Can a number be palindromic in more than 3 consecutive number bases?

_$2017:$ Was initially asked on [MSE](https://math.stackexchange.com/questions/2234587/can-a-number-be-palindrome-in-4-consecutive-number-bases)_ \- but wasn't solved or updated there since.

**Update $2019$:** I've returned to this problem, made some progress and updated the post here.
(I've basically rewritten this entire post here)

* * *

## **Introduction and problem**

An $k$-palindrome is a number palindromic in $k$ consecutive number bases. Here, I'm wondering about the existence of $k=4$. Note that if it does not exits, $k\gt 4$ can't exist by definition.

Also note that if a number is even length (number of digits is even) palindrome in base $b$, then it is divisible by $b+1$ and thus can't be palindromic in that number base. So we can look for odd length cases of digits in the first (last, depending on how you define it) base only.

> Can a natural number be _nontrivially_ [palindromic](https://en.wikipedia.org/wiki/Palindromic_number) in **more than** $3$ consecutive number bases?
>
> _Nontrivially_ means that I'm not counting one-digit palindromes.

Smallest number $N$ which is nontrivially palindromic in $k$ consecutive number bases:

$$ \begin{array}{|c|} \hline k& N & \text{Palindrome} \\\ \hline 1& 3 & 11_2 \\\ 2& 10 & 101_3=22_4\\\ 3& 178 & 454_6 =343_7 = 262_8\\\ 4& ?& \\\ \hline \end{array} $$

Where the index denotes the number base representation. For example $11_2$ is three in binary.

Note that $k=1$ is not special as those are just palindromes and easily can be constructed.

It is not hard to see $k=2,3$ have infinitely many examples. But finding all of them is hard.

> I conjecture solution for $k=4$ does not exist. That is, there are no $4$-palindromes.

My **question** here is:

> Can you provide arguments (help) or direction (on proving) why (why not) would this be true?

My ideas are presented below:

**First approach to proving this**

One idea to prove this is proving $(1)$ and $(2)$ which would imply this. The $(1)$ is now proven. The $(2)$ will be much harder to prove. The details are included below. Here are the claims:

  * $(1)$ A $3$ digit (when written in palindromic bases) number can't be a $4$-palindrome.

  * $(2)$ To be a $4$-palindrome, a number must have $3$ digits (when written in palindromic bases).

The second claim is based on the fact that _almost-all_ "almost $4$-palindromes" have $3$ digits. That is, numbers palindromic in three out of four consecutive bases (that is, for example in bases $b,b+3$ and either $b+1$ or $b+2$).

_Almost-all_ since the only two known "almost $4$-palindromes" that do not have $3$ digits are $71240, 1241507$ which have $5$ digits (when talking about digits, they are counted in the "pivot" number base among the consecutive bases in which the number is palindromic - "pivot" being either the smallest or largest base).

> Proving $(2)$ is equally hard as proving $(3)$ from alternate approach below, since all cases of $d$ must be resolved. The $(1)$ specifically asked to resolve $d=3$ which was accomplished.

**Second approach to proving this**

> $(3)$ I actually conjecture I have found all $3$-palindromes. If this is true, then a $4$-palindrome does not exist since neither of my $k=3$ solutions can be extended to a fourth consecuitve number base.

More details are linked at the end of the next section:

## **" $k$-problem system" results and conjectures**

What I was able to prove both computationally and by hand:

> Expanding $(1):$ In short, I have proven I have all $3$ digit $3$-palindromes, and nontrivial $d\le 3$ solutions don't exist. Now it is easy to see now that a $3$ digit $3$-palindrome can't be palindromic in fourth consecutive base, which means $4$-palindromes with $d\le 3$ digits do not exist. (Digits counted when written in palindromic number bases).
>
> More details about this are linked at the end of this section, as well.

Expanding $(3):$ What I was able to conjecture is presented below:

In short, we can write the problem of finding $k$-palindromes in form of solving linear diophantine equations whose order grows when solving for longer palindromes of length $d$ (number of digits). Lets call that system needed to be solved "$k$-problem system".

There is the "general", and there is the "non-general" case of this problem system.

The non-general $k$-problem system is related to finding $d=2l+1$ (WLOG $d$ is odd) digit solutions, where all representations in consecutive bases have exactly $d$ digits.

The general $k$-problem system allows different lengths of digits in those $k$ consecutive number bases, where "pivot" number base among these $k$ consecutive number bases, usually smallest or largest, is taken to have $d$ digits.

The $k=2$ case has infinitely many solutions for every case of $d$. It is hard to find them all.

The $k=3$ is what will be the case of the "problem-system" from now on:

> Expanded on $(3)$ in this context:
>
> For the non-general case of the "problem-system", I have computationally solved (proved) I have all solutions for two smallest cases, $d=3,5$. I also strongly conjecture same is true for $d=7$. I also conjecture based only on computation, that a $3$-palindrome solution for $d\ge 9$ does not exist. That is, I conjecture I have all solutions.
>
> For the general case of the "problem-system", I have computationally solved (proved) I have all solutions for smallest case, $d=3$. I also conjecture I have all solutions for $d\ge 5$ since another conjecture I strongly believe is that the general case is equivalent to the non-general case, except for having an extra finite set of solutions for sufficiently small bases $b$ which I have collected and computed all (conjectured).

* * *

_Summing this all up_ : the important thing is I have all $3$ digit $3$-palindromes and that they can't be $4$-palindromes. I have all conjectured solutions for $3$-palindromes, but unable to prove the fact that large $d$ can't have solutions for $k=3$.

* * *

Note that I was trying to find counterexamples to these conjectures for a long time now but couldn't.

The more details as promised: A post mainly focusing on non-general "problem-system" [can be read on MSE](https://math.stackexchange.com/questions/3305696/hard-system-in-integers-related-to-natural-number-representations) if you are interested in more details, or if you want to see the "all $3$-palindromes" solutions for the non-general case (the extra solutions from general case are easily re-computable if my conjectures are true).

I'll also include all _infinite families_ (the non-sporadic solutions) for $k=3$ given there, here:

Given $k\in\mathbb N_0 = \mathbb N \cup \\{0\\}$, "infinite families" giving $3$-palindromes:

$$ \begin{array}{l,l,l,l} d & (a_i)&(c_i)k & b\\\ d=3 & (2,6)&(1,1)k & 2k+8\\\ d=5 & (31,32,0)&(3,2,1)k & 4k+47 \\\ d=7 & (34,50,10,74)&(1,1,1,1)k & 2k+76 \\\ d=7 & (8,33,0,41)&(1,3,1,3)k & 6k+58 \\\ d=7 & (112,15,0,36)&(4,0,1,0)k & 6k+175 \\\ d=7 & (227,160,187,200)&(5,3,5,3)k & 6k+280 \\\ d=7 & (5,23,6,14)&(2,6,5,0)k & 12k+39 \\\ d=7 & (93,78,30,50)&(10,6,7,0)k & 12k+119 \\\ d=7 & (47,150,249,26)&(2,6,11,0)k & 12k+291 \\\ \end{array} $$

Where these give digits of $3$-palindromes in base $b$, which are also palindromic in $b-1,b-2$.

That is, for example, observing the second row (family), the only $d=5$ case infinite family, means $(31+3k,32+2k,0+1k,32+2k,31+3k)$ are digits of a $3$-palindrome in base $4k+47$, for every $k$. We can convert this to a decimal value easily.

> $(3):$ If we can prove these are all of the solutions for $k=3$ (and that sporadic solutions exist only for sufficiently small bases as I conjectured), we have solved the $k=4$ problem, the $4$-palindrome problem: There does not exist a number palindromic in four consecutive number bases.

The main thing needing proving is that $d\ge 9$ can't have solutions. Or can you find a counterexample, and actual solution? That would make this much more interesting.

The $d=9$ specifically for example will be very surprising if it had a solution (I found the period of solutions if they exist, must be larger than $500$ in this case, which is unlikely given $d=3,5$ have periods $2,4$ and $d=7$ has periods $2,6,12$). Period being families producing a solution every period amount of bases.

@GerryMyerson Regarding a relevant sequence, I think A279093 is updated with all known 3-palindromic solutions.

@joro Update: Now it is known that the $3$ digit solution does not exist. I will have to update this post.

@JoseBrox If we have $n$ and $n+3$, and we also include either $n+1$ or $n+2$ and "observe almost palindromic in four consecutive", then surprisingly, only 3 digit palindromes seem to have a chance to be palindromic in four consecutive bases ~ Updated the question

@Vepir Yes, sorry, I meant $n$ and $n+3$. I was just trying to understand if the obstruction for the existence of consecutive palindromes for $k=4$ (in case it exists) is of an algebraic/number-theoretic nature or rather of an analytic/combinatoric nature (which is what the existence of examples for $n$ and $n+3$ suggests).

@JoseBrox Why $n, n+4$? There are a lot of examples, smallest one being $7=111_2=11_6$. You can easily modify the linked python code to generate them. If you meant $n, n+3$ there are again a lot of examples, smallest being $31=11111_2=111_5$. I only found interesting the palindromes which share consecutive bases. Smallest example for $k=1,2,3,4,\dots$ consecutive bases: $3(=11_2), 10(=11_3=101_4),178(= 454_6 =343_7 = 262_8), ? (k=4)$... The solution for four is either very large or does not exist.

@Vepir Do you know of any example of palindrome simultaneously for bases $n$ and $n+4$?

@joro: That doesn't seem to be proven. At MSE, Vepir searched up to $10^7$ and found  a one parameter family which does not extend to $4$ bases, and a single sporadic example, but there is no proof these are the only $3$ digit solutions.

Is it known that three digit solution doesn't exist?

This is perfectly tweetable mathematics and thus certainly not offtopic on mathovertweet. I vote Reopen

@Andy, the question was asked at m.se two weeks ago. If it hasn't been answered to Vepir's satisfaction, no reason why it shouldn't be posted here. And no reason why anyone here has to accept Vepir's suggestion to answer there. I'm sure that if someone posts an answer here, Vepir will survive the trauma.

I voted to close.  MO is not intended to advertise questions on math.se, so if you want answers there, you should just ask there.


A Linear Order from AP Calculus

In teaching my calculus students about limits and function domination, we ran into the class of functions

$$\Theta=\\{x^\alpha (\ln{x})^\beta\\}_{(\alpha,\beta)\in\mathbb{R}^2}$$

Suppose we say that $g$ weakly dominates $f$, and write $f\preceq g$, if

$$\lim_{x\to\infty}\frac{f(x)}{g(x)} \hspace{3 mm} \text{is finite}$$

We can then readily see that $(\Theta,\preceq)$ is a total order isomorphic to the lexicographic order on $\mathbb{R}^2$.

But we can get more complicated total orders with, say

$$\Theta_n=(x^{\alpha_0}(\ln{x})^{\alpha_1}(\ln\ln{x})^{\alpha_2}\cdots(\ln^{n-1} x)^{\alpha_{n-1}})_{\vec{\alpha}\in\mathbb{R}^{n}}$$

$$\Phi=\\{e^{p(x)}\\}_{p(x)\in\mathbb{R}[x]}$$

which are isomorphic as total orders to the lexicographic orders on $\mathbb{R}^n$ and $\operatorname{List}\mathbb{R}$

All of these complicated orders live inside what I'd call "the AP Calc linear order" $(\Omega,\preceq)$ defined as:

$$\Omega_0=\\{f\in\mathscr{C}^0((\lambda,\infty))\\}_{\lambda\in\mathbb{R}}$$

$$f\preceq g\Longleftrightarrow \max\left\\{\left|\liminf_{x\to\infty} \frac{f(x)}{g(x)}\right|,\left|\limsup_{x\to\infty} \frac{f(x)}{g(x)}\right|\right\\}<\infty$$

$$\Omega=\Omega_0/\simeq \hspace{5 mm} \text{where} \hspace{5 mm} f\simeq g \Leftrightarrow \left[f\preceq g \text{ and } g\preceq f\right]$$

where the refinement on $\preceq$ is made so as to avoid problems with things like $\sin{x}$.

This seems to be a very complicated linear order, as it includes as a suborder things like

$$\Psi=\\{p_0(x)e^{p_1(x)}e^{e^{p_2(x)}}\cdots\exp^{n-1}(p_{n-1}(x))\\}_{p_i(x)\in\mathbb{R}[x]\forall i}$$

My question is the following: is there any combinatorial description or universal construction, i.e. as a colimit, of the isomorphism type of $(\Omega,\preceq)$?

This is essentially just a question involving Landau notation, no?

With Anthony Quas, I suggest you look up "Hardy field", and also "o-minimal structure" whose germs at infinity produce Hardy fields. The question as it stands suffers from problems noted by Eric Wofsey.

@DmitryV: That's not what I was talking about (I was also thinking like a category theorist and not caring about that).  If $f/g$ oscillates between approaching $0$ and approaching $\infty$, then $f\not\leq g$ and $g\not\leq f$.  Actually, if $f$ and $g$ don't have to be positive, you could just have an unbounded set where $f$ is zero and $g$ is not and an unbounded set where $g$ is zero and $f$ is not.

@EricWofsey you're right, I was being too much a category theorist and forgetting to mod out by isomorphisms, turning this pre-order into a total order. I'll edit it now.

A side comment: The class $\Theta$ is a very small sub-class of the logarithmico-exponential functions studied by Hardy (see also Boshernitzan's work). These have very nice properties including no oscillation: for any two distinct functions in a Hardy Field, eventually $f$ is bigger than $g$ or vice versa.

This is not a total order; $f$ and $g$ can oscillate between $f\gg g$ and $g\gg f$.


A question in Fontaine--Laffaille theory

Let $K$ be finite unramified extension of $\mathbf{Q}_p$ with ring of integers $W$. Let ${\rm MF}$ be the category of strongly divisible $W$-modules $M$ with ${\rm Fil}^0M=M$ and ${\rm Fil}^{p-1}M=0$. Let ${\rm Crys}(G_K)$ be the category of finite free $\mathbf{Z_p}$-representations of $G_K$ which become crystalline over $\mathbf{Q}_p$. There is a functor $$A:{\rm MF} \to {\rm Crys}(G_K), \qquad M \mapsto {\rm Hom}(M, A_{\rm crys}),$$ where the ${\rm Hom}'s$ are required to preserve the filtration and the Frobenius action, which is fully faithful by Fontaine-Laffaille theory.

Given $T$ in the image of the above functor, one can construct a finite free $W$-module with filtration and Frobenius, namely $B(T)={\rm Hom}_{G_K}(T, A_{\rm crys})$. Note that $B(A(M))[1/p]$ is isomorphic to $M[1/p]$ by the theory of crystalline representations. I believe that $B$ is not the inverse to $A$ in general, however.

Questions: 1) What is an example of a $T$ for which $B(T)$ is not the strongly divisible module associated to $T$? 2) For which $T$ is $B(T)$ the strongly divisible lattice associated to $T$?


The oriented homeomorphism problem for Haken 3-manifolds

Haken famously described an algorithm to solve the homeomorphism problem for [the 3-manifolds that bear his name](http://en.wikipedia.org/wiki/Haken_manifold) (fleshed out by many others, including Hemion and Matveev who fixed some gaps). But it's often natural to consider 3-manifolds _equipped with an orientation_ , and so my question is:

> **Question 1:** Is there an algorithm _in the literature_ that determines whether or not a pair of oriented Haken 3-manifolds M and N are _orientedly_ homeomorphic?

By Haken's Algorithm, we can reduce to the case in which M and N are (unorientedly) homeomorphic, and so the question can be reduced to:

> **Question 2:** Is there an algorithm _in the literature_ that determines whether or not an orientable Haken 3-manifold M admits an orientation-reversing self-homeomorphism?

**Notes:**

  1. Note the words 'in the literature'. I don't really doubt that this algorithm exists, but the details seem quite complicated, particularly in the non-geometric case, so I hope that someone will be able to supply a reference. A very short argument would also be nice. Answers of the form 'I don't know, but I'm sure it can be done if you work out each geometric case individually and then think hard enough about the JSJ decomposition', while appreciated, would be less useful.

  2. The hyperbolic case can be deduced from Sela's solution to the homeomorphism problem, which also computes the automorphism group. I've no doubt the other non-Haken cases can be handled similarly. Hence my interest in the Haken case.

  3. It may be that a simple modification of Haken's Algorithm handles the oriented case. This isn't at all clear to me, but I would be very interested to hear if it's true.

  4. **(added later)** I would also be happy to confirm that this question is `open' (in the sense that an answer is not in the literature, rather than that the answer is in doubt)---I very much suspect that this is the case. An authoritative pronouncement by someone who knows the literature very well would therefore be a valid answer.

Certain special cases can be deduced from the literature. If you look at Theorem 6.1.6 in Matveev's book, it says that there is an algorithm to tell if two Haken manifolds with boundary pattern are homeomorphic taking boundary pattern to boundary pattern. If you have two oriented knot complements, then you can tell if they are orientation preserving homeomorphic by choosing a boundary pattern which is not preserved under mirror image (such as a single closed (1,1) curve). However, this is also appealing to the knot complement problem. springerlink.com/content/978-3-540-45898-2

Actually, Ian, this is precisely my motivation for asking the question.  As far as I can tell, the treatments of the homeomorphism problem for reducible manifolds in the literature miss this point.

If you connect sum with a manifold that doesn't admit an orientation reversing isometry, then check homeomorphism, then this checks if the manifolds are orientation preserving homeomorphic. However, I'm not sure if the homeo. problem for reducible (or even connect sums of Haken) manifolds is in the literature explicitly.


Reference request: Parallel processor theorem of William Thurston

Sometime in the 1980's or 1990's, Bill Thurston proved a theorem regarding the existence of a universal parallel processing machine, using a certain class for such machines having finite deterministic automata at each node. There is no such published paper on MathSciNet, and I have no record of this theorem other than my own faulty memory.

I'd like to ask whether there exists a preprint somewhere, or some other reference.

Although I cannot quite reproduce the theorem's exact statement, here's what I roughly remember. Caveat: please take this all with a grain or three of salt, because I am sure my memory has gaps and/or errors.

The parallel processors of this class are built on an underlying framework consisting of a connected graph with finite valence at each vertex. At each node of this graph one places a finite deterministic automaton. This automaton has some special capabilities. After each computation, it can send one of finitely many outputs along each adjacent edge. Also, before each computation it can receive inputs along each adjacent edge (which have previously been sent by adjacent automata, or have been set up as part of an initialization) and it uses those inputs together with the current state to compute its next state and its next outputs.

I believe there are strong regularity constraints. The graph should have an automorphism group which is transitive on vertices. In fact the entire setup, including the automata themselves and how their inputs and outputs are associated to the adjacent edges, should have an automorphism group that is transitive on vertices. The reason for these constraints is so that the entire network can be "finitely described" or "finitely programmed", and perhaps that is the true constraint, however it may be formalized.

One programs or initializes the processor by choosing initial states and initial inputs. I believe there must be some finiteness condition, for example outside of some bounded region the initial states and initial inputs are perhaps in some default position. Then one lets the processor run until some point (I don't remember the halting conditions), at which time one reads off the terminal states and outputs.

I remember Thurston proved two theorems about universal machines in this class (subject to some restrictions about how to encode or program one machine within another, which I totally do not remember; but this argues for some strong regularity constraints in the class of machines).

One theorem is that a processor of this type whose underlying graph is the Cayley graph of a free group cannot be universal. Roughly speaking the reason one might expect this to be a universal processor is because of exponential growth of nodes, but the trouble is that there is too much bottlenecking at each node.

The main theorem is that there exists a processor of this type whose underlying graph is any Cayley graph of a lattice in the 3-dimensional solvable Lie group. For example, one such lattice is the Cayley graph of the semidirect product $$\mathbb{Z}^2 \rtimes_M \mathbb{Z} $$ using the action of $\mathbb{Z}$ on $\mathbb{Z}^2$ generated by the matrix $M = \pmatrix{2 & 1 \\\ 1 & 1}$. Here the rough idea is that one still has exponential growth of nodes, but the bottlenecking phemonenon has been avoided.

**Added:** I'm convinced now that so far this is just a description of computable functions in the ordinary sense. But this jogs a really dim memory that what's missing is a way to constrain or measure how the programming process is allowed to compress information. I'm still hoping that someone will have a more accurate record of this theorem.

Maybe you could try theoretical computer science,  cstheory.stackexchange.com

@GerryMyerson: No, it's not in that paper, although the issues of mesh partitions discussed in that paper remind me of the "bottlenecking" issues that are at the heart of this parallel processing model I'm asking about, and this may have been around the same time as well.

@IgorRivin: I did ask Matt, and although he remembers working on somewhat similar problems around the same (vague) time, this particular theorem is not something he remembers.

@GerryMyerson: I wondered the same, but it seems far from Cayley graphs and solvable Lie groups...

Is it possible we are talking about the paper, Miller, Teng, Thurston, and Vavasis, Automatic mesh partitioning, available at cs.cmu.edu/~glmiller/Publications/Papers/MiTeThVa93.pdf ?

@Algernon: You might be right, I could be overlooking some kind of space-or-time compression restrictions. But I cannot remember.

My recollection that this is actually due to Matt Grayson, so you might want to ask him...

Your model is what people call a cellular automaton.  It is easy to simulate a Turing machine with a "one-dimensional" cellular automaton (i.e., the underlying graph is a bi-infinite path, a Cayley graph of the one-generator free group!), with reasonable finiteness constraint.  On a free group with a larger generator set, you can simply ignore all but one of the generators and simulate the Turing machine along the remaining generator.  I suspect your memory is putting the emphasis of whatever Thurston's result was on the wrong place.


Coarse moduli spaces of stacks for which every atlas is a scheme

Let $X = [P/G]$ be a smooth finite type separated DM-stack over $\mathbb C$ given as the quotient of a smooth quasi-projective scheme $P$ by the action of a smooth (finite type separated) reductive group scheme $G$. Since $X$ has finite inertia, the coarse space $X^c$ of $X$ exists as an algebraic space.

At _atlas_ of $X$ is an étale morphism $U\to X$ with $U$ an algebraic space.

I sm interested in the converse of the following statement:

**Thm.** If the coarse moduli space of $X$ is a scheme,then every atlas $U$ of $X$ is a scheme.

_Proof._ The map from an atlas $U$ to the coarse moduli space $X^c$ of $X$ is quasi-finite separated. Therefore $U$ is a scheme by Knutson Cor. II.6.16., p. 138 QED

So the converse would read as follows:

**Q.** Suppose that every atlas $U$ of $X$ is a scheme. Is the coarse moduli space of $X$ a scheme?

I expect the answer to be negative, but can't find a good example. Note that a counterexample can not be an algebraic space, as an algebraic space with the property that every atlas is a scheme is itself a scheme (the identity morphism being an atlas).

Edit: I have an application in mind of the above question, and in the context of my application the stack $X$ is even generically a scheme. Not sure if that helps...

Yes it looks like the example from Corollary 6 does the job! After taking a quick look at the proof, it seems like the condition that the coarse space is a scheme should be equivalent to something like the following: for each point $x \in P$, there exists a linearized ample line bundle $L$ such that $x$ is $L$-semistable. I'm not sure how this interacts with your atlas condition though.

@DoriBejleri   I think (but I'd have to double-check) that Kollar's paper arxiv.org/pdf/math/0501294.pdf contains such examples. Do you agree? (BTW, this question is seven years old, and I somehow decided to revive it now, but I don't even remember the "application in mind" at the moment.)

Do you have an example, without assuming your condition on the atlases, where $P$ is quasi-projective and $G$ is reductive, $[P/G]$ is separated DM, but the coarse space of $[P/G]$ is not  a scheme?


The cofinality of $(\mathbb{N}^\kappa,\le)$ for uncountable $\kappa$?

For a partially ordered set $P$, a set $A\subseteq P$ is _cofinal_ if for each element of $P$ there is a larger element in $A$. The _cofinality_ of $P$, ${\rm cof}(P)$, is the minimal cardinality of a cofinal family in $P$.

Let $\kappa$ be an infinite cardinal number. Consider the set $\mathbb{N}^\kappa$ of all functions from $\kappa$ to $\mathbb{N}$ (equivalently, $\kappa$-sequences of natural numbers), with the pointwise partial order: $f\le g$ if $f(\alpha)\le g(\alpha)$ for all $\alpha<\kappa$.

Let $$f(\kappa):={\rm cof}(\mathbb{N}^\kappa,\le).$$ The cardinal $f(\aleph_0)$ is the well-understood dominating number $\mathfrak{d}$, which is consistently any cardinal of uncountable cofinality that is not larger than the continuum (see [Blass](http://www.math.lsa.umich.edu/~ablass/hbk.pdf)).

Basic facts include:

  1. For cardinal numbers $\kappa\le\mu$, we have that $f(\kappa)\le f(\mu)$: A projection of a cofinal family in $\mathbb{N}^\mu$ on the first $\kappa$ corrdinates is cofinal in $\mathbb{N}^\kappa$.
  2. $\kappa<f(\kappa)$: Given $\\{h_\alpha:\alpha<\kappa\\}\subseteq\mathbb{N}^\kappa$, define $h(\alpha):=h_\alpha(\alpha)+1$ for all $\alpha<\kappa$.
  3. The value of $f(\kappa)$ with "$\le$" replaced by eventual dominance remains the same (see [Comfort](http://dml.cz/bitstream/handle/10338.dmlcz/106682/CommentatMathUnivCarol_029-1988-4_5.pdf)).

Some deeper results:

  1. [Jech-Prikry](http://journals.cambridge.org/download.php?file=%2F4622_1176E3359F6DC1B50E531EBE349A0A5F_journals__PSP_PSP95_01_S0305004100061272a.pdf&cover=Y&code=f33a33db8489693def0a6e06d5ccf9fa): Assume that $2^{\aleph_0}$ is regular and smaller than $2^{\aleph_1}$. If $f(\aleph_1)=2^{\aleph_0}$, then there is an inner model with a measurable cardinal.
  2. If $\kappa^{\aleph_0}=\kappa$, then $f(\kappa)=2^\kappa$ (see [Hathaway](http://arxiv.org/abs/1401.7948)).

I did not conduct an extensive literature search. I would appreciate your pointing out relevant results that are not mentioned here.

**Question:** Are there additional cardinals $\kappa$ for which $f(\kappa)$ can be evaluated in ZFC?

(The question also applies to large cadrinals. I do not request that the cardinals provably exist.)

**Update.** Intially, I also asked whether $\kappa^{\aleph_0}=\kappa$ could be weakened to ${\rm cof}([\kappa]^{\aleph_0},\subseteq)=\kappa$. Todd Eisworth reminds in the comments below that the latter hypothesis is true in ZFC for $\aleph_1$.

Strangely, I couldn't find an explicit reference for the following.

**Question.** Is it consistent (modulo suitable large cardinals) that $f(\aleph_1)<2^{\aleph_1}$? Assuming that (See Assaf Rinot's comment below), is there an explicit reference for that?

**Motivation:** The values $f(\kappa)$ pop up whenever I study (usually, jointly with colleagues) the character and other cardinal invariants of topological groups. They are unavoidable.

(Comment: By no means do I mean that the symbol $f(\kappa)$ should be reserved for the function thus defined. This is just a short, temporary name for brevity.)

The Komjath-Shelah model does not involve large cardinals. Just force to add $\aleph_{\omega_1}$ many Cohen reals over a model of $2^{\aleph_0}=\aleph_1, 2^{\aleph_1}=\aleph_2, 2^{\aleph_2}=\aleph_{\omega_1+1}$ and $(\aleph_{\omega_1})^{\aleph_0}=\aleph_{\omega_1}$. E.g., force with $Add(\aleph_2,\aleph_{\omega_1+1})*Add(\omega,\aleph_{\omega_1})$ over a model of GCH.

@ToddEisworth I had such a feeling from what I did see in the literature. This is absurd that such a "simple" question is still open. I wonder if Shelah knows about it (probably not, for otherwise it should have been solved :) ). I'll have a chance to mention/remind/motivate it to him and others next week, in my lecture at the YST. Would appreciate any futher info about this problem that I can mention then.

@BoazTsaban:  Jech's Chapter 29 mentions that the question of whether $f(\aleph_1)<2^{\aleph_1}$ is consistent is open, and I seem to recall that the question was also mentioned in the original "green version" of his book from 1978.

@Todd: Oops. I should have considered these immediate cases. Thanks for pointing this out. Will update the question accordingly.

Should have said "stronger implication" as the assumption is weaker...

Just observe that the cofinality of $([\aleph_1]^{\aleph_0},\subseteq)$ is $\aleph_1$, as witnessed by initial segments.  Same holds for all the $\aleph_n$ by induction.

In the particular case where $\kappa=\aleph_1$, one may show that not only CH, but also $2^{\aleph_0}=\aleph_2$, is enough to imply that $f(\aleph_1)=2^{\aleph_1}$. See Jech's book for a reference (Chapter 29).

A wild guess concerning the consistency of $f(\aleph_1)<2^{\aleph_1}$: have a careful look at Komjath-Shelah model of [F1445].

@ToddEisworth could you provide some details, or a reference?

If the weakened implication in (2) holds then we would have $f(\aleph_1)=2^{\aleph_1}$ in ZFC.


Homotopy type of the affine Grassmannian and of the Beilinson-Drinfeld Grassmannian

The affine Grassmannian of a complex reductive group $G$ (for simplicity one can assume $G=GL_n$) admits the structure of a complex topological space. More precisely, the functor $$X\mapsto |X^{an}|$$ that associates to a scheme locally of finite type the underlying topological space of its analytification ([SGA1, XII, 1]) can be left-Kan-extended to a functor $PSh(Sch)\to Top$, and therefore applied to the ind-scheme $Gr_G$.

Some topological properties of this space are known: its connected components are in bijection with the topological fundamental group $\pi_1(G)$ ([Beilinson-Drinfel'd, _Quantization of Hitchin's integrable system and Hecke eigensheaves_ , Proposition 4.5.4]). The étale fundamental group (which is the profinite completion of the topological fundamental group) of each orbit in $Gr_G$ under the left action of $L^+G$ is trivial (Richarz, _Affine Grassmannians and the Geometric Satake Equivalences_ , proof of Proposition 3.1).

Also, $|Gr_G^{an}|$ is known to be homotopy equivalent to a polynomial loop space ([Zhu, An introduction to the affine Grassmannian and the Geometric Satake Equivalence] and [Pressley/Segal, Loop Groups, Sec. 8.3]).

However, my knowledge essentially stops here. For example, the homotopy groups of this polynomial loop space are unknown to me.

Has anyone ever written or thought about the topology of the affine Grassmannian in further detail? Relevant questions to me are the following:

  1. What are the fundamental groups of the connected components of $|Gr_G^{an}|$?
  2. The cohomology of the affine Grassmannian is much more studied and used than its homotopic properties. Is there any chance that the connected components of $|Gr_G^{an}|$ satisfy conditions like "being a simple space", which often allows homotopical issues to be translated into cohomological issues, by means of the homological Whitehead theorem?

Needless to say, the answer to such questions for the Beilinson-Drinfel'd Grassmannian will probably be more difficult. However, any ideas or references in this more sophisticated context would be highly appreciated.

Thank you in advance.


support of the coupling between two probability measures

Given two Borel probability measures $\mu$ and $\nu$ on $\mathbb{R}$, let $\Pi(\mu, \nu)$ denote all couplings between them, i.e., all Borel probability measures on $\mathbb{R}^2$ such that the marginal distribution of the first and second coordinate are $\mu$ and $\nu$ respectively. Can we describe the set of all possible support of the joint law:

$S(\mu, \nu) = \\{E \subset \mathbb{R}^2: \exists \lambda \in \Pi(\mu, \nu), s.t. \lambda(E) = 1\\}$.

[Strassen](http://projecteuclid.org/euclid.aoms/1177700153) (Theorem 11) characterized this collection: $C \in S(\mu, \nu)$ if and only if for all $U$ open in $\mathbb{R}$, $\nu(U) \leq \mu(U^C)$, where $U^C = \\{x: \exists y \in U, s.t. (x,y) \in C\\}$. But this condition seems not easy to verify! For instance, it is probably not true to only consider open interval $U$.

Let us focus on the normal distribution $\mu=\nu=N(0,1)$. Can we characterize $S' = S(N(0,1), N(0,1))$ **more explicitly** in this case, i.e., what kind of subset in the plane admits a probability measure whose marginals are standard normal? For instance we can ask the following question: for what pair of $a < b$ do we have $\\{(x,y): a \leq |x-y| \leq b\\} \in S'$, that is, can we construct a joint distribution on this strip that has standard normal marginals. It seems a non-trivial question. For $b = \infty$, one can show that the largest possible $a$ is between $\sqrt{\pi/2}$ and $3/2$. I am thinking is there a systematic way to do it in the simple case of real line. Note that Strassen's result work for any Polish space.


Does this variant of a theorem of Hasse (really due to Gauss) have an "elementary" proof?

BACKGROUND

Here are 3 theorems of varying difficulty. Let $M$ be the $Z/2$ subspace of $Z/2[[x]]$ spanned by $f^k$, with the $k>0$ and odd, and $f=x+x^9+x^{25}+x^{49}+\cdots$. For $g$ in $M$, let $S(g)$ consist of the primes, $p$, for which the coefficient of $x^p$ in $g$ is 1. Note that each $p$ in $S(f^k)$ is congruent to $k$ mod 8.

T1.----- If $k=3 {\rm\ or\ } 5$, $S(f^k)$ consists of the $p$ that are $k$ mod 8

T2.----- $S(f^7)$ consists of the $p$ that are 7 mod 16

T3.----- If $k=19 {\rm\ or\ } 21$, then $S(f^k)$ consists of the $p$ that are $k$ or $k+8$ mod 32.

To prove T1 when $k=3$, we write $f^k$ as $f*f^2$ and use the fact that if $p$ is 3 mod 8, then $p$ is uniquely the sum of a square and twice a square. When $k=5$ we argue similarly using Fermat's two square theorem.

As I indicated in a comment on a recent MO question of Joel Bellaiche, "[Primes and x^2+2y^2+4z^2](https://mathoverflow.net/questions/100701/primes-and-x22y24z2)" ,T2 follows from a result of Hasse on the class number of $Q(\sqrt{-2p})$, using Gauss' theorem that the number of representations of $2p$ as a sum of 3 squares is 12*(this class number). Hasse's proof is an application of the Gauss theory of genera and ambiguous forms.

T3 is thornier. Because $f$ is the mod 2 reduction of (the Fourier expansion of) the normalized weight 12 cusp form for the full modular group, each $g$ is the mod 2 reduction of a modular form of integral weight. A profound result of Deligne, relating Hecke eigenforms to Galois representations, then shows that $S(g)$ is a "Frobenian set". Nicolas, Serre and Bellaiche, continuing in this vein, developed a theory of level 1 modular forms in characteristic 2 that led to more precise results. Their investigations motivated me to try to determine $S(f^k)$ empirically for small $k$, and I was led to conjecture T3. Joel then applied his methods to give a proof. But this is very hard, and so I ask:

QUESTION

Does there exist an "elementary proof" of T3, using the theory of binary quadratic forms, along the lines of the Hasse-Gauss argument?

EDIT: Motivated by my recent simple proof of T2 (see my answer to the question of Joel cited above), I've found arguments that ought to reduce the proof of T3 to Sage calculations. The point is that forms of weight 2 are easier to deal with than forms of weight 3/2, so one should work with quadratic forms in 4 variables rather than in 3, even when the genera that arise have more than 1 class in them. Here's the idea of my argument for f21.

Let p be a prime that is 5 mod 8. Writing $f^{21}$ as $(f)(f^2)(f^2)(f^{16})$ we find that if $R$ is (1/16)*(the number of representations of $p$ by $G_1=x^2+2y^2+2z^2+16t^2$ with $x,y,z$ and $t$ all odd), then $p$ is in $S(f^21)$ if and only if $R$ is odd. Now since $p$ is 5 mod 8, in any representation of $p$ by $G_1$, $x,y$ and $z$ must be odd. So if we set $G_2=x^2+2y^2+2z^2+64t^2$ then $R=(N1-N2)/16$, where $N_1$ and $N_2$ are the numbers of representations of $p$ by $G_1$ and $G_2$ respectively. Now write $p$ as $a^2+4b^2$ with $a$ and $b$ congruent to 1 mod 4. Computer calculations indicate:

Conjecture 1. $N_1=p+1+2a$

Conjecture 2. $N_2=((p+1)/2)+a+4b$

If these conjectures hold then $R=(p+1+2a-8b)/32$. The numerator here is $4(b-1)^2 +(a+3)(a-1)$, which mod 64 is $(a+3)(a-1)$. So $R$ has the same parity as $(a+3)(a-1)/32$ and is odd just when $a$ is $5$ or $9$ mod $16$. Now mod $32$, $p=a^2+4$. So $R$ is odd just when $p$ is $29$ or $85$ mod $32$, and so the conjectures imply Joel's result for $S(f^21)$.

How does one attack the conjectures? The theta series attached to $G_1$ and $G_2$ are modular forms for $\Gamma_0 (64)$ and $\Gamma_0 (256)$ respectively. If the conjectures are to hold it seems that each of these theta series should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\mathbb Q(i)$. It should be possible, using Sage, to get an explicit formulation of this, and prove the conjectures.

My proposed treatment of $S(f^{19})$ is entirely similar. Suppose $p$ is $3$ mod $8$. Writing $f^{19}$ as $(f)(f)(f)(f^{16})$ and arguing as above we find that if we take $H_1$ and $H_2$ to be $x^2+y^2+z^2+16t^2$ and $x^2+y^2+z^2+64t^2$ respectively, and let $N_1$ and $N_2$ be the number of representations of $p$ by $H_1$ and $H_2$, then $p$ is in $S(f^{19})$ just when $R=(N_1-N_2)/16$ is odd. Now Jacobi's 4 square theorem, (see the argument in my answer to Joel's question), shows that $N_1$ is $2(p+1)$. Write $p$ as $a^2+2b^2$ with $a =1$ or 3 mod 8. The computer suggests:

Conjecture 3. $N_2=p+1+4a$

So if the conjecture holds, $R=(p+1-4a)/16$, and one sees easily that this is odd just when $p$ is 19 or 27 mod 32. Once again the theta series attached to H2 is a modular form for $\Gamma_0 (256)$. The conjecture indicates that it should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\mathbb Q(\sqrt{-2})$; all this should admit a proof using Sage.


Is there some way to see a Hilbert space as a C-enriched category?

The inner product of vectors in a Hilbert space has many properties in common with a hom functor. I know that one can make a projectivized Hilbert space into a metric space with the Fubini-Study metric and then follow Lawvere in considering it a $[0,\infty)$-enriched category; it's particularly nice because the distance between two vectors is a function of the inner product.

Is there a way of defining a monoidal structure on $\mathbb{C}$ such that the inner product is the hom so that we can think of a (perhaps projectivized) Hilbert space as $\mathbb{C}$-enriched?

You might also be interested in John Baez's paper on the subject: arxiv.org/abs/q-alg/9609018


are there high-dimensional knots with non-trivial normal bundle?

Does there exist a smooth embedding $\varphi\colon S^k\to S^n$ such that $\varphi(S^k)$ has non-trivial normal bundle? I looked at some of the old papers by Kervaire, Haefliger, Massey, Levine but I couldn't find an answer. There are various related results, for example it is shown by Kervaire et al that in many dimensions the normal bundle is trivial. Furthermore Kervaire showed that there exists an immersion, such that the pullback of the normal bundle is non-trivial.

@StefanFriedl Thanks for the corrections.

@archipelago   thanks, you put me on the right track!

Namely Danny Ruberman writes: "Jerry Levine's paper "A classification of differentiable knots" (Annals 82 (1965) 15-50) determines (see proposition 6.2) the possible normal bundles in some range of codimensions in terms of maps in various exact sequences. It's hard to summarize the results, but he gives a table at the end dealing with relatively low dimensions, where calculations can be made explicit. The case n=11 and k=6 contains Haefliger's example; in fact there are exactly 5 possible normal bundles among all embeddings."

more googling gives an answer in mathoverflow.net/questions/237384/….

@archipelago but for this conversation it's more important that by accident you misquoted their comment, they write "since Haefliger (unpublished) has shown that S^11 embeds in R^17 with a +++++non-trivial++++ normal bundle." That's a very good find and answers my question with a shaky reference. It would be great if there was a reference with details

@archipelago   Since Levine was my advisor I should point out that it's a paper by Hsiang, Levine and Szcarba

@archipelago: Of course, thanks for pointing that out.

@archipelago  thanks for your comments! But from you and I unearthed so far in the literature it could in principle be that the normal bundle is always trivial, isn't it? You'd think that this question was answered in the 60's.

According to Hsiang, Lee, and Szczarba, it is an unpublished result of Haefliger that $S^{11}$ embedds into $S^{17}$ with trivial normal bundle (see 'On the normal bundle of a homotopy sphere embedded in Euclidean space'). This would show that Kervaire's bound is sharp.

@MichaelAlbanese Any normal bundle arising from such an embedding is stably trivial, since the tangent bundle of $S^k$ is.

Do you know if Kervaire's example $\nu$ is stably trivial? Maybe I'm oversimplifying, but if you have an immersion $f : S^k \to S^n$ with $k < n$, then it factorises as $i_n\circ f_0$ where $i_n : \mathbb{R}^n \to S^n$ is the standard inclusion and $f_0 : S^k \to \mathbb{R}^n$. The map $h : S^k \to S^{n+k+1}$ given by $i_{n+k+1}\circ(f_0, \iota)$ is an embedding where $\iota: S^k \to \mathbb{R}^{k+1}$ is the standard embedding. If I'm not mistaken, the normal bundle of $h$ is $\nu\oplus\varepsilon^{k+1}$ which is trivial if and only if $\nu$ is stably trivial.

Using this, one sees that all embeddings with $k\le 6$ have trivial normal bundle. I believe it is an open question whether there is an embedding with nontrivial normal bundle for k=7 and n=11.

If k<2(n-k)-1 then the normal bundle is trivial by a result of Kervaire (Theorem 8.2 in his "An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant"). If n=k+1,k+2,k+3 it is trivial as well.


About the equivariant analogue of $G_n/O_n$

Let $BO_n$ and $BG_n$ be the classifying spaces for rank $n$ vector bundles and for spherical fibrations with fiber $S^{n-1}$, respectively, and let $G_n/O_n$ be the homotopy fiber of $BO_n\to BG_n$.

There is an interesting old fact, that the stabilization map $G_n/O_n\to G_{n+1}/O_{n+1}$ is $(2n-4)$-connected, about twice as good as the corresponding maps $BO_n\to BO_{n+1}$ and $BG_n\to BG_{n+1}$. Equivalently, the map $G_n/O_n\to G/O$ is $(2n-4)$-connected. I was wondering about the history of this result.

But what I was really wondering about was the following equivariant analogue. Let $\Gamma$ be a finite group. For a real representation $V$ of $\Gamma$, let $O(V)^\Gamma$ be the group of linear (isometric) automorphisms respecting the $\Gamma$-action and let $G(V)^\Gamma$ be the topological monoid of equivariant homotopy equivalences from the unit sphere $S(V)$ to itself. Write $G(V)^\Gamma/O(V)^\Gamma$ for the homotopy fiber of the map of classifying spaces, and write $G^\Gamma/O^\Gamma$ for the colimit of this over all $V$ in a complete universe (or the sequential colimit over direct sums of copies of the regular representation).

I think maybe with some work I could get a good connectivity statement for the map $G(V)^\Gamma/O(V)^\Gamma\to G^\Gamma/O^\Gamma$, but I wonder if someone has already done so, and more generally if anyone has any insights to offer about any of this.

Note that $BO^\Gamma$ is the product, over irreducible real representations $W$, of either $BO$, $BU$, or $BSp$ depending on the type of $W$, while $G^\Gamma$ is the union of the invertible components of $(\Omega^\infty S^\infty)^\Gamma$. In particular, $\pi_1(BG^\Gamma)$ is the group of units of the Burnside ring of $\Gamma$.


The free complete lattice on three generators, beyond ZF

_This was originally[asked at MSE](https://math.stackexchange.com/q/4261226/28111); although it is still under bounty it seems unlikely to be answered there._

[$\mathsf{ZF}$ proves that there is no free complete lattice on three generators](https://www.jstor.org/stable/1993166?seq=1#metadata_info_tab_contents) since any such lattice would have to surject onto the ordinals. This isn't a problem in the context of set theories which admit a universal set, such as [$\mathsf{NFU}$](https://plato.stanford.edu/entries/settheory-alternative/#NewFounRelaSyst) or [$\mathsf{GPK}_\infty^+$](https://plato.stanford.edu/entries/settheory-alternative/#SystGPKiOlivEsse). However, in such theories recursive constructions (which are utterly unproblematic in $\mathsf{ZF}$) become more complicated. So the status of the free complete lattice on three generators in such a theory is unclear to me:

> Is the existence of a free complete lattice on three generators consistent with either of $\mathsf{NFU}$ or $\mathsf{GPK_\infty^+}$?

Of course $\mathsf{NFU}$ and $\mathsf{GPK_\infty^+}$ are not the only two set theories with a universal set, and I'd be interested in an answer for any such theory. My focus on these two is due to $(i)$ my impression that $\mathsf{NFU}$ is the best-understood such set theory by far at the moment and $(ii)$ my personal liking of $\mathsf{GPK_\infty^+}$.

**EDIT** : I forgot to include exactly what "free complete lattice on three generators" means for me in this context (given that such notions are often more fragile in non-$\mathsf{ZF}$ contexts). While to a certain extent I'm interested in any reasonable interpretation of the phrase, I'm specifically interested in the following precisiation: a complete lattice $L$ with distinguished elements $a,b,c$ such that for every complete lattice $M$ and every $x,y,z\in M$ there is exactly one complete lattice homomorphism $L\rightarrow M$ sending $a$ to $x$, $b$ to $y$, and $c$ to $z$.

Given the results in the Crawley-Dean paper, I suspect that if the free complete lattice with $3$ generators exists in a model of $\mathsf{NFU}$ with choice, then it would actually have the same with any number of generators, including $|V|$.

@JamesHanson In fact, I can't even show that "the free complete lattice on $V$ generators" doesn't exist! (That is, a pair $(X,f)$ where $X$ is a complete lattice and $f:\{x:x= x\}\rightarrow X$ is such that for every other complete lattice $A$ and every map $g:\{x:x =x\}\rightarrow A$ there is exactly one complete lattice homomorphism $h: X\rightarrow A$ such that $h\circ f=g$.)

@JamesHanson Yes, I don't see how to get Dedekind-MacNeille completions to work in such settings either. (On that note I forgot to include my definition of freeness - fixed!)

I think you can show that in $\mathsf{NFU}$ there is no complete lattice generated by three elements that satisfies the conditions in Crawley and Dean's definition 2. The issue I'm having with applying this to the definition of freeness in your MSE question is actually that it seems likely that not all lattices have completions in $\mathsf{NFU}$. The Dedekind-MacNeille completion of a lattice naturally lives one type higher than the lattice itself, which makes it seem like in general you only know that completions exist for lattices on Cantorian sets.


Automorphic forms and coherent cohomology

Why is it (and what does it mean) that automorphic forms do not contribute in the coherent cohomology of Siegel modular varieties parametrizing abelian varieties of dimension $d>2$ (see section 7 of the [thesis](https://tel.archives-ouvertes.fr/tel-02940906/document) of Nguyen or section 1.4.1 of this [paper](https://arxiv.org/pdf/1812.09269.pdf) of Boxer, Calegari, Gee, and Pilloni)?

Apparently this comes from the fact that the Hodge-Tate weight of the associated motive is neither regular nor weakly regular, i.e. the same number appears more than twice in the Hodge-Tate weights, but why does this mean that we cannot find them in the coherent cohomology? Apparently this is also true for the Betti cohomology of the associated locally symmetric space. Does the theorem of Franke (generalization of Eichler-Shimura) have anything to say about this?

A good reference for coherent cohomology of Shimura varieties is the paper of Michael Harris, cited as [Har90b] in the BCGP paper. In particular, Proposition 4.3.2 there gives a formula for the infinitesimal character associated to the coherent cohomology of an automorphic vector bundle. You might also find the paper of Gross cited as [Gro16] in BCGM useful: it explains the role played by the endomorphism ring of the abelian variety.

I also agree that a short answer with some reference could be useful, not only for the OP but for other people interested in this question as well. Even if it is not perfect, it would be better than nothing.

@naf: I think I understand now why the motives of abelian varieties of dimension d>2 cannot come from the coherent cohomology of a Shimura variety: The former is never regular, since HT weights are d 1's and d 0's. The latter however are required to be regular. Could you give me a reference for coherent cohomology that discusses this?

@YemonChoi: Coherent cohomology actually corresponds to non-degenerate limits of discrete series. Without knowing anything about what the OP is willing to assume, it is not possible (for me) to write a short answer which would also be useful.

(Alternatively, is there a representation-theoretic explanation of why certain representations are not contained in the discrete series for the relevant group, without going through Hodge-Tate weights?)

@naf I am neither a number theorist nor an algebraic geometer, but perhaps the OP would find it useful to see an expanded version of the statement "Another way to see this..." that you refer to. I don't know if "computing the possible infinitesimal characters" is supposed to be a standard procedure in this setting, but perhaps an outline of how one does this for a toy example could be helpful for the OP?

@Tim: The explanation in the paper of BGGP is fairly detailed and seems quite clear to me. What I said corresponds to the statement in brackets beginning with "Another..." in the second paragraph there. If there is something specific you do not understand you should ask a more focused question.

@naf If you know the correct statement then do you also know an answer? It might be a bit more constructive to maybe expand on this a bit more :-)

Thank you, and sorry for getting it wrong. I would be happy to edit the question, but could you elaborate more or give me some references to learn more about this?

The question as written doesn't make sense (or is wrong...): the correct statement is that the motive of an abelian variety of dimension $>2$ with endomorphism ring the integers does not occur in the (motive associated to) the coherent cohomology of (automorphic vector bundles on) a Shimura variety.


Mysterious sum equal to $\frac{7(p^2-1)}{24}$ where $p \equiv 1 \pmod{4}$

Consider a prime number $p \equiv 1 \pmod{4}$ and $n_p$ denotes the remainder of $n$ upon division by $p$. Let $A_p=\\{ a \in [[0,p]] \mid {(a+1)^2}_p<{a^2}_p\\}$.

I Conjecture

$$\sum_{n \in A_p } n=\dfrac{7(p^2-1)}{24}$$

The question has already been asked in this [thread](https://math.stackexchange.com/questions/4639883/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4?noredirect=1#comment9793873_4639883). If it is a repetition, please vote to close this thread.

**Addition** After have [answer](https://math.stackexchange.com/questions/4639883/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4/4640608#4640608) Numerically, I find this equality : for every odd integer p ≥ 3 $$\sum_{n \in A_p} n=\dfrac{7p^2 -12p+5}{24}+\dfrac{1}{p}\sum_{n=1}^{p-1}{n^2}_p$$ Perhaps, it is useful to demonstrate in a simple way the fact that [the sum differs from 7(p^2-1)/24 by the class number of Q(sqrt(-p))](https://math.stackexchange.com/questions/4639930/even-more-mysterious-sum-equaling-to-frac7p2-124-h-p-where-p-equ).

In the same vein, see also in Franz Lemmermeyer's book "Reciprocity Laws", especially chapter 1 and exercise 1.32 (Gauss).

@Pascal I did some simple computations, this problem is so interesting feels like doing more, so I think we have $\sum_{a\in A_p}a^2=\frac{(p-1)(5p^2+7p-10)}{24}$. In general I think it is interesting to know if there is a degree $d$ polynomial $P_d$ such that $\frac{24}{p-1}\sum_{a\in A_p}a^d=P_d(p)$ for any prime $1$ modulo $4$ and if there are some patterns for the polynomials $P_d$.

I have added a numerical result that I think is interesting

Thank you very much , I received a proof in the other thread that I had reported here

This has been answered in your MSE post. Generally it's best practice to leave a reasonable amount of time (at least a week) between crossposts to avoid situations like this.

I believe this has nothing to do with primes. Try for instance powers of primes 1 mod 4 (the pattern is immediate for powers 2,3,4 for instance), or products of primes 1 mod 4, where the same formula holds (and probably also with class numbers).

Let $(a+1)^2=a^2+2a+1\equiv\alpha\pmod{p}$ and $a^2\equiv \beta\pmod{p}$. Then $\beta-\alpha\equiv-2a-1\pmod{p}$. Also for $a\in A_p$ we have $\beta-\alpha\geq1$. But according to the size of $a$, I think we should get $\beta-\alpha=2p-2a-1$. Not sure though.

To test that I understand correctly, for $p=5$ we are talking about $3+4$, because $0^2_5 < 1^2_5 < 2^2_5 = 3^2_5 > 4^2_5 > 5^2_5$, right?

. . . and as already noted at math.stackexchange.com/questions/4639930 when p is -1 mod 4 the sum differs from 7(p^2-1)/24 by the class number of Q(sqrt(-p)) !  This too is yet unproved.


Two curious series for $1/\pi$

On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have $$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$ where $(\frac{\cdot}p)$ denotes the Legendre symbol and $$R_p:=\frac1p\sum_{n=0}^{p-1}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}.$$ Motivated by this I believed in 2011 that $$\sum_{n=0}^{\infty}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}=\frac{c}{\pi}$$ for some algebraic number $c$, but I could not figure out the exact value of $c$ at that time.

On August 19, 2020 I learned from Dr. Deyi Chen that Maple has a function to justify the form of an algebraic number if we know enough digits of this number. Inspired by this, here I pose the following conjecture.

**Conjecture 1.** We have $$\begin{aligned}&\sum_{n=0}^{\infty}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}\\\&\qquad=\frac{24}{25\pi}\sqrt{375+120\sqrt{10}}. \end{aligned}\tag{1}$$

On June 16, 2011 I conjectured that for any odd prime $p\not=5$ we have $$\sum_{n=0}^{p-1}\frac{4n+1}{(-160)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-20)^{n-k}\equiv 0\pmod p.$$ Motivated by this I formulated the following conjecture on August 20, 2020.

**Conjecture 2.** We have $$\begin{aligned}&\sum_{n=0}^{\infty}\frac{4n+1}{(-160)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-20)^{n-k}\\\&\qquad=\frac{\sqrt{30}}{5\pi}\cdot\frac{5+\root 3\of{145+30\sqrt6}}{\root 6\of{145+30\sqrt6}}. \end{aligned}\tag{2}$$

I have checked $(1)$ and $(2)$ for the first $100$ decimal digits.

**Question**. How to prove the identities $(1)$ and $(2)$?

Note that my two other similar conjectural identities $$\sum_{n=0}^\infty\frac{357n+103}{2160^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}=\frac{90}{\pi}\tag{3}$$ and $$\sum_{n=0}^\infty\frac{n}{3645^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k486^{n-k}=\frac{10}{3\pi}\tag{4}$$ found in 2011-2012 and published in [this paper](http://maths.nju.edu.cn/%7Ezwsun/157b.pdf) also remain open.

You may look at my talk available from maths.nju.edu.cn/~zwsun/CNT-talk3.pdf and my recent paper available from dx.doi.org/doi:10.3934/era.2020070.

What is the relation between the congruence and the identity for pi? Is there an explanation for why we would expect such things?


Cycles in algebraic de Rham cohomology

Let $F$ be a number field, $S$ a finite set of places, and $X$ a smooth projective $\mathscr{O}_{F,S}$-scheme with geometrically connected fibers. For each point $t\in \text{Spec}(\mathscr{O}_{F,S})$, there is a Chern class map $$c_{1,t}:\text{Pic}(X_t)\to H^2_{dR}(X_t),$$ where $H^2_{dR}$ denotes algebraic de Rham cohomology. (See below for one construction of $c_1$.)

> Let $c\in H^2_{dR}(X_F)$ be an element so that there exists a non-zero integer $N$ such that $Nc$ specializes to an element of the image $c_{1,t}\otimes k(t)$, for almost all closed points $t\in \text{Spec}(\mathscr{O}_{F,S})$. Is $c$ in the image of $$c_{1, F}: \text{Pic}(X_F)\otimes F\to H^2_{dR}(X_F)?$$ Here $c_{1,F}$ means $c_{1, \eta}$, for $\eta$ the generic point of $\text{Spec}(\mathscr{O}_{F,S})$, and $k(t)$ is the residue field of $t$.

**Is the answer to the above question (and its evident analogue for higher degree Chern classes/cycle class maps) known in general? Does the question have a name, and is it discussed somewhere?**

In general, given a cohomology theory, there's some conjectural description of the image of the cycle class map (the Hodge conjecture for Betti/de Rham cohomology, the Tate conjecture for $\ell$-adic cohomology) -- my motivation for asking the question is that I realized I hadn't seen such a conjecture for algebraic de Rham cohomology.

* * *

I'll now give a brief description of $c_1$, for completeness. Let $\Omega^\bullet_{dR}$ be the algebraic de Rham complex of some variety $Y$. There's a natural map $$d\log: \mathscr{O}_Y^*[-1]\to \Omega^{\bullet}_{dR}$$ sending a nowhere-vanishing function $f$ to $df/f$. Applying $H^2$ gives the desired map $$c_1:\text{Pic}(Y)=H^1(Y, \mathscr{O}_Y^*)=H^2(Y, \mathscr{O}_Y^*[-1])\overset{d\log}{\longrightarrow} H^2_{dR}(Y).$$

By construction, it factors through $F^1H_{dR}^2(Y)$, the first piece of the Hodge filtration.

* * *

Some remarks, added September 6, 2022:

  * François Charles has explained to me how to give a positive answer to the above question for divisors on Abelian varieties, using Bost's work on foliations. I think a similar argument handles products of curves.
  * One natural strategy is to try to find a prime where the Picard rank does not jump. This will not work, unfortunately; there are varieties where the Picard rank jumps at _every_ prime, e.g. the square of a non-CM elliptic curve.
  * The question can be reduced to the case of surfaces, if I'm not mistaken.

@PeterScholze: Thanks! This does indeed seem quite related, and indeed the discussion on page 81 of "Une Introduction aux Motifs" is closely related to the argument François Charles sketched for the case of Abelian varieties.

Related in spirit, but I think a bit different, is the Ogus conjecture, discussed at mathoverflow.net/questions/389391/… (see in particular the reference to Yves Andre's "Une Introduction aux Motifs"). Ogus' conjecture is the only one I'm aware of in the neighborhood of what you are proposing.

@R.vanDobbendeBruyn: yes, I was hedging because otherwise specialization isn’t defined everywhere. But since it’s an almost-everywhere condition it’s fine to take N=1 if you don’t mind specialization not being defined somewhere.

You don't lose generality by taking $N = 1$, right? Because $N$ is invertible in $F$ and in $\kappa(t)$ for almost all $t$, and all maps are maps of vector spaces.

@DanPetersen: Good point. To be honest my actual motivation was: Simpson's conjecture says that a local system underlies an integral VHS iff it arises from geometry; this is the non-abelian analogue of the Hodge conjecture. The Fontaine-Mazur conjecture says a local system is arithmetic iff it arises from geometry; this is the non-abelian analogue of the Tate conjecture. Now there is a conjecture that a flat vector bundle has almost all p-curvatures nilpotent iff it comes from geometry (the p-curvature conjecture is closely related) -- I was wondering what the "abelian" analogue of this is.

An analogue of Hodge and Tate conjecture with respect to algebraic de Rham cohomology is the Grothendieck period conjecture: if $X$ is a smooth projective variety over $\overline{\mathbf{Q}}$, then conjecturally the image of the cycle class map is given by the intersection $ H^{2k}_{\mathrm{sing}}(X^{\mathrm{an}},\mathbf Q(k)) \cap H^{2k}_{\mathrm{dR}}(X/{\overline{\mathbf{Q}}})$.

Clearly, this is a special case of the Litt conjecture. :)


Čech functions and the axiom of choice

A **Čech closure function** on $\omega$ is a function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ such that (i) $X\subseteq\varphi(X)$ for all $X\subseteq\omega$, (ii) $\varphi(\emptyset)=\emptyset$, and (iii) $\varphi(X\cup Y)=\varphi(X)\cup\varphi(Y)$ for all $X,Y\subseteq\omega$; in other words, it obeys the Kuratowski closure axioms except possibly the idempotent law $\varphi(\varphi(X))=\varphi(X)$.

In 1947 E. Čech asked if there exists such a closure function (on any set, not necessarily $\omega$) which is also **surjective** and **nontrivial** (not the identity map). Čech's question has been answered in the affirmative [under various set-theoretic assumptions](https://www.sciencedirect.com/science/article/pii/0166864182900608), including ZFC. [The ZFC example](https://pdfs.semanticscholar.org/8870/c3c8a574f24aa1ee411f82283a1076c53303.pdf) is rather complicated and makes heavy use of the axiom of choice; it seems unlikely that one could construct such a thing without choice.

**Question.** Is it consistent with ZF that there is no nontrivial surjective Čech closure function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$?

That is, there is no function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ satisfying the conditions (i)-(iii) above and also (iv) for every $Y\subseteq\omega$ there exists $X\subseteq\omega$ such that $\varphi(X)=Y$, and (v) $\varphi(X)\ne X$ for some $X\subseteq\omega$. (Condition (ii) is now redundant, as it follows from (i) and (iv) that $\varphi(X)=X$ if $X$ is finite.)

@Haim Looks to me like they both use choice. My guess is, if you can construct such a function without choice, it will be by a completely different method.

@bof Out of Theorem 1 and Theorem 2 in the Galvin-Simon paper, which one is using AC in its proof?

@YCor: I got a partial answer to this in this question.

@YCor: Are you asking for a definable class (the class of functions such that blah blah, defined by formula) whose elements are individually not definable (so that you cannot write down such $F$'s to pick out individual elements)?

@bof I'd be happy with a weaker definition of "constructible". I'm just wondering whether the slogan "(exists in ZF)=(can be constructed)" has any foundation.

Is it correct to think as "in ZF there's such a function" as "one can construct such a function"? More precisely, is it true (from some general principle) that if "there is such a function" is a theorem of ZF, then there exists $f_0\in P(\omega)^{P(\omega)}$ satisfying the given condition, and a formula of set theory $F(x)$ with only parameter $x$, such that for every $f\in P(\omega)^{P(\omega)}$, $F(f)$ holds iff $f=f_0$?

I think it would be a good point in time to migrate this to MathOverflow. I don't have an obvious solution at hand, and I have quite a lot of things on my hands right now (which is why I might be missing something obvious).

Also a candidate: the Feferman–Levy model where the reals are a countable union of countable sets, or the Truss model (although if this holds there, I'd imagine the Solovay model would already capture this).

I could imagine several candidate models: Maybe Cohen's model is a suitable example with the Dedekind-finite set of reals "spoiling things", but more likely would be Solovay's model where there are no MAD families, etc.


Is the Frog game solvable in the root of a full binary tree?

_This is a cross-post from[math.stackexchange.com](https://math.stackexchange.com/q/3800570/318073)$^{[1]}$, since the bounty there didn't lead to any new insights._

* * *

For reference,

> The **Frog game** is the generalization of the _Frog Jumping (see it on[Numberphile](https://www.youtube.com/watch?v=X3HDnrehyDM))_ that can be played on any graph, but by convention, we restrict the game to tree graphs. The _Frog Jumping_ is equivalent to the case on "paths (tree graphs with exactly 2 leaves)" of the **Frog game** , which was resolved in the linked video.

The mechanics of the Frog game are in short,

> Given a tree graph and one frog on each vertex, the goal of the game is for all frogs to host a party on a single vertex of the graph. All of the $f$ frogs on some vertex can "jump" to some other vertex if and only if both vertices have at least one frog on them and both vertices are exactly $f$ edges apart from each other.
>
> If a sequence of "jumps" exists such that all frogs end up on a single vertex, then the party is successfully hosted on that vertex and we call that vertex a "lazy toad" (or just "lazy") because the frog that started there never jumped during the game.

Trivially, if $|V|=n$ is the number of vertices of some graph $G=(V,E)$, then every game on $G$ must end in $n-1$ moves (or less if the party isn't hosted, e.g. if the sequence of jumps is suboptimal and we can no longer make moves or if none of the vertices are lazy).

* * *

Here, I'm trying to resolve the Frog game on "full binary trees".

Let $T_h=(V,E)$ be a full binary tree of height $h$. Denote its layers with $0,1,2,\dots,h$ where the root $r\in V$ is on the layer $0$ ("bottom layer"), where layer $i=1,2,\dots,h$ contains vertices $v_{i,j}$ labeled with $j=1,2,\dots,2^i$.

That is, we have $|V|=2^{h+1}-1$. Let $h\in \mathbb N$ because $h=0$ is a single vertex which is trivial.

> **Conjecture.** For all $h\ge 4$, every vertex of $T_h$ is a "lazy toad" (is "lazy").

I couldn't prove the above, and for simplicity, my question is just about the root vertex:

> Let $h\in\mathbb N, h\ne 2$. Can we prove that the root $r$ of every such $T_h$ is a "lazy toad" (is "lazy")?

However, if the approach that can solve the root can be generalized to all other vertices, that's a big bonus.

* * *

**Solving the root**

**Reduction.** We say that $T_h$ can be reduced to $T_s,s\le h-2$, if there exists a sequence of moves that can bring all frogs from the top $s$ layers $h,h-1,\dots,h-s+1$ to the root vertex.

I've managed to reduce $T_h$'s for all $h\in(2,20)$ by hand pretty easily. At most, I needed to consider only $31$ vertices or less to find a pattern that allows a reduction of entire layers, which is very efficient if you consider that $T_{19}$ has over a million vertices. It looks like that maybe an inductive argument would be able to solve the problem, but I did not manage to find a sufficient set of move sequences.

I'll summarize some immediate results below.

> **Trivial reduction.** If $h=2^t+t-2,t\ge 2$ and $T_{t-1}$ is solvable in root, then $T_h$ can be reduced to $T_{h-t}$.

In the above reduction, we are simply (at the top layers) solving $T_{t-1}$ subtrees in root and then jumping from that root to the root of $T_h$.

> **Necessary minimal reduction condition.** Let $h\gt2,a\ge2$. If reduction from $T_h$ to $T_{h-a}$ is possible, then there exists $b\in[0,h-a+1]$ and a partition of the number $(2^a-1)2^b$ into the layers $H=\\{h,h-1,h-2,\dots,h-a+1\\}$ where $(h-l)\in H$ is used at most $(2^{a-l-1})2^b$ times.

In the above condition, we say that "a partition is solvable" if we can bring the corresponding amounts of frogs to the corresponding layers. I.e. the above condition simply says that we must be able to partition the total number of frogs from the top layers (that we are reducing) into the corresponding vertices from which they can jump to the root.

It remains to show (or find an explicit set of move sequences that will imply) that at least one partition from the above condition is solvable. Surprisingly, for all $h\in(2,20)$ that I've solved so far, the solution was always available in the smallest or one of the smallest partitions ($a\le 5,b\le 3$) of no more than $31$. The $h=20$ is the first case that requires considering more than $31$ vertices.

I still do not see how to construct explicit move sequences for larger $h$ without tackling each one individually. Alternatively, could there be a way to prove that the root is lazy, without finding explicit move sequences?


Orientation-reversing homotopy equivalence

If there is an orientation-reversing homotopy equivalence on a closed simply-connected orientable manifold is there an orientation-reversing homeomorphism?

It is not true, for instance, that if there is an orientation-reversing homeomorphism there must be an orientation-reversing diffeomorphism (consider exotic 7-spheres).

This can be heuristically reduced to finding manifolds that are homotopy equivalent but not homeomorphic, which should show you how to organize thinking about this. If $X$ does not admit an orientation-reversing homotopy equivalence and $Y$ is homotopy equivalent to $X$ but not homeomorphic, then the connected sum of $X$ and the opposite orientation of $Y$ is a manifold that admits an orientation-reversing homotopy equivalence, but no orientation-reversing homeo. $X=\mathbb C\mathbb P^4$ and $Y$ with different Pontrjagin classes. This isn't any easier than my previous example, though.

Correction: there is a smooth manifold homotopy equivalent to $S^4\times S^4$ with nonzero $p_2$ and thus no orientation reversing homeomorphism, but I don't see an easy way to construct it. However, linear sphere bundles have nontrivial $p_1$ and so do provide counterexamples to the more natural question of whether a particular homotopy equivalence is homotopic to a homeomorphism. That is, the homotopy equivalence which reverses orientation on the base sphere is not homotopic to a homeomorphism, even though the homotopy equivalence which reverses orientation on fiber is realized by a homeo.

@ConnorMalin I'm not really sure what you're asking, but the first topic is called surgery theory and the second topic is called smoothing theory (or triangulation theory). For most purposes, I recommend the two volumes "Surveys on Surgery Theory," ed Cappell-Ranicki-Rosenberg. But if you really want a foundational reference for the topological category, I think the only option is Kirby-Siebenmann.

@BenWieland Do you know of a standard reference for the obstructions to homotoping a homotopy equivalence to a homemorphism? And for isotopy between a homeomorphism and a diffeomorphism?

The difference between homeomorphism and diffeomorphism is subtle (torsion). The difference between homotopy equivalence and homeomorphism is crude (rational), the same as the difference between homotopy equivalence and diffeomorphism. Examples involving exotic spheres are difficult and overkill. There are easy smooth examples using Pontrjagin classes. There are linear $S^4$ bundles over $S^4$ homotopy equivalent to $S^4\times S^4$ with nonzero $p_2$. Obviously smoothly chiral. Given the difficult but blackbox theorem that $p_2$ is a rational homeomorphism invariant, topologically chiral.

According to arxiv.org/pdf/0907.5283.pdf (Section 2, 4th bullet point), the lens space $L_5(1,1)$ has an orientation reversing homotopy equivalence, but no orientation reversing homeomorphism.  Of course, it is not simply connected.


Are these local systems on $\mathscr{M}_{g,1}$ motivic?

Let $(\Sigma_g, x)$ be a pointed topological surface of genus $g$, and let $MCG(g,1)$ be the mapping class group of this pointed surface. Then $MCG(g,1)$ has a natural action on $\pi_1(\Sigma_g, x)$ $$MCG(g)\to \text{Aut}(\pi_1(\Sigma_g))$$ and hence acts on the set of irreducible characters of $\pi_1(\Sigma_g)$: $$\gamma: MCG(g,1)\to \text{Aut}(\text{Char}^{\text{irr}}(\pi_1(\Sigma_g))).$$

Suppose $$\rho: \pi_1(\Sigma_g)\to GL_n(\mathbb{C})$$ is an irreducible representation of $\pi_1(\Sigma_g)$ which is fixed by this action, i.e. for every element $\tau$ of $MCG(g,1)$, there exists a matrix $\varphi_\rho(\tau)$ conjugating $\rho^\tau$ to $\rho$ as a representation of $\pi_1(\Sigma_g)$. Because $\rho$ is irreducible, $\varphi_\rho(\tau)$ is well-defined up to scaling, and so we obtain a natural projective representation $$\varphi_\rho: MCG(g,1)\to PGL_n(\mathbb{C}).$$ Choosing a representation of $PGL_n(\mathbb{C})$ (say, the adjoint representation), we obtain a representation of $MCG(g,1)$ and hence a local system on the moduli space of pointed curves $\mathscr{M}_{g,1}$.

> Are the local systems obtained this way motivic?

I know how to produce several examples of such $\rho$ fixed by $MCG(g,1)$ (e.g. using the Parshin trick) and in these examples the local systems in question are indeed motivic.

Of course this question is too hard as stated in this generality -- so it would be better to ask:

> Do the local systems thus obtained underly $\overline{\mathbb{Z}}$-variations of Hodge structure?

This question is naturally divided into three parts:

>   1. Is $\varphi_\rho$ always defined over $\overline{\mathbb{Q}}$? I.e. is it conjugate to a representation into $PGL_n(\overline{\mathbb{Q}})$?
>
>   2. Is $\varphi_\rho$ always integral, i.e. is it conjugate to a representation into $PGL_n(\overline{\mathbb{Z}})$?
>
>   3. Does $\text{ad}_{PGL_n}\circ \varphi_\rho$ always underly a $\overline{\mathbb{Q}}$-variation of Hodge structure?
>
>

The only method of proof I can think of at the moment would be to show that $\varphi_\rho$ is _rigid_ (which would imply (1) and (3), using non-abelian Hodge theory) or _cohomologically rigid_ (which would imply (1-3) via recent results of Esnault and Groechenig). I haven't had any luck showing this thus far.

* * *

Some brief remarks on this old question (9/11/22):

  1. As written, the answer is "no": I know how to make counterexamples for $g\leq 2$. Indeed, for $g\leq 2$ one may construct such local systems which do not have quasi-unipotent mondoromy at infinity. It's possible that the answer is "yes" in $g\leq 2$ if one additionally assumes this.
  2. Such examples are impossible for $g\geq 3$, as any local system on $\mathscr{M}_{g,n}$ automatically has quasi-unipotent monodromy at infinity in this case by a result of Aramayona-Souto.
  3. The answer is "yes" for representations of rank $r<\sqrt{g+1}$ with finite orbit under $MCG(g,1)$ by one of the results of [this paper](https://arxiv.org/abs/2205.15352) of mine and Landesman. This is because such representations necessarily have finite image.

@BenWieland: (1) is an interesting idea -- I'll play with it a bit and see what I get. Thanks!

Oops, my (2) is not reducible as an $M_{g,1}$ representation. That was the whole point, yet it ruins it. The whole point is that it is the degeneration of an extension to a split extension, but such degenerations are trivial away from the split locus.

Yes, expand . . $M_2$ is virt $M_{0,6}$, so admits non-rigid reps. So ask $g\geqslant3$, but why stop there? Why did you think that 2 was enough? I don't really know any non-rigid reps, but here are two ideas. (1) a Burau-ish rep of $M_g$ on the homology of the abelian cover, as a $\mathbb Z[t_1,...t_{2g}]$-mod. (2) The rep of of $M_{g,1}$ on the Lie algebra of the 2-step nilpotent quotient of the fundamental group of the curve: shouldn't a 2-step nilpotent Lie algebra have a defm towards splitting, natural enough that it carries the rep? What happens when you restrict $M_{g,1}$ to $\pi_1S_g$?

@BenWieland: I should have assumed $g\geq 2$ -- the hope was that this construction rigidifies things, and it's motivated by some arithmetic considerations I omitted from the question, but which I can expand on if you'd like. Do you know of an example of a non-rigid representation of $\pi_1(M_{g,n})$ for some $g\geq 2$? Re: reps of $\pi_1(M_{g, 1})$, the Parshin trick gives examples, and there are others coming from TQFT's...

This seems very arbitrary to me. Why not ask if all representations of $M_{g,n}$ are motivic? But they're not all rigid. If you had a nonrigid representation of $M_{g,1}$, you could restrict to $\pi_1(\Sigma_g)$ and check if it remained irreducible. But I don't know any representations of $M_{g,1}$. For example, the Burau representation of $M_{0,n}^1$ is not rigid and probably remains irreducible when restricted to $\pi_1(P^1-n))$. (But it sounds motivic.)

@WillSawin: It seems to me that such reps will always be rigid as projective reps of the mapping class group, so at least (1) and (3) should be OK. And I think (2) should be OK by Esnault-Groechenig if the mapping class group action on the tangent space to $\rho$ has no invariants...

It might be an easier project to show that, if $\rho$ is rigid as a representation satisfying this additional condition, then $\varphi_{\rho}$ is motivic, and if $\rho$ is not rigid then $\varphi_\rho$ is not (always) motivic. Determining whether $\rho$ is / isn't always rigid could be a separate matter.

You’re right, fixed. Do you know how to make such a nontrivial family?

Your construction of a representation does not really work since you are trying to act via $Aut(\pi_1(\Sigma))$ instead of $Out(pi_1(\Sigma))$. What you are actually getting here are projective representations of $MCG(g,1)$. I am sure, these are non-motivic in general and there are nontrivial families of (equivalence classes ) such representations. One more thing: Each fixed point $[\rho]$ does get you a  linear representation of $MCG(g)$ but in a different dimension: You act on the Zariski tangent space to the character variety at the point $[\rho]$.

I agree with everything you’ve written—I tried the strategy in your second comment at some point but didn’t have any luck there either...some things are known in the case of 2-dimensional reps IIRC (due to Goldman and collaborators, for example), but not much.

With regards to showing at least the first couple of your three points, might it make sense to show that $\rho$ has these properties first? I think if $\rho$ is algebraic or integral then $\varphi_{\rho}$ is algebraic or integral as well. For algebraicity at least a plausible avenue of proof is to show that $\rho$ has no 1-parameter deformations that respect this invariance property.

In the $\ell$-adic world, if $\rho$, $\pi_1(\Sigma_g) \to GL_n( \overline{\mathbb Q}_\ell)$ factors continuously through the profinite completion of $\Sigma_g$, and this factorized representation is stable under the action of the arithmetic fundamental group of the moduli space of curves, then I think by the same construction we get a projective arithmetic representation. Of course by the grand theory of motives we expect variations of Hodge sructures and arithmetic representations should appear at the same time.


Is there a model category describing shape theory?

Is there a nice model structure on some category of topological spaces compatible with [shape theory](https://en.wikipedia.org/wiki/Shape_theory_\(mathematics\))? In particular, weak equivalences should induce isomorphisms on sheaf cohomology.

As an example, the [Polish Circle](https://en.wikipedia.org/wiki/Shape_theory_\(mathematics\)#Warsaw_Circle) has the same sheaf cohomology with constant coefficients as $S^1$, and both spaces also have equivalent categories of covering spaces. So one could ask if in such a model structure, the polish circle would be weakly equivalent to an ordinary one.

**Edit:** I would be happy with an answer for any nice category of topological spaces which contains the Polish circle, e.g. compactly generated weak Hausdorff spaces. Please ignore the following paragraph.

Also, sheaf theory is defined on rather general topological spaces, so it would be nice if the underlying category of topological spaces would be much more general than, say, compactly generated weak Hausdorff spaces.

@DenisNardin Thanks for the suggestion. I see the point now and did the edit.

@SebastianGoette Well, it is sort of the same definition, only you need to generalize it to a notion of the shape of a topos rather than just of topological spaces (you might have heard it under the name "étale homotopy type").The point is that the underlying space of an algebraic variety does not have much useful information (e.g. all curves over an alg. closed field are homeomorphic!),and you need more refined objects (e.g the étale topos)to recover the information present in the topology for manifolds. Still, your question is interesting, I would only remove the reference to alg. varieties.

@DenisNardin I was really thinking of the shape as a topological space, as in the wikipedia article. I was not even aware that there is a separate definition in algebraic geometry.

@SimonHenry It's not idempotent, but it becomes one if you consider the profinite shape instead (see the appendix to SAG and the Barwick-Glasman-Haine work on the stratified shape)

In Lurie's perspective shape theory is an adjunction between the $\infty$-category of $\infty$-toposes and the $\infty$-category of pro-spaces. I always had the feeling that this adjunction is idempotent ( not sure if this is really true though, is there someone who know ?) if it is the case "shapes" are a localization of the category of pro-spaces, as such it should be possible to construct a model structure representing them (maybe on pro-simplicial sets...). If this works one can start thinking about whether this transfer to a model structure on some category of topological spaces...

@SebastianGoette I'll have to admit I am a bit confused. What do you mean with the shape of an algebraic variety then? The notion I know is not encoded in the topology.

@DenisNardin I don't expect to see the 'algebraic' cohomology (in whatever sense) here, because the structure sheaf seems to be an extra datum that the topological space does not have. But maybe apart from constant sheaves, we also want to (maybe have to) consider constructible sheaves.  But then all information we need is still encoded in the topology.

@SimonHenry I must admit that I have no idea. The correct notion of equivalence seems to be part of the problem. Also I wonder whether it is possible to work on topological spaces, or whether one needs topoi.

What is the precise definition of 'shape equivalences' (it is not in the wikipedia link)  ? I'm not expert in shape theory but I believe there has been several non-equivalent definition... is it equivalent to the definition Lurie gave for infinity toposes ?

Hmm... the shape of algebraic varieties does not depend on the underlying topological space, unless I'm mistaken (rather, it is the shape of the étale topos), so I doubt a naive approach would work there.


Are simplicial finite CW complexes and simplicial finite simplicial sets equivalent?

**Edit** Originally the question was whether an arbitrary diagram of finite CW complexes can be approximated by a diagram of finite simplicial sets. In view of Tyler's comment, this was clearly asking for too much. I restricted the question from arbitrary diagrams to simplicial diagrams.

* * *

It is well-known that for any finite CW complex $K$ one can construct a finite simplicial set whose geometric realization is equivalent to $K$. But the construction is not functorial.

Suppose we have a simplicial object in finite CW complexes. Can one always construct a simplicial object of finite simplicial sets, whose levelwise geometric realization is connected to the original diagram by a zig-zag of levelwise weak equivalences?

One may ask a similar question in a more general context. Suppose we have an $\infty$-category $\mathcal C$. Suppose $A$ is a collection of (say compact) objects of $\mathcal C$. Let $CW^f(A)$ be the category of finite cellular objects generated by $A$. Roughly speaking, $CW^f(A)$ consists of objects that can be built as a finite homotopy colimit of objects of $A$. I am leaving the definition of $CW^f(A)$ a little vague - feel free to use any reasonable notion. Let $S_\bullet^f(A)$ be the category of simplicial objects in $\mathcal C$ that in each simplicial degree are a finite sum of elements of $A$, and are degenerate above some dimension. Geometric realization gives a functor $S_\bullet^f(A)\to CW^f(A)$. Now we have an obvious generalization of the question:

Suppose we have a simplicial object in $CW^f(A)$. Can we find a simplicial object in $S_\bullet^f(A)$ whose levelwise geometric realization is levelwise (weakly) equivalent to it? If it is not true in general, is there a reasonable sufficient assumption?

The point of the original question is that a model does not support every morphism. Tyler's example of a functor from $B\mathbb N$ is just the choice of an object and a morphism. If you allow Kan complexes or finite CW complexes, then they are cofibrant-fibrant and thus they support arbitrary morphisms. In the purely homotopical setting, the object supports the morphism. Tyler's example was $S^1$, which is as nicely cellular as you can imagine: represent it as the constant simplicial object. Maybe having to spell out the cells allows counterexamples, but not that one.

@BenWieland I am feeling a little slow. Why it is not a counterexample?

@BenWieland yes, absolutely. you can replace each object in a diagram with any weakly equivalent object and get an equivalent diagram at the cost of working coherently.

If the "more general" version is purely homotopical, Tyler's example isn't a counterexample, is it? (a counterexample to an even more general version) And a nitpick, just a problem for objects, if a "finite homotopy colimit" includes retracts, then you run into Wall finiteness problems, but @R.vanDobbendeBruyn's version doesn't.

A possible more general finiteness condition could be the assumption that every object in the diagram has finitely many maps out of it. (This has popped up in some inductive constructions I ran into; examples include finite diagrams and semisimplicial objects.)

@AchimKrause That is, indeed, a great deal simpler than what I had in mind.

In Tyler's example, can't we just observe that for any finite simplicial set $K$, only finitely many maps on $\pi_1$ are induced by endomorphisms $K\to K$? This is because there are only finitely many maps $K\to K$ in total.

Hi Tyler, thanks. This sounds right and suggests I should impose a local finiteness condition on the indexing diagram (such as, there can be only finitely many morphisms between any two objects). What I really wanted to know was about simplicial objects: can any simplicial object in finite CW complexes be approximated by a simplicial object in finite simplicial sets? I will edit the question.

Hi Greg, I suspect that the map $B\Bbb N \to CW$, representing $S^1$ with the endomorphisms $2^k$ for $k > 0$, can't be realized by a diagram of finite simplicial sets, but to prove it I'd want to come up with some explicit growth condition on the number of edges needed to represent elements of $\pi_1$ for a fixed finite model. This may be a little harder for s.sets than for $\infty$-categories because the mapping complexes aren't Kan complexes.


What is the logical complexity of the Hodge conjecture?

The Hodge conjecture seems to me the most mysterious among the Millennium problems (and many others). In particular, I am not sure about its logical complexity. It is not difficult to see that the conjecture is equivalent to a $\Pi^0_2$ statement if restricted to varieties over $\overline{\mathbb{Q}}$. (Basically, it is because these varieties form a recursive set, and the statement sounds like "for any Hodge class there is an algebraic cycle".) However the conjecture may be true over $\overline{\mathbb{Q}}$ but false in general ; if I am not mistaken, this possibility has not being ruled out yet. My guess is that the Hodge conjecture may be formulated as a $\Pi^0_3$ statement but, at present, not as $\Pi^0_2$. Is it true?

Some remarks. As is well known, the Hodge conjecture may be separated into two parts:

I. Hodge classes are absolute,

II. Absolute Hodge classes are algebraic.

The second part is $\Pi^0_2$. (It follows from results of Voisin, "Hodge loci and absolute Hodge classes", Compositio Mathematica, Vol. 143 Part 4, 945-958, 2007. ) So, the logically hard part seems to be the first one. In view of Proposition 0.5 of the Voisin's paper, it boils down to whether Hodge loci are defined over $\overline{\mathbb{Q}}$ or not. I have some ideas about how to translate this into $\Pi^0_3$ but nothing close to $\Pi^0_2$. Naturally, if it is true that the Hodge conjecture can _not_ be formulated as a $\Pi^0_2$ statement for now, there is no way one can actually prove this fact. In case it is not unlikely to be so, I am only asking for an expert opinion. Also, if possible, I would appreciate a slick and sound proof of the "upper bound" $\Pi^0_3$ (although I have some ideas about it, I do not really like them).


Is this Variation of the Continuum Hypothesis Inconsistent with ZFC or ZF?

It is a well-known fact that the Generalized Continuum Hypothesis is undecidable from ZFC. For similar sentences $\phi$, this is simply equivalent to ZFC having a model $M$ for which $M\models\phi$.

I wanted to see if there was any way to make all $\beth$-numbers limits. At first, this seemed dubious, but now, after creating a variation on GCH, I have realized it seems completely legitimate.

The variation I created is called the **Cofinal Continuum Hypothesis** , or CFCH. It is defined as follows:

$$\forall\kappa(2^\kappa=\aleph_{\kappa^+})$$

This immediately destroys the idea of the continuum hypothesis, as with this definition $2^{\aleph_0}=\aleph_\Omega$, which is very far off from $\aleph_1$.

The reason I chose specifically this definition is that its cofinality preserving properties when compared to GCH, namely that $\mathrm{cof}(2^\kappa)=\mathrm{cof}(\kappa^+)$. This property "piggybacks" off of GCH's undecidability in a sense; most proofs that a variation on GCH is inconsistent with ZFC (or similar theories) involve some failure of an inequality with $\mathrm{cof}(2^\kappa)$ and $\mathrm{cof}(\kappa)$. With this definition, it is clear to see that this problem no longer exists because in GCH $\mathrm{cof}(2^\kappa)=\mathrm{cof}(\kappa^+)$.

Because of this construction, one could generalize it to $2^\kappa$ is the $\kappa^+$-th aleph fixed point, or $2^\kappa=\aleph_{\aleph_\kappa}$. These, however, seem even more audacious than CFCH itself.

**However** , the reason I feel this is still dubious is because of it's interesting relation to the Singular Cardinal Hypothesis. It is clear that, with this definition, the first strong limit cardinal is a singular cardinal. This becomes a problem when introduced to the Singular Cardinal Hypothesis, as, assuming and CFCH and SCH:

$$\forall\alpha(\mathrm{cof}(\beth_{\omega\cdot\alpha})\neq\beth_{\omega\cdot\alpha}\rightarrow 2^{\beth_{\omega\cdot\alpha}}=\beth_{\omega\cdot\alpha}^+)$$ $$2^{\beth_\omega}=\beth_\omega^+$$ $$2^{\beth_\omega}=\aleph_{\beth_\omega^+}$$ $$\beth_\omega^+=\aleph_{\beth_\omega^+}$$

This is a contradiction, as it implies that a successor cardinal is an $\aleph$-fixed point. Thus, CFCH is inconsistent with SCH. Proving CFCH equiconsistent to ZFC would be a great feat, as it would prove $\neg$SCH equiconsistent to ZFC as well, thus showing that the existence of a measurable cardinal with Mitchell order $\kappa^{++}$ is equiconsistent to ZFC (as a result of Gitik), and every other large cardinal axiom implied by said axiom (which has not yet been done). This is why I doubt that CFCH will be proven equiconsistent with ZFC.

**Here are some other interesting facts implied by CFCH:**

  * $\beth_{\omega\cdot\alpha}$ is the $\alpha$-th $\aleph$-fixed point (and thus all strong limit cardinals are $\aleph$-fixed points). This is yet another doubt I have for CFCH.
  * $\mathrm{cof}(\beth_\alpha)=\mathrm{cof}(\aleph_\alpha)$ (and thus for regular cardinals $\aleph_\alpha$, $\mathrm{cof}(\beth_\alpha)=\aleph_\alpha$ and for limit ordinals $\alpha$, $\mathrm{cof}(\beth_\alpha)=\alpha$.)
  * The Konig's Theorem corollaries about cofinality seem are implied by it
($\kappa<\mathrm{cof}(2^\kappa)=\mathrm{cof}(\kappa^+)=\kappa^+$ and $2^\kappa<(2^\kappa)^{\mathrm{cof}(2^\kappa)})=2^{\kappa\cdot\mathrm{cof}(2^\kappa)}=2^{\mathrm{cof}(2^\kappa)}>2^\kappa$)
  * As for other variations of the Generalized Continuum Hypothesis and Continuum Hypothesis:
    * $\mathrm{CFCH}\rightarrow\neg\mathrm{GCH}$
    * $\mathrm{CFCH}\rightarrow\neg\mathrm{SCH}$
    * $\mathrm{CFCH}\rightarrow\neg\mathrm{CH}$
    * $\mathrm{CFCH}\rightarrow\neg\mathrm{NCH}$ (The Natural Continuum Hypothesis, which is found in another MathExchange link "[A New Continuum Hypothesis](https://mathoverflow.net/questions/148756/a-new-continuum-hypothesis-revised-version)")

**In general, it would be really useful if the reader of this question reply in the comments with either a link to a website listing many interesting models of ZF and ZFC or several Forcing methods or they send me the names of the models themselves.** Of course, you could reply to this question with comments pertaining to the question as well.

I mainly mentioned NCH because it seems like this is a "variant" of NCH (where cardinal exponentiation speeds up)

Just as a reminder, I would like to mention that despite its tempting appearance and intuitive background from natural numbers, the Natural Continuum Hypothesis (NCH) is inconsistent with ZFC due to Konig's lemma. However, it sounds interesting to me that you considered the cofinality version of CH independently.

Under CFCH, $2^{\aleph_n} = \aleph_{\omega_{n+1}}$ and therefore $2^{\aleph_{\omega}} = \aleph_{\omega_\omega}^\omega = \aleph_{\omega_{\omega + 1}}$. By the theory of pcf, we get that if $a$ is the set of all cardinals between $\aleph_{\omega}$ and $\aleph_{\omega_\omega}$, the possible cofinalities of $\prod a$ is the regular cardinals in the range between $\aleph_{\omega_\omega + 1}$ and $\aleph_{\omega_{\omega + 1}}$ which has cardinality $|a|^+$. A recent paper of Gitik claims that it is consistent, but it is open if one can obtain this type of pcf structure in accessible cardinals.

Easton's theorem implies that ZFC + "for all regular $\kappa$, $2^\kappa = \aleph_{\kappa^+}$" is equiconsistent with ZFC. Other variants, such as "the $\kappa^+$-th fixpoint" or $\aleph_{\aleph_{\kappa^+}}$ are possible, too.  - But it is well known that ZFC+ non-SCH is much stronger than ZFC, in terms of consistency strength.