In this post, every manifold is smooth unless otherwise stated. Given a closed manifold
, one old and very natural question is what kind of manifolds it bounds. The simplest form of this question just asks: Is
the boundary of some manifold
? This leads one to the theory of cobordisms: Two closed
and
are cobordant if
is the boundary of some compact manifold
. (Henceforth, we won’t talk about manifolds being closed or compact; they will be assumed to be compact, and whether or not they have boundary should be clear from context.) In fact, the set of manifolds of dimension
mod cobordism (that is, we set cobordant manifolds equal) is a group under the disjoint union operation, and the given problem was completely solved by Thom (you can tell whether or not a manifold is cobordant to the empty set, i.e. bounds a manifold, solely from its cohomology ring
).
Then one is led to consider manifolds with extra structure. In a topological direction: given an oriented manifold, does it bound another oriented manifold in a way compatible with the orientation? Or spin manifolds? More geometrically, what contact manifolds bound symplectic manifolds? Or what codimension 1 foliated manifolds bound codimension 1 foliated manifolds? (Which contact manifolds bound symplectic manifolds is still an active area of research.)
We will stay mostly in low dimensions here. So for this, it is important to note that every oriented 3-manifold bounds an oriented 4-manifold. Indeed, it will be important that there is an explicit construction of this. Because every oriented 3-manifold is surgery on a link in
, that same link diagram provides an oriented cobordism from
to the 3-manifold itself; filling in the
with a ball gives you the desired oriented 4-manifold.
Here we’ll look at a specific class of manifolds: homology 3-spheres. These are closed 3-manifolds with the same homology groups as
; equivalently, these are closed 3-manifolds with perfect fundamental group. Homology spheres are orientable; henceforth when we say homology sphere we mean a homology sphere with a chosen orientation. We’re generally interested in the class of 4-manifolds these can possibly bound; but specifically, we’ll often consider the question of whether or not they can bound homology balls: manifolds with trivial homology. As above with unoriented manifolds and cobordism, we can consider the set of homology 3-spheres, modulo homology cobordism. we set two homology spheres
to be the same if there is a manifold
with boundary
such that the inclusion maps
induce isomorphisms on every homology group. It is not hard to see that this forms a group under connected sum; this is called the
-dimensional homology cobordism group
. Being trivial in this group is the same as bounding a homology ball, because the identity element is
; if you bound a homology ball, just delete a small ball from the interior to get a homology cobordism to
.
Let’s think about a couple examples. There are no nontrivial homology 1- or 2-spheres. There are quite a lot in dimensions 3 and above; the simplest nontrivial example in dimension 3 is the Poincare sphere; it has finite fundamental group. Every other nontrivial example in dimension 3 has infinite fundamental group. The other simplest examples (and great test cases for anything you’re thinking about regarding homology 3-spheres) are the Brieskorn spheres
. It is not obvious how to say much about whether or not these are homology cobordant.
In dimension 4 and higher, things can only get more complicated – luckily, we’re only studying things up to homology cobordism. In fact, Kervaire proved that every homology 4-sphere bounds a contractible 4-manifold. The same theorem (Theorem 3) includes the fact that
is a quotient of the group of homotopy spheres (up to homeomorphism, group operation is connected sum). I believe this quotient should always be zero but didn’t verify it.
So what’s left is to study
. Before moving on, it is worth noting that if I’m willing to think about topological manifolds that bound homology 3-spheres, instead of smooth manifolds, this group is trivial: Freedman proved that every homology 3-sphere bounds a contractible topological 4-manifold. As we will see, this is far from true if we’re studying smooth 4-manifolds!
Rokhlin’s invariant
Before we say anything about Rokhlin’s invariant, we need to know a little about 4-manifolds and their intersection forms.
Given an oriented 4-manifold (not necessarily without boundary)
, consider
. There is then a cup product pairing
, given by representing every homology class by an embedded oriented surface and taking the intersection numbers of such classes. This is a bilinear, symmetric form on
called the intersection form; if
is closed or has boundary with
, this form is unimodular (has determinant 1). As an example,
has
, intersection form just multiplication;
has intersection form represented by
;
has intersection form given by
. Above, when we described how a surgery diagram of a 3-manifold gives a 4-manifold that bounds it, the link has associated to it a so-called linking form corresponding to how each of the knots link with each other; the associated 4-manifold’s intersection form is isomorphic to this linking form. (This will matter soon!)
Lastly, given any intersection form, we can find a subspace of
of maximal dimension such that the intersection form, restricted to the subspace, is positive (that is, has
for all
in the subspace). Call the dimension of this subspace
. You can find a complementary subspace on which the intersection form is negative; call the dimension of this
. Lastly, we say that the signature of our manifold is
.
For something not important to this post: signature is an oriented cobordism invariant; signature mod 2 is a cobordism invariant. Signature detects the fact that
does not bound an orientable manifold, though it does bound a non-orientable manifold.
Given an intersection form, we call it even if
is always even. It is a purely algebraic fact that the signature of an even intersection form is divisible by 8. What’s interesting is Rokhlin’s theorem that given a closed smooth manifold
with intersection form
, if
is even, its signature is divisible by 16. A modern proof for the case where
has no 2-torsion (in particular, if it’s zero) might go as follows: By the assumption that
is even, some characteristic class wizardry shows that
supports a spin structure, hence has a Dirac operator with even index, and the Atiyah-Singer index theorem shows that the index of this is
. This fact is very special to smooth manifolds: there is a closed topological 4-manifold with intersection form of signature 8.
Now we are in a position to define the Rokhlin invariant. Given a homology 3-sphere
, some fiddling with a surgery diagram for
shows that it bounds a smooth 4-manifold
with even intersection form
. Define
. The algebraic fact we invoked above means that this is indeed an integer mod 2; it’s well-defined because if
is another such 4-manifold,
is a smooth closed 4-manifold, hence has signature divisible by 16; but because its intersection form is
, its signatures thus have
divisible by 8.
Here are some basic facts about the Rokhlin invariant.
First, it’s a homology cobordism invariant. This is because, given a homology cobordism
, if
is a 4-manifold with boundary
and even intersection form, then
is a 4-manifold with the exact same even intersection form and boundary
.
Second, it’s additive under connected sum. This is because if
bound smooth 4-manifold with even intersection form
with intersection forms
, then we can form the “boundary sum” of
and
, a 4-manifold with intersection form
whose boundary is
. These two facts combined tell us the Rokhlin invariant descends to a well-defined homomorphism
.
Third,
. This is because
can be described as surgery on a link whose linking form is the
form, whose signature is 8. Thus, everything thus far gives us our first concrete result:
does not bound a homology ball! But the Rokhlin homomorphism doesn’t tell us anything about, for instance,
.
Now,
is a very mysterious group – some of the few things I can easily say are that it’s abelian and countable. Does it have any torsion? Is it free? As far as I know, these are both open. We can specify the first question to try to make it easier: Does the Rokhlin homomorphism find any 2-torsion? That is to say, is there a homology sphere
such that
bounds a homology ball and
? (Said another way, does the Rokhlin homomorphism split?) This turns out to be quite hard, and we’ll say more about it.
Improving the Rokhlin homomorphism
From one perspective, an invariant that takes a (homology cobordism class of) homology sphere(s) and spits out one of two numbers is not that good, especially if there are lots of homology spheres. So one would like integer-valued invariants. From another, to disprove the question above, we might want to find a homomorphism
that reduces mod 2 to the Rokhlin homomorphism – then if
, we have
, so
cannot bount a homology ball. Both of these perspectives lead us to find better invariants of homology 3-spheres.
The first such invariant was Casson’s invariant. It is hard to define rigorously, but roughly, it’s an oriented count of homomorphisms
mod conjugacy; if you’re lucky, there’s only finitely many of them. More precisely, define
to be the space of representations
mod conjugacy and denote the genus
handlebody as
. Given a Heegaard decomposition of
, we get two maps
. The intersection of the image of these is
. Inspired by this, we define the Casson invariant of
to be the intersection number of the two images of
. There are a number of subtleties I’ve ignored here, not least of which is the fact that we need to prove that this number is independent of the choice of Heegaard splitting.
The first thing to note is that the Casson invariant,
, reduces mod 2 to the Rokhlin invariant. I’m not going to prove this for you. The next thing to note is that it’s also still a homomorphism!
…from the monoid of homology 3-spheres up to homeomorphism. It is not, unfortunately, a homology cobordism invariant. So the Casson invariant is no good at showing that
doesn’t split, and it’s no good at showing that
doesn’t bound a homology sphere. We’ll move on momentarily, but it’s worth noting that the Casson invariant is still very powerful: for instance, it provided the first proof of the existence of non-triangulable 4-manifolds.
The Casson invariant soon received attention from gauge theorists. Here are two particular items of note. Given a homology 3-sphere
, we can define its instanton Floer homology
; this is a
-graded group, and it so turns out that its Euler characteristic (meaning
) is twice the Casson invariant. This is defined in an entirely differential geometric way, so we’ve now seen three different-looking ways of studying manifolds show up in studying similar invariants: for
, the topology of 4-manifolds it bounds; for
, the representation theory of its fundamental group; and for
, the differential geometry of its space of connections and the Chern-Simons functional.
Careful study of a certain technical issue in the instanton homology (factorization through the reducible connection) led Froyshov to define an invariant
. It is unrelated to the Rokhlin invariant, but is a homomorphism. Because one calculates that
, we’ve now proved one of our goals above: that
does not bound a homology ball.
Whether or not the Rokhlin homomorphism
splits is of much greater importance than it might initially seem. It turns out that the Rokhlin homomorphism admits a section if – and only if – every manifold of dimension 5 and up is triangulable. (For the details, see the excellent paper by Galewski and Stern that proves this fact. It is also, independently, due to Matumoto.) As far as I recall, it is still open whether or not
lifts to an integer-valued homomorphism. Luckily, we don’t need quite that much. Manolescu, in 2013, defined an invariant
(motivated by much the same ideas as Froyshov’s invariant) such that
, that reduces mod 2 to the Rokhlin invariant. So if
bounds a homology ball, then
is homology cobordant to
, and hence
; hence
. And, as a corollary, there are non-triangulable manifolds in dimension 5 (and then by other results in the Galewski-Stern paper, every dimension higher than that).
Bounding other kinds of manifolds
Some cleverness with intersection forms and gauge theory lets one construct minor, but additional, obstructions to bounding certain manifolds. Suppose
bounds a 4-manifold with even, positive definite intersection form. (Positive definite means that
.) Then
cannot possibly bound a negative definite 4-manifold (in particular, a homology ball). If it could, then gluing our two manifolds bounding
would give us a smooth 4-manifold with positive-definite nonstandard intersection-form, contradicting Donaldson’s theorem. This proves, in particular, that neither
nor
bound homology balls. It says nothing about
, because the intersection form of the “obvious” manifold it bounds is
, which is not positive definite (it has signature zero). Indeed,
does bound a homology ball.
is weird
To close out the post, some minor comments about
. First, we still really do know very very little. Here’s what I personally know about it, and I think it’s about the limit of what is known.
- It’s countable, because there are only countably many compact manifolds up to homeomorphism.
- It’s abelian, because the connected sum operation on the level of manifolds is abelian.
- The Rokhlin homomorphism
is a nonzero homomorphism that does not split. Indeed, it has a lift to a non-homomorphism with
.
- It’s infinitely generated! In fact, if you fix
relatively prime, the Brieskorn spheres
for
are linearly independent in
. So
has
as a subgroup. This was first proved by Furuta using what’s known as Atiyah’s
invariant (and studying – what else? – gauge theory on 3-manifolds); it was recently reproved by my friend Matt Stoffregen using Pin(2)-equivariant Seiberg-Witten Floer homology here.
- It has a nonzero homomorphism onto
(the Froyshov invariant), and hence has a
summand.
In particular, it’s still not known if the group has any torsion at all. Maybe there’s not! There’s lots to still figure out.