Is there a "direct" proof of the Galois symmetry on centre of group algebra?

Let $G$ be a finite group, and $n$ an integer coprime to $|G|$. Then we have the following map, which is clearly not a morphism of groups in general: $$g\mapsto g^n.$$

This induces a linear automorphism of $\mathbb{Z}[G]^G$, the algebra of $G$-invariant functions on $G$ under convolution, and surprisingly, this induced map is also an algebra automorphism, as can be seen by passing to $\mathbb{C}$ and noting that this is the Galois action on characters, which permutes the set of primitive idempotents of this algebra.

My question is whether this surprising fact can be explained directly, without using character theory? I would be interested in both a high-concept explanation of this symmetry, or a generators and relations argument for why there exists, for any $g,h$ in $G$, an $a,b\in G$ with: $$(gh)^n=ag^n a^{-1}bh^n b^{-1}.$$

By a generators and relations argument I would think some symbolic argument that holds in any torsion group of exponent coprime to $n$, building $a,b$ from $g,h$ using these assumptions and the assumed $n$th root function. If one could give a group of this type where the result fails, it would rule out such an argument, which would also be very interesting.

What do you mean by "a generators-and-relations argument for... $(gh)^n=ag^na^{-1}bh^nb^{-1}$"? I doubt there exist such $a,b$ in the free group over $g,h$. For $n=2$ there are none: any $x,y,z$ in a free group satisfying $x^2y^2=z^2$ must commute (it follows there are none whenever $n$ is even).

A slightly different (but definitely character theoretic) proof of this interesting fact uses the structure constants $a_{KLM}=\frac{|K||L|}{|G|}\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi(x_K)\chi(x_L)\overline{\chi(x_M)}}{\chi(1)}.$ If $K^\prime$ etc. are the classes containing the respective $n^{th}$ powers and if $\sigma$ is Chris' automorphism then $a_{K^\prime L^\prime M^\prime}=\frac{|K||L|}{|G|}\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi^\sigma(x_K)\chi^\sigma(x_L)\overline{\chi^\sigma(x_M)}}{\chi(1)}=a_{KLM}^\sigma,$ so $a_{K^\prime L^\prime M^\prime}=a_{KLM}^\sigma=a_{KLM}.$

There are a number of similar facts which seem non-obvious from a group-theoretic point of view, but are easy to prove with characters: for example, the number of times $ x \in G$ is expressible as a commutator is unchanged if we replace $x$ by a different generator of  $\langle x \rangle.$

@FedorPetrov: Note the assumption $\gcd(n, |G|) = 1$. So if $g^n = 1$, then $g = 1$.

The Frobenius group of order $20$ is an example, trivial outer automorphism group, with nonrational characters, take $n=3$.

Basic question: What is an example of a finite group $G$ and an integer $n$ prime to $|G|$ such that there is no automorphism of $G$ which induces the permutation $\text{Conj}(g) \mapsto \text{Conj}(g^n)$ on conjugacy classes?

Wow, that's a beautiful find. I am getting a slightly cohomological whiff from it (as, e.g., in the proof of the transfer's homomorphism property).

There is a theory of products and powers of conjugacy classes; see Beltrán, Felipe, and Melchor - Some problems about products of conjugacy classes in finite groups for what appears to be a recent survey.  It seems to be in a different direction, but I wonder if that might be the place to look.

Its true, and follows from the fact that $g\mapsto g^n$ is an algebra automorphism of $\mathbb{Z}[G]^G$, by expanding the product of the classes of $g$ and $h$. I don't know of a non character theoretic proof of why this map is actually an algebra automorphism however.

Is the quoted equality about $(g h)^n$ true, but you don't know how to prove it, or you're not sure if it's true?


monoidal (∞,1)-categories from weakly monoidal model categories

In _Higher Algebra_ section 4.1.7, Jacob Lurie proves that the underlying $(\infty,1)$-category of a monoidal model category is a monoidal $(\infty,1)$-category.

Dominic Verity and Yuki Maehara have (independently) defined the lax Gray tensor product for model categories $S$ and $T$ for two different models of $(\infty,2)$-categories and proved that their tensor product is "weakly monoidal" in various senses:

  * Maehara provides functors $\otimes_n \colon T^{\times n} \to T$ for each $n \geq 2$ that define the left adjoint parts of $n$-variable Quillen adjunctions. But these functors are associative in a weak sense, eg with a zig zag of natural weak equivalences

$$ X \otimes_2 (Y \otimes_2 Z) \to \otimes_3(X,Y,Z) \leftarrow (X \otimes_2 Y) \otimes_2 Z.$$

  * Verity constructs two binary tensor products $\otimes, \otimes' \colon S \times S \to S$ that are naturally weakly equivalent, one of which is a closed left Quillen bifunctor and the other of which is associative up to isomorphism.

Morally, this weakness should not matter for the underlying $(\infty,1)$-category, but has anyone verified that a monoidal structure on the underlying $(\infty,1)$-category can be produced under these weaker hypotheses?

Alternatively, would someone like to take this on? I'd be happy to help publicize a result along these lines whenever it appears.

(FYI: Andrea Gagna, Yonatan Harpaz, Edoardo Lanari have a third version of the lax Gray tensor product that defines a monoidal model category in the usual sense. But I'm still interested in the question of about weak monoidal model categories, which is why I've set this aside.)

There was further discussion in chat. To sum up: the suggestion from my second comment was misguided because Lurie really only deals in reflective localizations, and so can't say much about the localization from a model category to its associated $\infty$-category. But Rune points out that Hinich has results on how DK localization interacts with cocartesian fibrations which could do the trick if generalized to the locally cocartesian case.

@AlexanderCampbell Great, that's good to know! Regarding pre-complicial sets, Verity indeed showed that the Gray tensor product is monoidal biclosed on them, and Yuki, Chris Kaupulkin and I observed in Theorem 1.33 here that the usual model structures are monoidal with respect to it. Relatedly, in that same preprint, we construct a monoidal biclosed lax Gray tensor product on a category of marked cubical sets (without having to futz around with the category a la precomplicial sets) and show it's a monoidal model category.

@TimCampion Yuki's structure is indeed an example of a lax monoidal category (which is the standard name in the literature). Lurie's "corepresentable operads" correspond to (symmetric) colax monoidal categories. Of course, the opposite of a lax monoidal category is a colax monoidal category, so I agree that this a promising approach.

In the second situation, isn't the full subcategory of pre-complicial sets a genuine monoidal model category?

Hiya Emily, nice to see you here.  I think that this question was explored in the paper by Heuts, Hinich, and Moerdijk that fixes all of the mistakes from Moerdijk-Weiss.  I'm not sure if they verified transport of structure to the associated ∞-category, but it's another data point: arxiv.org/abs/1305.3658

Localization is a product-preserving functor from relative categories to $\infty$-categories. That means it's compatible with the enrichment of each category in itself, which should give you enough structure to say a functor of 2-categories to the 2-category of relative categories gives after localization one to the $(\infty,2)$-category of $\infty$-categories, which would give another approach to the first part. Unfortunately this kind of naturality for enrichments from module structures hasn't been worked out yet, however, though I believe Hadrian Heine is working on this.

It's conceivable that Lurie's framework might even be helpful here -- if you're interested mostly in the $\infty$-category presented by your model category, then maybe some of Lurie's general results on localization for $\infty$-operads could be used to say things like "there is a monoidal structure on the presented $\infty$-category".

Ignoring the model structure, I would like to call the situation of the first bullet a "lax monoidal category" (or maybe "oplax"?). I've only seen lax monoidal categories (or rather their dual) in one source, namely Higher Algebra Definition 6.2.4.3 where they're called "corepresentable ($\infty$-)operads" (of course, it's jazzed up to be $\infty$ everything). I think another example of a (op)lax monoidal category (interacting nicely with a model structure) would be Boardman's handicrafted smash products of spectra.


Is every positive integer the rank of an elliptic curve over some number field?

For every positive integer $n$, is there some number field $K$ and elliptic curve $E/K$ such that $E(K)$ has rank $n$?

It's easy to show that the set of such $n$ is unbounded. But can one show that _every_ positive integer is the rank of some elliptic curve over a number field?

The analogous question for a fixed number field is expected to have a negative answer (c.f. e.g., [this question](https://mathoverflow.net/questions/137970/is-it-expected-that-every-natural-number-is-the-rank-of-some-elliptic-curve-over)) but is still conjectural. But I wonder if one might be able to prove a positive answer to the question I asked above.

The "trick" here is we can use $q \to\infty$ results because the question lets us pass to any finite extension. A similar result would follow over number fields from heuristics for the distribution of $L$-functions. In particular, it requires our heuristics to be true only in the 100% setting, and not in the power savings setting rqeuired by the boundedness of rank arguments over the integers.

So for $q$ sufficiently large we can show the existence of an elliptic curve $E$ and discriminants $D_1,\dots, D_k$ such that of the $2^k$ twists generated by $D_1,\dots, D_k$, $n$ have analytic rank $1$and the rest have rank $0$. Then one just has to use the analytic rank $1$ part of the BSD conjecture (which I think is now known in full generality over function fields?) and deduce that $n$ of these twists have algebraic rank $1$, hence the pullback to the composite of quadratic extensions has rank $n$.

It should be possible to prove the analogous statement over function fields. For a quadratic twist of a fixed elliptic curve $E$, Katz proved that $1/2$ the twists have analytic rank $0$ and half the twists have analytic rank $1$, in the large $q$ limit. It should be possible to check by explicitly examining monodromy that the root numbers of different quadratic twists are suitably independent.

@VesselinDimitrov Yes it's also natural to consider the Tate-Shafaverich group, from the point of view of the BSD conjecture and analytic class number formula. Both are special cases of the Tamagawa number conjecture, which uses Galois cohomology. The class number arises as the boundary in the Poitou-Tate localization sequence. The ideal class group and the Mordell-Weil group are both Picard groups but for Dedekind domains which are of somewhat different nature.

@FrançoisBrunault: That rel Picard over $\mathbb{Q}$, though, kills the much more mysterious Shafarevich-Tate part; the Selmer group I think should be the true analogy. It is very easy to construct many $E/K$ of rank zero, much harder to construct number fields of class number one (apart from finitely many known cases).

@FrançoisBrunault: I didn't think of this, thanks for noting it! It makes sense, then. I assumed the class group was similar to a Selmer group, which includes only an upper bound on the rank. Still it looked to me that Stanley's question should be harder than the current one, but I could be off.

@VesselinDimitrov The Mordell-Weil group of $E$ is isomorphic to the relative Picard group of $E/\mathbb{Q}$, so this is a natural analogue. By the way it is  known that every abelian group is the Picard group of some Dedekind domain (Claborn 1966). One may ask the same thing while restricting the class of abelian groups and/or Dedekind domains considered.

@StanleyYaoXiao: Your question should be infinitely harder than this one. I don't think there is much of a similarity between a class number and a Mordell-Weil rank.

This other question I asked may be of relevance: mathoverflow.net/questions/88175/…

Sorry, I realized that my comment sounds less than clear, so let me add that apart from the remark in the first sentence, I do regard there both $K$ and $E$ as varying. Basically I am saying that a version of Silverman's specialization theorem can be used $r$ times over an $r$-dimensional char. $0$ base $T$ assuming that there exists an elliptic scheme $A/T$ such that $A(T) / \mathrm{tors} \cong \mathbb{Z}^n$. From this point of view the problem is geometric, and I am guessing that an explicit such $A/T$ is given by $A_{1,n+1}  / A_{1,n}$ (but any other construction would suffice).

@Wojowu: I am merely pointing out the existence of the phenomenon 'idle curiosity'. If you would label Goldbach's conjecture as such, that is up to you. Anyway, I wanted to give my reasons for my downvote, not enter into a discussion about the value of mathematics, so I will leave it here.

One may ask more generally: given a representation of a finite group $\rho : G \to \mathrm{GL}_n(\mathbb{Q})$, does $\rho$ arise as the Galois module $E(L) \otimes \mathbb{Q}$ for some elliptic curve $E/K$ and some extension $L/K$ with Galois group $G$. We can also fix the number field $K$. If I recall correctly, if one replaces "arises" by "appears inside" then for function fields $K=\mathbb{F}_p(t)$ these kinds of things are known.

@RP: What would be the use of knowing the answer of any question on this website? Unless you're only in it for cryptographic applications, this question is useful because of human curiosity.

@RP_ I would -1 your comment if I could. Since when does usefulness of the result dictate whether the question is good for this site? Which ranks are possible over a fixed number field is similarly useless, and yet it's a very interesting open problem.

-1. I think this question is a little silly. What could possibly be the use of knowing the answer? I mean, if you could reduce the Goldbach conjecture for $2n$ to the existence of an elliptic curve of rank $n$ over some number field,  then I would see the point. As the matter stands I'm afraid I don't.

I am not sure whether it is fair to say that the fixed number field case should admit a negative answer; of course that comes down to a famous unsolved problem. For the existence of $E/K$, an iterated application of Silverman's specialization theorem reduces us to proving that the group of sections of the elliptic scheme $A_{1,n+1} \to A_{1,n}$ has rank $n$, generated by the $n$ 'point doubling' sections $(x_1, x_2,\ldots,x_n) \mapsto (x_1,x_1;x_2,\ldots,x_n)$, etc. (Here $A_{g,n}$ is the space of $n$-pointed abelian varieties of dimension $g$). This should be true.


Non-rigid ultrapowers in $\mathsf{ZFC}$?

_Originally[asked and bountied at MSE](https://math.stackexchange.com/questions/4293034/non-rigid-ultrapowers-in-mathsfzfc):_

> **Question** : Can $\mathsf{ZFC}$ prove that for every countably infinite structure $\mathcal{A}$ in a countable language there is an ultrafilter $\mathcal{U}$ on $\omega$ such that the ultrapower $\mathcal{A}^\mathcal{U}$ has a nontrivial automorphism?

"Obviously" the answer is yes! _(... right?)_

Note that this is trivial in $\mathsf{ZFC+CH}$: $\mathsf{ZFC}$ alone proves that any nontrivial ultrapower is countably saturated, which in the presence of $\mathsf{CH}$ lets us run a back-and-forth argument to show that $\mathcal{A}^\mathcal{U}$ is non-rigid whenever $\mathcal{A}$ is countable and $\mathcal{U}$ is a nonprincipal ultrafilter on $\omega$. Separately, note that if we drop the requirement that the ultrafilter be on $\omega$ a positive answer follows from the Keisler-Shelah theorem, but [$\mathsf{ZFC}$ can't bring down KS to $\omega$ as one might expect](https://shelah.logic.at/files/130108/vive.pdf), so that ultimately doesn't seem like a promising direction.

I strongly suspect that I'm missing a very simple argument, but at present I'm not seeing it.

_As a minor aside, I'm especially interested in a proof which avoids using results about first-order logic (techniques from the model-theory of first-order logic are fine). This comes from the original motivation for this question, which was a weakness in[an MSE answer of mine](https://math.stackexchange.com/questions/41469/non-axiomatisability-and-ultraproducts/3569736#3569736) where the whole point was to avoid some standard model theory. However, in retrospect this seems to be jumping the gun here, so I'll relegate this aspect of the question from the question itself to this mere side comment._

@PaulLarson " An ultrapower of a structure of infinite Scott height by an ultrafilter on omega shouldn't have its points separated by its Scott process. Should that make it nonrigid?" I'm not certain, I'm not too familiar with Scott processes.

The proof of Harrington's theorem on the Scott ranks of counterexamples to Vaught's Conjecture should show that separating points characterizes rigidity for structures of size aleph_1 too.

A countable structure is rigid if and only if its Scott process separates points. An ultrapower of a structure of infinite Scott height by an ultrafilter on omega shouldn't have its points separated by its Scott process. Should that make it nonrigid?

@PaulLarson Ooh, fun question (although not directly related to this one unless I'm missing something?) - I can't think of anything off the top of my head, but this is well outside my normal sphere so that doesn't mean much.

Is there anything known about rigid structures whose countable elementary substructures are all non-rigid?

@FarmerS Good question! :P The best I can say is that Shelah's Vive la difference series probably has some relevant info, per Douglas Ulrich's comment, but I haven't worked through those papers to see if there's an answer there.

Maybe this is easy, but do we know there is a countably infinite structure and some non-principal ultrafilter on $\omega$ such that the ultraproduct is rigid?


A spin extension of a Coxeter group?

Consider a Coxeter group $W$ with simple generators $S$ and Coxeter matrix $\left( m_{s,t}\right) _{\left( s,t\right) \in S\times S}$.

Let $\mathfrak{M}$ be the set of all pairs $\left(s, t\right) \in S^2$ satisfying $s \neq t$ and $m_{s,t} < \infty$.

For every $\left( s,t\right) \in \mathfrak{M}$, let $c_{s,t}$ be an element of $\left\\{ 1,-1\right\\} $. Assume that:

  * We have $c_{s,t}=c_{s^{\prime},t^{\prime}}$ for any two elements $\left( s,t\right) $ and $\left( s^{\prime},t^{\prime}\right) $ of $\mathfrak{M}$ for which there exists some $q \in W$ satisfying $s^\prime = qsq^{-1}$ and $t^\prime = qtq^{-1}$.

  * We have $c_{s,t}=c_{t,s}$ for each $\left( s,t\right) \in \mathfrak{M}$.

Let $W^{\prime}$ be the group with the following generators and relations:

_Generators:_ the elements $s\in S$ and an extra generator $q$.

_Relations:_ \begin{align*} s^{2} & =1\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S;\\\ q^{2} & =1;\\\ qs & =sq\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S;\\\ \left( st\right) ^{m_{s,t}} & =1\ \ \ \ \ \ \ \ \ \ \text{for every }\left( s,t\right) \in \mathfrak{M} \text{ satisfying } c_{s,t}=1;\\\ \left( st\right) ^{m_{s,t}} & =q\ \ \ \ \ \ \ \ \ \ \text{for every }\left( s,t\right) \in \mathfrak{M} \text{ satisfying } c_{s,t}=-1. \end{align*}

There is clearly a surjective group homomorphism $\pi:W^{\prime}\rightarrow W$ sending each $s\in S$ to $s$, and sending $q$ to $1$. There is also a group homomorphism $\iota:\mathbb{Z}/2\mathbb{Z} \rightarrow W^{\prime}$ which sends the generator of $\mathbb{Z}/2\mathbb{Z}$ to $q$.

> **Question.** Is $\iota$ injective? Equivalently, is the sequence \begin{equation} 1\longrightarrow\mathbb{Z}/2\mathbb{Z}\overset{\iota}{\longrightarrow}W^{\prime }\overset{\pi}{\longrightarrow}W \longrightarrow1 \end{equation} exact? Equivalently, is $\left\vert \operatorname*{Ker}\pi\right\vert =2$ ?

**Background.** This would generalize at least one of the two "spin symmetric groups" to the situation of any Coxeter group. It would explain one of the results (Theorem 2.3 **(b)**) in [Alexander Postnikov and Darij Grinberg, _Proof of a conjecture of Bergeron, Ceballos and Labbé_](http://nyjm.albany.edu/j/2017/23-70.html), and prove a generalization of this result (Conjecture 6.1 **(b)**).

I have tried generalizing the [standard approach to constructing the spin symmetric groups by embedding them in the Hecke-Clifford algebra](https://arxiv.org/abs/1110.0263), but to no avail so far. Nor has the existing literature on central extensions of Coxeter groups been particularly helpful ([Howlett's _On the Schur multipliers of Coxeter groups_](http://onlinelibrary.wiley.com/doi/10.1112/jlms/s2-38.2.263/abstract) counts the extensions abstractly, but doesn't help understanding whether a given one exists; [Burichenko's _On extensions of Coxeter groups_](http://www.tandfonline.com/doi/abs/10.1080/00927879508825315?journalCode=lagb20) gives a criterion that I don't seem to properly understand, as it gives me wrong answers).

@DavidESpeyer: Thanks -- I've answered the Question positively in the meantime (a year ago), but only in a month or so will probably be able to write up my proof. Meanwhile, I did look into Howlett's and others' Schur-multiplier papers, but never found myself able to get something out of it that wasn't obviously wrong; it's too much of a foreign language to me.

Morris, "Projective representations of finite reflection groups. III.", Comm. Alg. (2004), is doing something very like this, although I haven't unwound his notation to see whether it exactly matches your formulation or not. The title of that paper has the word finite in it, but most of the results don't seem to need it. See also Howlett, "On the Schur multipliers of Coxeter groups", JLMS (1988), which is less concrete but is better about not including unneeded finiteness hypotheses.

This is a quotient of the Tits group (MSN) in a natural way.  It seems like maybe Théorème 2.5 there will give you what you want, although I can't see it right away.

@VictorProtsak: Thanks. I've reverted to the $\mathfrak{M}$ notation from the paper.

Sure, $s\ne t$. Actually, your last relation involving $q$ should also have restriction $s\ne t$.

@VictorProtsak: I take it you want $s \neq t$, since otherwise picking $s=t$ and $s' = t'$ to be two non-conjugate simple reflections would break it. Anyway, nice question! Perhaps what we did in the proof of Claim 1 in the proof of Lemma 4.0.3 in Alex's and my paper could be of use.

Darij, very interesting question! Clearly your two conditions are necessary: conjugate elements have the same order. But are they sufficient to guarantee that if $st$ and $s't'$ have the same order in $W$, then $c_{s,t}=c_{s',t'}$ ? More precisely, do you know whether the following statement is true for general $W$? $$ $$ If $st$ and $s't'$ are conjugate in $W$ then there is $q\in W$ simultateously conjugating either $s$ to $s'$ and $t$ to $t'$ or $s$ to $t'$ and $t$ to $s'$.

Jim, for any finite dihedral group (finite Coxeter group of rank 2) $\langle s,t\rangle\simeq D_m$ of type $I_2(m)$, in the nontrivial case $c_{s,t}=-1$ this construction produces the dihedral group $D_{2m}$ of type $I_2(2m)$ with double the size. So starting from $W(G_2)=D_6$ you would get $D_{12}$.

@Gro-Tsen: Interesting idea; but this would only give two "spin extensions" (and I have no idea how to precisely find their generators and relations), whereas my $c_{s,t}$ in general allow for much more freedom. (Not saying that the approach is broken; could be my extensions are all equivalent.)

@darij:  This is an intriguing area which I haven't gone into, but I wonder for example what would happen in the frequent case that -1 is in a given finite irreducible Coxeter group $W$?   (This is a special case of the longest element of $W$, which has been well studied using the classification.)    A small example is the $G_2$ case, though symmetric groups of rank > 1 aren't examples.

I had in mind that every Coxeter group can be realized as isometries of a certain "canonical" symmetric bilinear form (not necessarily definite), so that it embeds in $\mathit{O}(p,q)$: wouldn't its inverse image in $\mathit{Pin}^+(p,q)$ or $\mathit{Pin}^-(p,q)$ answer your question?  Or am I talking nonsense?


Isoperimetric inequality and geometric measure theory

The following version of the isoperimetric inequality can be easily deduced from the Brunn-Minkowski inequality:

> **Theorem.** _If $K\subset\mathbb{R}^n$ is compact, then $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}\mu_+(K), $$ where $\omega_n$ is the volume of the unit ball and
>  $$ \mu_+(K)=\liminf_{h\to 0} \frac{|\\{x:\, 0<{\rm dist}\, (x,K)\leq h\\}|}{h} $$ is the Minkowski content._

If $K$ is the closure of a bounded set with $C^2$ boundary, then $\mu_+(K)=H^{n-1}(\partial K)$ (Hausdorff measure) so we have $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K). $$ However, the above inequality is true for _any_ compact set without assuming anything about regularity of the boundary. My question is:

## Is there a simple proof of the following result?

> **Theorem.** _If $K\subset\mathbb{R}^n$ is compact, then $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K). $$_

This result can be proved using the machinery of the geometric measure theory. If $H^{n-1}(\partial K)=\infty$, the inequality is obvious. If $H^{n-1}(\partial K)<\infty$, then $K$ has finite perimeter and the isoperimetric inequality for sets of finite perimeter yields $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}P(K)= n^{-1}\omega_n^{-1/n} H^{n-1}(\partial^* K) \leq n^{-1}\omega_n^{-1/n} H^{n-1}(\partial K), $$ where $P(K)$ is the perimeter of $K$ and $\partial^*K\subset\partial K$ is the reduced boundary. For details see [1].

Unfortunately, this argument is very far from being elementary.

**[1] Ambrosio, L., Fusco, N., Pallara, D.:** _Functions of bounded variation and free discontinuity problems._ Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000.


Hahn-Banach and the "Axiom of Probabilistic Choice"

Stipulate that the Axiom of Probabilistic Choice (APC) says that for every collection $\\{ A_i : i \in I \\}$ of non-empty sets, there is a function on $I$ that assigns to $i$ a finitely-additive probability measure $\mu_i$ on $A_i$.

Then Hahn-Banach (HB) implies APC, since HB is equivalent to the existence of a finitely-additive probability measure on every boolean algebra, and hence implies the existence of such a measure on the direct sum of the powerset algebras $P(A_i)$.

APC is non-trivial in that it implies the Banach-Tarski paradox and the existence of nonmeasurable sets (the proof Foreman and Wehrung use to show that HB implies Banach-Tarski works).

**Question:** Does APC imply HB?

(When I think about this, I find I keep on wanting to use Stone representation, but of course to do that would be to assume Boolean Prime Ideal, which is stronger than HB or APC.)

1. D. Pincus, The strength of Hahn–Banach's Theorem, in: Victoria Symposium on Non-standard Analysis, Lecture notes in Math. 369, Springer 1974, pp. 203-248. Implies Hahn-Banach theorem is weaker than the ultrafilter lemma

@Fedor: You might have remembered that HB+SKM (Strong Krein-Milman) imply AC.

@Wojowu ah, indeed, I remembered not well

@FedorPetrov Not at all. It is very much weaker. It follows from the ultrafilter lemma (and at the same time doesn't imply it).

If I remember well, HB is equivalent to AC.

Oh, I wasn't implying it's trivial. I was hoping for someone to complete the proof. :)

Well, using Bartle integrals you can get linear functionals on $\ell^\infty(A_i)$ which do not come from $\ell^1(A_i)$. And HB is equivalent to "Every Banach space has a nontrivial functional" (continuous or otherwise). So...


Looking for an effective irrationality measure of $\pi$

Most standard summaries of the literature on [irrationality measure](http://mathworld.wolfram.com/IrrationalityMeasure.html) simply say, e.g., that $$ \left| \pi - \frac{p}{q}\right| > \frac{1}{q^{7.6063}} $$ for all sufficiently large $q$, without giving any indication of how large qualifies as "sufficiently large." It would occasionally be useful (for example, [for my answer here](https://math.stackexchange.com/a/1313626/30836)) to be able to come up with explicit bounds on the size of $q$.

I don't have access to Salikhov's paper where the exponent $7.6063$ is obtained, but Hata's paper which obtained an exponent of $8.0161$ is online [here](http://matwbn.icm.edu.pl/ksiazki/aa/aa63/aa6344.pdf). Looking through it, the proof seems to be effective (all the bounds he's using appear to be defined explicitly), but it is definitely not written to be clear about what those bounds are.

Has anyone extracted explicit bounds on the size of the denominator from any irrationality measure calculation for $\pi$ more recent than Mahler's proof (referenced in the Hata paper) that $$ \left| \pi - \frac{p}{q}\right| > \frac{1}{q^{42}} $$ unrestrictedly?

Salikhov's paper is available here: mathnet.ru/links/93c99ab6587bdc1bf0912ec96b1d9f4f/rm9175.pdf

I wasn't aware this was migrated, I'll delete my answer in that case.

The exponent -42 got by Mahler was quite extraordinary in his time ("Striking inequality", Alan Baker). It is valid for all rational p/q with q > 1 and also indicates that rationals "may not be very closed to $\pi$". I guess all improvement of the exponent -42 is valid just "for all sufficiently large q" in the ordinary meaning of this expression without explicity giving of bounds. Maybe possible but certainly quite difficult to get them.

@muaddib: Nope, you must be thinking of some other Micah...

I know this is random but by any chance did you go to New College?


Is the determinant of cohomology a TQFT?

If $M$ is an oriented $d$-manifold, let $D(M)$ denote the top exterior power of $H^*(M,\mathbf{C})$. Then $D(M_1 \amalg M_2) = D(M_1) \otimes D(M_2)$. Is there a good recipe for a map $D(M) \to D(N)$ induced by a cobordism from $M$ to $N$?

In some dimensions, there is a natural identification $D(M) = \mathbf{C}$ and you could take every cobordism to the identity map. But such a TQFT would not be completely trivial when $d = 4$, I think. For example it would distinguish $S^1 \times \mathbf{C}P^2$ from the mapping torus of complex conjugation.

If there's no problem defining it for $(d+1)$-manifolds, can it be "extended down" any distance, i.e. associate something to a $(d-1)$-manifold?

I would have thought that the $d$-dimensional theory (or maybe I mean $d\pm 1$) would assign $\mathrm{pt} \mapsto \mathbb C$-as-an-$E_d$-algebra.  But that defines the trivial (framed) theory.  Perhaps there's a nontrivial choice of how to descend from a framed theory to an oriented one, but I'm not seeing it.

I'm sure the answer is "yes, $D$ is part of an invertible TFT", but I'm not seeing totally immediately how to write it down.  From my point of view, you want to look at the "free topological boson on $M$", which is the theory whose field is a map $\phi: M \to \mathbb C$ and whose EOM is $\mathrm d \phi = 0$.  This has a "cotangent quantization" in the Costello-Gwilliam language, which up to some degree shift is $D$.  If you can realize it as a fully extended TFT, you win.

This TFT would presumably be invertible, so can in principle be classified using the cohomology of Madsen-Tillman spectra. There's a known invertible 2d TFT which assigns to a closed oriented surface the number $\lambda^{\chi(X)}$ where $\chi$ is the Euler characteristic and $\lambda \in \mathbb{C}^{\times}$, and this TFT is a categorification of that. Maybe this paper of Freed will  be helpful: arxiv.org/abs/1406.7278


Homeomorphisms of the sphere mapping a geodesic triangulation to another one

Consider the standard Riemannian 2-sphere $S$, equipped with a geodesic triangulation $T$. Let $L(S,T)$ be the space of homeomorphisms of $S$ which map $T$ to a geodesic triangulation. What is the homotopy type of $L(S,T)$? A similar question has been solved by [E. Bloch, R. Connelly, D. Henderson, Topology 23 (1984), 161-175](http://www.sciencedirect.com/science/article/pii/0040938384900375), when $S$ is a simplicial disc in the plane, and it is a "tour de force"!

This question is on behalf of Jean Cerf, my former advisor, who is interested in S. Cairns' work.

I agree with the change of title. It is better, but I thought of (S,T) as a structure, not an object; it was ambiguous.

The question in the edited title doesn't seem to be the same as the one in the body (and is much easier).


Polynomials with roots in convex position

Let $\mathcal P_n$ denote the set of all monic polynomials of degree $n$ with real or complex coefficients such that $P\in\mathcal P_n$ if for all $k\in\lbrace 0,1,\dots,n-2\rbrace$ the $n-k$ roots of $P^{(k)}=\left(\frac{d}{dx}\right)^kP$ are in strictly convex position (ie are the $n-k$ vertices of a convex polygon with $n-k$ extremal vertices).

The set $\mathcal P_n$ is clearly an open subset of all monic polynomials of degree $n$ over $\mathbb R$ or $\mathbb C$.

What is the geometry and topology of $\mathcal P_n$?

Facts:

(1) Over the reals, $\mathcal P_n$ has at least $2^{n-1}$ connected components: Indeed, consider a very fast decreasing sequence (I guess $n\longmapsto 1/((1+n)^{(1+n)^{1+n}})!$ will probably work) of strictly positive reals $\alpha_0 > \alpha_1 > \cdots$ and a sequence of signs $\epsilon_0,\epsilon_1,\ldots,\epsilon_{n-2}\in\lbrace \pm 1\rbrace$. Then
$$x^n+\sum_{k=0}^{n-2}\epsilon_k\alpha_k x^k\in \mathcal P_n(\mathbb R)$$ and different choices of signs correspond to different connected components. (Proof: Up to translation, we can assume that all roots sum up to $0$. If two polynomials of the above form are connected by a continuous path of real polynomials with roots summing up to $0$ , then all derivatives except the very last one do not have a root at $0$ (since it is otherwise in the interior of the convex hull of all roots). All coefficients up to degree $d-2$ are thus necessarily non-zero and different signs correspond to different connected components). Are there other connected components?

(2) Over the complex numbers, all the polynomials described in (1) are in the same connected component. Choosing $\epsilon_i$ on the complex unit circle suggests however that $\pi_1(\mathcal C_n)$ might be $\mathbb Z^{n-1}$ where $\mathcal C_n$ denotes the connected component of $x^n+\sum_{k=0}^{n-2}\alpha_kx^k$ in $\mathcal P_n(\mathbb C)$. Do we have $\mathcal C_n=\mathcal P_n(\mathbb C)$?

(3) We have the equalities $\mathcal P_{n-1}=\lbrace P'/n \ |\ P\in\mathcal P_n\rbrace$.

I have sketched a proof in order to adress the the comment of Denis Serre.

I am not sure of why all these real polynomials are in different CCs. Do you pretend that if a coefficient of degree $\le n-2$ is zero, then $P$ is not in ${\mathcal P)_n$ ? That seems wrong for the constant coefficient.

Ī̲ have some ideas how to prove $\mathcal C_n=\mathcal P_n(\mathbb C)$ and, similarly, to make a retraction of the real case to components described by the OP. Still searching for a suitable compactification of $\mathcal P_n$ because Ī̲ detest to construct ε–δ proofs.


On random Dirichlet distributions

Fix a dimension $d\ge2$.

  * Let $Q_d$ denote the positive quadrant of $\mathbb{R}^d$, that is, $Q_d$ is the set of points $\mathbf{x}=(x_i)_i$ in $\mathbb{R}^d$ such that $x_i>0$ for every $i$.
  * For every $\mathbf{x}$ in $Q_d$, let $|\mathbf{x}|=x_1+\ldots+x_d$.
  * Let $\Delta_d$ denote the set of points $\mathbf{x}$ in $Q_d$ such that $|\mathbf{x}|=1$.
  * For every $\mathbf{a}$ and $\mathbf{b}$ in $Q_d$, define $\mathbf{a}\cdot \mathbf{b}$ in $Q_d$ by $(\mathbf{a}\cdot \mathbf{b})_i=a_ib_i$ for every $i$.
  * For every $\mathbf{a}$ in $Q_d$, let $\mathrm{Dir}(\mathbf{a})$ denote the Dirichlet distribution of parameter $\mathbf{a}$.

**The problem in a nutshell**

Fix $\mathbf{a}$ and $\mathbf{b}$ in $Q_d$. Choose a random parameter $\mathbf{u}$ in $\Delta_d$ with distribution $\mathrm{Dir}(\mathbf{a})$. Then choose a random point $\mathbf{X}$ in $\Delta_d$ with distribution $\mathrm{Dir}(\mathbf{b}\cdot \mathbf{u})$.

> My aim is to understand the (absolute) distribution of $\mathbf{X}$.

**Some more notations**

For every $\mathbf{a}=(a_i)_i$ in $Q_d$, $\mathrm{Dir}(\mathbf{a})$ is the absolutely continuous probability measure on $\Delta_d$ whose density $f(\ |\mathbf{a})$ at $\mathbf{x}$ is proportional to $x_1^{a_1-1}\cdots x_d^{a_d-1}$. More precisely, $$ f(\mathbf{x}|\mathbf{a})=\Gamma(|\mathbf{a}|)\mathbf{x}^{\mathbf{a}-1}/\Gamma(\mathbf{a}), $$ with the following shorthands: $$ \Gamma(\mathbf{a})=\Gamma(a_1)\cdots\Gamma(a_d),\quad \mathbf{x}^{\mathbf{a}-1}=x_1^{a_1-1}\cdots x_d^{a_d-1}. $$ The density $f_{\mathbf{a},\mathbf{b}}$ of the distribution of $\mathbf{X}$ is $$ f_{\mathbf{a},\mathbf{b}}(\mathbf{x})=\int_{\Delta_d} f(\mathbf{x}|\mathbf{b}\cdot \mathbf{u})f(\mathbf{u}|\mathbf{a})\mathrm{d}u_1\cdots\mathrm{d}u_{d-1}. $$

**Some special cases**

If $a_i=b_i=1$ for every $i$, $\displaystyle f_{\mathbf{1},\mathbf{1}}(\mathbf{x})\propto\int_{\Delta_d} \frac{\mathbf{x}^{\mathbf{u}-1}}{\Gamma(\mathbf{u})}\mathrm{d}u_1\cdots\mathrm{d}u_{d-1}.$

The case $d=2$ yields $$ f_{\mathbf{1},\mathbf{1}}(x,1-x)\propto\int_0^1\frac{x^{w-1}(1-x)^{-w}}{\Gamma(w)\Gamma(1-w)}\mathrm{d}w=\frac1{\pi x}\int_0^1\left(\frac{x}{1-x}\right)^{w}\sin(\pi w)\mathrm{d}w, $$ hence $$ f_{\mathbf{1},\mathbf{1}}(x,1-x)=\frac1{x(1-x)}\frac1{\pi^2+(\log[x/(1-x)])^2}. $$ Writing $\mathbf{X}=(X_1,X_2)$ with $X_1\ge0$, $X_2\ge0$ and $X_1+X_2=1$, this can be rewritten as the fact that, for every $x$ in $(0,1)$, $$ P(X_1\le x)=P(X_2\le x)=\frac12+\frac1\pi\arctan\left(\frac1\pi\log\left(\frac{x}{1-x}\right)\right). $$ Are there other cases where the density $f_{\mathbf{a},\mathbf{b}}$ is (reasonably) explicit? Or, for example, where the moments $E(\mathbf{X}^\mathbf{n})$ of $\mathbf{X}$ with $\mathbf{n}=(n_1,\ldots,n_d)$ any $d$-uplet of integers, are (reasonably) explicit?


Etale fundamental group of a curve in characteristic $p$

Let $C$ be a connected, smooth, proper curve of genus $g$ over an algebraically closed field $k$ of characteristic $p>0$. Let $\pi_1(C)$ be the etale fundamental group of $C$ - I only care about this as an abstract profinite group, so I omit base points.

It is well known by Grothendieck that $\pi_1(C)$ is topologically generated by $2g$ elements. Formally, there is a surjection $\phi:\widehat{F_{2g}}\rightarrow \pi_1(C)$, where the first group is the profinite completion of the free group on $2g$ elements. Moreover, if we let $G^{(p)}$ be the prime-to-$p$ quotient of $G$, then $$\pi_1(C)^{(p)}\cong\widehat{F_{2g}/w}^{(p)},$$ where $w=\prod_{i=1}^g[x_i,y_i].$ I.e., you get what you expect from the characteristic 0 case.

Recall that the prime-to-p quotient is the quotient $G/K$, where $K$ is the intersection of the kernels of all the maps from $G$ to finite prime-to-$p$ groups.

**Question: Is there any case of a curve $C$ of genus at least 2 for which its etale fundamental group is known?**

I phrase the question in this way because the Etale fundamental group does depend on the curve in question. In fact, even its abelianization will depend on the $p$-rank of the Jacobian.

I think $\pi_1$ is not known for any hyperbolic curve in any reasonable sense. For instance, just knowing for each prime-to-$p$ cover the $p$-rank of its Jacobian seems very hard - it is not at all obvious that there is a usable finite description of this data.

Take $G_1$ an infinitely generated free $p$-group and $G_2$ the product of $G_1$ with $\mathbb Z/p$. Then they have the same finite quotients but are distinct. This is not so great an example, because group cohomology distinguishes them.

@Niels: Not to belabour the point, but i think knowing whcih groups $G$ arise suffices to figure out the cardinality of the set of $G$-corvers. The point being that if there are $k$ non-conjugate homomorphisms to $G$, then there is a homomorphism $\pi_1\to G^k$, and the image $H$ of this homomorphism has $k$ non-conjugate homomorphisms to $G$. So we can detect whether theres at least $k$ $G$-covers by asking whether such a group $H$ occurs as a finite quotient.  Probably even this is not enough to determine the profinite group. Perhaps I'll ask this a separate question. Thanks for your help!

@jacob: for an affine curve, we just know if a given finite group $G$ arises as a quotient of $\pi_1$ or not and nothing more. In particular the cardinality of the set of $G$-covers is not known for general $G$. And this data would be probably insufficient to recover the $\pi_1$ anyway. To sum up: no the $\pi_1$ is not known for any curve (complete or not) over any field of positive characteristic except for the trivial cases of $\mathbb P^1$ and elliptic curves.

@Niels: You said Joels observation of knowing the finite quotients does not tell us, for a fixed finite group, the cardinality of the set of $G$-covers.My third sentence was meant to say that,if we are working over a countable field, it does. Because the answer is either countable infinity,or some finite number,either of which can be detected on the level of finite quotients.  As you say, this is irrelevant if we are working with proper curves, but if we talk about affine curves this starts mattering. So, my question is: If the $\pi_1$ of a hyperbolic affine curve over a countable field known?

@jacob I am not sure I understand your third sentence. About the last sentence : the $\pi_1$ of a proper curve is invariant by algebraically closed field extension. So fixing a specific base field does not change the question for proper curves. And the answer is no, the $\pi_1$ is not known. By specialization theory we just know it is topologically of finite type. For affine curves the situation is even worse since the  $\pi_1$ is not topologically of finite type.

Aha! That is illuminating. But over a countable field like $\mathbb{F}_{p^{\infty}}$ we know that the set of $G$-covers is at most countably infinite. So in that context, at least, one can answer Niels's question simply by knowing the finite quotients. Am I correct that even in this restricted setting, we do not know the $\pi_1$?

@Joël I agree "completely wrong" was a bit strong :) but I disagree with your interpretation of the question. Clearly the question is : "is the $\pi_1$ known up to isomorphism ?". An answer to your weak version wouldn't answer the following question : "given a fixed finite group $G$, what is the cardinality of the set of $G$-covers" ?

@jacob I think I remember you can find an example in Fried-Jarden Field Arithmetic.

@jacob. The finite quotient of $\mathbb Z_p^I$ for $I$ infinite are all the abelian $p$-group. Now take two infinite sets $I$, one enumerable say and another say larger than the continuum, and you have two profinite groups non-isomorphic (because they don't have the same cardinality) having the same set of finite quotients. So Niels is right that there is more in understanding a $\pi^1$ than its understanding it's finie quotient. Now in algebraic geometry one could have the point of view that it is the finite quotients, corresponding to finite cover, that are what we want to understand.

@Niels, Ah okay, interesting. But "completely wrong" is perhaps not the right term (or possibly I am more wrong that you explain in your comment): one at least knows all the finite quotients of this profinite group (the $pi^1$ of a non-complete curve), which is already something important, and sufficient for many applications. Then it depends of what we mean by "know", and apparently the OP was not completely sure about that. But, do we know the complete list of finite quotient of the $\pi^1$ for a curve of genus $\geq 2$ without any point removed ?

@Niels: Actually, I was wondering about this. Is there an example of 2 non-isomorphic profinite groups with the same finite quotients? Perhaps this is worth posting as a separate question...

@Joël : what you wrote is completely wrong. Abyankhar's conjecture describes the set of finite quotients of the $\pi_1$, but since in the affine case the $\pi_1$ is not topologically of finite type, this is not enough information to get back the $\pi_1$. For instance, the $\pi_1$ of $\mathbb A^1$ is not known in positive characteristic, even if one knows that its finite quotients are the quasi-$p$ groups. In fact to my knowledge there is no example of an hyperbolic curve whose $\pi_1$ is "known" in positive characteristic.

Of course (you certainly know this), if you remove $t$ points to your curve, then the problem is solved, provided than $t>0$. It is Abyankhar's conjecture, now a theorem of Raynaud and Harbater.

Tamagawa proved that for all profinite groups $G$, there are only finitely many smooth proper curves over $\bar{\mathbb F_p}$ (up to isomorphism) with etale fundamental group isomorphic to $G$. Maybe you can find something useful in his article; see Finiteness of isomorphism classes of curves in positive characteristic with prescribed fundamental groups, J. Algebraic Geom. 13 (2004), 675–724.


Finiteness of etale cohomology for arithmetic schemes

By an _arithmetic scheme_ I mean a finite type flat regular integral scheme over $\mathrm{Spec} \, \mathbb{Z}$.

> Let $X$ be an arithmetic scheme. Then is $H_{et}^2(X,\mathbb{Z}/n\mathbb{Z})$ finite for all $n \in \mathbb{N}$?

Remarks: Let $j:U \to X$ be the open subset given by removing the fibres of $X \to \mathrm{Spec} \, \mathbb{Z}$ lying above those primes dividing $n$.

  1. I can easily show that $H_{et}^1(X,\mathbb{Z}/n\mathbb{Z})$ is finite. For example, it follows from the Leray spectral sequence that $H_{et}^1(X,\mathbb{Z}/n\mathbb{Z}) \to H_{et}^1(U,\mathbb{Z}/n\mathbb{Z})$ is injective, and the latter group is finite by a special case of [1, Proposition II.7.1].
  2. If $n$ invertible on $X$, then the answer is _yes_ and this is again a special case of [1, Proposition II.7.1]. So I'm interested in the case where $n$ is not invertible on $X$. (In particular, this shows that $H_{et}^2(U,\mathbb{Z}/n\mathbb{Z})$ is finite).
  3. If $\dim X = 1$ then the answer if _yes_ by [1,Theorem II.3.1].
  4. Another application of the Leray spectral sequence argument shows that is suffices to show that $H^0(X, R^1j_* \mathbb{Z}/n\mathbb{Z})$ is finite, but I don't see why this should be the case.
  5. The analogous question concerning finiteness of $H_{et}^i(X,\mathbb{Z}/n\mathbb{Z})$ is also interesting, but for my application I only need $i = 2$.

References:

[1] Milne - Arithmetic duality theorems.

@MartinBright: Thanks for the references! I have had a look at the paper of Sato, but it is written in the language of derived categories which are not my forte.... Could you please be more specific which result is most relevant? He seems to have many different purity results in his paper.

A couple of references: the purity result of Gabber referred to is here arxiv.org/abs/1207.3648 Exposé XVI, and indeed assumes the torsion order is invertible.  For p-torsion there is some kind of purity theorem in this article of Sato arxiv.org/abs/math/0610426 but I don't know whether it is any use to you.

@JasonStarr: as you know, there's always an excision sequence relating the cohomology of a scheme to that of an open subscheme with the extra term being cohomology with supports along the closed complement (rather than cohomology on that complement), so the Artin-Schreier stuff doesn't directly intervene (but there's likely no purity theorem for this situation with torsion-orders not invertible; with reasonable torsion-orders I think Gabber has proved purity theorems for abstract regular schemes).

I have one more comment.  For every flat, regular, affine $\mathbb{Z}$-scheme $X$ whose reduction $X_p$ over $\mathbb{Z}/p\mathbb{Z}$ is regular of dimension $\geq 1$, the cohomology $H^1_{\text{et}}(X_p,\mathbb{Z}/p\mathbb{Z})$ is infinite, because of the many different Artin-Schreier covers.  If there were a long exact sequence, presumably this would imply that $H^2_{\text{et}}(X,\mathbb{Z}/p\mathbb{Z})$ is also infinite.  That might be evidence that there is no long exact sequence, or it might be evidence that the cohomology is infinite.

The first paper of Esnault that I know including such a Gysin sequence (the one with the joint appendix with Deligne) is "Deligne's Integrality Theorem in Unequal Characteristic and Rational Points over Finite Fields", Ann. of Math. 164 (2006), pp. 715--730.  A second paper, where she uses alterations, is "Coniveau over $\mathfrak{p}$-adic Fields and Points over Finite Fields", C. R. Acad. Sci. Paris, Ser. I 345 (2007), pp. 73--76.

@Jason Starr: Which paper of Esnault are you referring to?

Esnault's uses alterations (in the strong form of a $G$-Galois cover that has a regular modification) and the map from the cohomology of the original scheme to the $G$-invariants of the cohomology of the $G$-cover to deal with the regularity properties.  You are completely correct that she needs to invert several integers.

@JasonStarr: The purity theorems (such as in Milne's notes that you mention) require the closed complement to satisfy regularity properties (which $X \bmod p$ may not) and more importantly assume torsion-orders for the sheaf are invertible on the scheme being considered (and avoiding the latter assumption is the main thrust of the question here).

@JasonStarr: In the papers of Deligne and/or Esnault surely the relevant torsion sheaves have torsion orders invertible on that base (and section 6 of Weil II assumes properness). With "invertible" torsion-orders one can use Deligne's "generic base change" expose from SGA 4.5 to see that the higher direct images on ${\rm{Spec}}(\mathbf{Z})$ are constructible, and it is a general theorem (perhaps due to some combination of Artin, Mazur, Zink?) that the cohomology of any constructible abelian sheaf (no hypothesis on torsion-orders) on the ring of $S$-integers of a number field is finite.

. . . Esnault cites Equation 6 in Section 3.6, p. 389 of "La Conjecture de Weil, II" by Pierre Deligne.  Deligne considers a finite type flat scheme over a DVR, but he seems to assume that it has equicharacteristic.  However, in the papers of Esnault (including the paper with a joint appendix with Deligne), the exact sequence is applied when the DVR has mixed characteristic . . .

The version in Milne is Theorem 16.1 of "Lectures on Etale Cohomology", but he only formulates there the theorem for finite type schemes over a field.  The purity theorem over a Dedekind domain base has been used by Esnault in her work on congruences for rational points . . .

There is a purity theorem that is really a Gysin sequence relating etale cohomology of $U$, etale cohomology of $X$, and etale cohomology with supports in the closed complement of $U$ (which then gets turned into etale cohomology of the complement via purity).  There is a reference in Milne.  I will write it in a moment.


Infinitely generated non-free group with all proper subgroups free

Is there any example of group $G$ satisfying the following properties?

  1. $G$ is non-abelian, infinitely generated (i.e. it is not finitely generated) and not a free group.
  2. $H< G$ implies that $H$ is a free group.

Clearly such an example should be torsion-free and countable.

* * *

(Added, from comments) For completeness, here's why it has to be countable. First, it is enough to construct a subgroup of index 2. Indeed, a theorem of Swan asserts that a torsion-free group with a free finite-index subgroup is free. Next, if a group $G$ has no subgroup of index 2 (i.e. every element is product of squares), it is easy to construct a nontrivial countable subgroup $H$ with the same property). In the current setting, $G$ is uncountable so $H$ is a proper subgroup, and hence has to be free, contradiction.

Since a partly equivalent question was just asked, I have added the short argument to discard the uncountable case.

@YCor You're probably right I should not have have said "clearly" but in my mind it was a very direct application of some known facts (to me at least) in infinite group theory. Swan Theorem says that if your group is torsion-free and it is a finite extension of a free group, then it must be free itself.

@W4cc0 thanks; I wouldn't have said "clearly"! I didn't know Swan's theorem, what does it say exactly? Here's an argument to show residual finiteness: it's enough to check that there is a proper subgroup of index 2. Indeed, the latter being free by assumption, it has to be res. finite and hence so is the whole group. Now, if by contradiction there is no subgroup of index 2, then every element is a finite product of squares. From this, it is easy to construct a nontrivial countable subgroup in which every element is a finite product of squares, and hence this is a non-free countable subgroup.

@YCor such a group must be countable because if it is not, then it would be residually finite, but then a very well known theorem of Swan would imply that it is free

@SantanaAfton -- Mark explained how to construct a non-free group in which every finitely generated subgroup is free. This contradicts your intuition that since "every finitely-generated subgroup is both proper and free" ... the group must itself be free.

@W4cc0 on the other hand do you maintain your claim that such a group has to be countable? or do you want to require the group to be countable? (I guess the countable case and uncountable case are quite distinct questions.)

@YCor, I meant $G$ non-abelian (even if it makes sense in this case, it is easy to see that there are no abelian such groups), of course it would have no sense in the other case; but unfortunately I always write the opposite of what I would like to, sorry.

Higman constructed non-free groups of cardinality $\aleph_1$ in which every countable subgroup is free.  But every proper subgroup being free sounds hard.

@SantanaAfton It's overly simplistic, if not senseless, to think of a free group as a group with no relations.

@W4cc0 no, you don't: there are always free abelian subgroups. So the question was correctly stated, and wouldn't if "non-abelian" were added. Also as already mentioned, every group with your property has a non-abelian free subgroup.

@SantanaAfton Every finitely generated group of the rationals/Hawaiian earring is free but these are not free

@HJRW Are you referring to the HNN construction? In that example, if you take the relation $a^t=ab$ and the group generated by all letters involved, you don’t get a proper subgroup. This never happens in OP’s situation.

@SantanaAfton -- see Mark Sapir's comment above.

It feels as though no such group exists. Since every finitely-generated subgroup is both proper and free by assumption, how could the group have any nontrivial relations?

I should have been more clear, I meant free non-abelian.

@BenWieland a torsion-free abelian group in which every proper subgroup is cyclic, is itself cyclic, by an easy argument.

@MarkSapir $\mathbf{Z}[3/2]=\mathbf{Z}[1/2]$, so it's not a proper subgroup. Actually, the only unital subrings of $\mathbf{Z}[1/2]$ are itself and $\mathbf{Z}$. Also, every subgroup of $\mathbf{Z}[1/2]$ is $\{0\}$, infinite cyclic, or has finite index and isomorphic to $\mathbf{Z}[1/2]$ (namely equals $k\mathbf{Z}[1/2]$ for some unique positive odd $k$.)

@MarkSapir Oh, yeah, but probably better to call the subgroup $3\cdot\mathbb Z[\frac12]$

@BenWieland: $\mathbb{Z}[1/2]$ is the derived subgroup of the Baumslag-Solitar group $BS(1,2)$, it contains non-free  subgroup $\mathbb{Z}[3/2]$

Sorry.  I missed the infinitely Generated.

About the countability issue, I think I'm not aware of a non-free group in which every countable subgroup is free. (By analogy, $\mathbf{Z}^\mathbf{N}$ is not free abelian but all its countable subgroups are. Also, the Hawaiian earring group is locally free but has non-free countable subgroups.)

@PaulPlummer: I do not think it is clear (or even true) that these groups must be countable,

@MarkSapir and YCor Thanks, for some reason my comment sounded like a better idea when typing it, even though I knew of locally free nonfree groups.

@PaulPlummer: Locally free but non-free groups are easy to construct. Take a proper ascending HNN extension of the free group $F_k$, and the normal subgroup $N$ generated by $F_k$. For example $\langle a,b,t \mid a^t=ab, b^t=ba>$. The normal closure $N$  of $a,b$ is locally free but not free. The OP does not want this example. He wants all proper subgroups to be free, not just finitely generated subgroups.

@PaulPlummer I don't really understand your first comment. A free group itself has plenty of generating families under which it is not free, so using non-freeness of a given generating family will not help much.

Also why must such groups be countable? I am not seeing it immediately

@BenjaminSteinberg To complement Mark's comment, groups with Olshanskii's property are generated by every non-commuting pair.


Characteristic subgroups and direct powers

Solved question: Suppose _H_ is a characteristic subgroup of a group _G_. Is it then necessary that, for every natural number _n_ , in the group $G^n$ (the external direct product of $G$ with itself $n$ times), the subgroup $H^n$ (embedded the obvious way) is characteristic?

Answer: No. A counterexample can be constructed where $G = \mathbb{Z}_8 \times \mathbb{Z}_2$ (here $\mathbb{Z}_n$ is the group of integers modulo $n$) with $H$ the subgroup

$$\\{ (0,0), (2,1), (4,0), (6,1) \\}$$

This subgroup sits in a weird diagonal sort of way and just happens to be characteristic (a quirk of the prime $2$ because there isn't enough space). We find that $H \times H$ is not characteristic in $G \times G$.

(ASIDE: The answer is _yes_ , though, for many important characteristic subgroups, including fully invariant subgroups, members of the upper central series, and others that occur through typical definitions. Since for abelian groups of odd order, all characteristic subgroups are fully invariant, the answer is yes for abelian groups of odd order, so the example of order $2^4$ has no analogue in odd order abelian groups.)

My question is this:

  1. Strongest: Is it true that if _H_ is characteristic in _G_ and $H \times H$ in $G \times G$, then each $H^n$ is characteristic in $G^n$ [NOTE: As Marty Isaacs points out in a comment to this question, $H \times H$ being characteristic in $G \times G$ implies _H_ characteristic in _G_ , so part of the condition is redundant -- as explained in (2)]?
  2. Intermediate: Is there some finite $n_0$ such that it suffices to check $H^n$ characteristic in $G^n$ for $n = n_0$? Note that if $H^n$ is not characteristic in $G^n$ for any particular _n_ , then characteristicity fails for all bigger $n$ as well. I'd allow $n_0$ to depend on the underlying prime of _G_ if we are examining $p$-groups.
  3. Weakest: Is there a test that would always terminate in finite time, that could tell, for a given _H_ and _G_ , whether $H^n$ is characteristic in $G^n$ for all _n_? The "try all _n_ " terminates in finite time if the answer is _no_ , but goes on forever if the answer is _yes_. In other words, is there a finite characterization of the property that each direct power of the subgroup is characteristic in the corresponding direct power of the group?

ADDED: My intuition, for what it's worth, is that those subgroups _H_ of _G_ that can be characterized through "positive" statements, i.e., those that do not make use of negations or $\ne$ symbols, would have the property that $H^n$ is characteristic in $G^n$. On the other hand, those whose characterization requires statements of exclusion (_not a ..._) would fail because negative statements are difficult to preserve on taking direct powers. But I don't know how to make this rigorous.

Do you have a reference for "if $H$ is fully invariant in $G$ then $H \times H$ is fully invariant in $G \times G$? Thanks in advance.

In the case of finite abelian groups, is it true that $\operatorname{Aut}(G \times G \times G)$ is generated by automorphisms stabilizing one copy of $G$ ? This holds at least when $G=(\mathbf{Z}/p^n\mathbf{Z})^k$.

@Marty Isaacs: I have added this note to Question 1. I had already incorporated a similar observation into Question 2, as you probably noticed.

There is a redundancy in Question 1 since if $H \times H$ is characteristic in $G \times G$, then $H$ is automatically characteristic in $G$. This is because every automorphism of $G$ extends to an automorphism of $G \times G$ stabilizing the first component.

No reference to a paper or book, but here's an online proof I wrote up: groupprops.subwiki.org/wiki/…  It's rather long. What fails for the prime 2 is the first half of the proof: "it is the direct sum of its intersections with the direct summands" because in that part we use that the doubling map is an automorphism. The second half is bookkeeping that takes homomorphisms between cyclic subgroups and converts them to automorphisms by adding the identity map.

That odd prime statement sounds handy, do you have a reference for "characteristic subgroups of abelian groups of odd order are fully invariant"?


Large values of characters of the symmetric group

For $g$ an element of a group and $\chi$ an irreducible character, there are two easy bounds for the character value $\chi(g)$: First, the bound $|\chi(g)|\leq \chi(1)$ by the dimension of the representation and, second, the bound $|\chi(g) |\ \leq \sqrt{ |Z(g)|}$ by the centralizer arising from the formula $\sum_{\chi} |\chi(g)|^2= |Z(g)|$.

I am interested on upper bounds for characters of the symmetric group that improve slightly on the second bound.

More specifically, fix $\delta> 0$ small. I want to know for which $g\in S_n$ there exists an irreducible character $\chi$ with $|\chi(g)| > \sqrt{ |Z(g)|} e^{- \delta \sqrt{n}}$.

Recall that for a permutation $g\in S_n$ with $m_k$ cycles of size $k$ for all $k$, so that $\sum_k m_k k =n$, we have $|Z(g)| = \prod_k (k^{m_k} m_k!)$.

For example, known upper bounds for the dimension of representations of the symmetric group imply there does not exist such a character for $g$ the identity. On the other hand, such a character does exist if $g$ is an $n$-cycle, since the right side is less than $1$. I suspect and hope that this can only happen for $g$ containing some relatively large cycles.

I found works in the literature giving bounds for $\chi(g)$ that improve on the $|\chi(g)|\leq \chi(1)$ by a multiplicative factor ([Asymptotics of characters of symmetric groups related to Stanley character formula by Féray and Śniady](http://doi.org/10.4007/annals.2011.173.2.6)) or a power ([Characters of symmetric groups: sharp bounds and applications by Michael Larsen and Aner Shalev](https://doi.org/10.1007/s00222-008-0145-7)) but it's not obvious if it's possible to transform them into a bound of the form I need.

But now suppose we consider a permutation with $n/{3k}$ $k-1$-cycles, $k$-cycles, and $k+1$-cycles, or something like that. The square root of the centralizer is now $\sqrt{ (n/3k)!^3 (k^3-k)^{n/3k} }$ which is smaller by a factor of roughly $\sqrt{ 3^{n/k} e^{ - \pi \sqrt{2n/3}}}$. So the upper bound you desire should imply that slightly changing the size of the strips gives us vastly fewer border strip tableaux than before, which seems unlikely (instead of merely more cancellation, which is more feasible.)

@TimothyChow Actually, consider the following. If we consider a permutation with $n/k$ $k$-cycles, its centralizer has size $(n/k)! k^{n/k}$. This is the sum of the squares of the character values and the number of character values is $e^{ \pi \sqrt{2n/3}}$ so one character value has to be at least $\sqrt{ (n/k)! k^{n/k}  e^{ - \pi \sqrt{2n/3}}}$. So the number of border strip tableaux is at least that number as well.

@SamHopkins Yes, good point, and an "unsigned combinatorial description for the square of the characters" as ruled out by this paper would certainly be very convenient for proving the bound I want if it was possible.

Of course this does not in any way preclude bounds of the form you are interested in, but it might also be worth pointing out on the negative side of things that there have been recent dramatic advances which show that symmetric group character values are computationally hard: arxiv.org/abs/2207.05423

@TimothyChow For $n=4$ if $g$ is a $3$-cycle and a $1$-cycle and the character corresponds to the partition $3+1$ then $\sqrt{ | Z(g)|} =\sqrt{3}<2$ and the number of border-strip tableaux is either $2$ or $0$ depending on the order of the cycles. So maybe the truth or falsity depends on the order. Another evidence is that this bound is true is if all the cycles are the same size so that there is no choice of order, by Corollary 9 of the paper A bijection proving orthogonality of the characters of $S_n$ by Dennis E White mentioned in a deleted answer.

My first instinct is to try Murnaghan-Nakayama, but I don't know if that will work. Best-case scenario would be that the number of border-strip tableaux is already $\le \sqrt{|Z(g)|}$ (so that you don't even have to worry about sign-cancellation), but maybe this is false?

@PerAlexandersson Thanks! It seems to me that one of these formulas will be helpful if there exists a relatively simple combinatorial proof of the bound $\chi(g)^2 \leq |Z(g)|$, because then one can try to study the proof more carefully to see when it can be sharp. So far I don't see how to do that for any of them, but it could be possible...

Some various combinatorial formulas for Sn-characters can be found here: symmetricfunctions.com/murnaghanNakayama.htm


What algebraic properties are preserved by $\mathbb{N}\leadsto\beta\mathbb{N}$?

Given a binary operation $\star$ on $\mathbb{N}$, we can naturally extend $\star$ to a semicontinuous operation $\widehat{\star}$ on the set $\beta\mathbb{N}$ of ultrafilters on $\mathbb{N}$ as follows:

$$\mathcal{U}\mathbin{\widehat \star}\mathcal{W}:=\\{A:\\{k: \\{a: a\star k\in A\\}\in\mathcal{U}\\}\in\mathcal{W}\\}.$$

It's a standard (and [quite useful!](https://web.williams.edu/Mathematics/lg5/Hindman.pdf)) fact that associativity is preserved under this transformation: if $\star$ is associative, then so is $\widehat{\star}$. On the other hand, commutativity [is _not_ preserved](https://math.stackexchange.com/questions/87109/addition-on-ultrafilters-is-non-commutative). Say that an equational sentence $\sigma$ involving a single binary function symbol is a **$\beta$ -property** iff we have $$(\mathbb{N};\star)\models\sigma\quad\implies\quad (\beta\mathbb{N};\widehat{\star})\models\sigma$$ for every binary operation $\star$ on $\mathbb{N}$.

> **Question** : Which equational sentences are $\beta$-properties?

Of course there is no need to restrict attention to equational properties per se _(which is why I've used "$\implies$" rather than "$\iff$" in the definition of $\beta$-property)_, and we can similarly transform arbitrarily many arbitrary-arity operations on $\mathbb{N}$; however, already this question seems difficult. I recall seeing a partial(?) answer to this question which involved the way variables were/were not _reordered_ in the left vs. right hand sides of $\sigma$, but at the moment I can't track it down.

* * *

Incidentally, the following question is at least of a related _flavor_ , and may involve relevant ideas (and be easier to tackle at first): what is $(\beta\mathbb{N};\widehat{+})$ _(or $(\beta\mathbb{Z};\widehat{+})$ using the analogous definition)_ like from a universal-algebraic perspective? For example, is it congruence modular?

@BenjaminSteinberg xy=zw is also preserved. :P

One thing that is preserved of course is either xy=x or xy=y

Yes.  I think chapter 6 talks about free subsemigroups.  Of course it doesn't looks at nonassociative magmas.  My impression is that few general semigroup identities are  preserved.

@BenjaminSteinberg Is that book Algebra in the Stone-Cech compactification?

There are some results  in Hindman's book on conditions guaranteeing the Stone-Cech compactification of a semigroup contains a free semigroup on 2 generators and hence satisfies no semigroup identity.  If is of course possible for the original semigroup to satisfy an identity and so this identity is not preserved. Commutativity is such and identity and so are Malcev's nilpotent identities.

Notice also many quasiidentities like cancellativity are not preserved.


Mumford-Tate conjecture for mixed Tate motives

Let $X$ be a (not necessarily smooth or proper) variety over a number field $k$. Suppose we are given

  1. A subquotient $V_{dR}$ of the algebraic de Rham cohomology $H_{dR}^i(X)$ (defined in the non-smooth case via a thickening, as in e.g. Hartshorne's algebraic de Rham cohomology paper),

  2. For each embedding $\iota: k\to \mathbb{C}$, a subquotient $V_\iota$ of $$H^i(X_{\mathbb{C}}^{an}, \mathbb{Q})$$ respecting the $\mathbb{Q}$-mixed Hodge structure, and

  3. For some fixed prime $\ell$ and algebraic closure $\bar k$ of $k$, a subquotient $V_\ell$ of $$H^i(X_{\bar k, \text{ét}}, \mathbb{Q}_\ell)$$ respecting the action of $\text{Gal}(\bar k/k)$, such that

  4. For each $\iota$, the corresponding comparison isomorphism $$H^i_{dR}(X)\otimes_k \mathbb{C} \to H^i(X_{\mathbb{C}}^{an}, \mathbb{Q})\otimes \mathbb{C}$$ sends $V_{dR}$ to $V_\iota$, and

  5. For each $\iota$, the corresponding comparison isomorphism $$H^i(X_{\mathbb{C}}^{an}, \mathbb{Q})\otimes \mathbb{Q}_\ell \to H^i(X_{\bar k, \text{ét}}, \mathbb{Q}_\ell)$$ sends $V_\iota$ to $V_\ell$, and

  6. The $V_\iota$ is of mixed Tate type (i.e. all $h^{p,q}$'s vanish for $p\not=q$). Or equivalently $V_\ell$ is an iterated extension of powers of the cyclotomic character.

(If you'd like, you can think of $V$ as the realization of a mixed Tate motive over $k$ under any of the various formalisms for mixed Tate motives over number fields -- I think that a priori the conditions above are slightly weaker but expected to be equivalent.)

> I'd like to know -- is the analogue of the Mumford-Tate conjecture known to be true for such $V$? Explicitly, one expects that the Lie algebra of the image of $$\text{Gal}(\bar k/k)\to GL(V_\ell)$$ is the extension of scalars to $\mathbb{Q}_\ell$ of the Lie algebra of a $\mathbb{Q}$-group, given as the Tannaka dual of the subcategory of mixed Tate Hodge structures over $k$, generated by the Hodge realization $(V_{dR}, V_{\iota_1}, V_{\iota_2}, \cdots)$.

My feeling is that one can probably extract this from the literature by comparing ranks of various Ext groups (in one's favorite category of mixed Tate motives and in the $\ell$-adic and Hodge settings) but I am hoping it is written explicitly somewhere.

**EDIT:** Fixed some errors in the formulation of the Hodge realization.

@DonuArapura It seems like I might have to, since I have an application in mind!

Hi Daniel. I'm not sure it is written down anywhere, so it would be nice if you did.


Is there a classification of reflection groups over division rings?

I asked a [version](https://math.stackexchange.com/questions/3491953/reflection-groups-of-division-rings) of this question in Math StackExchange about a week ago but I've received no feedback so far, so following the advice I received on [meta](https://meta.mathoverflow.net/questions/4416/would-this-question-be-appropriate-for-mathoverflow) I decided to post it here.

* * *

## Details

The irreducible finite [complex reflection groups](https://en.wikipedia.org/wiki/Complex_reflection_group) were classified by Shephard and Todd. The list consists of a three-parameter family of imprimitive groups and 34 exceptional cases. If $\mathbb{K}$ is any field of characteristic zero, it is known (see e.g. Section 15-2 [here](https://www.springer.com/gp/book/9780387989792)) that if a group has a representation as a $\mathbb{K}$-reflection group then it also has a representation as a complex reflection group, so we get no new examples for fields. I'm interested to know if an analogue of this classification exists when we allow $\mathbb{K}$ to be a division ring of characteristic zero.

I have only found partial results. The irreducible finite quaternionic reflection groups were classified in [this paper](https://www.sciencedirect.com/science/article/pii/0021869380901489) by Cohen. Finite $\mathbb{K}$-reflection groups of rank $1$, or equivalently finite subgroups of division rings of characteristic $0$, were classified in [this paper](https://www.jstor.org/stable/1992994?seq=1) by Amitsur. The classification for rank $2$ can perhaps be extracted from the [classification of finite subgroups of $GL(2,\mathbb{K})$](https://core.ac.uk/download/pdf/82740228.pdf) by Banieqbal using a case-by-case check.

In view of the complex and quaternionic lists, I would expect the full classification (if there is one) to follow a form similar to this:

  * An infinite family $G_n(M,P,\alpha)$ of imprimitive reflection groups of rank $n$, where $M$ is a finite subgroup of a division ring $\mathbb{K}$ (i.e. an Amitsur group) and $[M,M]\le P \trianglelefteq M$, possibly with some extra data $\alpha$ in low rank. Algebraically it should correspond to something like the group of generalized permutation matrices with entries in $M$ whose determinant is in $P$, like in the cases of fields and quaternions (note that order doesn't matter when computing the determinant, since $[M,M]\le P$).

  * A family or families of examples in rank $2$ (or perhaps in rank $\le m$ where $m^2$ is the dimension of $\mathbb{K}$ as a division algebra over its center). In the quaternionic case they are the primitive reflection groups whose complexification is imprimitive, and are all constructed from certain $2$-dimensional primitive complex reflection groups; it's not clear to me how this construction could generalize to arbitrary $\mathbb{K}$.

  * A number of exceptional cases of small rank. These are the ones I'm most interested in. In the quaternionic case these are precisely the primitive reflection groups whose complexification is also primitive; in the general case they might correspond to primitive reflection groups which remain primitive after tensoring the representation with a splitting field.

* * *

## Question

My main question is thus:

> Is there a classification of groups representable as a $\mathbb{K}$-reflection group over some division ring $\mathbb{K}$ of characteristic zero?

I would also appreciate any references dealing with this problem or with particular cases. If the classification turns out to be intractable, or currently out of reach, I would ask if at least an example can be found of a new exceptional reflection group of rank $\ge 3$ (see details above).

* * *

### Update

After much searching, I finally found a brief reference to this problem in the literature. The mention occurs at the end of Section 3 in [this 1981 paper](https://darkwing.uoregon.edu/%7Ekantor/PAPERS/GenLinearGps.pdf) by Kantor, which I quote here for convenience:

> [...] For example, consider the problem of determining all finite primitive reflection groups $G$ in $GL(n,D)$, for $D$ an arbitrary noncommutative division ring of characteristic $0$. If $n=1$, this is just the famous problem solved by Amitsur (1955) (and independently and almost simultaneously by J. A. Green). If $n=2$ and $G$ is solvable, the problem seems to involve even more difficult number theory than Amitsur used. But if $n\ge 3$, and if simple group classification theorems are thrown at the problem, no new nonsolvable examples arise. [...]

If the last sentence is true, it means that the new examples of "exceptional groups" I asked for must necessarily be solvable. However, the author does not provide any in-text citation for that statement, and I haven't been able to find which result is being alluded to.


xkcd's "Unsolved Math Problems", straight lines in random walk patterns

STEM student's favourite source of amusement posted a comic titled "Unsolved Math Problems" one of which looks like something that could actually be tackled.

> If I walk randomly on a grid, never visiting any square twice, placing a marble every N steps, on average how many marbles will be in the longest line after NK steps?

[Original comic (with ilustration of the problem)](https://xkcd.com/2529)

I know there are some results in form of power laws (critical exponents) like the 0.587597... constant for SARW in 3D.

Can we say anything interesting about the process described in the comic?

@Keba if you get boxed in discard that walk and begin again. A better way to think of the question might be: consider ALL possible SAWs of NK steps with a marble placed at every Nth step. What is the average number of marbles on the longest line?

@ZachTeitler, re, a (literally) bold new notation for the $5$-adics?

@Gro-Tsen It's a "conjection" whatever that is. (Also: I wonder what $\mathbb{5}$ is supposed to mean.)

@SamHopkins A horizontal or vertical line through the origin will have $\sqrt{K/N}$ marbles on average and I would wildly guess that the correct answer is within $(KN)^\epsilon$ of this. It looks conceivable to prove this by bounding for each line the expectation of the $\ell$th power of the number of marbles on the line and summing over all lines passing through at least a few points within a distance $NK$ of the origin.

If we consider the analogous process for a random walk which is not self-avoiding, is this understood? (SAWs are much harder than regular random walks, from what I understand…)

@Gro-Tsen: $\epsilon$ cannot be $\lt 0$.

I believe deeplinking to images on XKCD is discouraged, so I changed the link to point to the comic.

What I very much want to know is: is the Euler field manifold hypergroup isomorphic to a Gödel-Klein meta-algebreic $\epsilon<0$ quasimonoid connection under Sondheim calculus? (Also, what in Apollo's name is going on with this curve?)

@MartinHairer Is this 'dropping a marble' part redundant then? Is it roughly equivalent to how many collinear gridpoints were visited?

@Keba either seems interesting but I think the original intention was point-like marbles.

The self-avoiding random walk with $N$ steps is usually simply defined as the uniform measure on the set of all self-avoiding walks of $N$ steps. It doesn't have any kind of Markov property, but is conjectured in $2D$ to converge to an SLE curve with self-similarity exponent $3/4$. Rigorously, almost nothing non-trivial is known about its large-$N$ behaviour.

I feel like we are getting nerd sniped again...

Also, how large are these marbles? Do they fill the whole square? Are they just points?

I'm no expert on random walks so I probably miss something easy, but why is this process well-defined? I might corner myself in?


Is there a simpler proof of the key lemma in the paper by Hiroshi Iriyeh and Masataka Shibata on the 3D Mahler conjecture?

In this [remarkable paper](https://arxiv.org/abs/1706.01749) 30 pages are occupied by the proof of the following innocently looking lemma:

Let $K$ be an origin-symmetric convex body in $\mathbb R^3$. There exist three planes through the origin splitting $K$ into $8$ parts of equal volume and such that each two of these planes split the cross-section of $K$ by the third one into $4$ parts of equal area.

I cannot shake off the feeling that there must be a half-page proof of this statement though I don't have one yet. I also know that MO is swarming with good topologists. Anybody up to the challenge?

it might be of interest: in the first part of the proof of Theorem 1 keithmball.files.wordpress.com/2014/11/… Ball constructs a similar map to simplex as Makeev does but he uses Browers Fixed Point Theorem (BFPT) unlike Makeev. So it quite might be that one needs to apply BFPT in a proper way which I don't see yet how.

@fedja yes, the claim is indeed bit different, but the topological argument is still the same

@FedorPetrov Except in our case the measures to partition depend on the planes. However the idea that if the solution is unique in some generic position, then there is an odd number of solutions in every generic position is amazing and certainly very promising. It should be rather standard, of course, but the good side of ignorance (mine) is the possibility to get surprised with the facts everybody else considers routine :-).

(continued) Makeev partitions two measures which are absolutely continuous w.r.t. Lebesgue measure, but actually this absolute continuity is not essential.

Roman Karasev (private communication) suggests that this essentially follows from the results of  [V.V. Makeev. Equipartition of a continuous mass distribution. Journal of Mathematical Sciences, 140:4 (2007), 551--557, mathnet.ru/links/a1722f058ed63ed36cb13d9e67fd255f/znsl299.pdf, theorems 5 and 6]

Now one can hope that geometric considerations will show that (in this "generic" situation) the intersection class in $H^6((\mathbb{RP}^2)^3,\mathbb{Z}/2\mathbb{Z})$ will be nontrivial, and that furthermore that there is no contribution from some degenerate case e.g. two planes coinciding. Then it probably still remains necessary to pass to some perturbation of $K$, and to run this argument there instead. So even if this sketch can be made to work, I doubt that a fully rigorous proof would fit in a half-page...

Some naive thoughts: We can try and derive this from an intersection theory statement on $(\mathbb{RP}^2)^3$. Points correspond to triples of planes, and the condition that three planes split $K$ into $8$ parts of equal volume will give a codimension $3$ cycle for "generic" $K$. (We get codimension $3$ instead of codimension $7$ because diametrically opposite regions have the same area, by reflection symmetry of $K$). Similarly each of the conditions that two planes split a cross-section into four parts of equal area should lead to a codimension $1$ cycle.


Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space

Say you want to check that $|\sum_{n\leq x} \mu(n)|\leq \sqrt{x}$ for all $x\leq X$. (I am actually interested in checking that $\sum_{n\leq x} \mu(n)/n|\leq c/\sqrt{x}$, where $c$ is a constant, and in finding the best $c$ such that this is true in a given range, such as $3\leq x\leq X$, say; all of these problems are evidently related.)

The natural algorithm (the same you can find in section 4 of [1], say) has running time $O(X \log \log X)$ and takes space $O(\sqrt{X})$ (where we think of integers as taking constant space). I've coded it with some optimizations (for $\sum_{n\leq x} \mu(n)/n$, using interval arithmetic) and should have a result for $x=10^{14}$ in about a fortnight; $x=10^{12}$ takes an afternoon. (When I first coded this, less carefully and on worse hardware, it took a week.)

Is it possible to do things in either less time or less space? Space is important here in practice - ideally you would want to keep everything in cache, and it is rare to have more than 4MB per processor core - that is, enough for $1.6\cdot 10^7$ values of $\mu$, or $5\cdot 10^5$ large integers.

Notice that I am asking for a check for all $x\leq X$, and not just for a single $x=X$.

* * *

Further remarks: I am aware that there are algorithms for computing a single value of $\sum_{n\leq x} \mu(n)$ in time $O(x^{2/3} \log \log x)$ [2], or even (in what looks like a more sophisticated but less practical way that may have never been coded) in time $O(x^{1/2+\epsilon})$ ([3]; see also [Mertens' function in time $O(\sqrt x)$](https://mathoverflow.net/questions/95726/mertens-function-in-time-o-sqrt-x)). Once can of course use such an algorithm to compute the sum for values of $x$ that are about $c\sqrt{x}$ apart, and then use what I've called the "natural" algorithm to deal with intervals in which such a computation shows that the statement to be verified could be violated. This results in a running time of $O(x^{7/6} \log \log x)$, or $O(x^{1+\epsilon})$, and while memory usage could be decreased by applying the "natural" algorithm on intervals of length smaller than $\sqrt{x}$, this would result in a longer running time.

The same is true, of course, of the "natural" algorithm itself: one could store a list of primes $p\leq \sqrt{x}$ in the main memory in $\sqrt{x}$ bits, and compute $\mu(m)$ in blocks of size $M$ at a time; this would require working in memory $O(M)$ (to be stored in the cache) for the most part, but the running time would be $O(x^{3/2}/M)$. Or is there a way around this? Can one, say, select the primes $p\leq \sqrt{x}$ that might divide integers in an interval of length $M \lll \sqrt{x}$, and do so in time less than $\sqrt{x}$ or $\sqrt{x}/\log x$? Or is there any other way to take space substantially less than $O(\sqrt{x})$ while still having essentially linear running time? Or a way to have better than linear running time?

PS. I could also ask about parallelising this, but I would not like to risk making the discussion too hardware-specific here.

[1] _François Dress_ , MR 1259423 [**Fonction sommatoire de la fonction de Möbius. I. Majorations expérimentales**](http://projecteuclid.org/euclid.em/1048516214), _Experiment. Math._ **2** (1993), no. 2, 89--98.

[2] _Marc Deléglise and Joël Rivat_ , MR 1437219 [**Computing the summation of the Möbius function**](http://projecteuclid.org/euclid.em/1047565447), _Experiment. Math._ **5** (1996), no. 4, 291--295.

[3] _J. C. Lagarias, and A. M. Odlyzko_ , MR 0890871 [**Computing $\pi(x)$: an analytic method**](http://dx.doi.org/10.1016/0196-6774%2887%2990037-X), _J. Algorithms_ **8** (1987), no. 2, 173--191.

Well, numbers of size about $X$ without prime factors between $X^(1/3)$ and $X^{1/2}$ are very roughly as common as the primes - but the sum of the reciprocals of the primes diverges. Or rather - the sum $\sum_n 1/n$ over all $X\leq n<2X$ without such prime factors is about a constant times $1/\log X$, i.e., much larger than $1/\sqrt{X}$. The same happens if you work over much smaller intervals. So I'm afraid this cannot work as stated.

You might consider "faulty" versions of the natural algorithm, where one computes the wrong value on some of the values, or forgets to store the primes above cube root of X.  You might still be able to show the error is small enough that the inequalities you want to check are preserved.  Gerhard "Faulty Computing Isn't Computationally Wrong" Paseman, 2016.04.25.

And yes, the running time is as you said and I said. I'm looking for something better.

Gerhard: that's exactly what I mean by "the natural algorithm".

the simplest algorithm for checking $M(x)$ not for every $x \le X$ but for every $x = \lfloor X/n\rfloor$ is by using $\sum_{n=1}^x M(x/n) = 1$, and computing $M(k)$  for every different $k \in \{ \lfloor X/n \rfloor \}$, it takes $\mathcal{O}(X)$ additions and $\mathcal{O}(\sqrt{X}\log X)$ in space

It may be that I am talking about the "natural algorithm" already, which is a variant of the one at mathoverflow.net/a/50691 for determining distinct prime factors of integers up to X in a serial fashion.  However I still wonder what parameter you are talking about for space usage.  Gerhard "Space Usage: The Final Frontier" Paseman, 2016.04.25.

Do you mean $\sqrt{x}$ or $\sqrt{X}$?  If the latter, there is a simple sieving process that computes the sum on the fly, and you can save extremal values for later processing and check that the inequality is satisfied.  You can even sift just the squarefree values for processing.  The memory storage is $O(\log X)$ bits for each prime at most $\sqrt{X}$.  The runtime should be not much longer than $O(X\log\log X)$.  Gerhard "Sometimes Has Xtreme Case Sensitivity" Paseman, 2016.04.25.


Which manifolds decompose into pants?

In this [nice paper](http://arxiv.org/abs/math/0205011) Mikhalkin uses certain (more geometrical than algebraic) aspects of tropical geometry to prove that every complex projective hypersurface in $\mathbb C \mathbb P ^n$ decomposes as a smooth manifold into some _$(n-1)$-dimensional pair-of-pants_.

Following Mikhalkin, a $k$-dimensional pair-of-pants is a particular real $(2k)$-manifold with corners. It is obtained by removing $k+2$ generic hyperplanes from $\mathbb C \mathbb P^k$. For instance, a $1$-dimensional pair-of-pants is a sphere minus 3 points (as everybody knows), while a $2$-dimensional pair-of-pants is $\mathbb C \mathbb P^2$ minus 4 generic lines.

Boundaries and corners arise when you take the compact version of these pair-of-pants, i.e. you remove an open regular neighborhood of the $k+2$ generic hyperplanes:

  * When $k=1$ you get the genuine compact pair-of-pants $P$ with 3 boundary components.
  * When $k=2$ you get a real 4-manifold with boundary and corners: the boundary consists of four compact 3-manifolds homeomorphic to $P\times S^1$ (corresponding to the four lines), glued together along six tori (corresponding to the six intersections between the lines). The tori are the corners.

The 4-dimensional manifold with corners we get with $k=2$ is a nice object, which may look intriguing to a low-dimensional topologist.

As everybody knows, not only complex hypersurfaces (i.e. curves) in $\mathbb C \mathbb P^2$ decompose into pair-of-pants: every oriented surface of negative Euler characteristic does! It would be then natural to ask the following:

> Which compact 4-manifolds decompose into pair-of-pants?

The set of course includes all complex hypersurfaces in $\mathbb C \mathbb P^3$.

**Edit:** It is not true that every curve in $\mathbb C \mathbb P^2$ decomposes into pants: a line or a cubic give $S^2$ and $T^2$ and they of course do not decompose into pants. Reading more carefully Mikhalkin's paper, it seems to me that he also allows to collapse the natural fibering of the boundaries of the blocks: for instance, by collapsing the boundaries of a two-dimensional pair-of-pants we get $S^2$.

I think that the following holds: let $X$ be obtained by gluing $n$ Pairs-of-Pants, and suppose that $X'$ is obtained by the same gluing rules, except that you add a Dehn twist in one of the gluing faces. Then, in analogy with the Lickorish-Wallace theorem, $X'$ should be obtained from $X$ by a single surgery along a (probably homologically essential, zero-square) torus.

Thinking a bit more about your comment, I realized that Mikhalkin probably also admits the collapsing of the natural fibrations of the boundaries of the pants: the situation is slightly more complicated than I guessed, I have edited the text accordingly.

Hi Marco. I would put no restriction at all (I can't see any reasonable restriction...). You glue along diffeomorphisms of blocks of type $P\times S^1$: every such diffeomorphism must preserve the fiber, but there are infinitely many non-isotopic choices. You are absolutely right about $\chi$, I must admit I didn't think about it. It might be that you get only finitely many manifolds for each $\chi$, but since you might choose among infinitely many gluing maps between the $P\times S^1$ faces, this is not clear (to me).

Hi Bruno! What sort of restrictions do you put on the gluing maps between (not necessarily different?) Pairs-of-Pants? For example, how would you obtain $\mathbb{CP}^2$ in this picture? Is it explained in Mikhalkin's paper? In any case, it looks like a necessary condition is to have $\chi>0$, since your building blocks all have $\chi=1$, and both the submanifolds you glue them along and the corners have $\chi=0$. In particular, $\chi$ of your manifolds seems to determine how many PoPs you need.