5
$\begingroup$

$n = 4$ is the smallest dimension such that the fundamental group of a closed $n$-manifold can be any finitely-presentable group (leading e.g. to various undecidability results stemming from the undecidability of the word problem in finitely-presentable groups).

$n = 4$ is also the “wild” dimension where low-dimensional smoothing techniques fail but the Whitney trick doesn’t work smoothly, the smooth Poincaré conjecture is open, etc.

Question: Are these two facts related, or is it just a coincidence of small numbers?

If these facts are related, then the relation is not immediate — even simply-connected 4-manifolds are “wild” (and heck, uncountably many smooth structures on $\mathbb R^4$ exist without a fundamental group in sight!). But if you told me, for example, that the closed 4-manifolds you can build without 1-handles are not “wild”, I might take it as evidence that there might be some connection between these two facts. (Sanity check: apparently 4 is the only dimension where simply-connected closed manifolds are not known to admit handle decompositions without 1-handles.)

Improve this question
$\endgroup$
3
  • 8
    $\begingroup$ I don't see how your question can avoid the tautological answer: manifolds in dimensions $n \geq 4$ all have the same set of fundamental groups, but when $n \geq 5$ there's nothing "wild" going on. $\endgroup$
    – Ryan Budney
    Commented Apr 16 at 3:14
  • $\begingroup$ @RyanBudney Yeah, that’s a good point. Here’s one kind of thing which might happen (with apologies for talking out of my ass): my understanding is that Freedman showed how to do the Whitney trick continuously but not smoothly in dimension 4. If the obstruction to smoothing an instance of the Freedman-Whitney trick is fundamentally a 1-skeleton / $\pi_1$ /1-handle phenomenon, then I could imagine that there might be some connection. $\endgroup$
    – Tim Campion
    Commented Apr 16 at 3:20
  • 7
    $\begingroup$ I view the Whitney trick as saying you can mimic the language of group presentations geometrically, with handle-decomposition manipulations. So the failure of a smooth whitney trick amounts to saying there's a gap between 4-manifold fundamental group presentations and handle-decompositions of 4-manifolds. Not at the level of group isomorphisms, but at the level of group presentations (and relator manipulations) vs. handle decompositions (and handle manipulations). So the class of objects are the same but their isomorphisms are not faithfully represented. $\endgroup$
    – Ryan Budney
    Commented Apr 16 at 4:12

1 Answer 1

Reset to default
2
$\begingroup$

Here is a not-quite-answer. Note that $\pi_1$, but not $\pi_{>1}$, can be nonabelian. Higher homotopy groups can interact in a nonabelian, aka nonstable, way, but still they fit together solvably; $\pi_1$ can be insolvable. In terms of representation theory, whether a group is abelian translates into whether its simple representations are invertible. Indeed, for this reason physicists use the term "(non)abelian anyon" to refer to a (simple) quasiparticle which is (non)invertible under fusion, whether or not the cause is a charge under a (non)abelian group. Extended objects (quasistrings, etc.) can be nonabelian in this sense, but it seems (and it is known in sufficiently finite contexts) that any nonabelian nature is kind of fake, in the same way that solvable groups are only fake-nonabelian. Note for dimension counting: particles are 1-dimensional and strings are 2-dimensional in spacetime.

Manifolds, for example the spacetime of a quantum system, have one primary extra feature that homotopy types lack: Poincaré duality. I don't just mean this stably as a pairing on cohomology. I also mean to refer to the nonstable/nonabelian phenomenon of knotting and linking of submanifolds. The upshot is that, whereas $\pi_k X$ and $k$-dimensional bordism groups of $X$ (like $\Omega_k^O X$) are about $k$-manifolds in $X$ that may self-intersect, if $X$ is an $n$-manifold, then there are more "quantum" things like $\pi_k X$ having to do with non-self-intersecting loops or $k$-manifolds. I'm intentionally being imprecise. The point is that this more "quantum" thing has $k$ units of one type of (co)multiplication, if you are looking at $k$-manifolds in your $n$-manifold, and $(n-k-1)$ units of a different type of (co)multiplication, corresponding to the ways that $k$-manifolds can wiggle and fuse in $n$-space. Well, let me be slightly less imprecise by supplying an example. Suppose you have a 3-manifold $X$, perhaps framed (or I'll need to put some other structure somewhere else) and a braided monoidal category $\mathcal{B}$. Then you can look at the space of all skeins in $X$ labeled by objects and morphisms in $\mathcal{B}$. I want you to think that this is some sort of (space of) representation(s) of some "quantum" object that sees some of the "quantum homotopy theory" of $X$. Poincaré duality then becomes a sort of duality between $k$-manifolds and $(n-k-1)$-manifolds, given by linking them. (Or, if you want, fill your manifolds, and study intersection pairings.)

Recall that monoidal $1$-categories, but not higher (except for fake aka solvable cases), can be "nonabelian" in the physicists' sense of having noninvertibility. This supplies a richness to the skein theory of $1$-submanifolds that isn't present for higher submanifolds. On the other hand, Poincaré duality says that $1$-submanifolds are dual to $(n-2)$-submanifolds. So $(n-2)$-submanifolds also collect an extra richness that other dimensions don't have. The richest case is when $n-2 = 1$: knots and links in 3-manifolds are much richer than in any other dimension. Specifically, they have a lot more "nonabelian" nature than the general case.

I like to think of this "doubly nonabelian" nature of knots in 3-manifolds as the quantum version of insolvability of $\pi_1$. Hurewicz tells you that solvable objects with trivial homology are trivial — but there are perfect groups, and you can kill their higher homology without killing them, to produce homology-points that are not points. Pick up a manifold (with vanishing L-theory obstruction) and try to surger it to a sphere. Surgery kills homology in a Poincaré-duality-type way — it kills the abelian (quadratic) version of the "quantum homotopy" of your manifold. It can hang if your manifold has too much insolvable/nonabelian/perfect quantum homotopy theory. This only happens in specific dimensions. After more carefully counting dimensions, you find that surgery hangs in dimension 4, exactly due to nonabelian knotting in dimension 3.

There's other parts of the story I don't quite understand. For reasons I do not have a picture for, you can continue to surger at the cost of smooth structure. Specifically, Freedman supplies an interesting topological but non-smooth filling of the Poincaré 3-sphere. Recall that only the reason there even is a Poincaré 3-sphere is that there is a (super)perfect group which acts without fixed points a 3-manifold (specifically, on a 3-sphere). It is the first (super)perfect group: $A_5 = \mathrm{PSL}_2(5) = \dots$. That perfect groups exist at all is already a surprise — if you just start listing groups in increasing order, or if you randomly finite sample groups, it takes a long time to find an insolvable one. I don't really have a good story for why a perfect group can act on a 3-sphere. It requires some coincidences, the most fundamental being that there exist $a,b,c \in \mathbb{N}$ all distinct with $\frac1a + \frac1b + \frac1c = 1$. I tried to tell this story for an advanced high school audience in http://categorified.net/MathcampColloquium-novideo.pdf. And I really don't have a good story for why, at the cost of smoothness, you can evade this perfect group.

All together, I'm saying that the reason that $\mathbf{4}$-dimensional manifolds are wild is that $\mathbf{4} - 2 = 2\times \mathbf{1}$, where $\mathbf{1}$ is the categorical dimension in which perfectness (essential nonabelianness) occurs. The existence of a 4-manifold with arbitrary $\pi_1$, on the other hand, is that $\mathbf{4} - 2 \geq 2\times \mathbf{1}$, as Ryan Budney points out in the comments. In both cases, the factor of $2$ and the shift by $2$ are about Whitney embedding statements (but they're not all the same "$2$"s).

Improve this answer
$\endgroup$
4
  • $\begingroup$ I think in your second-to-last paragraph there you want $SL(2,5)$, not $A_5$. $\endgroup$
    – Steve D
    Commented Apr 18 at 1:55
  • $\begingroup$ @SteveD SL(2,5) is the spin double cover of A_5 = PSL(2,5). It is quasisimple (=perfect central extension of simple), although not simple, and yes, it is the one that acts on S^3. Usual Poincaré duality implies that if a perfect (=trivial abelianization) group acts on S^3, then it must be superperfect (=trivial abelianization and trivial Schur multiplier). Each quasisimple group has two canonically associated groups: its "adjoint form" (quotient by the centre) and its "simply connected form" (Schur cover of adjoint form). For the purpose of general intuition, it is fine to conflate them. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 19 at 11:16
  • $\begingroup$ Yes, good points. IMO it makes sense to discuss $SL(2,5)$ here too. For example, the fact it's superperfect follows from its balanced presentation, coming from a Heegaard splitting. Whether every finite superperfect group has a balanced presentation is an open problem. Or $SL(2,5)$ is the unique perfect Frobenius complement, where a Frobenius complement acts fixed-point-freely on some vector space $\mathbb{C}^n$ (here on $\mathbb{C}^2$). That such an action comes from an action on $S^3$ is essentially the spherical space form conjecture. ... $\endgroup$
    – Steve D
    Commented Apr 19 at 22:40
  • $\begingroup$ So to get a homology $3$-sphere with finite fundamental group, you need at least a perfect Frobenius complement. Only one exists. But it just so happens it's the right dimension and superperfect, so it all works out. $\endgroup$
    – Steve D
    Commented Apr 19 at 22:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .