as a linear combination
$$y=c\_1y\_1+\ldots+c\_ny\_n.\quad\quad\quad\quad\quad\quad\quad\quad (1)$$
Solving this by Cramer's rule we obtain $c\_k$ as a ratio of two determinants made of coordinates of $y$ and $y\_j$. Since the coordinates of $y$ make one column in the numerator, it is clear that this ratio is a linear function of coordinates of $y$, so we can write
$$c\_j=a\_j(t)y,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (2)$$
for some row vector $a\_j$ which depends on $t$.
Now, since all elements of $E$ satisfy your non-linear equation
we must have for every $y\in E$ of the form (1):
$$y'=\sum\_{j=1}^n c\_jy\_j^\prime=\sum\_{j=1}^nc\_jf(t,y\_j),$$
Substituting our formula for $c\_j$ we obtain that
$$y'=A(t)y,\quad\mbox{where}\quad A(t)=\sum\_jf(t,y\_j(t))a\_j(t)$$
(column vectors $f(t,y\_j(t))$ times row vectors $a\_j$ give you
$n\times n$ matrices.)
There is a little problem with the denominator in the Cramer Rule:
one has to show that it cannot be zero at some point.
If all functions $f,y\_j$ are analytic, this can only happen at isolated
points. Away from these isolated points $t\_k$ both equations
$y'=f(t,y)$ and $y'=A(t)y$ have the same set of solutions, therefore
$f(t,y)=A(t)y$ on some open set of $(t,y)$, and since all functions
are analytic they must coincide.
| 5 | https://mathoverflow.net/users/25510 | 422499 | 171,813 |
https://mathoverflow.net/questions/422497 | 4 | Let $M$ be a non-empty subset of $\mathbb R^n$, $n \geq 2$.
Recall that a vector $v$ is tangent to $M$ at the point $m \in M$ if it exists a differentiable curve $\gamma : I \to M$ such that $\gamma(0) = m$ and $\gamma'(0) = v$, where $I \subset \mathbb R$ is an interval that contains a neighborhood of $t=0$.
Suppose that it exists an integer $k \geq 1$ such that, for every $m \in M$, the set of vectors that are tangent to $M$ at $m$ is a linear space of dimension $k$ (the same $k$ for every $m$).
Is it true that $M$ is a differential manifold of dimension $k$?
| https://mathoverflow.net/users/80602 | If tangent vectors are a vector space of same dimension at every point, does one has a manifold? | I found that the answer is no.
For $n=2$: take $M$ as the union of the two circles of radius 1 centered at $(\pm 1,0)$, at any point one has a tangent space of dimension $k=1$ but it is not a manifold (double point at $(0,0)$).
This idea should be generalizable to arbitrary $n$ and $k$.
| 1 | https://mathoverflow.net/users/80602 | 422503 | 171,815 |
https://mathoverflow.net/questions/422507 | 6 | Let $M$ be a compact orientable $n$ dimensional manifold. Assume that $M$ has trivial cobordism class.
Is there an embedding of $M$ in some Euclidean space $\mathbb{R}^m$ such that the convex hull of $M$ is a $n+1$ dimensional manifold whose boundary is $M$?
Here the image of $M$ under the embedding is denoted again by $M$.
**Note:** One can pose the same question in the following geometric manner:
Let $M$ be a compact Riemannian manifold with trivial cobordism class. Is there an isometric embedding of $M$ in some Euclidean space such that the convex hull of $M$ is a manifold whose boundary is $M$?
| https://mathoverflow.net/users/36688 | The convex hull of a manifold whose cobordism class is trivial | There are exotic spheres (which are null cobordant) which do not bound a parallelisable manifold. Since the convex hull is contractible, it would be parallelisable if it were a manifold, so these guys do not admit embeddings like you want.
| 8 | https://mathoverflow.net/users/10839 | 422508 | 171,816 |
https://mathoverflow.net/questions/182989 | 3 | Is there a lower bound on the number of perfect matchings in a $k$-regular bipartite graph?
One can use Hall's marriage theorem and induction on $k$ to derive the lower bound of $k$. I can't come up with an example where this bound is actually tight. Is there a better lower bound than this?
| https://mathoverflow.net/users/39492 | Minimum number of perfect matchings in a regular bipartite graph | as already mentioned in the comments,
that is answered by Alexander Schrijver in his publication "**Counting 1-factors in regular bipartite graphs**":
>
> any $k$-regular bipartite graph with $2n$ vertices has at least $$\left(\frac{(k-1)^{k-1}}{k^{k-2}}\right)^n$$ perfect matchings.
>
>
>
| 2 | https://mathoverflow.net/users/31310 | 422525 | 171,820 |
https://mathoverflow.net/questions/422510 | 5 | The early seminal result of Bernstein in 1914 for $n=2$ is the well-known Bernstein theorem:
>
> The only entire solutions to the minimal surface equation in $\mathbb R^3$ are the affine functions
> $$u(x,y)=ax+by+c,$$
> where $a, b, c\in\mathbb{R}$.
>
>
>
Actually, Bernstein obtained his result as an application of the so called Bernstein’s geometric theorem:
>
> If the Gauss curvature of the graph of $u\in C^\infty(\mathbb R^2)$ in $\mathbb R^3$ satisfies $K\leq 0$ everywhere and $K<0$ at some point, then $u$ cannot be bounded.
>
>
>
As a corollary, Bernstein proved a very general Liouville theorem:
>
> Suppose $u$ is a solution to the elliptic equation
> $$\sum\_{i,j=1}^2 a\_{ij}u\_{ij}=0\quad\text{in }\mathbb R^2$$
> with $a\_{ij}\in C^\infty$ such that
> $$|u(x)|=o(|x|) \text{ as }|x|\to+\infty.$$
> Then $u$ is a constant.
>
>
>
Note that in the above Liouville theorem, the equation doesn't need to be uniformly elliptic, hence it is a very powerful result. What I want to know if this result has a half space version, which is like harmonic functions. More precisely, I want to obtain the following proposition:
>
> Suppose $u\geq 0$ is a solution to the elliptic equation
> $$\left\{\begin{aligned}\sum\_{i,j=1}^2 a\_{ij}u\_{ij}&=0\quad\text{in }\mathbb R^2\_+,\\
> u(x,0)&=0\quad \text{on }\mathbb R,\end{aligned}\right.$$
> where $a\_{ij}\in C^{\infty}$.
> Then $u$ is a linear function of form
> $$u(x,y)=Ay,\quad A\geq 0.$$
>
>
>
Note that in the question, $a\_{ij}$ could be degenerate or sigular at $\infty$.
This question is motivated by seeing Mooney's notes: The Monge-Ampère equations. He used partial Legendre transform to investigate the Liouville theorem for Monge-Ampère equation in half space, and one of steps in his proof used the similar proposition for harmonic functions, and which can be proved by boundary Harnack inequality and odd extension of $u$. But it is failed for the case without uniform ellipticity.
I have searched on the internet, and I didn't find any references about this proposition. Is this propersiton true? And if there are some references that I missed? Thanks in advance.
| https://mathoverflow.net/users/140934 | Bernstein's corollary for the case of half space | Here is a counterexample: let
$$u(x,y) = e^{-x^2}\sinh(y).$$
Then
$$\det D^2u = -2e^{-2x^2}(\sinh^2(y) + 2x^2) < 0 \text{ ơn } \mathbb{R}^2 \backslash \{0\},$$
and the equation
$$u\_{xx} + (2-4x^2)u\_{yy} = 0$$
is uniformly elliptic in a neighborhood of the origin.
| 5 | https://mathoverflow.net/users/16659 | 422529 | 171,822 |
https://mathoverflow.net/questions/422531 | 13 | Let $f(z)$ be an entire holomorphic function in $\mathbb{C}$, and consider the real-valued function
$$g\_f(z)=\frac{|f'(z)|}{1+|f(z)|^2}.$$
If $f(z)$ is a polynomial, then it is easy to prove that $\lim\_{|z|\rightarrow \infty}g\_f(z)=0$.
When $f$ is transcendental, say $f(z)=e^z$, then $g\_f(z)=\frac{e^x}{1+e^{2x}}$, which goes to zero when $Re(z)$ is going $\infty$, but remains a constant when $z$ is moving on any vertical line. So far I have not found any example such that the limit goes to zero when $f$ is transcendental.
**My question** is, can we rigorously prove that there is no transcendental entire function $f$ such that $$
\lim\_{|z|\rightarrow \infty}g\_f(z)=0\; ?
$$
| https://mathoverflow.net/users/51546 | Behavior of $|f'(z)|/(1+|f(z)|^2)$ as $|z| \rightarrow \infty$? | This is not true. The optimal estimate from below for transcendental entire functions is
$$\limsup\_{z\to\infty}\frac{|z||f'(z)|}{\log|z|(1+|f(z)|^2)}=\infty,$$
and this is best possible,
J. Clunie and W. Hayman, The spherical derivative of integral and meromorphic functons, Comment Math. Helv., 40 (1966) 117-148.
More precisely, for every function $\phi(r)\to+\infty$, they constructed an example of transcendental entire function
for which
$$\frac{|z||f'(z)|}{\phi(|z|)\log|z|(1+|f(z)|^2)}$$ is bounded. The function in this example is of the form
$$f(z)=\prod\_{n=1}^\infty\left(1-\frac{z}{2^{k\_n}}\right)^{k\_n},$$
where $k\_n$ is an increasing sequence of integers, which is choosen, depending on $\phi$.
| 16 | https://mathoverflow.net/users/25510 | 422543 | 171,825 |
https://mathoverflow.net/questions/421946 | 4 | Consider the following discrete-time population model. We start with a single "good" individual who reproduces asexually into $k$ children and dies in the process. At generation $t=2$, those children themselves each asexually reproduce into $k$ children of their own and die in the process. That is, at generation $t=n$, there are $k^n$ individuals.
Good individuals' children are good with probability $1-q$ and evil with probability $q$.
Evil individuals' children are evil with probability $1-q$ and good with probability $q$.
In all of this, $q<1/2$, with the idea that good individuals are more likely to beget good individuals and evil individuals are more likely to beget evil individuals.
---
I am most interested in the asymptotic behavior of the probability that the good individuals outnumber the evil individuals. That is, even if we start with a good individual with a moderately high probability of begetting good individuals, might this asymptotic probability tend to $1/2$, with good and evil equally matched?
Calling the number of good individuals $G\_n$ and the number of evil individuals $E\_n$, I am curious about $\lim\_{n \to \infty} Pr(G\_n > E\_n) = f(q)$. In particular, one might expect that if $q$ is close enough to $1/2$ that the asymptotic probability $f(q)=1/2$, while if $q$ is close enough to $0$ that $f(q) > 1/2$.
Indeed, $f(q)$ can be shown through Chebyshev's inequality to be strictly greater than $1/2$ for $q$ small enough, and $f(1/2) = 1/2$.
My question is, is there some $0<q\_c<1/2$ for which
$$f(q)=\begin{cases}
>1/2 & q<q\_c \\
1/2 & q\_c< q \leq 1/2
\end{cases}? $$
Note that the $k$ dependence is implicit in the above; I expect $q\_c$ is a function of $k$. An explicit form for $f(q)$ would also be very interesting.
| https://mathoverflow.net/users/153549 | Population growth with good and evil children - probability good outnumbers evil | The process you describe has been studied extensively as a mutation model [5], as a model of broadcasting on trees [2], as a representation of the Ising model on the Bethe lattice [1]. A very general relevant analysis in the context of mutitype branching processes is in [3], some of the results there were refined in [4].
Your intuition was correct. There is a phase transition, and the critical value $q\_c$ on the $k$-ary tree is given by the equation
$$(1-2q\_c)^2=1/k \,,$$
see [1] or Theorem 1.1 in [2]. For $q \ge q\_c$, one has (with your notation) $f(q)=1/2$, while for $q<q\_c$ one has $f(q)>1/2$.
[1] BLEHER, P. M., Ruiz, J. and ZAGREBNOV, V. A. (1995). On the purity of the limiting Gibbs state for the Ising model on the Bethe lattice. J Statist. Phys. 79 473-482.
[2] Evans, William, Claire Kenyon, Yuval Peres, and Leonard J. Schulman. "Broadcasting on trees and the Ising model." Annals of Applied Probability (2000): 410-433. <https://www.jstor.org/stable/2667156>
[3] KESTEN, H. and STIGUM, B. P. (1966). Additional limit theorems for indecomposable multi-dimensional Galton-Watson processes. Ann. Math. Statist. 37 1463-1481.
[4] Mossel, Elchanan, and Yuval Peres. "Information flow on trees." The Annals of Applied Probability 13, no. 3 (2003): 817-844.
[5] STEEL, M. (1989). Distribution in bicolored evolutionary trees. Ph.D. thesis, Massey Univ., Palmerston North, New Zealand.
| 5 | https://mathoverflow.net/users/7691 | 422555 | 171,831 |
https://mathoverflow.net/questions/422180 | 2 | Let $f:[0,1]^2 \rightarrow \mathbb{R}^2,$ where $f\_1(x,y) = g(y)-x$ and $f\_2(x,y) = g(x)-y.$ Here $g(\cdot)$ is a strictly decreasing polynomial function such that $g(0)=1$ and $g(1)=0.$ I am interested in analyzing the asymptotic behavior of the following system of differential equations:
\begin{equation}
\dot{x} = f\_1(x,y) \ \text{and} \ \dot{y} = f\_2(x,y).
\end{equation}
I want to show that the above system's limiting behavior will be to one of the stationary states i.e., rule out limit cycles and other complicated behavior. Since $\frac{\partial f\_1}{\partial x} + \frac{\partial f\_2}{\partial y} = -2 \neq 0,$ can I use the Bendixson–Dulac theorem to rule out limit cycles and conclude that the above system converges to one of the stationary states?
The domain in the statement of Bendixson–Dulac's theorem is usually an open set. However my domain of interest $[0,1]^2$ is not open and so am confused. I would really appreciate help on this.
| https://mathoverflow.net/users/151054 | Asymptotic behavior of system of differential equations | **Preliminary remark.** I am not certain whether Bendixon-Dulac grants the *global attractiveness* of the equilibria. However, via sheer leveraging on the (strict) monotonicity of $g$ and symmetry of the ODE, we can prove that the equilibria is indeed the *global attractor* which leaves no room for limit cycles or other nontrivial attractors.
The *right* region in the phase space to be studied is the invariant set $\mathcal{I}=\left\{\left(x,y\right)\in \left[0,1\right]^2\,:\,g(x)\leq y \leq g^{-1}(x)\mbox{ or }g^{-1}(x)\leq y \leq g(x)\right\}$, i.e., the region *between* the graphs $g$ and $g^{-1}$. This is the *right region* in the sense that trajectories necessarily accumulate onto it (exponentially fast). Further, observe that $\mathcal{I}$ contains the equilibria given by $\mathcal{E}=\left\{(x,y)\in\left[0,1\right]^{2}\,:\,y=g(x)=g^{-1}(x)\right\}$.
Let $\left(x(t),y(t)\right)\_{t\geq 0}$ be the solution to the ODE with initial condition $(x(0),y(0))\in\left[0,1\right]^2$. Define $T$ as the hitting time to hit the set $\mathcal{I}$, i.e., $T\overset{\Delta}=\inf\left\{T\geq 0\,:\, (x(T),y(T))\in \mathcal{I}\right\}$. Let ${\sf d}(w,z)\overset{\Delta}=\|w-z\|\_2$.
**Claim $1$.** If $T=\infty$, then ${\sf d}((x(t),y(t)),\mathcal{E})\overset{t\rightarrow\infty}\longrightarrow 0$ (exponentially fast).
>
> *Proof.* Define $f\_1(t)=\frac{1}{2}\left(x(t)-g(y(t))\right)^2$ and $f\_2(t)=\frac{1}{2}\left(x(t)-g^{-1}(y(t))\right)^2$. Assume that $(x(0),y(0))$ is in the *lower triangular part* of the phase space, i.e., $y(0)<g(x(0))$ and $y(0)<g^{-1}(x(0))$ -- everything that follows applies similarly if we assume an initial condition at the upper triangular part. If $t<T$, then we have that
> $$f'\_1(t)=(x(t)-g(y(t)))(\dot{x}(t)-g'(y(t))\dot{y}(t))=(x(t)-g(y(t)))(g(y(t))-x(t)-g'(y(t))(g(x(t))-y(t)))\leq -(x(t)-g(y(t)))^2,$$
> where the inequality follows since $-g'(y(t))(x(t)-g(y(t)))(g(x(t))-y(t))<0$ as $(g(x(t))-y(t))>0$, $-g'(y(t))>0$ and $(x(t)-g(y(t)))<0$ since $y(t)<g^{-1}(x(t))$.
> Therefore, from Grönwall's inequality, $f\_1(t)\leq f\_1(0) e^{-2t}$. If $T=\infty$, then $f\_1(t)\overset{t\rightarrow \infty}\longrightarrow 0$ exponentially fast. Similarly, if $T=\infty$, we conclude that $f\_2(t)\overset{t\rightarrow \infty}\longrightarrow 0$ exponentially fast. This is equivalent to ${\sf d}((x(t),y(t)),\mathcal{E})\overset{t\rightarrow \infty}\longrightarrow 0$.
>
>
>
**Claim $2$.** If $T<\infty$, then ${\sf d}((x(t),y(t)),\mathcal{E})\overset{t\rightarrow\infty}\longrightarrow 0$.
>
> *Proof.* Now, $(x(T),y(T))\in \mathcal{I}$. Let $g^{-1}(x(T))\leq y(T) \leq g(x(T))$ -- the other case $g(x(T))\leq y(T) \leq g^{-1}(x(T))$ can be dealt with similarly. If $g^{-1}(x(T)) = y(T) = g(x(T))$, then $(x(T),y(T))$ is already at equlibrium. Assume $g^{-1}(x(T))\leq y(T) < g(x(T))$. Remark that $g^{-1}(x(t))\leq y(t) \leq g(x(t))$ for all $t\geq T$. Let us refer to this invariant set as $\mathcal{I}\_1$. Further, let $(x^{\star},y^{\star})$ be the equilibrium that lies in the left part of $\mathcal{I}\_1$, i.e., $x^{\star}=\sup\limits\_{(w\_1,w\_2)\in \mathcal{E}} w\_1 < x(T)$. Then, $V((x,y))\overset{\Delta}= \frac{1}{2}\left(\left(x-x^{\star}\right)^2+\left(y-y^{\star}\right)^2\right)$ conforms to a Lyapunov function granting attractiveness to the equilibrium for any $(x(T),y(T))\in\mathcal{I}\_1$: $V$ is definite positive and $\dot{V}(x,y)<0$ for all $(x,y)\in \mathcal{I}\_1\setminus \left\{{\sf eq}\_{{\sf right}} \right\}$, where ${\sf eq}\_{{\sf right}}$ is the equilibrium on the right side. In words, $g^{-1}(x(T))\leq y(T) < g(x(T))$ implies that the solution will acumulate onto the left closest equilibrium, whereas $g(x(T))\leq y(T) < g^{-1}(x(T))$ will imply convergence to the right closest equilibrium.
>
>
>
Claims $1$ and $2$ combined yield the global attractiveness of the equilibria.
**Theorem $3$. [$\mathcal{E}$ is the global attractor]** ${\sf d}((x(t),y(t)),\mathcal{E})\overset{t\rightarrow\infty}\longrightarrow 0$ regardless of the initial condition $\left(x(0),y(0)\right)\in\left[0,1\right]^2$.
| 1 | https://mathoverflow.net/users/138242 | 422572 | 171,836 |
https://mathoverflow.net/questions/422556 | 3 | Suppose that I have a collection of known $\gamma\_1, \dots, \gamma\_{2N} \in \mathbb{C}$. Is there a known method to compute $\zeta\_1, \dots, \zeta\_N, \alpha\_1, \dots, \alpha\_N \in \mathbb{C}$ that solve the following system of equations?
$$
\begin{pmatrix}
\zeta\_1&\zeta\_2&\dots&\zeta\_N \\
\zeta\_1^2&\zeta\_2^2&\dots&\zeta\_N^2 \\
\vdots&\vdots&\ddots&\vdots \\
\zeta\_1^{2N}&\zeta\_2^{2N}&\dots&\zeta\_N^{2N}
\end{pmatrix}
\begin{pmatrix}
\alpha\_1 \\
\alpha\_2 \\
\vdots \\
\alpha\_{N}
\end{pmatrix}
=
\begin{pmatrix}
\gamma\_1 \\
\gamma\_2 \\
\vdots \\
\gamma\_{2N}
\end{pmatrix}
$$
| https://mathoverflow.net/users/170939 | A system of $2N$ equations resembling a Vandermonde matrix | Your system is
$$\sum\_{k=1}^N\alpha\_k\zeta\_k^j=\gamma\_j, \quad 1\leq j\leq 2N.\quad\quad\quad(1)$$
Putting $x\_k=\alpha\_k\zeta\_k$ we obtain
$$\sum\_{k=1}^Nx\_k\zeta\_k^j=\gamma\_{j+1},\quad 0\leq j\leq 2N-1.\quad\quad\quad(2)$$
This system of equation was studied much,
and it is called the Sylvester-Ramanujan system.
A necessary and sufficient condition of solvability is given
in the paper of Y. I. Lyubich,
The Sylvester-Ramanujan system of equations and the complex power moment problem.
Ramanujan J. 8 (2004), no. 1, 23–45.
It is somewhat complicated to be reproduced here; they also give a uniqueness statement.
Let me sketch Ramanujan's method of solving it.
Consider the rational function
$$f(z)=\sum\_{k=1}^n\frac{x\_k}{1-z\zeta\_k},$$
then our system is equivalent to
$$f(z)=\sum\_{j=0}^{2N-1}\gamma\_{j+1}z^j+O(z^{2N}),\quad z\to 0.$$
On the other hand, the rational function $f=A/B$ is a ratio
of two polynomials of degree $N$. So the problem is to find
two polynomials of degree $N$ whose ratio has prescribed $2N-1$
coefficients at $0$. This is not difficult, and nowadays this
is called the "diagonal Pade approximation". The problem is also equivalent to the finite "moment problem", but the moment problem is usually studied when $x\_k$ are positive while $\zeta\_k$ and $\gamma\_j$ are real.
Remark. The trouble may arise when some $\zeta\_k$ are equal or
some $\alpha\_k$ are zero. Lyubich gives a (generic) necessary and sufficient condition on the RHS for these degenerations not to happen. For generic RHS, the system has unique solution.
| 7 | https://mathoverflow.net/users/25510 | 422584 | 171,841 |
https://mathoverflow.net/questions/422537 | 1 | **Motivation.** The following has a real-life (!) inspiration from a discussion about how to connect lamps and switches in an efficient way.
**Question.** Let $n\in\mathbb{N}$ be a positive integer and let $\{1,\ldots,n\}$ represent $n$ lamps, each of which is in exactly one of the states OFF or ON. Let $E\subseteq {\cal P}(\{1,\ldots,n\})$. For every $e\in E$ we have an "*$e$-button*", such that if that button is pressed, every element of $e$ switches its state (either from OFF to ON, or vice versa). We say that $E$ is *state-complete* if the following condition holds:
>
> If all lamps are OFF and if $k\in \{1,\ldots, n\}$ is given, there is a finite button-sequence $e\_1, \ldots, e\_m \in E$ such that after all the buttons have been pressed, lamp $k$ is ON and all other lamps are OFF.
>
>
>
For instance, $\big\{\{k\}: k\in \{1,\ldots,n\}\big\}$ is state-complete, and $\big\{\{1,\ldots,n\}\big\}$ is not state-complete for $n\geq 2$.
Given $n\in \mathbb{N}$, what is the least cardinality that a state-complete set $E\subseteq {\cal P}(\{1,\ldots,n\})$ can have?
| https://mathoverflow.net/users/8628 | "Lamp-switch set-up number" of $n$ | This problem is a sort of generalization of the game 'Lights out'.
There are a few paper's e.g. by M. Zaidenberg
(for example "Periodic harmonic functions on lattices and points count in positive characteristic.")
on the underlying mathematics.
| 1 | https://mathoverflow.net/users/4556 | 422585 | 171,842 |
https://mathoverflow.net/questions/422587 | 0 | Given a (separable) Hilbert space **H** and an unbounded densely defined linear operator $T:{\cal D}(T) \to $**H** such that ${\cal D}$ is **diagonalizable** (it means $\exists$ an O.N.B. of **H** such that all basis elements are eigenvectors of $T$). As normal, take the point spectrum of $T$ to mean those $\lambda \in \sigma(T)$ such that $\lambda$ is an eigenvalue of $T$. Will the cloure of the point spectrum of $D$ be equal to the spectrum of $D$?
| https://mathoverflow.net/users/128876 | Closure of the point spectrum of an unbounded diagonalizable operator | From your assumption you can easily see that $T$ is unitarily similar to a multiplication operator on $\ell^2$ (and thus, $T$ is normal, by the way). This shows that the answer is "yes" (as it is easy to analyse the spectrum of multiplication operators on $\ell^2$).
| 1 | https://mathoverflow.net/users/102946 | 422591 | 171,844 |
https://mathoverflow.net/questions/422582 | 39 | When I was an undergrad, the field of [spherical trigonometry](https://en.wikipedia.org/wiki/Spherical_trigonometry) was cited as a once-popular area of math that has since died. Is this true? Are the results from spherical trigonometry relevant for contemporary research?
| https://mathoverflow.net/users/128876 | Is spherical trigonometry a dead research area? | It is not. As a proof, I will mention three relatively recent papers where I am a co-author:
M. Bonk and A. Eremenko, [Covering properties of meromorphic functions, negative curvature and spherical geometry](https://arxiv.org/abs/math/0009251), Ann of Math. 152 (2000), 551-592.
A. Eremenko, Metrics of positive curvature with conic singularities on the sphere, Proc. AMS, 132 (2004), 11, 3349--3355.
A. Eremenko and A. Gabrielov, [The space of Schwarz--Klein spherical triangles](http://jmag.ilt.kharkov.ua/jmag/pdf/16/jm16-0263e.pdf), Journal of Mathematical Physics, Analysis and Geometry, 16, 3 (2020) 263-282.
As you see, they are all published in mainstream math journals. All contain some new results on spherical triangles.
And I am not the only person who is involved in this business:
Feng Luo, [A characterization of spherical polyhedral surfaces](https://doi.org/10.4310/jdg/1175266233),
J. Differential Geom. 74(3): 407-424.
Edit. To address one comment: here is a forthcoming [conference on spherical geometry](http://www1.mat.uniroma1.it/people/mondello/ricerca/spherical2022/index.html)
| 60 | https://mathoverflow.net/users/25510 | 422593 | 171,845 |
https://mathoverflow.net/questions/422611 | 2 | Consider a function
$$
f:\mathbb{R}^n\rightarrow\mathbb{R}^m
$$
given by $m$ functions $f\_i:\mathbb{R}^n\rightarrow \mathbb{R}$ that we can assume to be polynomials in $x\_1,\dots,x\_n$.
Does there exist any formula expressing $f$ as a linear combination of $\frac{\partial f}{\partial x\_1}(x\_1,\dots,x\_n),\dots,\frac{\partial f}{\partial x\_n}(x\_1,\dots,x\_n)$, $f(x\_1,0,\dots,0), \frac{\partial f}{\partial x\_1}(x\_1,0,\dots,0),\dots,\frac{\partial f}{\partial x\_n}(x\_1,0,\dots,0)$, where the coefficients of the linear combination are functions of $x\_1,\dots,x\_n$? Here $\frac{\partial f}{\partial x\_i} = \left(\frac{\partial f\_1}{\partial x\_i},\dots,\frac{\partial f\_m}{\partial x\_i}\right)$.
Basically I am asking for an analogue of Euler's formula for homogeneous polynomials which says that if $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a homogeneous polynomial of degree $d$ then $d\cdot f = x\_1\frac{\partial f}{\partial x\_1}(x\_1,\dots,x\_n) + \dots + x\_n\frac{\partial f}{\partial x\_n}(x\_1,\dots,x\_n)$.
Thank you.
| https://mathoverflow.net/users/482323 | Expressing a vector valued function in terms of its derivatives | $\newcommand{\pa}{\partial}\newcommand{\R}{\mathbb R}$The answer is no. Indeed, suppose the contrary: that for each polynomial $f$ there are functions $a\_j,b,c\_j$ such that
\begin{equation}
f(x\_1,\dots,x\_n)=\sum\_{j\in[n]}a\_j(x\_1,\dots,x\_n)(\pa\_j f)(x\_1,\dots,x\_n) \\
+b(x\_1,\dots,x\_n)f(x\_1,0\dots,0) \\
+\sum\_{j\in[n]}c\_j(x\_1,\dots,x\_n)(\pa\_j f)(x\_1,0\dots,0) \tag{1}\label{1}
\end{equation}
for all $(x\_1,\dots,x\_n)\in\R^n$, where $[n]:=\{1,\dots,n\}$ and $\pa\_j f$ denotes the partial derivative of $f$ with respect to its $j$th argument.
Let now
\begin{equation}
f(x\_1,\dots,x\_n):=3x\_2^2-2x\_2^3.
\end{equation}
Then $(\pa\_j f)(1,1,0,\dots,0)=f(1,0\dots,0)=(\pa\_j f)(1,0\dots,0)=0$ for all $j\in[n]$, so that the right-hand side of \eqref{1} for $(x\_1,\dots,x\_n)=(1,1,0,\dots,0)$ is $0$, whereas the left-hand side of \eqref{1} for $(x\_1,\dots,x\_n)=(1,1,0,\dots,0)$ is $f(1,1,0,\dots,0)=1\ne0$, which contradicts \eqref{1}. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 422621 | 171,857 |
https://mathoverflow.net/questions/422622 | 2 | I would like to ask a question with possibly a reference. If we have a Schrödinger operator $-\Delta+V$ on an interval $[0,L]$ with $V$ continous and Dirichlet conditions, can we state that the eigenfunctions of such operator are uniformly bounded, i.e. there exists $M>0$ such that the eigenfunctions $\{\phi\_n\}\_n$ satisfy
\begin{equation\*} \sup\_n \lvert\lvert \phi\_n \rvert\rvert\_\infty\leq M \end{equation\*}
?
| https://mathoverflow.net/users/482337 | Uniform boundedness Schrödinger operator eigenfunctions with Dirichlet conditions | Yes, this follows because asymptotically, as $|z|\to\infty$, the solutions of $-y''+Vy=zy$ look like those of the free equation $V\equiv 0$, and the eigenfunctions of $-y''=zy$, $\phi\_n=(2/(L\pi ) )^{1/2}\sin n\pi x/L$, are uniformly bounded.
In fact, they are uniformly bounded not just in $n$, but also in the potential $V$ as long as we impose a uniform bound on $\|V\|\_1$.
(I'm assuming here that you normalize your eigenfunctions as usual, $\|\phi\_n\|\_2=1$, and then you are asking about $\|\phi\_n\|\_{\infty}$.)
| 2 | https://mathoverflow.net/users/48839 | 422625 | 171,858 |
https://mathoverflow.net/questions/422631 | 8 | Let $X\_1,\ldots,X\_n$ be $\{0,1\}$-valued random variables drawn from some joint distribution. Let $\tilde X\_1,\ldots,\tilde X\_n$ be their *independent version*: $\mathbb{E}X\_i=\mathbb{E}\tilde X\_i$ for each $1\le i\le n$ and the $\tilde X\_i$ are mutually independent (as well as being independent of the $X\_i$'s). I would like to compare
$M:=\mathbb{E}\max\_{1\le i\le n}X\_i$
and
$\tilde M:=\mathbb{E}\max\_{1\le i\le n}\tilde X\_i$.
It's obvious that $M$ can be made arbitrarily small, while $\tilde M$ is arbitrarily close to $1$, so in general, there can be no bound on $\tilde M/M$. But it seems that in the reverse situation, one should be able to say something. Is some bound of the form
$$ M \le c\tilde M $$
known, where $c$ is an absolute constant? If such a thing is impossible (what's the counterexample?), what's the best dependence on $n$ in the bound?
| https://mathoverflow.net/users/12518 | Max decoupling inequality | Yes, this inequality holds with
$$c:=\frac e{e-1}.$$
Indeed, let $A\_i:=\{X\_i=1\}$. Then
$$M=P\Big(\bigcup\_i A\_i\Big)\le\min(s,1)\le c(1-e^{-s}),$$
where $s:=\sum\_i P(A\_i)$. On the other hand,
$$\tilde M=1-\prod\_i(1-P(A\_i))
\ge1-\prod\_i\exp(-P(A\_i))=1-e^{-s}.$$
So, $M\le c\tilde M$.
| 11 | https://mathoverflow.net/users/36721 | 422636 | 171,862 |
https://mathoverflow.net/questions/422518 | 2 | Consider an uncountable perfect $K\_\sigma$ set $X\subseteq \omega^\omega$, where $K\_\sigma$ means countable union of compact sets, perfect means that $X$ has no isolated points and $\omega^\omega$ is the Baire space. Suppose now that there exists a subset $A\subseteq X$ which is dense in $X$ (wrt the subspace topology).
My question is:
* Can we prove that there exists a Cantor set $\mathcal{C}\subseteq X$ such that $\mathcal{C}\cap A$ is dense in $\mathcal{C}$?
Since $X$ is uncountable it must contain a Cantor set, as all analytic sets have the perfect set property, but here we are looking for a Cantor set that preserves the property of $A$ of being dense.
Ideas?
Thanks!
| https://mathoverflow.net/users/141146 | Strong form of $\mathtt{PSP}$ for $K_\sigma$ sets | Here's a counterexample. Let $X$ be the set of bounded sequences, and let $A$ be the set of sequences which have only finitely many nonzero terms and achieve a strict maximum at the last nonzero term. Clearly $A$ is a dense subset of $X.$
Let $P \subset \omega^{\omega}$ be a perfect set such that $P \cap A$ is dense in $P.$ We will show $P \not \subset X.$
Enumerate $P \cap A.$ Let $p\_0$ be the first sequence, and recursively define $p\_{i+1}$ to be the first sequence such that the nonzero part of $p\_{i+1}$ strictly extends the nonzero part of $p\_i$ (here we are using that $p\_i$ is a limit point of $P$). Clearly $\langle p\_i \rangle$ converges to an unbounded sequence $p,$ so $p \in P \setminus X.$
| 4 | https://mathoverflow.net/users/109573 | 422644 | 171,863 |
https://mathoverflow.net/questions/422660 | -1 | Let $ A(x)=(a\_{ij}(x)):\mathbb{R}^d\to\mathbb{R}^{d\times d} $ be a matrix satisfying ellipticity condition, that is
\begin{align}
\mu^{-1}|\xi|^2\geq \sum\_{i,j=1}^da\_{ij}(x)\xi\_i\xi\_j\geq\mu|\xi|^2
\end{align}
with $ \mu>0 $ being a positive constant and $ \xi\in\mathbb{R}^d $. Since the spectrum of $ L=-\operatorname{div}(A(x)\nabla) $, for $ \zeta\in\mathbb{C}\backslash\mathbb{R}\_+ $, $ (L-\zeta I)^{-1} $ exists. I read some paper and know that for some special $ A $ and $ p $, we can derive that
\begin{align}
\left\|(L-\zeta I)^{-1}\right\|\_{L^p\to L^p}\leq \frac{C}{1+|\zeta|}.
\end{align}
where $ C $ may depends on $ \arg\zeta $. I want to know what is the motivation to study such estimates and where can I use such estimates.
| https://mathoverflow.net/users/241460 | Applications and motivations of resolvent for elliptic operator | To begin with, the ellipticity condition is useless if you don't ask also that
$$\sum\_{i,j}a\_{ij}\xi\_i\xi\_j\le M|\xi|^2$$
for some finite constant $M$.
Now the resolvant estimate is used to define an operator $e^{-zL}$ for $\Re z>0$. In particular $S\_t=e^{-tL}$ is the semi-groups associated with the evolution equation
$$\partial\_tu+Lu=f$$
where $f(t,x)\in L^1(0,T;L^p)$ is given, as well as an initial value $u\_0\in L^p$.
| 2 | https://mathoverflow.net/users/8799 | 422662 | 171,869 |
https://mathoverflow.net/questions/422672 | -1 | Can one prove for a sequence of positive random variable $X\_{n}$ such that $\lim\_{n\to \infty}E[x\_{n}] = 0$ and $\lim\_{n\to \infty}E[x\_{n}x\_{n}]= 0$ all the cumulants go to zero once $n\to \infty$ ?
| https://mathoverflow.net/users/482383 | Cumulants of a sequence of variables with zero mean and variance | Counter example: probability distribution of $X\_n$ given by
$$P\_n(x)=\frac{3n}{(1+nx)^4},\;\;x\geq 0,$$
properly normalized to unity. Then $\mathbb{E}[X\_n]=1/2n$ and $\mathbb{E}[X\_n^2]=1/n^2$ both vanish in the limit $n\rightarrow\infty$, but higher moments and cumulants diverge.
Alternatively, for an example where the moments do not diverge for finite $n$, let $X\_n$ take the value $1/n$ with probability $1-1/n$ and the value $n^{1/3}$ with probability $1/n$. Then $\mathbb{E}[X\_n]$ and $\mathbb{E}[X\_n]^2$ both vanish in the limit $n\rightarrow\infty$, while $\mathbb{E}[X\_n^3]$ tends to unity.
| 1 | https://mathoverflow.net/users/11260 | 422674 | 171,872 |
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