Boundaries of noncompact contractible manifolds

It is known that a manifold $B$ bounds a compact contractible topological manifold if and only if $B$ is a homology sphere. The "only if" direction follows by excising a small ball in the interior of the contractible manifold, and noting that its boundary sphere has the same homology as $B$ by Poincare duality. The "if" direction is due to Freedman in dimension $3$ and to Kervaire in dimensions $>3$.

**Question.** Is there a characterization of boundaries of _noncompact_ contractible manifolds?

Note that if $B$ is a boundary component of a contractible $n$-manifold $W$, then the following holds.

  1. If $B$ is compact, then $W$ is compact and $\partial W=B$.
(If $V$ denotes the union of $B$ and the interior of $W$, then the long exact sequence of the pair gives $H_{n-1}(B)\cong H_{n}(V,B)$ and Poincare duality gives implies that $H_{n-1}(B)$ is nontrivial, and hence so is $H_{n}(V,B)\cong H^0_c(V)$ which is only possible if $V$ is compact.)

  2. $B$ is stably parallelizable (because $W$ is parallelizable).

  3. If $U$ is a proper open subset of $B$, then $U$ bounds a noncompact contractible manifold, namely, $U\cup\mathrm{Int}(W)$.

  4. The product of $B$ with any open contractible manifold bounds a noncompact contractible manifold.

  5. If $B'$ bounds a contractible manifold $W'$, then the connected sum $B\\# B'$ bounds a contractible manifold, namely, the boundary connected sum of $W$, $W'$.

The above question may be too hard, so I would be happy with any addition to 1--5.

@User See theorem 11.3 in Massey "Homology and cohomology theory", or the section on Poincare duality in Bredon's "Sheaf theory".  There two references involve (co)homology of general spaces and may be hard for a beginner. Alternatively, solve exercise 35 in section 33 in Hatcher's "Algebraic topology".

Could you please tell me a source of the proof of $H^k_\textbf{c}(M)\cong H_{n-k}(M,\partial M)$, where $n=\dim M$?

Igor, of course,  one has to construct a Gromov-hyperbolic metric on $W$ "relative" to $B$, so that $B$ embeds. The difficulty with my approach, I think, is in two absolute cases: (1) constructing a Gromov-hyperbolic metric on $W$ so that the entire boundary is a homology sphere; (2) Given a compact contractible manifold $W$ with boundary $B$, construct a metric on $W$ so that $B$ is the ideal boundary. Part (2), I think, is quite doable and mildly interesting.   Part (1), I think, is difficult, if not impossible.

Misha, thank you for bringing geometry into the game. Yet I do not see how this gives an embedding of the boundary of $W$ into the ideal boundary of (a hyperbolic metric on) the interior of $W$. The ideal boundary is not necessarily the largest compactification. Say start with a compact contractible manifold whose boundary is a homology sphere $B$, remove a proper closed subset of $B$, and try to use your idea to embed what is left in a homology manifold; this looks hard.

A remark: The approach I suggested could work only for $n\ge 4$, since every homology 2-sphere is homeomorphic to the usual sphere. But for $n\ge 4$ I do not see any obvious obstructions.

Recall that Bestvina proved in 1996 that a hyperbolic group $G$ is a $PD(n)$ group if and only if its ideal boundary is a homology manifold which is also a homology sphere (of dimension n-1 of course). One can also ask an absolute version of your question in the hyperbolic context: Suppose $W$ is contractible $n$-manifold. Does $W$ admit a bounded geometry hyperbolic metric so that the ideal boundary is a homology manifold/homology sphere (of dimension n-1)? No group is involved here, just pure geometry.

I do not have an answer, but only a vague suggestion: Instead of trying to embed $B$ in a homology (n-1)-manifold, one can try to embed it in a homology sphere which is also a homology (n-1)-manifold. This might be easier. This situation could be similar to Mladen Bestvina's theorems on ideal boundaries of hyperbolic PD(n) groups. One approach would be to construct a suitable Gromov-hyperbolic metric on $W$ so that $B$ is a part of the ideal boundary. Alternatively, one can try to imitate Bestvina's arguments without having a metric (or, instead, using a weaker structure a la John Roe).

Tom's comment on intersection product together with classification of noncompact 2-manifolds implies that if $B$ has dimension $2$, then it is homeomorphic to an open subset of $S^2$, and clearly any open subset of $S^2$ occurs as $B$. Indeed, if the surface is not planar, then it contains a handle, and hence there are two curves that only intersect once.

@Igor: I meant the $Q/Z$ valued linking form on the torsion homology classes.  I was thinking a warm up problem is to determine which $B=$interior$(M)$, $M$ compact, are boundaries, and suggesting maybe all such embed as codim 0 submanifolds of a homology sphere.

@Paul, linking numbers do not vanish, not even in the sphere bounding a disk. I am not sure whether "open subsets of homology spheres" is the answer. Certainly, $B\times\mathbb R^n$ is homeomorphic to an open subset of $\mathbb R^{2n-1}$: just embed $B$ properly in $\mathbb R^{2n-1}$ and note that the normal bundle is trivial by 2. But I cannot see why $B$ itself would be homeomorphic to an open subset of $\mathbb R^{2n-1}$. Would not it be like showing that any open contractible manifold has a manifold compactification (which is surely false)?

Maybe the answer is any open subset of a homology sphere? Linking numbers vanish also.  I can't see how to find a compact $n$-manifold with boundary $M$ whose interior bounds a contractible manifold but which doesn't embed in a homology $n$-sphere. For $n=3$, it seems like you can add 2-handles to $M$ to get a homology ball, and perhaps in higher dimensions you could add handles to get a homology ball.

@Tom, oh, and the intersection number is trivial because the $0$-cycle is a boundary, so its points come in pairs with opposite signs, and cancel. Thank you for suggesting this!. I think this "intersection product" is Poincare dual to the cup product, as discussed in Dold's "Algebraic Topology", section VIII.13.

In particular I am referring to intersection numbers, not to the ring structure on homology.

Every cycle in $B$ is the boundary of a chain in $W$. So basically given a $p$-cycle and an $(n-p-1)$-cycle in $B$ their intersection is the boundary of the intersection of a $(p+1)$-chain and an $(n-p)$-chain in $W$, so the boundary of a $1$-chain, so a homologically trivial $0$-chain. To make this rigorous I guess you would need to say something about Poincare duality and compactly supported cohomology.

@Tom, I am struggling to make the notion of an intersection product precise. Is triviality of the intersection product based on the fact that $B$ has trivial normal bundle in $W$, or does the contractibility of $W$ get used?

By 3, $B$ can have the homotopy type of any finite  complex, by taking a thickening of some embedding in  Euclidian space. So for example any $k$ dim'l complex has  the homotopy type of a $n=2k+1$  manifold which is the boundary of a contractible $n+1$ manifold.

6. Intersection products in $B$ are trivial.


Riemannian manifolds etc. as locally ringed spaces?

There are, among others, three general ways of equipping a "space" (which for the purposes of this question could be a topological space or a differentiable manifold, according to the case) with further structure:

(1) "Specifying regular funcions", which leads to locally ringed spaces, e.g. real-analytic manifolds or holomorphic manifolds.

(2) "Choosing a section (perhaps with some nondegeneracy properties) of some (usually tensor) bundle", which leads for example to Riemannian manifolds, quasi-symplectic manifolds and quasi-holomorphic manifolds.

(2') A variant of the latter: "Choosing a nondegenerate section _with some integrability properties_ of some (usually tensor) bundle", which leads e.g. to symplectic geometry and holomorphic geometry.

(3) "Choosing a sub-bundle (with some properties) of some bundle" , and this leads for example to foliated manifolds and contact structures.

There are certainly some overlaps between the above approaches. For example, one can think of a complex manifold as a topological space equipped with a sheaf of local rings (the sheaf of holomorphic functions of the complex manifold), which is an instance of (1), or as an even dimensional smooth manifold equipped with an "integrable" field of endomorphisms of its tangent bundle which square to $-\mathrm{id}$, and that's an instance of (2), or rather (2'). Another example is that of a foliated manifold: it's usually seen as an instance of (3), but you can construct the sheaf of locally-constant-on-the-leaves smooth functions, and that reduces, in some sense, to the point of view (1).

Furthermore, smooth (in the sense of differentiable) maps between complex manifolds that preserve the tensor $J$ giving the complex structure correspond exactly to locally ringed space maps between the corresponding complex analytic spaces. And I would imagine that some foliation-preserving map in the case of foliated manifolds translate fully faithfully into maps for the locally ringed structure mentioned above.

In either case, if I'm not mistaken, you get a fully faithful embedding of some "geometric category" into the category of locally ringed spaces (such that, say, the forgetful functor to $\mathrm{Top}$ is respected). So it seems that the approach (1) is quite general. It's then spontaneous to ask:

> Is the category of symplectic manifolds (with symplectomorphisms as morphisms) fully faithfully embeddable (say, preserving the forgetful functor to $Top$) into the category of locally ringed spaces?

and

> Is there a good notion of morphisms between Riemannian manifolds (e.g. local isometries? Riemannian submersions? compositions thereof?) such that the resulting category has a fully faithful embedding into locally ringed spaces as above?

* * *

The first (perhaps useless?) construction that comes to my mind is the following. Given a smooth manifold $X$ with sheaf of differentiable functions $\mathcal{O}_X$, let $\mathcal{T} \\!en_X^{\bullet}:=\oplus_{i\geq 0} (T_X^{\; *})^{\otimes i}$ be the sheaf of covariant tensors on $X$. If $g$ is a Riemannian metric on $X$, one could consider the sheaf of (commutative) local rings generated by $\mathcal{O}_X$ and $g$:

$\mathcal{O}_X[g]\subseteq \mathrm{Sym}^{\bullet}(T_X^{\; *})\subseteq \mathcal{T} \\!en_X^{\bullet}$

and then consider the locally ringed space $(X,\mathcal{O}_X[g])$.

One can do an analogous thing with a symplectic manifold $(X,\omega)$ viewing $\omega$ as an anti-symmetric tensor:

$\mathcal{O}_X[\omega]\subseteq \Omega_X^{\mathrm{even}} \subseteq \mathcal{T} \\!en_X^{\bullet}$

to get a locally ringed space $(X,\mathcal{O}_X[\omega])$.

Can this construction be of any help to answer the above questions?

From scheme theoretic perspective, how do wish to 'deal' with additional structures like metrics, symplectic forms, hermitian or kahler structures?

I'm quite interested in this question. Made any progress since Feb?

@JohanesEbert: recovering $g$ up to conformal equivalence would already be nice. And what about LRS morphisms: would they correspond to conformal mappings?

@Zack: ya.. it looks like so. Perhaps my definition is just useless. But I don't know how morphisms of symplectic/Riemannian manifolds induce homomorphisms on that sheaves.

Isn't $\mathcal{O}[\omega]$ just $\mathcal{O}[x]/x^{n+1}$?

You cannot recover $g$ from $\mathcal{O}[g]$, only up to conformal equivalence.

@André: In the theory of "supermanifolds" people already use sheaves of $\mathbb{Z}/2\mathbb{Z}$-graded rings. And there are some recent theories of "derived manifolds" (of which I don't know anything). But my question was perhaps more down to earth. Anyway, why do you suggets using differential graded rings? Is it to keep track that the "generator" $g$ in $\mathcal{O}_X[g]$ is placed in (tensor) degree 2?

Maybe it could be useful to generalize the notion of locally ringed space, and allow "structure sheaves" with values in categories other than rings (e.g. differential graded rings).


When does a representation admit a spin structure?

Let $G$ be a finite group, and let $V$ be an $n$-dimensional _real_ representation of $G$. Think of $V$ as given by a homomorphism $$ \rho_V\colon G\to O(n).$$ Write $\chi_V$ for the character of $V$.

Here are two problems.

  1. Using _only_ the character $\chi_V$ of $V$, determine whether $\rho_V(G)\subset SO(n)$.

  2. Using _only_ the character $\chi_V$ of $V$, and assuming $\rho_V(G)\subset SO(n)$, determine whether $\rho_V$ admits a factorization through a homomorphism $\widetilde{\rho}_V:G\to Spin(n)$.

Here's one answer to 1: the identity of formal power series $$ \sum_{k\geq0} \chi_{\Lambda^kV}(g)\,T^k = \exp\Bigl( -\sum_{k\geq1} \frac{1}{k}\chi_V(g^k)\, (-T)^k \Bigr)$$ where $\Lambda^kV$ is the $k$-th exterior power representation of $V$, gives $\chi_{\Lambda^nV}(g)$ as a polynomial in $\chi_V(g),\dots,\chi_V(g^n)$, and $\rho_V(G)\subset SO(n)$ if and only if all $\chi_{\Lambda^nV}(g)>0$.

Is there a better answer for 1? Is there any answer in a similar spirit for 2?

For 1, it depends what you mean by "using the character ONLY". You can't tell the number of elements of each order from the character alone. If you do know the order of each element, then 1 requires just that each element of $G$ has the eigenvalue $-1$ with even (possibly zero) multiplicity, and this is an easy calculation, given the character values ( in fact, you only need to check elements whose order is a power of $2$).

Charles, I can see at least a necessary condition on the character (that you probably know of). It is that for any involution $s\in G$, $\chi_V(s)\equiv\chi_V(1) \mod 8$ (this comes from restriction to cyclic subgroups). Given the "miracles" of Brauer theory, might it be sufficient also ? A perhaps relevant reference is this Deligne's 1976 [paper][1], related to Marty's first remark about $\epsilon$ factors.     [1]: digizeitschriften.de/de/dms/img/?PPN=GDZPPN002092654

Marty, your observation about pairs of conjugacy classes seems convincing to me.

@BS:  Ah yes.. that was dumb on my part.  I had in mind the lifting of subgroups isomorphic to $Z$, not $Z / kZ$, when I wrote that.  The remainder of my comment -- that one needs information about pairs, not just individual elements -- should hold (though it's vague).

@marty : are you saying that any morphism $Z/kZ \to SO(n)$ lifts $Spin(n)$ ? This seems not true if $k$ is even.

Here's another reason why such an answer shouldn't exist for (2).  Let's say you know all the eigenvalues for $\rho(g)$, $\rho(g^2)$, etc., information given by the character.  Well, that information tells you nothing about whether $\rho$ factors through the 2-fold Spin cover, because every element of $SO(n)$ lifts (compatibly with its powers) to an element of $Spin(n)$.  Instead, you need information about $\rho(g_1), \rho(g_2), \rho(g_1 g_2)$ for all pairs $(g_1, g_2)$ -- information depending probably on the conjugacy class of the pair $(g_1, g_2)$ (not individual conjugacy classes).

Okay.  But I don't see why, a priori, having a cohomological obstruction means you can't compute it from the character; my question 1 is really about the vanishing of $w_1(\rho_V)$, after all.    On the other hand, I could certainly believe that such a problem could be intractible.

(I suppose that the $SU(2)$ and $SO(3)$ case is just the case $n = 3$ of the question...)

I don't think that any such answer exists for (2).  The obstruction to lifting is called the second Steifel-Whitney class of the orthogonal real representation $\rho_V$, written $w_2(\rho_V)$.  It's difficult to compute and quite important for Galois representations (related to $\epsilon$-factors).  Such results lie beyond simple character calculuations, I think.  Compare, for instance, the problem of deciding whether a rep. into $SO(3)$ arises as $Sym^2$ of a rep into $SU(2)$.  Can you tell by the character?

I guess there is a cohomological obstruction  for 2 in $H^2(G,\mathbb{Z}/2\mathbb{Z})$. Although, I have no idea how to express this in character theory.


Is the set of integers of the form $a/(b+c)+b/(a+c)+c/(a+b)$ computable?

The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying

$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$$

are [absurdly high](https://web.archive.org/web/20170813162605/https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM), namely $$(154476802108746166441951315019919837485664325669565431700026634898253202035277999,$$ $$36875131794129999827197811565225474825492979968971970996283137471637224634055579 ,$$ $$ 4373612677928697257861252602371390152816537558161613618621437993378423467772036) .$$ This leads to the following general question: Is the set $C\subseteq {\mathbb N}$ defined by $$ C = \\{n\in\mathbb{N}\setminus\\{0\\}: (\exists a,b,c \in\mathbb{N}\setminus\\{0\\}):\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\\}$$ computable? (As user [Watson](https://mathoverflow.net/users/84923/watson) points out in the comment section below, $C$ contains no odd numbers. It would also be great to see an even number $\geq 6$ not contained in $C$.)

The goo.gl link in a previous comment points to Quora.

Also discussed at mathematica.stackexchange.com/questions/184956/…

(continued) Note that there are odd $n$ for which the associated elliptic curve has positive rank, but does not have rational points on the component that contains the positive points. (IIRC, $n = 19$ is an example.) This gives an additional twist to the question.

This is somewhat similar to the question which natural numbers $n$ are congruent numbers (i.e. the area of a right-angled triangle with rational sides), which comes down to asking whether the elliptic curve $E_n \colon y^2 = x^3-n^2x$ has positive rank over $\mathbb Q$. In this case, there is a conjectural answer (Tunnell's Theorem, conditional on the BSD conjecture), which relies on the fact that the $E_n$ are all quadratic twists of a fixed curve. The question asked here is likely to be harder, since the resulting curves are not twists, and there is the positivity condition. -->

Here is explained why $C$ doesn't contain any odd number.

See also math.stackexchange.com/questions/402537/… and Bremner and Macleod, An unusual cubic representation problem, Annales Mathematicae et Informaticae 43 (2014) 29-41, ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf

I don't have the time to look into the details, but from a cursory glance at this quora answer it looks like the problem amounts to deciding whether a certain elliptic curve has solutions over $\mathbb{Q}$. Now IIRC there is an algorithm for that (find solutions locally and try to globalize) provided Ш is finite, which is conjecturally always the case. So I think conjecturally your set is indeed computable.

A quick heuristic comment: for each $n$, the existence of such $a, b, c$ corresponds to the existence of a solution to a degree-$3$ Diophantine equation in $3$ variables; and this is, I believe, a bit beyond what is generally known to be decidable. So I suspect that there will be no general reason why this set is computable; rather, if it is computable (which I extremely strongly suspect it is), the proof is likely to be a corollary of a complete characterization of these $n$s, which will not use any computability theory.


A mysterious paper of Stallings that was supposed to appear in the Annals

In Stallings's paper

  * Stallings, John, _[Groups with infinite products](https://doi.org/10.1090/S0002-9904-1962-10817-9)_ , Bull. Amer. Math. Soc. **68** (1962), 388–389.

he briefly discusses how to prove "several generalizations" of Brown's theorem saying that monotone union of open $n$-cells is an open $n$-cell; however, he never formulates a precise statement of what he proves.

For the details, he refers to the paper

  * J. Stallings, On a theorem of Brown about the union of open cones, Ann. of Math, (to appear).

But as far as I can tell this paper never appeared. Does anyone know what happened and what the theorem is?

I feel the pain. There's more than one of these in category theory from the 70s-80s, announced to appear in JPAA, but which never turned up.

With regards to the publication of this paper, it looks like the journal may have been... stalling

@CarloBeenakker: Thanks for noticing that reference!  I think that the lack of a journal might just be the Annals's style guide -- there are several papers in that bibliography that are "to appear", and none of them specify a journal.

in a 1963 Ann. of Math. article by Stallings that paper is still cited as "to appear", however now without a journal being mentioned.


The multiplication game on the free group

Fix $W\subseteq\mathbb F_2$ and consider the following two-person game: Player 1 and Player 2 **simultaneously** choose $x$ and $y$ in $\mathbb F_2$. The first player wins, say one dollar, iff $xy\in W$. Does this game admit Nash equilibria?

Of course, for very particular choices of $W$ it does, for instance if $W$ is either finite or cofinite. But I am not interested in these cases, I am interested in _solvability_ of the _multiplication game_ for all $W$.

One possible mathematical reformulation (game theorists would say that the mixed extension of this game is not uniquely defined and then there are _several_ different mathematical reformulations) is the following:

> **Question.** Do there exist, for all $W\subseteq\mathbb F_2$, finitely additive probability measures $\overline\mu,\overline\nu$ (of course depending on $W$) on the power set of $\mathbb F_2$ such that for all other finitely additive probability measures $\mu,\nu$ one has $$ \int\int\chi_W(xy)d\mu(x)d\overline\nu(y)\leq\int\int\chi_W(xy)d\overline\mu(x)d\overline\nu(y)\leq\int\int\chi_W(xy)d\overline\mu(x)d\nu(y) $$ ?

**Motivation.** If we play the same game on a (countable) amenable group, one can show that a Nash equilibrium exists and is given by a right-invariant mean $\overline\mu$ and a bi-invariant mean $\overline\nu$. I am wondering what can happen for non amenable groups... I suspect (motivated by some possibly wrong speculative consideration) that the _multiplication game_ is actually solvable for all $W$ only on amenable groups, giving an hopefully nice new characterization of amenability. However, any approach to proving that solvability implies amenability so far got stuck.

Thanks in advance,

Valerio

I've edited, saying explicitly that the choices of $x$ and $y$ are simultaneous.

Why should P2 win? Notice that the choices are done independently (simultaneously, if you want), so that P2 cannot choose a suitable $y$ such that $xy\in W$, because (s)he does not know $x$.

I'm confused. If $W$ is the whole group, player 1 wins, otherwise player 2 wins. Right?


Characterising critical points of $E(f)=\int_{M}| \bigwedge^2 df|^2 \text{Vol}_{M}$

$\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\TM}{\operatorname{TM}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{T^*\M}$

Let $\M,\N$ be $d$-dimensional oriented Riemannian manifolds. Let $\tilde E:C^{\infty}(M,N) \to \R$ be defined by

$$\tilde E(f)=\frac{1}{2}\int_{\M} | \bigwedge^2 df|^2 \text{Vol}_{\M}.$$ $\tilde E(f)$ measures the mean action of $f$ on 2$D$-cubes. (How $f$ affects areas of surfaces, locally).

Note that $\bigwedge^2 df\in \Omega^2\big(\M,\Lambda_2(f^*{\TN})\big)$. Let $\nabla^{\Lambda_2(f^*{\TN})}$ be the induced connection on $\Lambda_2(f^*{\TN})$ and let $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} $ be the adjoint of the covariant exterior derivative $d_{\nabla^{(f^*{\TN})}}$.

> **Question:** Does every critical point of $\tilde E$ satisfy $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$?

Of course, by counting degrees of freedom, this doesn't look reasonable.

The converse direction follows from the Euler-Lagrange equation of $\tilde E$: I proved [here](https://mathoverflow.net/a/271264/46290) that the E-L equation is $$h_{f^*\TN} \big( \tr_{\TM}\big( df \otimes \delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)\big) \bigg)=0,$$

where $h_{f^*\TN}:f^*\TN \otimes \Lambda_2(f^*\TN) \to f^*\TN$ is the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2.$$

Thus $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$ implies $f$ is critical.

**Edit 1:**

A natural way to produce critical points of a functional is to look at its symmetries. $\tilde E$ is conformally-invariant exactly at dimension $4$. So, in dimension $4$ conformal maps are critical. However, I proved that they also satisfy the stronger equation $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$.

(In fact I proved that a conformal map $\M^d \to \N^d$ satisfies $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ if and only if $d=2k$ or it is a homothety).

**Edit 2:**

As I shall prove below (see "Edit 3"), in dimension $d=2$ the answer is positive. (Note that in this case the "number of constraints" is the same). Focusing on the simplest next case, we consider $d=3,\M=\N=\R^3$ with the flat metrics.

**First, let's try to answer a more degenerate version of the question (forgetting $h_{f^*\TN}$):**

> Is there a smooth map $f:\mathbb{R}^3 \to \mathbb{R}^3$, which satisfy
>
> $$\delta \big( df \wedge df \big) \neq 0, \, \text{and } \, \tr \big( df \otimes \delta(df \wedge df) \big)=0, $$

If there is such an $f$ then it cannot be a local diffeomorphism; its Jacobian must vanish somewhere.

Here $ \delta \big( df \wedge df \big)\in \Omega^1\big(\R^3;\Lambda_2(\R^3)\big)$ is a one-form on $\R^3$ with values in $\Lambda_2(\R^3)$, and

$$ \delta \big( df \wedge df \big)(X)=\sum_i \nabla_{e_i}(df \wedge df)(e_i,X)=\sum_i (\nabla_{e_i} df)(e_i) \wedge df(X) + df(e_i) \wedge (\nabla_{e_i} df)(X)$$ $$ =\Delta f \wedge df(X)+\sum_i df(e_i) \wedge (\nabla_{e_i} df)(X).$$

We can try to look first for harmonic counter-examples, but so far I failed doing even that.

**Edit 3:**

> The answer is positive for $d=2$, so we need to restrict our attention to $d \ge 3$.

Indeed, in this case $E(f)=\int_{\M}| \bigwedge^2 df|^2 \text{Vol}_{\M}=\int_{\M}(\Det df)^2 \text{Vol}_{\M}$. We shall prove a map $f$ is $E$-critical if and only if its determinant is constant.

Since $\Det df$ is constant $\iff$ $\nabla (\bigwedge^2 df)=0 \iff \delta_{\nabla^{\Lambda_2(f^*\TN)}} \big( \bigwedge^2 df \big)=0$, we are done.

The Euler-Lagrange equation of $E$ can be written as

$$ \delta (\Det df \cdot \Cof df) =0, \tag{1}$$ where $\Cof df:\TM \to f^*(\TN)$ is the _cofactor map_ of $df$ defined by $$ \Cof df= \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1, $$

and $\delta$ is the adjoint of the pullback connection on $f^*\TN$.

(The Euler-Lagrange equation of the Jacobian functional $E(f)=\int_{\M} \Det df \text{Vol}_{\M}=\int_{\M} f^* \text{Vol}_{\N}$ is $\delta (\Cof df)=0$. Details can be found in lemma 2.9 in my paper [here](https://arxiv.org/abs/1701.08892)).

Expanding equation $(1)$ we get $$ 0=\delta(\Det df \cdot \Cof df )= \Det df \delta(\Cof df ) - \tr_{g}(d \Det df \otimes \Cof df). $$

We now use the fact the Jacobian functional is a null-Lagrangian, i.e. every smooth map is a critical point, or equivalently $\delta (\Cof df)=0$. (This is essentially Stokes theorem, you can see lemma 2.5 [here](https://arxiv.org/abs/1701.08892)).

So, the E-L equation $(1)$ reduces to $$\tr_{g}(d \Det df \otimes \Cof df)=0. \tag{2}$$

Let $f$ be a map satisfying equation $(2)$. We shall prove $\Det df$ is constant; suppose that $\Det df_p \neq 0$ for some $p \in \M$. This implies $ \Cof (df_p)$ is invertible, so by equation $(2)$ $d\Det df_p=0$.

Now [we observe](https://math.stackexchange.com/a/2489068/104576) that any $C^1$ function $g : \M \to \mathbb R$ on a connected manifold, satisfying $g(p) \ne 0 \implies dg_p = 0$ is constant.

For the local problem, it may be useful to assume that $M,N$ are both flat in which case you can get rid of the geometry and concentrate on the multilinear algebra.


Smooth thickenings of non-smoothable manifolds

It is known that any closed topological manifold is homotopy equivalent to an open smooth manifold.

**Question 1.**_What can be said about the**smallest** dimension of a smooth manifold that is homotopy equivalent to a given closed topological manifold?_

The following somewhat heavy-handed argument yields a smooth manifold of roughly twice the dimension, namely, use

  * [ West's solution ](http://www.maths.ed.ac.uk/~aar/papers/west.pdf) of Borsuk's conjecture that any compact ANR of dimension $n$, and in particular any closed $n$-manifold, is homotopy equivalent to a finite polyhedron of dimension $\max\\{n, 3\\}$;
  * [ Stallings-Dranishnikov-Repovs's ](http://www.pef.uni-lj.si/repovs/clanki/1993/BullAustralMathSoc.pdf) embedding up to homotopy type theorem that any finite $n$-dimensional polyhedron is homotopy equivalent to a finite $n$-dimensional subpolyhedron of $\mathbb R^{2n}$, so its regular neighborhood is the desired open smooth manifold.

Here is a specific question that shows the state of my ignorance on this matter:

**Question 2.** _Is there a closed $n$-manifold which is**not** homotopy equivalent to a smooth $(n+1)$-manifold? _

The naive idea to look at the product of a non-smoothable manifold of dimension $\ge 5$ with $\mathbb R$ fails, because such a product is also non-smoothable (by the topological product structure theorem of Kirby-Siebenmann).

**Edit:** Misha kindly corrects me that a $5$-manifold is smoothable if and only if its Kirby-Siebenmann invariant vanishes; in particular, this apples to products of a $\mathbb R$ and a closed $4$-manifold $M$. Thus $M\times\mathbb R$ is smoothable iff $M$ has zero KS invariant. Smooth $4$-manifolds have zero KS invariant, but amazingly so do some non-smoothable ones.

@Misha: certainly, not all $(n+1)$-dimensional thickenings of closed $n$-manifolds are proper homotopic to $I$-bundles. One can start with an $I$-bundle, and then attach a one-sided $h$-cobordism along the boundary; there are general methods to construct those in higher dimensions. One can also take boundary connected sums with (possibly noncompact) contractible manifolds with boundary, and this way one gets tons of examples when $n\ge 3$ whose ends are quite complicated.

@Igor: Concerning KS invariant for 4-manifolds: It vanishes iff $M^4\times {\mathbb R}$ is smoothable. In dimension 4 there are non-smoothable manifolds with both zero and non-zero KS invariants since there are other, gauge-theoretic, obstructions to smoothability.

@Igor: Actually, I realized that even for n=3, a tame 4-dimensional thickening of a hyperbolic 3-manifold need not be properly homotopy-equivalent to an interval bundle.

@Misha: I am familiar with examples of closed non-smoothable aspherical manifolds by Davis-Januszkiewicz and Davis-Hausmann, in fact, my question was motivated by the desire to see how much the action dimension of arxiv.org/abs/math/0010141 would change if one defines it using smooth (!) properly discontinuous actions.

@Mark: corrected; I hope you won't insists on accents in Repovs. @Misha: Do you know a $4$-dimensional example? In higher dimensions there is an old preprint of Galewski-Hollingworth that implies that there is no $(n+1)$-dimensional compact smooth thickening $V$ of an $n$-dimensional Poincare complex $X$, provided both of them are orientable; compactness of $V$ can be replaced by the assumption that an image of $X$ separates a neighborhood in $V$. Showing that $V$ is properly h.e. to an $\mathbb R$-bundle seems tricky.

@Igor: 1. Sometimes, the product with ${\mathbb R}$ is smoothable. 2. You can avoid Stallings' theorem if you use immersion to ${\mathbb R}^{2n}$ instead of an embedding. 3. Try Davis-Januszkiewicz argument, see the reference  here  and see if this gives a counter-example at least for a tame smoothing. (Try to argue that your open smooth (n+1)-manifold is properly homotopy-equivalent to an open interval bundle over a non-smoothable n-manifold.)


On certain representations of algebraic numbers in terms of trigonometric functions

Let's say that a real number has a _simple trigonometric representation_ , if it can be represented as a product of zero or more rational powers of positive integers and zero or more (positive or negative) integer powers of $\sin(\cdot)$ at rational multiples of $\pi$ (the number of terms in the product is assumed to be finite, the empty product is taken to be $1$).

_Examples:_

  * $\sqrt[3]{\sin\\!\left(\frac\pi3\right)}$ has a simple trigonometric representation, because $\sqrt[3]{\sin\\!\left(\frac\pi3\right)}=\frac{3^{1/6}}{2^{1/3}}$.
  * $\sqrt{\sin\\!\left(\frac\pi{10}\right)}$ has a simple trigonometric representation, because $\sqrt{\sin\\!\left(\frac\pi{10}\right)}=\frac{2^{1/2}}{5^{1/4}}\,\sin\\!\left(\frac\pi5\right)$.
  * $\pi$ does not have a simple trigonometric representation, because it is not an algebraic number.

_Questions:_

  * Do $\sqrt{\sin\\!\left(\frac\pi5\right)},$ $\sqrt{\sin\\!\left(\frac\pi8\right)},$ $\sqrt{\sin\\!\left(\frac\pi{12}\right)},$ $\sqrt{\sin\\!\left(\frac\pi{15}\right)},$ $\sqrt{\sin\\!\left(\frac\pi{20}\right)},$ $\sqrt{\sin\\!\left(\frac\pi{24}\right)}$ have simple trigonometric representations?
  * Is there an algorithm that, given a rational power of $\sin(\cdot)$ at a rational multiple of $\pi$, would determine if it has a simple trigonometric representation? If so, could you give (or outline) a concrete example of such an algorithm (efficient, if possible)?
  * More generally, is there an algorithm that, given a real algebraic number (in some explicit form, e.g. as its minimal polynomial and a rational isolating interval), would determine if it has a simple trigonometric representation? If so, could you give (or outline) a concrete example of such an algorithm (efficient, if possible)?

You mean solve all minimal polynomials with $\sin(\pi\times rational)$?

This is related I guess: mathworld.wolfram.com/TrigonometryAngles.html


Infinitely many planets on a line, with Newtonian gravity

(I previously asked essentially this [on physics.stackexchange](https://physics.stackexchange.com/questions/56843/infinitely-many-planets-on-a-line-with-newtonian-gravity), but was actually
hoping for answers with something closer to a proof than what I got there.)

Suppose we have a unit mass planet at each integer point in 1-d space. $\:$ As described in that answer, the sum of the forces acting on any particular planet is absolutely convergent. $\;\;$ Suppose we move planet_0
to point $\epsilon$, where $\: 0< \epsilon< \frac12 \:$. $\;\;$ For similar reasons, those sums will still be absolutely convergent.
Now we let Newtonian gravity apply. $\:$ What will happen?

If it's unclear what an answer might look like, you could consider the following more specific questions:

Will there be a positive amount of time before any collisions occur?
(As opposed to, for example, a collision at time $\frac1n$ for each positive integer $n$.)

"Obviously" (at least, I hope I'm right), planet_0 will collide with planet_1. $\:$ Will that be the first collision?

planet_0 will start out moving right, and all of the other planets will start out moving to the left.
Will there be a positive amount of time before any of them turn around?

How long will it be before there are any collisions? $\:\:$ (perhaps just an approximation for small $\:\epsilon\:$)

@Anthony Quas and Jon: I think this is not what he means. He does not "live in 1D world" but in 3-space. Just the planets happen to be on the line. The gravity law must be inverse squares. Otherwise the questions make no sense.

@AnthonyQuas is correct. The Poisson equation in one dimension in this case is $\phi''(x)=\delta(x)$ that has as a solution $\theta(x) x$. So, the potential between two bodies is not decaying in this case.

Excellent question! Perhaps  You should start as in thermodynamics with a large number $N$ of particles located at $-N,\dotsc,-1, \epsilon, 1,\dotsc, N$ and see what happens then. At least in this case the total energy is finite, which is not the case  when infinitely many  particles are present.

Are the planets point masses, or some radius $r \lt \frac{1}{2}$?

Yes, particles clump together, typically forming smaller systems first. This is studied extensively in cosmology, both analytically and numerically. Gravitational effects are easy to model, and affect dark matter. youtube.com/watch?v=8C_dnP2fvxk However, dissipative effects such as the inelastic contraction of gas clouds are important, too.

I think that as long as each planet is at most $\delta$ from its nearest integer, the total force on each planet is $O(\delta)$. This can be used to prove rigorously that there's a positive $\tau>0$ before any collision can occur.

Of course if you really live in a 1D world, gravitational force presumably doesn't decay at all?

I read recently about a very similar problem that appeared in a 1949 letter from Ulam to von Neumann. (In that case the particles started at points of $\mathbb Z$ with each node being occupied with probability 1/2). He showed(?) that something analogous to the universe happens: nearby groups of particles come together; and then those "solar systems" form galaxies etc.


Are there "chain complexes" and "homology groups" taking values in pairs of topological spaces?

Throughout this question, notation of the form $(X,A)$ denotes a sufficiently nice pair of topological spaces. I think for most of what I'm saying here, it is enough to assume that the inclusion $A \hookrightarrow X$ is a [cofibration](http://ncatlab.org/nlab/show/Hurewicz+cofibration), or that $(X,A)$ is a neighborhood deformation retract ([NDR](http://ncatlab.org/nlab/show/neighborhood+retract)) pair. All maps $f:(X,A) \to (Y,B)$ are continuous and satisfy $f(A) \subset B$.

Consider a sequence $\mathcal{S}$ given by: $$\ldots \to (X_{n-1},A_{n-1}) \stackrel{d_{n-1}}{\to} (X_n,A_n) \stackrel{d_n}{\to} (X_{n+1},A_{n+1}) \to \ldots$$ so that $d_n \circ d_{n-1}(X_{n-1}) \subset A_{n+1}$ for each $n$.

> Has this object been defined and studied? If so, where?

One can associate to this sequence of space-pairs a "homology" which takes values in the category of space-pairs. That is, define $$\mathcal{HT}_n(\mathcal{S}) = \left(d_n^{-1}(A_{n+1}),d_{n-1}(X_{n-1})\right).$$ The construction is functorial if one considers the obvious analogue of "chain maps" in the category which contains objects like $\mathcal{S}$. I'm wondering if basic definitions and properties, etc. of this or related constructions have been set down somewhere.

@ViditNanda This is two years old by now. Did you make any progress?

@Vidit: I've sent an email on this. One point is that a pair $(X,S)$ defines filtrations $E_n(X,A)$ which are a base point in dim $0$, $A$ in levels $1$ to $n-1$ and $X$ thereafter. So if $X$ already has a filtration $X_*$ then we get two filtrations for a given $n$.

Ronnie, I'm not sure I see an "obvious" bifiltration anywhere in this question. Could you explain why that would be the natural thing to do? I am happy to discuss this via email if things get too intricate for comment boxes.

@Vidit: Reading your question again, I would be inclined to go for defining invariants of bifiltered spaces, rather than invariants in spaces.  Such bifiltered spaces in truncated form can be regarded as special cases of $n$-cubes of spaces, to which my work with Loday applies. I'm slowly writing up something on the bifiltered case, and my exposition  [60]  ``Triadic Van Kampen theorems and Hurewicz theorems'', Algebraic  Topology, Proc. Int. Conf. March 1988, Edited M.Mahowald and S.Priddy,  Cont. Math. 96 (1989) 39-57. (pdf on my web site) gives an idea. Also I missed the word "discrete"!

Thank you, Ronnie. I already have your book(s), being a fan if the groupoid point of view, so I will take a look.

@Vidit: You mention a "filtered cell complex".  The book "Nonabelian algebraic topology" (2011) see my web page, pages.bangor.ac.uk/~mas010/nonab-a-t.html , builds algebraic topology directly from filtered spaces, via homotopically defined functors. Problem 16.1.17 is about relating these methods to Morse Theory.

Johannes: somewhere in between. I'd call it concretely-motivated curiosity. My thesis work involved simplifying a filtered cell complex via discrete Morse theory in a way that preserved algebraic topological invariants (persistent homology groups). As such, I find the process of constructing massive chain groups and then taking a huge quotient somewhat inefficient, and was wondering if there is a "compact representation" (both words used non-technically) of some object in the category of topological spaces itself from which one could derive the invariants whenever necessary.

Where did you meet this structure? In a concrete problem or just out of curiosity?


Fake CM elliptic curves

Suppose one has an elliptic curve $E$ over $\mathbb{Q}$ with conductor $N < k^3$ for some (large) positive $k$, with the property that its Fourier coefficients satisfy $$ a_p=0, \; \mbox{ for all } \; p \equiv -1 \mod{4}, \; \; \; k/3 < p < k. $$ Can one prove unconditionally that $E$ has CM? This follows from work of Serre (under GRH) or Elkies (under something like Szpiro's conjecture), since otherwise we'd have a surplus of supersingular primes. It does not appear to be a consequence of, say, Serre's argument without additional hypotheses (though I'd be happy to be wrong on this score).

Dear BCrd : I've pondered that, but I really don't know. My admittedly uneducated guess would be "probably not" with the level of precision I'm after....

Dear Mike: Can one make "effective" (in the spirit of your question) any information coming from the knowledge that Sato-Tate holds in the non-CM case? (Sorry to answer a question with another question...).


Is the equivariant cohomology an equivariant cohomology?

Suppose a finite group $G$ acts piecewise linearly on a polyhedron $X$. Then there are two kinds of equivariant cohomology (or homology).

$\bullet$ With coefficients in a $\Bbb Z G$-module $M$. A reference is K. Brown's "Cohomology of groups". Namely, $H^\ast_G(X;M)=H^\ast(Hom_{\Bbb Z G}(C_\ast,M))$, where $C_\ast$ is the simplicial chain complex of a suitable invariant triangulation of $X$. Note that $H^\ast_G(X;M)\simeq H^\ast_G(X\times EG;M)$. Generalized cohomology (including stable cohomotopy) of $X$ with coefficients in a module can be defined via generalized cohomology of $X\times_G EG$ with local coefficients, which in turn is defined in the last chapter of the Buoncristiano-Rourke-Sanderson book.

$\bullet$ Representation-graded. A reference is "Equivariant homotopy and cohomology theory" by J. P. May et al. (but to simplify matters, I'm thinking of trivial coefficients, as in [Segal](http://www.mathunion.org/ICM/ICM1970.2/Main/icm1970.2.0059.0064.ocr.pdf) and [Kosniowski](http://resolver.sub.uni-goettingen.de/purl?GDZPPN002309874) rather than coefficients in a Mackey functor). Instead of defining $H^*_G$ let me only say that there is a Hurewicz homomorphism $\omega^*_G(X)\to H^*_G(X)$, and the equivariant stable cohomotopy $\omega_G^{V-W}(X)$ can be defined as the equivariant homotopy set $[S^{W+U}\wedge X,S^{V+U}]_G$, where $U$ is a sufficiently large (with respect to the partial order by inclusion) finite-dimensional $\Bbb RG$-submodule of $\Bbb R G\oplus\Bbb RG\oplus\dots$, and the "representation sphere" $S^U$ is the one-point compactification of $U$ (viewed as a Euclidean space with an action of $G$).

What relations are there between these two kinds of (e.g. ordinary) equivariant cohomology? More specifically,

> is cohomology with coefficients in a $\Bbb ZG$-module a special case of representation-graded cohomology?

A related (equivalent?) question is whether cohomology with coefficients in a (nontrivial!) module is representable in some reasonable sense.

For instance, by reducing both sides to non-equivariant cohomology, I know that $H^m_{\Bbb Z/2}(X;I^{\otimes m})\simeq H_{\Bbb Z/2}^{mT}(X_+)$, where $\Bbb Z/2$ acts freely on $X$, $X_+=X\sqcup$(basepoint), $I$ is the augmentation ideal of $\Bbb Z[\Bbb Z/2]$ (so $I^{\otimes m}$ is $\Bbb Z$ if $m$ is even and $I$ if $m$ is odd) and $T$ is the augmentation ideal of $\Bbb R[\Bbb Z/2]$ (so $mT$ is $\Bbb R^m$ with the antipodal involution). Something of this kind can also be done for stable cohomotopy. However this all depends on a twisted Thom isomorphism which I don't know for more general modules.

When the action is free, $H^n_G(X;M)\simeq [X,K(M,n)]_G$; the relative case looks a bit more complicated, $H^n_G(X,Y;M)\simeq [(X,Y),(K(M,n)\times BG,BG)]_G$ (see VI.3.11 in the Goerss-Jardine book and projecteuclid.org/euclid.ijm/1256052280). On the other hand, for every $G$-spectrum $E$, $h^{V-W}_G(X):=[S^W\wedge X,S^V\wedge E]_G$ is an equivariant cohomology theory. So we want to find for an $M$ e.g. a $V$ such that $H^{dim V+1}_G(S^V\wedge B|M|)\simeq M$, where $|M|$ is $M$ with the trivial action of $G$. This is indeed possible for $G=\Bbb Z/2$ but not in general.

Another reference for generalized cohomology with local coefficients is K. Brown's paper, ams.org/journals/tran/1973-186-00/S0002-9947-1973-0341469-9


bound on the genus of a fiber of the Albanese map of a surface with $h^1({\mathcal O})=1$?

This is maybe more an open problem than a question, since I have seriously thought about it and asked several people working on algebraic surfaces with no success. I hope somebody here can suggest an approach different from the standard arguments in surface theory.

BACKGROUND: let $X$ be a smooth minimal complex projective surface of general type. An _irrational pencil_ is a morphism with connected fibers $f\colon X\to B$, with $B$ a smooth curve of genus $b>0$.

For $b>1$, $X$ has at most finitely many pencils of genus $b$, having such a pencil is a topological property and it is possible to bound explicitly the genus of a general fiber of $f$ in terms of $K^2_X$ (Arakelov' theorem).

For $b=1$, namely for _elliptic pencils_ , things are very different in general: a surface can have infinitely many such pencils, the genus of the general fibers of these pencils can be unbounded, and it is possible that a surface with an elliptic pencil deforms to a surface without elliptic pencils.

However, if $h^1({\mathcal O}_X)=1$, then the Albanese map $a\colon X\to Alb(X)$ is an elliptic pencil, and for fixed $K^2$ the genus of a general fiber of $a$ is bounded, since the moduli space of surfaces with fixed $K^2$ is quasiprojective.

QUESTION: can one give a bound for the genus of the general fiber of the Albanese pencil of a minimal surface of general type $X$ with $h^1({\mathcal O}_X)=1$ in terms of $K^2_X$? Such a bound would be very interesting in the fine classification of surfaces of general type.

If $K_S-F$ is effective, than of course I have a bound since $K_S(K_S-F)\ge 0$ and $K_SF=2g(F)-2$. And  by the same argument, I would have a bound if I could determine explicitly an  $m$ such that $mK_S-F$ is effective, but that's precisely what I don't know how to do.

I GUESS in some sense you can make it. For example, you can consider the canonical map of the surface, which will work if $K_S^2$ is not so small. Assume that the fiber of the Albanese map is $F$, then consider whether $\mathscr{O}(K_S-F)$ has global section or not. If it has a global section, then after computing some intersection number, maybe you can have a bound of $g(F)$.


Row of the character table of symmetric group with most negative entries

The row of the character table of $S_n$ corresponding to the trivial representation has all entries positive, and by orthogonality clearly it is the only one like this.

Is it true that for $n\gg 0$, the row of the character table of $S_n$ corresponding to the sign representation has the most negative entries of any row (about half of them)?

Note that for $S_4$, in addition to the sign representation row there are two other rows which also have 2 negative entries.

Possibly the fact that row sums are positive is relevant here.

**EDIT** : see also <https://realopacblog.wordpress.com/2025/05/02/maximizing-negative-entries-in-symmetric-group-character-tables/>

@PerAlexandersson: I don't see any obvious connection to that conjecture.

Is this related perhaps to the conjecture that about half of the partition numbers, p(n) are even, as n goes to infinity?

@DenisSerre: the sum of the row corresponding to the sign representation is oeis.org/A000700. This quantity grows much slower than $p(n)$, hence why I said “about half.”

Is it clear/known that "about half of conjugacy classes have negative signature" ?

Some questions of a somewhat similar flavor that have been considered lately include whether most entries in the character table are highly divisible, or in fact are equal to $0$ (see e.g. arxiv.org/abs/2010.12410).

@FedorPetrov: each conjugacy class. The character table is a $p(n)\times p(n)$ table, where $p(n)$ is the number of partitions of $n$.

Do we count each permutation once or each conjugacy class once?


Can a 4D spacecraft, with just a single rigid thruster, achieve any rotational velocity?

_(Copied[from MSE](https://math.stackexchange.com/questions/4472850/can-a-4d-spacecraft-with-just-a-single-rigid-thruster-achieve-any-rotational-v). Offering four bounties over time, I got no response, other than twenty-nine upvotes.)_

It seems preposterous at first glance. I just want to be sure. Even in 3D the behaviour of rotating objects can be surprising (see the Dzhanibekov effect); in 4D it could be more surprising.

A 2D or 3D spacecraft (with no reaction wheels or gimbaling etc.) needs at least two thrusters to control its spin. See my [answer](https://space.stackexchange.com/a/59430) to [Could this three-thruster spherical spacecraft de-tumble itself with zero final velocity?](https://space.stackexchange.com/questions/53590) on space.SE.

* * *

For any two $4\times4$ matrices $X$ and $Y$, define the commutator $[X,Y]=XY-YX$, the anticommutator $\\{X,Y\\}=XY+YX$, and the Frobenius inner product $\langle X,Y\rangle=\operatorname{tr}(X^\top Y)$, where $\operatorname{tr}$ is the trace and $\top$ is the transpose.

Let $M$ be a symmetric positive-definite $4\times4$ matrix, and $T=\mathbf f\,\mathbf r^\top-\mathbf r\,\mathbf f^\top$ an antisymmetric $4\times4$ matrix. ($M$ describes the distribution of mass in the spacecraft, $\mathbf r$ is a vector locating the thruster relative to the centre of mass, $\mathbf f$ is the force produced by the thruster, and $T$ is the torque produced by the thruster. More details at [Chiral Anomaly's answer](https://physics.stackexchange.com/a/471197/247726) to [Rigid body dynamics derivation from Newton's laws for higher dimensions](https://physics.stackexchange.com/questions/471106). Everything is described in the rotating reference frame.)

The angular velocity $\Omega(t)$, an antisymmetric $4\times4$ matrix, changes with time $t$ according to Euler's equation

$$\\{M,\Omega'(t)\\}+[\Omega(t),\\{M,\Omega(t)\\}]=f(t)T$$

where $f(t)\geq0$ is a function (continuous, piecewise-constant, or just integrable) describing when and how strongly the thruster is used.

**Question:** Can $M$ and $T$ be chosen such that, for any two antisymmetric $4\times4$ matrices $\Omega_0$ and $\Omega_1$, there exist $t_1>0$ and $f$ such that the solution $\Omega$ to Euler's equation with initial value $\Omega(0)=\Omega_0$ has final value $\Omega(t_1)=\Omega_1$?

To simplify things, we may take $f$ to be a combination of Dirac deltas instead of an ordinary function. Then the angular velocity satisfies $\\{M,\Omega'(t)\\}+[\Omega(t),\\{M,\Omega(t)\\}]=0$ except at a finite set of times when $\\{M,\Omega(t)\\}$ changes by a positive multiple of $T$.

* * *

Here is Euler's equation in terms of the components $\omega_{ij}$ of the angular velocity. Assume that $M$ is diagonal, with components $m_i>0$.

$$(m_1+m_2)\,\omega_{12}'(t)+(m_2-m_1)\big(\omega_{13}(t)\,\omega_{32}(t)+\omega_{14}(t)\,\omega_{42}(t)\big)=f(t)\,\tau_{12}$$

(and permute the indices to get 6 equations like this).

If $m_1=m_2$, then $\omega_{12}'$ has constant sign; $\omega_{12}$ is either always non-increasing, or always non-decreasing. So, if the difference between initial and final values of $\omega_{12}$ has the wrong sign compared to $\tau_{12}$, then there is no solution. Thus, we must take $m_1\neq m_2$, and similarly $m_i\neq m_j$ for $i\neq j$.

* * *

The Frobenius inner product of two antisymmetric matrices has every term duplicated: $\sum_{i,j}x_{ij}y_{ij}=2\sum_{i<j}x_{ij}y_{ij}$ (since $x_{ji}=-x_{ij}$ and $y_{ji}=-y_{ij}$). So it's natural to take half of this.

The angular momentum is $L(t)=\\{M,\Omega(t)\\}$. Its squared magnitude is $\tfrac12\langle L(t),L(t)\rangle$; the derivative of this is $\langle L(t),f(t)T\rangle$. So the angular momentum has constant magnitude whenever $f(t)=0$. (The angular momentum itself would be constant in an inertial reference frame, but here we're using a rotating reference frame.)

The rotational energy is $\tfrac14\langle\Omega(t),L(t)\rangle$; the derivative of this is $\tfrac12\langle\Omega(t),f(t)T\rangle$. So the energy is constant whenever $f(t)=0$.

All of that applies in any dimension. But in 4D there's another constant, the angular momentum bivector's exterior square, or equivalently its inner product with its Hodge dual.

Here are those constants in component form. Squared magnitude of angular momentum:

$$(m_1+m_2)^2\omega_{12}^2+(m_1+m_3)^2\omega_{13}^2+(m_2+m_3)^2\omega_{23}^2\\\\+(m_1+m_4)^2\omega_{14}^2+(m_2+m_4)^2\omega_{24}^2+(m_3+m_4)^2\omega_{34}^2.$$

Doubled energy:

$$(m_1+m_2)\omega_{12}^2+(m_1+m_3)\omega_{13}^2+(m_2+m_3)\omega_{23}^2\\\\+(m_1+m_4)\omega_{14}^2+(m_2+m_4)\omega_{24}^2+(m_3+m_4)\omega_{34}^2.$$

Halved exterior square of angular momentum:

$$(m_1+m_2)(m_3+m_4)\omega_{12}\omega_{34}-(m_1+m_3)(m_2+m_4)\omega_{13}\omega_{24}+(m_2+m_3)(m_1+m_4)\omega_{23}\omega_{14}.$$

(Its derivative is the exterior product of angular momentum and torque: $(m_1+m_2)\omega_{12}\,f\,\tau_{34}+\dotsb$ .)

These three constants define two ellipsoids and another quadratic surface (with signature $+^3-^3$) in the 6D space of antisymmetric matrices. When $f=0$, $\Omega$ must remain on the intersection of these three quadratic surfaces. In terms of $L$ instead of $\Omega$ (with $\ell_{ij}=(m_i+m_j)\omega_{ij}$), those constants are

$$ \ell_{12}^2+\ell_{13}^2+\ell_{23}^2+\ell_{14}^2+\ell_{24}^2+\ell_{34}^2, $$$$ \left(\tfrac1{m_1+m_2}\right)\ell_{12}^2+\left(\tfrac1{m_1+m_3}\right)\ell_{13}^2+\left(\tfrac1{m_2+m_3}\right)\ell_{23}^2+\left(\tfrac1{m_1+m_4}\right)\ell_{14}^2+\left(\tfrac1{m_2+m_4}\right)\ell_{24}^2+\left(\tfrac1{m_3+m_4}\right)\ell_{34}^2, $$$$ \ell_{12}\ell_{34}-\ell_{13}\ell_{24}+\ell_{23}\ell_{14}. $$

We can assume $m_1>m_2>m_3>m_4>0$. Taking the squared magnitude of $L$, and subtracting $(m_1+m_3)$ or $(m_2+m_4)$ times the doubled energy, we get

$$ \left(\tfrac{m_2-m_3}{m_1+m_2}\right)\ell_{12}^2-\left(\tfrac{m_1-m_2}{m_2+m_3}\right)\ell_{23}^2-\left(\tfrac{m_3-m_4}{m_1+m_4}\right)\ell_{14}^2-\left(\tfrac{m_1-m_2+m_3-m_4}{m_2+m_4}\right)\ell_{24}^2-\left(\tfrac{m_1-m_4}{m_3+m_4}\right)\ell_{34}^2, $$$$ \left(\tfrac{m_1-m_4}{m_1+m_2}\right)\ell_{12}^2+\left(\tfrac{m_1-m_2+m_3-m_4}{m_1+m_3}\right)\ell_{13}^2+\left(\tfrac{m_3-m_4}{m_2+m_3}\right)\ell_{23}^2+\left(\tfrac{m_1-m_2}{m_1+m_4}\right)\ell_{14}^2-\left(\tfrac{m_2-m_3}{m_3+m_4}\right)\ell_{34}^2, $$

both being constant. Notice the signs of the coefficients; the first expression has only one positive term, so (if the expression is $0$) it defines a cone, or a pair of opposite cones, around the $\ell_{12}$ axis in 6D. And the second expression defines a pair of opposite cones around the $\ell_{34}$ axis. If the torque $T$ is inside one of these cones, and the initial angular momentum $L_0$ is in the same cone, then $L$ stays in that cone, and there's no solution with the final angular momentum $L_1$ outside of the cone. Thus, we must take $T$ to be outside of all four of these cones. That is, the first expression (with $\tau_{ij}$ in place of $\ell_{ij}$) must be negative, and the second expression must be positive.

* * *

Notice that Euler's equation has a time symmetry: Given a solution $\Omega(t)$, we can construct another solution $\tilde\Omega(t)=-\Omega(t_1-t)$, so that $\\{M,\tilde\Omega'(t)\\}+[\tilde\Omega(t),\\{M,\tilde\Omega(t)\\}]=f(t_1-t)\,T$, and the initial and final values are swapped (and negated): $\tilde\Omega(0)=-\Omega_1$ and $\tilde\Omega(t_1)=-\Omega_0$. Therefore, any angular velocity can be reached from any other, if and only if any angular velocity can be reached from $0$. (We can reverse a path from $0$ to $-\Omega_0$ to get a path from $\Omega_0$ to $0$, and concatenate that with a path from $0$ to $\Omega_1$, to get a path from $\Omega_0$ to $\Omega_1$.)

Also, Euler's equation has a scale symmetry: Given a solution $\Omega(t)$, for any $k>0$ we can construct another solution $\tilde\Omega(t)=k\,\Omega(kt)$, so that $\\{M,\tilde\Omega'(t)\\}+[\tilde\Omega(t),\\{M,\tilde\Omega(t)\\}]=k^2f(kt)\,T$, and the final value is $\tilde\Omega(t_1/k)=k\,\Omega(t_1)=k\,\Omega_1$. Therefore, any angular velocity can be reached from $0$, if and only if any angular velocity _in a small neighbourhood of $0$_ can be reached from $0$.

(It looks like these two symmetries are related by $k=-1$, but we should keep a distinction between positive time and negative time.)

* * *

This has the form of a quadratic differential equation, $\mathbf x'(t)=\mathbf x(t)\odot\mathbf x(t)$ where $\odot$ is some bilinear function. (Specifically, for antisymmetric matrices $X$ and $Y$, define $X\odot Y=-\\{M,\\}^{-1}([X,\\{M,Y\\}])$, so Euler's equation is $\Omega'=\Omega\odot\Omega$, as long as $f=0$. Alternatively, define $X\odot Y=-[\\{M,\\}^{-1}(X),Y]$, so Euler's equation is $L'=L\odot L$.)

Fix a norm on the space, and find some constant $\lVert\odot\rVert>0$ such that $\lVert\mathbf x\odot\mathbf y\rVert\leq\lVert\odot\rVert\lVert\mathbf x\rVert\lVert\mathbf y\rVert$ for all $\mathbf x,\mathbf y$.

Given initial value $\mathbf x(0)=\mathbf a$, the equation $\mathbf x'=\mathbf x\odot\mathbf x$ has the power series solution

$$\mathbf x(t)=\sum_{n=0}^\infty t^n\frac{\sum^{n!}\mathbf a^{n+1}}{\sum^{n!}1}$$

where the coefficient of $t^n$ is the average of all possible ways of evaluating $\mathbf a^{n+1}$ using the product $\odot$. For example, the coefficient of $t^3$ is $\tfrac16$ of

$$\sum^6\mathbf a^4=((\mathbf a\mathbf a)\mathbf a)\mathbf a+(\mathbf a(\mathbf a\mathbf a))\mathbf a+2(\mathbf a\mathbf a)(\mathbf a\mathbf a)+\mathbf a((\mathbf a\mathbf a)\mathbf a)+\mathbf a(\mathbf a(\mathbf a\mathbf a))$$ $$\newcommand{\aaaa}[3]{\mathbf a\underset{#1}\odot\mathbf a\underset{#2}\odot\mathbf a\underset{#3}\odot\mathbf a} =\aaaa{1}{2}{3}+\aaaa{2}{1}{3}\begin{matrix}{}+\aaaa{1}{3}{2} \\\ {}+\aaaa{2}{3}{1}\end{matrix}+\aaaa{3}{1}{2}+\aaaa{3}{2}{1}.$$

(In the $1$-dimensional algebra, the product is associative, and the power series simplifies to $\sum_nt^na^{n+1}=a/(1-at)$.)

This converges absolutely, for small $t$:

$$\sum_{n=0}^\infty\left\lVert t^n\frac{\sum^{n!}\mathbf a^{n+1}}{\sum^{n!}1}\right\rVert\leq\sum_{n=0}^\infty|t|^n\lVert\odot\rVert^n\lVert\mathbf a\rVert^{n+1}=\frac{\lVert\mathbf a\rVert}{1-|t|\lVert\odot\rVert\lVert\mathbf a\rVert},$$ $$|t|<\frac{1}{\lVert\odot\rVert\lVert\mathbf a\rVert}.$$

And at certain times (when the thruster is used) an impulse may be applied to $\mathbf x$, with a fixed direction $\mathbf b$ but an arbitrary magnitude $c>0$, thus: $\mathbf x(t+)=\mathbf x(t-)+c\mathbf b$. These impulses, and the time intervals between them, are many variables that we can control. If the number of variables is at least $6$ (or the dimension of the space $\mathbf x$ is in), then there is hope of surrounding $0$ in an open set, and thus reaching everywhere in the space (according to the previous section).

$$ \mathbf x(0+)=0+c_0\mathbf b $$$$ \mathbf x(t_1-)=\sum_{n=0}^\infty\frac{t_1^n}{n!}\sum^{n!}\mathbf x(0+)^{n+1} $$$$ \mathbf x(t_1+)=\mathbf x(t_1-)+c_1\mathbf b $$$$ \mathbf x(t_1+t_2-)=\sum_{n=0}^\infty\frac{t_2^n}{n!}\sum^{n!}\mathbf x(t_1+)^{n+1} $$$$ \mathbf x(t_1+t_2+)=\mathbf x(t_1+t_2-)+c_2\mathbf b $$$$ \mathbf x(t_1+t_2+t_3-)=\sum_{n=0}^\infty\frac{t_3^n}{n!}\sum^{n!}\mathbf x(t_1+t_2+)^{n+1} $$$$ c_0,t_1,c_1,t_2,c_2,t_3\geq0 $$$$ \mathbf x(t_1+t_2+t_3-)\approx0\quad? $$

Re, I had not meant to make any adverse changes, and didn't even realise that it made a difference in the rendering.  My apologies.  I have changed back to back-to-back $$ $$ environments.

@LSpice - Now the displayed equations are too closely spaced.


Reference request: deforming a G-local system to a variation of Hodge structure

Let $X$ be a smooth connected quasiprojective variety over $\mathbb{C}$ and let $G$ be a complex reductive group. Let $$\iota: G\to GL_N$$ be a representation and let $$\rho: \pi_1(X(\mathbb{C}))\to G(\mathbb{C})$$ be a homomorphism. I am looking for a reference for the following fact:

> $\rho$ is deformation-equivalent to a homomorphism $$\rho': \pi_1(X(\mathbb{C}))\to G(\mathbb{C})$$ such that the local system on $X$ associated to $\iota\circ \rho'$ underlies a polarizable complex variation of Hodge structure.

If $X$ is projective, this is due to Simpson [1, Theorem 3]; in the quasiprojective case some special cases have been written down by T. Mochizuki, e.g. if $G=GL_N$ [2, Theorem 10.5] or if $\rho$ is rigid with Zariski-dense image [2, Lemma 10.13].

I think this can probably be extracted with some difficulty from existing literature, but I'm hoping the statement has been written down explicitly somewhere.

EDIT: If it helps, I'm primarily interested in the case $G=PGL_n$.

[1] _Simpson, Carlos T._ , [**Higgs bundles and local systems**](http://dx.doi.org/10.1007/BF02699491), Publ. Math., Inst. Hautes Étud. Sci. 75, 5-95 (1992). [ZBL0814.32003](https://zbmath.org/?q=an:0814.32003).

[2] _Mochizuki, Takuro_ , [**Kobayashi-Hitchin correspondence for tame harmonic bundles and an application**](http://www.arxiv.org/abs/math/0411300), Astérisque 309. Paris: Société Mathématique de France (ISBN 978-2-85629-226-6/pbk). viii, 117 p. (2006). [ZBL1119.14001](https://zbmath.org/?q=an:1119.14001).

@SashaP: actually I think this should probably be false over infinite finitely generated fields; I have an idea for a counterexample conditional on Fontaine-Mazur, but it’s maybe too involved for an MO comment.

@SashaP: This is a nice question! I think I see an argument for the analogous statement over finite fields (i.e. working in a component of the rigid generic fiber of the deformation ring of a given residual rep of the geometric fundamental group). Plausibly I see an argument over p-adic fields as well, but I need to check details...send me an email and let's discuss further? I don't yet see how to do it over finitely generated infinite fields, which is I think the most interesting case...

Apologies for an off-topic comment, but do you know if the analogous statement (say, at least for $GL_n$) is true for 'arithmetic' in place of 'underlies a polarizable VHS'?

I see, I just did not read the OP carefully.

@MoisheKohan: As I remark in the question, Mochizuki has proven a quasiprojective version, just not for arbitrary $G$...

I remember that Simpson proved this for projective varieties. I do not remember seeing a quasiprojective version.


A multiple integral

Let us consider the multiple integral $$I_{n}=\int_{-\infty }^{\infty }ds_{1}\int_{-\infty}^{s_{1}}ds_{2}\cdots \int_{-\infty }^{s_{2n-1}}ds_{2n}\;\cos {(s_{1}^{2}-s_{2}^{2})}\;\cdots \cos {(s_{2n-1}^{2}-s_{2n}^{2})}.$$

Using an indirect method (Landau-Zener formula from the theory of molecular scattering), we had shown in <http://arxiv.org/abs/1201.1975> that $$I_n=\frac{2}{n!}\left(\frac{\pi}{4}\right)^{\,n}.$$ I'm very interested how to get this result by direct mathematical (preferably simple) methods, without using the Landau-Zener formula. Can anyone help?

Actually, I asked similar question an year ago [Multiple Integral (American Mathematical Monthly problem 11621 and its generalization)](https://mathoverflow.net/questions/125237/multiple-integral-american-mathematical-monthly-problem-11621-and-its-generali) but have not got an answer.


Bounding failures of the integral Hodge and Tate conjectures

It is well know that the integral versions of the Hodge and Tate conjectures can fail. I once heard an off hand comment however that they should only fail by a "bounded amount". My question is what this means and whether this can be made more precise.

> > Let $\pi: X \to B$ be a smooth projective morphism of smooth varieties over $\mathbb{C}$. Assume that the (usual rational) Hodge conjecture holds for the fibre $B_x$ over every point $x \in B(\mathbb{C})$. Does there exist an integer $d$ such that for any integral Hodge class $h$ on $B_x(\mathbb{C})$, the class $dh$ is represented by an algebraic cycle for all $x \in B(\mathbb{C})$?

Of course the point of this question is whether $d$ can be chosen uniformly with respect to the family.

As a brief remark, it is quite easy to prove this when $B={\mbox{Spec}}\ \mathbb{C}$. Indeed, by assumption the algebraic classes form a subgroup of finite index inside the Hodge classes!

I am also similarly interested in the analogous question for the integral Tate conjecture for families of varieties over number fields and finite fields.

It seems unlikely that any of the known counterexamples to the integral Hodge conjecture would give a negative answer to your question. On the other hand, there seems to be no reason to expect a positive answer since the dimension of the space of Hodge classes could jump over infinitely many subvarieties of $B$. Interesting!


Is the mapping class group of $\Bbb{CP}^n$ known?

In his paper ["Concordance spaces, higher simple homotopy theory, and applications"](https://pi.math.cornell.edu/~hatcher/Papers/ConcordanceSpaces.pdf), Hatcher calcuates the smooth, PL, and topological mapping class groups of the $n$-torus $T^n$. This requires an understanding of the surgery structure set of the $n$-torus, as well as the space of self-equivalences (easy, because the torus is aspherical and a Lie group).

At least $\pi_0$ and $\pi_1$ of $\text{hAut}(\Bbb{CP}^n)$, the space of self-homotopy equivalences, are known: the former is $\Bbb Z/2$ because $[\Bbb{CP}^n, \Bbb{CP}^n] = [\Bbb{CP}^n, \Bbb{CP}^\infty] = \Bbb Z$, and the only invertible elements are $\pm 1$; further the homology of this space seems to have been calculated by Sasao, who gives that $\pi_1 \text{hAut} = \Bbb Z/(n+1)$.

The best references I know for the surgery structure set of $\Bbb{CP}^n$ are Wall's book on surgery and the Madsen-Milgram book on Top/PL cobordism groups "The classifying spaces for surgery and cobordism of manifolds", neither of which seem to give a completely explicit description.

From this, it's not immediately clear to me whether enough is known to run Hatcher's argument for $T^n$ on complex projective space.

Has anybody computed the various mapping class groups of $\Bbb{CP}^n$ when $n \geq 3$? Maybe in any specific examples, or when $n$ is very large? If not, is it out of reach with current technology?

(I would also be interested in other infinite families of high-dimensional manifolds, like real projective spaces and lens spaces, but I imagine those require strictly more work to understand, given the presence of non-trivial fundamental group without being aspherical like the torus is.)

Regarding your last comment: Wall (MR0156354, MR0177421) and Kreck (MR0561244) have computed various mapping class groups of highly connected manifolds up to extension problems.

@skupers Kreck points out to me that the oriented smooth mapping class group $\pi_0 \text{Diff}^+(\Bbb{CP}^3)$ was calculated by Brumfiel to be $\Bbb Z/4$, all of whose elements are represnted by diffeomorphisms supported in a ball. I did not check but presumably $\pi_0 \text{Diff}$ is the dihedral group on 8 elements.

Kreck and Su have announced a paper containing the case n=3, see Remark 1.4 of arxiv.org/abs/1907.05693. You could try asking one of them.

Only small homotopy groups seem relevant in Hatcher's computation, though, so I hope that doesn't cause too much trouble. Thanks for the reference!

There are spectral sequences developed by Schultz, and Becker-Schultz that converge to homotopy groups of the identity component of the space of self-maps of $CP^n$. See section 6 in arxiv.org/abs/0912.4874 and references therein. The computational difficulties are substantial, e.g., in the linked paper we only manage to get partial information on $\pi_7$.


Cauchy matrices with elementary symmetric polynomials

$\newcommand{\vx}{\mathbf{x}}$ Let $e_k(\vx)$ denote the [elementary symmetric polynomial](http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial), defined for $k=0,1,\ldots,n$ over a vector $\vx=(x_1,\ldots,x_n)$ by

\begin{equation*} e_k(\vx) := \sum_{1 \le l_1 < l_2 < \cdots < l_k \le n} x_{l_1}x_{l_2}\cdots x_{l_k}. \end{equation*}

To avoid messy conjugates in what follows, let me limit $x$ to be a real vector.

[**EDIT**]: Changed notation below to highlight that size of $M_k$ is independent of size of the vectors being used to define it.

> **Definition** (GC-matrix) Let $(\vx^1,\ldots,\vx^m)$ be vectors in $\mathbb{R}^{n}$ such that each component $|x^p_j| < 1$ (for $1\le p \le m$, $1\le j \le n$). Denote by $\vx^i \circ \vx^j$ the vector formed by taking elementwise products; also, let $\mathbf{1}$ denote the length $n$ vector of all ones.
>
> Define now the $m\times m$ _Generalized Cauchy (GC) matrix_ $M_k$, for $1\le k\le n$, as \begin{equation*} M_k := \left[\frac{1}{e_k(\mathbf{1}-\vx^i\circ \vx^j)}\right]_{i,j=1}^m. \end{equation*}

**Example** For $e_1(\vx)=\sum_lx_l$, the GC matrix above essentially (up to scaling) boils down to a standard [Cauchy matrix](http://en.wikipedia.org/wiki/Cauchy_matrix), of the form \begin{equation*} \left[\frac{1}{1-\lambda_i\lambda_j}\right], \end{equation*} for suitable scalars $\lambda_i$; while for $e_n(\vx)=\prod_l x_l$, the GC matrix reduces to a Hadamard product of ordinary Cauchy matrices.

> **Question.** The above example shows that $M_1$ and $M_n$ are positive semidefinite (since the involved Cauchy matrices are). So my question is there a slick representation for $e_k$ that I could use to conclude positive semidefiniteness of $M_k$?

* * *

**PS:** I suspect the answer follows immediately from well-known (in the correct circles) results; but I need some MO help here to put me rapidly in the correct direction!

Unfortunately, this question has seems to have a negative answer (as proved to me very recently by one of the mathematicians with whom I discussed it; will update this question once I get a chance.)


p-groups as rational points of unipotent groups

Is it true that every finite p-group can be realized as the group of rational points over $\mathbb{F_p}$ of some connected unipotent algebraic group defined over $\mathbb{F_p}$? For abelian p-groups, the answer is yes via Witt vectors, but is it true in general?

Sorry, after a bit more thought, it occurs to me that, since $p$-groups have centres, it's OK to have the Abelian group as sub-object.  Then Theorem 1.8(c) of the paper by Kumar and Neeb shows that $\operatorname{Ext}^1_{\text{alg}}(G, \mathbb G_a) \cong H^2(\mathfrak g, \mathfrak{gl}_1)^{\mathfrak g}$; but I still can't see my way through to showing that the necessary map is surjective.

L. Spice, do you want to replace $GL_1$ with $\mathbb{G}_a$?

By using a filtration of your $p$-group with all successive quotients $\mathbb F_p$, you can reduce the problem to showing that $\operatorname{Ext}^1_{\text{alg}}(\operatorname{GL}_1, G) \to \operatorname{Ext}^1_{\mathbb Z}(\mathbb F_p, G(\mathbb F_p))$ is surjective for all connected, unipotent $\mathbb F_p$-groups $G$.  A quick Google search turned up “Extensions of algebraic groups” by Kumar and Neeb (#48 at math.unc.edu/Faculty/kumar), but the Abelian group by which you're extending there is the subobject, not the quotient.


Deciding whether a given power series is modular or not

The degree 3 modular equation for the Jacobi modular invariant $$ \lambda(q)=\biggl(\frac{\sum_{n\in\mathbb Z}q^{(n+1/2)^2}}{\sum_{n\in\mathbb Z}q^{n^2}}\biggr)^4 $$ is given by $$ (\alpha^2+\beta^2+6\alpha\beta)^2-16\alpha\beta\bigl(4(1+\alpha\beta)-3(\alpha+\beta)\bigr)^2=0, $$ where $\alpha=\lambda(q)$ and $\beta=\lambda(q^3)$. This has a very simple rational parametrization $$ \alpha=\frac{p(2+p)^3}{(1+2p)^3}, \qquad \beta=\frac{p^3(2+p)}{1+2p} $$ (with $p$ ranging from 0 to 1 as as $q$ changes in the range).

By certain heuristical reasons (which are hidden behind analysis in my [recent joint work](http://arxiv.org/abs/1012.3036)), I expect that modularity can occur in some other similar parameterizations. In particular, the expansions $$ \mu(q) = 4096q - 294912q^2 + 12238848q^3 - 379846656q^4 $$ $$ \+ 9737920512q^5 - 217011585024q^6 + 4333573472256q^7 - 79091807551488q^8 $$ $$ \+ 1337378422542336q^9 - 21157503871942656q^{10} + 315428695901356032q^{11} $$ $$ \- 4455786006742302720q^{12} + 59885350975571779584q^{13} + O(q^{14}) $$ and $$ p = 4q + 12q^2 - 48q^3 + 156q^4 - 12q^5 - 6576q^6 + 78144q^7 - 607812q^8 $$ $$ \+ 3017364q^9 + 156q^{10} - 208502832q^{11} + 2876189520q^{12} - 24837585384q^{13} + O(q^{14}) $$ (which, of course, can be further extended) satisfy $$ \mu(q)=\frac{p(4+p)^5}{(1+4p)^5} \quad\text{and}\quad \mu(q^5)=\frac{p^5(4+p)}{1+4p}. $$ (These relations define $p$ and $\mu$ in a unique way.)

Is there any way to identify $\mu(q)$ with a known modular function, or to show that $\mu(q)$ is not modular at all?

Thanks! Best wishes already from 2011.

@Kevin, thanks for this hint. Because I have no guess about the index of the underlying group in $\Gamma(1)$, I am not sure that $O(q^{1000})$ would be enough. I didn't try to expand so far (the coefficients grow extremely fast), but what I did (up to $O(q^{50})$) was verifying a possible algebraic relation between $\mu(q)$ and the classical $j$-invariant. None was found...

If $\mu$ were modular then presumably there would be an algebraic relation between $\mu(q)$ and $\mu(q^n)$ for all positive integer values of $n$, the relation being of degree something like the index of $\Gamma_0(n)$ in $SL(2,Z)$ (but perhaps this isn't precisely right---the exact degree would depend on the level of $\mu$). So you could expand $\mu$ out to $O(q^1000)$ and then it would be easy to search for these relations. If it doesn't work out, i.e. if $n=5$ is OK but the others don't seem to come out, then this is evidence to suggest that $\mu$ isn't modular.


Schemification (schematization?) of locally ringed spaces

**Motivation:**

Say $F: D \to Sch$ is a diagram in the category of schemes, and we're interested in whether it has a [colimit](http://en.wikipedia.org/wiki/Limit_\(category_theory\)#Colimits_2) (gluings, pushouts, and "categorical" quotients are all examples of colimits). Its colimit $Q$ in the category of locally ringed spaces always exists, and a scheme $Y$ is the scheme-colimit of $F$ iff there is a map $Q\to Y$ that is "initial among maps to schemes", i.e. any other map from $Q$ to a scheme factors through it.

Thus the problem of when colimits of schemes exists can be answered if we know when a locally ringed space admits a "schemification".

**Examples:**

1) The process of turning a classical variety $V$ into a "variety with generic points" $V^s$ is a schemification. Proving this boils down the classical affine case. If $R$ is a finite-type reduced $k=\overline{k}$-algebra, and we map $V=mSpec(R)$ to a scheme $Y$, first cover $Y$ by affines $Spec(B_i)$, and then pull back this cover to $V$ and refine it by principal opens $mSpec(R_{f_{ij}})$, so we have maps $B_i \to R_{f_{ij}}$ that determine the map $V\to Y$ (by the adjunction of $Spec({\cal O}(-))$ to the inclusion $AffSch\hookrightarrow LRS)$. But these define the desired map $V^s\to Y$ because $V^s$ can be obtained by gluing $Spec(R_{f_{ij}})$.

2) For $k=\overline{k}$, if we let $\mathbb{G_m}=\mathbb{A}^1_k\setminus 0$ act on $\mathbb{A}^1_k$, then the coequalizer in schemes of the action and the projection $G_m \times_k \mathbb{A}^1_k \rightrightarrows \mathbb{A}^1_k$ (i.e. the "categorical quotient") is $Spec(k)$, hence the locally ringed space coequalizer $Q$, which has two points and is not a scheme, has $Spec(k)$ as its schemification.

3) Some locally ringed spaces have no schemification; for example, two affine lines glued along their generic points. This is precisely because the gluing diagram has no coequalizer in schemes, as per BCnrd and Anton Geraschenko's [answer here](https://mathoverflow.net/questions/9961/colimits-of-schemes/23966#23966).

> When does a locally ringed space $X$ admit a "schemification", i.e. a map to a scheme $X^s$ through which any other map to a scheme must factor?

**EDIT:** (Response to Martin's answer)

4) Any locally ringed space $X$ with exactly one closed point $x$ has its "affinization", $X^a:=Spec({\cal O}_X(X))$, as a schemification. This is because in a map $F:X\to Y$ with $Y$ a scheme, every point maps to a generization of $f(x)$, so and since open sets are closed under generization, $f$ factors through an affine neighborhood of $f(x)$. But any map from $X$ to an affine factors uniquely through $X^a$, so we're done.

5) Schemification commutes with disjoint unions, in that if $X= \coprod X_i$ then $X^s$ exists iff all $X_i^s$ exist, and in that case $X^s = \coprod X_i^s$. So _if $X$ is locally connected_ , since it then decomposes into a disjoint union of connected clopen components, we might as well assume it is connected.

@Andrew: I'm thinking of something like a "generalized GAGA", but I'm not really sure what conditions you need on $(X,A)$. I think something like what I'm talking about is covered in chapter 8 of that book,  but I really don't know the specifics.

@Buschi, Chapter IV seems to be about forming locally ringed topoi (or spaces) from ringed topoi (or spaces).  If I understand, this construction is right adjoint to the inclusion $LRT \to RT$.  @Harry, Could you say more specifically what result you're talking about?  I'm dubious, because I'd expect any ringed topos to be a relative scheme over itself...

@Andrew: There is some stuff in the next chapter (chapter 5) that might be useful to you (if you can show that your locally ringed topos is a relative scheme for some other ringed topos $(X,A)$, you have a universal scheme associated to it by way of a crazy adjunction induced by the universal map $(X,A)\to (Set,Z)$), but I don't think it really addresses your question in any sort of concrete way.

I haven't. Should I?  I just skimmed through a copy and didn't see anything about turning espaces annelés en anneaux locaux into schemas.

HAve you read CAP.IV of  "Topos Annelés et schemas relatifs" by M. Hakim , Ergebn. Math. Grenzgeb. , 64  ?


Closed connected additive subgroups of the Hilbert space

It is a classical result that a closed and connected additive subgroup of $\mathbb{R}^n$ is necessarily a linear subspace. However, this is no longer true in infinite dimension: a very easy example is the subgroup $L^2(I,\mathbb{Z})$ of the real Hilbert space of all $L^2$ real valued functions on the unit interval $I:=[0,1].$

Indeed, any element $\phi$ of $L^2(I,\mathbb{Z})$ is connected to the origin by the path $\gamma:I\ni t\mapsto \phi\chi_{[0,t]}\in L^2(I,\mathbb{Z}),$ where $\chi_{[0,t]}$ is the characteristic function of the interval $[0,t].$ Actually, up to a reparametrization, this path is also $1/2$-Hölder continuous. (Indeed, if $\sigma:[0,1]\to\big[0,\ 1+ \|\phi\| _2^2\, \big]$ is the strictly increasing, surjective continuous map $t\mapsto t+\int_0^t \phi^2dx$, then $\|\gamma(t)-\gamma(t')\|_2\le|\sigma(t)-\sigma(t')|^{1/2}$, meaning that $\gamma\circ\sigma^{-1}$ is $1/2$-Hölder continuous).
So we may say that $L^2(I,\mathbb{Z})$ is even $1/2$-Hölder-path-connected, though it is certainly not a linear subspace.

It is also not hard to see that the Hölder exponent $1/2$ is critic: any closed subgroup $G$ of a Hilbert space $H$, which is connected by $\alpha$-Hölder paths, with $\alpha > 1/2,$ is necessarily a linear space. (Reason: as a consequence of the generalized parallelogram identity, it turns out that the lattice generated by $n$ vectors $g_1,\dots,g_n$ in $H$, with norms $\|g_k\|\leq r,$ is a $rn^{1/2}$-net in their linear span. In particular, if $\gamma:[0,1]\to G$ is an $\alpha$-Hölder path, for any $n\in\mathbb{N},$ the $n$ elements $g_{k,n}:=\gamma(\frac{k+1}{n})-\gamma(\frac{k}{n})\in G,\quad k=0,\dots,n-1$ are a $Cn^{1/2 - \alpha }$-net in their linear span. Since $G$ is closed this implies that it is a cone, hence a linear subspace).

I find this quite nice, but at this point some questions arise quite naturally. Let $H$ be the infinite dimensional real separable Hilbert space.

  * Let $0 < \alpha < 1/2.$ Are there closed additive subgroups of $H$ which are connected by $\alpha$-Hölder paths, but not by $\beta$-Hölder paths for any $\beta >\alpha \, $?

  * More generally: connected / non-connected w.r.to paths with given modulus of continuity? Are there closed, connected, not path-connected additive subgroups?

  * Are these objects just pathologies/curiosities of the mathematical Zoo, or did anybody gave an application of them to functional analysis?


Almost everywhere differentiability for a class of functions on $\mathbb{R}^2$

A while ago, I came across the following problem, which I was not able to resolve one way or the other.

> Let $f,g\colon\mathbb{R}^2\to\mathbb{R}$ be continuous functions such that $f(t,x)$ and $g(t,x)$ are Lipschitz continuous with respect to $x$ and increasing in $t$. Is it necessarily true that the partial derivative $f_x=\partial f/\partial x$ exists almost everywhere with respect to the measure $\iint\cdot\,d_tg(t,x)dx$?

The notation $\int\cdot\,d_tg(t,x)$ refers to the Lebesgue-Stieltjes integral with respect to $g(t,x)$, for each fixed $x$. At first sight, this looks like a basic real-analysis problem for which it should be possible to either prove true or construct a simple counterexample. However, after playing around with this for a while, I was none the wiser.

The reason for the interest in this question comes from attempts to construct a particular decomposition of stochastic processes. Supposing that $X$ is a continuous martingale, it can be useful to decompose $f(t,X_t)$ into a martingale term and a 'drift' component, $$ \begin{align} f(t,X_t)=\int_0^tf_x(s,X_s)\,dX_s+A_t.&&{\rm(1)} \end{align} $$ For this expression to be well-defined, it is necessary that $f_x$ exists outside of a null set. That is, $\int1_{\\{f_x(s,X_s)\rm\ doesn't\ exist\\}}dX_s$ should be zero. However, letting $g(t,x)=\mathbb{E}[\vert X_t-x\vert]$, it can be shown that the expected value of the square of this integral is $\iint1_{\\{f_x\rm\ doesn't\ exist\\}}d_tg(t,x)dx$. So a positive answer to my question would mean that decomposition (1) is well-defined. Furthermore, it can then be shown that $A$ has zero [quadratic variation](http://en.wikipedia.org/wiki/Quadratic_variation), so that $f(t,X_t)$ is what is sometimes referred to as a _Dirichlet process_ (although this term does also have a completely different meaning). Decomposition (1) was used to prove some results about the distribution of martingales, but I also published it as a stand-alone paper ([here](http://dx.doi.org/10.1214/09-AOP476)). However, not knowing an answer to the question above, this did require imposing the annoying additional condition that the left and right hand derivatives of $f(t,x)$ with respect to $x$ exist everywhere.

Some further points are in order.

  * Lebesgue's differentiation theorem tells us that the partial derivative $f_x$ exists almost everywhere with respect to the Lebesgue measure. That does not help here though, because $\iint\cdot d_tg(t,x)dx$ need not be absolutely continuous. In fact, even restricting to the case where $g(t,x)$ is convex in $x$, it is possible to construct nontrivial examples where $\iint\cdot d_tg(t,x)dx$ is supported on the graph $\\{(t,\gamma(t))\colon t\in\mathbb{R}\\}$ of a (discontinuous) function $\gamma\colon\mathbb{R}\to\mathbb{R}$. One such example is given by $g(t,x)=\mathbb{E}[(M_t-x)_+]$ where $M$ is the martingale constructed in [this blog post](https://almostsure.wordpress.com/2016/10/05/a-martingale-which-moves-along-a-deterministic-path/). Another (similar) example is the function $f(t,x)$ constructed [here](https://almostsure.wordpress.com/2016/10/12/do-convex-and-decreasing-functions-preserve-the-semimartingale-property-a-possible-counterexample/).
  * If the left and right hand partial derivatives of $f(t,x)$ with respect to $x$ exist everywhere, then the question has a positive answer.
  * If the left and right hand partial derivatives of $g(t,x)$ with respect to $x$ exist everywhere, then $f_x$ exists in a weak sense. Given any bounded measurable $\theta\colon\mathbb{R}^2\to\mathbb{R}$ with compact support,$$\iint h^{-1}(f(t,x+h)-f(t,x))\theta d_tgdx\to\iint f_x\theta d_tgdx$$as $h\to0$.
  * Rather than monotonicity in $t$, I am really interested in the weaker condition that $\iint\cdot\vert d_tf(t,x)\vert dx$ is locally finite ($\int\cdot\vert d_tf(t,x)\vert$ is the variation of $\int\cdot d_tf(t,x)$), and similarly for $g$. If I find a positive answer to this, then I think I would update my papers on arXiv to reflect this.


Is the Dieudonne module actually a cohomology group?

One often times thinks of the Dieudonne module $M(X)$ of a $p$-divisible group (say over $k$, a perfect characteristic $p$ field) as being some sort of cohomology theory

$$M:\left\\{p\text{- divisible groups}/k\right\\}\to \left\\{F\text{-crystals }/k\text{ with slopes in }[0,1]\right\\}.$$

This is supported by the famous observation of Mazur-Messing-Oda that if $A/k$ is an abelian variety, then

$$M(A[p^\infty])=H^1_\text{crys}(A/W(k))$$

as $F$-crystals.

There are two natural questions that come from this, in my opinion. First, can one make sense of $M(X)$ as a crystalline cohomology group (in a literal sense) if $X$ 'comes from a scheme' (e.g. is the $p^\infty$-torsion of some group scheme). I've asked this [here](https://mathoverflow.net/questions/219015/extension-of-messing-mazur-oda-to-general-groups) as well as asking when $M$ (an $F$-crystal) 'comes from a scheme'.

But, the other question, and the one I am asking here is the following. Is there some way to interpret the understanding of the Dieudonne module 'as cohomology' in a rigorous way? For example, is there a natural site over $X$ (perhaps with underlying category $p$-divisible groups over $X$) such that $M(X)$ is cohomology on this site? I would even be interested in knowing if there is a rigorous way of considering $M(X)/pM(X)$ as a cohomology theory (in the Mazur-Messing-Oda setup it's $H^1_\text{dR}(A/k)$).

The obvious guess is to create a crystalline/infinitesimal site on $X$, but I don't really know how to make this precise.

Doesn't Mazur-Messing interpret the Dieudonne module of a p-divisible group G as rigidified extensions? This suggests that defining a category of G-invariant D-modules should be possible. I would naively guess that the resulting cohomology would be "generated in degree 1", just as for abelian varieties, though.

One wants not a site on $X$ but a site whose "fundamental group" is in some way "dual" to $X$. The simplest picture of this seems like it should be the quotient of a point by the Cartier dual of $X$.


Density of first-order definable sets in a directed union of finite groups

This is a generalization of the following [question](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) by John Wiltshire-Gordon.

Consider an inductive family of finite groups: $$ G_0 \hookrightarrow G_1 \hookrightarrow \ldots \hookrightarrow G_i \hookrightarrow \ldots $$

We may view each group as a subgroup of the next. Let $G$ be the directed union of the $G_i$: $$ G := \bigcup_{i=0}^\infty \ G_i $$

Now, $G$ is a countable group. Suppose we're asked a question like, "What proportion of the elements of $G$ are commutators?", or "What proportion of pairs $(g,h) \in G^2$ satisfy $g^2h^2=1$ but $g^2 \neq 1$?" We try to make sense of such questions in terms of _density_ on $G^n$, which is not defined for all subsets. For $E \subseteq G^n$, let $$ d(E) := \lim_{i \to \infty} \frac{|E \cap G_i^n|}{|G_i^n|}, $$ if the limit exists.

My question is: If $E$ is a subset of $G^n$ that is first-order definable in the language of groups, does the density of $E$ necessarily exist? John's [question](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) covers the (still unresolved) special case of when $E$ is defined by an atomic formula.

A first-order definable subset $E \subseteq G^n$ is the set of $n$-tuples of group elements where a particular first-order formula in $n$ free variables is true. Such a formula is a finite string of symbols and may involve multiplication, inversion, the identity element, equality, logical connectives ($\wedge$, $\vee$, $\neg$, $\implies$), quatifiers ($\exists$ and $\forall$), and parentheses. For example, the following first-order formula defines the set of commutators $g \in G$:

$$ (\exists x)(\exists y)(g = xyx^{-1}y^{-1}). $$

On the other hand, if you try to define the commutator subgroup by such a formula, you run into trouble. You might want to say, "$g$ is a commutator or $g$ is a product of two commutators or $g$ is a product of three commutators or ...," but infinite disjunctions are not allowed.

Please note that the counterexample to a similar question in a comment by Vipul Naik [here](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) is _not_ a counterexample to this question. In the inductive family $$ A_3 \hookrightarrow S_3 \hookrightarrow \ldots \hookrightarrow A_{2i-1} \hookrightarrow S_{2i-1} \hookrightarrow A_{2i+1} \hookrightarrow S_{2i+1} \hookrightarrow \ldots, $$ the groups alternate between having all the elements as commutators and half the elements as commutators. However, in the directed union, which is $A_\infty$, all the elements are commutators. The upshot is that quantifiers in a first-order formula may demand that you "look ahead" in your inductive family.

It may also not be true that you will get a density, i.e. that there will be a limit law.  I definitely don't know, but Vardi might.  Gerhard "Ask Me About System Design" Paseman, 2010.11.02

@Gerhard:  It's not true that you'll always get density zero or one, if that's what you mean.  For the union $S_\infty$ of the symmetric groups $$ S_1 \hookrightarrow S_2 \hookrightarrow \ldots \hookrightarrow S_n \hookrightarrow \ldots, $$ the set of commutators is $A_\infty$, and its density is $1/2$.  An even more basic example would be the set of elements squaring to one in the union of the groups $\mathbb{Z}/3\mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z})^i$, which has density $1/3$.

This shouts "zero-one law" at me.  Perhaps looking at limit laws, and works of Vardi, Fagin, and others on finite structures may give you what you want.  Gerhard "Ask Me About System Design" Paseman, 2010.10.27


If $X\times Y$ is homotopy equivalent to a finite-dimensional CW Complex, are $X$ and $Y$ as well?

Is there a space $X$ that is not homotopy equivalent to a finite-dimensional CW complex for which there exists a space $Y$ such that the product space $X\times Y$ is homotopy equivalent to a finite-dimensional CW complex? If so, how might we construct an example?

A first consideration could be where $X$ has infinitely many nontrivial homology groups, in which case $X$ is not homotopy equivalent to a finite-dimensional CW complex. Yet, in this case it follows from the Künneth Formula that no suitable $Y$ exists. Of course, this only rules out spaces with infinitely many nontrivial homology groups.

Another consideration could be where $X$ is an Eilenberg–Maclane space for which $\pi_1(X)$ has non-trivial torsion, in which case $X$ is, again, not homotopy equivalent to a finite-dimensional CW complex. In this case, if $\pi_1(X)$ is abelian then $X$ is homotopy equivalent to $L\times Z$ for some other space Z and an infinite-dimensional lens space $L$, and hence $X$ has infinitely many nontrivial homology groups, which implies, again, that no suitable $Y$ exists. This has led me to wonder if there exists a space $Y$ and a noncommutative group $G$ with non-trivial torsion such that $Y\times K(G,1)$ is homotopy equivalent to a finite-dimensional CW complex, where $ K(G,1)$ denotes an Eilenberg–Maclane space.

Any insight into approaching this problem is greatly appreciated.

There are examples by R. Bing and many other mathematicians when a product of a non-manifold B and a real line gives a euclidean space, see e.g. the following papers and references there: ams.org/journals/bull/1958-64-03/S0002-9904-1958-10160-3/… ams.org/journals/proc/1961-012-01/S0002-9939-1961-0123303-2/… However, I do not know whether such space B has the homotopy type of a CW-complex.

A postscript to Igor Belegradek's comment:  The argument in Proposition A.11 of my book shows that a space dominated by a CW complex of dimension $n$ is homotopy equivalent to a CW complex of dimension $n+1$, where the extra dimension arises from a mapping telescope construction.  The argument, which is due to J.H.C.Whitehead if I remember correctly, uses only elementary homotopy theory.  Using cohomology and Wall's work one obtains the stronger result that the increase in dimension is not really necessary, at least in dimensions 3 and greater.

Just to add references to Tom's  answer. The last sentence it is  theorem E (p.63) of C.T.C. Wal, "Finiteness conditions for CW complexes", see  math.uchicago.edu/~shmuel/tom-readings/…. The statement that a homotopy retract of a CW-complex is a CW complex is e.g. in Hatcher's "Algebraic Topology", proposition A.11.

If $Y$ is nonempty then $X$ is a homotopy retract of a CW complex, therefore homotopy equivalent to a CW complex. If $X$ is a homotopy retract of a finite-dimensional CW complex, then $X$ has no cohomology above some dimension $d$, for any coefficient system. This in turn implies (if $d\ge 3$) that $X$ is weakly, therefore strongly equivalent, to a $d$-dimensional complex.

Not exactly what you wanted, but there are pointed CW complexes $X$ that are not finite, but such that $S^1\wedge X$ is finite, for example $BG$ for $G$ an infinite acyclic group.


Are the eigenvalues of the Laplacian of a generic Kähler metric simple?

It is a [theorem of Uhlenbeck](http://www.jstor.org/stable/2374041) that for a generic Riemannian metric, the Laplacian acting on functions has simple eigenvalues, i.e., all the eigenspaces are 1-dimensional. (Here "generic" means the set of such metrics is the complement of a [meagre set](http://en.wikipedia.org/wiki/Meagre_set) in the space of all metrics on a given manifold.)

I would like to know whether this property is also true for a generic Kähler metric. More precisely, given a compact complex manifold with a fixed choice of Kähler class, is it true that for the generic representative of this class, the Laplacian acting on functions has simple eigenvalues?

(To be honest, on a first reading it seems as if one might be able to apply similar arguments to those of Uhlenbeck to prove this result. I guess I would like to know if someone has already done this, or if there is a cunning counter-example that I'm missing, before I commit the time and energy to try!)


Homotopy flat DG-modules

A right DG-module $F$ over an associative DG-algebra (or DG-category) $A$ is said to be homotopy flat (h-flat for brevity) if for any acyclic left DG-module $M$ over $A$ the complex of abelian groups $F\otimes_A M$ is acyclic. Homotopy projective and injective DG-modules are defined in the similar way, e.g., a left DG-module $P$ over $A$ is h-projective if for any acyclic left DG-module $M$ over $A$ the complex of abelian groups $Hom_A(P,M)$ is acyclic. Homotopy adjusted DG-modules are used to define derived functors on the derived categories of DG-modules.

My question is: can one describe the class of h-flat DG-modules in any way alternative to the definition? Here are examples of the kind of description I have in mind:

  1. A DG-module is h-projective if and only if it is homotopy equivalent to a DG-module obtained from the free DG-module $A$ over $A$ using the operations of shift, cone, and infinite direct sum. (Similarly, a DG-module is h-injective if and only if it is homotopy equivalent to a DG-module obtained from the cofree DG-module $Hom_{\mathbb Z}(A,\mathbb Q/\mathbb Z)$ using the operations of shift, cone, and infinite product.)

  2. A module over a noncommutative ring is flat if and only if it is a filtered inductive limit of projective (or even projective and finitely generated, or free and finitely generated) modules (the Govorov-Lazard theorem).

The class of h-flat DG-modules is closed with respect to the operations of shift, cone, filtered inductive limit, and passage to a homotopy equivalent DG-module. It also contains the free DG-module $A$ over $A$. Can all h-flat DG-modules be obtained from that one DG-module using these four operations?

References: 1. Keller's paper "Deriving DG-categories"; 2. my preprint arXiv:0905.2621, section 1.

I think the paper arxiv.org/abs/1607.02609 answers your question.

The following paper solves a closely related problem: L.W. Christensen and H. Holm, "The direct limit closure of perfect complexes", Journ. of Pure and Appl. Algebra 219, #3, p.449-463, 2015.  arxiv.org/abs/1301.0731