Supercompact and Reinhardt cardinals without choice

A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:

_Definition._ A cardinal $\kappa$ is _supercompact_ if for all ordinals $\lambda$, there exists a transitive inner model $M_\lambda$ of $V$ and an elementary embedding $j_\lambda: V\rightarrow M_\lambda$ such that $crit(j_\lambda):=\min\lbrace\alpha\in ON: j(\alpha)\not=\alpha\rbrace=\kappa$ and $M_\lambda$ is closed under $\lambda$-sequences (that is, $^\lambda M_\lambda\subseteq M_\lambda$).

Now in the definition of supercompactness, the inner models $M_\lambda$ are allowed to vary with $\lambda$. We could define a large cardinal notion, _uniform supercompactness_ , by removing this possibility: $\kappa$ is uniform supercompact if there is a single $M$ and $j$ such that $M$ is closed under all sequences of ordinal length and $j: V\rightarrow M$ is an elementary embedding with critical point $\kappa$. But in ZFC, this is not an interesting notion. First, assuming choice, $M$ is closed under sequences of ordinal length if and only if $M=V$ (since two models with the same ordinals and sets of ordinals are equal), so $ZFC\models$ "$\kappa$ is uniformly supercompact $\iff$ $\kappa$ is Reinhardt;" second, Kunen showed that Reinhardt cardinals are inconsistent with $ZFC$. (EDIT: as Joel points out, this statement doesn't actually make sense, as Reinhardt-ness is not a first-order property. One needs to pass to some extension of ZFC which can talk about proper classes directly, like Morse-Kelly set theory, which is the setting Kunen used for his proof.)

However, assuming $\neg$Choice, both of these points seem suspect. It is still open whether Reinhardt cardinals are consistent with $ZF$, and it is unclear to me that $ZF$ alone proves that uniformly supercompact cardinals are Reinhardt. So my questions are the following:

Question 1: Does $ZF\models$ every uniformly supercompact cardinal is Reinhardt?

Question 2: What is the consistency strength of $ZF+$ there exists a uniformly supercompact cardinal?

Obviously $ZF$ proves that every Reinhardt cardinal is uniformly supercompact; also obviously, Question 2 is really only interesting in lieu of a positive answer to Question 1. The most basic obstruction to a positive answer to Question 1 is the fact that given an elementary embedding $j: V\rightarrow M$ with $M$ closed under ordinal-length sequences and $crit(j)=\kappa$, the restriction of $j$ to $M$, $j\upharpoonright M: M\rightarrow M$ is not obviously (to me at least) an elementary embedding.

The point here is that we require $j(\kappa)$ to be the same in all of these embeddings. Clearly if there is a class embedding its restrictions satisfy this property. And now we can quantify over all $\lambda$ if need be.

I know that this is quite old by now, and I'm sure that Noah and several others here would have no problem tackling @Joel's last comment. Here is my "starting suggestion" to modify Woodin's definition: $\kappa$ is $\lambda$-supercompact* if there is $\lambda'\geq\lambda$ s.t. for every large enough $\alpha$ there is an elementary $j\colon V_\alpha\to N$, where $N$ is transitive, the critical point is $\kappa$, $j(\kappa)=\lambda'$, $V_\lambda\subseteq N$, and $N^\lambda\subseteq N$. (Some indices might need a small tweak, e.g. $V_{\lambda+1}$ or so to be "faithful to the idea".)

Hm, good point. Is there a "uniformized" version of the existence of normal fine measures that might be to the existence of normal fine measures what the second-order definition of uniform supercompact is to the second-order definition of supercompact? (If that makes any sense.) Looking at it now, I don't see one . . .

I noticed that the definition you give of supercompact is explicitly second-order. With AC, this is equivalent to the existence of normal fine measures, but without AC, one no longer has the Los theorem and so it isn't clear that your second-order definition has a first-order equivalent.

Thanks for your thoughtful response, Noah. Elliott Mendelson dispenses with global AC by positing, in place of Limitation of Size, AC for sets plus a form of Replacement. In that case, ZFC only holds for sets, not for proper classes ("pc's") such as V and M (for M an inner transitive model of V). Hence, ZFC doen't rule out the existence of a suitable pc M, i.e. a pc that witnesses a negative answer to Question 1. Also, a negative answer to 1 is compatible with the existence of Reinhardts; it simply says that the class of Reinhardts doesn't include all of the uniform supercompacts.

will yield a theory which still proves ZFC; in this case, there will be no such suitable model. Specifically, I believe that given any (I suppose wellfounded) model $W$ of this (admittedly vague) version of NBG, it will still be the case that any submodel of $W$ with the same ordinals and sets of ordinals has the same sets as $W$ itself.  If, on the other hand, we weaken Limitation of Size so much that even normal choice is not provable, then we can have such an $M$, but we no longer no that no Reinhardts exist; so we still can't answer 1 negatively.

Oh, I see what you mean. Sorry, I missed the "global" in your comment and didn't understand what you were saying. Actually addressing the point you made: The way I'm familiar with it, NBG doesn't actually have global choice as an axiom; instead, it follows from Limitation of Size. So there are two ways I could understand "dropping global choice" from NBG: weakening Limitation of Size so that only normal AC is provable, or weakening Limitation of Size so that not even normal AC is provable. Both are vague, but some things can be said about them. Any suitable way of doing the former (cont'd)

I agree that the mere existence of an M with certain properties doesn't mean we have an elementary embedding. Question 1, however, doesn't ask for an e.e.; it asks if all uniform supercompact cardinals are Reinhardt (given ZF). I.e., it assumes that uniform supercompacts and their e.e.'s exist (including a suitable M), and asks if this entails an e.e. of V into V. In NBG without global AC, there exists (I claim) a suitable M distinct from V, thus answering Question 1 negatively. I may be wrong, of course; but even if I am, I don't think it's due to not showing that a certain e.e. exists.

@wmitt, I don't think that works - uniform supercompactness (and, for that matter, Reinhardtness) doesn't just ask for an inner model with some properties, it asks for an elementary embedding. So even given an $M\subsetneq V$ closed under sequences of ordinal length, that doesn't mean that we have an elementary embedding $V\preccurlyeq M$. (I think what I've said is right; somebody who knows more, please correct me if I'm wrong!)

Thanks. If we do drop global AC from NBG, doesn't the resulting lack of a bijection between ON and V entail the existence of an inner model M that's (i) closed under sequences of ordinal length, and (ii) not equal to V? If so, we would seem to have a negative answer to Question 1, making the idea of uniform supercompactness interesting.

wmitt, Yes, it does not require any AC or global AC to formulate the Reinhardt cardinal concept in second-order Godel-Bernays set theory (without AC). So you are correct, wmitt, that we should speak of formalizing it in GB rather than in GBC. The Kunen inconsistency, which rules out Reinhardt cardinals, is a theorem of GBC, with choice.

Is there a way to formalize Reinhardt cardinals and uniform supercompactness without using Choice? The above examples of using NBG or KM to formalize these notions both involve the use of Choice. But we need to work with a Choiceless universe in order to answer Questions 1 and 2.

We are jumping over each other's comments!  Yes, by "second-order GBC", I just meant that GBC was a second-order theory. Sorry to confuse...

Yes, NBG and GBC are the same; just an alternative acronym. N = von Neumann. It is conservative over ZFC, because you can force to add a global choice class and then just take the definable classes.  So every model of ZFC is the first-order part of a model of GBC. This is not true for KM, since KM proves the existence of satisfaction classes for first-order truth, and hence proves Con(ZFC).

(Not to keep spamming comments, but:) Your most recent comment confuses me somewhat. Is GBC a theory in second-order logic, or by "second-order GBC" did you just mean something similar to "second-order arithmetic?"

The super compactness of $\kappa$ is expressible in first-order as the assertion:  for every ordinal $\lambda$, there is a $\kappa$-complete normal fine ultrafilter on $P_\kappa\lambda$.

Also, you say that it is also possible to formalize Kunen's inconsistency argument in GBC. I'm not familiar with GBC; is it the same as NBG, and if not, is it a conservative extension of ZFC?

Noah, while we can prove that Reinhardt cardinals are not first order expressible (otherwise let $\kappa$ be the least one, and then consider $j(\kappa)$...), it seems that we should similarly expect the same for uniform supercompactness. But it can be formalized in second-order GBC, as an assertion about a class $j$ and a class $M$. One replaces the assertion that $j$ is fully elementary with the assertion that $j$ is $\Sigma_1$-elementary and cofinal, which implies $\Sigma_n$-elementarity by a meta-theoretic induction.

@Joel: good points. I've corrected the question. Also, building off of your first comment, is it obvious that uniform supercompactness is first-order expressible? For that matter, how is supercompactness itself first-order expressible?

I believe (from your remark about ordinal lengths) that you intend that the assertion "for all $\lambda$..." in the definition is meant to be interpreted as, "for all ordinals $\lambda$...". The difference matters, as you likely know, because if you have $j:V\to M$ and $j"A\in M$ for all sets $A$, then it follows easily that $M=V$ even in ZF, since if $A$ is transitive, then $A$ is the Mostowski collapse of $j"A$.

Very nice question. One quibble: it isn't really correct to say that ZFC refutes Reinhardt cardinals, since there is no first-order characterization of Reinhardt cardinals; that is, you can't say "$\kappa$ is a Reinhardt cardinal" in a first-order way. Kunen proved his inconsistency in the second-order Kelly-Morse set theory, but it is also possible to formalize the result in Goedel-Bernays set theory GBC. See jdh.hamkins.org/generalizationsofkuneninconsistency for a further discussion of this point.

This is an interesting question. The contradiction in ZFC comes from the fact that models with the same sets of ordinals are isomorphic. In ZF it is not true. Monro proved that there is a chain of $\omega$ models, where $V_n\subseteq V_{n+1}$ and they both have the same sets-of-sets-..$n$-times of ordinals, but they are not isomorphic (Jech The Axiom of Choice, Ch. 5, Problem 17). This, however is "going up" while inner models are "going down" so there might be some difference there.

Like you say in the last paragraph, it is not clear that the restriction of j to M is going to be an elementary embedding into M in general.   Instead you could consider the restriction of j to smaller models, like the class N consisting of all sets constructible from sequences of ordinals.  Then j does restrict to an embedding N -> N but you still have might have the problem that the embedding is not amenable to the model.  This seems like the bigger issue to me.


Orders in number fields

Let $K$ be a degree $n$ extension of ${\mathbb Q}$ with ring of integers $R$. An order in $K$ is a subring with identity of $R$ which is a ${\mathbb Z}$-module of rank $n$.

Question: Let $p$ be an unramified prime in $K$. Is it true that the number of orders in $R$ of index equal to $p^r$, for some natural number $r$, is less than or equal to the number of subrings with identity of ${\mathbb Z}^n$ of index equal to $p^r$?

Nathan Kaplan and I need this fact for $n=5$ in a project where we are trying to find asymptotic formulae for the number of orders of bounded index in a given quintic field.

We've been staring at Jos Brakenhoff's [thesis](https://openaccess.leidenuniv.nl/bitstream/handle/1887/14539/proefschrift-brakenhoff.pdf?sequence=2) for a while, but I haven't gotten anywhere. Any advice will be greatly appreciated. Thanks.

Added in edit: Here is an elementary reformulation of this problem. Let $r(x)$ be a polynomial with integer coefficients. Then show that for any natural number $a$ in order to maximize the number of subrings of $({\mathbb Z}/p^a {\mathbb Z})[x]/(r(x))$ of a given index, the polynomial $r(x)$ has to be a product of linear factors modulo $p$.

Update: In a soon-to-be-posted preprint Nathan Kaplan, Jake Marcinek, and I have proved an asymptotic formula for the number of orders in a given quintic field of bounded discriminant, using an argument independent of the above question. We can also give non-trivial upper bounds for a general number field. I'm now convinced the answer to the question is yes, but I still don't know how to prove it.

Thanks everyone. You can actually count the number of subrings of index $p$. Namely, if $r(x) \mod p$ has $u$ factors of degree $1$ and $w$ factors of degree $2$, and the rest of higher degrees, then the number of subrings of ${\mathbb Z}/p{\mathbb Z}[x]/(r(x))$ of index $p$ is ${u \choose 2} + w$.

The $r=1$ case can be handled as follows. Each order in $R$ of index $p$ is uniquely determined by its image in $R/pR$. Now $R/pR$ is an \'etale $k := \mathbf{F_p}$-algebra of rank $n$ as $p$ is unramified. Now the set of such rings injects into the set of index $p$ $k$-subalgebras of $R/pR \otimes_k \overline{k} \simeq \overline{k}^n$. The latter set is a combinatorial object (it's the set of surjective maps from an $n$-element set to an $(n-1)$-element set), and can be bounded above by the set of index $p$ subrings of $\mathbf{Z}^n$.

I found a paper by Melanie Wood which might be relevant, although I did not check it carefully: arxiv.org/pdf/1007.5508v1.pdf

Have you tried (probably you have) writing down a binary $n$-ic form and playing the same game as in Theorem 9 to Lemma 13 of Bhargava, Shankar, and Tsimerman's paper arxiv.org/pdf/1005.0672.pdf ? This doesn't "just work", the details are much messier, but since you need something weaker I wonder if it is possible to salvage the result you are looking for?

In our work we only need this for unramified primes.

It is clear that this question can be studied locally at $p$. In other words, this depends only on $R \otimes \mathbb Z_p$, so it depends only on $K \otimes \mathbb Q_p$, which is just a product of $p$-adic fields.

Is it obvious that the elementary reformulation is the same, given that one is restricted to $p$ unramified and the other is not?


Where do uncountable models collapse to?

Suppose $T$ is a complete first-order theory (in an finite, or at worst countable, language). Given any model $\mathcal{M}\models T$ of cardinality $\kappa$, we can ask whether $\mathcal{M}$ can be made isomorphic to any countable model $\mathcal{N}\models T$ in the ground model; that is, whether there is some $\mathcal{N}\models T$, which is countable, such that $V[G]\models \mathcal{M}\cong\mathcal{N}$ for $G$ $Col(\kappa,\omega)$-generic over $V$. **For simplicity, we'll say "$\mathcal{M}$ collapses to $\mathcal{N}$".**

In case $T$ has _countably many_ countable models, the answer is yes: the set of models of $T$ is a countable Borel set, and so is absolute. The uncountable models of such a $T$ are thus partitioned into $\omega$-many classes. My question is about what the nonempty classes are. For example, suppose $T$ is $\aleph_1$-categorical. Then the countable models of $T$ form an elementary chain, with a "top" model $\mathcal{N}_\omega$, and it is easy to see that this is the only countable model which uncountable models can collapse to.

However, there are plenty of non-$\aleph_1$-categorical theories with only countably many countable models, and beyond the $\aleph_1$-categorical setting things are much less clear. Some examples:

  * The language has predicates $P_i$ ($i\in\omega$); $T$ asserts that the $P_i$ name infinite disjoint sets. Then the countable models of $T$ are classified by how many elements they have not in $\bigcup_{i\in\omega} P_i$, and so there are countably many countable models. For each countable model of $T$, there is an uncountable model of $T$ which collapses to it.

  * The language has a single binary relation $E$; $T$ asserts that $E$ is an equivalence relation and that there is exactly one class of cardinality $n$, for each $n$. Countable models of $T$ are determined by how many infinite classes they have; and every countable model of $T$ **except the prime** is collapsed to by some uncountable model of $T$.

I'm curious what we can say in general about the set of models which can be collapsed to. There's a lot of questions around here that one can ask; let me focus on:

> Is there a theory $T$ with countably many countable models, including $\mathcal{M}_0\prec\mathcal{M}_1$, such that $\mathcal{M}_0$ can be collapsed to by some uncountable structure but $\mathcal{M}_1$ cannot?

That is, can there be "gaps" in the range of the collapse?

* * *

EDIT: As Paul Larson pointed out in the comments below, a countable model is collapsed to by some uncountable model iff it is **extendible** (see Definition 2.6 in <http://shelah.logic.at/files/1003.pdf>) - that is, iff it has an uncountable $L_{\omega_1\omega}$-elementary extension. One direction is trivial. To show that every extendible model is collapsed to, let $\mathcal{M}$ be extendible with $\mathcal{M}\prec_{\omega_1\omega}\mathcal{N}$ for $\mathcal{N}$ uncountable. Let $\varphi$ be the Scott sentence of $\mathcal{M}$, and let $V[G]$ be a forcing extension in which $\mathcal{N}$ is made countable. Then $\mathcal{N}\models\varphi$ in both $V$ and $V[G]$; moreover, the statement "$\varphi$ is the Scott sentence of $\mathcal{M}$" is $\Pi^1_2$ (actually, better, but this is enough) and so absolute between $V$ and $V[G]$. So $V[G]\models\mathcal{M}\cong\mathcal{N}$.

It's worth keeping in mind the broader result, due to Barwise and Karp if I recall correctly, that $\mathcal{M}\equiv_{\infty\omega}\mathcal{N}$ iff $\mathcal{M}\cong\mathcal{N}$ in some forcing extension.

I asked Saharon my version of your question, and it turns out to have a relatively easy positive answer. Here is a write-up, which could probably be improved : users.miamioh.edu/larsonpb/extendible.pdf

I've been wondering about this too. I don't see that it matters. By the way, Su Gao's "On automorphism groups of countable structures" shows that a countable structure is extendible if and only if its automorphism group is not closed in the group of permutations of its domain. So one  could rephrase the problem using this. Here's a variation I've wondered about : can you have countable $M \prec N$ such that $M$ is extendible and $N$ is rigid?

@PaulLarson Coming back to this a year later, it occurs to me that we can ask the same question of theories with uncountably many countable models. That is, are there $\mathcal{M}_0\prec \mathcal{M}_1$ with $\mathcal{M}_0$ extendible but $\mathcal{M}_1$ not? It's not obvious to me that this makes the question any easier - I don't see how to get such a pair, even if I allow lots of countable models - but maybe you know an easy answer in this case?

For instance, it seems that the model-theoretic forcing approach gives : if $T$ is a complete first order theory having an uncountable model and only countably many countable models, then there is a countable extendible model of $T$ having a copy of each countable model of $T$ as an elementary submodel. Again, I don't know how much insight this gives into your question.

@PaulLarson That looks really cool, thanks!

You might also get some mileage out of Shelah's theory of model-theoretic forcing, from his publication #88r, depending on what you're interested in.

John Baldwin used "extendible" in our paper with Saharon (shelah.logic.at/files/1003.pdf). I assume that it was used previously, but I don't really know.

@PaulLarson I didn't know about extendible models - yes, it looks like a model is extendible iff it is collapsed to. But googling around I can't find much about them - can you point me to a good source?

It seems that you could naturally rephrase this in terms of extendible models (i.e., countable models which have the same Scott sentence as some uncountable model) and then drop the mention of forcing.


Which sets of roots of unity give a polynomial with nonnegative coefficients?

> **The question in brief:**   When does a subset $S$ of the complex $n$th roots of unity have the property that $$\prod_{\alpha\, \in \,S} (z-\alpha)$$ gives a polynomial in $\mathbb R[z]$ with nonnegative coefficients?

Some trivial necessary conditions include $1\not\in S$ and also that $S$ is self-conjugate so that the coefficients will at least be real.

A sufficient condition that is not obvious (to me) is that the roots of unity in $S$ are precisely those lying outside of some wedge-shaped region of the complex plane, i.e., those with $\left|\arg(z)\right| > t$ for some fixed $t > 0$. This follows from the main theorem of

> Barnard, R. W., W. Dayawansa, K. Pearce, and D. Weinberg. "Polynomials with nonnegative coefficients." _Proceedings of the American Mathematical Society_ 113, no. 1 (1991): 77-85. [Journal link](http://www.ams.org/journals/proc/1991-113-01/S0002-9939-1991-1072329-2/S0002-9939-1991-1072329-2.pdf)

which says that, given any polynomial with nonnegative coefficients, dividing it by the linear factors corresponding to exactly those of its roots lying in such a region produces a polynomial with only positive coefficients. Applying this to $z^{n-1}+\cdots+z+1$ gives the desired result.

But that theorem is not number-theoretic; it doesn't care at all that we are dealing with roots of unity. So maybe there is a nicer proof in this special case?

More generally, what other necessary and/or sufficient conditions exist? Is there some reasonable number-theoretic or combinatorial characterization of such sets? What tools seem likely to shed some light on this question?

**ADDED LATER:** A couple of very interesting suggestions have been made in the comments so far, and I'm eager to see where they lead. However, I remain quite curious as to what can be said from a number-theoretic perspective. For example, if $n=p^r > 1$ for some prime $p$, then the cyclotomic polynomial $\Phi_n(x)=\Phi_p(x^{n/p})$ has only nonnegative coefficients. So a sufficient condition that I did not mention above is that $S$ is the set of primitive $n$th roots of unity for some prime power $n$. I believe this is the only way to obtain such a polynomial that is a _cyclotomic_ polynomial, but that doesn't imply that nothing else can be gained from a number-theoretic perspective on this question – however I myself lack the expertise necessary to make the most of such a perspective.

I should also mention that (ironically, given my desire for a number-theoretic perspective on this) the cyclotomic result actually follows from a more general, not number-theoretic result of

> Evans, Ronald, and John Greene. "Polynomials with nonnegative coefficients whose zeros have modulus one." _SIAM Journal on Mathematical Analysis_ 22, no. 4 (1991): 1173-1182. [Author's link](http://math.ucsd.edu/~revans/PolynomialsGreene.pdf)

which gives that for any proper divisor $d$ of $n$, $(x^n-1)/(x^d-1)$ has only nonnegative coefficients. So that generalizes the cyclotomic case, but not via number theory!

"I believe this is the only way to obtain such a polynomial that is a cyclotomic polynomial": yes, you're correct. The easiest way to see this is plugging in x=1: it's easy to show that $\Phi_m(1) = \begin{cases}0 & m=1 \\ p & m=p^k \\ 1 & \text{otherwise}\end{cases}.$ This essentially says there must be cancellation outside of the prime-power case.

@Josep: The reference you gave is extremely interesting.  I was totally unaware of Suffridge's result.

In fact the Suffridge polynomials cover the more general case when $1\not\in S,$ S self-conjugate and the arguments of the roots in S are separated by the same angle, except for a pair.

In the wedge case you can compute explicitly the coefficients and see they are positive. It's a particular case of Suffridge's extremal polynomials- you have the expression deduced in Sheil-Small, Complex polynomials, p. 251-252.

Maybe, the following related result of Kellog is helpful: Let $A$ be a complex $n\times n$ matrix. If all its elementary symmetric functions are positive (so that the characteristic polynomial has alternating signs), then the spectrum of $A$ lies in the set $\{z : |\text{arg}z| \le \pi - \pi/n\}$....


Chromatic Spectra and Cobordism

I apologize in advance, if some of the things I've written are incorrect.

The cobordism hypothesis states that $\mathbf{Bord}^\mathrm{fr}_n$ is the free symmetric monoidal $(\infty,n)$-category with duals generated by a point. There is a geometric realization functor $|-|:\text{Cat}_{(\infty,n)}\rightarrow \text{Gpd}_\infty$, where $\text{Cat}_{(\infty,n)}$ is the $\infty$-category (I mean a quasicategory, but you can think of it as a model category, or whatever you like) of (small) $(\infty,n)$-categories, and $\text{Gpd}_\infty$ is $\infty$-category of $\infty$-groupoids, otherwise known as spaces. This functor is a left adjoint (in the homotopical sense) to the inclusion functor $\text{Gpd}_\infty\hookrightarrow \text{Cat}_{(\infty,n)}$. Therefore, $|-|$ preserves (homotopy) colimits. If we "restrict" ourselves to the symmetric monodical $(\infty,n)$-categories, we would get a functor $|-|:\text{SymMon}_{(\infty,n)}\rightarrow \text{Alg}_{\mathbb{E}_\infty}$, where $\text{SymMon}_{(\infty,n)}$ is the $\infty$-category of symmetric monoidal $(\infty,n)$-categories and $\text{Alg}_{\mathbb{E}_\infty}$ is the $\infty$-category of infinite loop spaces. This should also preserve colimits. It manifests itself in the fact that $|\mathbf{Bord}^\text{fr}_n|$ is the infinite loop space of the sphere spectrum.

There is a universal characterization of the bordism $(\infty,n)$-category of manifolds with singularities [Theorem 4.3.11 in [Lurie's paper](http://arxiv.org/abs/0905.0465)]. The type of singularities considered are those of Baas-Sullivan theory, which produce a wide array of chromatic spectra, such as the Brown-Peterson spectrum $\text{BP}$, the Johnson-Wilson spectra $\text{E}(n)$, and, of course, the Morava K-theory spectra $\text{K}(n)$. I am wondering:

> Is it possible to give these spectra a universal characterization using the fact that geometric realization preserves colimits?

**EDIT:** When I think about it, I am not sure whether there is a universal description of $|\mathbf{Bord}^{(X,\zeta)}_n|$, where $X$ is topological space, $\zeta$ a vector bundle on it of dimension $n$, other than the case when $X$ is point. We _can_ write it down. It is simply $\Omega^{\infty-n}M\zeta$, where $M\zeta$ is the Thom spectrum of $\zeta$.

> Do we get a different characterization of these infinite loopspaces from the cobordism hypothesis?


Planar minor graphs

The theorem of Robertson-Seymour about graph minors says that there exists no infinite family of graphs such that none of them is a minor of another one.

Apparently, it came as a generalization of the Kruskal's theorem that states that there exists no infinite family of rooted ordered trees such that none is a minor of another one. Here, _rooted ordered_ means that the tree has a root, and that the edges escaping from a vertex are ordered. In other words, the trees are assumed to be embedded in the plane, and the minor operation has to respect this embedding.

Here comes the question: is Robertson-Seymour theorem true for planar graphs, when we add the condition that the minor operation respects the embedding? (_i.e._ we not only ask $G_i$ to be a minor of $G_j$ as an abstract graph, but also as an embedded graph.)

It is not clear to me that this should be a direct corollary of the original theorem, because of the amount of possible embeddings into the plane for a given planar graph.

@PierreDehornoy: I think it would improve the question if "true for planar graphs" was replaced with 'true for plane graphs', because what you write immediately afterwards suggests that you are thinking about planar graphs together with a specified embedding into the plane, and that is called a plane graph.

@Pierre Dehornoy: Pierre, sorry, you a right, this argument does not answer your question.

@ Misha: This looks almost convincing, but I do not understand the end of the argument. One of the embedding of $G_m$ appears infinitely offen, but how does it says that this particular embedding is the one that was given at the beginning, i.e. that corresponds to the given embedding of $G_m$?

I guess, my understanding of R-S theorem is correct: If a sequence of graphs were to contain infinitely many minimal elements $G_m$ then the new sequence of graphs $(G_m)$ would violate R-S theorem.

@Pierre Dehornoy: I thought that the conclusion of Robertson-Seymour theorem could be strengthened to "A certain graph $G_m$ occurs as a minor infinitely many times in a given infinite sequence" ($G_m$ would be one of the finitely many minimal elements of the sequence, with respect to the "minor" order). If my understanding is correct, then the answer to your question is positive since $G_m$ has only finitely many planar embeddings, up to isotopy, so one of them would be repeated (infinitely often).


Is every $p$-group the $\mathbb{F}_p$-points of a unipotent group

Let $\Gamma$ be a finite group of order $p^n$. Is there necessarily a unipotent algebraic group $G$ of dimension $n$, defined over $\mathbb{F}_p$, with $\Gamma \cong G(\mathbb{F}_p)$?

I have no real motivation for this question, it just came up in conversation and no one knew the answer. There does not appear to be any sort of uniqueness to $G$; both the groups $\mathbb{Z}/p^2 \mathbb{Z}$ and $(\mathbb{Z}/p \mathbb{Z})^2$ have infinitely many lifts to unipotent groups.

* * *

I am bumping this question to the top because I now suspect I have a counterexample, namely the dihedral group of order $16$, also known as $C_2 \ltimes C_8$, where $C_n$ is cyclic of order $n$. The "obvious" lift of $C_8$ is the additive group of the $3$-truncated Witt vectors, $\mathcal{W}_3$. So the obvious thing to try is a semidirect product $\mathbb{G}_a \ltimes \mathcal{W}_3$. But $\mathbb{G}_a$ has exponent $2$, so an action of $\mathbb{G}_a$ on $\mathcal{W}_3$ must live in the $2$-torsion elements of $\mathrm{Aut}(\mathcal{W}_3)$. I compute that the space of $2$-torsion elements in $\mathrm{Aut}(\mathcal{W}_3)$ has two connnected components, one of which maps $1$ to things that are $1 \bmod 4$ and one of which maps $1$ to things which are $-1 \bmod 4$. Since $\mathbb{G}_a$ is connected, it must land in one component, but the two elements of $C_2$ live in different components.

I've made some partial progress thinking about the structure of connected unipotent groups whose $\mathbb{F}_2$ points are $C_2 \ltimes C_8$, but I've gotten stuck, so I am putting up a bounty for progress either on this case or the questions as a whole.

@DrorSpeiser the Zariski closure of a finite set is itself (so not unipotent). I don't see any reason why we should expect any given embedding of D_8 in a unipotent group to come from the F_2-points of a connected subgroup.

Take $a=1+E_{12}+E_{34}+E_{56}$, $b=1+E_{23}+E_{45}+E_{67}\in Mat_{7\times7}(\mathbb{F}_2)$, where $E_{ij}$ has a unique nonzero element at row $i$ and column $j$ equal to 1. Set $c=ab$. Let $V$ be $\overline{\mathbb{F}}_2^7$, and let $D_{16}\cong<x, y|x^2=y^8=xyxy=1>$ act on $V$ by sending $x$ to $a$ and $y$ to $c$. With this action $V$ is an indecomposable module for $D_{16}$. Let $\overline{D_{16}}$ be the Zariski closure of $<a, c>$ in $GL(V)$, which is unipotent. Does it satisfy $\overline{D_{16}}(\mathbb{F}_2)=<a,c>\cong D_{16}$?

@DrorSpeiser I decided to put the bounty on my own version, rather than the duplicate because I'd rather edit my own question and because it has a bunch of useful comments. If I had (two years ago) seen Gjergji's link to the duplicate before a bunch of those comments showed up, I probably would have self-closed, but it doesn't seem like a big deal either way. Of course, if someone posts a good answer, I'll link it on that question as well to make sure searchers find it.

Given that this is a duplicate of a question six years earlier, how is this not closed?

Okay, I'll take a look at Abhyankar's conjecture and see what that does for me. In the mean time, I'll record that the formula was meant to be $(x_1, y_1, z_1) \ast (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2, z_1+z_2 +(x_1 y_2)^p-x_1 y_2)$, sorry for the incorrect subscripts in the previous one.

I am specifically suggesting to use the faithful representations of $p$-groups arising from Abhyankar's conjecture.  If we realize $\Gamma$ as the Galois group of a finite etale extension $\mathbb{F}_q[t] \subset R$, then for every subgroup $\Lambda \subset \Gamma$, then $R$ is a finite etale extension of the $\Lambda$-invariant subring $S$, with Galois group $\Lambda$.  This is a "complete reducibility" result that is missing for arbitrary representations.  I suggest we projectively complete these rings at $\infty$, and quotient by the maximal ideal of $\infty$ in $\mathbb{F}_q[t^{-1}]$.

Suppose there were such a $G$, and let $Z$ be the center $\{ (0,0,z) \}$ of $U$. Case 1: $G \supset Z$. Then $G(\mathbb{F}_p) \supset \{ (0,0,z) : z \in \mathbb{F}_p \}$, which is not in $\Gamma$. Case 2: $G \cap Z$ is finite. In that case, the projection $G \to U/Z$ is surjective. But any subvariety of $U$ which surjects onto $U/Z$ generates $U$, contradicting that $G$ is two dimensional.

I think there is a fatal flaw in @JasonStarr's approach. It seems that Jason's method would show that, if $U$ is a unipotent group, and $\Gamma$ is a subgroup of $U(\mathbb{F}_p)$, then there is a unipotent subgroup $G$ of $U$ with $G(\mathbb{F}_p) = \Gamma$. This isn't true. Let $U$ be the group structure on $\mathbb{A}^3$ given by $(x_1,y_1,z_1) \ast (x_2,y_2,z_2) = (x_1+x_2, y_1+y_2, z_1+z_2+x_1^p y_1^p - x_1 y_1)$ and let $\Gamma = \{ (x,y,0) : x,y \in \mathbb{F}_p \} \subset U(\mathbb{F}_p)$.  Then $\Gamma$ is not $G(\mathbb{F}_p)$ for any $2$-dimensional subgroup of $U$. (continued)

@DavidSpeyer. I should not have called this the unique smallest subgroup.  I am saying that for a vector space $V$ and a finite subgroup $Z$ of the group of $\mathbb{F}_p$-points of the scheme $\text{Spec}\ \text{Sym}_{\mathbb{F}_p}(V)$, the kernel of the restriction map $V\to \prod_{z\in X} \kappa(z)$ is an $\mathbb{F}_p$-linear subspace of $V$ generating an ideal whose associated closed subscheme is a closed subgroup scheme whose associated group of $\mathbb{F}_p$-points equal to $Z$.

@JasonStarr It seems like you are saying that, given an connected abelian group $H$ (in your setting, $Z(U)$), and a discrete subgroup $B$, there is a smallest connected group $A$ containing $B$. But that isn't true: consider $H = \mathbb{G}_a^2$ and $B = \{ (x,0) : x \in \mathbb{F}_p \}$. Then $B$ is contained in $\mathbb{G}_a$ embedded in $H$ by either $t \mapsto (t,0)$ or $t \mapsto (t,t^p-t)$. What am I missing?

@JasonStarr: I was wondering about that connectedness issue too but thought I might be overlooking something.

@nfdc23.  You are correct.  I realize now the problem with the approach I suggest: why should the centralizer of $Z(\Gamma)$ in $U$, much less the center of the centralizer, be a connected group?

@JasonStarr: In view of your inductive suggestion it seems that what you call "$n$" (from SL$_n$) is not "$n$" as in the statement of the question, just some unknown integer, in which case such $\rho$ is provided by the regular representation of $\Gamma$ on itself over $\mathbf{F}_p$ (that gives $\rho$ into ${\rm{GL}}_N(\mathbf{F}_p)$, necessarily landing inside SL$_N({\mathbf{F}}_p)$ and even conjugating into $U_N(\mathbf{F}_p)$ since the latter is a $p$-Sylow subgroup).

. . . So there is a unique minimal, connected subgroup $A\subset Z(U)$ such that $A(\mathbb{F}_p)$ equals $Z(\Gamma)$.  Now replace $U$ by $U/A$, and use induction on the rank.

Here is how I would try to prove this.  There exists a faithful homomorphism, $\rho:\Gamma\to \textbf{SL}_n(\mathbb{F}_p)$.  This follows, for instance, from the solution of the Abhyankar conjecture by Raynaud and Harbater.  By counting $\mathbb{F}_p$-points of the flag variety of $\textbf{SL}_n$, there exists a Borel subgroup $B$ that contains the image of $\rho$.  Thus, there exists a faithful homomorphism from $\Gamma$ to a unipotent group.  Replace that unipotent group by the centralizer $U$ of $Z(\Gamma)$.  Thus $Z(U)(\mathbb{F}_p)$ is an additive group that contains $Z(\Gamma)$ . . .

A step towards an inductive proof:  if $G$ is a unipotent $k$-group, and $c \in Z^2(G(k), Z / p Z)$ is a 2-cocycle (for trivial action), does there exist a polynomial 2-cocycle $\gamma \in Z^2(G, G_a)$ such that $\gamma$ gives $c$ upon taking $k$-points?  I want to attack this one inductively with inflation-restriction with respect to $G$ and $[G,G]$.  But no time for more than a sketchy comment.

This was asked a while ago, as well, but there was no answer. mathoverflow.net/questions/69397/…

If $U$ is smooth connected unipotent of dimension $n$ over $k=\mathbf{F}_p$ then $\#U(k)=p^n$ (use composition series with successive quotients $\mathbf{G}_a$).  Thus, if $\Gamma=G(k)$ for a unipotent $k$-group $G$ of dimension $n$ then by passing to $G_{\rm{red}}$ so that $G$ is smooth we see that if $G$ is not connected then $G^0(k)$ of size $p^n$ must exhaust $G(k)$. In other words, we lose nothing by assuming $G$ is smooth and connected. So it is the same as asking for smooth connected unipotent $k$-groups $U$ if there is any constraint on $U(k)$ beyond its order being $p^n$. Good puzzler.


Does the Tate construction (defined with direct sums) have a derived interpretation?

Any abelian group M with an action of a finite group $G$ has a Tate cohomology object $\hat H(G;M)$ in the derived category of chain complexes. There are several ways to define this. One is as the tensor product $W \otimes_G M$ with a particular (typically unbounded) complex of projective $\Bbb Z[G]$-modules which happens to have trivial homology. One is to instead use $Hom^G(W',M)$ for $W'$ a similar complex. One is as the mapping cone of a norm map $$ M_{hG} \to M^{hG} $$ from derived coinvariants to derived invariants, where we view $M$ as a complex concentrated in degree zero.

We can attempt to generalize this construction to the derived category of $\Bbb Z[G]$-modules, or of $R[G]$-modules for $R$ a commutative ring. If $M_*$ is a chain complex of $\Bbb Z[G]$-modules, these give us three definitions of the Tate construction on $M_*$: $$ \begin{align*} Tate^\oplus(M)_n &= \bigoplus_{p+q=n} W_p \otimes_G M_q\\\ Tate^\prod(M)_n &= \prod_{p+q=n} W_p \otimes_G M_q\\\ Tate(M)_n &= \bigcup_N \prod_{p+q = n, p \leq N} W_p \otimes_G M_q \end{align*} $$ (There are natural maps $Tate^\oplus \to Tate \to Tate^{\prod}$ which are isomorphisms on bounded complexes.)

The first construction is nice because it commutes with filtered colimits, and the second with filtered limits. The third is neither, but it does have a norm-cofiber sequence and a Tate cohomology spectral sequence $$ \hat H^p(G; H^q(M)) \Rightarrow H^{p+q}(Tate(M)). $$ The third construction is also the only one that preserves quasi-isomorphisms, and thus the only one that (naturally) extends to the _unbounded_ derived category.

Are there derived-category interpretations for the constructions $Tate^\oplus$ or $Tate^\prod$?

Such an interpretation probably needs to take in more information than an object of the derived category of $\Bbb Z[G]$-modules. The main motivation for asking is that I'd like to see what is necessary in more general contexts (e.g. modules over a differential graded algebra).

Another characterisation of Tate cohomology $\hat H^n(G, M)$ is as $Hom(\mathbf Z, M[n])$, where the Hom-set is in the category of singularities, defined as the quotient of the usual derived category $D^b(Mod_{\mathbf Z G})$ modulo its subcategory of perfect complexes. A natural extension thus seems to be (I remember this is also mentioned somewhere in Gaitsgory's work, but don't seem to find it right now) to consider the Hom-set or -space in the quotient of Ind-Coherent sheaves modulo Quasi-coherent sheaves (in the $\infty$-sense).

@Jesper Both of those are good suggestions. I haven't asked either yet, and it looks like I'll probably be doing so soon.

Dave Benson or John Greenlees would be my go-to people for this. Maybe you've already asked them...


Are amenable groups topologizable?

I've learned about the notion of topologizability from "On topologizable and non-topologizable groups" by Klyachko, Olshanskii and Osin (<http://arxiv.org/abs/1210.7895>) - a discrete group $G$ is topologizable iff there exists a topology on $G$ which makes it into a Hausdorff non-discrete topological group.

> **Main question:** Is every infinite amenable group topologizable?

* * *

Main motivation for this question is that perhaps naively non-topologizability seemed to me such a strange property that I hoped it could be used to show existence of non-sofic groups.

> **Question:** Is there an infinite non-topologizable sofic group?

After failing to prove that sofic groups are topologizable I thought it would be still interesting to prove that infinite "elementary sofic" groups are topologizable. Elementary sofic groups are for the purpose of this discussion the class of "groups which are provably sofic by current methods", i.e. it contains all amenable groups, is closed under taking free products amalgamated over amenable groups, extensions with sofic kernel and amenable quotient, /any other results which are in the literature/, and with the property that if G is residually elementary sofic then G is elementary sofic.

> **Question:** Is there an infinite non-topologizable elementary sofic group?

Unfortunately my plan to answer the above question negatively failed at step 1, and hence the Main question.

I essentially stopped thinking about this, because I don't know how to proceed, but I thought I'd share one idea which at one point I thought was hopeful: G acts on a certain metric space X by isometries - namely fix a mean m on G and define X to be the set of subsets of G up to sets of mean 0. The metric on X is d(A,B) = m(A-B \cup B-A). There are various topologies on Isom(X) but I failed to prove any of them gives a Hausdorff non-discrete topology on G.

@Simone: AFAI understand, Bohr compactif'n B(G) is in particular compact, so by Peter-Weyl thm if G embeds in B(G) then G is res. linear, and so by Malcev thm if it is fin. generated then it is res. finite. So to produce example of an amenable group G which doesn't embed into B(G) take a fin. generated simple amenable group, or easier take a fin.gen. solvable non-res. finite group (e.g. BS(1,n)). Taking a simple group is more convincing though, because the image in B(G) is trivial, whereas if the image in B(G) is infinite one could still hope for inducing a (Hausdorff) topology on G somehow...

probably Corollary 3 on page 213 in mscand.dk/article.php?id=2006    will help...

unfortunately it seems that the bohr topology is T2 only for MAP groups... is there any known example of amenable non-MAP group (with the discrete topology)?

Any infinite abelian group is topologizable (e.g. by its Bohr topology).

Just to clarify,  Łukasz's first sentence actually characterizes the pro-p topology for fixed p; the profinite topology on $\mathbb{Z}$ is generated by all the pro-p topologies.

@Ben: you can use profinite topology, e.g. fix a prime number $p$; then integer $n$ is "near" to $0$ if large power of $p$ divides $n$. Other way to topologize integers is - take the action of integers $Z$ on the circle such that the generator $t$ of $Z$ acts by irrational rotation, and define $t^k$ to be "close" to the identity element iff $t^k$ is a rotation by a "small" angle. This is special case of "find an infinite cyclic subggroup of a compact group and induce the topology"

Wouldn't the evenly spaced integer topology work?

Are the integers topologizable?  What does that topology look like?


Caramello's theory: applications

In [this](https://www.laurentlafforgue.org/math/TheorieCaramello.pdf#page=9) text, the author says (well, he says it in French, but I am too lazy to fix all the accents, so here is a Google translation):

_In any case, contemporary mathematics provides an example of extraordinarily deep and highly studied equivalence which, without having so far been formulated as an equivalence of Morita, still seems very close to the general framework of Caramello's theory: it is the correspondence from Langlands._

I am trying to understand this remark. The entire mathematical world has produced few, if any, correspondences which unify lots of non-trivial mathematics as elegantly as the Langlands program does (other than the Langlands, that is). Several genuises (including L. and V. Lafforgue, A. Wiles, P. Scholze) have contributed to its study and our understanding of it is still far from complete.

> I wonder if there were any papers confirming the overall sentiment of the author, i.e. papers where Caramello's theory is applied to Langlands. Were there at least some long-standing problems that historically were not believed to have any serious connection with logic or topoi, and were solved using Caramello's techniques?

There were some similar questions, but [this](https://mathoverflow.net/q/330087/140765) one seemed to be largely about understanding an incorrect translation of a quote of another distinguished mathematician (hopefully, our translation is correct!), and [this](https://mathoverflow.net/q/29232/140765) one is pretty similar but not the same (it asks about producing theorems without any creative effort, while we allow additional creative input but just want to see theorems that rely in an essential way on ideas of Caramello, among other things). The second question did receive a response that might have qualified as an answer to our question, but that response also received [criticism](https://mathoverflow.net/questions/29232/the-unification-of-mathematics-via-topos-theory?noredirect=1&lq=1#comment70456_29826) from BCnrd (which was not addressed AFAICT).

P.S. The kind of response we are not looking for in this question:

  * some philosophical argument why topoi are great that does not give a specific resut. While such arguments can be of interest in some situations, not right now.
  * something along the lines of "One can produce thousands of deep theorems easily, so easily in fact that I won't even bother producing a single example".

The kind of response we are looking for:

  * a link to a paper proving a result outside of logic or set theory and whose statement does not involve topoi, and an explanation of how Caramello's techniques enter the proof as an important ingredient.

@TimCampion OK, maybe somebody who actually read those papers will comment on them. But who knows, maybe logic will give us the motivic t-structure!

The first two publications listed under "Publications" on her website are both papers in English written jointly with Lafforgue. In particular, the second appears to do something reasonably concrete in algebraic geometry / number theory. The first appears to be more category-theoretic, but perhaps closer inspection would reveal some algebraic geometry / number theory.

@TimCampion I browsed through some of them and did not manage to find any results of geometric or topological interest (though those that I saw were in French, so I did not try super hard).

Caramello's website lists 3 papers joint with Lafforgue by my count. Presumably they contain some kind of working-out of her theoretical framework, if not directly in the area of the Langlands program, then at least somewhere in that area.


$\infty$-topos and localic $\infty$-groupoids?

It's known that every classical (Grothendieck) topos is equivalent to the topos of sheaves on a localic groupoid (a groupoid in the category of locales).

For the record, this is proved by, starting form a topos $T$, constructing a locale $L$ and a surjection $L \rightarrow T$ 'nice enough' (like a proper surjection, or an open surjection depending on the proof). Then $(L, L \times_T L, L \times_T L \times_T L)$ is a truncated simplicial locale, which can be seen as a localic groupoid. There is a canonical geometric morphism from the topos of sheaves on this groupoid to $T$, and if the surjection $L \rightarrow T$ was nice enough it's an isomorphism.

My question is : Can we hope for a similar result for $\infty$-toposes ? for example by replacing localic groupoids by a localic $\infty$-groupoids, defined as a simplicial locale satisfying some form of Kan complex condition and defining the associated $\infty$-topos as the colimits of the corresponding simplicial diagram of localic $\infty$-toposes.

@SimonHenry In classical topos theory, the Diaconescu localic cover theorem gives a connected locally connected geometric morphism. Often people only stress on the fact that it is open.

@IvanDiLiberti  the stability under pullback of localic morphisms might be useful at some point yes. But the crucial point would be some form of Diaconescu Localic open cover theorem for $\infty$-toposes (though "open" should probably be replaced by locally $\infty$-connected or something like that...).

@SimonHenry I think that would be a very good starting point to prove the result you want.

@IvanDiLiberti Probably - but as far as I remember it hasn't been written (from what I remember Lurie in HTT talks about n-localic toposes but does not study the relative version)

Do we have a hyperconnected-localic fact system?

So, you're asking if every classical $\infty$-topos is equivalent to the category of sheaves on a localic $\infty$-topos, correct?

If I found any things about this, I will. But for now I have a lot of thing to learn before...

This is something I have thought about in the back of my head for several years. If you come up with something, or would just like to brainstorm, let me know.


Status of the Euler characteristic in characteristic p

In the introduction to the Asterisque 82-83 volume on `Caractérisque d'Euler-Poincaré, Verdier writes:

> Enfin signalons que la situation en caractéristique positive est loin d'être aussi satisfaisante que dans le cas transcendant. On aimerait pouvoir disposer d'un groupe de Grothendieck `sauvage' des faisceaux constructibles tels que deux faisceaux ayant même rang et même comportement sauvage en chaque point aient même classe dans ce groupe de Grothendieck et tel que ce groupe soit foncteur covariant. Des resultats récents de Laumon donnent des indications dans ce sens. On aimerait aussi disposer d'une théorie des classes d'homologie de Chern qui sereait une transformation naturelle de ce groupe de Grothendieck sauvage dans les groupes d'homologie étale.

My question is:

> > Has the situation improved in the intervening 32 years?

I recall in detail below some background and ask some more precise questions.

For a map of complex algebraic varieties $f: X \to Y$, one can define the pushforward of a constructible function $\xi$:

$$(f_* \xi)(p) = \sum_n n \chi(\xi^{-1}(n) \cap f^{-1}(p) )$$

This is a relative notion of Euler characteristic: $f_* (1_X)$ is the function on $Y$ whose value at $y$ is the Euler characteristic $\chi(X_y)$.

This notion is functorial: $(gf)_* = g_* f_*$.

The (naively) analogous notion in characteristic $p$ is not functorial. The standard counterexample is the Artin-Schreier map

$g:\mathbb{A}^1 \to \mathbb{A}^1$

$z \mapsto z^p - z$

Note $g_* 1 = p$. On the other hand, taking $f: \mathbb{A}^1 \to \mathrm{Spec}\,k$ the structure map, one has $f_* (g_* 1)) = p \ne 1 = (fg)_* 1$. The problem seems to be that the map is wildly ramified at infinity. The beginning of the above paragraph of Verdier is a request that this situation be remedied by introducing some intermediary between constructible sheaves and their stalkwise Euler characteristics which would record the possibility of wild ramification and thus be functorial under pushforward.

> Is there now some such object?

The second part of Verdier's paragraph presumes that the answer to the question above is yes, and requests an analogue of Macpherson's Chern class transformation. Over $\mathbb{C}$, this is a natural transformation $c_{SM}: Con(\cdot) \to H_*(\cdot)$ between the functors (covariant with respect to proper pushforward) which respectively assign to a variety its constructible functions and homology; it has the property that if $X$ is proper and smooth then $c_{SM}(X) = c(TX) \cap [X]$.

> Is there an analogue of the Macpherson Chern class transformation in characteristic $p$?

The Macpherson Chern class is defined in terms of something called the 'local Euler obstruction'. This is a constructible function $Eu_V$ attached to a variety $V$ which, roughly speaking, at a point $v\in V$ records the virtual number of zeroes of any extension of the radial vector field from the boundary of ball around $v$ to its interior. (One passes to the Nash blowup in order to make sense of this vector field.)

The Euler obstruction enjoys the following hyperplane formula: locally embedding $(V, v)$ in a smooth variety $Y$, taking some sufficiently small $\epsilon$ ball $B_\epsilon(v)$, and a generic projection $l: B_\epsilon(v) \to \mathbb{C}$, one finds that $l_* Eu_V$ is locally constant near $l(v)$. In other words, the Euler obstruction is annihilated by taking vanishing cycles with respect to a general linear function.

> Is there an analogue of the local Euler obstruction in characteristic $p$, and is it annihilated by taking vanishing cycles with respect to a general linear function?

You've probably seen this already, but I'm adding it here anyway for others: arxiv.org/abs/1510.03018 by Takeshi Saito is relevant.


0's in 815915283247897734345611269596115894272000000000

> Is 40 the largest number for which all the 0 digits in the decimal form of $n!$ come at the end?

Motivation: My son considered learning all digits of 40! for my birthday. I told him that the best way to remember them would be to come up with a mnemonic. He asked me what word he should come up with if the digit is 0. The rest is history.

Maybe it will easy to solve in binary first. Where satisfied for 0!, 1!, 2!, 3! and 4!.

The point is that the question is not about the zeros in this number (which is a trivial question) but about the zeros in infinitely many other numbers.

I changed the title to "$0$'s in 40!=...", that is, added "40!". This makes the title less cryptic within the list of questions. I regret you reverted it.

The title would be improved (and less puzzling) if it were replaced by the sentence that is the actual question.

@Malkoun I did a similar check before seeing your post, and ran it up to 40,000.  OEIS A182049 mentioned by Jeppe Stig Nielsen implies it's been checked up to 100,000.  I guess I could check somewhat further by letting a computer run through the weekend, but I doubt there's much point.

$41!$ seems to be the last factorial number with no digit 9 in it. It looks like for $n\ge 42$, the decimal representation of $n!$ will contain all ten digits, even after trailing zeros have been removed. Edit: Found OEIS entry.

Ok, it seems you're Hungarian. Well, finding such words should not be too much of an issue :-)

As for the mnemonic, he can try to use 10 letters words, if they're not too uncommon in your native language.

Interesting... I could not find another example for $41 \leq n \leq 16000$. I wrote a small Python program for that purpose, which checked the $n$s in the previous range one by one.

Then I suppose this answers my question as 'known open problem'.

One could call it the Zumkeller conjecture.

@domotorp: oh, that's what I feared... :D

I would learn a cyclic number instead, that he can use for magic tricks en.wikipedia.org/wiki/Cyclic_number

Oh, I understood it the wrong way! My bad.

@JukkaKohonen: I mean that 40 probably is the largest number $n$ for which all the 0's in $n!$ appear at the end.

@Will, did your back-of-envelope calculation provide any estimate on when it would happen?

Using the heuristic that the digits of $n!$ are just random, followed by some zeroes at the end, you can calculate without too much trouble that the answer to your question is: probably yes. My (very crude) estimate puts it at a better than 99% chance. I don't know whether this heuristic is any good, and I doubt this way of thinking will lead to a real answer . . . but there you have it.

What follows doesn't answer your question, and probably won't contribute to an answer either, but a result by John Edward Maxfield -- A note on $N!$, Mathematics Magazine 43 #2 (March 1970), pp. 64-67 -- might be of interest: Given any positive integer $n,$ there exists a positive integer $N$ such that the decimal digits of $N!$ begin with all the decimal digits of $n,$ in their correct order. For more about this, see my answer to Short papers for undergraduate course on reading scholarly math.

Is there an integer n which does not divide 10 such that for any k integer with 10 does not divide a = n * k and a has a zero in its decimal places ?

Interesting question, but very difficult. I am not sure we even know if there are an infinite number of 0s in the decimal expansion of pi. (But it's believed this is the case)


What's the smallest $\lambda$-calculus term not known to have a normal form?

For Turing Machines, the question of halting behavior of small TMs has been well studied in the context of the Busy Beaver function, which maps n to the longest output or running time of any halting n state TM. The smallest TM with unknown halting behavior has 5 states. It takes $5\cdot2\cdot\log_2(5\cdot4 + 4)$ or roughly 46 bits to describe, 50 bits if we assume a straightforward encoding.

In comparison, $\lambda$-calculus terms have a [simple binary encoding](http://tromp.github.io/cl/Binary_lambda_calculus.html): $00$ for lambda, $01$ for application, and $1^n0$ for variable with de Bruijn index $n$. It's natural to define a $\lambda$-calculus analog of the busy beaver function as the maximum normal form size of any size $n$ closed lambda term.

As the smallest closed lambda term is $\lambda\,1$, with encoding $0010$, we determine $$ BB_{\lambda}(4) = 4 $$ The next smallest ones, $\lambda\,\lambda\,1$ and $\lambda\,\lambda\,2$ are similarly already in normal form, and give $$ BB_{\lambda}(6) = 6,\qquad BB_{\lambda}(7) = 7 $$ $BB_{\lambda}(n)$ will have to remain undefined for $n < 4$ or $n = 5$.

The first enlarged normal form shows up at $\lambda\,(\lambda\,1\,1)\,(1\,(\lambda\,2))$ which gives $$ BB_{\lambda}(21) = 22 $$ $BB_{\lambda}$ starts to grow rapidly at $n \geq 30$, since tripling Church numeral two, $(\lambda\,1\,1\,1)\,(\lambda\,\lambda\,2\,(2\,1))$ with normal form Church numeral $2^{2^2}= 16$, gives $$ BB_{\lambda}(30) \geq 5 \cdot 16 + 6 = 86 $$ and quadrupling/quintupling give $$ BB_{\lambda}(34) \geq 5 \cdot 2^{16} + 6 $$ $$ BB_{\lambda}(38) \geq 5 \cdot 2^{2^{16}} + 6 $$ which exceed the TM Busy Beavers for 4 and 5 states.

An Ackermann-like function takes a mere [29 bits](https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/ackermann.lam). A twisted application to Church numeral $2$ yields a $BB_{\lambda}(51)$ exceeding 2↑↑↑↑5.

Graham's number is exceeded in at most [49 bits](https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam), giving $$ BB_{\lambda}(49) \geq 5 \cdot G + 6 $$ (compared with a 16 state TM that needs over 192 bits to describe).

What's the smallest n for which $BB_{\lambda}(n)$ is unknown in ZFC?

One upper bound is [213 bits](https://codegolf.stackexchange.com/questions/79620/laver-table-computations-and-an-algorithm-that-is-not-known-to-terminate-in-zfc).

Let's try to narrow it down some more.

Function $BB_{\lambda}$ has been added to the [Online Encyclopedia for Integer Sequences](https://oeis.org/A333479).

Yes, I meant unknown in ZFC. I'll make that explicit.

Does "unknown" mean "unknown in ZFC"? I infer that this is the case from the upper bound being a reference to a program which is known to halt under the assumption of a rank-into-rank cardinal.

I mean "unknown halting behaviour". We know that (\ 1 1)(\ 1 1) is the smallest nonhalting term, of size 18.

By “not known to halt”, do you mean “unknown halting behavior” or “known to not halt or unknown halting behavior”?

I also have an upper bound of 247 bits for finding an odd perfect number at github.com/tromp/AIT/blob/master/oddperfect.lam

Yes; Goldbach is 267 bits, as can be seen at github.com/tromp/AIT/blob/master/goldbach.gif which is compiled from goldbach.lam; larger than the 213 bit Laver statement above.

For an upper bound, do you know what a lambda calculus formulation of twin primes / Goldbach / similar number-theoretic statements might look like?

Curiously, triple Church_2 is not the best 30 bit beaver. A very minor tweak gives a 160 bit normal form!

It seems that application is also the other way round... quoting yourself :-)

When i changed from 0-based to 1-based variables, I forgot to update the encoding. It's fixed now.

Shouldn't $\lambda 1$ have encoding 00110 (of length five), or am I missing something in the notation here?

Yes, they were. I replaced them by actual lambdas now.

Are all those slashes supposed to represent lambda?


conjectures regarding a new Renyi information quantity

In a recent paper <http://arxiv.org/abs/1403.6102>, we defined a quantity that we called the "Renyi conditional mutual information" and investigated several of its properties. We have some open conjectures, and I would like to pose them as open questions to the mathematics community on this forum. Perhaps there is a simple resolution, but we have considered known approaches and it appears that new tools are necessary for solving these problems (or maybe known tools could work?).

## Background

Let $\mathcal{H}$ denote a Hilbert space, and let $\mathcal{D}(\mathcal{H})$ denote the set of [density operators](https://en.wikipedia.org/wiki/Density_matrix#Formulation) acting on this Hilbert space (positive semi-definite operators with trace one). (It suffices for our purposes to consider finite-dimensional Hilbert spaces, but of course the infinite-dimensional case is interesting as well.) We are interested in "three-party" density operators $\rho_{ABC}$ acting on the tensor-product Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C$. Let $\rho_A$ denote the "marginal" density operator of $\rho_{ABC}$, obtained by taking a partial trace over the spaces $\mathcal{H}_B \otimes \mathcal{H}_C$: $$ \rho_A = \operatorname{Tr}_{BC} \\{ \rho_{ABC}\\} $$ In a similar way, we can define $\rho_{B}$, $\rho_{C}$, $\rho_{AB}$, $\rho_{BC}$, and $\rho_{AC}$.

We define the "Renyi conditional mutual information" of order $\alpha \geq 0 $ as follows: $$ I_{\alpha}(A;B|C)_{\rho} \equiv \frac{1}{\alpha - 1} \log \operatorname{Tr} \\{ \rho_{ABC}^{\alpha} \rho_{AC}^{(1-\alpha) / 2} \rho_C^{(\alpha-1)/2} \rho_{BC}^{1-\alpha} \rho_C^{(\alpha-1)/2} \rho_{AC}^{(1-\alpha) / 2} \\} $$ In the above, identity operators are implicit, so that, e.g., $\rho_C^{(\alpha-1)/2} = I_{AB} \otimes \rho_C^{(\alpha-1)/2}$, where $I_{AB}$ is the identity operator acting on $\mathcal{H}_A \otimes \mathcal{H}_B$. For more motivation for the above quantity, please consult our paper.

We can easily prove that the above quantity is "monotone under a local quantum operation acting on system $B$" for all $\alpha \in [0,2]$, i.e., that the following inequality holds $$ I_{\alpha}(A;B|C)_{\rho} \geq I_{\alpha}(A;B|C)_{\omega}, $$ where $\omega_{ABC} \equiv (\operatorname{id}_A \otimes \mathcal{M}_B \otimes \operatorname{id}_C)(\rho_{ABC})$, $\operatorname{id}$ denotes the identity map, and $\mathcal{M}_B$ is a completely positive trace preserving linear map acting on system $B$. A proof follows by applying a standard approach with the [Lieb concavity theorem](https://en.wikipedia.org/wiki/Trace_inequalities#Lieb.27s_concavity_theorem) and the [Ando convexity theorem](https://en.wikipedia.org/wiki/Trace_inequalities#Ando.27s_convexity_theorem) (proof is detailed in the paper).

## Conjectures/Questions

We have several conjectures which (to fit the MathOverflow format) could be converted into questions in obvious ways (e.g., are the statements true? please provide an argument or counterexample, etc.).

Conjecture 1:

> The quantity $I_\alpha(A;B|C)$ is "monotone under a local quantum operation acting on system $A$" for all $\alpha \in [0,2]$, i.e., the following inequality holds $$ I_{\alpha}(A;B|C)_{\rho} \geq I_{\alpha}(A;B|C)_{\tau}, $$ where $\tau_{ABC} \equiv (\mathcal{N}_A \otimes \operatorname{id}_B \otimes \operatorname{id}_C)(\rho_{ABC})$ and $\mathcal{N}_A$ is a completely positive trace preserving linear map acting on system $A$. (Numerical evidence indicates that this conjecture should be true.)

Conjecture 2:

> The Renyi conditional mutual information of order $\alpha$ should be monotone increasing in the Renyi parameter $\alpha$, i.e., if $0 \leq \alpha \leq \beta$, then $$ I_{\alpha}(A;B|C)_{\rho} \leq I_{\beta}(A;B|C)_{\rho}. $$ (Numerical evidence indicates that this conjecture should be true as well, and we furthermore have proofs that it is true in some special cases (for $\alpha$ in a neighborhood of one and when $\alpha + \beta = 2$).)

These are the basic conjectures, but please consult our paper for more general forms of them. Solving the first conjecture would be very interesting for us, as it would establish the Renyi conditional mutual as a true "Renyi generalization" of the "von Neumann style" quantum conditional mutual information. Solving the second conjecture would have widespread implications throughout quantum information theory and even for condensed matter physics ([topological order](http://dao.mit.edu/~wen/topartS3.pdf) \- pdf).

There has now been even more progress on Conjecture 2 that is useful enough for applications in quantum information theory. The idea was to make use of the Hadamard three-line theorem (in particular, Riesz-Thorin interpolation). This is detailed in the following paper: arxiv.org/abs/1505.04661 . See also arxiv.org/abs/1506.00981 for follow-up work.

This is just to say that there has been what I would consider siginificant progress on Conjecture 2 since it was posted. Fawzi and Renner have proved a variation of the case when $\beta = 1$ and $\alpha = 1/2$, in the paper arxiv.org/abs/1410.0664 . My coauthors and I have also generalized the notion of Renyi conditional mutual information Renyi relative entropy differences in the paper arxiv.org/abs/1410.1443 and we generalized the methods of Fawzi and Renner in the paper arxiv.org/abs/1412.4067 . Of course, the conjectures stated here still remain open.

In response to earlier comments (now deleted), I have reformatted the post in order to better highlight the apparent questions. With that, let's please focus on the mathematics.

I really like this question --- I hope I manage to get some time to think about it!


Is A276175 integer-only?

The terms of the sequence [A276123](https://oeis.org/A276123), defined by $a_0=a_1=a_2=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-3}}\;,$$ are all integers (it's easy to prove that for all $n\geq2$, $a_n=\frac{9-3(-1)^n}{2}a_{n-1}-a_{n-2}-1$).

But is it also true for the sequence [A276175](https://oeis.org/A276175) defined by $a_0=a_1=a_2=a_3=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)(a_{n-3}+1)}{a_{n-4}} \;\;?$$

**Remark** : This question has been [asked previously on math.SE](https://math.stackexchange.com/questions/1905063/is-a276175-integer-only) ; one participant gave an interesting answer, but partial.

@YCor the sequences are examples of Y-frieze patterns, which are naturally associated to Fock and Goncharov's $\mathcal{X}$-cluster varieties.

More precisely, it seems that $a^{(5)}$ is still integral (checked for $n<35$), and $a^{(k)}$ is non-integral for $k>5$.

Initially I thought all of the recurrences $a^{(k)}_n = \prod_{i=1}^{k-1} (a^{(k)}_{n-i}+1)/a^{(k)}_{n-k}$ (with initial values all $1$) might have all terms integral - where $a^{(3)}$ is oeis.org/A276123 and $a^{(4)}$ is oeis.org/A276175.  But it seems that $a^{(6)}_{12}$ is non-integral.

@YCor The first sequence (the one with the explicit quasi-linear recurrence) is one of the period-1 Laurent phenomenon sequences from Alman, Cuenca and Huang's paper, thus presumably related to cluster algebras (not sure about this).

@CarloBeenakker The proof relies on a computer calculation that no one but mercio seems to have seen let alone checked. And from the sound of his comments, mercio doesn't seem 100% sure of it either.

@FedorPetrov what's the connection with cluster algebras?

@darij grinberg : perhaps in view of today's edit it would be helpful to indicate why the MSE proof is not complete (I note that also OEIS lists this as a proven statement).

@FedorPetrov they are actual polynomials in the first eight $a_{n-1}a_{n+1}/(a_n(a_n+1))$ but unfortunately for this particular sequence, those quantities are not all integers.

At least a priori it could be Laurent polynomial not literally in $a_0,\dots,a_3$, but in something different, like, say, also $(a_1a_2+1)/(a_2+a_3)$ or whatever.

When using generic $a_0,a_1,a_2,a_3$, one does not get Laurent polynomials (starting with $a_8$).

I feel that 'cluster algebras' tag may be appropriate.


Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property, without Choice

While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then wondered whether this was unique. When I got back to my computer, I looked on google and came up with [this](http://camoo.freeshell.org/cfield.isor.html), which indicated that is not. However, the proof given there makes heavy use of Choice, and based on similar things, I am guessing that this cannot be avoided.

Define DC($\omega_1$) as:

For all [trees](http://en.wikipedia.org/wiki/Tree_%28set_theory%29) $T$, $\quad$ $T\:$ has a branch $\ $ or $\ $ $T\:$ has a chain of length $\omega_1$ $\quad$.

which, if I haven't messed up the simplification, ZF proves is equivalent to the definition given at shelah.logic.at/files/446.ps .

Does $\ ZF+DC(\omega_1)\ $ prove that there is a unique subfield of $\mathbb{C}$ which is both isomorphic to $\mathbb{R}$ and has Baire property?

Just a comment - The first link should be camoo.freeshell.org/cfield.isor.html instead of camoo.freeshell.org/cfield.isor to make it more readable.

@Ricky: Your translation of DC_omega1 to trees is not correct.  The above statement is obviously false, for instance since clearly the one element tree has no chain of length omega1.

@Harry, What does there seem to be wrong with the formatting?

Ah, thanks for the explanation Tobias. I found the result very counterintuitive until you pointed out that there was no reason to have that intuition to begin with :)

There seems to be something wrong with the formatting?

@Zev: The fact that $\mathbb C$ has uncountably many field automorphisms does not seem so surprising to me. After all the only obstruction that keeps $\mathbb R$ from having many automorphisms is the fact that its order can be described algebraically, hence is preserved. Since there is no such obstruction on $\mathbb C$, there should be plenty of automorphisms, and in fact there are.

+1 just for blowing my mind with the fact that there are subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ other than $\mathbb{R}$!


Example of a quasi-Bernoulli measure which is not Gibbs?

Let $X=\\{0,1\\}^{\mathbb{N}}$. For simplicity I consider measures on $X$ only.

A measure $\mu$ is **quasi-Bernoulli** if there is a constant $C\ge 1$ such that for any finite sequences $i,j$, $$ C^{-1} \mu[ij] \le \mu[i]\mu[j] \le C\mu[ij]. $$

(Here as usual $ij$ is the juxtaposition of $i$ and $j$ and $[k]$ is the cylinder of all infinite sequences starting with $k$.)

Let $f:X\to \mathbb{R}$ be continuous. The measure $\mu$ is a **Gibbs measure** with potential $f$ if there are $C>0$ and $P\in\mathbb{R}$, such that for every infinite sequence $i_1 i_2\ldots$ and all natural $n$, $$ C^{-1} \le \frac{\mu[i_1\ldots i_n]}{\exp(-nP+f(i)+f(\sigma i)+\cdots+f(\sigma^{n-1}i))} \le C, $$ where $\sigma$ is the left shift.

Of course, there are other definitions of Gibbs measure, but they all agree if the potential $f$ is Hölder. In this case, it follows readily that a Gibbs measure is quasi-Bernoulli.

If a measure is quasi-Bernoulli, there is an equivalent measure (mutually absolutely continuous with bounded densities) which is invariant and ergodic under the shift.

* * *

**Question** : Are all quasi-Bernoulli measures Gibbs? (for some continuous potential, not necessarily Hölder). If not, what is a counterexample?

**Motivation** : Gibbs measures (with Hölder potentials) enjoy many nice statistical properties. Sometimes I have a measure that is quasi-Bernoulli or satisfies some similar but weaker property. I would like to understand if and to what extent good statistical properties continue to hold in that setting.

Dear Vaughn, dear John, could you give us references about both of your arguments? That would be very useful!

Let me point out that there is a standard procedure associating to each almost additive sequence of functions a measure on the algebra up to each level (say $n$ if you consider the element $\varphi_n$ of the sequence of functions). Moreover, any weak sublimit of an almost additive sequence with bounded variation is a Gibbs measure. This implies that your question has a positive answer at each level $n$, that is, indeed any quasi-Bernoulli measure (in your sense) is Gibbs at the algebra of level $n$. Moreover, any of its weak sublimits is also Gibbs, now on the whole $\sigma$-algebra.

Just a thought that doesn't necessarily go anywhere.  If $\mu$ is a Gibbs measure, then you can recover the corresponding potential $f$ (up to a constant) by $f = \log(\frac{d\mu}{d(\mu\circ\sigma)})$.  So if we take an arbitrary quasi-Bernoulli measure and take $f$ to be its log-Jacobian, can we deduce anything about the regularity of $f$ and/or a Gibbs relationship based on the quasi-Bernoulli property?


Is the Poset of Graphs Automorphism-free?

For $n\geq 5$, let $\mathcal {P}_n$ be the set of all isomorphism classes of graphs with n vertices. Give this set the poset structure given by $G \le H$ if and only if $G$ is a subgraph of $H$.

> Is it true that $\mathcal {P}_n$ has no nontrivial automorphisms?

**Remarks:**

This follows if one can recognize a graph from the set of isomorphism classes of its edge-deleted subgraphs. However, since recognizing a graph from the set of isomorphism classes of its edge-deleted subgraphs is stronger than edge reconstruction, I'm wondering if there is an alternative way of proving this.

Probably has something to do with: math.stackexchange.com/questions/3052384/…

Of course there is in fact no need to check, as it anyhow follows from the set edge reconstruction conjecture but I could not find anything about for what values that has been checked.

$\mathcal P_5$ can pretty easily be checked by hand. As for anything larger, I'm not sure.

Maybe it is useful to add that $\cal P_4$ has a nontrivial automorphism. Did you ckeck for higher values, e.g. using the tables from maths.uq.edu.au/~pa/research/posets4to8.html?


An $n \times n$ matrix $A$ is similar to its transpose $A^{\top}$: elementary proof?

A famous result in linear algebra is the following.

> An $n \times n$ matrix $A$ over a field $\mathbb{F}$ is similar to its transpose $A^T$.

I know one proof using the Smith Normal Form (SNF). However, I want to find an elementary proof avoiding any concepts related to the SNF. My question is: is there an elementary way to prove this?

The requirements are:

  * Do NOT use the structure theorem over PID.

  * Do NOT use the Smith Normal form (nor Jordan canonical form).

  * Do NOT use the concept of invariant factors.

  * Provide an explicit invertible matrix $P$ such that $A=PA^T P^{-1}$.

After many years this question was first asked, I cannot find anything meeting all of my requirement. Actually, finding Smith Normal Form can be done through elementary row/column operations. So, it is already elementary enough, and I do not have any reason left for avoiding such a simple and powerful tool. Also, SNF provides not only the existence of $P$, but also an explicit way to write down a such $P$.

@user46896 The method is interesting, but not very far from Jordan canonical form (see questions V.B.1. and V.B.2.). One dreams about an algorithm providing $P$ from scratch ...

@i707107 So, I hope you can work with this : sujets-de-concours.net/sujets/centrale/2003/tsi/math2.pdf , look at Partie V, it's an elementary approach. The solution here : pomux.free.fr/corriges-2003/index.html

@Julien I know very little french, but there is google translate.

@i707107 Can you read some french papers ?

See projecteuclid.org/euclid.pjm/1103039127. Maybe math.stackexchange.com/questions/62497/… is related.

@BenjaminDickman The first link shows that the solution space of $XA=A^T X$, $X=X^T$ has at least dimension $n$. However, the paper does not discuss how to obtain a "nonsingular" matrix in the solution space. That was actually a motivation of my question.

Is the proof given here msp.org/pjm/1959/9-3/pjm-v9-n3-p25-p.pdf sufficiently "elementary"? Incidentally, note the remark on p.895 (pdf 4/7) comes from M. Newman; there was an earlier MO question answered using Smith's work (which you wish to avoid here...) that turned out to have a more elementary solution from M. Newman (mathoverflow.net/questions/151166/…). If the linked proof here does not suffice, perhaps Newman is the fellow to look to...

@MarkSapir : Yes, of course the fact that $A$ and $A^{T}$ have the same minimum polynomial is obvious: it was the use of rational canonical form ( which is really more general that Jordan Normal Form) that seemed to be outside the spirit of the question.

This is slightly related to my question <a href="http:mathoverflow.net/questions/42072

@Geoff: the fact that $A$ and $A^T$ have the same minimal polynomial is obvious: if $p(A)=0$, then $p(A^T)=p(A)^T=0$.

In the two by two case $P$ can always be chosen to be symmetric. Is this true in general?

It's easy to prove using normal forms, but you can reframe it as the statement that if $A:V\to V$ is linear then there is a nondegenerate bilinear form on $V$ such that $\langle Av,w\rangle=\langle v, Aw\rangle$ identically. Is there any chance of a proof of this that is quite different from the normal-form proofs?

There is a proof using rational canonical form, which basically just needs the fact that $A$ and $A^{t}$ have the same minimum polynomial, but I don't think that will be considered explicit .enough

@Mariano: Yeah, it might not be possible to avoid those to prove it. The matrix P coming from Jordan canonical form proof is already complicated enough. Maybe requiring to provide explicit P is not possible without those. The generalized eigenvectors are involved and an inversion that transforms Jordan block to its transpose.

This is a lot like looking for a proof which does not need the letter a in order to be written down :-) What exactly do you  mean by elementary? The Jordan canonical form is basic linear algebra, really.


What is the symmetric monoidal functor from Clifford algebras to invertible K-module spectra?

There ought to be a symmetric monoidal functor from the symmetric monoidal $2$-groupoid whose objects are Morita-invertible real superalgebras (precisely the Clifford algebras), morphisms are invertible superbimodules, and $2$-morphisms are invertible superbimodule homomorphisms to the symmetric monoidal ($\infty$-)groupoid whose objects are invertible $KO$-module spectra, morphisms are invertible morphisms of $KO$-module spectra, etc. Everything I'm about to say should generalize to the complex case and $KU$ but let me only address the real case.

On $\pi_0$ this functor ought to induce a map from the graded Brauer group of $\mathbb{R}$ to the Picard group of $KO$, both of which turn out to be $\mathbb{Z}_8$; this directly relates Bott periodicity for Clifford algebras and for K-theory.

Recently I tried to write down this functor "explicitly" (to prepare for a talk and to confirm that I wasn't lying when I wrote [this](https://mathoverflow.net/questions/87345/brauer-groups-and-k-theory/177321#177321)) and found that I couldn't quite get some details to work out the way I wanted them to. The idea should be, starting with a real (resp. complex) superalgebra $A$, to

  1. Construct the topological groupoid of finitely generated projective $A$-supermodules (automorphism groups have their usual topologies as Lie groups),
  2. Pass to the algebraic K-theory spectrum of this groupoid, regarded as a symmetric monoidal groupoid under direct sum,
  3. At some point during this process, kill off "trivial objects."

The correct version of this process should, when given $A = \mathbb{R}$ itself, reproduce $KO$, and more generally, when given the Clifford algebra $\text{Cliff}(p, q)$, reproduce a shift of $KO$ (as a $KO$-module spectrum) by $p - q$ either up or down depending on some [sign conventions](https://mathoverflow.net/questions/185645/what-are-the-correct-conventions-for-defining-clifford-algebras).

I ran into two problems getting this picture to work out:

  1. The construction I know of the algebraic K-theory spectrum of a symmetric monoidal groupoid produces a connective spectrum. For example, applied to $\text{Vect}_{\mathbb{R}}$ it reproduces $ko$ rather than $KO$.

  2. It seems like there are two inequivalent ways to kill off trivial objects, and they don't produce the same answer.

In detail, to fix the sign conventions, let $\text{Cliff}(p, q)$ be the Clifford algebra generated by $p$ anticommuting square roots of $1$ and $q$ anticommuting square roots of $-1$, so that $\text{Cliff}(1, 0) \cong \mathbb{R} \times \mathbb{R}$ and $\text{Cliff}(0, 1) \cong \mathbb{C}$ (as ungraded algebras). Let $K(\text{Cliff}(p, q))$ be the algebraic K-theory of the category of finite-dimensional $\text{Cliff}(p, q)$-supermodules (so some connective spectrum). There are two natural forgetful functors inducing maps

$$K(\text{Cliff}(p + 1, q)) \to K(\text{Cliff}(p, q)), K(\text{Cliff}(p, q + 1)) \to K(\text{Cliff}(p, q))$$

of connective spectra; as maps on categories, they pick out the supermodules admitting an additional odd automorphism squaring to $1$ or an odd automorphism squaring to $-1$ respectively. For example, when $p = q = 0$, these are two ways of identifying the supervector spaces you want to regard as trivial for the purposes of defining $KO$ in terms of supervector spaces.

I computed that the homotopy cofibers of these maps (in connective spectra) are $\Omega^{p-q} ko$ and $B^{p-q} ko$ respectively; that is, we get the infinite loop spaces making up $ko$, but in two different orders.

> How can this construction be fixed so that it outputs $KO$-module spectra rather than $ko$-module spectra, and what's the "correct" way to kill trivial objects?

One desideratum for "correct" is that it should generalize cleanly to the case where we replace $\mathbb{R}$ with a discrete field $k$ and replace $KO$ with the algebraic K-theory spectrum $K(k)$ of $k$. The end result of the "correct" process is that it should naturally give a symmetric monoidal functor inducing a map from the graded Brauer group of $k$ to the Picard group of $K(k)$. But here there are not only two choices for what an odd automorphism could square to: instead the set of choices is parameterized by

$$k^{\times}/(k^{\times})^2 \cong H^1(\text{Spec } k, \mu_2).$$

So I don't know what to do. Maybe always pick squaring to $1$?

One thing that made me uncomfortable about the above is that it's conceptually important that the supermodule categories should be regarded as enriched and tensored over supervector spaces, but I ignored their enrichment over supervector spaces when I passed to algebraic K-theory spectra. The enrichment seems like it ought to be important for defining the "correct" way to kill of trivial objects.

Update: awhile back I asked Lurie this question and his sense is that the functor I want isn't symmetric monoidal but at best lax / oplax.

Have you worked out the complex version (graded complex central simple algebras and invertible $KU$-module spectra) of this story? If so, perhaps you can go down to $KO$-module spectra by taking $\mathbb{Z}/2$-fixed points.


Base change for $\sqrt{2}.$

This is a direct follow-up to [Conjecture on irrational algebraic numbers](https://mathoverflow.net/questions/173414/conjecture-on-irrational-algebraic-numbers).

Take the decimal expansion for $\sqrt{2},$ but now think of it as the base $11$ expansion of some number $\theta_{11}.$ Is there an easy (or, failing that, hard) proof that $\theta_{11}$ is transcendental? Of course, same question stands for $\theta_k,$ for your favorite $k>10.$

@NikitaSidorov I, for one, am completely agnostic on the subject, and have absolutely no clue why anyone would believe what you say everyone believes. I am, of course, just a caveman.

@NikitaSidorov And of course, we all speak for ourselves.

Of course, we all believe it is transcendental. And of course, we all know that this is way beyond our reach at present. Why ask, then?

@AnthonyQuas I cannot actually understand your comment...

Since the digits are the same but the base is larger, in a certain sense, $\theta_{11}$ has a quicker rational approximations than  $\sqrt 2$. Can this fact affect the irrationality measure of $\theta_{11}$?

@AnthonyQuas: Are there reasons to believe in this independency, besides mere numerical observations?

Algebraic irrationality "is independent of" base expansions; there are countably many algebraic irrationals; Therefore "with probability 1", all algebraic irrationals are normal to all bases.

@GeraldEdgar What is the justification for the well-known conjecture?

$\theta_{11}$ is certainly irrational.  There is a well-known (but still far from proved) conjecture that all algebraic irrationals are normal in all bases.  So (of course) your number is transcendental.  But your question (whether there is an easy proof of it) is not answered by this.


Do all possible trees arise as orbit trees of some permutation groups?

## I.Motivation from descriptive set theory

(Contains some quotes from Maciej Malicki's paper.)

The classical theorem of Birkhoff-Kakutani implies that every metrizable topological group G admits a compatible left-invariant metric, that is, a metric d inducing the group's topology, and such that d(zx,zy)=d(x,y) for all x,y,z in G. However, even if G is completely metrizable, it is not always true that G admits a metric that is left-invariant and complete at the same time.

Polish groups admitting a complete left-invariant metric are called CLI groups. We only focus on CLI subgroups of $S_{\infty}$, the symmetry group of natural numbers, for now. Malicki proved that a permutation group G is CLI iff the _orbit tree_ of G is well-founded, i.e. every branch of the tree terminates at some finite level, where the concept _orbit tree_ is defined as following.

* * *

## II.The problem

Let $G\leq S_{\infty}$ be a permutation group acting on $\mathbb{N}$ (starts from 1 for convenience). Define $G_n$ to be the n-point stabilizer, $G_n=\\{g\in G|g(i)=i,1\leq i\leq n\\}$, and $G_0=G$. We represent an infinite orbit $O$ of $G_n$ by a node $N$ on level $n$ (finite orbits do not count).

If $O_1$ is an infinite orbit of $G_k$, $O_2$ an infinite orbit of $G_{k+1}$ and $O_2\subset O_1$ (possibly the same), we correspondingly draw $N_2$ as a one-step extension of $N_1$.

Assuming $G$ is transitive, this process defines a map from $G$ to a rooted tree $T_G \subset\omega^{<\omega}$. We say it is the _orbit tree_ of $G$ and it roughly describes how the orbits of $G$ split when we keep fixing more and more points. The natural inverse question is

> For every tree $T\subset\omega^{<\omega}$, is there a group $G$ such that the orbit tree $T_G$ is isomorphic to $T$? Can we construct it?

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## III.Some attempts

Since I do not know too much about group theory, especially not about infinite permutation groups, I started with some concrete examples. The construction was hard even for very simple finite trees, though. For instance, $\mathbb{Z}$ acting on $\mathbb{Z}$ realizes the tree of a single node. $\mathbb{Z}_{wr}\mathbb{Z}_2$(wr for wreath product) on $\mathbb{Z}\times 2$ realizes a "chain" of any finite depth, etc. But these seem to be too _ad hoc_.

I also received a generous hint from an expert, saying that

> 1\. If $G$ is a permutation group on $\Omega$ and $\alpha\in\Omega$ and the orbits of $G_{\alpha}$ are $\Omega_{\lambda}(\lambda\in\Lambda)$ then $G_{\alpha}$ is a subdirect product of the restrictions $H_{\lambda}:=G_{\alpha}^{\Omega_{\lambda}}$. Specifically $G_{\alpha}$ can be embedded in the Cartesian product $\prod_{\lambda}H_{\lambda}$ such that the projections onto each factor are surjective.
>
> 2\. Now let $G$ be a free group of countably infinite rank. We want to find a corefree subgroup $H$ of $G$ of countable index such that $H$ is free of infinite rank and has a specified number of double cosets $HxH$ for which $|H:x^{-1}Hx|$ is infinite ($H$ is going to play the role of $G_{\alpha}$). We know that every subgroup of a free group is free and many (most of them?) have infinite rank when $G$ has infinite rank. I don't think it is that hard to satisfy the condition about the number of cosets. If this is true then we can go from level 0 to level 1.
>
> 3\. To go to level 2 we know that $H$ is a subdirect product of the $H_{\lambda}$. Suppose that all the infinite $H_{\lambda}$ are free of infinite rank and that the embedding of $H$ induces an isomorphism onto each of these. Now for each infinite $H_{\lambda}$ choose (corefree, infinite rank) $K_{\lambda}$ so that $H_{\lambda}$ and $K_{\lambda}$ play the roles of $G$ and $H$ in 2 to provide the correct number of infinite orbits at the next level. Can we choose the isomorphisms $H\rightarrow H_{\lambda}$ and a subgroup $K$ of $H$ such that $K$ maps onto $K_{\lambda}$ for each $\lambda$?
>
> 4.If the answers to these questions are yes, I think that you can then prove what you want.

It is not easy for me to comprehend his thought since I have no training in free groups. I have some vague doubts that

  1. What is the action of $G$ on countably many things? $G$ naturally acts on itself by concatenation but this seems not correct.

  2. Why must such an $H$ be a one-point stabilizer?

  3. Moreover, the free group of countably infinite rank is countable itself. It is known that every countable group is CLI, hence its orbit tree must be well-founded. That is to say, $G$ may never realize ill-founded trees. I suspect we need to use other methods for non-CLI groups, taking inverse limit for example.

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## IV.Possible modifications of the problem (not strict)

From the viewpoint of descriptive set theory, it is not necessary to realize every tree, "a subcollection of trees" which has the same Borel complexity as "all trees" will suffice. For example, we may properly "stretch" the tree so that there is at most one "event" (a node terminates or a node splits into more than one nodes) happens on each level. we may also require every splitting node to split infinitely, etc. As long as the transformation is Borel, we may choose the easiest class of trees to realize.

To realize the tree, we may also choose to fix finitely many points at one time if necessary.

If we only look at the well-founded trees, we may construct a big tree recursively. We know that the one-point stabilizer is isomorphic to a subdirect product of the restrictions on each orbit. But

> Given transitive permutation groups $H_1$, $H_2$... (finitely or countably many) acting on $\mathbb{N}$, is there a transitive group $G$ such that $G_{\alpha}$ is isomorphic to the direct product of these $H_{\lambda}$?

My greedy hope (possibly incorrect!) is, there is such a group. If this is true, noticing that factors of direct product are "independent" of each other, we may fix points on orbits "one by one" and let other orbits "wait". Say the orbit tree of $H_{\lambda}$ is $T_{\lambda}$, then the orbit tree of $G$ will be of the shape that each subtree is a "stretched" $T_{\lambda}$. It will work since we only need to realize all "stretched" trees.

Please do not laugh at my silliness if I am miserably wrong :-P Thank you so much for your time and effort.

a rooted tree is a subset(well, my fault) in \omega^{<\omega} that is closed under taking initial segment. For example, the set {(0),(0,0),(0,1),(0,2)} is the tree that has one root and three "branches", each terminates at level 1. But we usually think of it intuitively(as drawn on paper).   As we fix more and more points in N, the orbits become smaller and smaller as a set. We intuitively draw a node for each orbit with infinite size. If O2 is a subset of O1, then N2(on the level below) is linked to N1. Thus we may draw the orbit tree.

Thank you for your comment. G_1 could have infinitely many orbits. But your construction will not obtain the wanted direct product. This is because g(a1,a2...)g^{-1}=(g(a1),g(a2)...) and it will introduce a lot more permutations.

How are you relating a tree to an element in \omega^{<\omega}? Does this mean all finite sequences in natural numbers? If so why does the tree need to be finite? Probably I am misunderstanding the notation. Can you clarify?

Can't G1 have infinitely many orbits? Construct your group in the following fashion. Divide N into {1} and countably more disjoint infinite sets Sis. Let G be generated by transpositions between the 1 and the first element of Sis and the full permutation group on Si. I was not sure how G corresponded to the rooted tree