What is the three-dimensional hyperbolic volume of a four-manifold?

Every smooth closed orientable 4-manifold may be constructed via a handle decomposition. Before asking a couple of questions, I recall some well-known facts about handle-decompositions of 4-manifolds.

We can of course order the handles according to their index. Handles of index 0 and 1 form a connected 4-dimensional handlebody, whose boundary is a closed 3-manifold, diffeomorphic to a connected sum $\\#_g(S^2\times S^1)$ of some $g$ copies of $S^2\times S^1$ (if $g=0$ we get $S^3$). Handles of index 2 are attached to some framed link $L\subset \\#_g(S^2\times S^1)$. Since 3- and 4-handles also form a 1-dimensional handlebody, after the attaching of the 2-handles we must necessarily obtain a 4-manifold whose boundary is again diffeomorphic to $\\#_h(S^2\times S^1)$, for some $h$ which is not necessarily equal to $g$. The new $\\#_h(S^2\times S^1)$ is obtained by _surgery_ along the framed link $L$.

Therefore, in some sense, constructing closed orientable 4-manifolds reduces to constructing framed links $L\subset \\#_g(S^2\times S^1)$ that produce some $\\#_h(S^2\times S^1)$ via surgery. Let us define the _3-dimensional hyperbolic volume_ of a handle decomposition as the Gromov norm of $(S^2\times S^1) \setminus L$ (which is in turn the sum of the hyperbolic volumes of its pieces according to geometrization, whence the name). Let us then define the _3-dimensional hyperbolic volume_ of a closed orientable 4-manifold as the infimum of all the hyperbolic 3-dimensional volumes among all its handle decompositions. The infimum is actually a minimum because the set of 3-dimensional hyperbolic volumes is well-ordered.

The general question is:

> What can we say about the 3-dimensional hyperbolic volume of a closed 4-manifold?

A more specific one:

> Which closed 4-manifolds have zero 3-dimensional hyperbolic volume?

which is equivalent to the following:

> Which closed 4-manifolds admit a handle decomposition such that $\\#_g(S^2\times S^1)\setminus L$ is a graph manifold?

Complex projetive plane belongs to this class, and also many doubles of 2-handlebodies: if you take any link $L\subset S^2\times S^1$ whose complement is a graph manifold, you can attach 2-handles to it, and then make the double of the resulting bounded 4-manifold. The resulting double has volume zero.

Finally, we have the following very specific question:

> Is there a 4-manifold with positive 3-dimensional hyperbolic volume?

I would expect that most (all?) aspherical 4-manifolds have positive volume, and maybe also many simply connected ones, but I don't know the answer.

I think this might be related to this question: mathoverflow.net/q/19974/1345 In particular, if the 3-dimensional volume is zero, I suspect that the manifold admits an F-structure, and the simplicial volume is zero (and likely minimal volume =0). The point is that the circle action on a Seifert manifold M extends over dehn fillings and attaching 2-handles along MxI.  If all of the level sets of a Morse function on a 4-manifold are graph manifolds (and the complements of the 2-handles attaching maps), then I think that these local circle actions may be assembled to get an F-structure.

This question sounds a bit similar to what is written in the last lines of a recent article of Gromov and Guth : arxiv.org/abs/1103.3423


Two-convexity ⇒ Lefschetz?

Assume that

  * $\Omega$ is an open simply connected set in $\mathbb R^n$
  * _(two-convexity)_ if 3 faces of a 3-simplex belong to $\Omega$ then whole simplex in $\Omega$.

Is it true that any component of intersection of $\Omega$ with any 2-plane is simply connected?

**Comments:**

  * Note that an open connected set $\Omega\subset\mathbb R^n$ is convex (in the usual sense) if together with two sides of triangle $\Omega$ contains whole triangle.

  * If the boundary of $\Omega$ is a smooth hypersurface then the answer is YES. [The above property implies that at most one principle curvature of the boundary is negative. Then the statement follows easily from a Morse-type argument; see Gromov's "Sign and geometric meaning of curvature", Lefschetz theorem in Section $\tfrac12$. In fact this argument, shows that in this case any component of intersection of $\Omega$ with any affine subspace (not necessary 2-dimensional) is simply connected.]

  * The answer is YES if $n=3$. [In this case I can mimic the Morse-type argument.]

  * The statement would follow if one could approximate any $\Omega$ by two-convex domains with smooth boundary. BUT the example below shows that such approximation does not exist for $n\ge 4$.

  * I was surprised that my question has a lot in common with [this question](https://mathoverflow.net/questions/29373/minimal-length-embeddings-of-braids-into-r3-with-fixed-endpoints/)

**Example:** We will construct two-convex simply connected open set $\Omega$ in $\mathbb R^4$ such that intersection $L_{t_0}$ of $\Omega$ with some hyperplane is not simply connected. [This NOT a counterexample, but it shows that it is impossible to prove it using smoothing, as indicated in the comments.]

Set $$\Pi=\\{\,(x,y,z,t)\in\mathbb R^4\mid\,y< x^2\\}.$$ Let $\Pi'$ be the image of $\Pi$ under a generic rotation of $\mathbb R^4$, say $(x,y,z,t)\mapsto(z,t,x,y)$.

Note that $\Pi$ is open and two-convex. Therefore $\Omega=\Pi\cap \Pi'$ is also open two-convex set. One can choose coordinates so that $\Omega$ is an epigraph for a function $f\colon\mathbb R^3\to\mathbb R$ like $$f=\max\\{\alpha_1-\beta_1^2,\alpha_2-\beta_2^2 \\},$$ where $\alpha_i$ and $\beta_i$ are linear functions. In particular $\Omega$ is contactable.

Let $L_{t_0}$ be the intersection of $\Omega$ with hyperplane $t=t_0$; it is a complement of two convex parabolic cylinders in general position. If these cylidners have a point of intersection then $\pi_1 L_{t_0}=\mathbb Z$.

[In particular, the function $f$ can not be approximated by smooth functions which Hessian has at most one negative eigenvalue value at all points.]

@WillSawin, I guess your question is about last par, starting with "Let $L_{t_0}$...". We are in hypereplane, i.e., in $\mathbb R^3$. Imagine that your parabolic cylinders have exactly one point of intersection. Then the complement is homotoipically equivalent to $\mathbb{S}^1$.

I don't understand the example. I'm visualizing pairs parabolic cylinders in general position and all of them have simply-connected complements. Could you describe a specific hyperplane that does the job?

Here's an article of mine where (a slight variant of) higher convexity is used: staff.science.uu.nl/~henri105/PDF/advgometry.pdf

For an open set $\Omega$, why is [ if 3 faces of a 3-simplex belong to $\Omega$ then whole simplex in $\Omega$ ] not the same as convex?


Orthogonal vectors with entries from $\{-1,0,1\}$

Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \\{-1,0,1\\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $n \in \\{1,2\\} \cup \\{4k : k \in \mathbb{N}\\}$?

A few notes are in order:

  1. If we do not allow "$0$" entries in the vectors then this is a well-known necessary condition for the existence of an $n \times n$ [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix), so the answer to my question is "yes" if we replace $\\{-1,0,1\\}$ with$\\{-1,1\\}$.

  2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is "yes" when $n \leq 6$.

**Edit (May 19, 2023):** In the comments, Max Alekseyev has shown that the answer is "yes" when $n \leq 12$, and Ilya Bogdanov has shown that the answer is "yes" when $n$ is an odd prime.

I've computationally verified the conjecture $n=15$ and together with Ilya's and Mikhail's arguments that makes it proved for all $n\leq 20$. For $n=21$, I can tell for now that if a counterexample exists, it should have at least 6 vectors with exactly 14 nonzero components.

@MaxAlekseyev Same argument that establishes that Hadamard matrices are of order $1$, $2$, or $4k$. If there are three mutually orthogonal ±1-vectors $\mathbf{1}$, $v_1$, $v_2$, then the number of indices $i$ where $v_{1i} = s_1$, $v_{2i} = s_2$ is the same for any of the four combinations of $s_1, s_2 \in \{+1, -1\}$.

@MikhailTikhomirov: What does "invoke Hadamard" stand for?

Further, $n = 2p$ for odd prime $p$ is also a no-go. $1 = 1/2p + \ldots$ requires at least one more $1/2p$, and their sum $1 / p$ requires an extra (even) denominator divisible by $p$ (thus, another $2p$). We now have three full rows, which is enough to invoke Hadamard.

@IlyaBogdanov's argument also seems to apply when $n = p^k$ for $p$ odd prime. For the diagonal element we have $1 = 1/p^k + \ldots$, which has to involve at least $p - 1$ more $1/p^k$'s. But two odd-sized full rows are already a contradiction.

@kodlu Yes, even many such weighing matrixes exist. But the all-1 column changes a lot.

@IlyaBogdanov - You're quite right, thank you.

@NathanielJohnston The life seems more complicated for me. E.g.,  it could happen that the sum involves $1/15$,  $2\times 1/6$ and  $6\times 1/10$ (,the other denominators should be even!). However, this may fasten algorithms searching such matrices.

@IlyaBogdanov - If you agree with my reasoning, please feel free to write this up as an answer and I will accept it. Your denominators-cancelling argument is what blew this open. Thank you!

@IlyaBogdanov - And a similar argument works to rule out the $n \equiv 2\pmod{4}$ cases, thus proving the entire thing. Suppose $n = 2m$ for some odd number $m$. Then (again, by your denominators-cancelling argument) $A$ must have at least $2$ columns with all entries non-zero (one of which is the all-ones column). If it had $3$ or more such columns we get the Hadamard contradiction, so assume it has exactly $2$. The remaining columns each have their number of non-zero entries each equal to factors of $m$, which are all odd. Such columns cannot be orthogonal to $\mathbf{1}$. Contradiction.

Very interesting. If the all 1 vector is removed there are solutions with odd lengths.

@IlyaBogdanov - Very nice, thank you! I think your argument can be extended to all odd numbers, not just odd primes. For example, if $n = 15$ then (by your denominators-cancelling argument) there must be at least 3 columns of $A$ that have all $15$ entries with no $0$s (since then you can get $1/15 + 1/15 + 1/15 = 1/5$ to reduce the denominator). But the standard proof of the Hadamard necessary condition says that you can't have $3$ orthogonal columns of $\pm 1$ entries unless $n \in \{1,2\} \cup \{4k\}$.

@NathanielJohnston Each diagonal element of $BB^T$ is a sum of $n$ njmbers each of them is either 0 or  of the form $1/k$ with $k\leq n$. At least one summand is $1/n$; if there are less than $n$ such, then $n$ in the denominators cannot cancel.

@Ilya - Can you expand upon the statement "But the diagonal elements in $BB^T$ may equal $1$ only if $D = nI$"? I expect that this is the step that is using the hypothesis of $n$ being an odd prime, but I'm having a hard time seeing how.

As a first step, $n$ cannot be an odd prime. This is because the matrix $A$ composed from those columns should be such that $D=A^TA$ is diagonal, so $B=AD^{-1/2}$ should be orthogonal. But the diagonal element s in $BB^T$ may equal $1$ only I’d $D=nI$ which is impossible by Hadamard. It seems that one may investigate $BB^T$ or, equivalently, the relation $AD^{-1}A^T=I$  further…

@domotorp - Only when there's a Hadamard matrix.

Do you have a construction for such a matrix for every $n$ that is divisible by 4, or only for those $n$ for which there exists a Hadamard matrix?

The question seems related to weighing matrices except that the all-ones vector kind of messes things up.

@PeterMueller: I've also checked with cliquer and it confirmed no solutions for $n=11$ within a couple of hours. So, the "yes" answer holds for all $n\leq 12$.

Maybe linear algebra modulo $4$ is relevant, as these vectors miss $2 \bmod 4$? There is also a natural combinatorial interpretation -- with each $\mathbf{v}_i$ associate disjoint subsets $A_i^+$, $A^-_i$ of $\{1,2,\ldots,n\}$ defined by $A^{\pm}_i=\{j: (\mathbf{v}_{i})_j=\pm 1\}$. Then $|A^+_i|=|A^-_i|$ by orthogonality of $\mathbf{v}_i$ with $\mathbf{1}$. The rest of the orthogonality conditions are encoded by $|A^+_i \cap A^+_j| + |A^-_i \cap A^-_j| = |A^+_i \cap A^-_j| + |A^-_i \cap A^+_j|$ for $i \neq j$.

@MaxAlekseyev ... and so did I using cliquer. Presently running $n=11$ which might take a while ...

I confirm the affirmative answer for all $n\leq 10$ with ILP.

@Peter: Thanks, I do need that condition so I'll add that in. (I'm reminded of a time when I asked my linear algebra students to produce some orthogonal vectors satisfying certain conditions, and lots of them got easy marks since I forgot to specify "non-zero"...)

@NathanielJohnston Which condition prevents all $\mathbf{v_i}$ to be the $0$-vectors?

@Amir: If you just require linear independence then you can always (for all $n$) find such a set of vectors. Furthermore, if you just require linear independence then whenever $n$ is even you can find such a set of vectors with entries in $\{-1,1\}$.

Interesting! Would the answer change if we replace the "orthogonality" requirement with "independence"?


Is there a Mathieu groupoid M_31?

I have read something which said that the large amount of common structure between the simple groups $SL(3,3)$ and $M_{11}$ indicated to Conway the possibility that the Mathieu groupoid $M_{13}$ might exist.
Indeed, it exists, and the group of permutations of the other points generated by paths in which the hole starts and ends at the same point is the Mathieu group $M_{12}$.

It is natural to ask whether or not something analogous exists for the Mathieu group $M_{24}$. The corresponding linear group seems to be $GL(5,2)$.
$2^{4}:A_{8}$ is a vector (or hyperplane) stabilizer in $GL(5,2)$ and an octad stabilizer in $M_{24}$. $2^{6}:(GL(2,2)\times GL(3,2))$ is likewise the stabilizer of a 2-dimensional (or 3-dimensional) space in $GL(5,2)$, and the stabilizer of a 'trio' (to use the term from SPLaG) in $M_{24}$.

The natural setting, therefore, in which to look for something analogous to $M_{13}$ is $(\mathbb{Z}/(2))^{5} \backslash 0$. (This can also be regarded as $4$-dimensional projective space over $\mathbb{Z}/(2)$, but I will here think of it as $5$-dimensional space over $\mathbb{Z}/(2)$ with the origin removed.)
Then, to have $24$ points left over when the 'holes' are put in place, it would be natural to have a 'missing $3$-dimensional space' for a hole.
A move should consist of moving the $3$-dimensional 'hole' to some other $3$-dimensional subspace (minus the origin) of $(\mathbb{Z}/(2))^{5}$. The intersection of these spaces can have dimension $1$ or $2$, and there should be different rules for moving the hole to another location depending on the dimension of its intersection with the new location.

Who has tried this before? Are there any papers on it?


Subalgebras of von Neumann algebras

In the late 70s, Cuntz and Behncke had a paper

H. Behncke and J. Cuntz, _Local Completeness of Operator Algebras_ , Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100

the about the following question. Let $A$ be a $C^\star$-algebra and let $B \subset A$ be some $*$-subalgebra. Let $B$ be dense and such that every maximal abelian subalgebra of $B$ is norm-closed. Is it true that $B=A$? They proved this in various cases.

I want to ask a related question in the von-Neumann-setting. More precisely: Let $A$ be a von Neumann algebra and $B \subset A$ be some ultra-weakly dense $\star$-subalgebra such every MASA of $B$ is ultra-weakly closed in $A$. Is it true that $B=A$? A result of Gert Pedersen implies that once $B$ is a $C^\star$-algebra, then $B=A$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that $B=A$ if $A$ has no $II_1$-part.

> **Question:** Is it true for every von Neumann algebra?

or a little more modest

> **Question:** Is it true for the hyperfinite $II_1$-factor?

Another strenghtening of the the assumption (which could help) would be the following:

> **Question:** Let $A$ be a $II_1$-factor and $B \subset A$ be a ultra-weakly dense $\star$-subalgebra such that for every hyperfinite subalgebra $R \subset A$ one has that $R \cap B$ is ultra-weakly closed in $A$. Is it true that $B=A$?

If I understand correctly, you need to add a density hypothesis in your question?


Isometric embeddings of finite subsets of $\ell_2$ into infinite-dimensional Banach spaces

Question: Does there exist a finite subset $F$ of $\ell_2$ and an infinite-dimensional Banach space $X$ such that $F$ does not admit an isometric embedding into $X$?

There are some results of the type: each finite ultrametrics admits an isomeric embedding into any infinite dimensional Banach space. See Shkarin, S. A. Isometric embedding of finite ultrametric spaces in Banach spaces. Topology Appl. 142 (2004), no. 1-3, 13-17 for this, and also the papers: Dekster, B. V. Simplexes with prescribed edge lengths in Minkowski and Banach spaces. Acta Math. Hungar. 86 (2000), no. 4, 343-358; Faver, Timothy; Kochalski, Katelynn; Murugan, Mathav Kishore; Verheggen, Heidi; Wesson, Elizabeth; Weston, Anthony Roundness properties of ultrametric spaces. Glasg. Math. J. 56 (2014), no. 3, 519-535.

There is a related question on MathOverflow: [Almost isometric embeddability implies isometric embeddability](https://mathoverflow.net/questions/209435/almost-isometric-embeddability-implies-isometric-embeddability)

Added on 9/29/2016: Related results were obtained by James Kilbane in <https://arxiv.org/abs/1609.08971>

Added on 4/3/2017: In another [recent paper](https://arxiv.org/pdf/1704.00319.pdf) James Kilbane proved that the set of possible counterexamples (if they exist) is small in a certain sense.

The last result I know in this direction is in Petrov, F. V.; Stolyarov, D. M.; Zatitskiy, P. B. On embeddings of finite metric spaces in ln∞. Mathematika 56 (2010), no. 1, 135–139

Is much known when $X$ is $\ell^n_{\infty}$ for $n<\#F$?


Peano Arithmetic and the Field of Rationals

In 1949 Julia Robinson showed the undecidability of the first order theory of the field of rationals by demonstrating that the set of natural numbers $\Bbb{N}$ is _first order definable_ in $(\Bbb{Q}, +, \cdot$).

It is not hard to see that Robinson's result can be reformulated in the following symmetric form.

**Theorem A.** _The structures_ ($\Bbb{N}, +, \cdot$) _and_ $(\Bbb{Q}, +, \cdot$) _are bi-interpretable._

The following generalization of Theorem A is considered folkore (I am not aware of a published reference).

**Theorem B.** _If $(M, +, \cdot)$ is a model of $PA$ (Peano arithmetic), then the field of rationals $\Bbb{Q}^M$ of $(M, +, \cdot)$ is bi-interpretable with $(M, +, \cdot )$._

Let $EFA$ denote the _exponential function arithmetic_ fragment of $PA$, a fragment also known as $I\Delta_{0}+exp$.

Based on _a posteriori_ evidence _classical_ theorems of Number Theory do not require the full power of $PA$ since they can be already verified in $EFA$ (indeed Harvey Friedman has conjectured that even FLT can be verified in $EFA$, with a proof that would be very different from Wiles').

This suggests that in Theorem B one should be able to weaken $PA$ to $EFA$, hence my question:

**Question**. _Is there a published reference for the strengthening of Theorem B, where $PA$ is weakened to $EFA$?_

P.S. The following paper provides an excellent expository account of Robinson's theorem (and related results).

D. Flath and S. Wagon, _How to Pick Out the Integers in the Rationals: An Application of Logic to Number Theory_ , American Mathematical Monthly, Nov. 1991.

@Samuel Ried: No I have not (sorry for the tardy reply, I have been "away" for a long while).

@AliEnayat: Have you found an answer to your question yet? I would be very interested to know.

SJR: The answer to your first question is positive, i.e., nothing fancier is going on.  Regarding the second one: the arguments of Robinson's paper should go through in $EFA$ (based on $EFA$'s track record in handling "elementary" number theory); my question is whether anyone has actually shown - in a publishd source - that this is indeed the case.

Ali: Suppose $M\models PA$. To interpret $M$ in $\mathbb{Q}^M$ does Julia Robinson's formula with the operations of $\mathbb{Q}^M$ do the job, or is there something fancier going on? Is your question about how much of the argument in Robinson's paper goes through in models of $I\Delta_0+exp$?


$f\circ f=g$ revisited

This may be related to [solving $f(f(x))=g(x)$](https://mathoverflow.net/questions/17614). Let $C(\mathbb{R})$ be the linear space of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$, and let $\mathcal{S}:=\\{g\in C(\mathbb{R}) ; \exists\; f\in C(\mathbb{R})$ s.t. $f\circ f=g \\}$ . Is there some infinite dimensional (or, at least, bidimensional) linear subspace of $C(\mathbb{R})$ contained in $\mathcal{S}$ ?

P.S. As a remark, there is a [maybe] interesting connection between [How to solve $f(f(x)) = \cos(x)$?](https://mathoverflow.net/questions/17605) and [Borsuk pairs of Banach spaces ](https://mathoverflow.net/questions/11675). Namely, let $E$ be the closed subspace of $C[-1,1]$ consisting of all even functions, and let $K$ be the closed unit ball of $E$. Then the continuous mapping $\Psi:$ $K$ $\rightarrow$ $E$ expressed by $\Psi(f)$ $:=$ $f\circ f$ $+$ $\left(\left\Vert f\right\Vert _{\infty}-1\right)\cdot\cos$

is odd on $\partial K$, and has no zeros in $K$.

In the previous comment, the full stop was taken as a part of the url - however, the intention was clearly to link to this post: solving $f(f(x))=g(x)$.

Unfortunately, not even two vectors in that cone generate a subspace in ${\mathcal S}$ since ${\mathcal S}$ contains no decreasing functions (see mathoverflow.net/questions/17614.)  Any solution to this problem is a linear space containing no injective functions.

All increasing bijections are ok, so there is an infinite-dimensional cone.

ad least there is a 2-dimensional cone (a space which is closed under linear combinations with positive coefficients) given by the space of all nondecreasing linear maps.


The central insight in the proof of the existence of a class of Kervaire invariant one in dimension 126

I understand from [a helpful earlier MO question](https://mathoverflow.net/questions/101031/kervaire-invariant-why-dimension-126-especially-difficult) that the techniques leading to the celebrated resolution of the Kervaire invariant one problem in the other candidate dimensions yield no insight on dimension 126, to the extent that until recently there was not consensus as to which way it was likely to go.

For a few weeks now, I've been aware of [a manuscript of Minami](https://arxiv.org/pdf/2106.14604) which, if I'm reading correctly, asserts the existence of such an element in dimension 126 as a consequence of Wang and Xu's 2017 paper on the stable 61-stem. Presumably they would not have failed to remark on the resolution of the Kervaire invariant one problem if such was readily visible from what they had done, so it would seem that there is something that is not _obviously_ a consequence of it, but which can nevertheless be extracted in eight pages if one does somehow know what to look for. It's common that when an old conjecture like this is resolved, it is as an application of some newly discovered techique or heretofore-ungleaned insight into the inner workings of the encompassing theory.

**What is the new insight here?**

The work on the stable stems is forbiddingly technical work I am unlikely to ever grasp the full details of, but it's still possible sometimes to get the flavor of the ideas involved, and that's what I would hope for here.

Probably you aware by now but just in case, there is also this preprint  "On the Last Kervaire Invariant Problem" by Weinan Lin, Guozhen Wang, and Zhouli Xu (arxiv.org/abs/2412.10879) which also claims to resolve the final case of dimension 126 in the affirmative.

You're most welcome. "Suitable for MathOverflow" is among my favorite genres of MathOverflow question!

Thank you for asking this question in a way that is suitable for MO! (For those wondering: it's asking about mathematical content/insight, not correctness of a preprint)


On the definition of regular (non-noetherian, commutative) rings

All rings are commutative with unit. A ring $R$ is called _regular_ if it satisfies

**(Reg)** Every finitely generated ideal of $R$ has finite projective dimension.

Clearly this gives the usual definition if $R$ is noetherian. In the non-noetherian world, valuation rings are regular (this is the reason why I got interested), as are polynomial rings in arbitrary sets of variables over a field.
I am wondering about the reason for this definition, as opposed in particular to:

**(Reg')** Every finitely presented $R$-module has finite projective dimension.

Clearly (Reg') implies (Reg), and they are equivalent if $R$ is _coherent_ (every finitely generated ideal is finitely presented).
To me, (Reg') looks much more natural than (Reg), and from my limited knowledge of the literature it seems that the association "coherent+regular" is quite popular. So, my question is

> What are examples of regular rings not satisfying (Reg')?

Since my post is really about motivating the definition, I am of course more interested in "natural" examples (not just constructed for the purpose).
On the other hand, all examples are welcome since I don't know any.

If I understand correctly, to show that regular Noetherian is regular in this sense, you have to use the fact that for Noetherian rings being proj dimension <n at a point implies it in a neighborhood?

@KestutisCesnavicius: Thanks for the reference. Both properties are stable under any localization, and are Zariski-local on the spectrum. I doubt that they are local in the stronger sense of localizing at primes, but this should be true if the ring is assumed coherent, which is the case for valuation rings and finitely generated flat algebras over them. See Knaf, J. of Alg. Appl. 7 (2008) #5, 575-591.

Are these notions local on $R$? Perhaps it would be better to impose these conditions on the local rings of $R$? Anyway, I just wanted to mention that for a valuation ring $V$, not only $V$ itself is regular (in the stronger sense) but also so is every finite type, flat $V$-algebra $R$ with regular $R$-fibers (e.g., $R$ could be $V$-smooth), at least if one takes the local definition I have alluded to. This is a result of Gabber--Ramero, Prop. 11.4.1 in arxiv.org/abs/math/0409584


Todd class as an Euler class

Let $X$ be a relatively nice scheme or topological space.

In various physics papers I've come accross, the Todd class $\text{Td}(T_X)$ is viewed as the Euler class of the normal bundle to $X\to LX$. Here $LX$ is the loop space of $X$. They usually call it $e_{reg}$, the regularised Euler class.

**Question:** Is there a way of making this notion of regularised Euler class rigorous?

Whatever the correct definition is, it should work for

  * finite dimension vector bundles, where it reduces to usual notion of Euler class,
  * loop spaces, where it reduces to the Todd class.

You can define $i_*$ for any closed embedding between smooths actually, and $i^*i_*=e$ is still true.

@Pulcinella The two questions are indeed related: in finite dimensions in the equivariant setting, after localization both $e(N_i)$ and $i^*$ become invertible, so that one can define $i_*$ from the formula $i^*i_*=e(N_i)$. This is another of the reasons inducing me to think that the equivariant setting is the correct one to handle the Euler class of the normal bundle to $i\colon X\to LX$. For Gysin I admit I have never thought to that: it is a very good question!

Also I'd be curious whether you think it's necessary to work equivariantly to get an answer. The Atiyah paper worked non equivariantly, but a lot of their intermediate algebra makes it look like equivariant setup is the one to use.

@domenicofiorenza That's a good question! It was part of what I was getting at originally. A question I might have for you is if you know whether 1. this regularised Euler class might fit into a Gysin sequence, or 2. if it satisfies a formula like $i^*i_*=e(N_i)$ for $i:X\to LX$. Obviously $i_*$ is undefined at the moment, but ignore that for now...

@Pulcinella Well, in the finite dimensional case that is the definition of equivariant Chern classes. In the infinite dimensional case the finite dimensional formulas become infinite products and one has to make sense of them. So the real question, I guess, is what singles out the zeta-regularization among the possible ways of giving a meaning to those infinite products? Surely zeta-regularization has good properties (see, e.g., the paper by Quine,Heydari and Song), but I do not know how to completely avoid an element of choice making it absolutely canonical.

@domenicofiorenza I've had a look and I'm a bit worried by the fact that your top chern classes (of the normal bundle of $X$ inside $LX$) are defined by hand like in Atiyah's original paper, rather than follow from some more general notion of euler class, from which the formal Weierstrass-regularisation type formulas also fall out.

I'd say Sections 2 and 6. In Section 6 you'll have to replace every occurrence of $\mathbb{C}/\Lambda$ with $\mathbb{R}/\Lambda$ or ask Mattia for a copy of his thesis, where you can find the computation spelled out in detail

@domenicofiorenza Thanks! Could you point to the most relevant section(s)?

Always unpleasant to mention a paper one has contributed to, but something very close to an answer to the original question can be found in arxiv.org/pdf/2106.14945.pdf . There, only the Witten class case is treated in detail as we considered Atiyah's treatment in "Circular symmetry and stationary-phase approximation" to be complete for the Todd class. Yet, the details on the Todd class case in the spirit of our paper can be found in Mattia Coloma's thesis.

You might also want to look at the closely related literature deriving the Witten and A^ genus from [rigorous mathematical] quantum field theory - see Costello on the Witten genus arxiv.org/abs/1006.5422, and Grady-Gwilliam and Grady on the A^ genus arxiv.org/abs/1110.3533 and arxiv.org/abs/1211.6816

There's a version of this question in derived algebraic geometry (explaining the Todd genus and Grothendieck-Riemann-Roch via derived versions of loop spaces) that's well understood - the idea is due to a visionary but hard to understand paper of Markarian arxiv.org/abs/math/0610553, a related story is explained in arxiv.org/abs/1305.7175 and the final derivation of the Todd genus is due to Kondyrev and Prikhodko arxiv.org/abs/1906.00172 --- see also closely related arxiv.org/abs/1511.03589 and arxiv.org/abs/1804.00879.

I suspect this might still be unknown: see Atiyah's paper "Circular symmetry and stationary-phase approximation", where it's apparent that these sorts of questions are still not rigorously solved (and Brylinski's more recent book on loop spaces where he repeats this).

@kiran Of course. This comes up a lot in papers of Rongmin Lu, e.g. "Regularized Equivariant Euler Classesand Gamma Functions", section 6 digital.library.adelaide.edu.au/dspace/bitstream/2440/50479/8/…

Could you include a reference or give a little more context for this "regularized euler class"  ?


When are two C*-algebras isomorphic as Banach spaces?

We may consider each $C^*$-algebra as a Banach space (by forgetting the multiplication and adjoint). I wonder how drastic this step is, i.e., which properties of the $C^*$-algebra are reflected by its Banach space structure.

Clearly, finite-dimensionality and other cardinality properties can be detected. Thus, to avoid these cases, let us restrict to separable, infinite-dimensional $C^*$-algebras.

For commutative $C^*$-algebras the problem has been solved by Milutin. In particular, $C(X)$ and $C(Y)$ are isomorphic as Banach spaces for any uncountable, compact, metric spaces $X$ and $Y$.

In the paper
Hamana. **On linear topological properties of some $C^*$-algebras** , Tohoku Math. J., II. Ser. 29, 157-163 (1977).
it is shown that the Banach space structure also reflects if all irreducible representations of the $C^*$-algebra are finite-dimensional.

Let us ask what happens at the 'opposite' end of the scale:

> Question: Are all simple, separable, infinite-dimensional $C^*$-algebras isomorphic as Banach spaces?

This has been answered many times in the comments but is the pair $c_0$ (the null sequences) and the algebra of continuous functions on the closed unit interval not a simple counter example to your initial question?

It is an open problem whether $C^∗_r(F_2)$ is Banach isomorphic to a nuclear $C^*$-algebra (say the CAR algebra $B$ as a model). It is known that (1) the dual of $B$ has the bounded approximation property (because it's AFD), (2) $B$ has a Schauder basis (Junge--Nielsen--Ruan--Xu, Adv. Math. 2004), and (3) $B\cong B\otimes_\alpha B$ for some reasonable tensor product (in this case the spatial tensor product). None of these properties is known for $C^∗_r(F_2)$.

@HannesThiel, so maybe you could organize all this long conversation into the single answer?

@Tomek: Thank you for pointing out that reference, I was not aware of it. It deals exaclty with the topic I asked about.

As Taka pointed out, among the simple, separable, non-type I C*-algebras, all the nuclear ones are isomorphic as Banach spaces (even much more is true) but beyond the nuclear case this is no longer the case. This certainly answers my question, thanks.

Hannes, this ma.utexas.edu/users/rosenthl/pdf-papers/93.pdf survey article of Rosenthal gives a nice overview of the Banach/operator-space structure of C*-algebras.

@BillJohnson: HRS proved that none of Type II and III noncommutative $L^1$ spaces Banach embed into a type I noncommutative $L^1$ space. If $A$ is a type I $C^*$-algebra, then $A^*$ is a predual of type I von Neumann algebra.

Thanks, Nik.  Taka, can you give a lazy old man a hint as to why the HRS classification theorem yields that Type I is invariant under Banach space isomorphisms?

By Szankowski (Acta Math 1981), there is a separable (simple,  unital, etc.) $C^*$-algebra which does not have the approximation property. Such a $C^*$-algebra cannot be Banach isomorphic to a nuclear one.

A complement to Caleb's remark: Being type I is invariant under a Banach space isomorphism (at least in the separable case). This follows from Haagerup--Rosenthal--Sukochev's classification of noncommutative $L^1$-spaces up to Banach isomorphism.

@Caleb: Thanks for pointing out the result of Kirchberg. This clarifies the picture a lot. The question now seems to be whether the Banach space structure can detect nuclearity of the C*-algebra.

@Qiaochu: Thanks for clarifying. Here, two Banach spaces E and F are called isomorphic if there exists a bounded linear map from E to F that is bijective. (The inverse map will automatically be bounded, too). I do not want to assume isometric isomorphism.

Hi Hannes, Kirchberg showed in his subalgebras of CAR algebra paper that all separable nuclear, non Type I C*-algebras are isomorphic as operator spaces (much stronger than just isomorphic as Banach spaces).  I think Christensen and others also did some things in this direction for von Neumann algebras: those results are in Pisier's operator space book, but my copy isn't with me.

@Bill: the "simplest" example of a simple C*-algebra (hard not to make puns here) is the algebra $K(H)$ of compact operators on a separable Hilbert space. The irrational rotation algebras $A_\theta \subset B(L^2({\bf T}))$ generated by multiplication by $e^{2\pi i t}$ and rotation by the angle $2\pi\theta$ for irrational $\theta$ are also simple. The CAR algebra is obtained by embedding $M_{2^n}$ in $M_{2^{n+1}}$ via the map $A \mapsto \left[\matrix{A&0\cr 0&A}\right]$ and completing the union $\bigcup_{n=1}^\infty M_{2^n}$; it is also simple.

I assume "isomorphic as Banach spaces" means linear homeomorphism. It's well known that there are non-isomorphic C*-algberas that are isometric as Banach spaces. (Google "C*-algebra not isomorphic to its opposite algebra.) The correct statement is: A and B are isomorphic as C*-algebras if and only if they are completely isometric as operator spaces.

What are some simple and complicated examples of simple separable $C^*$-algebras?

Is an isomorphism of Banach spaces a bounded linear map with bounded linear inverse or an isometric isomorphism?


"Three great cocycles" in Complex Analysis as cohomology generators

In his [lecture notes](http://www.math.harvard.edu/~ctm/home/text/class/harvard/213a/00/html/course/course.pdf), C. McMullen discusses "the three great cocycles" in Complex Analysis: the derivative $$f\mapsto\log f',$$ the non-linearity $$f\mapsto (\log f')'dz$$ and the Schwarzian derivative $$f\mapsto \left(\frac{f'''}{f'}-\frac32\left(\frac{f''}{f'}\right)^2\right)(dz)^2$$ I would like to understand what "cocycle" means here precisely, and in which sense these maps are "the only" cocycles.

For diffeomorphisms of the circle, this is very clearly explained, e. g., in the [book](http://www.math.psu.edu/tabachni/Books/BookPro.pdf) by Ovsienko and Tabachnikov. Namely, $\mathrm{Diff}(S^1)$ is a group acting naturally on each of the spaces $F_\lambda(S^1)$ of tensor densities of degree $\lambda$. The corresponding (group) 1-cohomology is trivial unless $\lambda$ equals $0$, $1$, or $2$, where it is one-dimensional and generated by the above cocycles.

Question: is there an equally simply formulated statement in the context of complex variable, e. g. treating the above cocycles as generators of a group (or perhaps groupoid?) cohomology? This [MO answer](https://mathoverflow.net/a/38133/56624) suggests that the group is $\mathrm{Aut}$(meromorphic functions), but, at least naively, this seems to coincide with a subgroup of the Mobius group.

**Update:** Let me restate the question in simpler terms.The maps above satisfy the cocycle equation $$ D(g\circ f)(z)=(f'(z))^\lambda D(g)(f(z))+D(f(z)), $$ for $\lambda=0,1,2$, respectively. For a tensor density $\psi$, one can define the coboundary $$ D_\psi:f\mapsto (f')^\lambda(\psi\circ f)-\psi, $$ satisfying the same equation. So, the question is: what are the explicit conditions guaranteeing that the three maps above are the only, up to coboundaries, solutions to the cocycle equation?

I'm confused by the notation, especially $D$ and $\circle$.

See perhaps "Explicit formulae for cocycles of holomorphic vector fields with values in $ \lambda$ densities" by Wagemann (arxiv.org/pdf/math-ph/0003035.pdf).

@WillSawin, well, in the case of the circle the space of tensor densities of degree $\Lambda$ is just the vector space of smooth functions on the circle, endowed with the diffeomorphism group action $g\mapsto (f')^\lambda \cdot g\circ f$. In the complex variable case, it's a part of my question what this space should be...


Freedman's $E_8$-manifold is nontriangulable, as proved on page (xvi) of the Akbulut-McCarthy 1990 Princeton Mathematical Notes "Casson's invariant for oriented homology 3-spheres". Kirby showed that a compact 4-manifold has a handlebody structure if and only if it is smoothable: [1](http://www.manifoldatlas.him.uni-bonn.de/index.php/Questions_about_surgery_theory) and [2](https://mathoverflow.net/questions/7921/failure-of-smoothing-theory-for-topological-4-manifolds/38739#38739). When is a compact topological 4-manifold a CW complex?

I have good reason to believe that it is an open question! Apologies - I hadn't seen the earlier posting  mathoverflow.net/questions/36838/…

I take it this is an open problem?  Small note, but your question was effectively asked in another form, here: mathoverflow.net/questions/36838/…  I usually think of CW-complexes as being a tool for describing homotopy-types rather than homeomorphism types, so my answer was to a weaker question than the one asked.


Cohomology of symmetric groups and the integers mod 12

When $n \ge 4$, the third homology group $H_3(S_n,\mathbb{Z})$ of the symmetric group $S_n$ contains $\mathbb{Z}_{12}$ as a summand. Using the universal coefficient theorem we get $\mathbb{Z}_{12}$ as a summand of the cohomology group $H^3(S_n,\mathbb{Z}_{12})$.

**What's a nice 'formula' for a $\mathbb{Z}_{12}$-valued 3-cocycle on $S_n$ that generates this summand of the third cohomology?**

In more concrete terms, I'm looking for a nontrivial recipe to get an element $c(g,g',g'') \in \mathbb{Z}_{12}$ from three elements of $S_n$, such that

$$ c(g',g'',g''') - c(gg',g'',g''') + c(g,g'g'',g''') - c(g,g',g''g''') + c(g,g',g'') = 0 $$

for all $g,g',g'',g''' \in S_n$. Here "nontrivial" means that for all nonzero $\alpha \in \mathbb{Z}_{12}$, we do not have

$$ \alpha c(g,g',g'') = f(g',g'') - f(gg',g'') + f(g,g'g'') - f(g,g') $$

for some $f \colon S_n \times S_n \to \mathbb{Z}_{12}$.

@DevSinha - can you explain that class and how it works in more detail, or point me toward an explanation?   I wanted a "formula", but an explicit geometrical description like this would also be nice.

@PhillippLampe - another typo!  I'll fix it, thanks.

@JohnBaez Why do the second and the third summand in the cocycle condition have the same sign? Should the sign not alternate?

From a number theoretic perspective, the third homology group of the general linear group is related to algebraic $K$-theory, namely $K_3$ (this seems relevant because $S_n$ embeds into $\mathrm{GL}_n$). In general the definition of $K$-group is highly non-constructive but $K_3$ is related to the Bloch group which is very explicit and some people have studied its torsion. The group $K_3(\mathbb{Z})=K_3(\mathbb{Q})$ is cyclic of order 48. By conjectures of Quillen-Lichtenbaum and Bloch-Kato the size of $(K_3)_\mathrm{tors}$ is linked with $\zeta'(-1)=-1/12$.

@JohnBaez, I can tell you about the geometry of this cocycle, using Fox-Neuwirth cochains in the unordered configuration space model for $BS_n$: this class is represented by having four points which share a coordinate, and three points which all share a coordinate, with two sharing another coordinate.  There is probably a prettier subvariety which represents this, but I don't have any techniques/ideas to get at those.

@JohnBaez -- I guess you've seen the formulae for 3-cocycles on dihedral groups in the math-phys literature?  It doesn't directly help, since you assume $n \geq 4$... but it gives some idea of what might be possible.

It would be equivalent (and also interesting) to demonstrate a $\Bbb Z/12$-valued invariant for bordism classes of oriented  3-manifolds equipped with $n$-fold covering space, $n \geq 4$. It's not so clear to me that this is any easier, though.

This doesn’t really address your question, but I once saw Fernando Rodriguez-Villegas give a talk called “All 12s are the same.” The theme was exactly your motivation here - that of describing deep connections that underlie various occurrences of the number 12 throughout mathematics. I don’t recall if this occurrence was discussed, but he might be a good guy to hit up as a curator of 12s.

@YCor Yes thank you, I guess I constructed an element of $H_3(S_n)$ coming from the generator of $H_3(A_n)$ but maybe I am confused here too... At any rate I did not think at all about how to move to $H^3$

@მამუკაჯიბლაძე $H_3$ is not cyclic in this case, so I'm confused what you want to call a generator in $H^3(S_n,Z/12Z)$. You maybe mean an element of order 12.

On the afterthought - I think I had (at least) one thing imprecise: $S_5$ embeds in $O(3)$, and its pullback to $Pin(3)$ is actually a double cover of $S_5$, so you get a 3-cycle for that double cover, which must then be descended back to $S_5$ via transfer/induction/Gysin/whateveritscalled...

@JohnBaez Yes I think one might also obtain a generator of $H_3(A_n)$ this way. I said a little bit about it in an answer here

Our comments crossed paths.  Okay, so your idea (in the case $S_5$) is based on embedding $S_5$ as a subgroup of $\mathrm{Pin}(3)$.  This is closely related to Epa and Gantner's method embedding of the double cover of $A_n$ in $\mathrm{Spin}(n)$.

I didn't see that recipe in there.  I like it but I don't quite understand it. It seems there should be a direct way to see how 3-simplices in the triangulation you describe give 3-chains in some chain complex that computes the homology  of $S_5$.  But I'm not seeing it.

Hmmm I don't actually see clearly a proof that for larger $n$ the cycle I describe will continue to be a generator - in fact that it will not even bound. In detail one should do this: realise $S_5$ as a subgroup in $Pin(3)$; for larger $n$, find an $m$ such that there is an embedding of $Pin(3)$ into $Pin(m)$ that (1) induces isomorphism on $H_3$ and (2) carries $S_5$ into a subgroup of a $S_n$ subgroup. While this will give what is needed, what I don't see is whether you can choose such a thing in such a way that $S_5$ will land in a subgroup as indicated in my previous comment...

And I think that answer linked by @QiaochuYuan contains also a recipe for an explicit 3-cycle representing a generator in $H_3$. Take the Cayley graph for some presentation of $S_5$ by generators and relations - e. g. the Coxeter one. It gives a polyhedral subdivision of the 3-sphere (for the Coxeter presentation it consists of all (hyper)faces of a permutahedron). Triangulate it, and take the sum of all obtained 3-simplices (with signs according to the orientation). For larger $n$, just choose any $S_5$ subgroup on a 5-element subset of $\{1,...,n\}$

Yes, it needs that summand.  Fixed, thanks!

I believe the last (non-coboundary) condition needs one more summand, $-f(g,g')$

@MinseonShin - yes, I'll fix that.  Thanks!

Should the domain of $f$ be two copies of $S_{n}$?

@QiaochuYuan - that's a great explanation of what's secretly motivating my question! However, I'm looking for something maximally concrete: a 'formula', as explicit as possible, that computes an element of $\mathbb{Z}_{12}$ from 3 permutations, obeying the conditions I listed. If you have to use conformal nets to get that element, okay... but I'm hoping there's a formula that combinatorists or finite group theorists could more easily enjoy.  Sort of like the sign of a permutation, only a lot subtler.

Here's a thing I wrote which is maybe relevant to your question: mathoverflow.net/a/196130/290 In the frame of that answer you are looking for something like a suitably nontrivial action of $S_n$ on an object in a suitably nice 3-category, and you can try to get such a thing by taking powers of an invertible object in a symmetric monoidal 3-category (for example, as suggested in that answer, conformal nets). The action of $S_n$ on such powers factors through the $3$-truncation of the sphere spectrum which naturally introduces a $\mathbb{Z}_{24}$.

OK. Note that this is equivalent to independently asking for (a) an explicit 3-cocycle $c_4$ valued in $Z/4Z$, such that $2c_4$ is not a coboundary; (b) an explicit 3-cocycle $c_3$ valued in $Z/3Z$, which is not a coboundary. [The solutions $c_3$ and $2c_4$ are unique modulo coboundaries, but not $c_4$.]

@ChrisGerig - an explicit 4-term exact sequence would be a big step in the right direction and in some ways even better than the cocycle.

Since I'm trying to get to the bottom of why the numbers 12 and 24 show up in many contexts in mathematics, I greatly prefer a $\mathbb{Z}_{12}$-valued cocycle and suspect there will be a good answer to my question, even if technically speaking it involves some noncanonical choices.  However, beggars can't be choosers!  So, a $\mathbb{Z}_6$-valued cocycle would also be okay.

The homology is isomorphic to $C_{12}\oplus C_2(\oplus C_2)$ for $n\ge 4$ (without the second $\oplus C_2$ for $n=4,5$ groupprops.subwiki.org/wiki/…) but there is no reason that the $C_{12}$ summand to be canonical a priori. If you're looking for a nice formula it's likely that you don't want non-canonical choices and hence have to accept these few $C_2$ terms (on the other hand, modding out the 2-torsion should give something canonical, in $C_6$).

Would you accept the following rephrasal? The generator of your summand in $H^3(S_n,\mathbb{Z}/12)$ corresponds to a 4-term exact sequence (a crossed module extension) $0\to \mathbb{Z}/12\to N\to E\to S_n\to 0$, from which you can write down an explicit 3-cocycle using a choice of cross-section $S_n\to E$. I am not sure if this is due to MacLane (could check later).


Is there an Ehrhart polynomial for Gaussian integers

Let $N$ be a positive integer and let $P \subset \mathbb{C}$ be a polygon whose vertices are of the form $(a_1+b_1 i)/N$, $(a_2+b_2 i)/N$, ..., $(a_r+b_r i)/N$, with $a_j + b_j i$ being various Gaussian integers. For any Gaussian integer $u+vi$, let $h_P(u+vi)$ be the number of lattice point in $(u+vi) \cdot P$. I would like to say that $h_P(u+vi)$ is in some sense "quasi-polynomial modulo $N$". To emphasize: **The reason that this does not follow from the ordinary Ehrhart polynomial theorem is that I am multiplying by a Gaussian integer, not just an ordinary integer.**

The case $N=1$ can be studied using [Pick's theorem](http://en.wikipedia.org/wiki/Pick%27s_theorem). The area of $P$ transforms simply under multiplication by Gaussian integers, the number of lattice points on the boundary is only a little messier. In particular, it is easy to show that there is some non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $h_P(u+vi) = (u^2+v^2) \mathrm{Area}(P) + \mbox{constant}$.

A result that would satisfy me is that there is some non-zero non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $$h_p(u,v) = (u^2+v^2) \mathrm{Area}(P) + a(u,v) u + b(u,v) v + c(u,v)$$ where $a$, $b$ and $c$ are periodic modulo $N$ in each input.

Motivation: This is enough to give an elementary deduction that the [quartic residue](http://en.wikipedia.org/wiki/Quartic_reciprocity#Quartic_residue_character) symbol $\left[ \frac{z}{u+vi} \right]$ is periodic as a function of $u+vi$.

Motivation behind the motivation: There are a series of papers: [Habicht](http://www.ams.org/mathscinet-getitem?mr=113867), [Kubota](http://www.ams.org/mathscinet-getitem?mr=921585), [Hill](http://www.ams.org/mathscinet-getitem?mr=1324545) which promise to prove $m$-th power reciprocity laws roughly by counting lattice points. I can't follow any of them, so I thought I would true to reverse engineer what might be true in order to make the $\mathbb{Z}[i]$ proof come out.

For this reason, I would also be interested in statements for other number fields, but I don't have a precise question worked out in that setting.

Dear David. Sorry for the late question, but, did you eventually found some place (in addition to the articles you mention) where this is explained/worked, hopefully for other number fields?

Yeah, right, I didn't think straight. Sorry for noise.

I do not see this. A rotation can change the lattice in a quite messy way. What am I missing?

naively, at least if $|a+ib|\in\mathbb{Z}$, this should follow from the classical case; indeed, by multiplying by $a+ib$ you apply to your $\mathbb{C}$-plane the composition of a rotation and the scalar matrix $C=|a+ib|I$, and so the integer points would behave in the same way as by scaling with $C$ alone.

@DimaPasechnik Because I'm multiplying by Gaussian integers, not ordinary integers. Edited to clarify.

I was under impression that the quasiperiodicity of Ehrhart polynomial holds for any lattice $\Lambda$ in $\mathbb{R}^d$ and a polytope with vertices in $\Lambda$. How does your setting differ?


Nontrivial tangent bundle that is diffeomorphic to the trivial bundle

Is there an example of a smooth $n$-manifold $M$ whose tangent bundle is nontrivial as a bundle but is nonetheless (abstractly) diffeomorphic to the trivial bundle $M \times \mathbb{R}^n$?

(This question was inspired by [Trivial fiber bundle](https://mathoverflow.net/questions/58685/trivial-fiber-bundle).)

$T^*G\cong G\times \mathfrak g^*\cong G\times \mathbb R^{n}$

@Dylan and @Somnath: Tangent bundles over spheres never give such examples. Moreover, in general, if the homotopy-equivalence $M\to M$ induced by a diffeomorphism $TM\to M\times {\mathbb R}^n$ is homotopic to a diffeomorphism $M\to M$, then $TM$ is trivial. See "Diffeomorphism of total spaces and equivalence of bundles" by De Sapio and Waldschap. This still leaves open the case when $M$ is a homotopy-sphere though.

@Dylan - That was the example I was thinking of too! It is clear that $TS^5$ is not the trivial rank $5$ bundle over $S^5$. It only has $3$ linearly independent nowhere vanishing vector fields (for lack of anything elementary one can use Adams' vector fields on spheres result). Although I don't know how to do it myself, if you can prove (extend your sphere bundle result to vector bundles) your claim, we're done!

I think I can show that the unit tangent bundle over $S^5$ is diffeomorphic to the trivial one... but I'm having trouble otherwise, since most of the diffeomorphisms I can think of rely on theorems about compact spaces.

@David, is no reason for the diffeomorphism $TM\to M\times \mathbb R^n$ to cover a diffemorphism of $M$. @Faisal, the answer to your question is almost certainly yes, but I do not see an explicit example  at the moment, cf. my answer to mathoverflow.net/questions/58685/trivial-fiber-bundle.

A trivial observation: such an isomorphism would cover an automorphism of $M$. A possibly unhelpful observation: One could try to put a Riemannian metric on $M$ and see what constraints this gives in the light of my first remark.


Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$?

> Is there a field $F$ such that $F \cong F(X,Y)$ as fields, but $F \not \cong F(X)$ as fields?

I know only an example of a field $F$ such that $F$ isomorphic to $F(x,y)$ : this is something like $F=k(x_0,x_1,\dots)$. But in this case we have $F \cong F(x)$.

This is related to [this](https://mathoverflow.net/questions/225896) question. This was previously [asked](https://math.stackexchange.com/questions/2000333) on MSE but after 2 months and a bounty, no answer was provided.

Thank you very much!

Oops, I forgot to mention that I've posted a related question here: mathoverflow.net/questions/259117/…

I have an example where I know $F\cong F(X,Y)$ but I don't know whether $F\cong F(X)$. There are torsion-free abelian groups $G$ such that $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$ but $G\not\cong G\oplus\mathbb{Z}$ (see mathoverflow.net/questions/218113/… for example). Let $\mathbb{Q}(G)$ be the field of fractions of the group algebra $\mathbb{Q}[G]$. Then $\mathbb{Q}(G)\cong\mathbb{Q}(G)(X,Y)$, but I don't see why it should be isomorphic to $\mathbb{Q}(G)(X)$.

Related on MSE: math.stackexchange.com/questions/388291


The field of fractions of the rational group algebra of a torsion free abelian group

Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.

> Are there non-isomorphic torsion free abelian groups $G$ and $H$ such that $\mathbb{Q}(G)\cong\mathbb{Q}(H)$? If so, are there easy examples?

This is motivated by [this question](https://mathoverflow.net/questions/258647/is-there-a-field-f-which-is-isomorphic-to-fx-y-but-not-to-fx). There are [examples](https://mathoverflow.net/questions/218113/a-is-isomorphic-to-a-oplus-mathbbz2-but-not-to-a-oplus-mathbbz) of torsion free abelian groups $G$ where $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$, but $G\not\cong G\oplus\mathbb{Z}$. Since $\mathbb{Q}(G\oplus\mathbb{Z}\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X,Y)$ and $\mathbb{Q}(G\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X)$, a negative answer to my question would give a positive answer to the motivating question.

I've found a little about this construction in the literature, but not much. For example, in

_Gervasio G. Bastos and T. M. Viswanathan_ , MR 942063 [**Torsion-free abelian groups, valuations and twisted group rings**](http://dx.doi.org/10.4153/CMB-1988-021-x), _Canad. Math. Bull._ **31** (1988), no. 2, 139--146.

it is proved that $\mathbb{Q}(G)$ is a purely transcendental extension of $\mathbb{Q}$ if and only if $G$ is free abelian.

Any other relevant references would be welcome. Also, there are obvious variants: e.g., using a different field, such as $\mathbb{F}_2$ instead of $\mathbb{Q}$.

Lovely question Jeremy. I only wish I could say something useful about it!

Just a remark: if $G$ has no primitive element (in the sense that for every $g\in G$ there exists $n\ge 2$ such that $n^{-1}g\in G$) then I guess that $\pm G$ is characterized within $\mathbb{Q}(G)^*$ as those elements with roots of arbitrarily high order. (This holds if $G$ is a $\mathbb{Z}[1/p]$-module, or if $G$ is noncyclic of rank 1.) Since $G$ is isomorphic to $(\pm G,\cdot)$ modulo its torsion subgroup, it follows that $\mathbb{Q}(G)$ recognizes $G$ in this case.


Linking formulas by Euler, Pólya, Nekrasov-Okounkov

Consider the formal product $$F(t,x,z):=\prod_{j=0}^{\infty}(1-tx^j)^{z-1}.$$ (a) If $z=2$ then on the one hand we get Euler's $$F(t,x,2)=\sum_{n\geq0}\frac{(-1)^nx^{\binom{n}2}}{(x;x)_n}t^n,$$ on the other we get Pólya's formula (the "cycle index decomposition", see Stanley's EC2, p.19) $$F(t,x,2)=\sum_{n\geq0}Z(S_n,(1-x)^{-1},\dots,(1-x^n)^{-1})t^n.$$ (b) If $t=x$ then we get Nekrasov-Okounkov's $$F(x,x,z)=\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\left(1-\frac{z}{h_{\square}^2}\right);$$ where $h_{\square}$ is the hook-length (for notations and references you may see _[Theorems, problems and conjectures](http://dauns.math.tulane.edu/%7Etamdeberhan/conjectures.html)_).

> **My Question still waiting for an answer.** Is there a unifying combinatorial right-hand side in $$\prod_{j\geq0}(1-tx^j)^{z-1}=?$$

**Remark 1.** One possibility is perhaps a refinement in line with (b) involving hook-lengths.

**Remark 2.** Even special cases, such as evaluation for specific numerical values of $t$ and/or $z$, would be interesting.

**Remark 3.** Yet another direction is to find a hook-length expansion for $F(t,x,2)$. This could be fascinating because the result will connect the cycle index polynomial with hooks.

never mind -- NO formula generalizes the cycle index

does Nekrasov-Okounkov formula really fall out of cycle index for permutation groups??


Non-linear expanders?

Recall that a family of graphs (indexed by an infinite set, such as the primes, say) is called an _expander family_ if there is a $\delta>0$ such that, on every graph in the family, the discrete Laplacian (or the adjacency matrix) has spectral gap $|\lambda_0-\lambda_1|\geq \delta$. (Assume all graphs in the family have the same degree (= valency) $d$.)

If, for every $p$, we specify some maps from $\mathbb{Z}/p\mathbb{Z}$ to itself, then we are defining a graph with vertex set $\mathbb{Z}/p\mathbb{Z}$ for each $p$: a vertex is adjacent to the vertices it is mapped to.

If we consider the maps $x\to x+1$, $x\to 3x$ (say), then we do not get expanders. On the other hand, if we take $x\to x+1$, $x\to x^{-1}$, then, as is widely known, we do get an expander family.

Consider the maps

$x\mapsto x+1$, $x\mapsto x^3$.

Do they (and their inverses, if you wish) give an expander family? (Let $p$ range only over primes $p\equiv 2 \mod 3$, so as to keep the map $x\mapsto x^3$ injective.)

What about the maps $x\mapsto x+1$, $x\mapsto 3 x$, $x\mapsto x^3$? Do they, taken together, give an expander family?

(Is there a way to relate these maps to the action of a linear group? Are there examples of sets of not neessarily linear maps giving rise to expander families?)

UPDATE: What if these were shown _not_ to be expanders? What would be some interesting consequences?

Can you say anything on the structure of those $A\subset{\mathbb F}_p$ with $|(3A)\setminus(A+1)|<c|A|$ (or smaller than $|A|^\delta$ at least)? If so, it might be possible to use this information to show that such $A$ cannot be stable under the map $x\mapsto x^3$.

Yes, as I said it should do something like expand $|A|$ to $|A| + |A|^\delta$ for some small $\delta$.

That would give only a very, very weak expansion result, wouldn't it?

I was thinking something more along the lines of Proposition 4.3 of my paper with Ben: arxiv.org/abs/0709.3562 , which should show enough equidistribution in the function $x \mapsto x^3$ on medium-sized intervals in ${\bf Z}/p{\bf Z}$   (maybe cutting out the interval of size $p^{1/3}$ or so around the origin, and maybe removing other "major arcs" also) to show that the image of that interval has small intersection with other medium-sized intervals.

I can see that $|A\cup (A+1)|\leq (1+\delta) |A|$ implies that $\sum_{x\in A} e(\xi x)$ is small for $\epsilon p\leq \xi \leq (1-\epsilon) p$ (say), and that one can get from that (by Weyl differencing) that $\sum_{x\in A} e(\xi x^2)$ must be small for all $\xi$ in a range. But how would one show from that that $|A\cup A^3|$ is large?

Actually, thinking about it more, Bourgain-Konyagin (which has to do with how arithmetic progressions and geometric progressions interact) is more relevant for the $x \mapsto x+1, x \mapsto 3x$ case.  For $x \mapsto x+1, x \mapsto x^3$, it is indeed Weyl differencing which is the key: to show that if A is basically the union of a small number of intervals, then $A^{\cdot 3}$ does not behave like the union of a small number of intervals.

It looks like Bourgain-Konyagin type exponential sum estimates may be able to prove a very weak expansion result of the form $|A \cup (A+1) \cup A^{\cdot 3} | \geq |A| + |A|^\delta$ when $p^\varepsilon \leq |A| \leq p^{1-\varepsilon}$, where $A^{\cdot 3} = \{ a^3: a \in A \}$, which would almost give a non-trivial diameter bound, but the improvement is slight and one also would have to treat the case of very small or very large A somehow.  [Maybe a Weyl differencing argument would work?]

Agreed - yes, these are my reasons for asking! But what about diameter bounds? I actually want expansion, but that might already be somewhat interesting.

... proving expansion rigorously, though, would probably require substantial new ideas since there does not appear to be any natural group involved other than the free group.  But one could imagine that sum-product theorems could at least give some non-trivial bounds on diameter and mixing time, even if they fall short of true expansion.

As for the last question, as you well know, it was proven long ago by Pinsker that any pair of random permutations will form an expander on $\{1,\dots,n\}$ with probability $1-o(1)$.  Absent any obvious nilpotent or solvable structure (as with the $x \mapsto x+1, x \mapsto 3x$ case) I would conjecture that $x \mapsto x+1, x \mapsto x^3$ behaves like two random maps for the purposes of expansion, since there do not appear to be any local relations between these maps (in contrast to the former case in which we have $3(x+1) = 3x + 1 + 1 + 1$).


Can every 3-dimensional convex body be trapped in a tetrahedral cage?

Although the question is fairly unambiguous, I give all relevant definitions:

$\bullet$ A subset $C$ of $\mathbb{R}^n$ is an $n$-dimensional _**convex body**_ if $C$ is convex, compact, and has non-empty interior.

$\bullet$ A _**polyhedral cage**_ $P^{(1)}$ in $\mathbb{R}^3$ is the union of all edges (_i.e._ , the 1-skeleton) of a convex $3$-dimensional polyhedron $P$. In particular, a _**tetrahedral cage**_ is the union of the six edges of some tetrahedron.

$\bullet$ A convex $3$-dimensional body $C$ is _**trapped**_ by the tetrahedral cage $T^{(1)}$, that is, by the $1$-skeleton of the tetrahedron $T$, if the cage is fixed (motionless), and if for every continuous rigid motion (rotations allowed) $f_t(C);\ 0\le t\le 1$, either $f_t(C)$ intersects $T$ for every $t\in [0,1]$ or $T^{(1)}$ contains an interior point of $f_t(C)$ for some $t_0\in [0,1]$.
In other words, $C$ cannot be moved arbitrarily far from $T$ while, during the entire motion, avoiding the cage's bars penetrating $C$'s interior.

$\bullet$ Remark. In dimension $n>3$, one can consider analogous questions, with a variety of types of a simplicial cage, by taking the $i$-skeleton of a convex $n$-dimensional simplex, with $1\le i\le n-2$.

A trivial example: A ball is trapped in the cage consisting of the $1$-skeleton of the regular tetrahedron edge-tangent to the ball. Also, obviously, every convex body can be trapped in some polyhedral cage.

For an interested reader, I suggest a few of somewhat less trivial exercises: each of the following convex bodies can be trapped in a tetrahedral cage: the cube, the circular cylinder of any (finite) height, the circular cone, the regular tetrahedron.

@Anton  The half-circle, for example, and many other, related.

In 2-dimensional case, one can ask which convex figures can be trapped by three nails. Rectangle cannot be trapped, but I do not see other examples.

@DanRomik Romik : Yes, it is quite a bit different.

The paper How to cage an egg by Oded Schramm (Inventiones Math. 107 (1992), 543-560) comes to mind as potentially relevant, although the question he answers is a bit different than yours.

@JosephO'Rourke: Yes. More precisely, some of the traps for the cone also trap the spindle, no matter how thin it is.

Does your trap for a cone also trap a thin spindle? Say, two tall, base-to-base cones?


Mathieu group $M_{23}$ as an algebraic group via additive polynomials

An elegant description of the Mathieu group $M_{23}$ is the following: Let $C$ be the multiplicative subgroup of order $23$ in the field $F=\mathbb F_{2^{11}}$ with $2^{11}$ elements. Then $M_{23}$ is the group of additive maps of $F$ to itself which permute the set $C$. The restriction to $C$ then is the natural action of $M_{23}$ of degree $23$.

Any additive map of $F$ to itself has the form $x\mapsto\sum_{i=0}^{10}a_ix^{2^i}$ for $a_i\in F$. The condition that such maps preserve $C$ can be expressed easily as an equivalent set of polynomial conditions on the coefficients $a_0,\dots,a_{10}$. For instance, for a variable $t$, compare coefficients of $$\prod_{c\in C}(t-\sum_{i=0}^{10}a_ic^{2^i})=t^{23}-1.$$ In addition, we need the equations $a_i^{2^{11}}=a_i$ to ensure that the $a_i$ are in $F$.

However, the polynomials we obtain this way (or some other way, there are somewhat smarter possibilities) are extremely complicated.

Of course, any finite subgroup of a linear group is an algebraic set, that is a solution of a system of polynomial equations. And in general such a system will be messy.

Here, I somehow expect that there should be a simple system of polynomials discribing $M_{23}$ as an algebraic set. The obvious attempt would be to first take a system as above, and then hope that a Groebner basis looks better. After some naive attempts it seems that the systems are too complex for Magma or Singular to compute the Groebner bases.

Before trying to refine the approach, here is my question: Has anyone seen this version to describe $M_{23}$?

@DaveBenson Indeed, I had also tried something along this idea. The upper bound of the Galois group of $Y^{23}+XY^3+1$ is achieved by showing that this polynomial divides an additive polynomial of degree $2^{11}$. So actually this additive polynomial has Galois group $M_{23}$, and its groups acts on an $11$-dimensional $\mathbb F_2$-space.

This is a very old question, but I can't help adding the following comment, that may or may not be relevant. I wonder whether Abhyankar's paper, "Mathieu group coverings in characteristic two" is relevant here? He demonstrates that there are very innocuous looking degree 23 trinomials over ${\mathbb F}_2(X)$ with Galois group $M_{23}$. His example is $Y^{23}+XY^3+1$.

The additive subgroup generated by $C$ is a subfield, is not reduced to $F_2$, and hence equals $F_{11}$. Hence the fixator of $C$ in $GL_{11}$ is reduced to the trivial subgroup. Hence the stabilizer $G$ of $C$ in $GL_{11}$ is finite (as an algebraic group). Since $|C|=23$ and the dimension is 11, there exists a 3-element subset $F$ of $C$ such that $C-F$ generates $F_{11}$ additively. Hence a 3-cycle on $F$ cannot belong to $G(K)$ for any extension $K$ of $F_2$. So the (faithful) action of $G(K)$ on $C$ is reduced to $M_{23}$, and hence $G(K)$ is reduced to $G(F_2)=M_{23}$ for every $K$.

If I understand correctly, you don't know the answer (whether $G$ has positive dimension, where $G\subset\mathrm{GL}_{11}$ is the stabilizer of $C$).

@YCor: I tried to answer that in my first comment above.

Why are you not happy with just defining the $G$ as the subgroup $\mathrm{GL}_{11}$ that preserves $C$? then $G(\mathbf{F}_2)$ is reduced to the Mathieu group. No need to incorporate the conditions that entries are in $\mathbf{F}_2$: this is part of the definition of $G(\mathbf{F}_2)$. If $G$ is defined as I just did, I don't even know if it's positive-dimensional.

one natural way to produce an explicit set of equations is to identify a polynomial $p$ left invariant by $M_{23}$, but not by any overgroup of it in GL(11,2), and write down equations for the matrix entries saying that  $p$ is invariant.

@DimaPasechnik Yes, I know that. But I believe that has nothing to do with my question.

It is known how to express generators of $M_{23}$ as a permutation group in terms of rational univariate maps over $\mathbb{F}_{23}$. (Not sure about the original reference, I saw it in a book by D. Gorenstein ams.org/mathscinet-getitem?mr=698782). More precisely, it gives such representation for $M_{24}$, as the maps in PSL(2,23) and one extra binomial of relatively high degree.

@DimaPasechnik Right, one could try some smaller toy examples. The group $PSL(3,2)$ maybe is not that interesting though, because it equals $GL(3,2)$. So all that we polynomially encode would be bijectivity of the linear maps.

@JimHumphreys I hope to use it for certain combinatorial questions about $M_{23}$, for instance in the study of sharply transitive sets of permutations. Such question have a polynomial formulation, so if the $M_{23}$ has a nice polynomial formulation too, I have some hope.

E.g. one may ask for such a description for PSL(3,2): I suppose you have to replace 11 by 3 and 23 by 7, i.e. $|C|=7$ and $F=\mathbb{F}_{2^3}$, and consider PSL(3,2) as the group of additive maps of $F$ preserving $C$. (Disclaimer---I didn't check all the details here :-)).

the natural object preserved by the group here is dual of the perfect binary Golay code en.wikipedia.org/wiki/Binary_Golay_code One might ask for similar descriptions of smaller codes to begin with.

This looks like a reasonable (though maybe quite difficult) question.    But I wonder what consequences such a description would have?


Derivative of Class number of real quadratic fields

Let $\Delta$ be a fundamental quadratic discriminant, set $N = |\Delta|$, and define the Fekete polynomials $$ F_N(X) = \sum_{a=1}^N \Big(\frac{\Delta}a\Big) X^a. $$ Define $$ f_N(X) = \frac{F_N(X)}{X^N-1}. $$ Canceling common factors it can be shown that $$ f_N(X) = \frac{H_N(X)}{\Phi_N(X)}, $$ where $\Phi_N$ is the $N$-th cyclotomic polynomial and $H$ has integral coefficients. (These definitions seem rather ad hoc, but in fact the objects above are all quite "natural".)

If $\Delta < 0$, then $f_N(1) = \frac{2h}w$, where $h$ is the class number of $\Delta$ and $w$ the number of roots of unity in the corresponding quadratic number fields. In fact we get $$ f_N(1) = - \frac{\sum_{a=1}^{N-1} a \Big( \frac{\Delta}a\Big)}N . $$

If $\Delta > 0$, then $f_N(1) = 0$, which suggests looking at the first derivative. Here we find that $f_N'(1)$ is always even, so we set $f_N'(1) = - 2 h$ and get $$ h = \frac{\sum_{a=1}^{N-1} \frac{a(a-1)}2 \Big( \frac{\Delta}a\Big)}N . $$ Here is a small table:
$ $ $ \begin{array} {c|ccccccccccc} N & 12 & 13 & 17 & 21 & 24 & 28 & 29 & 33 & 37 & 41 & 44 \\\ \hline h & 1 & 1 & 2 & 2 & 3 & 4 & 3 & 6 & 5 & 8 & 7 \end{array} $

It is quite remarkable that $h$ seems to possess, for positive discriminants, properties analogous to class numbers. In particular, $h$ is odd if and only if $\Delta = p > 5$ is a prime $\equiv 5 \bmod 8$, and if $\Delta$ is divisible by $n$ distinct primes, then $h$ is divisible by $2^{n-2}$ etc.

In the case $\Delta < 0$, Hasse and Bergstrom have shown how to extract the "genus factor" from the analytic class number formula, but the necessary calculations were horrible. Thus if we want to avoid going through similar technicalities in the case $\Delta > 0$ then we would need a natural interpretation of $h$ as the order of an analogue of a "class group".

So I guess my question is whether there is such a mysterious group (connected to real quadratic number fields) with order $h$. Even a proof that $h \ne 0$ would be welcome, since in the case of negative discriminants this is equivalent to the nonvanishing of Dirichlet's L-function at $s = 1$.

**Edit.** As Rene Schoof has pointed out, the numbers in question are essentially generalized Bernoulli numbers $B_{2,\chi}$ for (quadratic) Dirichlet characters with conductor $N$. These numbers are defined by $$ B_{2,\chi} = \sum_{a=1}^N \chi(a) B_2(a/N), $$ where $B_2(X) = X^2 - X + \frac16$ is the second Bernoulli polynomial. The constant term $\frac 16$, in which this expression differs from the one I have used above, is irrelevant since $\sum \chi(a) = 0$.

These numbers $B_{2,\chi}$ show up as factors of eigenspaces of cuspidal divisor class groups as studied in the book by Kubert-Lang (more exactly, the $p$-part of the group is reflected in the divisibility by $p$ of the Bernoulli numbers). The Birch-Tate-Lichtenbaum conjecture also predicts that the $B_{2,\chi}$ are factors of $K_2({\mathcal O})$, where ${\mathcal O}$ is the ring of integers of the maximal real subfield of the $N$-th roots of unity (Kubert-Lang, p. 151).

Although these numbers have such a prominent place in number theory, their elementary properties apparently have not been investigated (I've looked into Washington's and Lang's books on cyclotomic number fields). Numerical experiments suggest e.g. the following:

Let $\chi$ be an even Dirichlet character with conductor $N$. Then $h_2(N) = \frac{N}4 B_{2,\chi}$ is an integer for all $N > 8$.
Moreover, $h_2(N)$ is odd if and only if

  1. $N = p$ for primes $p \not \equiv 1 \bmod 8$;
  2. $N = 4p$ for primes $p \equiv 3 \bmod 8$;
  3. $N = 8p$ for odd primes $p \not \equiv \pm 1 \bmod 8$.

In absence of any kind of genus theory for cuspidal divisor class groups it seems that the only option for proving such results is working directly with the definition of these numbers. As mentioned above, this becomes technical for integers with many prime factors.

There are similar results for $h_3(N) = \frac{N^2}6 B_{3,\chi}$, and there can be no doubt that most of this generalizes in some form to all generalized Bernoulli numbers $B_{n,\chi}$.

This whole set of question came out of my approach to quadratic reciprocity in Chapter 7 of my notes on [Pell conics](http://www.rzuser.uni-heidelberg.de/~hb3/pell.html).

Hi!  I think these are generalized Bernoulli numbers $B_{2,\chi}$ associated the quadratic character $\chi$ of conductor $\Delta$. Apart from a power of $2$ perhaps, they are equal to the order of the $\chi$-part of the cuspidal divisor class group of the modular curve $X_1(\Delta)$. See the book by Kubert-Lang for more details.


derived category of equivariant coherent sheaves and fixed points

The K-group $K^T(X)$ of $T$(torus)-equivariant coherent sheaves on a variety $X$ is isomorphic to $K^T(X^T)$, that of the fixed point locus via the inclusion homomorphism, when we tensor the quotient field of the representation ring $R(T)$ of $T$.

Is there a similar result for $D_T(Coh X)$ and $D_T(Coh X^T)$, derived categories of $T$-equivariant coherent sheaves ? I do not know even how to formulate `the quotient field of $R(T)$' in the derived category case.

@Ben, you're absolutely right, somehow I missed the salient word "coherent" in the question.  But even for constructible sheaves, is there a localization statement at the level of derived categories?

It doesn't seem like that to me at all.  Is there anything about coherent sheaves in GKM?

This seems like exactly the sort of thing that ought to be one of the main results in the paper of Goresky-Kottwitz-MacPherson.  Unfortunately, I haven't been able to extract the kind of statement you're asking for; it would be very interesting if someone can give a precise answer.  They do give a localization-type theorem for an element of the equivariant derived category in (7.8), but only after taking cohomology.


Computational complexity of topological K-theory

I am a novice with K-theory trying to understand what is and what is not possible.

Given a finite simplicial complex $X$, there of course elementary ways to quickly compute the cohomology of $X$ with field coefficients. However, it is known that computing the rational homotopy groups of $X$ is at least $NP$-hard (in fact, at least $\\# P$-hard), simply because it is possible to reduce NP-hard problems to the computations of some rational homotopy groups.

For any of: (1) Real K-theory, (2) Complex K-theory, (3) p-adically completed K-theory, is there an algorithm to compute $K^0$ of a finite simplicial complex? Is this algorithm polynomial-time?

If there is no known algorithm, is there at least evidence that any theoretical algorithm, should it exist, must be at least $NP$-hard? In other words, is there any way to reduce an $NP$-hard problem to the calculation of some $K$ group?

Just thought I'd add a citation for the fact that rational homotopy groups of simply-connected spaces are NP hard here. In particular, this is not just some $\pi_1$ issue.

Have you looked at Gunnar Carlsson's stuff? They have this software for computing homology; maybe it could be made to work with K-theory, or maybe they've addressed the computational complexity and that could help you bound the complexity of K-theory. Here is a link: comptop.stanford.edu. In a similar vein, Rob Ghrist has done work on algorithms for cohomology theory, but I'm not sure if he's addressed computational complexity or not. Last thought: at the JMM last year there was a special session about K-theory and electrical engineering. Maybe those guys have looked into this question

Dominique Arlettaz has a result regarding the order of differentials in the Atiyah-Hirzebruch spectral sequence, see "The order of the differentials in the Atiyah-Hirzebruch spectral sequence". I would imagine that one might be able to make certain conclusions if you assumed enough about the torsion of the homology of the space. However, on rereading your question I realize this is not so much what you were interested in. Thought I'd mention the paper of Arlettaz just in case though.

I have thought about this occasionally.  There is some reason to think that there might be a tractable combinatorial construction of a minimal Kan complex of homotopy type $\Omega^\infty(KU/2)$.  From that it should be possible to determine the computational complexity of $(KU/2)^*(X)$.  Odd primes might be possible as well.  I have various kinds of calculations related to this, but no real conclusion.

Probably not. The first differential is easy (for KU), it's determined by a cohomology operation, but I think the higher ones are algorithmically intractable and would describe higher-order cohomology operations. Maybe one should try to describe vector bundles on finite simplicial complexes combinatorially instead?

Yes, but only as a black-box.  I do not know an algorithm to compute differentials or solve extension problems.  Is this possible?

Are you familiar with the Atiyah-Hirzebruch SS?


Finite-dimensional subalgebras of $C^\star$-algebras

Let $A$ be a unital $C^\star$-algebra and let $a_1,\dots,a_n$ be a finite list of normal elements in $A$ which (together with their adjoints) generate a norm-dense $\star$-subalgebra $B \subset A$. Clearly, if $A$ is finite-dimensional, then every element in $A$ (and hence $B$) has finite spectrum. I am asking for the converse.

> **Question:** Assume that each element of $B$ has finite spectrum. Is it true that $A$ has to be finite-dimensional?

The existence of finitely-generated infinite torsion groups shows that this might be a highly non-trivial problem. In this case one would consider the reduced group $C^\star$-algebra and note that all monomials in the generators of the group and their inverses (which are equal to the adjoints of the generators) would have finite spectrum. However, the generated algebra would still be infinite-dimensional. I do not know of any simpler way to come up with such an example. In this case it is conceivable that the random-walk operator associated with the generating set (which is an element in the real group ring) has infinite spectrum, even though I did not prove this.

Maybe there is also need to consider the spectra of elements in matrices over $B$ (which of course follow to be finite if $A$ is finite-dimensional.) In view of this, I am not only asking for an answer to the question but also for the right question (or a better one) if the answer to the original question is negative.

A stronger assumption would be to assume that $A$ itself consists only of elements with finite spectrum. This case seems much easier to approach and the answer seems to be positive. In fact every infinite-dimensional $C^\star$-algebra should contain an element with infinite spectrum.

Just to get started, a more concrete instance of the question above is:

> **Question:** Let $p_1, p_2$ and $p_3$ be three projections in a $C^*$-algebra with the property that every (non-commutative) polynomial in $p_1, p_2$ and $p_3$ has finite spectrum. Is it true that the projections generate a finite-dimensional algebra?

Every finitely generated *-algebra is generated by a finite list of normal elements (the real and imaginary parts of its generators) so the assumption that the $a_i$ are normal is superfluous.

@AliTaghavi: Yes, in this case, the spectrum of $A$ is equal to the joint spectrum of $a_1,\dots,a_n$ (which is contained in the product of the spectra of $a_1,\dots,a_n$).

Is the  answer to the original qustion obvious  for  commutative  C*  algebras?

Let $G$ be a finitely-generated Group, $H$ a Hilbert space, $h : G \mapsto \mathbb{B}(H)$ a unitary representation of $G$ such that $h(G)$ is linear independent and $B = span(h(G))$. Then G is finite. For a simple proof see Proposition 2 in ON THE SPECTRUM OF THE SUM OF GENERATORS OF A FINITELY GENERATED GROUP, II, matwbn.icm.edu.pl/ksiazki/cm/cm65/cm65111.pdf

To "In fact every infinite-dimensional C⋆-algebra should contain an element with infinite spectrum." :  For a proof see ALAN W. TULLO, CONDITIONS ON BANACH ALGEBRAS WHICH IMPLY FINITE DIMENSIONALITY, journals.cambridge.org/action/…

Could be; I do not know. I was just talking with Joachim Cuntz and George Elliott about the problem, but they did not mention Richard Kadison. There is a related problem to find an element with infinite spectrum in an infinite dimensional $C^\ast$-algebra, which is a lot easier. Where did you hear that Kadison asked the question?

@Andreas,  Why don't you go ahead and ask this as a separate question. I was thinking about this as an approach to your question. One next step (which would be much harder) would to say if the spectra of these elements isn't the solution of a polynomial equation, then one of these elements has an infinite spectrum. I have no idea how to prove this, or even if it's true, but it seems to work for a few examples.

I do not know. Beautiful question! Do you want to (or may I) ask this as a separate question?

Suppose you have a C$^*$ algebra generated by $g_1$, $g_2$ and $g_3$. Consider the elements $x_1g_1+x_2g_2+x_3g_3$, where the $x_i$ are variables. If the spectra of these elements are the solutions of a polynomial equation in $x_1$, $x_2$ and $x_3$, is the algebra finite dimensional?

I have just learned that Choi-Elliott proved in (ams.org/mathscinet-getitem?mr=1081290) that the set of self-adjoint elements with finite spectrum in certain rotation algebras are dense in the set of all self-adjoint elements. (Interestingly, this result is not explicit, in the sense that there is no concrete $\theta$ for which it happens.) However, from what I understand, it seems unclear whether there is a finitely generated dense sub-algebra such that each element has finite spectrum.


Can one divide by the cardinal of an amorphous set?

This question arose in a discussion with Peter Doyle.

It is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \times \\{1,\ldots,k\\} \simeq Y \times \\{1,\ldots,k\\}$ then $X \simeq Y,$ where I use $\simeq$ to denote the existence of a bijection between two sets.

Assuming the Axiom of Choice, the positive finite cardinals are precisely the cardinals that one can divide by since $1\cdot\mathfrak{m} = 2\cdot\mathfrak{m}$ for any infinite cardinal $\mathfrak{m}$. The ultimate question is whether this is actually provable in ZF:

> If $\mathfrak{m}$ is an infinite cardinal, must there be two cardinals $\mathfrak{p} \neq \mathfrak{q}$ such that $\mathfrak{p}\cdot\mathfrak{m} = \mathfrak{q}\cdot\mathfrak{m}$?

As a candidate for a counterexample, I thought it might be possible to divide by the cardinal of a (strongly) amorphous set. Specifically, in the First Fraenkel model:

> Is it true that if $X \times A \simeq Y \times A$ then $X \simeq Y$, where $A$ is the set of atoms?

@Lorenzo: A hint would imply that this is easy... it's not! Peter Doyle has written up division by three and division by four - math.dartmouth.edu/~doyle

Can someone give me a hint on how to prove that in $\text{ZF}$ we can divide by finite cardinals?

There is Lovasz's unique factorization and unique roots for finite structures, and related results.  It might be prudent (and useful if not tangential for this question) to see in which logical domains his arguments can be repeated.  Gerhard "Forward Engineering For Reverse Mathematics" Paseman, 2016.04.09.


Unital $C^{*}$ algebras whose all elements have path connected spectrum

A unital $C^{*}$ algebra is called a "Path connected algebra" if the spectrum of all its elements is a path connected subset of $\mathbb{C}$.

> **What is an example of a non commutative path connected $C^*$ algebra? What is an example of a simple path connected $C^{*}$ algebra? In particular does $C^{*}_{red} F_{2}$ satisfiy this property?**
>
> As another question: Is the tensor product of two path connected algebra, a path connected algebra?(For spatial norm).

@JohannesHahn  Thanks  for  your  comment. As you pointed out a minimal dynamical system gives  a simple  C*  algebra. But what  about the spectrum of its  elements(Path  connected spectrum)?

@HannesThiel  Thank you very much for your  comment  and your attention. The motivation for the second part of  my question, about the reduced $C^*$ algebra, is obvious. Regarding the first part: the motivation is that the product topology of two path connected space is  a path connected space. But according to your comment, one can ask that: Is the tensor product of two idempotent less C* algebras, an idempotent less C* algebra?

@AliTaghavi What is your motivation to consider path connected and not just connected spectra?

Example of a connected C∗ algebra :  See B.E. Blackadar. A simple unital projectionless C∗-algebra. J. Operator Theory, 5:63–71, 1981.  By the comments of Hannes Thiel and Sam Evington this algebra is an example of a connected C* algebra.  But is it also path connected (remark by Ali Taghavi) ?

Hannes' observation can be improved: all elements of a C*-algebra with no non-trivial projections have connected spectrum. Indeed given an element with disconnected spectrum the holomorphic functional calculus gives a non-trivial idempotent. This will be similar to some self-adjoint idempotent (see Blackadar's K-Theory book) which will be non-trivial.

@HannesThiel  Thank you for the comment. So my second question leads to the following: What is  an example of  a projectionless  simple  $C^{*}$ algebra for which the spectrum of all elements are path connected.

The spectrum of a nontrivial projection (i.e., nonzero and not the unit) is disconnected. Thus, a unital C*-algebra with the property you consider cannot contain any nontrivial projections. Conversely, if a C*-algebra contains a normal element with disconnected spectrum, then it contains a nontrivial projection. Thus, in a unital, simple C*-algebras without nontrivial projections (such as the Jiang-Su algebra), at least the spectrum of every normal element is connected.

@JohannesHahn Thanks for your  comment. If I am not mistaken, this algebra is simple iff the action of $G$ on $X$ does not have  a nontrivial compact invariant set. Now how you controle the spectrum of elements?

Well $\mathbb{C}$ is simple. But I assume you want something less trivial. A stab in the dark: One could imagine putting some group action on $C(X)$ such that none of its non-trivial ideals are $G$-invariant and hope that $C(X)\rtimes G$ is simple.

@JohannesHahn yes I mean without closed two sided ideal.

Oh you meant simple as in simple not as in easy...

@JohannesHahn  But $C(X)$ is not simple.

An easy example: $C(X)$ for some path connected compactum $X$ is path connected in this sense because the spectrum of $f\in C(X)$ coincides with the range of $f$.


A question on simultaneous conjugation of permutations

Given $a,b\in S_n$ such that their commutator has at least $n-4$ fixed points, is there an element $z\in S_n$ such that $a^z=a^{-1}$, and $b^z=b^{-1}$? Here $a^z:=z^{-1}az$.

Magma says that the answer is yes up to $n\leq 11$ (computed by M. Zieve). The answer is also yes if $a$ and $b$ commute via the argument below. It would be great to have an answer in case a,b do not commute, even under the assumption that $\langle a,b\rangle$ contains the alternating group $A_n$. If $\langle a,b\rangle$ contains $A_n$ with $n\geq 4$, then any $z$ as above is unique.

The commutative case: If $a^b=a$, the orbits (and cycles) of $a$ are permuted by $b$. Say $b$ permutes under conjugation cyclicly the orbits $A_1,A_2,\ldots,A_r$ of $a$. Let $a_i\in S_n$ be the cycle induced by $a$ on $A_i$, for $i=1,\ldots,r$. Choose $c\in S_n$ to be element of order $r$ such that $b=c$ on $A_i$ for $i=1,...,r-1$. (This forces $a_r^c=a_1$). Since $b=c$ outside $A_1$ and $a_1^{bc^{-1}}=a_1$, we have $b=a_1^e c$ for some integer $e$.

Choose $v$ to be an involution on $A_1$ such that $a_1^v=a_1^{-1}$ on $A_1$, and trivial outside $A_1$. Choose $w$ to be the involution in the symmetric group on $1,\ldots,r$ such that $(1,\ldots,r)^w=(r,\ldots,1)$ and $1^w=2$. Define $u$ to be a permutation such that if $(i,j)$, $i\leq j$, is an orbit of $w$, then $\alpha^u = \alpha^{c^{j-i}}$ for every $\alpha\in A_i$, $i=1,\ldots,r$. This way $c^u =c^{-1}$. Let $z:=vv^c\cdots v^{c^{r-1}}u$, so that $c^z=c^{-1}$, and $a^z=a^{-1}$. (Moreover, $a_i^z=a_{i^w}^{-1}$ for $i=1,\ldots,r$). It follows that $b^z=a_{1^w}^{-e}c^{-1} = a_2^{-e}c^{-1}=c^{-1}a_1^{-e} = b^{-1}$.

Some might consider this proof (in the commutative case) to be simpler in the language of wreath products.

For the record: This is proved in Theorem 4.9 of arXiv:1810.08971v4 by Junyao Pan. (I have not checked the proof.)

Consider the polygonal mesh with vertices $\{1,\dots,n\}$, directed edges $i\to a(i)$ and $i\to b(i)$ for each $i$, and for each (possibly trivial) cycle $(k\;\cdots\;k+m-1)$ of $[a,b]$ a face with boundary $[a,b]^m$ starting at $k$. The genus depends on the cycle type of $[a,b]$. Lifting to the universal cover gives a tessellation of the Euclidean or hyperbolic plane. It suffices to find a symmetry of the plane sending each $a$ (resp $b$) edge to an $a$ (resp $b$) edge with opposite orientation. In the commutative case this is realized by a $180^\circ$ rotation around any square face.

Have you tried using Rutilo Moreno, Luis Manuel Rivera, Blocks in cycles and $k$-commuting permutations, SpringerPlus, December 2016, 5:1949? (A preprint also appears as arXiv:1306.5708.) Corollary 1 characterizes when the commutator of two permutations has at least $n-k$ fixed points (the authors say that these two permutations $k$-commute).

@SamHopkins I don't think so. I think a reason why many fixed points matter is as follows: Suppose $a^{-1}b^{-1}ab=x$. If $z$ inverts $a$ and $b$, then $x^z=a^{-z}b^{-z}a^zb^z=aba^{-1}b^{-1}$, so $x^{zab}=x$. Thus the $z$ we are looking for is in the coset $C_G(x)b^{-1}a^{-1}$. Note that $C_G(x)$ is $S_{n-k}$ times a subgroup of $S_k$, where $k\le4$. I believe the reason for the existence of $z$ is the hugeness of $C_G(x)$.

Totally naive comment, but would the 4 vs 5 thing have anything to do with when A_n is simple?

Note also that it is very easy to find examples in which $[a,b]$ has $n-5$ fixed points and there is no element $z$ inverting both $a$ and $b$.

After doing lots of random experiments I am starting to believe that this might be true. If so, then I am afraid that it might not be easy to prove!