Need explicit formula for certain "$q$-numbers" involving gcd's

The question is motivated by yet another possible approach to a combinatorial problem formulated previously in ["Special" meanders](https://mathoverflow.net/q/146802/41291). I'm not giving details of the connection as I believe the question is sufficiently motivated by itself.

I've got a vector space with basis $\left\\{e_n\mid n\geqslant0\right\\}$ and scalar product $$ \left\langle e_m,e_n\right\rangle=q^{\gcd(m,n)} $$ (with the convention $\gcd(0,n)=n$ for all $n\geqslant0$).

What I need is a maximally explicit expression for an orthogonal basis $\left\\{o_n\mid n\geqslant0\right\\}$ with respect to this scalar product. I do not mind if the $o_n$ are not of unit norm (and this is clearly a minor point anyway).

Here are the first few values (calculated with MAPLE). Up to arbitrary rescalings, \begin{align*} o_{{0}}&=e_{{0}}\\\ o_{{1}}&=e_{{1}}-qe_{{0}}\\\ o_{{2}}&=e_{{2}}- \left( q+1 \right) e_{{1}}+qe_{{0}}\\\ o_{{3}}&=e_{{3}}-qe_{{2}}-e_{{1}}+qe_{{0}}\\\ o_{{4}}&=\left( q+1 \right) e_{{4}}-{q}^{2}e_{{3}}- \left( {q}^{2}+q+1 \right) e_{{2}}+{q}^{2}e_{{1}}+{q}^{2}e_{{0}}\\\ o_{{5}}&=\left( {q}^{2}+q+1 \right) e_{{5}}- \left( {q}^{2}+1 \right) qe_{{4}}- \left( {q}^{2}+1 \right) qe_{{3}}+ \left( {q}^{3}-{q}^{2}-1 \right) e_{{1}}+ \left( {q}^{2}+1 \right) qe_{{0}}\\\ o_{{6}}&=\left( {q}^{3}+{q}^{2}+2\,q+1 \right) e_{{6}}- \left( {q}^{3} +q-1 \right) qe_{{5}}- \left( {q}^{3}+q-1 \right) qe_{{4}}\\\&\- \left( {q} ^{2}+1 \right) \left( {q}^{2}+q+1 \right) e_{{3}}- \left( {q}^{3}+{q} ^{2}+2\,q+1 \right) e_{{2}}+ \left( 2\,{q}^{4}+{q}^{3}+3\,{q}^{2}+1 \right) e_{{1}}+ \left( {q}^{3}+q-1 \right) qe_{{0}}\\\ o_{{7}}&=\left( {q}^{2}+1 \right) e_{{7}}- \left( {q}^{2}-q+1 \right) qe_{{6}}- \left( {q}^{2}-q+1 \right) qe_{{5}}- \left( {q}^{2}-q+1 \right) qe_{{4}}\\\&\+ \left( {q}^{2}-q+1 \right) qe_{{2}}+ \left( {q}^{3} -2\,{q}^{2}+q-1 \right) e_{{1}}+ \left( {q}^{2}-q+1 \right) qe_{{0}} \end{align*}

The inverse transformation (again up to rescalings) does not look any more enlightening (except that everything is positive):

\begin{align*} e_{{0}}&=o_{{0}}\\\ e_{{1}}&=o_{{1}}+qo_{{0}}\\\ e_{{2}}&=o_{{2}}+ \left( q+1 \right) o_{{1}}+{q}^{2}o_{{0}}\\\ e_{{3}}&=o_{{3}}+qo_{{2}}+ \left( {q}^{2}+q+1 \right) o_{{1}}+{q}^{3}o_ {{0}}\\\ e_{{4}}&=o_{{4}}+{\frac {{q}^{2}}{q+1}}o_{{3}}+ \left( {q}^{2}+1 \right) o_{{2}}+ \left( q+1 \right) \left( {q}^{2}+1 \right) o_{{1}} +{q}^{4}o_{{0}}\\\ e_{{5}}&=o_{{5}}+{\frac { \left( {q}^{2}+1 \right) q}{{q}^{2}+q+ 1}}o_{{4}}+{\frac {q \left( {q}^{2}+1 \right)}{q+1}} o_{{3}}+q \left( {q}^{2} +1 \right) o_{{2}}+ \left( {q}^{4}+{q}^{3}+{q}^{2}+q+1 \right) o_{{1}} +{q}^{5}o_{{0}}\\\ e_{{6}}&=o_{{6}}+{\frac { \left( {q}^{3}+q-1 \right) q}{{q}^{3}+ {q}^{2}+2\,q+1}}o_{{5}}+{\frac { \left( {q}^{3}+q-1 \right) q}{{q}^{2} +q+1}}o_{{4}}+{\frac { \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right)}{q+1}} o_{{3}}\\\ &\+ \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+ 1 \right) o_{{2}}+ \left( q+1 \right) \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right) o_{{1}}+{q}^{6}o_{{0}}\\\ e_{{7}}&=o_{{7}}+{\frac { \left( {q}^{2}-q+1 \right) q}{{q}^{2}+ 1}}o_{{6}}+{\frac { \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right) q}{{q}^{3}+{q}^{2}+2\,q+1}}o_{{5}}+ \left( {q}^{2}-q+1 \right) qo_{{4}}\\\ &+{\frac {q \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right)}{q+1}}o_{{3}} +q \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right) o_{{2}} \+ \left( {q}^{6}+{q}^{5}+{q}^{4}+{q}^{3}+{q}^{2}+q+1 \right) o_{{1}}+{q}^{7}o_{{0}}\\\ e_{{8}}&=o_{{8}}+{\frac { \left( {q}^{2}+1 \right) {q}^{4}}{ \left( {q}^{2}+q+1 \right) \left( {q}^{3}+q+1 \right) }}o_{{7}}+{\frac {{q} ^{4}}{{q}^{2}+q+1}}o_{{6}}+{\frac { \left( {q}^{2}+1 \right) {q}^{4}}{{q}^{3}+{q}^{2}+2\,q+1}}o_{{5}}\\\ &+{\frac { {q}^{6}+{q}^{4}+{q}^{2}+q+1 }{{q}^{2}+q+1}}o_{{4}} +{\frac {{q}^{2} \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right)}{q+1}}o_{{3}}+ \left( {q}^{2}+1\right) \left( {q}^{4}+1 \right) o_{{2}}+ \left( q+1 \right) \left( {q}^{2}+1 \right) \left( {q}^{4}+1 \right) o_{{1}}+{q}^{8}o_{ {0}} \end{align*}

There are some patterns, but I cannot even guess any statement that I could try to prove about these coefficients. Also the above was computed under the assumption that the transformation matrix is triangular, i. e. that $o_n$ only depends on $e_k$ with $k\leqslant n$. It is quite possible that there is a nicer basis which violates this assumption, but again I cannot think of any natural alternative form of the transformation matrix.

What I know is an explicit orthogonal basis for a similar scalar product "without $q$": if I would have just $\left\langle e_m,e_n\right\rangle=\gcd(m,n)$, then an orthogonal basis would be given by $e_n=\sum_{d|n}o_d$, so $o_n=\sum_{d|n}\mu\left(\frac nd\right)e_d$. Strangely enough, I obtain this from the above by substituting $q=0$, and again I have no idea why.

As @alpoge points out in a comment, if one removes $e_0$, then this actually works "with $q$" too.

PS As suggested by @Wolfgang in a comment, I've looked at polynomials resulting from the substitutions $q=\pm1$, $e_n=x^n$. Here:

$q=1$: \begin{align*} &1\\\ &x-1\\\ & \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 2\,x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 3\,{x}^{2}+x+2 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 5\,{x}^{3}+4\,{x}^{2}+8\,x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( {x}^{2}+x+1 \right) \left( 2\,{x}^{2}-x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 9\,{x}^{5}+7\,{x}^{4}+14\,{x}^{3}+10\,{x}^ {2}+6\,x+2 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 11\,{x}^{6}+8\,{x}^{5}+16\,{x}^{4}+10\,{x} ^{3}+15\,{x}^{2}+9\,x+3 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 7\,{x}^{5}+6\,{x}^{4}+5\,{x}^{3}+4\,{x}^{2 }+10\,x+1 \right) \left( {x}^{2}+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 16\,{x}^{8}+11\,{x}^{7}+22\,{x}^{6}+12\,{x }^{5}+18\,{x}^{4}+3\,{x}^{3}+9\,{x}^{2}-6\,x+5 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 21\,{x}^{9}+19\,{x}^{8}+38\,{x}^{7}+34\,{x }^{6}+51\,{x}^{5}+45\,{x}^{4}+39\,{x}^{3}+33\,{x}^{2}+6\,x+2 \right) \left( x-1 \right) ^{2} \end{align*} $q=-1$: \begin{align*} 0&:1\\\ 1&:1+x\\\ 2&:\left( x-1 \right) \left( 1+x \right)\\\ 3&:\left( x-1 \right) \left( 1+x \right) ^{2}\\\ 4&:- \left( x-1 \right) \left( 1+x \right) ^{2}\\\ 5&:\left( x-1 \right) \left( {x}^{2}+x+2 \right) \left( 1+x\right) ^{2}\\\ 6&:-\left( x-1 \right) \left( {x}^{3}+2\,{x}^{2}+2\,x+3 \right) \left( 1+x \right) ^{2}\\\ 7&:\left( x-1 \right) \left( {x}^{2}+x+1 \right) \left( 2\,{x}^{2}- x+3 \right) \left( 1+x \right) ^{2}\\\ 8&:-\left( x-1 \right) \left( {x}^{4}+2\,{x}^{2}+2 \right) \left( 1 +x \right) ^{3}\\\ 9&:\left( x-1 \right) \left( {x}^{5}+{x}^{4}+{x}^{3}+3\,{x}^{2}+3 \right) \left( 1+x \right) ^{3}\\\ 10&:- \left( x-1 \right) \left( {x}^{2}+1 \right) \left( {x}^{5}+2\, {x}^{4}+{x}^{3}+2\,{x}^{2}+2\,x+3 \right) \left( 1+x \right) ^{2}\\\ 11&:\left( x-1 \right) \left( 4\,{x}^{8}+{x}^{7}+8\,{x}^{6}+2\,{x}^{ 5}+12\,{x}^{4}+3\,{x}^{3}+11\,{x}^{2}+4\,x+5 \right) \left( 1+x \right) ^{2} \end{align*}

@OfirGorodetsky Thanks for the link! Reading that - seems an interesting idea to give it a combinatorial interpretation in terms of the Möbius function of some poset. In fact this might be important for the original problem which is of combinatorial nature.

@alpoge But this is great! I was sticking to $e_0$ so much it never occurred to me that dropping it gives such a nice orthogonal basis! It is true that I still need $e_0$ essentially (it is the unit of certain algebra), but your version does almost all of it. Could you please post this as an answer? It's worth it in any case, and if nobody will find a way to incorporate $e_0$ I will go for it

Just a remark\restatement, nothing fancy: Let $A_{i,j} = q^{(i,j)}$ be a $(n+1) \times(n+1)$ matrix. Decompose $A$ as $B^T B$ (Cholesky-Decomposition; $B$ can be upper-triangular). The condition that $o_i = \sum_{j} o_{i,j} e_j$ are orthonormal w.r.t to the given inner product is the same as requiring that $B o_i$ are orthonormal w.r.t to the ordinary inner product, so we can take $o_i = B^{-1} e_i$. Hence the problem reduces to Cholesky-decomposing $A$, and inverting.   Bruce Sagan's talk here about GCD-matrices might provide useful input: users.math.msu.edu/users/sagan/Slides/mfp5.pdf

Formally doesn't $o_n := \sum_{d\vert n} \mu(d) e_{n/d}$ for $n > 0$ and $o_0 := e_0 - \sum_{n\geq 1} o_n$ work? (Not that that helps much, since the expression for $o_0$ is nonsense.)

@Wolfgang Well $q=-1$ looks somehow less hopeless but still I have no clue :) I'll add it too

and I suppose the same with $q=-1$ doesn't reveal more either?

I had skipped $n=6 $, so that was a wrong conclusion of mine! :(.

@PeterMueller I use $\gcd(0,0)=0$, I think it is not senseless in view of $\gcd(0,n)=\gcd(n,n)=n$ for all $n$.

@OP: What is $\langle e_0,e_0\rangle$? It seems to me that the problem is not well formulated, as $gcd(0,0)=\infty$. In particular, I don't know how to interpret your assertion $0=\langle o_0,o_1\rangle=\langle e_0,e_1-qe_0\rangle=q-q\cdot q^{gcd(0,0)}$.

@Wolfgang I've looked at it, what happens that seeminglu each of these polynomials (well, starting from 3) is divisible by $(x-1)^2(x+1)$, but the quotient seems impenetrable to me. I'll add it to the question though

@Wolfgang Many thanks for the suggestion, sounds promising! I'll definitely try it

Have you tried putting $q=1$ and replacing $e_i$ by $x^i$ in the $o_j$ expressions? The resulting polynomials seem to split into linear and quadratic factors, and maybe you'll find some patterns in those factors, which might give some ideas to start with.


Bunnity of multilinear maps

Is there a way to compute the following nullity of multilinear maps? As it is different from any nullity I know of, I call it _bunnity_ after myself:-)) If it already has a name, it be nice to know it. References are particularly appreciated.

Let $V$ be a $d$-dimensional vector space over a field $K$, $f:V^{\otimes n}\rightarrow K$ a multilinear map. We define the bunnity by

$Bun(f) = sup \\{ \sum_{k=1}^n dim(U_k) \ | \ f(U_1\otimes U_2 \ldots \otimes U_n)=0\\}$.

If $n=2$, the bunnity is nullity plus dimension. It is more complicated if $n\geq 3$. I expect that for $n=3$ or $n=4$, there should be an algorithm because the 3-subspace problem is finite, while the 4-subspace problem is tame. I would be pleasantly surprised if there is an algorithm for a general $n$.

**EDIT** I forgot to say what $U_i$ are. They are **non-zero** vector subspaces of $V$.

The dimension of the null-space: think of $f$ as a map $V\rightarrow V^\ast$ and take the dimension of its kernel. Similarly, the usual multinullity is defined for any $n$: use position $i$ to write $f$ as $V\rightarrow (V^{\otimes (n-1)})^\ast$ and take the dimension of its kernel.

Could you remind what definition of "nullity" do you use (e.g. when $n=2$)?

You should offer a bounty to whoever manages to frame the answer in terms of Wascal's triangle.

Inequality. In the usual tensor rank/nullity, you choose each $U_i$ separately. I ask you to choose them simultaneously.

What's the connection to multilinear tensor rank?


Question about combinatorics on words

Let $\\{a_1,a_2,...,a_n\\}$ be an alphabet and let $\\{u_1,...,u_n\\}$ be words in this alphabet, and $a_i\mapsto u_i$ be a substitution $\phi$.

Question: Is there an algorithm to check if for some $m,k$ some prefix of the word $\phi^m(a_1)$ coincides with some suffix of the word $\phi^k(a_2)$?

This is related to a question about the R. Thompson group $F$.

@domotorp The equality version is decidable, by Theorem 4 of [1]. Unfortunately, this doesn't seem very helpful for the original question, because it goes via saying that if the substitution homomorphism is not injective, then it factors through a smaller alphabet and we are happy by induction. [1] A. Ehrenfeucht, G. Rozenberg, Simplifications of homomorphisms, Information and Control, Volume 38, Issue 3, 1978 (sciencedirect.com/science/article/pii/S0019995878900955)

With the below simple trick we can ensure that if a prefix equals a suffix, then the whole words are equal. I'm thinking that deciding equality might be already undecidable.  Trick: $\phi(a_1)=a_{start} a_1' a_{end}$ and $\phi(a_2)=a_{start} a_2' a_{end}$, where $\phi(a_{start})=a_{start}$ and $\phi(a_{end})=a_{end}$, and none of $a_1, a_2, a_{start}, a_{end}$ are in the image of any other $a_i$.

@NaradRampersad: You can ask him and post his answer here, if he is not on MO himself.

I don't have an actual solution to the problem, but I know the best person to ask about it: Juha Honkala of the University of Turku. If anyone knows the answer it would be him.

This is a decision problem. I need to decide whether or not the property holds. The second $n$ is a $k$.

Assuming the words are finite, there is the brute force method with no guarantee of termination.  Are you looking for an algorithm with such a guarantee, or do you want even more?  Also, is the n in the question the same as the alphabet size? (Probably not.) Gerhard "Just Checking On Some Details" Paseman, 2016.09.19.


The topos for forcing in computability theory

My understanding is that forcing (such as Cohen forcing) can be described via a topos. For example this [nlab article on forcing](http://ncatlab.org/nlab/show/forcing) describes forcing as a "the topos of sheaves on a suitable site."

My question concerns forcing in computability theory, for example as described in Chapter 3 or these [lecture notes of Richard Shore](http://www.math.cornell.edu/~shore/papers/pdf/SingLect2NS.pdf). The idea is that the generics are those which meet all _computable_ dense sets of forcing conditions. (Computable can mean a few things. Often it is taken to mean a $\Sigma^0_1$ set of forcing conditions. Also, usually the forcing posets are countable.) Since there are only countably many such dense sets, such effective generics exist.

> Is there a known/canonical type of topos corresponding to the forcing in computability theory?

Any references would be appreciated.

_FYI: My background is in computability theory, proof theory, and computable analysis. I know little about topos theory, but I am willing to learn a bit. I am mostly asking this question because I want to compare some ideas I have about effective versions of Solovay forcing with some work by others about the topos corresponding to Solovay forcing. Also, it is always nice to learn new things._

Have you, since asking the question two years ago, learned about a topos-theoretic treatment of forcing in computability theory?

@BjørnKjos-Hanssen, I would be happy to know the answer in that case.  (The application I have in mind is a bit more general, but I think knowing what is going on in the case of a computable forcing poset is sufficient for me to get a general idea of what is going on.)

Did you also want the forcing partial order to be a (close to) computable relation?


Elements of finite fields with many powers of trace zero

Let $p$ be an odd prime number, $n>1$ be an integer, and $\mathrm{tr}$ be the trace map of the field extension $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$. For which pair $(p,n)$ does there exists $x\in\mathrm{GF}(p^{2n})$ such that $x^{2(1+p^n)}=1$ and $\mathrm{tr}(x^{1+p^j})=0$ for $j=0,1,\dots,n-1$?

After computing the gcd of the corresponding $n+1$ polynomials over $\mathrm{GF}(p)$ for small $p$ and $n$, I believe that the answer should be ``if and only if $p$ divides $n$" and moreover, when $p$ divides $n$, all such $x$ would form a subgroup of $\mathrm{GF}(p^{2n})^\times$ of index $2(1+p^{n/p})$.

**Update.** If $(p,n)$ is such a pair, then denoting by $X$ the $(2n)\times(2n)$ matrix with $(i,j)$th entry $x^{p^{i+j-2}}$ we deduce from the conditions that $$ X^2= \begin{pmatrix} 0&2naI_n\\\ 2naI_n&0 \end{pmatrix}, $$ where $a=x^{1+p^n}=\pm1$. This yields two necessary conditions:

(1). If $n$ is not divisible by $p$ then $n$ is odd.

_Proof._ Suppose on the contrary that $n$ is even. Then $$ |X|^2= \begin{vmatrix} 0&2naI_n\\\ 2naI_n&0 \end{vmatrix} =(-1)^n(2na)^{2n}=(2na)^{2n} $$ is a square in $\mathrm{GF}(p)$, which implies that $|X|\in\mathrm{GF}(p)$. However, changing columns shows that $-|X|=(-1)^{2n-1}|X|$ is the determinant of the matrix with $(i,j)$th entry $x^{p^{i+j-1}}$. Hence $-|X|=|X|^p$ and so $|X|\notin\mathrm{GF}(p)$, a contradiction.

(2). If $n$ is not divisible by $p$ then $x,x^p,\dots,x^{p^{2n-1}}$ is a normal basis of $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$.

_Proof._ $|X|^2=(-1)^n(2na)^{2n}\neq0$ implies that $|X|\neq0$ and so $x,x^p,\dots,x^{p^{2n-1}}$ are $\mathrm{GF}(p)$-linearly independent.


Groups generated by 3 involutions

Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the class transposition $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$ and which fixes everything else.

  1. Is it algorithmically decidable whether all orbits on $\mathbb{Z}$ under the action of a group generated by 3 given class transpositions are finite?

  2. Is it algorithmically decidable whether a group generated by 3 given class transpositions acts transitively on the set of nonnegative integers in its support (i.e. set of moved points)?

**Added on Dec 11, 2013:** This question will appear as Problem 18.47 in:

Kourovka Notebook: _Unsolved Problems in Group Theory_. Editors V. D. Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.

Remarks:

Ad 1.: A hard case is $G = \left\langle\tau_{0(2),1(2)}, \tau_{0(5),4(5)}, \tau_{1(4),0(6)}\right\rangle$. For example we have $|32^G| = 6296$ and $|25952^G| = 245719352$, and the largest point in the latter orbit is about $10^{5759}$. The finiteness of the orbit $173176^G$ is not known to the author so far.

EDIT: In the meantime, the length of the cycle of $g := \tau_{0(2),1(2)} \cdot \tau_{0(5),4(5)} \cdot \tau_{1(4),0(6)}$ containing 173176 has been computed by Jason B. Hill (<http://mail.gap-system.org/pipermail/forum/2012/003948.html>) -- it is 47610700792, and the largest point in the cycle is about $10^{76785}$. An earlier attempt of the same computation ended without success after one week of CPU time (<http://mail.gap-system.org/pipermail/forum/2012/003915.html>). Possibly the cycle length is also the orbit length $|173176^G|$.

Ad 2.: A hard case is $G = \left\langle\tau_{1(2),4(6)}, \tau_{1(3),2(6)}, \tau_{2(3),4(6)}\right\rangle$. As the question whether this group acts transitively on $\mathbb{N} \backslash 0(6)$ is equivalent to Collatz' $3n+1$ conjecture (cf. <http://en.wikipedia.org/wiki/Collatz_conjecture>), likely only a negative answer might be given without solving a known open problem. This will likely also not be easy -- at least unless someone has e.g. a good idea on how to encode arbitrary computations with just 3 class transpositions.

**Added on Apr 24:** An easy case is $G = \left\langle\tau_{0(2),1(2)}, \tau_{0(3),2(3)}, \tau_{1(2),2(4)}\right\rangle$. By means of computation it can be checked that this group acts at least 5-transitively on $\mathbb{N}_0$.


Is there a n/2 version of the Erdős-Hanani conjecture?

This question comes out of REU research from this past summer. Unfortunately weeks of thought led to only trivial observations and the conclusion that the problem is quite hard.

Fix $k,t$. Let $F$ be a set of $k$-subsets of $[n] := \\{1,\ldots,n\\}$ of minimal cardinality such that $F$ covers all $t$-subsets of $[n]$ (covers in the sense that any $t$-subset of $[n]$ is a subset of an element of $F$.) Let $\kappa_n := |F|$. The Erdős-Hanani conjecture states that

> $\kappa_n = \binom{n}{t} / \binom{k}{t}(1 + o(1))$.

Of course $\binom{n}{t} / \binom{k}{t}$ is a lower bound on $\kappa_n$, so the EH conjecture is saying that the obvious necessary condition is asymptotically sufficient. Rödl proved the EH conjecture in 1985.

This question is about what happens when $k$ and $t$ are not fixed. Specifically, take $k = \lfloor n/2 \rfloor$ and $t = \lfloor n/2\rfloor - 1$. Define $F$ and $\kappa_n$ as above. Is it true that

> $\kappa_n = \frac{1}{\lfloor n / 2 \rfloor} \binom{n}{\lfloor n/2 \rfloor}(1 + o(1))$?

# Background

The EH conjecture lead to the study of what is called "packing in a hypergraph." See <http://en.wikipedia.org/wiki/Packing_in_a_hypergraph>. Rödl's proof introduced what is now called the "Rödl nibble" and is pseudo-random in nature. Spencer gave a lovely proof using branching processes. There are a lot of results from the late 80s to 90s that say, as Kahn puts it in "Asymptotics of Hypergraph Matching, Covering and Coloring Problems", that hypergraphs are asymptotically well-behaved _as long as their edge sizes are bounded_! Unfortunately the $n/2$ version of EH involves hypergraphs of unbounded edge size and the existing methods appear useless.

# Some ideas

A straightforward application of the method of alterations (or equivalently, some easy analysis of the greedy algorithm) gives that $\kappa_n \leq \log n \frac{1}{\lfloor n / 2 \rfloor} \binom{n}{\lfloor n/2 \rfloor} (1 + o(1))$, so the whole question is whether we can eliminate this log factor.

A set of $\lfloor n/2 \rfloor$-subsets has maximum coverage of $(\lfloor n/2 \rfloor - 1)$-subsets when all its elements have pairwise symmetric difference of at least 4. So really this is a coding theory problem. The paper "Lower bounds for constant weight codes" by Graham and Sloane shows that we can find a set $H$ of $\lfloor n /2\rfloor$-subsets of $[n]$ such that $|H| \geq \frac{1}{2}\binom{n}{\lfloor n/2 \rfloor}$ and the hamming distance between elements is at least 4. Let $G$ be the set of $(\lfloor n/2 \rfloor - 1)$-subsets covered by $H$. $G$ is half the size we want it to be, but we only used half as many elements are we are allowed. So we might be optimistic that by allowing some small overlap we can cover everything we want. If we take a permutation $\sigma \in S_n$ and look at $\sigma(H)$ (i.e. apply the permutation to the elements of the elements of $H$) it covers $\sigma(G)$. Of course $|\sigma(G)| = |G|$. We could hope that a good choice of $\sigma$ gives $|G \cup \sigma(G)| \approx 2|G|$ and we have found an appropriate set $F := H \cup \sigma(H)$. I asked the question of whether such a $\sigma$ must exist before: [Size of union of a set of subsets and its permutations](https://mathoverflow.net/questions/101886/size-of-union-of-a-set-of-subsets-and-its-permutations). That question is interesting in its own right, but this EH conjecture is really why I wanted an answer.

@Gerhard Paseman: it looks like the repository has $n < 100$, $k \leq 25$ and $t \leq 8$. With such small values it is not possible to detect the presence or absence of a log factor.

Some structural results on this part of the boolean lattice can be found with the search phrases "middle levels" and "boolean" together.  For example, see the references in arxiv.org/pdf/math.CO/0608485.pdf .

I should mention also that the Graham-Sloane construction means that the EH-conjecture for $k = \phi(n)$ and $t = k - 1$ has a positive answer when $\phi(n) = o(n)$, so $n/2$ is in fact the "hard case."

@Brendan McKay: Yes, thank you. It is corrected.

Does your description have $t$ and $l$ mixed up?  If so, please edit.

Have you checked out the La Jolla Covering Repository online?  That might give a nice suggestion of what to expect.  Gerhard "Ask Me About System Design" Paseman, 2013.02.15


Katz--Mazur for abelian varieties

Over $\mathbb Z$, there is a smooth DM stack $A_g$ classifying abelian varieties.

Over $\mathbb Z[\frac 1N]$, there is finite etale cover $A_g(N)_{\mathbb Z[\frac 1N]}\to A_g\otimes\mathbb Z[\frac 1N]$ classifying abelian varieties with $N$-level structure.

> Is there a finite proper map $A_g(N)\to A_g$ over $\mathbb Z$ with a _nice moduli interpretation_ , which recovers the usual notion of $N$-level structure over $\mathbb Z[\frac 1N]$?

**Remark:** In fact, I believe that such an $A_g(N)\to A_g$ is determined canonically from $A_g(N)_{\mathbb Z[\frac 1N]}\to A_g\otimes\mathbb Z[\frac 1N]$ by "normalizing in the function field", so the content of the question is really to have a good moduli interpretation for this canonical extension.

In more informal terms, I'm just asking:

> What is the "right" notion of $p$-level structure on an abelian variety in characteristic $p$?

The impression I get from reading in various places is that the answer to this is well-known, but I haven't been able to find a good reference. I don't have a specific type of level structure in mind. I'd just like to know what is known (and where to find it) for some common notions of level structure. **EDIT** : Actually, for level structures consisting of a choice of _subgroup(s)_ , there is a natural choice of moduli space over $\mathbb Z$ classifying abelian varieties with a choice of subgroup scheme(s) (or equivalently, isogenies) and this is valid in "bad" characteristics -- see, e.g. the end of [Chai](http://www.ams.org/mathscinet-getitem?mr=1124633). So, let me state my question for level structures given by choices of _point(s)_ , e.g. "a point of exact order $N$", or full level $N$ structure.

For elliptic curves, [Katz--Mazur](http://www.ams.org/mathscinet-getitem?mr=772569) showed that the answer is provided by the notion of a _Drinfeld level structure_ (first considered by [Drinfeld](http://www.ams.org/mathscinet-getitem?mr=384707)). For example, the "right" notion of a $p$-torsion point on an elliptic curve $E\to S$ (for _any_ scheme $S$) is a point $P:S\to E$ so that:

  1. $pP=1$.
  2. The Cartier divisor $\sum_{i=0}^{p-1}[iP]$ is a subgroup scheme.

As Katz--Mazur write, it is not clear how to generalize this notion to abelian varieties since points are no longer divisors.

Dear Daniel Litt:  Thanks for tracking it down; that is indeed the right paper.  It is also worth noting (in view of the question posed) that this paper provides a sense in which mere normalization in higher level (which doesn't have positive-dimensional fibers) is the "wrong" thing to consider, suggesting that perhaps the question posed about normalization is not a "useful" point of view for applications with $g > 1$.

The paper of Chai and Norman user76758 alludes to is likely this one:  jstor.org/stable/2374734

Note that $A_g$ classifies principally polarized abelian schemes (of relative dimension $g$); the polarization aspect comes along for free uniquely when $g=1$ but not otherwise, as you undoubtedly know.  I was once told that there is a paper by Chai and Norman (title I do not know) which shows in some sense that there isn't a good version of Drinfeld's idea beyond relative dimension 1.  Perhaps email your question to Chai (or Kai-Wen Lan) if you don't receive a satisfactory answer here.


Souslin trees and weakly compact cardinals

In [Souslin trees on the first inaccessible cardinal](https://mathoverflow.net/questions/143530/souslin-trees-on-the-first-inaccessible-cardinal) it is asked if it is consistent that there are no $\kappa-$Souslin trees at the least inaccessible cardinal $\kappa$.

In this question I would like to ask an apparently simpler question:

**Question.** Is it consistent that there are no $\kappa-$Souslin trees for some inaccessible not weakly compact cardinal $\kappa$?

**Remarks.** (A) If $\kappa$ is weakly compact, then the tree property holds at $\kappa$, in particular there are no $\kappa-$Souslin trees,

(B) If $V=L,$ then $\kappa$ is weakly compact iff there are no $\kappa-$Souslin trees.

(C) $\Diamond_\kappa+GCH$ does not imply the existence of a $\kappa-$Souslin tree, for $\kappa$ inaccessible. For example if $\kappa$ is ineffable and $GCH$ holds, then $\Diamond_\kappa+GCH$ holds and there are no $\kappa-$Souslin trees.

Its about your comment above (since in the situation of your comment $S$ is non reflecting at every $\alpha$, as witnessed by $\text{acc }C_\alpha$).

Your comment is about my above comment for a weaker sufficient condition?

@Mohammad: you don't actually need a $\square (\kappa )$ sequence, all you need is a non reflecting stationary set $S$ with diamond on it.

@Yair: Your last comment seems like a familiar topic all of a sudden! :-)

If there is a $\kappa$-closed $\kappa$ saturated ideal on $\kappa$ and $\kappa$ is not measurable then $\kappa$ is inaccessible and there is a $\kappa$-Suslin tree (by splitting the positive sets in a partial binary tree). Actually, in this case every normal $\kappa$-tree contains a Suslin subtree.

Does the existence of a $\kappa-$Souslin tree for $\kappa$ inaccessible have any effects on saturated ideals on $\kappa$ or something else? Maybe this way we can reduce the problem to something that we know more about it.

I think it does not give the answer. Note that a sufficient condition for having a $\kappa-$Souslin tree for $\kappa$ inaccessible, is square-with-built-in-diamond at $\kappa$, as is implicit in [Sh:624]. A weaker sufficient conditions is, e.g.,: there exists a $\square(\kappa)-$sequence $< C_\alpha: \alpha<\kappa >$, and a stationary set $S$ in $\kappa$, which is disjoint from $acc(C_\alpha)$ for all $\alpha<\kappa$, and $\Diamond(S)$ holds. To force the failure of these principles, we need more than weakly compact cardinals.

What happens if we start with a weakly compact cardinal $\kappa$ in $V$, and then force to kill the Mahloness of $\kappa$ by adding club $C\subset\kappa$ avoiding the regular cardinals below $\kappa$? Does this create a $\kappa$-Suslin tree?


Maximum automorphism group for a 3-connected cubic graph

The following arose as a side issue in a project on graph reconstruction.

**Problem:** Let $a(n)$ be the greatest order of the automorphism group of a 3-connected cubic graph with $n$ vertices. Find a good upper bound on $a(n)$.

There is a [paper of Opstall and Veliche](http://arxiv.org/abs/math/0608645) that finds the maximum over all cubic graphs, but the maximum occurs for graphs very far from being 3-connected.

I foolishly **conjecture** : for $n\ge 16$, $a(n) < n 2^{n/4}$.

When $n$ is a multiple of 4 there is a vertex-transitive cubic graph achieving half the conjectured bound, so if true the bound is pretty sharp.

The maximum does not always occur for vertex-transitive graphs. There is no vertex-transitive counterexample to the conjecture with less than 1000 vertices.

Any bound for which the exponential part is $2^{cn}$ for $c<\frac12$ is potentially useful.

**Added:** As verret pointed out in a comment, a [paper of Potočnik, Spiga and Verret](http://arxiv.org/abs/1010.2546), together with the computation I mentioned, establishes the conjecture for vertex-transitive graphs, so the remaining problem is whether one can do better for non-transitive graphs. For 20, and all odd multiples of 2 vertices from 18 to at least 998 (but not for 4-16 or 24 vertices) the graph achieving the maximum is not vertex-transitive.

In the cubic vertex-transitive case and n twice an odd number, it immediately follows from the same paper that you get a polynomial rather than exponential upper bound. For example Corollary 4 yields that, for large enough n, we have |G|<n^2. This is not best possible, but is not far off, at least for some values of n. If you need more precise estimates, I can show you a few more references that deal with this. (By the way, the link in your "added" section has a typo.)

@verret: Any idea what the maximum is for transitive cubic graphs of order an odd multiple of 2?

Great! I vaguely remembered such a paper but hadn't managed to find it today.

The family of vertex-transitive graphs you have in mind are in fact best possible among large enough cubic vertex-transitive graphs. (n=100 or so should already suffice.) This is shown in the paper : "Bounding the order of the vertex-stabiliser in 3-valent vertex-transitive and 4-valent arc-transitive graphs", arxiv.org/abs/1010.2546.  Therefore, a counter-example to your conjecture will necessarily be not vertex-transitive.


Actions on ℍⁿ generated  by torsion elements

Let $n$ be a large integer.

I am looking for a cocompact properly discontinuous isometric action on $n$-dimensional Lobachevky space which is generated by elements of finite order.

Or equivalently, I need a cocompact properly discontinuous isometric action $\Gamma\curvearrowright\mathbb{H}^n$ such that $\mathbb{H}^n/\Gamma$ is simply connected, here $\mathbb{H}^n/\Gamma$ stays for quotient space (NOT for orbifold).

**Comments.**

  * I am aware that for large $n$ these is no cocompact action generated by reflections (Vinberg, 1984).
  * Consider the group of matrices with integer coefficients from $\mathbb{Q}[\sqrt{5}]$ which preserve the quadratic form $$\tfrac{1+\sqrt{5}}{2}\cdot x_0^2-x_1^2-\cdots-x_n^2.$$ This gives cocompact properly discontinuous isometric action, say $\Gamma\curvearrowright\mathbb{H}^n$. (For $n=2$, the group $\Gamma$ contains the Coxeter group of regular right-angled pentagon.)
_I would be very happy if the subgroup generated by the elements of finite order in $\Gamma$ would have finite index. Or, equivalently, if_ $$|\pi_1(\mathbb{H}^n/\Gamma)|<\infty.$$

**P.S.** It seems that examples of such actions are not known. (In addition to Agol's comment, I've got a letter from Vinberg stating this.)

@ Anton: I mean he computed the reflection subgroup for $n$ such that the subgroup is finitely generated. The full group is then generated by the reflection subgroup together with the symmetries of its fundamental domain, so in some sense he computed $O(n,1;Z)$. All that I'm pointing out is that these are actually generated by involutions.

@Agol; What do you mean by "up to the dimensions that Vinberg had computed them"? Did Vinberg gave a list of generators of $O(n,1;\mathbb{Z})$ for small $n$?

At one point, I checked that all of the groups $O(n,1;Z)$ are generated by rank one involutions (reflections through planes and points), up to the dimensions that Vinberg had computed them. In fact, the even unimodular Lorentzian lattice in dimension 25 is also generated by such involutions (Daniel Allcock checked this using properties of the Leech lattice and Conway groups). I don't see why this might not be true in all dimensions, but I would have no clue how to approach proving it. Are you interested in non-uniform lattices too?

I suspect this is an open question, but I'm not aware of a reference explicitly stating it.


Does a symplectic group act on a tensor power of a spin representation?

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}$More specifically, let $S_k$ be the spin representation of $\Spin(2k+1)$. Then is there are action of $\Sp(2r-2)$ on $\bigotimes^{2r}S_k$ which commutes with the action of $\Spin(2k+1)$?

I am not asking for a duality. That is, I do not require that each isotypic vector space is an irreducible representation of $\Sp(2r-2)$.

The motivation is that the space of invariant tensors in $\bigotimes^{2r}S_k$ looks like the irreducible representation of $\Sp(2r-2)$ with highest weight $k\omega_{r-1}$. For instance, for $k=1$, both vector spaces have dimension given by the Catalan numbers $C_r = \frac{1}{r+1}\binom{2r}{r}$. More generally, for arbitrary $k$ the dimension of both of these spaces is $$ \prod_{1\leq i \leq j \leq r-1} \frac{i+j+2k}{i+j} $$ which is the number of so-called "_$k$ -fans of nested Dyck paths of semilength $r$_" (see, e.g., Section 3.1.6 of [Federico Ardila - Algebraic and geometric methods in enumerative combinatorics](https://arxiv.org/abs/1409.2562)).

(For an explanation of a similar, well-known numerical "coincidence," but for $SL_n$ and $GL_m$, using Howe duality, see: sbseminar.wordpress.com/2007/08/10/the-ubiquity-of-howe-duality/…)

As mentioned in the bounty message, I would be interested in any algebraic explanation of the numerical coincidence of the dimensions of the spaces discussed in Bruce's question. Does some version of Howe duality apply here?


Antichains of Cardinals in ZF Without Choice...

With the Axiom of Choice, the cardinals form a nice linearly ordered "set". In the absence of the Axiom of Choice, the cardinals form a partially ordered "set". Broadly, I am wondering what properties these posets can have.

A specific question I am interested in is the following.

**Is there, for each (infinite) subset $S \subseteq \mathbb{N}$ containing $1$ and not containing $0$, a model of ZF in which there is a maximal antichain of cardinals of size $n$ if and only if $n \in S$?**

Though I have a mild interest in knowing how such a model would be constructed (assuming a positive answer), my primary interest is in knowing that it is (is not) possible. Hence, if such a result exists in the literature, a citation would be all that I ask for.

Joel: My motivation for requiring infinite $S$ was the initial context: How complicated can the theory of cardinals in a model of $ZF$ (in the language of posets) be?  That said, now my interest goes beyond the initial context, so I am curious about sets such as $\{1,2\}$.  Thanks for elaborating!  <p>  For comparison, the theory of the cardinals of a model of $ZFC$ (as a poset or under addition) is decidable.  <p>  Note that, so the underlying universe is a set rather than a proper class, I am restricting the "model" of set theory to any downward closed subset of cardinals within the "model".

Asher, suppose that $\{\a\}$ was a maximal antichain (again, I'm ignoring finite sets) then $a$ is comparable with its Hartog number and thus well orderable. So everyone below $a$ can be well ordered as well. It's clear that $W_{\aleph_0}$ implies $1\in S$; on the other hand, if $W_{\aleph_0}$ does not hold, pick $a$ to be a D-finite set, every cardinal above it cannot be well ordered and thus incomparable with its Hartog number. Therefore no one above $a$ can be a maximal antichain of size $1$.

Asher, the connection is that I believe that whenever there is an antichain of size 2, then one can make larger finite antichains, but it wasn't clear to me how to make larger finite maximal antichains. But if the answer to my question was that it always held, then one could not attain $S=\{1,2}$, for example. Meanwhile, you had asked about infinite $S$, but it isn't clear to me why you insist on that.

Joel: Naively, I would expect models of ZF with finite antichains having no finite maximal antichain extension.  Is there a connection between this phenomena and the primary question?

Trevor: Thanks, I was not even certain that (nontrivial) finite maximal antichains were consistent with ZF.  Asaf and Joel: Indeed, the only reason I required $0 \not \in S$ and $1 \in S$ was to avoid such trivialities with the finite cardinals.  Excepting them, why is $1 \in S$ if and only if $W_{\aleph_0}$ holds?  Naively, it seems plausible to have a set an infinite set $S$ with no countable subset, with some other cardinal (with countable subsets and size $S$ subsets) comparable to all cardinals.

Asaf, I don't think your embedding fact settles the question I raise in my comment, since perhaps the finite antichain can be extended to a maximal antichain that does not lie in the copy of the poset in the cardinals.

@Joel: Sure, if you allow finite cardinals then $1\in S$ is a requirement. I just don't think it is very interesting that way. Considering only infinite cardinals, on the other hand... :-) As for your second question, I'd not think so. Jech has a theorem [Axiom of Choice, Thm 11.1] which embeds every poset into the cardinals, take a poset with only infinite maximal antichains, then its embedding will produce a model in which there are finite antichains that cannot be extended into finite maximal antichains.

Do we have any reason to hope that every finite antichain of cardinals can be extended to a finite maximal antichain?

Asaf, I don't think what you say is correct. We need $1\in S$ always because, for example,  {7} is a maximal antichain, since every set either has fewer than 7 elements or at least 7. Similarly for {0} or {k} for any finite k.

To answer a very small part too, $1\in S$ if and only if $W_{\aleph_0}$ holds, that is every infinite set has a countable subset.

To answer a very small part of your question, $\lbrace \omega_1, \mathbb{R} \rbrace$ is a maximal antichain in all known natural models of the Axiom of Determinacy.  See A.E. Caicedo and R. Ketchersid, \emph{A trichotomy theorem in natural models of AD}, available online at sites.google.com/site/richardketchersid/home/research/adcf-final.pdf.


"extended TQFT" versus "TQFT with defects"

There are two ways in which higher categories appear in topological field theory: in extended TFTs and in TFTs with defects. How are these appearances related?

According to the Atiyah-Segal axioms, a d-dimensional TFT is a symmetric monoidal functor

\begin{equation*}\text{Bord}_d\rightarrow\mathcal{C}\end{equation*}

where the target category $\mathcal{C}$ is a symmetric monoidal category, typically $\text{Vect}$. The objects of $\text{Bord}_d$ are closed $(d-1)$-manifolds and map to the vector spaces of $\text{Vect}$ under the TFT. The morphisms are d-dimensional bordisms of the closed $(d-1)$-manifolds and map to the linear maps of $\text{Vect}$.

This description can be "extended down" to define a d-dimensional n-extended TFT as a symmetric monoidal $n$-functor

\begin{equation*}\text{Bord}_d^n\rightarrow\mathcal{C}\end{equation*}

where $\mathcal{C}$ is now an symmetric monoidal $n$-category. The objects of $\text{Bord}_d^n$ are $(d-n)$-manifolds (for a "fully extended" TFT, _i.e._ $n=d$, the bordism objects are points) and map to the objects of $\mathcal{C}$, which can be thought of as $(n-1)$-categories. One should really specify $\mathcal{C}$ as some extension of $\text{Vect}$, but I am ignoring this technicality by thinking of the n-category of all (small) $(n-1)$-categories. More generally, $k$-morphisms are $(d-n+k)$-dimensional bordisms between $(d-n+k-1)$-manifolds and map to $(n+k-1)$-categories. There are also technicalities surrounding how one manages "manifolds with corners" in $\text{Bord}_d^n$, but allow me to gloss over them here.

Meanwhile, one uses higher categories to decorate manifolds with defects. In this picture, $k$-dimensional manifolds are decorated with $k$-dimensional (extended?) TFTs called "$k$-defects." Lower dimensional submanifolds can be decorated with "defects within defects." The $k$-defect assigned to a $k$-dimensional "boundary" between two $(k+1)$-dimensional regions, each with attached $(k+1)$-defects, amounts to a morphism of $(k+1)$-defects. Fusion of defects and sub-defects endows the set of $k$-defects with the structure of a $k$-category. (For details, see <http://arxiv.org/abs/1002.0385>.) If we again imagine the $k$-category of (small) $(k-1)$-categories, we can understand decoration as an assignment of $(k-1)$-categories to $k$-manifolds.

This construction of TFT with defects feels "upside down" compared to extended TFT: $k$-manifolds are decorated by $(k-1)$-categories and map to $(d-k-1)$-categories under the TFT functor. I realize that decoration is not a functor from a bordism category, but is it a functor in some other sense (from some category where higher degree morphisms are lower dimensional submanifolds)? Defects may have an interpretation as inserted operators (Wilson loops, surface operators, etc); can these two formalisms be combined to compute path integrals with operator insertions in an extended theory? In general, I am curious about the relation between how higher categories are used to define an extended TFT and how they are used to characterize TFTs with defects.

Here is a particular problem from <http://arxiv.org/abs/1309.1489>. Consider a fully extended 4D TFT. The TFT functor assigns $1$-categories to Riemann surfaces $\Sigma$. Meanwhile, in the defect description, 2D TFTs form a $2$-category. Restricting ourselves to a single object leaves us with a symmetric monoidal category of $1$-dimensional domain walls (essentially boundary conditions) of the 2D theory. Is this category of boundary conditions identified with the category assigned by the TFT functor (as is claimed)? It seems that this identification is only possible since $\dim\Sigma=2=\text{codim }\Sigma$ in dimension four and does not reflect a general relation between extended TFT and TFT with defects.

One approach to this is discussed in detail at arxiv.org/abs/1009.5025 .  See sections 2 and 6, and the remark about defects near the start of section 6.7.

Perhaps the following question needs to be answered first: a defect theory (an object of the k-category of defects that decorates a k-fold) is itself a TFT. How is this object realized as a TFT? In particular, given a d-dim theory Bord->C, how does one characterize the (d-1)-dim TFTs on boundaries (or domain walls)? And how are these boundary theories related to the defect decoration?

@KevinWalker Following your hint, the upside-down can be turned rightside-up, and the degrees and dimensions work out; however, I am still unclear about how to identify the data of the TFT with the defect data. For a d-dim theory Bord->C, the defects on a k-fold M form a k-category, the objects (defect theories) of which might be interpreted as k-morphisms of C. Endomorphisms in C of a defect theory form a (n-k-1)-category of boundary conditions (or an (n-k)-category with one object) of the defect theory. Somehow this might be identified with the (n-k-1)-category assigned to M by the TFT.

Forget about TQFTs for a moment and consider an $n$-category $C$.  To a $k$-morphism $x$ of $C$ we can, of course, associate the $k$-morphism $x$ itself.  We can also associate to $x$ the $(n-k)$-category of endomorphisms of $x$ in $C$.  The "upside-down-ness" you are noticing is closely related to this familiar example.

Yes, representations of cobordisms with singularities encode TFTs with boundaries and defects. For "pre-quantum" field theory this is discussed in sections 3.9.14.4 to 3.9.14.6 of arxiv.org/abs/1310.7930 (improved version in preparation at ncatlab.org/schreiber/show/Local+prequantum+field+theory). For quantization of this: arxiv.org/abs/1402.7041 . This is based on discussion with Domenico Fiorenza and Alessandro Valentino that recently appeared as arxiv.org/abs/1409.5723

I think you want the cobordism hypothesis with singularities (Theorem 4.3.11 in arxiv.org/abs/0905.0465).


The Grothendieck Ring of Higher Stacks

The Grothendieck ring of varieties is defined to be the free abelian group spanned by isomorphism classes of varieties modulo the _cut & paste_ (or scissor) relations, which say that $[X] = [U] + [Y]$, for any closed $Y \subset X$ and where $U$ is the open complement.

Now, we can equally speak of a Grothendieck ring of schemes, or algebraic spaces, and they all turn out to be isomorphic via the obvious inclusions. (for the scheme case one sees this by noticing that $[X] = [X_\text{red}] - 0$) There is also a Grothendieck ring of Artin stacks (everything here is of finite type over a field of characteristic zero, possibly algebraically closed, and with affine diagonal). This ring turns out to be a localisation of the previous one.

My question is: what happens if we include _higher_ stacks?

I think there is work by Toën on the Grothendieck ring for derived stacks, but here I'm only asking about higher (underived) stacks. Do we have a similar phenomenon to the case of varieties VS schemes? In the sense that $[X] = [\pi_0(X)] - 0$, where $\pi_0$ of a derived stack is its underived truncation?

are there any interesting phenomena you expect to happen? or just for fun?


Vector Bundles on the Moduli Stack of Elliptic Curves

As is well known, there is classification of line bundles on the moduli stack of elliptic curves over a nearly arbitrary base scheme in the paper _The Picard group of $M_{1,1}$_ by Fulton and Olsson: every line bundle is isomorphic to a tensor power of the line bundle of differentials $\omega$ and $\omega^{12}$ is trivial.

I am now interested in a similar classification scheme for higher dimensional vector bundles (i.e. etale-locally free quasi-coherent sheaves of finite rank). I am especially interested in the prime 3, so you may assume, 2 is inverted, or even that we work over $\mathbb{Z}_{(3)}$. I found really very little in the literature on these questions. I know only of two strategies to approach the topic:

1) I think, I can prove that every vector bundle $E$ on the moduli stack over $\mathbb{Z}$ localized at $p$ for $p>2$ is an extension of the form $0\to L \to E \to F$ where $L$ is a line bundle and $F$ a vector bundle of one dimension smaller than $E$ (this may hold also for $p=2$, but I haven't checked). In a paper of Tilman Bauer ([Computation of the homotopy of the spectrum tmf](http://arxiv.org/abs/math/0311328)) Ext groups of the so called Weierstraß Hopf algebroid are computed, which should amount to a computation of the Ext groups of the line bundles on the moduli stack of elliptic curves if one inverts $\Delta$. It follows than that every vector bundle on the moduli stack of elliptic curves over $\mathbb{Z}_{(p)}$ is isomorphic to a sum of line bundles for $p>3$ if I have not made a mistake. But for $p=3$, there are many non-trivial Ext groups and I did not manage to see which of the occuring vector bundles are isomorphic.

2) One can try to find explicit examples of a non-trivial higher dimensional vector bundles. A candidate was suggested to me by M. Rapoport: for every elliptic curve $E$ over a base scheme $S$ we have an universal extension of $E$ by a vector bundle. Take the Lie algebra of this extension and we get a canonical vector bundle over $S$. As explained in the book _Universal Extensions and One Dimensional Crystalline Cohomology_ by Mazur and Messing, this is isomorphic to the deRham cohomology of $E$. This vector bundle is an extension of $\omega$ and $\omega^{-1}$ and lies in a non-trivial Ext group. But I don't know how to show that this bundle is non-trivial.

I should add that I am more a topologist than an algebraic geometer and stand not really on firm ground in this topic. I would be thankful for any comment on the two strategies or anything else concerning a possible classification scheme.

The fact that vector bundles on $\mathcal M_{1,1}$ split as sums of line bundles in characteristic larger than 3 can also be seen by the classical description of $\mathcal M_{1,1}$ as an open substack of the weighted projective stack $\mathbb P(4,6)$, coming from the Weierstrass form. Any locally free sheaf on $\mathcal M_{1,1}$ extends to a reflexive sheaf on $\mathbb P(4,6)$, which is locally free, because $\mathbb P(4,6)$ is regular of dimension 1. It is not hard to prove that any locally free sheaf on a weighted projective stack $\mathbb P(m,n)$ splits as a direct sum of line bundles.

@Charles: Good point, I should have noticed the link with the non-liftability to char. 0 for the Hasse invariant when $p = 3$ (and its square when $p=2$). Thanks.

@Tyler Lawson: you use the covering $\mathcal{M}(4) \to \mathcal{M}$ and some flatness criterion. I can send you a pdf file with some details if you like.

You can also deduce that $Ext^1(O, \omega^2)$ is non-trivial from the fact that there are non-trivial "mod p" modular forms of weight 2 when p=2 or 3; these are computed (for instance) in Deligne's note on "formulas, after Tate" in Antwerp 4.

@Brian: The computation of the Ext-groups of tensor powers of $\omega$ on this moduli stack is written up in the Bauer paper that was linked to under part (1).

Since the coarse space is affine and the automorphism groups of the geometric points have order a power of 2 times a power of 3, and the  Ext-group (or $\omega^{-1}$ by $\omega$, to be precise) is identified with degree-1 coherent cohomology (of $\omega^2$), its nontriviality is also not visible over $\mathbf{Z}[1/6]$. So can you also briefly indicate how you know it is nontrivial?

How did you prove that every vector bundle is an extension?


Combinatorial identity involving the Coxeter numbers of root systems

The setup is:

$R$ = irreducible (reduced) root system;

$D$ = connected Dynkin diagram of $R$, with nodes numbered $1,2,...,r$;

$\hat D$ = extended Dynkin diagram, nodes numbered $0,1,2,...,r$;

$\alpha_k$ = $k^{th}$ simple root ($1\le k\le r$); $\alpha_0$ = -(highest root);

label $n_k$ ($1\le k\le n$) = multiplicity of root k in the highest root; $n_0=1$;

$n^\vee_k$ = dual label = multiplicity of co-root $\alpha_k^\vee$ in the highest _short_ co-root;

$\quad\quad=n_k$ ($\alpha_k$ long) or $n_k/c$ ($\alpha_k$ short), where $c=2$ (types $B_n,C_n,F_4$) or $3$ (type $G_2$) (also $n_0^\vee=1$);

$h(D)$ = Coxeter number of D = $\sum_{k=0}^rn_k$ (for example $h(E_8)=30$);

Now, delete node $k$ from $\hat D$ where $n_k>1$. Write $\hat D-\\{k\\}$ as a union of connected Dynkin diagrams $D_1,\dots, D_s$. Let $h_1,\dots, h_s$ be the Coxeter numbers of $D_1,\dots, D_s$.

**Lemma** : If $\alpha_k$ is long:

$$ \sum_{i=1}^{s}\frac1{h_i}\sum_{j\in D_i}n_j=n_k $$ If $\alpha_k$ is short: $$ \sum_{i=1}^s\frac1{h_i}\sum_{j\in D_i}n_j^\vee=n_k^\vee $$

Note that the sum over $D_i$ involves the labels $n_j$ coming from $D$, which have no obvious relation to those coming from $D_i$. On the other hand $h_i$ is the Coxeter number for $D_i$, so this is a mix of data from $D$ and $\\{D_i\\}$.

The proof is case-by-case.

> **Question** : is there a conceptual proof of the Lemma?

**Examples:** (Bourbaki numbering)

1) $D=E_8$, $k=2$, $n_2=3$, $\hat D-\\{2\\}$ = type $A_8$, $s=1$, $h(A_8)=9$: $$ \frac19(2+4+6+5+4+3+2+1)=3=n_2 $$ (As mentioned above, the numbers $2,4,6,5,4,3,2,1$ come from $E_8$, not $A_8$.)

2) $D=E_8$, $k=4$, $n_4=6$, $\hat D-\\{6\\}=A1+A2+A5$, $s=3$, $h(A_i)=i+1$: $$ \frac12(3) + \frac13(2+4) + \frac16(5+4+3+2+1) = \frac32 + 2 + \frac{15}6 = 6 = n_4 $$ Note that the terms in the sum are not all integers.

3) $D=E_8$, $k=8$, $n_8=2$, $\hat D-\\{8\\}=E_7+A_1$, $s=2$, $h(E_7)=18$, $h(A_1)=2$: $$ \frac1{18}(2+3+4+6+5+4+3)+\frac12(1)=\frac{27}{18}+\frac12=2=n_8 $$

4) Here is an example involving $n_k^\vee$: $D=C_8$, $k=3$, $\alpha_k$ is short, $n_k=2$, $n^\vee_k=1$ (in fact all $n^\vee_j=1$). $D-\\{3\\}=C_3+C_5$, $h(C_n)=2n$: $$ \frac16(1+1+1) + \frac1{10}(1+1+1+1+1)=\frac12+\frac12=1=n_3^\vee $$

5) By popular demand, $G_2$:

a) $k=2$ (long root), $n_2=2$, $n_2^\vee=2$, $\hat D-\\{2\\}=A_1+A_1$. Note that $n_1=3,n_1^\vee=1$.

$$ \frac12(1)+\frac12(1)=1=n_1^\vee $$

b) $k=1$ (short root), $n_1=3$, $n_1^\vee=1$, $\hat D-\\{3\\}=A_2$, $h(A_2)=3$. Note that $n_2=n_2^\vee=2, n_0=n_0^\vee=1$: $$ \frac13(2+1)=1=n_1^\vee $$

**Remark:** Write $H$ for the subgroup of $G$ corresponding to $\hat D-\\{k\\}$. This identity plays a role in some computations related to embedding elliptic elements of the Weyl group of $H$ in those of $G$ (the Coxeter element is an example).

@Jeff: One small nit-pick about your formulation is that the Coxeter number $h$ only depends on the Weyl group, not on $D$; for example, Lie types $B_r, C_r$ yield the same $h$.    Coxeter originally found the concept in the framework of finite real groups generated by reflections, which is independent of Lie theory and root systems but of course has a lot of overlap.     Your question, however, definitely comes from Lie theory.

Maybe a proof of this result can be furnished via Vinberg's classification of Dynkin diagrams in terms of the existence of a subadditive function? I think the coefficients $n_k$ you are describing are just the coefficients of the unique (up to scale) subadditive function...

This is an intriguing observation, which may not have a convincing explanation outside representation theory or the combinatorial geometry of affine Weyl groups.   It might for example come from Langlands duality in some context not so visible in classical Lie theory   (The tag weyl-group and/or lie-algebra might be appropriate.   Also, it makes the dualization process clearer if you include the case $G_2$.)


How many Hecke operators span the Hecke algebra?

This is a generalisation of my [earlier question](https://mathoverflow.net/questions/42809/how-many-hecke-operators-span-the-level-1-hecke-algebra) about generators for the level 1 Hecke algebra.

Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbb{Z})$, and $k \ge 1$ an integer. There's an explicit bound $S$ (the "Sturm bound") known such that if $f \in M_k(\Gamma)$ (the space of modular forms of weight $k$ and level $\Gamma$) satisfies $a_n(f) = 0$ for $0 \le n \le S$, then $f = 0$, where $a_n(f)$ denote the coefficients of the $q$-expansion of $f$ (at the cusp $\infty$).

Since $M_k(\Gamma)$ is finite-dimensional and doesn't contain any nonzero constant functions, there must actually exist some $S'$ such that if $a_n(f) = 0$ for $1 \le n \le S'$ (but with no assumption on $a_0(f)$), then in fact $f = 0$. Can one give an effective upper bound on $S'$ (in terms of $k$ and the index $[\operatorname{SL}_2(\mathbb{Z}) : \Gamma]$)?

I'd be interested to know the answer to this even for the special case $\Gamma = \Gamma_0(N)$ or $\Gamma_1(N)$. For these groups it is equivalent to asking that the Hecke algebra acting on $M_k(\Gamma)$ is spanned by the operators $T_1, \dots, T_{S'}$ (hence the title of the question).

Joel: see e.g. wstein.org/books/modform/modform/newforms.html#sturm-s-theorem.

David, do you know a good reference for sturm bounds?

A data point: for $\Gamma_0(p)$ in weight 4, $p$ prime, the Sturm bound $S$ is $(p + 1)/3$, but the Eisenstein series $E_4(pz)$ has $a_n = 0$ for $1 \le n < p$, so $S'$ must be at least $p$.

Julian, if you read the question you'll see that I've carefully explained why what I'm looking for is not the same as the Sturm bound.

Hmmm... isn't it called a "Sturm bound"?

I've decided to undelete this old question -- I did delete it because I wanted to set it as a student project, but the student didn't solve it.

@David: if you don't want anyone to answer the question you should delete it.


monomorphisms and epimorphisms of local rings

I want to understand the structure of monomorphisms/epimorphisms in the category of local rings (with local homomorphisms), or dually in the category of local schemes. Let $LR$ denote this category.

  * Every monomorphism is injective.

Proof: Let $R \to S$ be a monomorphism and $a,b \in R$ mapped to the same element. Regard $a,b$ as ring maps $R[t] \to R$. Let $\mathfrak{p},\mathfrak{q}$ the preimages of the maximal ideal $\mathfrak{m}$. Then we get two morphisms of local rings $R[t]_{\mathfrak{p}} \times_S R[t]_{\mathfrak{q}} \to R$ which agree when composed with $R \to S$, thus they are equal. Evaluating at $(t,t)$ gives $a=b$. There is a shorter proof which uses $R[t]_{(t)+\mathfrak{m}}$.

  2. Is every monomorpism regular? (i.e. the equalizer of some morphism pair)

  3. Is every monomorphism extremal? (i.e. $i=ep$ with $p$ epi implies $p$ iso)

  4. Is every epimorphism, which is also injective, and isomorphism?

  5. Is every epimorphism surjective?

Clearly we have 4 => 3 <=> 2 and 1 => 2. Since in $LR$ every morphism can be factorized as $ep$ where $p$ is surjective and $e$ is injective, we also have 3 => 4.

In the category of rings, 1,2,3,4 are false (see [this discussion](https://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like)). However, the counterexamples don't carry over to $LR$. In the "category theory bible" (joy of cats!) you can find nothing about local rings. The questions are partially motivated by Anton's problem about [coequalizers](https://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective) in the category of (local) schemes.

In Lazards article about flat epimorphisms ([pdf link](http://archive.numdam.org/ARCHIVE/SAC/SAC_1967-1968__2_/SAC_1)), there is an example (1.6) in which 4 fails. I have not understood it yet. What is $Z$ there? Perhaps $B$ is a localization of $k[T,Z]$ and not $k[T]$? But I don't believe that $f$ is well-defined.

If the answers turn out to be "no":

  * Is there a nice description for the regular monomorphisms? What about field extensions?

If $L/K$ is a finite galois extension, then $K \to L$ is a regular monomorphism iff $L/K$ is cyclic (see Brians comments).

Oh yes, this is the example. I'll take a look at Paul Balmer's explanation.

The link to Lazard's article seems to be broken; is this (mathoverflow.net/questions/19282/…) the counterexample you were referring to?  If so, I don't see why $C$ maps to $D$ either.

@Martin: here's proof of non-existence when Gal. group G for K/k  is non-cyclic. Suppose $s,t:K \rightrightarrows A$ is equalizer pair for $k \rightarrow K$ in category $LR$. Then $A$ is naturally a $K \otimes_k K$-algebra, and by locality of $A$ and structure of $K \otimes_k K$ (using $K/k$ is Galois), the algebra structure factors through projection to a factor field. Thus, $t=s \circ g$ for some $g \in G$. Hence, by univ. property of $A$ and inj. of $s$, any $f:B \rightarrow K$ s.t. $g \circ f=f$ factors through $k$. Thus, $K^g=k$, so $G=\langle g \rangle$. Contradiction.

I'm thinking about that, but it's pretty difficult to start. Even for nice small extensions ...

@Martin: sorry, for #1 I got disoriented by Spec and was thinking of co-equalizers. But how about a Galois extension whose Galois group doesn't admit a pair of generators? By restricting to local rings can you find an equalizer pair description for that?

If $K/k$ is a cyclic galois extension with generator $\sigma$ of the galois group, then $k \to K$ is regular since it is the equalizer of $id,\sigma : K \to K$.

1 fails for quadratic Galois ext'n of fields $K/k$ (due to non-locality of $K \otimes_k K$ and how its isom with $K \times K$ is defined). In category of local noetherian rings, map from ring to completion is an epimorphism (due to Krull intersection theorem, hence the noetherian condition), and in non-complete case provides counterexamples to 2, 3, 4. Doesn't show 2,3,4 fail in $LR$, but kills a reason to care, since affirmative results then seem useless. Category theory is like point-set topology: can get wrapped up in "weird examples"; not necessarily a good use of time.


Does every connected set that is not a line segment cross some dyadic square?

A _dyadic square_ is a subset of $R^2$ of the form $x + 2^{-n} [0,1]^2$ with $x \in 2^{-m} Z^2$, for integers $m,n \geq 0$. We say that a set $A$ _crosses_ a square $S$ if there exists a connected subset of $A \cap S$ which intersects two opposite sides of the square $S$. Clearly, the 45 degree line $\\{ (\pi + t, t) : t \in R \\}$ does not cross any dyadic square. Does every non-trivial closed, connected set that is not a line segment cross some dyadic square?

Since you @mirko don't mind rephrasing-reposting the question I'd repost as a dimension/measure-theoretical question on curves in the plane, someone will spit some result at you really quick if you're convinced it's true.  But I think it's false, you should be able to get good control of limiting behavior with a rectilinear construction and finish using basic results about direct limits of continua.

Really, some zero-span curve can't be constructed with just a uniform direct limit by oscillating off corners in Koch curve style with rectilinear paths by like a scaled factor of $> \frac{1}{4^n}$?  In my head it works, the argument should go like 'what's the biggest dyadic square it crosses' and then induct.  Has anyone tried to write it down?  In my head it works.  If there is no example then a dimension theorist probably can handwave why, you should tag dimension theory and GMT.

@IanAgol No, I don't have a proof for the pseudoarc, though this is usually what I think of, as a totally path-disconnected continuum (i.e. path-components are points). If $C$ is say hereditarily indecomposable plane continuum I have considered (something elementary), take $\frac1n$-neighborhood (union of all $\frac1n$-balls with centers in $C$) this is path-connected, try to use it somehow (then let $n\to\infty$), but haven't had much success. Your comment reminded me that the pseudoarc is arc-like(chainable), so perhaps use this to modify the path-connected case, but I have to think about it

@Mirko: do you think that you can prove it for pseudo-arcs?

I believe I have a positive answer under the additional assumption that the set in question is path-connected. I posted a related question which eventually got a bit long, but see the definition of hsb plane continuum there, at the beginning, along with the Result after the "Edit May 29-31, 2023" here: mathoverflow.net/q/446092 . For completeness, here are a couple of other, related questions that I posted recently mathoverflow.net/q/446317 and math.stackexchange.com/q/4694709 . (I don't feel like posting my "path-connected" solution here, it is only a special case.)

Is it pure curiosity? In other words --- do you have some motivation for this question?

Hello Kevin, this is a nice question and seems difficult, though I came up with some ideas that I need to develop. Do you know if it is an open question, and if anyone has done work on it, any references? Did you find it somewhere in a book or a paper, or did you come up with it on your own? I am inclined to believe that there is no example, though I changed my mind about it several times yesterday and today. I will email a colleague whose area is closer to this question, to see if he knows anything or has any advice (just sending him the link to your question as posted here). Thank you,

Hi Gunter, I pointed out that there were two mistakes in this answer. A day later it was removed by somebody.

There used to be an interesting answer posted here. Where is it? Was it wrong?

Gunter Rote, I do not mean the condition $m\leq n$. E.g. the square $[1/8,5/8]\times [1/4,3/4]$ is a dyadic square with my definition. Perhaps you would like to call such a square a translated dyadic square.

I think that every path-connected component of such a non-crossing set should be a line segment.

I would have expected the condition $m\le n$. Is that what you mean? Or is this another version of the problem? Every line segment that is not $\pm45$ degrees, and every sufficiently smooth curve that is not a $\pm45$ degree line segment must cross a dyadic square (even with the strong definition of dyadic square). So if there are counterexamples they are pretty pathological.


Rational equivalence of smooth closed manifolds

All spaces below will be assumed simply connected. A continuous map is a _rational equivalence_ if it induces an isomorphism of the rational homology groups. Two spaces are _rationally equivalent_ if they can be connected by a sequence of rational equivalences (not necessarily all going in the same direction). Now suppose $X$ and $Y$ are rationally equivalent smooth compact manifolds without boundary. Can one choose a sequence of rational equivalences from $X$ to $Y$ that only passes through smooth compact manifolds of dimension $\dim X=\dim Y$?

Note: if $X$ and $Y$ are CW-complexes, or more generally, compactly generated Hausdorff spaces, then $X$ is rationally equivalent to $Y$ if and only if the $\mathbb{Q}$-localizations $X_\mathbb{Q}$ and $Y_\mathbb{Q}$ are homotopy equivalent, see e.g. Felix, Halperin, Thomas, Rational homotopy theory, Proposition 9.8. If this is the case, then there is a roof $X\to X_\mathbb{Q}\gets Y$ of rational equivalences. The middle of the roof is of course an infinite complex (except when it is a point).

To add to Denis T's comment, if the rational homotopy type in question is formal, then $X$ admits endomorphisms of arbitrarily divisible degree (see Infinitesimal Computations in Topology Theorem 12.2, and F. Manin's "Positive weights and self-maps" for a detailed discussion, Corollary 1.1 arxiv.org/abs/2108.02173v3) and so, upon precomposing with an appropriate endomorphism one can lift the map $X \to X_{Q}$ through $Y \to X_{Q}$ to get a direct rational equivalence $X \to Y$.

...And SW classes are functorial along any maps which are $\Bbb Z/2$ cohomology equivalence, so one can maybe just use them instead of more complicated things like cokernel of Hurewitz map for oriented bordism provided our cohomology rings are "sturdy" enough in sense I defined above.

Well, yeah, that's why I'm writing it as a comment and not an answer, but I think my non-example can be promoted. By methods of Manuel Amann I think one can produce "sturdy" PD rational types, i. e. ones without self-maps of degree $\neq 1$ and retaining that property under connected sums. Then one can use some secondary obstructions (i. e. homology classes not realizable as manifolds) to distinguish two realisations of that type. Sturdiness will prevent positive degree things from killing our obstructions, which will forbid any nonzero degree maps in either direction.

.. also, the tangent Stiefel-Whitney classes are preserved under genuine homotopy equivalences of manifolds, but not in general under rational ones.

Denis: this is a good example, but rational equivalences may have degree $\neq 1$. For example, any map $S^n\to S^n$ of degree $\neq 0$ is a rational equivalence.

I think there's no way to connect $S^2 \times S^3$ with the unique other 5-mfd with $H_2 = \Bbb Z$ (double of nontrivial $D^3$-bundle over $S^2$) through degree 1 maps because one has trivial $w_2$ and other one does not. They are rationally equivalent, since they are both formal.


Transcendence of sum of reciprocals of factorials

For $A \subseteq \mathbb{N}$, define $\displaystyle x_A = \sum_{n \in A} \frac{1}{n!}$. It is easy to see that for every infinite $A$, $x_A$ is irrational.

**Question** : Is there an infinite $A \subseteq \mathbb{N}$ for which $\displaystyle x_A = \sum_{n \in A} \frac{1}{n!}$ is algebraic?

At least we can prove that if $x_{A}$ is an algebraic number then $A$ is not a cofinite subset of $\mathbb{N}$. :)

For other results along these lines, see Kumar and Vance and the references therein. I'm not familiar with them all but at first glance they all require faster growth rates than $n!$. And for an arbitrary $A$ it seems to me that there's not much you can say about $\sum_{n\in A} 1/n!$ other than the general rate at which the summands shrink.

I suspect the answer is no but that a proof is beyond current technology. The proof that $e$ is transcendental relies on special properties of the function $e^x$, not just on the rate of growth of $n!$. There are transcendence results that rely only on how fast the summands shrink but they require a faster growth rate than $n!$. For example, Nyblom proves that if for some fixed $\lambda>2$ we have $\liminf_{n\to\infty} a_{n+1}/a_n^{λ+1}>1$ then $\sum_n 1/a_n$ is transcendental.

Exactly this class of numbers is proved irrational (and some examples given, which I think all happen to be transcendental) in Martin Griffiths, "Irrational sums from reciprocals of factorials", Math. Gaz. 2015, jstor.org/stable/24496963


Is "Escherian metamorphosis" always possible?

$\DeclareMathOperator\int{int}\DeclareMathOperator\diam{diam}\DeclareMathOperator\area{area}\DeclareMathOperator\cl{cl}\DeclareMathOperator\ran{ran}\DeclareMathOperator\dom{dom}$_This is a tweaked version of a question which was asked and bountied[at MSE](https://math.stackexchange.com/q/4555946/28111); that question still hasn't quite been answered, but suggestions in the comments convinced me that this was the "right" version to ask instead:_

This question is motivated by Escher's series of **Metamorphosis** woodcuts (see e.g. [here](https://www.escherinhetpaleis.nl/story-of-escher/metamorphosis-i-ii-iii/?lang=en)), where one tesselating tile is gradually transformed into another. Basically, this is a precise way of asking the following: can we always "metamorphose" between any pair of tesselating shapes?

First, some definitions (everything is in $\mathbb{R}^2$ with the usual metric):

  * A _shape_ is a compact connected set $X$ with $X=\cl(\int(X))$.

  * A _tiling_ is a pair $(\mathscr{S},F)$ where $\mathscr{S}$ is a finite set of shapes and $F$ is a set of functions such that

    * each $f\in F$ is an isometric embedding of some $S\in\mathscr{S}$ into $\mathbb{R}^2$,

    * $\bigcup_{f\in F}\ran(f)=\mathbb{R}^2$, and

    * if $f,g\in F$ are distinct with domains $S,T$ respectively and $x\in \int(S)$, then $f(x)\not\in \ran(g)$ ("shapes only meet at their boundaries").

  * A _tile_ is a shape $X$ such that $(\\{X\\},F)$ is a tiling for some $F$.

  * $d_H$ is [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance). _(Note that we could replace $d_H$ with the modified version $d_{H*}(U,V)$ = the infimum over planar isometries $p$ of $d_H(p(U), V)$ without changing the question substantively; all that would change is that the number of tiles needed would shrink, but I'm not looking at that here.)_

Now with apologies to Escher, given tiles $A,B$ and $\epsilon>0$ let a **strong $\epsilon$-metamorphism** from $A$ to $B$ be a tiling $(\mathscr{S},F)$ such that

  * if $f\in F$ and $(x,y)\in \ran(f)$ with $x<0$ then $\dom(f)=A$,

  * there is some $N$ such that if $g\in F$ and $(x,y)\in \ran(g)$ with $x>N$ then $\dom(g)=B$ (call the least such $N$ the _length_ of the $\epsilon$-metamorphism), and

  * if $f,g\in F$ and $\ran(f)\cap \ran(g)\not=0$ then $$d_H(\dom(f), \dom(g))<\epsilon\cdot {\area_f+\area_g\over 2},$$ where $\area_j$ is the area of the domain of $j$ for $j\in F$. (This "scaling" of the Hausdorff distance means that we can't trivialize things by shrinking. This should have been included in the original problem, but I forgot it.)

Finally, say that an **$\epsilon$ -metamorphism** is as above but with the final inequality having right hand side merely $\epsilon$.

> **Question 1** : Is there always a strong $\epsilon$-metamorphism between $A$ and $B$ for any tiles $A,B$ and any positive $\epsilon$?

I recall seeing a theorem that the answer is **yes** , and that moreover we can always find metamorphisms with length "close to" the naive guess $${\max\\{\diam(A),\diam(B)\\}\cdot d_H(A,B)\over \epsilon},$$ but I haven't been able to track it down or prove it myself. I'm also interested in whether the situation changes as we appropriately tweak things to work in $\mathbb{R}^n$ for $n>2$.

> **Question 2** : Assuming for the moment that $\epsilon$-metamorphisms always exist (which seems plausible based on the comments at the MSE version of this question), what can we say about the length function $$\mathscr{l}_{A,B}: \epsilon\mapsto\inf\\{N: \mbox{an $\epsilon$-metamorphism from $A$ to $B$ with length $N$ exists}\\}$$ for a pair of tiles $A,B$? In particular, what is $\mathscr{l}_{\mbox{unitsquare, unithexagon}}$?

@MattF. Metamorphosis II has a portion that goes squares-lizards-hexagons, I believe.

Which of the Escher images do you find most suggestive for an answer about unit squares and unit hexagons?


A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?

**Background.** If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof of this fact I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem. It is worth mentioning that Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$.
T. Coquand and H. Lombardi have found a surprisingly elementary, first-order characterization of the Krull dimension that does not use prime ideals at all.

T. Coquand, H. Lombardi, A Short Proof for the Krull Dimension of a Polynomial Ring, The American Mathematical Monthly, Vol. 112, No. 9 (Nov., 2005), pp. 826-829 (4 pages)

You can read the article here. Here is a summary.
For $x \in R$ let $R_{\{x\}}$ denote the localization of $R$ at the multiplicative subset $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have
$$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \label{1}\tag{$\ast$}$$
It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that
$$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$
You can use this to define the Krull dimension.
This new characterization of $\dim(R) \leq k$ can be seen as a statement in first-order logic (whereas the usual definition with prime ideals uses second-order logic): Use two sorts $N,R$, the usual ring operations for $R$, but also the "mixed" operation $N \times R \to R$, $(n,x) \mapsto x^n$. Every ring becomes a model of this language.
A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.

Question. Can we use the characterization \eqref{1} of the Krull dimension by Coquand-Lombardi to prove the formula $$\dim(R[T])=\dim(R)+1$$ for Noetherian commutative rings $R$?

Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$. I suspect that this can only work if we find a first-order property of commutative rings (with powers) which is satisfied in particular by Noetherian rings and prove the formula for these rings.
Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization. If you post an answer that doesn't mention the characterization, then it's not an answer to my question and therefore offtopic. As of writing this, 20 answers have been posted, all of which have been deleted.

@Alexey Back then I removed the remark about first-order because of your comment. But I have just added this back, because we can just change the language to make it first-order. I have added an explanation for this.

Why this question has no answer or answers were deleted?

This may not be a direct answer to your question, but it is relevant to the problem and too long to share in the comments. You can use valuation overrings and valuative dimension instead of prime ideals and Krull dimension, respectively. Polynomial rings are well-behaved with respect to valuative dimension, and Jaffard rings are the natural context in which to study the condition you seek. All Noetherian rings and all Prufer rings are Jaffard, for example, and the class of Jaffard rings is much larger than that of the Noetherian rings. See, for example, reader.elsevier.com/reader/sd/pi

@WernerGermánBusch Of course $T$ is just one variable.

@moderators: Please make this question CW.

This question has been referred to at math.meta.stackexchange.com/questions/33453/…

@Foobanana I know them. There is no such proof.

This might not be too helpful, but have you tried searching through Gathmann's old notes on algebraic geometry? We're going through his notes in my algebraic geometry this year and he gives a lot of alternative proofs to commutative algebra facts :D

Might I ask why the dimension definition using prime ideals isn't elementary enough?   The proof isn't short, but it isn't terrible to see why the dimension increases by 1.   Is there a reason this other construction is the one you'd prefer to use?

For my own education, what do you mean when you write "first order property"? What is the definition of a first order property?

I guess T is a set of polynomial variables, like $T= \{x\}$ gets you $R[x]$ polynomials in $x$ with coefficients in the ring $R$, $T= \{x_1, x_2, ... , x_n\}$ gets you the polynomials in $n$ variables, and so on. I'm not sure if OP specificies T to be numerable or even finite.  +1 means just that, +1, the dimension is an integer number, so it makes sense to add 1 to it.

If you can explain in a few words, what is T and what does it mean +1 in your equation?

bourbaki's proof of this theorem uses only general results, not more partiuclar results such as krull IT

MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1

Indeed. By the compactness theorem, it can be seen that ${\rm dim}R\le l$ is not a first order statement for any finite $l$.

This characterization of $\dim R\le l$ does not look like first-order to me, at least in the language of rings: it uses quantifiers over $\mathbb N$ and exponentiation with the exponent as an argument, which is not a ring operation.

Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.

At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?


Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?

Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ is connected.
Question: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?

Full disclosure: I've now crossposted the question on MO.

Various remarks

If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = \sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.

With the bijection, it holds true in $n = 1$. But this is using the order structure of $\mathbb{R}$: a bijection that preserves connectedness on $\mathbb{R}$ must be monotone.

As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)

Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:

Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?

Some properties of $\mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $\mathbb{R}^n$ with the standard topology to itself, self-maps of some arbitrary topological space, it is easy to make the answer go either way.

For example, if the topological space $(X,\tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by the cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.

On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = \mathbb{Z}$ be equipped with the topology generated by:
$$ \{\mathbb{N}\} \cup \{ \{k\} \mid k \in \mathbb{Z} \setminus \mathbb{N} \} $$
Then the map $k \mapsto k+\ell$ for any $\ell > 0$ maps connected sets to connected sets, but its inverse $k\mapsto k-\ell$ can map connected sets to disconnected ones.
Working a bit harder one can construct in similar vein examples which are Hausdorff.

@HelloWorld: thank you for the reminder. The cash prize is now posted on X.

Will there be a cash price in twelve days? ;)

@Carlyle: nothing immediate comes to mind. You can have highly discontinuous functions whose graph still remain connected.

How does this definition of a connected function relate, if at all, to the property of the graph of a function (in the product Topology) being connected?

@FurdzikZbignew: the question is asking about mappings from $\mathbb{R}^n$ to itself, not from $\mathbb{R}^n\to\mathbb{R}^m$; I am not sure how your observation is relevant?

A counterexample could be a space-filling Peano curve, which is a bijection from the unit interval to the unit square but does not preserve dimensions/connectedness.  Therefore, the answer to the question is "no" for dimensions $n>1$. The total order on $\mathbb{R}$ was crucial in the proof for $n=1$, but does not generalize to higher dimensions. Additional regularity like continuity would be needed.

@user65023: the function $f$ maps $(s,t) \mapsto (s, t + \sin(\pi/s))$. And $J$ maps $(s,t) \mapsto (-s,t)$. You want to show $J\circ f \circ J \circ f$ is the identity, so just write it out $$ (s,t) \mapsto (s,t + \sin(\pi/s)) \mapsto (-s, t + \sin(\pi/s)) \mapsto (-s, t + \sin(\pi/s) + \sin(\pi / (-s))) = (-s,t) \mapsto (s,t) $$

@AlanU.Kennington: It seems like the example you gave above has the following properties: (1) $f$ is connected, (2) $f$ is discontinuous, (3) $f^{-1}$ is not connected, (4) $f$ is a bijection.  $f(s,t) = (s,t+\sin{({\pi}/{s})})$ for $s\neq 0$ and $f(0,t)=(0,t)$.

@PietroMajer: Do you have a proof of why AlanU.Kennington's example satisfies $f^{-1}=J\circ f \circ J$?  I'm not seeing it.  Also, it looks like for AlanU.Kennington's example $f^{-1}$ is not connected.  I think $f^{-1}$ maps the connected set: $\{(s,sin(\frac{\pi}{s})): s>0\} \cup \{(0,t): 0 < t < 1\}$ to the unconnected set: $\{ (s,0): s>0\} \cup \{ (0,t): 0 < t < 1\}$.  To see this set is unconnected, it is covered respectively by two open half planes: $\{ (s,t): t < s\}$ and $\{ (s,t): t > s\}$.

@WillieWong: Why do you think this would be embarrassing? Almost all of important branches of mathematics have been born just by such questions. For example think about Graph theory. Or think about one of my favorites unsolved problem in math i.e. square peg problem! a simple conjecture but so beautiful one. :). My suggestion is to offer this challenge to undergraduate and graduate students first. then history will find its way.

@C.F.G It would be embarrassing if, 30 years later, this question becomes my one legacy in mathematics. :-) That said, I'll perhaps save that (announcement of a cash prize) for the tenth anniversary of this post (which is only 1.5 years away).

@WillieWong: You Should (as the author of this problem :) ) offer a prize for any progress on this problem.

To be thorough: the statement is also true for $n = 0$. :-)

@AlexHeuman your $f$ is not a bijection. $f(x_1, x_2, \ldots) = f(x_1, x_2 + 2\pi, x_3,x_4 ,\ldots)$.

@GerryMyerson: Now 17 answers, all deleted.

@WillieWong might it be possible to send borromian rings in $\Bbb R^3$ to trefoil knots?  I can see the "rope" would need to be spliced into possibly three strands, with twists in such a way that it continuously morphs between knots and rings. I've no idea if this can hang together rigorously.

Ok I think I see the issue. The piecewise assignment is a priori and causes us to voluntarily disconnect the domain. So the pre-image and the image are both disconnected. But the problem asks for a bijection with a connected pre-image but disconnected image, for example.

For n=1 with f(x) = x, if x is rational; and x+1, if x is irrational, f is a bijection that is nowhere continuous. The usual topology on R begins its life connected. What is the definition of connected that lets us discard this f and conclude n > 1 for R^n?

@GerryMyerson: though 6 of them are from the same user....

Currently, 12 answers have been posted – all deleted!

@MiloBrandt: for $S^2$ your suspicion is sound. See this answer on the MathOverflow version of this question. Thanks!

I suspect that there is no such map in the analogous problem for $S^2$ instead of $\mathbb R^2$; the reason being that if we have a separating set $X$ (i.e. one such that $S^2\setminus X$ is disconnected), then its preimage has to be a separating set as well. Separating sets are kinda circle-like and I suspect have properties like "there are connected subsets $A,B,C\subseteq X$, no pair disjoint, such that $A\cap B\cap C=\emptyset$". Given this lemma or a similar ones, one can show the non-existence of such a forward-but-not-backwards connected map $S^2\rightarrow S^2$.

@WillieWong I think I can prove this if I succeed to prove that f maps closed connected sets onto closed connected sets.

@RobertFrost: this particular question I am asking about is turning out to be very subtle. This particular answer of the cross post to MathOverflow shows how to answer the question if we change some small number of conditions. But the original question is still, as far as I know, open.

@WillieWong I'm not competent with the subtleties of the concept of continuity but this reminds me of a function I'm doing some work with.  The branch cuts of its inverse are all superimposed on top of each other so for any $x+\epsilon$ there are $f^{-1}(x+\epsilon)$ not necessarily nearby to $f^{-1}(x)$ but I'm not totally clear how the concept of continuity treats this and whether it would fit your conditions.

@RobertFrost: mostly for my curiosity. But I note that the answer to a related question is known, and that question is what inspired me to ask this one.

@RobertFrost: you are half right. See this previous comment

@AndréLevy: Can you justify that the bijection is connected from the 2 spheres to the 1 sphere? Each of the paradoxical pieces is dense in the sphere, this suggests that the corresponding bijection cannot be connected. (Let $P,Q\subset \mathbb{S}^2$ be two disjoint dense sets. And let $\rho$ be a non-trivial rotation. Then for any $\epsilon > 0$ there exists $p\in P$ and $q\in Q$ such that $d(p,q) < \epsilon d(p,\rho(q))$.)

Well, @WillieWong, isn't the disassembling and reassembling of the pieces in B-T a bijection, isn't this bijection connected from the 2 spheres to the 1 sphere, but not the reverse?


Does every ring of integers sit inside a ring of integers that has a power basis?

Given a finite extension of the rationals, $K$, we know that $K=\mathbb{Q}[\alpha]$ by the primitive element theorem, so every $x \in K$ has the form
$$x = a_0 + a_1 \alpha + \cdots + a_n \alpha^n,$$
with $a_i \in \mathbb{Q}$.
However, the ring of integers, $\mathcal{O}_K$, of $K$ need not have a basis over $\mathbb{Z}$ which consists of $1$ and powers of a single element (a power basis). In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.

Question: Can every ring of integers $\mathcal{O}_K$ that does not have a power basis be extended to a ring of integers $\mathcal{O}_L$ which does have a power basis, for some finite $L/K$?

The difficulty in finding a general approach to construct such an extension arises from the fact that the structure of the ring of integers can be quite complicated, and power bases may not always exist for every number field.

@Jacob Wakem, if you take a field extension, let's say $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, then the ring of integers (of $\mathbb{Q}(\sqrt{2})$) refers in some sense to the 'analogon' of what $\mathbb{Z}$ is to $\mathbb{Q}$. In this case the ring of integers would be $\mathbb{Z}[\sqrt{2}]$. But I hope my comment is not misleading, because if $\mathbb{Q}(\alpha)/\mathbb{Q}$ is a field extension, then the ring of integers of $\mathbb{Q}(\alpha)$ is not necessarily(!) $\mathbb{Z}[\alpha]$.

What I should say is this: The number of generators of a ring of integers, as an algebra, can be 1 (monogenic) or arbitrarily large.

I honestly don't understand your remark "In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.": rings of integers are $\mathbb Z$-modules, so all their $\mathbb Z$-bases have the same number of elements (namely the degree of the corresponding field extension).

What is a ring of integers? Merely a ring containing only integers?

Note a similar result holds for the ordinals. See: en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form

@Gro-Tsen and sadly it's not had much luck there either!

I think this question should really be reposted on MathOverflow. Also, before tackling the number field case, it's probably worth thinking about curves over a field: given a base curve $C$ and an affine open set $U$, say that $C_1\to C$ (ramified covering) is "monogenic" for these data when there is $f$ regular on $U_1:=U\times_C C_1$ such that $\mathcal{O}(U_1)=\mathcal{O}(U)[f]$. Is it true that there is always $C_2\to C_1$ that is monogenic? (Maybe this is stupidly true or false, I'm not sure. But it could be easier to think about.)

Just a remark on terminology: If $\mathcal O_K=\mathbb Z[\alpha]$, then $\mathcal O_K$ is sometimes called monogenic.

Every $abelian$ number field $K$ sits inside $\mathbb Q(\zeta)$ for some root of unity $\zeta$ and has a ring of integers $\mathbb Z[\zeta]$. So your question is answered in the positive for abelian $K/\mathbb Q$.


If polynomials are almost surjective over a field, is the field algebraically closed?

Let $K$ be a field.  Say that polynomials are almost surjective over $K$ if for any nonconstant polynomial $f(x)\in K[x]$, the image of the map $f:K\to K$ contains all but finitely many points of $K$.  That is, for all but finitely many $a\in K$, $f(x)-a$ has a root.
Clearly polynomials are almost surjective over any finite field, or over any algebraically closed field.  My question is whether the converse holds.  That is:

If $K$ is an infinite field and polynomials are almost surjective over $K$, must $K$ be algebraically closed?

(This answer to a similar question gave a simple proof that $\mathbb{C}$ is algebraically closed from the fact that polynomials are almost surjective over $\mathbb{C}$.  However, this proof made heavy use of special properties of $\mathbb{C}$ such as its topology, so it does not generalize to arbitrary fields.)

I guess the problem with the original reasoning is that it relies on the fact that the image of $\mathbb{C}$ has to be simply connected, but this is hard to use for abitrary fields, the only topology you could use would be zariski but knowing zariski would give you info about the algebraic closeness. Maybe studying homology would give some results? The only nice thing is that being infinite gives you irreducibility

I do not know whether OP knew this but this was conjectured by Reinke in their 1975 paper "Minimale Gruppen", and as far as I know is still open. A similar conjecture in model theory is the so-called "minimal fields conjecture" of Podewski, asserting that infinite "minimal" fields are algebraically closed (where "minimal" means that every field formula with one free variable either defines a finite subset of the field or a cofinite one). Wagner solves the latter in his paper "Minimal fields" in positive characteristic, where he also solves a weaker variant of OP's question.

@student91: That's why I restrict $K$ to be infinite in the precise statement of the question.

I am slightly confused. If $K$ is finite, then any polynomial is "almost surjective" in this sense, because there are only finitely many points that are not in the image of $f$ (even if the image of $f$ consists of only one point). Furthermore, finite fields are not algebraically closed.  Do you assume your characteristic to be 0?

Ooh you are right. I thought it was strange to have such a regular topology for arbitrary fields without being known.

@AndreaMarino: I think it is extremely unlikely that you can get such topology at the end of an infinite chain.  Local compactness is also not enough (then you could have the discrete topology!), and you need $T_2$, not just $T_0$, since you need to know the limit of $y_n$ is unique to conclude $p(x)=s$.  Note then that no countably infinite compact $T_2$ space is homogeneous, so you can never have such a topology on a countably infinite field.

Sorry for double comments but I can't edit. It is enough for the topology to be locally compact, so that we recover the cases of $\mathbb{Q}, \mathbb{C}$.

How I would build such a topology: start from $\mathbb{Q}$ and $\mathbb{F}_p$, then show that the topology extends by transcendent, algebraic extensions and by infinite chains. By Zorn, this gives the topology on whatever field. Don't know if this route map can work!!

Here it is my 1 cent also. Suppose there exist a $T_0$, compact and first countable topology on K such that polynomials are continuous. There are no isolated points x, otherwise $x+k$ would be also isolated because polynomials are continuous, thus the topology is discrete. But being compact it should be finite. Now take a polynomial p with non taken values $s_1,\ldots, s_k$. Take $y_n$ in the $U_n $ (given by first countability) of $s_1$ , WLOG $x_n$ different by $s_i$. Take $x_n$ st $p(x_n)=y_n$. The space is sequentially compact, so WLOG $x_n\to x$. Then $p(x)=s$ and p is surjective.

Or wait, can your question be rephrased as 'If $K$ is an infinite field and polynomials are almost surjective over $K$, must polynomials be fully surjective over $K$?' ?

I find the concept of almost surjective polynomials a bit confusing. Can you give an example of an almost surjective polynomial over $\mathbb{C}$ that is not surjective?

Here's my 5 cents. $K$ must contain an $n^{th}$ root for each of its elements (take $f:=X^{n}$; for $\alpha\in K^{*}$, $f$ cannot miss the infinitely many elements $\alpha\cdot\beta^{n}$ with $\beta\in K^{*}$). At least when $char(K)=0$, the roots of unity are expressible as radicals. This implies that $K$ does not admit any finite Galois extensions with solvable Galois group. I.e., any minimal non-trivial Galois extension of $K$ must have a simple non-abelian Galois group.

I believe this is open. At least, it was posted on MO with no definitive answer: mathoverflow.net/questions/6820/…