Is the universality of the surreal number line a weak global choice principle?

I'd like to consider the principle asserting that the surreal number line is universal for all class linear orders, or in other words, that every linear order (including proper-class-sized) linear orders, order-embeds into the surreal number line.

This principle is the class-analogue of the corresponding result for countable linear orders, proved by Cantor, namely, that every countable linear order embeds into the rational line $\newcommand\Q{\mathbb{Q}}\Q$. Using the "forth" part of Cantor's back-and-forth method, one enumerates a given countable linear order and then maps each new point to a rational number filling the corresponding cut in the images of the previous points.

A similar transfinite extension of this argument works with the surreal numbers. Namely, if $\langle A,\leq_A\rangle$ is any class-sized linear order and $A$ is $\newcommand\Ord{\text{Ord}}\Ord$-enumerable, then we may map each new point to the first-born surreal number filling the corresponding cut in the images of the previous elements of $A$, and thereby construct an embedding of $A$ to the surreal numbers.

What this argument shows is that, provided the elements of the linear order $A$ are $\Ord$-enumerable, then we may recursively build up an embedding of the linear order into the surreal number line. In other words, if the principle of global choice holds, which is equivalent to the assertion that every class is $\Ord$-enumerable, then every class-sized linear order embeds into the surreal line. So global AC implies that the surreals are universal for class linear orders.

Similar uses of global choice underlie many other assertions of universality for the surreal numbers, with respect to class-sized structures, as ordered fields and so on. It is fundamental in all the arguments that I know for such universality that one should be able to enumerate the elements of the domain structure in order type $\Ord$, in order to carry out the transfinite embedding procedure.

Since this use of the $\Ord$-enumerations has always struck me as essential, I have long believed that the global axiom of choice was required in order to prove that the surreal number line is universal for class linear orders.

But upon recent reflection, I realized that I don't actually have a proof that global AC is required. I simply don't have a proof that global AC is required for universality, and at the same time, I don't know how to omit it.

Hence these questions:

  * Is the global axiom of choice required to prove that the surreal number line is universal for all class-sized linear orders?

  * Can one prove the universality of the surreal number line merely in GB+AC, without the global axiom of choice?

  * Or is the universality of $\text{No}$ an intermediate principle between AC and global AC?

To answer these questions, it would seem to be useful to settle the issue of the surreal-numbers-universality principle (which I propose we denote by $\text{UNo}$), in some of the known models where global AC fails. For example, in my answer to the question [Does ZFC prove the universe is linearly orderable?](https://mathoverflow.net/a/110823/1946), I showed that there are models of ZFC with no definable linear order of the universe. If we could settle the universality of the surreals in that model, it would answer at least one of the questions above. For example, if the model had surreal-universality UNo, then we would know that universality is not equivalent to global AC; if it is didn't have it, then we would know that UNo isn't provable in GB+AC.

(This topic is related to some ideas I'll be speaking about at my talk on [The hypnagogic digraph](http://jdh.hamkins.org/the-hypnagogic-digraph-jmm-2016/), which I am giving this Friday at the JMM in Seattle.)

Or to put it the other way round: the fact that a class-sized linear order A is embeddable in No apparently does not give a way to Ord-enumerate A. If one wants to somehow use the property `all class-sized linear orders are embeddable in No', then again, we trivially know that all class-sized linear orders isomorphic to a subclass of No are embeddable...so you would have to come up with a class-sized linear order B defined outside of No, and from B's embeddability in No deduce that No is Ord-enumerable. [Well, I hope I'm making some sense here, feel free to disregard my question/comments].

The question intrigues me although I have extremely little knowledge of such matters (but I'm pondering on variations of surreal numbers for constructive math). If I understand correctly, No itself is Ord-enumerable iff Global AC holds? (I don't know the first thing about this, but read it in another post). If that is correct, then intuitively I'd say: global AC is not necessary for UNo, because one can embed No in No without global AC... This may seem a strange argument, but it is meant to illustrate that one can have weaker conditions which already give embeddability.

Keep in mind that the surreal numbers No themselves are definable in L[G], and as elements these define every object in L[G].

Also, when you appeal to UNo, do you have to shout UNo when you have one line left in your proof, or else you have to take four more lines to finish?

Joel, what I'd try to do is to show that every linear ordering comes from a bounded part.

If you think about $L[G]$ in the model in my answer to your question about linear orders of $V$, then the question is whether UNo holds there. We know there is no linear order of all of $L[G]$ there, so perhaps there is a paucity of definable linear orders there (always allowing parameters). If so, then we might expect UNo to hold. But I can't quite prove it.

I just opened my eyes ten minutes ago, it's my pleasure to have choice related questions to think about when I wake up. :-)

@AsafKaragila But let me say that I'm very glad you're looking at this question. I'd appreciate any insight you might have.

I think of global-AC as an assertion in second-order set theory, as in Goedel-Bernays set theory, so one doesn't need to worry about definability. That is, one simply asserts that there is a proper class choice function on all nonempty sets, or equivalently, that there is an Ord-enumeration of V. In particular, global AC definitely holds after adding a Cohen real.

When you say global choice, does it come with parameters? Specifically, if you add a Cohen real to $L$, you have global choice with parameters, but not global choice itself. It stands to reason that in such model you might have what you were looking for.

I'll post my slides on Friday, and you can read all about it!

Hypnagogic digraph!  Quite an intriguing title.  Will there be forthcoming discussions of the hypnopompic digraph as well? :-)


Does the "holomorphic spheres-to-continuous spheres" forgetful function respect the mixed Hodge structures on homotopy groups?

For each smooth, projective, complex variety $X$ that is simply connected, John Morgan constructed a natural mixed Hodge structure on the homotopy group $\pi_k(X,x)\otimes \mathbb{Q}$. This was extended to (possibly singular) quasi-projective varieties independently by Hain, by Deligne, and by Navarro Aznar. For ~~any~~ the basepoint connected component of the double loop space $\Omega^2 X_0\sim [(S^2,0),(X,x)]_0$, each homotopy group $\pi_k(\Omega^2 X_0)\otimes \mathbb{Q}$ is canonically isomorphic to $\pi_{k+2}(X)\otimes \mathbb{Q}$, $k>0$. Thus these also have natural mixed Hodge structures.

Grothendieck constructed a Hom scheme $$H=\text{Hom}_{\mathbb{C}}((\mathbb{CP}^1,0),(X,x))$$ parameterizing algebraic maps (i.e., holomorphic maps in case $X$ is projective). Each quasi-compact, open and closed subset $H_\beta$ of the Hom scheme is quasi-projective. Assuming that $H_\beta$ is connected, there is a well-defined homotopy class of maps $$\Phi_{\beta}:H_\beta \to \Omega^2 X_0,$$ that associates to each holomorphic map from $\mathbb{CP}^1 = S^2$ the underlying continuous map. There are associated pushforward maps on homotopy groups, $$\Phi_{k,\beta}: \pi_k(H_\beta)\otimes \mathbb{Q} \to \pi_{k+2}(X,x)\otimes \mathbb{Q}.$$ Assume now that $H_\beta$ is simply connected so that the groups $\pi_k(H_\beta)\otimes \mathbb{Q}$ have mixed Hodge structures. (There are many cases of $X$ where the associated spaces $H_\beta$ are connected and simply connected. Also, with no hypothesis on $H_\beta$, one can also ask about compatibility with mixed Hodge structures of the induced homomorphism $H_1(H_\beta;\mathbb{Q}) \to \pi_3(X,x)\otimes \mathbb{Q}.$) Also, as Dan Petersen points out, it is necessary to form the Tate twist of the target, i.e., $\pi_{k+2}(X,x)\otimes \mathbb{Q}(1)$.

**Question**. Is the pushforward map $\Phi_{k,\beta}$ a morphism of mixed Hodge structures?

Of course there are many complex varieties $X$ for which $H$ is empty, and then the answer is trivially yes. Yet there are many other complex projective manifolds, e.g., Fano manifolds, where the maps $\Phi_{k,\beta}$ are nontrivial.

@AknazarKazhymurat No, unfortunately I have not found an answer to this question.

. . . This reduces to the smooth, projective case.  In that case, the Sullivan minimal model of the De Rham complex agrees with the Sullivan minimal model of the De Rham cohomology (with zero differentials).  Since the Sullivan minimal model of a dga is (roughly) a multilinear algebra operation on the dga, the work is to check that these multilinear operations respect the weight and Hodge filtrations.

@VivekShende.  "Is this mixed hodge structure constructed by actually making a simplicial scheme ..."  Whether it can be made that way, I do not believe that is how Morgan first constructed these mixed Hodge structures.  Morgan uses a single log resolution and an algebraic construction (basically by spectral sequences) of the rational homotopy type of the open stratum in terms of those of the (projective, smooth) boundary strata.  Deligne proved that this spectral sequence respects mixed Hodge structures.  Morgan checks that is also respects Sullivan minimal models . . .

Is this mixed hodge structure constructed by actually making a simplicial scheme (which?) whose cohomology groups are the homotopy groups of the original $X$?

@DanPetersen.  You are correct, there should be a Tate twist.  I will add it now.

I'd expect there to be a Tate twist at least. If $k=0$ you get a map from $H_0(H_\beta)=\mathbf Q(0)$ to $H_2(X)$, which has weight $-2$.

"Is it easy to see?"  Yes; this is part of adjointness of the space of continuous functions with the compact-open topology.  Let $C$, $X$, and $H$ be topological spaces with $H$ Hausdorff. Let $F:H\times C \to X$ be continuous.  The claim is that the induced function $\Phi:H\to [C,X]$ is continuous, where $[C,X]$ has the compact-open topology.  For an open subset $U$ of $X$, for a compact subset $K$ of $C$, the subset $(H\times K)\cap \Phi^{-1}(X\setminus U)$ is a closed subset of $H\times K$, hence it is proper over $H$.  So it has closed image in $H$, and the complement of the image is open.

Yes, I did mean that topology. Is it easy to see? Are the $H_\beta$s actually homeomorphic to subspaces of $\Omega^2X$?

"Is it easily seen that there is a continuous $\Phi_\beta$?"  Certainly $\Phi_\beta$ is not continuous for the Zariski topology on $H_\beta$.  It is continuous for the analytic topology on the associated analytic space of $H_\beta$.  Similarly, the notation $\pi_{k+2}(X,x)$ means the homotopy group for the associated analytic space of $X$.

@TomChurch.  You are correct.  I am eliding the issue that the function $\Phi_\beta$ is only well-defined up to homotopy.  There is a canonical function $\Phi_\beta$ to the connected component of $\Omega^2 X$ corresponding to the homotopy class $\beta\in \pi_2(X)$.  The homotopy equivalence of this component with the basepoint component is only well-defined up to homotopy.

What does the subscript 0 mean in $\Omega^2 X_0\sim [(S^2,0),(X,x)]_0$? If it means "basepoint component", then why do you say "for any connected component"? If not, why is the isomorphism between $\pi_k(\Omega^2 X_0)\otimes \mathbb{Q}$ and $\pi_{k+2}(X)\otimes \mathbb{Q}$ canonical?


Deforming a basis of a polynomial ring

The ring $Symm$ of symmetric functions in infinitely many variables is well-known to be a polynomial ring in the elementary symmetric functions, and has a $\mathbb Z$-basis of Schur functions $\\{S_\lambda\\}$, with structure constants $c_{\lambda\mu}^\nu$ called Littlewood-Richardson numbers.

I have been studying a commutative deformation of these structure constants, i.e. lifting $c_{\lambda\mu}^\nu$ to a 2-variable polynomial $c_{\lambda\mu}^\nu(a,b)$, for which I have a very explicit formula (much too complicated to describe here). However, since $Symm$ is a polynomial ring (i.e. free), any commutative deformation of it must be trivializable: there must be a way to deform each $S_\lambda$ to a function $S_\lambda(a,b)$ whose ordinary multiplication has my structure constants.

> Given the $\\{c_{\lambda\mu}^\nu(a,b)\\}$, how can I trivialize this family, finding such deformations $S_\lambda(a,b)$?

It seems like the answer should be nonunique (hence harder to find); it's easy to imagine deformations of the basis that don't change the structure constants at all.

> What extra conditions should one put on the deformation in order to make the trivialization unique?

I think $a=0$ is triangular one way, $b=0$ triangular the other way. The one-box rule is here: mathoverflow.net/questions/88569/…

Is it possible that the deformation is upper-triangular (i.e. that $S_\lambda(a,b) = S_\lambda + \textrm{lower order terms}$)? How complicated are the structure constants when $\mu$ is 1 box?


An algebraic strengthening of the Saturation Conjecture

The Saturation Conjecture (proved by Knutson-Tao) asserts that $c_{n\mu,n\nu}^{n\lambda}\neq 0\Rightarrow c_{\mu,\nu}^{\lambda} \neq 0$, where $c$ denotes a Littlewood-Richardson coefficient and $n$ is a positive integer. Let $\mathfrak{o}$ be a (commutative) discrete valuation ring with a finite residue field $k=\mathfrak{o}/\mathfrak{p}$. Let $M$ be a finite $\mathfrak{o}$-module. (For instance, we can take $\mathfrak{o}$ to be the $p$-adic integers $\mathbb{Z}_p$, in which case $M$ is a finite abelian $p$-group.) Thus there is a unique partition $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_r)$ such that $$ M \cong \bigoplus_{i=1}^r (\mathfrak{o}/\mathfrak{p}^{\lambda_i}). $$ We call $\lambda$ the _type_ of $M$. A fundamental result of Philip Hall is that if $M$ has type $\lambda$, then there exists a submodule $N$ of type $\mu$ and cotype $\nu$ (i.e., $M/N$ has type $\nu$) if and only if $c^{\lambda}_{\mu\nu}\neq 0$. See Macdonald, _Symmetric Functions and Hall Polynomials_ , Chapter II, (4.3).

Now suppose that $M$ has type $n\lambda$, and $N$ has type $n\mu$ and cotype $n\nu$. Is there is a submodule $L$ of $M$ of type $\lambda$, such that $L\cap N$ has type $\mu$ and cotype (with respect to $L$) $\nu$?

@DavidSpeyer You are right about the definition of Hall algebra. I have fixed this.

A note on terminology (and then I will think about this interesting question). I usually understood the Hall algebra to refer to the ring whose elements were formal sums of isomorphism classes of $\mathfrak{o}$-modules, and where multipication was given by counting submodules of given type and co-type. See, for example, arxiv.org/abs/math/0611617 . Also, as I imagine you know, Derksen and Weyman gave a far reaching generalization of the saturation conjecture in terms of this sort of Hall algebra -- see ams.org/journals/jams/2000-13-03/S0894-0347-00-00331-3

Is there a combinatorial interpretation for some special case (similar to Yamanouchi tableaux for the LR-coeffs)?  The saturation conjecture implies the analogous statement for Kosktka and skew Kosktka coefficients, so is there a "Kostka"-analogue of this question, which should be easier to prove?


Smooth curves on smooth varieties

Let $X$ be a smooth, proper algebraic variety over a field $k$, of positive dimension.

Is it true that $X$ contains a smooth Zariski-closed curve?

If it is projective, this is true by Bertini. But is it true in general?

@algori : yes, indeed !  The answer is positive if $X$ is separably rationally connected (over an algebraically closed field). Indeed, if $X$ is of dimension $\leq 2$, it is projective so there is no problem. And if it is of dimension $\geq 3$ it contains (lots of) smooth rational curves by Theorem IV 3.9 in Koll'ar's book "Rational curves on algebraic varieties".

Olivier -- it's worse than you think: I'm not sure how to prevent the curve from intersecting itself.

@algori : The problem with this strategy is that it is really possible to have a divisor $E$ on $X'$, exceptional over $X$, such that every curve meeting $E$ has singular image in $X$. This happens for instance if $X$ is a smooth surface, and $X'$ is constructed by blowing up a point, then blowing up a point on the exceptional divisor, then blowing up the intersection of the two exceptional divisors, and if you choose $E$ to be the third exceptional divisor. The problem is that the differential of $X'\to X$ is identically zero at every point of $E$.

Here is a strategy for showing that the answer is positive: blow up a proper smooth variety, call it $X$, to get a projective smooth $X'$ such that $X'\setminus D'$ is isomorphic to $X\setminus D$ with $D\subset X$ closed and $D'\subset X'$ a divisor with normal crossings; then take a curve in $X'$ that intersects transversally all the components of $D'$ and project back again.


An "exercise" on von Neumann algebra tensor product

The following problem appears to be an easy exercise on von Neumann algebra tensor products, but since I've been failing to find a rigorous proof, I'd like to make sure it's not that trivial. Suppose $M$ and $N$ are von Neuamann factors such that $L^\infty[0,1] \mathbin{\bar\otimes} M \cong L^\infty[0,1] \mathbin{\bar\otimes} N$. Does it follow $M \cong N$? It's true under the assumption of separability by the disintegration theory, but not so sure if $M$ (and hence $N$) does not have separable predual.

Obviously when $M$ or $N$ are Type $\mathrm{I}$ it can done in a different way and the result is well-known, but I'm unsure how to proceed if the factors in question are Type $\mathrm{II}_\infty$ or Type $\mathrm{III}$, even assuming separability.

This is obviously achievable if $M$ and $N$ are $\mathrm{II}_1$ factors, with $H$ and $K$ being their standard representations. In which case $L^2(0, 1) \otimes H$ and $L^2(0, 1) \otimes K$ are standard representations of $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$, respectively, so a spatial isomorphism of the above form just comes from any isomorphism between $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$. But without trace can this still be done?

I'm late here, but I'm curious as to how the question is resolved even in the separable case. All the treatments I could find on direct integral theory always needs a spatial isomorphism between $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$. More precisely, suppose $M$ acts on $H$ and $N$ acts on $K$, it seems to always require a unitary between $L^2(0, 1) \otimes H$ and $L^2(0, 1) \otimes K$ that conjugates $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$ onto each other.


What is the Hochschild cohomology of the Fukaya-Seidel category?

Let $(Y, \omega)$ be a compact symplectic manifold and let $Fuk(X,\omega)$ be its Fukaya category. The Hochschild cohomology of this category should be given by $HH^\bullet(Fuk(Y,\omega))=H^\bullet(Y, \mathbb{C})$, at least at the level of vector spaces (the product structure should be the quantum cohomology on $H^\bullet(Y,\mathbb{C})$: see for example [Hochschild (co)homology of Fukaya categories and (quantum) (co)homology](https://mathoverflow.net/questions/11081/hochschild-cohomology-of-fukaya-categories-and-quantum-cohomology) ). As $Fuk(Y,\omega)$ should be a Calabi-Yau category (see for example [Are Fukaya categories Calabi-Yau categories?](https://mathoverflow.net/questions/13114/are-fukaya-categories-calabi-yau-categories) ), the Hochschild homology $HH_\bullet(Fuk(Y,\omega))$ coincides with the Hochschild cohomology $HH^\bullet(Fuk(Y,\omega))$ up to a shift of the grading.

Let me deform the preceding setting: let me assume that $(Y,\omega)$ is a Kähler manifold and that $W \colon Y \rightarrow \mathbb{C}$ is an holomorphic function on $Y$ (in particular $Y$ is noncompact if one wants $W$ non-constant). One should be able to define the Fukaya-Seidel category $FS((Y,\omega),W)$ (I probably need to assume that $W$ has only isolated non-degenerate critical points given the current technology). The Hochschild homology of this category should be given by:

$HH_\bullet(FS((Y,\omega),W)=H^\bullet(Y,Y_{-\infty},\mathbb{C})$,

where we take the cohomology relative to the fiber $Y_{-\infty}=W^{-1}(z)$ of a point $z\in\mathbb{C}$ with $Re(z)<<0$. But now $FS((Y,\omega),W)$ is no longer Calabi-Yau and so the Hochschild cohomology is no longer simply related to the Hochschild homology. So my question is:

What is (or should be) the Hochschild cohomology $HH^\bullet(FS((Y,\omega),W))$ of the Fukaya-Seidel category $FS((Y,\omega),W)$?

If $(Y,\omega),W)$ is the mirror of a smooth projective Fano variety $X$ then

$FS((Y,\omega),W)=D^b Coh(X)$

and one knows by the Hochschild-Kostant-Rosenberg theorem that

$HH_\bullet(D^bCoh(X))=\oplus_{p-q=\bullet}H^p(X,\Omega_X^q)$

and

$HH^\bullet(D^bCoh(X))=\oplus_{p+q=\bullet}H^p(X,\wedge^q T_X)$.

In particular, Hochschild homology and cohomology are indeed very different in general.

This question is motivated by the general principle saying that the deformations of a category are governed by $HH^2$. In particular, a subquestion could be: what are the deformations of the Fukaya-Seidel category?, and this should be equivalent to understand what $HH^2$ is.

Disclaimer: when I write that something "should be" true, it is what I think is the general expectation but as I am not an expert in symplectic geometry, I don't know the technical state of the art of the rigorous proofs. As my question is itself of the "should be" form, I hope it is fine.

Another important paper which is related to your question is Seidel's symplectic homology as a Hochschild homology. If you denote by $\mathscr{B}$ the Fukaya category of the fiber and $\mathscr{A}$ the Fukaya-Seidel category, then $HH_\ast(\mathscr{A}\oplus t\mathscr{B}[[t]])$ gives the symplectic homology of the total space.

The deformation issues concerning fiberwise compactification of a Lefschetz fibration are discussed in II1/2. The key conceptual piece is the fixed point Floer cohomology.

You probably need to read Seidel's papers entitled Fukaya $A_\infty$ structures associated to Lefschetz fibrations I, II, II1/2.


A cohomology class associated with a complex representation of a group

$\newcommand\CC{\mathbb C}\newcommand\ZZ{\mathbb Z}\newcommand\ad{\mathsf{ad}}\newcommand\Ext{\operatorname{Ext}}$ Suppose that $G$ is a finite group and that it acts on a finite dimensional complex vector space $V$.

For each $g\in G$ we may consider the subspace $V_g$ of $V$ spanned by the eigenvectors of $g$ corresponding to eigenvalues different from $1$, and put $d(g)=\dim V_g$. We pick any nonzero linear map $\omega_g:\Lambda^{d(g)}V_g\to\CC$.

If $h\in G$ then the map $x\in \Lambda^{d(g)}V_{hgh^{-1}}\mapsto \omega_g(h^{-1}v)\in\CC$ makes sense, and we write it $h\cdot\omega_g$. Since it is not zero, there is a nonzero scalar $\lambda(h,g)\in\CC^\times$ such that $h\cdot\omega_g=\lambda(h,g)\omega_{hgh^{-1}}$. Associativity of the group action implies that $$\lambda(gh,k)=\lambda(h,k)\lambda(g,hkh^{-1})$$ for all $g$, $h$, $k\in G$. This means that the function $\lambda:G\times G\to\CC^\times$ is a $1$-cocycle in the complex which computes $\Ext_{\ZZ G}^\bullet(\ZZ G^\ad,\CC^\times)$, when we use a bar resolution of the $G$-module $\ZZ G^\ad$ (which is the permutation $G$-module constructed from the conjugation action of $G$ on itself) We may therefore take its class $[\lambda]$ in $\Ext^1_{\ZZ G}(\ZZ G^\ad,\CC^\times)$.

This class depends only on the representation and not on the choice of the $\omega_g$'s, so we write it $c(V)$.

The long exact sequence corresponding to the exponential sequence $0\to\ZZ\to\CC\to\CC^\times\to0$ allows us to identify that $\Ext^1$ with $\Ext^2_{\ZZ G}(\ZZ G^\ad,\ZZ)$. If $CG$ is the set of conjugacy classes and for each $c\in CG$ we let $\ZZ c$ be the permutation module corresponding to the conjugation action on $c$, we have $\ZZ G^\ad=\bigoplus_{c\in CG}\ZZ c$. If $G_c$ denotes the centralizer of an element of $c$, some form of Shapiro's lemma allows us to identify $\Ext^2_{\ZZ G}(\ZZ c,\ZZ)$ with $\Ext^2_{\ZZ G_c}(\ZZ,\ZZ)$ and then our class $[\lambda]$ is an element of $$\bigoplus_{c\in CG}H^2(G_c,\ZZ).$$ This can be written a bit more canonically as $$\left(\bigoplus_{g\in G}H^2(C_g,\ZZ)\right)^G$$ with now $C_g$ the centralizer of $g\in G$ and $G$ acting on the direct sum in the only sensible way. If $G$ is abelian, this can be identified further to $\ZZ G\otimes H^2(G,\ZZ)$.

> **Question:** What is this cohomology class $c(V)$?

This reminds me of Burghelea's description of the cyclic homology of the group algebra, so it might be some form of Chern class (if some vector bundle constructed from V on the free loop space of BG?).

This class shows up when one computes the Hochschild cohomology of the cross product algebra $S(V)\rtimes G$, but explaining how would make this post too much longer. It appears as an obstruction to making this nice :-/

**On triviality:** suppose there is a central element $z$ in $G$ which acts by multiplication by a scalar different from $1$. In that case $c=\\{z\\}$ is a conjugacy class, $G_c=G$, $\lambda(g,z)=\det g$ for all $g$, and the component of $c(V)$ in $\Ext^1_{\ZZ G}(\ZZ c,\CC^\times)=\hom(G,\CC^\times)$ is the determinant map. If the representation does not land in $SL(V)$, then this is not zero. Examples of representations like this are the geometric representation of Weyl groups which have nontrivial center, in which case the longest element is central and acts as $-1$ —see [this question and its answers](https://mathoverflow.net/questions/81858/does-i-belong-to-weyl-group) for the list of which types work— or abelian groups acting with some element not having $1$ in its spectrum.

In the other direction, if the representation respects a symplectic form $\omega$ on $V$, then all the $d(g)$ are even and we may take $\omega_g$ to be the restriction of the $\tfrac12d(g)$th power $\omega^{d(g)/2}$ to $V_g$. In this case, $\lambda(h,g)=1$ identically, so $c(V)=0$.

**Note.** If $e$ is the exponent of $G$ and $\Omega$ is any set of $e$th roots of unity, we can consider the subspace $V_g^\Omega$ spanned by eigenvectors of $g$ corresponding to eigenvalues in $\Omega$, and proceeding as above we get a class $c_\Omega(V)$ in the same direct sum. The class above corresponds to taking $\Omega$ the set of all $e$th roots of unity different from $1$. The assigment $V\mapsto\bigoplus_{g\in G}V_g^\Omega$, for $c$ a conjugacy class of $G$, is a nice functor into Yetter-Drinfeld modules, by the way.

(Here LBG is the free loop space on BG) The first Chern class of this descended bundle is an element of $H^2(LBG,\mathbb Z)$, which is a direct sum of cohomologies of centralizers. Is that class the same as the one described? That would be an answer to "what is c(V)?".

@Andreas, well, I know that the class is a bunch of homomorphisms (which are coherent in some way): that is how it is defined! What I am asking is what the class is, in the sense of is it something? For example, from V you can construct a vector bundle $\tilde V=EG\times_GV$ on $BG$, which you can pull back along the evaluation map $LBG\times S^1\to BG$ to get a vector bundle on that product which descends onto $LBG\times_{S^1}S^1$, which is $LBG$.

@MarianoSuárez-Alvarez: Your question was "What is this cohomology class?". Maybe I do not understand what a possible answer would be, but didn't appear the class in $H^1$ naturally from the representation. What are you looking for?

$\beta$ will be a cocycle here. An easy way to prove this is the following: in the twisted group algebra $\mathbb{C}^{\alpha}G$ we have the equality, if $g$ and $h$ commute in $G$, $U_h^{-1}U_gU_h = \beta(g,h)U_g$. Now it is quite easy to prove that the resulting map $C_g\rightarrow \mathbb{C}^{\times}$ is a homomorphism, and thus a one cocycle, since the action on $\mathbb{C}$ is trivial.

I honestly do not understand what «that is all» means in this context.

The identification $H^2(C_g,\mathbb Z)=H^1(C_g,\mathbb C^{\times})$ makes it look more obscure than it is. For each $g$, the centralizer $C_g$  fixes the subspace spanned by non-fixed eigenvectors and hence there is a determinant homomorphism $C_g \to \mathbb C^{\times}$, i.e., a class in $H^1(C_g,\mathbb C^{\times})$, that's all.

@EhudMeir, indeed, my example with the determinant was an example of something else! :-| I'll correct it. if  I start with with  a $2$-cocycle $\alpha:G\times G\to\CC^\times$, fix $g\in G$ and let $\beta:h\in G_g\mapsto \alpha(g,h)/\alpha(h,g)\in\CC^\times$, I am only being able to prove that $d\beta=\beta^2\smile\beta^2$, no that it is a $1$-cocycle on $G_g$; is that construction treated somewhere? (I can do it if $\alpha:G\times G^{\ad}\to\CC^\times$ is a $1$-cocycle giving an element of $\Ext^1(\ZZ G^\ad,\CC^\times)$, though: is that what you meant?)

Another question: if $[\alpha]\in H^2(G,\mathbb{C}^{\times})$, then we get another element in the same group $\oplus_{c\in CG}H^2(G_c,\mathbb{Z})\cong \oplus_{c\in CG}H^1(G_c,\mathbb{C}^{\times})$, namely if $h$ commutes with $g$ then we have the one cocycle $h\mapsto \alpha(g,h)/\alpha(h,g)$. Is it possible to see any sort of connection between elements arising from two cocycles and from representations?

Shouldn't $\lambda(g,1_G)=1$? That is: for the identity element $e=1_G$ it holds that $d(e) = 0$ because the only eigenvalue of $e$ is 1. I guess that in this case the interpretation of $\wedge^0 V_e$ should be $\mathbb{C}$, and the induced map should be the identity?


Cohomological characterization of CM curves

In his 1976 classical Annals paper on $p$-adic interpolation, N. Katz uses the fact that if $E_{/K}$ is an elliptic curve with complex multiplications in the quadratic field $F$, up to a suitable tensoring the decomposition of the algebraic $H_{\rm dR}^{1}(E,K)$ in eigenspaces for the natural $F^\times$-action coincides with the Hodge decomposition of $H_{\rm dR}^{1}(E,{\Bbb C})$ and (for ordinary good reduction at $p$) with the Dwork-Katz decomposition of $H_{\rm dR}^{1}(E)\otimes B$ for $p$-adic algebras $B$.

Then, he asks for a converse statement. Namely, is it true that if the Hodge decomposition of $H_{\rm dR}^{1}(E,{\Bbb C})$, where $E_{/K}$ is an elliptic curve, is induced by a splitting of the algebraic de Rham, then $E$ has complex multiplications?

The question is left unanswered in that paper. Does anyone know if the question has been answered since?

Ok, I guess I was a bit lazy myself too..... :-)  $K$ is a finite extension of $\Bbb Q$ over which the complex multiplications are defined, that is $F\subseteq K$. The comparison between algebraic $H^1$ and the Rham over $\Bbb C$ is allowed by choosing (and fixing) an embedding $K\rightarrow{\Bbb C}$.

What is $K$ in this question? Too lazy to find Katz' paper in the mess behind me :-) To compute the de Rham cohomology with coefficients in $C$ you'll need a map $K\to C$ right? Or are we considering all such maps? Or have I misunderstood the question?


Lipschitz constant of a homotopy

Let $n>0$, $\mathbb S^n$ be $n$-sphere and $1\in \mathbb S^n$ be its north pole.

A am looking for an example of compact manifold $M$ with a continuous $n$-parameter family of maps $h_x\colon M\to M$, $x\in \mathbb S^n$ such that

  * $h_1=\mathop{\rm id}_M$
  * For any Riemannian metric $g$ on $M$ and any family of maps $h'_x\sim h_x$ (i.e., such that the maps $h,h'\colon \mathbb S^n\times M\to M$ are homotopic) there is $x\in\mathbb S^n$ such that the Lipschitz constant of $h'_x\colon (M,g)\to(M,g)$ is $\ge 10000$.

**Comments.**

  * I think about this problem for few years; I do not think it is an easy one, but I want to make sure that I did not miss a well known trick. A positive answer might have interesting consequences for collapsing with lower curvature bound.
  * If $n=0$, so $\mathbb S^0=\\{+1,-1\\}$ then $M=\mathbb T^k$ and $h_{-1}=$ a hyperbolic linear map with a big eigenvalue will do.
  * Denote by $F$ the space of all maps $M\to M$. If $\pi_n(F,\mathop{\rm id}_M)$ is finitely generated then for every $h_x\colon (M,g)\to (M,g)$ there is $h'_x$ with a fixed Lipschitz constant $L=L(M,g)$ for all $x$.
  * I know few examples of families $h_x$ for which one can not find a metric $g$ and homotopic family $h'_x$ with Lipschitz constant $=1$ (in other words the family can not be made isometric).
  * I do not know an answer even if instead of 10000 I would have $1+\varepsilon$ for a fixed $\varepsilon> 0$.

If it were me, I'd try to start : Let $\Sigma$ be an exotic $n$-sphere.  By conjugacy through the round sphere, this carries a very-transitive action by homeomorphisms of $S O (n+1)$; this action cannot be by diffeomorphisms, or else one could use it to give $\Sigma$ an $S O (n+1)$-equivariant Riemann metric...  But I can't see whether this is helpful after perturbing away from the subspace of group actions; still, I do expect exotic spheres to be useful somewhere, especially since it is known that some carry obstructions to global curvature positivities.

@Henrik: it is almost correct, you should change "any" to "some".

The paper "Homotopical effects of dilation" by Gromov considers similar question, but I haven't checked the exact relationship. Links: projecteuclid.org/euclid.jdg/1214434601 , or (non-paywalled) seven.ihes.fr/~gromov/PDF/4%5B19%5D.pdf

So is the following rephrasing correct: The question is whether any element in $\pi_k(Map(M,M))$ lies in the image of $\pi_k(Lip_{10000}(M,g))$ for any Riemannian metric $g$ on $M$, where $Lip_C(M,g)$ denotes the space of self maps of $(M,g)$ Lipschitz constant $C$.


local equivalence of loop group representations

Let $G$ be a compact, simple, connected, simply connected (cscsc) Lie group, and let its smooth loop group $LG:=C^\infty(S^1,G)$. Given an interval $I\subset S^1$, we have the **_local loop group_** $$ L_IG := \\{\gamma\in LG \ | \ \forall z\not\in I \ \gamma(z)=e\\} $$ which is a subgroup of $LG$. Let $k\ge 1$ be an integer. The level $k$ central extension of $LG$ is denoted $\mathcal{L}G_k$. It restricts to a central extension of the local loop group that we denote $\mathcal{L}_IG_k$.

A representation of $\mathcal{L}G_k$ on a Hilbert space is called _**positive energy**_ if it admits a covariant action of $S^1$ (i.e., the action should extend to $S^1\ltimes \mathcal{L}G_k$) whose infinitesimal generator has positive spectrum. Here, the center of $\mathcal{L}G_k$ is required to act by scalar multiplication.

> **Definition 1:**
>  _Two level $k$ positive energy representation of the loop group are called **locally equivalent** if they become equivalent when restricted to $\mathcal{L}_IG_k$._

The follows is believed to be true:

> **Claim 2:**
> _Let $G$ be a cscsc group and let $V$ and $W$ be any two positive energy representations of $\mathcal{L}G_k$. Then $V$ and $W$ are locally equivalent._

* * *

~~I know a paper that proves the following:~~

> **Theorem 3:**
> _Let $G$ be a **simply laced** cscsc group and let $V$ and $W$ be two positive energy representations of $\mathcal{L}G_k$. Then $V$ and $W$ are locally equivalent._

_**Edit:** The argument in [[GF]](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-155/issue-3/Operator-algebras-and-conformal-field-theory/cmp/1104253398.full) seems to contain a mistake (on lines -4 and -3 of page 600)_

~~The basic ingredients that are needed (see page 599 of [Gabbiani & Fröhlich _[Operator algebras and conformal ﬁeld theory](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-155/issue-3/Operator-algebras-and-conformal-field-theory/cmp/1104253398.full)_] for the proof) are the following two facts about positive energy representations of simply laced loop groups:
• Every level 1 rep can be obtained from the vacuum rep by precomposing the action by an outer automorphism of $\mathcal{L}G_1$ that is the identity on $\mathcal{L}_IG_1$.
• Every level $k$ rep appears in the restriction of a level 1 rep under the map $\mathcal{L}G_k\to \mathcal{L}G_1$ induced by the $k$-fold cover of $S^1\to S^1$. ~~

There are proofs in the literature, due to A. Wassermann ([here](https://arxiv.org/abs/math/9806031) p23) and V. Toledano-Laredo ([here](https://arxiv.org/abs/math/0409044) p82) respectively, for the cases $LSU(n)$ and $LSpin(2n)$, that are based on the theory of free fermions -- actually, Toledano only treats half of the representations of $LSpin(2n)$.

* * *

Is there a proof of Claim 2 in the literature?
How does one prove Claim 2?

After having a further look at his preprint, it is now my opinion that Antony Wassermann does not provide a complete proof of the desired claim. So the question is open.

Many thanks to the person who sent me Wassermann's preprint.

There seems to be a proof of Claim 2 in some unpublished notes, namely in Böckenhauer and Evans "Modular Invariants, Graphs and $\alpha$-induction for Nets of Subfactors. II" on page 66 is claimed that local equivalence hold for any compact connected simple $G$ with reference  to the unplished notes of Wassermann "Subfactors arising from positive energy representations of some infinte dimensional groups", 1999.

probably not really helpful, but there seems to be a paper in preparation by Toledano Laredo which maybe could shed some light about this question in the B and C case


Steenrod algebra at a prime power

Let $n=p^k$ be a prime power.

When $k=1$, the algebra of stable operations in mod $p$ cohomology is the [Steenrod algebra](http://en.wikipedia.org/wiki/Steenrod_algebra) $\mathcal{A}_p$. It has a nice description in terms of generators and relations. Its dual (as a Hopf algebra) is also well understood by work of Milnor.

What about when $k>1$? Has any work been done on trying to understand the algebra of stable operations in mod $n$ cohomology? For instance, is there a description of $\mathcal{A}_4$ in terms of generators and relations? What about the dual Hopf algebra?

Of course I am just asking about the cohomology ring $H^\ast (H\mathbb{Z}/n;\mathbb{Z}/n)$, where $H\mathbb{Z}/n$ denotes the mod $n$ Eilenberg--Mac Lane spectrum. So I fully expect the answer to be either one of "yes, this is well-known and classical" or "no, this is known to be a big mess".

@elidiot: I don't know anything about mod p^k Dyer-Lashof operations, but that's probably just my own ignorance. It seems worthy of an MO question for sure!

I know that this is another question but as it is related I'm taking the liberty to ask it here: has anything been written about mod p^k Dyer Lashof operations ? I would especially be interested in mod 4 ones. Please tell me if this deserves a new thread.

@Sean: No. Shortly after asking this question I started working on other things. I did check out Adams' Berkeley notes, but couldn't find anything therein. Tyler's answer to mathoverflow.net/questions/119604/computation-of-hz-4-hz-4 seems encouraging though.

@Mark: Did you figure out a satisfactory answer? I had a thought about one possible way to "reverse engineer" it.

@Tyler: Thanks for the tip. I've ordered Adams' Berkeley notes from our library, I'll let you know if I find anything!

@Mark: I do remember reading somewhere in one of Adams' texts - possibly the Berkeley lectures? - about looking at cohomology operations into or out of mod $2^n$ cohomology.  I might start looking there.  The method is to filter $H\mathbb{Z}/p^n$ by $H\mathbb{Z}/p^k$ and look at the two associated Bockstein sequences that you get.  Sorry for the indeterminacy in references.

@Tyler: This sounds promising. Was this a self-made calculation? If not, is there any hope that you could remember where you are remembering it from? Thanks.

Err, a lot of the multiplications become zero.

If I remember correctly, this can be calculated because the mod-$p$ Steenrod algebra has a relatively simple structure when considered as a bimodule over the subalgebra generated by the Bockstein.  (I'm less sure at odd primes.)  For $p=2$: in degrees zero and one you get $\mathbb{Z}/2^k$, generated by the Bockstein; I believe that in higher degrees it's abstractly isomorphic to the Steenrod algebra as a module except that admissible monomials starting with $Sq^1$ start with the Bockstein $\beta$ and admissible monomials ending with $Sq^r$, $r$ odd, end with $\beta Sq^{r-1}$.  A lot of the mul

@Sean: I am perhaps not so interested in the answer for finite fields, but I would be interested to know why it is easier (presumably issues with the Kunneth Theorems and duality)?

This is not directly related, but it is not hard to compute the Steenrod algebra corresponding a finite field. I am guessing you are not interested in that though.


Elliptic $\infty$-line bundles over $B G$

Theorem 5.2 in Jacob Lurie's "Survey of Elliptic Cohomology" ([pdf](http://www.math.harvard.edu/~lurie/papers/survey.pdf)) states the equivalence of two maps

$$ B G \longrightarrow B \mathrm{GL}_1(A) $$

for $A$ an $E_\infty$-ring carrying an oriented derived elliptic curve $E \to \mathrm{Spec}(A)$.

The theorem is stated for $G = \mathrm{Spin}$. That's the case of relevance in the paragraph directly following its statement. But, in view of the whole 2-equivariance story, the statement of the theorem as such would make sense for more general $G$ equipped with a homomorphism $G \to O$, as in the discussion on the preceding page.

**First question:** Might this statement still be true for more general $G\to O$? (What's the idea of the proof, anyway?)

More in detail, one of the two maps is the composite

$$ B G \to B O \stackrel{J}{\longrightarrow} B \mathrm{GL}_1(\mathbb{S}) \longrightarrow B \mathrm{GL}_1(A) $$

where $J$ denotes the $J$-homomorphism and $\mathbb{S}$ the sphere spectrum. (Here I am giving a nominally different but hopefully straightforwardly equivalent description of what is indicated on p. 38 of Lurie's note.) The other one is the restriction of the derived theta-line bundle

$$ \theta \;\colon\; \mathrm{Loc}_{\mathrm{Spin}}(E) \longrightarrow \mathbf{B} \mathbb{G}_m $$

along the global point inclusion $\mathrm{Spec}(A)\to \mathrm{Loc}_{\mathrm{Spin}}(E)$ that picks the trivial local system/flat connection (that's again my slight refomulation which I believe is straightforwardly equivalent, but check -- my $\mathrm{Loc}_{\mathrm{Spin}}(E)$ is Lurie's "$M_{\mathrm{Spin}}$").

Regarding the first map, for $G = \mathrm{Spin}$ and for the special case $A = \mathrm{tmf}$ this is discussed in section 8 of Ando-Blumberg-Gepner [arXiv:1002.3004](http://arxiv.org/abs/1002.3004) and crucially equivalent to the $\mathrm{tmf}$-line bundle which is associated via the String-orientation $\sigma$ to the "Chern-Simons 3-bundle" classified by $\tfrac{1}{2}p_1$, i.e. to the map

$$ B \mathrm{Spin} \stackrel{\tfrac{1}{2}p_1}{\longrightarrow} B^4 \mathbb{Z} \stackrel{\tilde \sigma}{\longrightarrow} B \mathrm{GL}_1(\mathrm{tmf}) \,. $$

I suppose the argument there straightforwardly generalizes from $A = \mathrm{tmf}$ to any $A$ as above. But as a sanity check:

**Second question** : is that right?

In conlusion I am wondering:

**Third question** : In which generality in $E_\infty$-rings $A$ carrying oriented derived elliptic curves and in compact simply connected simple Lie groups $G$, is it true that the restriction of the derived theta line bundle along the $\mathrm{Spec}(A)$-point inclusion is equivalent to the $A$-line bundle classified by

$$ B G \stackrel{c_2}{\longrightarrow} B^4 \mathbb{Z} \stackrel{\tilde \sigma}{\longrightarrow} B \mathrm{GL}_1(A) \,. $$

?

(Where now $c_2$ denotes the generator of $H^4(B G, \mathbb{Z}) \simeq \mathbb{Z}$.)


Computation of low weight Siegel modular forms

We have these huge tables of elliptic curves, which were generated by computing modular forms of weight $2$ and level $\Gamma_0(N)$ as N increased.

For abelian surfaces over $\mathbb{Q}$ we have very little as far as I know. The Langlands philosophy suggests that every abelian surface should be attached to a Siegel modular form of weight $(2,2)$ on $GSp_4$, but the problem is that this weight is not cohomological, which has the concrete consequence that it's going to be tough to compute such things using group cohomology. In particular one of the reasons that modular symbols work for computing elliptic curves, fails in this situation.

I guess though that one might be able to somehow use the trace formula to compute the trace of various Hecke operators on Siegel modular forms of weight $(2,2)$ and various levels, because presumably the trace formula translates the problem into some sort of "class group" (in some general sense) computation, plus some combinatorics.

[EDIT: from FC's comment, it seems that my guess is wrong.]

Has anyone ever implemented this and tabulated the results?

[NB I know that people have done computations for low level and high weight, for example there's a lovely paper of Skoruppa that outlines how to compute in level 1; my question is specifically about the weights that are tough to access]

@David: in which case my answer to your aside is then simply "yes".

@Kevin: I am aware of the low-dimensional coincidences SO(2,1)=PGL(2) and Spin(2,3)=Sp(4).  I suppose n > 2 was implicit in my query. :)

@David: there is a low-degree coincidence: the dual group of GSp_4 is GSp_4 (think about the Dynkin diagram of Sp_4). I agree that the pattern should not continue for Sp_6 (think about the Dynkin diagram again).

Aside: Shouldn't the automorphic representation associated to an abelian n-fold live on SO(2n+1), rather than a symplectic group? (Look at the L-groups)

Thanks for the reference FC. What you say about Sp_6 is terrifying!

@David: I think that trying to describe the moduli space is not what I am after. I am after an algorithm to compute traces of Hecke operators on spaces of cusp forms that takes a level and a prime (or some slightly more precise data; a diagonal matrix with powers of p on the diagonal) as an input and gives a number as an output and doesn't care less about whether a certain moduli space is general type or not. The same way as modular symbols would do the job for classical modular forms. But from FC's comment it seems that the trace formula won't do it either :-/

@FC: I am pretty sure one can't use the trace formula to compute forms of weight 1 but I have so little understanding of the trace formula that I don't know why this is the case. Seems to me that you're saying that the same obstruction stops me computing low weight Siegel forms :-(

@Kevin: Curves of genera up 6 have a very pleasant description, and you can comfortably (depends on your standard of comfort of course) live with the description of curves up to genus 15; I'd buy a curve of genus up to 15 as a moduli space any day. I have yet to see a general type three-fold with a nice description (unless you cooked it for this purpose). Disclaimer: I don't know the method involved, you may well be right, it's just my vague intuition speaking here.

I don't buy this. Isn't that like saying "wouldn't computational efforts with elliptic curves of conductor 100 or more be hard because the moduli space has genus at least 2", isn't it? The point is that you don't compute the moduli space---that's the last thing you want to do! You use the trace formula applied to a carefully-chosen function which will give you essentially the trace of Frobenius on the cohomology in terms of some much more algebraic/combinatorial/number-theoretic data and compute that instead. Well, that's my suggestion, but a lot of thought needs to go in to making it work.

Wouldn't any computational efforts with N > 3 be very difficult simply because the moduli space is general type ?


Quotients of residually finite groups by amenable normal subgroups

My questions are:

> Is there any group, which cannot be written as the quotient of a residually finite group by an amenable normal subgroup? Is it possible for large classes of groups?

and

> Is there a group, such that every extension by an amenable group is split?

@Andreas: Me too! Can you send me the text? This sounds very interesting!

Have you imagined a way to answer the first question considering only the one-relator groups?

@Andreas: Could you please send me a copy of this paper, also? I am very interested in this.

@Andreas: Can you send me the text? I suggest you publish the result. Do you need advice about a journal?

@Mark: I proved it and it is not published, since I do not yet know what to do with it.

Also, 'if the answer is "yes", it would provide a new way to construct groups because old ways don't seem to work' is exactly the sort of motivation I was asking for.

Mark - I agree that if the answer to the first question were 'no' then it would be remarkable!  I was wondering what motivates Andreas to ask the question.  It's always useful when the asker gives all the information that s/he has about a problem.

@Andreas: who proved the result of measured group theory that you mentioned, and where?

I meant that no motivation is required because if the answer to the first question is "no", then it would be the first non-trivial property of all groups (which would disprove Gromov's thesis). And if the answer is "yes", it would provide a new way to construct groups because old ways don't seem to work.

̯@Henry: If the second question has a positive answer, then this would (most likely) help to answer the first. My motivation is the study of sofic groups. In the sense of measured group theory, every sofic group is the quotient of a residually finite group by an amenable subgroup (but that takes some time to explain). I was wondering what kind of condition one gets in the more rigid setup of ordinary group theory. Burnside groups are not known to be sofic, so Mark's comment is very interesting.

@Henry: None of these questions requires an additional motivation. @Andreas: I suspect that the free Burnside group of large enough exponent is an example. That would solve mathoverflow.net/questions/37344/… (see the discussion there). But that is a difficult question.

Could you give some motivation, and explain what (if anything) the two questions have to do with each other?


What is known about module categories over general monoidal categories?

All of the literature I have seen on module categories over monoidal categories has been in the rigid $k$-linear semisimple case, more or less in the spirit of Ostrik's paper,

  * Ostrik, V. _Module categories, weak Hopf algebras and modular invariants_ , Transformation Groups **8** (2003) 177–206, doi:[10.1007/s00031-003-0515-6](https://doi.org/10.1007/s00031-003-0515-6), arXiv:[math/0111139](https://arxiv.org/abs/math/0111139).

However, the basic definition of module category makes perfect sense in general, so I am wondering if there has been any work done in more general settings.

In particular, I am interested in the following sort of question. Let $\mathcal{C}$ be a braided monoidal category (maybe ultimately with some added assumptions, such as rigidity). Any left $\mathcal{C}$-module category is automatically a $(\mathcal{C}, \mathcal{C})$-bimodule category via the braiding. For bimodule categories there should be a good notion of "tensoring over $\mathcal{C}$," which makes the 2-category of bimodules (and hence the 2-category of left modules) into a monoidal 2-category. If we extract invertible objects and morphisms at all levels, we should obtain a sort of Brauer 3-group for $\mathcal{C}$. In the fusion category setting, this object is considered in the recent preprint

  * Pavel Etingof, Dmitri Nikshych, Victor Ostrik, _Fusion categories and homotopy theory_ , Quantum Topol. **1** (2010), 209–273, [journal](https://ems.press/journals/qt/articles/2876), arXiv:[0909.3140](https://arxiv.org/abs/0909.3140).

I'd like to know how this is related to the "internal" Brauer 3-group of $\mathcal{C}$, whose objects are Azumaya algebras in $\mathcal{C}$, morphisms are invertible bimodules, and 2-morphisms are invertible bimodule morphisms. In the fusion category setting, the connection is furnished by the main theorem in Ostrik's paper, which states that any semisimple indecomposable module category is equivalent to the category of modules over some algebra in $\mathcal{C}$. (I think this implies that the two notions of Brauer 3-group are equivalent in the fusion category setting.) Is any result along these lines known in more generality? It's not even clear to me that the module category of left modules over an Azumaya algebra in $\mathcal{C}$ is invertible, or even what the tensor product of two module categories of modules is.

The "module categories are categories of modules" yoga is probably very special to the finite rigid setting. We tried to be careful about what exactly was needed for this here:arxiv:1406.4204. An interesting example is the following (non-rigid) tensor category C. It is finite, semisimple and has two simple objects 1 and x. The object 1 is the unit and we have $x \otimes x = 0$. This determines the monoidal structure by linearity. If I remember right, the subcategory generated by x is a module category which is not of the form A-mod(C) for an algebra A in C.

The Deligne tensor product doesn't exist for arbitrary abelian categories. But if you relax "abelian" to instead mean "has finite colimits" then there is a very general tensor product that always exists called the "Kelly tensor product". A similar statement for locally presentable categories is in Cor 2.2.5 of (arXiv:1105.3104). I think the categorical machinery underlying these results can be adapted to the case of tensor products of module cats in the generality you are considering, but I don't know of a reference that does it.

An Azumaya algebra is one in which the tensor actions $X \mapsto A \otimes X$ and $X \mapsto X \otimes A$ are equivalences between $\mathcal{C}$ and the categories of left $A \otimes A^{\text{op}}$-modules and right $A^{\text{op}} \otimes A$-modules, respectively. This definition comes from the paper by Van Oystaeyen and Zhang, "The Brauer Group of a Braided Monoidal Category."

Could you recall the definition of Azumaya algebra in a general braided monoidal category?


Almost complex 4-manifolds with a "holomorphic" vector field

**Main question.** What is the class of smooth orientable 4-dimensional manifolds that admit an almost complex structure $J$ and a vector field v, that preserves $J$?

_The following sub question is rewritten thanks to the comment of Robert Bryant:_

Is it true that if $(M,J)$ admits a vector field that preserves $J$ then there is $J'$ on $M$ homotopic to $J$ that is preserved by an $S^1$-action on $M$?

_After three years I don't think indeed that the following is such a reasonable motivation_

POSSIBLE MOTIVATION. Claire Voisin gave a construction of the Hilbert scheme of points for every almost complex 4-fold by an analogy with the Hilbert scheme of points of a complex surface. The first calculation of the Euler charactericstics of Hilbert scheme of points of complex surfaces was done via localisation techniques, for $CP^2$. Now, if we have a "holomorphic" vector field on an almost complex manifold, this could potentially help to reduce the calculation of the Euler charachteristics of its Hilbert scheme to the study of fixed points of the manifold. So the question is how flexible this notion is... But in its nature this question seems to be more a question (maybe not a hard one) on dynamical systems.

Robert, thanks for this remark, you are right that this second phrase makes no sense. I rewrote so that I it makes sense.

@Dmitri:  Perhaps I don't understand your terminology, but it seems to me that $S^4$ admits a smooth $S^1$ action, but it doesn't admit any almost complex structure, let alone one with a (nontrivial) symmetry vector field.  Thus, your second sentence is mysterious to me.

Yes, I refer to the flow associated to the vector field. I don't want to impose that all zeros are isolated.

When you say the vector field preserves the almost-complex structure, are you referring to the flow associated to the vector field?  And do you want the vector field to be everywhere nonzero?  Nonzero at all but finitely many points?

Tim, thanks for the comment! In fact the approach I propose is exactly to give a justification of your words:)) How do we calculate the Euler charecteristics of a manifold - count the number of zeros of a vector field. I want to say, that for a Hilbert scheme this should be a result of the same caclulation. I think it would be cool to have a "simple" calculation of the Euler charateristics. By the way this speculation can also be applied to DT invariants -- when you count 0-dim subschemes on 3 dimensional CY manifolds. This calculation was done in 2005 and was one of MNOP conjectures.

It's an intriguing question, but it seems to me a surprisingly delicate way to calculate the Euler characteristic of the Hilbert scheme, which one expects to be a universal function of the Betti numbers of the 4-manifold, as in the integrable case. Perhaps one can prove this using the Cech spectral sequence coming from the open cover of Hilb arising from a good cover of the 4-manifold?  [BTW, I know you're quoting Voisin, but the following rule is worth insisting on: a 4-fold is an algebraic or analytic variety of dimension 4. A 4-manifold has real dimension 4.]

Thanks :) Maybe this sound cooler than it should be. He is the reference, it was published in 2002 people.math.jussieu.fr/~voisin/Articlesweb/almost.pdf

This sounds really cool. Is this recent work of Voisin?


$G$ a group, with $p$ a prime number, and $|G|=2^p-1$, is it abelian?

During my research I came across this question, I proposed it in the chat, but nobody could find a counterexample, so I allow myself to ask you :

$G$ a group, with $p$ a prime number, and $|G|=2^p-1$, is it abelian ?

@YemonChoi  Well, I would wait 500 years for the math community to find a complex 600-page solution, like for Fermat.

If you are claiming that you can answer this question, then you can either add an answer below, or write up your work as a preprint in the normal academic way, or write it in a blog post and leave a link. Saying "I can answer this, can you?" is not a constructive use of this site, nor is it collegial.

I have rolled back the edits to this old question, which seem to be purely cosmetic and done for promotion. Yes the question is appropriate for MO, but that doesn't mean you need to add some kind of "RL" label

Indeed, your suggestion mathoverflow.net/questions/273976/… is the same as @spin's.

I did a quick check for $p$ at most $800$ or so and found no examples. In the end this is a question about number theory and not group theory, and it might be difficult. For example it is an old open problem to determine whether $2^p - 1$ is squarefree for all primes $p$. If the answer to that old problem is yes, then you are asking whether every group of order $2^p - 1$ is cyclic.

One can use the Cunningham tables to do a quick check on the factors for more than 50 prime exponents. Gerhard "Not Computationally Motivated This Morning" Paseman, 2017.07.08.

@PéterKomjáth: No, we have then $2^p \equiv 2$ (mod $q^3$), respectively, $2^{p-1} \equiv 1$ (mod $q^3$).

@Glorfindel, it's hard to believe that never happens, but a quick test in Mathematica shows that (I screwed up in Mathematica or) it doesn't happen for the first 50 primes.

Yes, but probably easier to solve. I doubt there is a prime $q$ for which $q^3 | 2^p - 1$, but the other 'case', two primes $q_1, q_2$ in the factorization for which $q_1 | q_2 - 1$ might very well be possible.

It depends on the factorization of $2^p-1$, see groupprops.subwiki.org/wiki/…


Symmetries of local systems on the punctured sphere

Let $X=S^2\setminus D$, for $D\subset S^2$ some finite set of points, say with $|D|=n\geq 1$. The category of locally constant sheaves of $\mathbb{C}$-vector spaces on $X$ (equivalently, complex representations of $\pi_1(X)=F_{n-1}$), $\text{LocSys}(X)$, has many natural autoequivalences.

For example, if $f: S^2\to S^2$ is a homeomorphism fixing $D$, then $\mathbb{V}\mapsto f^*\mathbb{V}$ is an autoequivalence of $\text{LocSys}(X)$. Two homeomorphisms fixing $D$, homotopic through homeomorphisms fixing $D$, induce naturally equivalent autoequivalences, so one obtains an action of the mapping class group $\text{Mod}(S^2, D)$ on $\text{LocSys}(X)$. Since $\text{Mod}(S^2, D)$ is the pure spherical braid group on $n$ strands, I'll denote it by $B_n$. My question is:

> What are the auotequivalences of $\text{LocSys}(X)$ commuting (up to natural isomorphism) with the actions of $B_n$?

I know of three sources of examples. The first is invertible objects in $\text{LocSys}(X)$: if $\mathbb{L}$ is a rank one local system on $X$, the functor $\mathbb{V}\mapsto \mathbb{V}\otimes \mathbb{L}$ is $B_n$-equivariant. The second is the evident action of the group $\text{Aut}(\mathbb{C}/\mathbb{Q})$ on the structure constants of any given local system. Both of these operations preserve the rank of a local system.

The other is Katz's middle convolution operation, which does not.

Let $\Delta\subset X\times X$ be the diagonal, $\pi_1, \pi_2: X\times X\setminus \Delta\to X$ the two projections, $a: X\times X\setminus \Delta\to \mathbb{C}^\times$ the map $(x, y)\mapsto x-y$ (where here we choose an isomorphism $S^2\simeq \mathbb{CP}^1$ sending some point of $D$ to $\infty$ to coordinatize $S^2$), and $j: X\times X\setminus \Delta\hookrightarrow S^2\times X$ the evident inclusion. Then given a non-trivial rank one local system $\chi$ on $\mathbb{C}^\times$, the middle convolution $$MC_{\chi}(\mathbb{V}):=R^1\pi_{2\ast}j_{\ast}(\pi_1^{\ast}\mathbb{V}\otimes a^{\ast}\chi).$$ It is not obvious but Katz shows it is true that $MC_\chi$ is (up to equivalence) inverse to $MC_{\chi^{-1}}$. Moreover this operation (not entirely obviously) commutes with the action of $B_n$.

Of course one can compose these operations in various ways to get more autoequivalences commuting with the $B_n$-action.

> Is that everything?

I'm also curious about the analogous question with $S^2\setminus D$ replaced by a surface of higher genus.


Does the Ackermann function count something?

Let $\mathrm{FinSet}$ be the category of finite sets.

A _finite set structure_ is a faithful functor $F\colon C\to \mathrm{FinSet}$ such that, for any $n\geq 1$, there are only finitely many isomorphism classes of objects $F$ maps to $\\{1, \dots, n\\}$.

> **Question.** Is there a natural finite set structure realizing the Ackermann function (or some other computable function growing faster than any primitive recursive function)?

So, I do not know  what the Ackermann function counts, but the TREE function does count something explicit and combinatorial, and the Ackermann function is tiny in comparison with TREE. mathoverflow.net/questions/93828/how-large-is-tree3

The inverse Ackermann function appears in some algorithms, e.g. using a "soft heap" you can find the minimum spanning tree in time $O(m\alpha(m, n))$. A decade ago I probably knew why and whether something was being counted.

Interesting question. Maybe some nested set-partition-like structures? A stupid answer would be to take a deterministic rewrite system that takes $A\left(n,m\right)$ steps to terminate (I think there should be one), and just say "the intermediate states of this rewrite system".

There is a whole "school" of enumerative combinatorics done using Joyal's species! Relevant names are Joyal himself, Labelle, Leroux, Viennot, Mendez, Rajan...

Not sure why this (good) question has a category theory tag, but there is a kind of "categorical number theory / combinatorics" that was studied by Steve Schanuel. I don't know any more than that.

@SimonHenry even though the question isn't phrased in this way, I suspect it is asking about a species of structure with the Ackermann function as generating species. I have no idea what the sequence $a_n=A(n,n)$ could be counting, though, and throwing the very few terms of the sequence that can be computed at OEIS doesn't seem a viable choice: is there a combinatorial interpretation for those numbers?

@fosco : yes but,I am not aware of anything that forces the exponential generating function of combinatorial species to converge, so why would the convergence brings anything to the question ? But maybe you have in mind some result proving the convergence for some class of "nice" species ?

@SimonHenry yes you can associate some category but I wondered if there was a natural (in the vague sense of the word) choice. Maybe a structure arising independently elsewhere in mathematics.

I'm not sure how to understand this question. For any function $f:\mathbb{N} \to \mathbb{N}$ I can chose a sequence of sets $P_n$ such that $\# P_n = f(n)$ and take $C$ to be a naively defined category with an object for each $i \in P_n$ mapping to a set with $n$ elements... so any functions can be obtained this way.

hi @SimonHenry ! To a combinatorial species you can associate various formal power series, and especially from the exponential series you can understand a lot about the species itself --although not everything. Ad yes, that series it's not convergent, so there is no associated function you can study. This makes the problem elusive, even provided you find a meaning for what kind of structure $A(n,n)$ is counting.

@fosco, surely not; $f(n) \ge (n!)^2$ eventually.

@fosco : why is the convergence of the series relevant to the question ?

oh, fair enough; well, is the series $\sum \frac{f(n)}{n!}X^n$ convergent in at least a disk $B(0,\epsilon[$?

The Ackermann function $A(m,n)$ takes two inputs; apart from this, if I understand well, you're asking about the existence of a species of structure whose associated exponential series (in the variables X,Y) is $\sum \frac{A(m,n)}{n!m!}X^nY^m$. First of all, is this series convergent?


Picture of Lambert's proof that $\pi$ is irrational?

With a suitably generous notion of "picture proof" or "proof without words" or "geometric proof," there do exist such proofs of [the irrationality of square roots](https://www.cut-the-knot.org/proofs/GraphicalSqRoots.shtml) and even of the [irrationality of $e$](https://arxiv.org/pdf/0704.1282.pdf). However, I am not aware of anything that could plausibly be called a [geometric proof of the irrationality of $\pi$](https://math.stackexchange.com/q/1658174). Given that the original definition of $\pi$ was geometric, it seems a pity that there is no such thing. Or maybe there is?

Perhaps one can try to draw pictures to accompany [Lambert's irrationality proof](https://math.stackexchange.com/a/895728). For example, is there a way to draw a picture of the following fact? $$\tan(a/b) = {a \over\displaystyle b - {a^2 \over \displaystyle 3b - {a^2 \over \displaystyle 5b - {\displaystyle a^2 \over 7b - \cdots}}}}$$ And if so, is there any way to draw a picture of the fact that such a continued fraction is irrational when $a$ and $b$ are positive integers?

I don't have a specific suggestion, but of possible related interest is a collection of references I assembled 3 years ago in my answer to Irrationality of $\pi^2$ and $\pi^3$.

Correction: The above formula should have $\tan(x^{-1})$ on the right hand side. I also think there might be a sign error somewhere.

I'll think more about this later but: Let $SL_2(\mathbb{R})$ act on $\mathbb{R}$ by Mobius transformations in the usual way. It looks to me like your continued fraction says that $\begin{bmatrix} 0 & 1 \\ -1 & x \\ \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 3x \\ \end{bmatrix}  \begin{bmatrix} 0 & 1 \\ -1 & 5x \\ \end{bmatrix}  \cdots (0) = \tan(x)$. So maybe draw a sequence of lines given by partial products of this, and see that their slopes are approaching $\tan(x)$ or, equivalently, their angles are approaching $x$? (Here $x = a/b$.)


When is the determinant an $8$-th power?

I am working over $\mathbb{R}$ (though most of the story goes over any field). I am looking for linear spaces of matrices such that the restriction of the determinant to this spaces can be written (non-trivially) as the power of another polynomial. Let me give some examples where such a phenomenon appears, before asking my question in more details.

  * let $n = 2m$ be an even integer. Then, the restriction of the determinant to $\bigwedge^2 \mathbb{R}^n \subset \mathrm{End}(\mathbb{R}^n)$ can be written as the square of a non-zero polynomial : the pfaffian.

  * let $n = 2m$ be again even. Let $\mathcal{H}_{m}(\mathbb{C})$ be the set of $m \times m$ Hermitian matrices with complex coefficients. Using the matrix representation of $i$ (square root of $-1$) as: $$ i = \begin{pmatrix} 0 & -1 \\\ 1 & 0 \\\ \end{pmatrix},$$ one can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{C})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{C})} \right)^2.$$

  * let $n = 4m$. Let $\mathcal{H}_{m}(\mathbb{H})$ be the set of $m \times m$ Hermitian matrices with quaternionic coefficients. Using the $4 \times 4$ matrix representation of $i,j,k$ (square roots of $-1$ in $\mathbb{H}$), One can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is again easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{H})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{H})} \right)^4.$$

**Question** : Is there a similar story for $\mathcal{H}_{3}(\mathbb{O})$?

More precisely, one can define a good notion of determinant for Hermitian $3 \times 3$ matrices with octonionic coefficients. I was wondering if there is an embedding $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{S}^{2} \mathbb{R}^{24} \subset \mathrm{End}(\mathbb{R}^{24})$, such that the following hold on $\mathcal{H}_{3}(\mathbb{O})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})} = \left( \mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})} \right)^8?$$

The algebra $\mathcal{H}_{3}(\mathbb{O})$ being non-associative, it can not be embedded **as an algebra** into a matrix algebra. On the other hand, what I am asking is (probably) considerably weaker : I just want an embedding of vector spaces $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{End}(\mathbb{R}^{24})$ such that the restriction of $\mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})}$ to $\mathcal{H}_{3}(\mathbb{O})$ is the $8$-th power of $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}$.

Of course, since $\mathcal{H}_{3}(\mathbb{O})$ can not be embedded into $\mathrm{End}(\mathbb{R}^{24})$ as an algebra, it will probably be harder to check such an equality of determinants on $\mathcal{H}_{3}(\mathbb{O})$. Indeed, we can't use the representative of the Hermitian comatrix in $\mathrm{End}(\mathbb{R}^{24})$ and just multiply it with the orginal matrix to get $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}.Id_{24}$. But I thought that it was perhaps possible to find another trick for $\mathcal{H}_{3}(\mathbb{O})$?

Thanks for your help.

But I may be wrong and I would be delighted to learn that there is such an embedding!

@DenisSerre : Yes sure, I tried a lot of stuff related to the Jordan algebra theory. In fact I am now able to prove that there is an embedding of $\mathcal{H}_{3}(\mathbb{O})$ in $\mathbb{R}^{24}$ such that $\mathrm{det}_{\mathbb{R}^{24}}$ is the $4$-th power of a certain sextic related to $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{0})}$. However, I did a lot of computations with Macaulay2, and I start to believe that there is no embedding such that the restriction of $\mathrm{det}_{\mathbb{R}^{24}}$ to $\mathcal{H}_{3}(\mathbb{O})$ is a $8$-th power.

Have you tried using the fact that ${\cal H}_3({\mathbb O})$ is the exceptional Jordan algebra, that called the Albert algebra ?

@MarekMitros : Thanks! Yes I have seen this question a bit after having posted mine. It is very much related. Though, I guess mine is more focused on the equation lining the two dterminants while the other one focuses more to the existence of a determinant for hermitian matrices with octonionic coefficients.

The first suggestion of embedding octonions into matrices is to map octonion $x$ to $8\times 8$ matrix $L_x$ or $R_x$ of left and right multiplication by octonion $x$. Similarly octonion matrix $2\times 2$ can be mapped to block matrix consisting of four blocks $L_{a_{i,j}}$ where $a_{i,j}$ is octonion in original matrix. Do we have formula for determinant of block matrix ?

What is that first map? The multiplication table should be neither symmetric nor skew-symmetric. I see how given an $\mathbb{R}$-linear map $f:\mathbb{O}\hookrightarrow\mathrm{End}(\mathbb{R}^8)$ satisfying $f(z)f(\bar z)=|z|^2I_8$ one could extend to $\mathcal{H}_2(\mathbb{O})$ with the right determinant relation. That would require 7 anticommuting matrices $e_1,\dots,e_7$ in $\mathrm{End}(\mathbb{R}^8)$ that square to $-I_8$. How are you building these out of the multiplication table?

@MTyson In the case of $\mathcal{H}_{2}(\mathbb{O})$ I think there is something going on. Let $\mathbb{O} \hookrightarrow \mathrm{S}^2 \mathbb{R}^8$ which associates to an octonion its multiplication table. This gives rise to an embedding $\mathcal{H}_{2}(\mathbb{O})$ in $\mathrm{S}^2 \mathbb{R}^{16}$.  Using the block computation of a determinant for a $2*2$ block matrix (with the bottom right being a translation), one sees that $\mathrm{det}_{\mathrm{End}(\mathbb{R}^{16})} = (\mathrm{det}_{\mathcal{H}_{2}(\mathbb{O})})^3$

Do you know anything about the $\mathcal{H}_{2}(\mathbb{O})$ case?


Kan's simplicial formula for the Whitehead product

In his article on _Simplicial Homotopy Theory_ (Advances in Math., 6, (1971), 107 –209) Curtis quotes a formula (on page 197) for the Whitehead and Samelson products in a simplicial group $G$. The formula for the Samelson product $\langle x,y\rangle$ of elements $x\in \pi_p(G)$ and $y\in \pi_q(G)$ is in terms of an ordered product of commutators of pairs of certain degenerate elements. The order of multiplication is determined by an antilex total order on the set of $(p,q)$-shuffles. I tried, and think I succeeded in filling in the details of this sketch from Curtis. I understand that the result was due to Dan Kan but that he never published it.

My questions are

(i) is there a published proof of this formula and if so where? (As I said I have a proof that it seems to work, but do not yet have one that it actually deserves to be called the Samelson product and that the related Whitehead product corresponds to the classical form, e.g. involving a 'universal example'.)

and

(ii) I have so far failed to prove that this formula gives a bilinear pairing on homotopy groups. (It is simple to prove a related formula which incorporates an action as in the Witt-Hall identities for commutators but I do not see why the action involved should be trivial as would seem to be required for the bilinearity result to hold.)

Can anyone suggest a reference to a classical proof of bilinearity of the (topological) Whitehead product that is fairly categorical and hence adaptable to my simplicial setting. Most sources seem to leave it as an exercise and Whitehead's original approach although clear is very topological in nature.

(Edit: I note that Adams in his _Student's Guide to Algebraic Topology_ , stated: The Whitehead product is bilinear and anticommutative; this can be proved by diagram chasing with the universal example'. Unfortunately I have yet to find the diagram through which I have to chase, although I have tried several ones that initially seemed to give some hope of being good for the task in hand.)


Almost monochromatic point sets

There are many sort of equivalent theorems about monochromatic configurations in finite colorings, such as [Van der Waerden](https://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem), [Hales-Jewett](https://en.wikipedia.org/wiki/Hales%E2%80%93Jewett_theorem) or [Gallai's theorem](https://arxiv.org/abs/1411.1038), the latter of which states that in a finite coloring of $\mathbb Z^d$ or $\mathbb R^d$, there is a homothetic (i.e., scaled and translated) copy of any finite configuration $S$. Motivated by [this problem](https://mathoverflow.net/questions/299616/bichromatic-pencils), I wonder if similar statements hold if instead we require that in a configuration $(S,s_0)$ all points in $S$ are monochromatic, while $s_0$ has a different color from the rest.

Obviously, we need to impose some conditions on the coloring and the configuration. About the coloring, I only want to demand that it is non-monochromatic, i.e., not all points of the space are colored with the same color. About $(S,s_0)$, I want to require that $s_0\notin conv(S)$, i.e., $s$ is not in the convex hull of some points from $S$, as then we might not have a solution if the "first" half of the space is red, while the "second" half is blue.

> > Is there always an almost monochromatic copy (homothetic or isometric) of any finite $(S,s_0)$ with $s_0\notin conv(S)$ in a non-monochromatic finite coloring of $\mathbb R^d$?

Note that the answer is no for $\mathbb Z$ if $s_0=0$ and $S=\\{1,2\\}$ as shown by coloring odd numbers red and even numbers blue. This particular configuration, however, is easy to find in $\mathbb Q$.

@fedja I've just discovered that essentially the same problem was posed by some dude called Erdős and his pals in '75, and they made the same conjecture, so you might want to spell out your argument for their sake; see Conjecture 4 in old.renyi.hu/~p_erdos/1975-12.pdf.

Didn't I spell it out? $\tau$ is a field automorphism of $\mathbb C$ preserving $\mathbb Q$.

@fedja I'm not sure, but it feels like your reasoning goes the other way. Let me try. We know that there is a bad coloring (of even $\mathbb C$) for $(0,1,s_0)$ whenever $0<s_0<1$ and from this we want to show that for some $\tau$ there is also a bad coloring (of $\mathbb R$) for $(0,1,\tau(s_0))$ where $1<\tau(s)$. And then what do you require of $\tau$?

If I understand things right, you can design a field automorphism $\tau$ of $\mathbb C$ over $\mathbb Q$ so that $\tau(s_0)<0$. Your configuration is then $0,1,\tau(s_0)$ and in your coloring $\tau(z)$ is blue if $\Re z\ge 0$ and red if $\Re z<0$. The point is that the "similar" configurations are obtained by linear transformations and $a\tau(s_0)+b=\tau(\tau^{-1}(a)s_0+\tau^{-1}(b))$

@fedja Sorry, but my field automorphisms would need some polishing, though your idea certainly seems good. Could you give me an example of how you want to derive a coloring?

You might be interested in mathoverflow.net/q/3322/806, which was a problem in a similar spirit in $[n]$.

What prevents you from taking the "convex" configuration $(0,s_0,1)$ with some transcendental $s_0\in(0,1)$ and the "half-plane" coloring and apply a field automorphism of $\mathbb C$ over $Q$ that moves $s_0$ to a negative number to create a non-convex configuration and some crazy coloring that just encode the old ones?


On manifolds which do not admit (smooth) actions of finite groups

**Question** : Assume a smooth manifold $M$ does not admit any effective smooth group actions of finite groups $G \neq 1$, does it follow that $M$ also admits no _continuous_ effective group actions of finite groups $G \neq 1$?

Manifolds which do not admit actions of finite groups are interesting because if $M$ is compact and we choose a Riemannian metric $g$ on $M$, then $\text{Isom}(M,g)$ is a compact Lie group by Steenrod-Myers and must be trivial, because otherwise it contains a non-trivial finite group acting on $M$.

For this nice property, it is enough to consider only smooth actions.

However, there are some interesting [articles](https://arxiv.org/pdf/math/0606714v1.pdf) which construct manifolds which do not admit any _continuous_ actions of finite groups. And so I'm wondering if its enough to eliminate smooth actions if the manifold is smooth to get a manifold of this type.

The linked article of V. Puppe seems to say "manifold" for "(connected) closed manifold". Is the same implicit assumption made in the question?

Restatement of the question using group theory language: If $\mathrm{Diff}^\infty(M)$ is torsion-free, does it follow that $\mathrm{Homeo}(M)$ is torsion-free?


Jets of sections of vector bundles expressed by symmetrized iterated covariant derivatives - who did it first?

The (non-unique) bundle isomorphism between the bundle $J^r E$ of $r$-th order jets of sections of a _vector_ bundle $\pi:E\rightarrow M$ and the direct sum $$\bigoplus^r_{k=0}\vee^kT^*M\otimes E\rightarrow M$$ of the vector bundles of $E$-valued totally symmetric rank-$k$ covariant tensors for all $k=0,\ldots,r$ (here $\vee^1 T^*M=T^*M$ and $\vee^0 T^*M\rightarrow M=$ the trivial line bundle $\mathrm{pr}_1:M\times\mathbb{R}\rightarrow M$ of scalars, so that $\vee^0T^*M\otimes E=E$) may be chosen at the level of $r$-th order jet prolongations $j^r s$ of smooth sections $s\in\Gamma(E)=\\{s'\in C^\infty(M,E)\ |\ \pi\circ s'=\mathrm{id}_M\\}$ of $E$ by means of symmetrized iterated covariant derivatives of $s$. This is used to write linear partial differential operators between spaces of smooth sections of vector bundles in a global form (see below).

To do so, we start by fixing a covariant derivative on $E$ and a (say, torsion-free) covariant derivative on $M$, and then extend them to a covariant derivative on $E$-valued covariant tensor fields on $M$ by means of the Leibniz rule for tensor products of smooth sections. We denote this common extension by $\nabla$. Given $s\in\Gamma(E)$, we define the $k$-th order iterated covariant derivative $\nabla^k s$ of $s$ with respect to $\nabla$ recursively as $$\nabla^0s=s\ ,\,\nabla^1 s=\nabla s\ ,\,\nabla^{k+1}s=\nabla(\nabla^k s)\ ,$$ so that $\nabla^k s\in\Gamma(\otimes^k T^*M\otimes E)$. The totally symmetric part of $\nabla^k s$ (which, of course, belongs to $\Gamma(\vee^k T^*M\otimes E)$) will be denoted by $D^k_\nabla s$.

> One can show that the map $$j^r s \mapsto i_\nabla(j^r s)\doteq(s,D_\nabla s=\nabla s,\ldots,D^r_\nabla s)$$ is a $C^\infty(M)$-linear isomorphism between the $C^\infty(M)$-modules $\Gamma(J^r E)$ and $\Gamma(\oplus^r_{k=0}\vee^kT^*M\otimes E)$, thus establishing an instance of the bundle isomorphism stated in the first paragraph. Now, given two vector bundles $\pi:E\rightarrow M$, $\pi':F\rightarrow M$, a linear partial differential operator $P$ of type $E\rightarrow F$ and order $r\geq 0$ is the composite of the $r$-th order jet prolongation map $\Gamma(E)\ni s\mapsto j^r s\in\Gamma(J^r E)$ with a $C^\infty(M)$-linear map $p$ from $\Gamma(J^r E)$ into $\Gamma(F)$ (the "symbol map" of $P$). Using $i_\nabla$, one may write $$Ps=p(j^r s)=(p\circ i_\nabla^{-1})(s,D_\nabla s,\ldots,D^r_\nabla s)=\sum^r_{k=0}a_k D_\nabla^k s\ ,$$ where $a_k\in\Gamma(\vee^k T M\otimes E^*\otimes F)=$ the $k$-th order coefficient of $P$ with respect to $\nabla$ is uniquely determined by $P$ and $\nabla$ for each $k=0,\ldots,r$ (_remark:_ uniqueness of $a_k$ no longer holds if the symmetry requirement is dropped). Moreover, $a_r$ (the "principal symbol" of $P$) does not depend on $\nabla$.

The earliest reference I could find on this statement is Chapter IV, Section 9 (more precisely, the Corollary to Theorem 7, pp. 90-91) of the book edited by Richard S. Palais, _Seminar on the Atiyah-Singer Index Theorem_ (Princeton University Press, 1965). This chapter was written by Palais himself, and it has no references.

> **Question:** Is this really the first published proof of this result? As stated is it due to Palais, or does it go back even farther?

Conversely, given $P$ as defined in the question above, it is clear that $[P,f]=$ commutator of $P$ and (multiplication by) a smooth function $f$ on $M$ is a linear differential operator of order $k-1$. (Remark: one has to add to your definition the "initial condition" that linear differential operators of order $-1$ are identically zero)

@AliTaghavi Not really - such a definition is essentially the same as the one in terms of jet bundles. To see that, it suffices to know that the latter's fiber over $p\in M$ may be defined as the quotient of the space $E(p)$ of germs of smooth sections of $\pi$ over $p$ modulo the subspace of such germs vanishing to order $k+1$ at $p$. As such, it can be seen by induction on $k$ that given smooth functions $f_1,\ldots,f_{k+1}$ on $M$, one always has $D((f_1-f_1(p))\cdots(f_{k+1}-f_{k+1}(p))s)(p)=0$ for all $p\in M$, $s\in\Gamma(\pi)$, hence $D$ is necessarily of the form stated for $P$ above.

I confess I did not understand your terminologies completely. But what would be happen if one define a diffm operator of order k as a linear map $D$ such that $s\mapsto D(fs)-fD(s)$ would be of order k-1.(inductive definition). Is there a problem of GLOBAL definition of differentoal operatores in this manner?

Pohl's results were derived independently by Libermann (1963) (a student of Ehresmann) and Feldman (1963), but one also cannot find any explicit isomorphism there. There is a couple of references by Ehresmann (1955) and Libermann (1961) on higher-order connections I couldn't get access to (they are both in rather obscure proceedings volumes) and maybe there is something there closer to what I ask, but I really don't know.

@DmitriZaitsev That's true, but that's not what I am asking. I've also tracked Ehresmann's pioneer work on jets and (arbitrary-order) connections, and couldn't find any trace of the result I've stated - to wit, relating jets to iterated (first-order) covariant derivatives. The book by Jafarpour and Lewis quoted by Umberto Lupo above traces related results to a Trans. AMS paper by Pohl (1966) whose preprint actually  dates back to 1963, but there (as the authors themselves state) no such formula can be found explicitly.

The jets are usually attributed to Ehresmann

That is actually a very nice reference (despite lacking a bit of historical care, as you noticed and most people using/quoting the result do), thanks!

I had not known that the result dates back to Palais' work. I learned it from engineering.ucsb.edu/~saber.jafarpour/time_varying.pdf (see Lemma 2.1) and apparently those authors did not come across your reference either.


Trying to reconcile two facts about the Appell-Lerch sum learned from Polishchuk and Zwegers

One of the key characters in the [thesis](https://arxiv.org/abs/0807.4834) of Zwegers is the modular correction $\tilde\mu(u,v;\tau)=\mu(u,v;\tau)+\frac i2R(u-v;\tau)$ of the Lerch sum $\mu(u,v;\tau)=\frac{e^{\pi i u}}{\vartheta(v;\tau)}\sum_{n\in\mathbb Z}(-1)^n\frac{e^{\pi i(n^2+n)\tau}e^{2\pi inv}}{1-e^{2\pi in\tau}e^{2\pi iu}}$, where $$ R(u;\tau)=\sum_{\nu\in\frac12+\mathbb Z}(-1)^{\nu-\frac12}\left\\{\operatorname{sgn}(\nu)-E\left(\left(\nu+\frac{\Im(u)}{\Im(\tau)}\right)\sqrt{2\,\Im(\tau)}\right)\right\\}e^{-\pi i\nu^2\tau}e^{-2\pi i\nu u}, $$ with $E(x)=2\int_0^xe^{-\pi t^2}dt$.

In an illuminating (for me at any rate) paper "[M. P. Appell's function and Vector Bundles of Rank 2 on Elliptic Curves](https://pdfs.semanticscholar.org/4aef/7841fc86e1e2024c020094f75a1e6e9ef8f9.pdf)", Polishchuk explains that the function $\kappa_a(z;q)=\sum_{n\in\mathbb Z}\frac{q^{n^2/2}}{q^n-a}z^n$ can be uniquely characterized by the fact that $(\kappa_a(z;q),\theta(z;q))$ defines a holomorphic section of the rank 2 vector bundle on the elliptic curve $\mathbb C^*/q^{\mathbb Z}$ given by the quotient $(z,v_1,v_2)\sim(qz,av_1+v_2,q^{-1/2}z^{-1}v_2)$ of $\mathbb C^*\times\mathbb C^2$.

What confuses me is this - the way $\kappa$ is involved in the section of a bundle on the universal elliptic curve, it must have certain modularity properties; and it is clearly holomorphic (well, for $a\notin q^{\mathbb Z}$). On the other hand $\mu$, which is obviously closely related to it, although holomorphic, is not modular, while $\tilde\mu$, being modular, is not holomorphic. So seemingly $\tilde\mu$ must have a counterpart in the context considered by Polishchuk; but the latter does not mention any possible nonholomorphic modular corrections.

How exactly are $\mu$, $\tilde\mu$ and $\kappa$ related to each other?


Are there two non-equivalent exotic structures on $\mathbb{R}^4$ coming from topologically slice, non-slice knots?

For a knot $K \subset S^3$, which is topologically slice but not slice (in a smooth way), there's a four manifold $\mathbb{R}^4_K$, homeomorphic but not diffeomorphic to standard euclidean $\mathbb{R}^4$.

There are ways to find such knots, particularly knots with the Alexander polynomial equal to 1, are topologically slice (Freedman). Among these knots one can find non-slice knots using obstructions like $\tau$-invariant (coming from Knot Floer Homology) or Rasmussen $s$-invariant (coming from Khovanov Homology). Knots with non-zero $\tau$- or $s$- invariant are not slice.

The negative Whitehead double of left trefoil is an example of these knots.

Question. The above argument proves the existence of an exotic structure on $\mathbb{R} ^4$, but this method proves the existence of only one exotic structure on $\mathbb{R} ^4$. Can you show if there are two different knots, $K$ and $K'$ (both topologically slice, non-slice knots), which they give us two non-equivalent exotic $\mathbb{R} ^4$s? Or if all of $\mathbb{R}_K^4$ coming from these knots are the same, i.e. you can get to only one exotic structure through this method.