What are the simply connected surfaces?

There is a full classification of open surfaces (noncompact surfaces without boundary); see here. This is perhaps a bit more daunting than the classification of compact surfaces, as the latter needs only two simple invariants: the number of boundary components and the Euler characteristic. The added complexity in the noncompact case should not be surprising, as one can have truly wild-looking spaces (imagine making an infinite binary tree and for each edge, inserting a copy of the twice-punctured torus). But it does provide the following nice fact: the only simply connected surfaces without boundary are the sphere and plane. (If you allow boundary, you in addition gain the closed disc and any subset of the closed disc obtained by deleting closed subsets of the boundary.) We don’t have so nice a classification in higher dimensions; there are uncountably many pairwise non-homeomorphic noncompact, contractible 3-manifolds. (However, if you’re “simply connected at infinity” and contractible, then in dimensions at least 5, you’re homeomorphic to $\Bbb R^5$.)

The purpose of this post is to provide two proofs of that, one unnecessarily difficult and the other elementary.

Given an open subset of the plane, somehow the most obvious approach is to apply the Riemann mapping theorem, which provides a conformal map to the disc. To generalize this to other surfaces, we need to invoke something for more general Riemann surfaces: the uniformization theorem. This is annoyingly difficult, since you need to start by having a complex structure on your open surface. To do this, start by finding an almost complex structure on it; this is a smoothly varying automorphism $J_x: T_x \to T_x$ such that $J_x^2 = -I$. It’s straightforward to see than an orientation and Riemannian metric determine such an automorphism: let $J_x(v)$ be the unique tangent vector orthogonal to $v$ with $\{v, J_x(v)\}$ a positively oriented basis. Now invoke the Newlander–Nirenberg theorem that a manifold carries an honest complex structure inducing the almost complex structure $J$ if and only if its Nijenhuis tensor $NJ$ vanishes, and one can check by hand that this is always true for surfaces (rather miraculously; this is wildly false in higher dimensions). This is known in dimension 2 as “the existence of isothermal coordinates”. This gives an honest complex structure on our open manifold, and the uniformization theorem then literally gives us what we want. If you want to take this approach, you probably want both a proof of Newlander-Nirenberg in two dimensions and the uniformization theorem; the first is reasonably classical (due originally to Gauss in the real analytic case), and can be solved by solving what’s known as the Beltrami equation, and my favorite proof of the second (since we’re already doing PDE) is in Mazzeo-Taylor, Curvature and Uniformization (thanks to Willie Wong on StackExchange for pointing this paper out to me here).

Here’s a less preposterous approach (in that it involves no more than basic algebraic topology); in flavor, it is very similar to the link in the first sentence. First, note that every open smooth manifold is the increasing union of a sequence of compact submanifolds. To prove this, you should first prove the existence of a proper smooth function $f: M \to [0,\infty)$; then pick a sequence of regular values $\lambda_k \to \infty$, and let $M_k = f^{-1}([0,\lambda_k])$. One unfortunate fact is that these $M_k$ might always be disconnected. (Imagine a disc with lowest point at height $k$ centered at $(k,0)$, and a height-increasing “bridge” from one to the next, sitting inside $\Bbb R^3$, and extend these discs more-or-less vertical so that this is a closed submanifold of $\Bbb R^3$; this is diffeomorphic to a plane, but if you set $f$ to be the above height function, and pick $\lambda_k = k+\varepsilon$, we run into this problem. To resolve this, we just force the first stage to be connected: if $M_1$ is eg two components for convenience of discussion, pick a path from the boundary of one to the other, and take a tubular neighborhood of it; rounding out the corners, $M_1$ union this bridge is a connected compact manifold. Using the same idea, we may further demand that all of the $M_k$ are connected.

Henceforth for convenience all homology will be taken with $\Bbb Z/2$ coefficients.

Fact 1: If $N$ is a $(k+1)$-dimensional noncompact manifold and $C$ a $k$-dimensional compact boundary component, then $H_k(C) \to H_k(N)$ is injective. This follows from the relative long exact sequence, because $H_{k+1}(N,C) \cong H_{k+1}(N/C) = 0$, because $N/C$ is a noncompact manifold. A similar proof shows that if $N$ (now possibly compact) has at least two compact boundary components, then $H_k(N)$ has positive rank.

Fact 2: All of the $M_k$ have genus zero, and hence (this is just what it means for a compact surface with boundary to be genus zero!) are given as a sphere minus the interiors of finitely many closed discs in the sphere. To see this, apply Mayer-Vietoris to $M_k \cup (M \setminus \text{int } M_k)$; on the way, you should prove that the genus zero assumption is equivalent to $H_1(\partial M_k) \to H_1(M_k)$ being surjective.

Fact 3: The complement of $\text{int } M_k$ has all but one component compact. If it didn’t, Apply Fact 1 and Mayer-Vietoris to derive a contradiction.

Fact 2 implies that these compact components must be discs by the classification of compact surfaces. So define $N_k$ to be the union of $M_k$ and these discs, so $N_k$ is a disc for all $k$. Thus $N_{k+1} \setminus \text{int } N_k$ is an annulus. A union of successive annuli $S^1 \times [i,i+1]$ is diffeomorphic to $S^1 \times [0,\infty)$, which is diffeomorphic to a punctured disc; we conclude that $M$ is diffeomorphic to gluing this collar to the boundary of the disc $N_1$, which is indeed a plane.

What is homology?

When motivating the notion of homology, you’ll frequently see people saying “it counts holes!” They then present some unconvincing pictures (“check out this torus!”) and use a lot of exclamation points without giving any real argument why this algebraically defined thing does what they claim it does.

We’ll start out with some history to see the geometric origin of homology, travel through the necessary modifications to make it work, and then come back and see why homology actually does count holes.

History

In Poincare’s first paper on topology (Analysis Situs), he works with objects he calls $C^1$ manifolds – connected subsets of Euclidean space, cut out by the zero set of some set of equations; nowadays we would say that they’re cut out as regular values. Poincare is interested in submanifolds of these. Suppose we have $n$ oriented submanifolds, $M_1, \dots, M_n$; Poincare will say that these are a complete boundary if there is some submanifold $W$ such that the disjoint union of the $M_i$ form its oriented boundary. Poincare will say that this submanifold, $W$, is a homology between the $M_i$. We will write, then, $k_1 M_1 + \dots + k_n M_n ~ 0$ if we can find a set of homologies between the $M_i$ that add up to give us the left hand side of the above relation.

He then defines the $q$-dimensional Betti number to be the number $P_q$ such that $P_q -1$ is the maximum number of distinct, compact connected $q$-dimensional submanifolds, between which there is no homology $\sum k_i M_i$. If we are to eventually take our modern “homology groups” to be the class of submanifolds, modulo homologies, these $P_q-1$ are evidently the same thing as the ranks of the homology groups $H_q$. (The $-1$ in the above formula is just there so that his numbers match up with the numbers Betti had previously defined in a similar manner.)

Later in his paper, Poincare writes down an argument for his now-famous duality theorem: that if the ambient manifold we are working in is compact, $n$-dimensional, and oriented without boundary, then $P_k = P_{n-k}$ for all $k$.

As noted in Dieudonne’s book, much of the paper is on extremely shaky mathematical ground, and many arguments are unconvincing. A few years after Analysis Situs, Heegaard wrote a paper pointing out some obscurities in Poincare’s paper. First and foremost: if the submanifolds $M_i$ above are not disjoint, how is the notion of a homology defined? In addition, Heegaard gave an argument which effectively showed that Poincare’s and Betti’s numbers were not the same, which led Poincare to redevelop the whole theory from scratch, as he noticed that his sketchy proof seemed to also prove the duality theorem for Betti’s orders of connectedness (which, as Heegaard showed, was simply not true). Whence came the idea we now call simplicial homology (though Poincare did not yet introduce homology groups, he nonetheless was very close to doing so; it was a student of Noether’s who later introduced the groups.)

Bordism

Instead of combinatorializing Poincare’s original idea, let’s embrace it (as Thom did in 1950). Suppose $X$ is some (reasonably nice) topological space. Instead of running into trouble with intersections, as we do with submanifolds, let’s consider the set of maps $f: M \to X$, where $M$ is an oriented closed $n$-manifold (not necessarily connected). We now identify two pairs $(M,f)$ and $(N,g)$ if there is some compact oriented $(n+1)$-manifold $(W,h)$, such that the boundary of $W$ is $M$ and $N$ with the opposite orientation, and the restriction of $h$ to the boundary are the maps above. This forms a group under disjoint union of (equivalence classes of) manifolds (and maps); we call this group the $n$th bordism group $\Omega_n(X)$.

Why should we not define Poincare’s connectedness numbers $P_q - 1$ to be the ranks of these abelian groups? Because for a point, $\Omega_n(*)$ needn’t be trivial! That is to say, not every oriented compact manifold is the boundary of an oriented compact manifold one dimension higher. The simplest example is the complex projective plane $\Bbb{CP}^2$. Indeed, Thom proved that the rank of $\Omega_n(*)$ is zero when $n$ is not a multiple of 4, but if $n=4k$, then the rank is the number of partitions of $k$. Can Poincare’s original attempt at defining connectedness numbers be recovered?

Singular homology and singular bordism

Onto a modern definition of homology. I’m going to work with $\mathbb Z/2$ coefficients for convenience, but rest assured this works if you throw in the word “oriented” a lot.

We first define the singular chain complex by setting $C_k(X)$ the free $\Bbb Z/2$-vector space, generated by the set of continuous maps from the simplex $\Delta^n$ to $X$. The boundary operator $\partial_k: C_k(X) \to C_{k-1}(X)$ is given by sending a “singular simplex” $\Delta^n \to X$ to the formal sum of the restrictions to each face. We define the singular homology groups $H_k(X)$ to be $\text{ker}(\partial_k)/\text{im}(\partial_{k+1})$.

What, actually, is an element of that kernel? Precisely, it’s a collection of (maps from) simplices such that each face of a simplex has the same map as the face of some other simplex (possibly the same one). Using this fact, we can construct a $\Delta$-complex  by gluing together matching faces, one by one. By construction, each codimension 1 face in the resulting $\Delta$-complex is contained in precisely two (possibly the same) simplices. That is, the resulting complex is a manifold away from a subset of codimension 2.

As in Kreck, we say that a $\Delta$-complex is “$\Bbb Z/2$-orientable” if it has the above combinatorial condition (and thus that it is a manifold away from a codimension 2 subset). Playing the above game again, we see that an element of $\text{im}(\partial_{k+1})$ is the boundary of a $\Delta$-complex, one dimension higher, whose codimension 1 simplices either are the face of precisely 1 simplex (in which case we say that they are in the boundary of the complex) or precisely 2 simplices.

What have we discovered? That we can define homology as bordism groups of singular manifolds: $\Delta$-complexes which are manifolds away from a codimension 2 subcomplex. We immediately see from this description why homology groups of the point in this context are trivial: if $X$ is a $\Delta$-complex of dimension greater than zero, then its cone $CX$ is a $\Delta$-complex of dimension one greater, whose only singularities are either the previous singularities cross the interval, or the cone point. (This is also “why” the point itself is not trivial in degree 0 homology – the cone on the point is the interval, which has two boundary points; if you prefer, you might think of it having one boundary point, and one codimension 1 singularity.)

Kreck has a similar perspective in his beautiful and highly recommended book. Instead of moderately-singular $\Delta$-complexes, he works with what he calls stratifolds: again a form of singular (smooth) manifold, modulo bordism thereof. The difference between this approach and ours here is not simply a matter of taste; Kreck’s construction allows him to make Poincare duality a triviality, make geometrically clear what Mayer-Vietoris, Thom classes, and the cup product are… and so on. (His groups tend to only be the same as the ones we know well when they’re CW complexes; his groups disagree with singular homology for, say, the one-point compactification of an infinite-genus surface.) The smooth structure on his stratifolds allow him to perform plenty of useful geometric constructions, at the (in my mind, usually minor) cost of having the algebra be somewhat more difficult because he never uses chain complexes.

This leads one to what I think is a natural question: How singular do homology classes have to be? It is straightforward to see that any $\Bbb Z/2$-orientable 2-dimensional closed $\Delta$-complex is bordant to a closed surface, where you can add in the word ‘oriented’ throughout if you like. (Corollary: the abelianization of $\pi_1(X)$ is $H_1(X)$. Can you see why?) Thom studied this question in great detail in the 50s, working rather hard and using the bordism groups to prove that every $\Bbb Z/2$-oriented closed $\Delta$-complex is bordant to a manifold. (In modern language: Every class in $H_k(X;\Bbb Z/2)$ can be represented by a smooth manifold.) When working with oriented $\Delta$-complexes (that is, classes in $H_k(X;\Bbb Z)$, this isn’t true, but we can still do rather well; up to degree 6, every class can be represented by a manifold, and so can every class of codimension 1 or 2. But there are 7-dimensional classes in a certain 10-manifold that cannot be represented as a map out of a manifold.

Homology and holes

As promised, we’ll explain why homology counts some sort of “holes” like we claimed it does.

The following argument essentially comes from Kreck’s book. Given a space $X$, and a subspace $L$, we say that the space $Y = X \setminus L$ has the hole $L$; we call this an “extrinsic hole” because we can’t possibly see this hole from looking at $Y$ itself, but rather by looking at the larger space $X$. Now let’s suppose that $L$ is the interior of a compact $\Bbb Z/2$-oriented $\Delta-complex$ $K$. As before, we can notice the hole from within $X$, but from just looking at $Y$, how do we see that there was a hole there? Well, we could try to plug the hole. We don’t know what was originally deleted, so if we have the boundary of $K$ (without knowing about $K$ itself), we would have to check if we can extend this to a map to some $\Delta$-complex, but we don’t know which. And now we’ve more or less come to the definition of homology: we say that this boundary is actually zero if we can extend it to a map from some $\Delta$-complex. If not, then we can convince ourselves that yes, this subcomplex does correspond to the boundary of some hole – something we got by deleting a subcomplex. This is why nonzero homology classes can, in some way, represent holes, and why we test them by trying to make them bound any $\Delta$-complex, and not just, say, trying to make maps from spheres bound maps from balls.

A different classification of surfaces

There are a couple ways of giving the classification of (compact without boundary) surfaces. One standard one approach is as follows. Take the surface, and triangulate it (it’s of course a theorem that one can do so, but it’s a theorem I’m willing to hide behind the curtains). Cut it open along certain 1-simplices to get a disc; remembering how you cut it open, you get certain boundary identifications of the disc that give you your surface. Now cut the disc apart, and repaste, ad nauseum, until you’ve gotten to a standard form, at which point you say “Aha! This is one of the surfaces on the following list: …”

While I like the combinatorics of the proof, it somehow doesn’t quite satisfy me, since the fact that every surface is a connected sum of tori and real projective planes is a result, rather than input – somehow what I find to be the key property of surfaces is just a bonus instead of something we see. I don’t really feel “why” the theorem is true.
(As a side note, before I move on, the triangulation manipulations become both cleaner and clearer when you change them into handlebody manipulations instead. The proof, once you believe the basic rules of handleslides and birth/death, is almost a complete triviality. It’s an exercise in Gompf and Stipsicz’s wonderful book, “Kirby Calculus and 4-Manifolds”.)

I’d like to present here an alternative proof inspired by ideas from 3-manifolds that makes the appearance of tori and projective planes obvious. It’s probably longer, but I find it more comprehensible and satisfying. We’ll still need to be able to triangulate surfaces, so let’s just assume off the bat that we can do so. We’ll use some basic algebraic topology; let’s start by writing out some homological lemmas. Henceforth we will use $H_1(X)$ to mean the first singular homology group with $\Bbb Z/2$ coefficients, and $\Sigma$ is a compact surface without boundary.

Lemma 1. $H_1(\Sigma)$ is a finitely generated vector space over the field $\Bbb Z/2$.
Proof. This is an immediate corollary of the fact that singular homology is isomorphic to simplicial homology, and a compact surface has finitely many simplices.

Lemma 2. If $\Sigma^\circ$ is $\Sigma$ with a point removed from its interior, then $H_1(\Sigma^\circ) \cong H_1(\Sigma)$.
Proof. Apply the Mayer-Vietoris sequence (with reduced homology for convenience) to $U$ a small open ball around the deleted point and $V = \Sigma^\circ$ its complement. Because $U$ is contractible, it has trivial homology; and $U \cap V$ is homotopy equivalent to a circle, so the exact sequence is precisely
$H_2(\Sigma) \to H_1(S^1) \to H_1(\Sigma^\circ) \to H_1(\Sigma) \to 0$.
The key point is that the map from $H_1(S^1)$ is zero. But the image of the circle bounds a compact submanifold on both sides (possibly non-orientable); and because compact manifolds are triangulable, a chase through the definitions shows that this means that the generator of $H_1(S^1)$ maps to a boundary, proving that $0 \to H_1(\Sigma^\circ) \to H_1(\Sigma) \to 0$ is exact, as desired.

Lemma 3. If $\Sigma = M \# N$ ($\#$ the connected sum operation), then $H_1(\Sigma) = H_1(M) \oplus H_1(N)$.
ProofRun the Mayer-Vietoris argument again, this time with $U = M^\circ, V = N^\circ$, recalling from lemma 2 that $H_1(M^\circ) = H_1(M)$.

Now let’s do a little bit of classification – let’s classify surfaces with $H_1(\Sigma) = 0.$ (Actually, this is the hardest part of the whole proof – it had to happen somewhere, by the law of Conservation of Effort, since there’s some difficulty in the standard proof.)

Theorem 1. If $H_1(\Sigma) = 0$, then $\Sigma \cong S^2$.
Proof. Start by deleting the interior of a 2-simplex in $\Sigma$ to get a compact manifold with boundary $S^1$, which we will denote $\Sigma'$. Now I claim that $\Sigma'$ deformation retracts onto a 1-complex. How do you do this? Start at the boundary, and start pushing the 2-simplices in. The picture you have here is that you take a solid triangle and deformation retract it to two of its sides by pushing the third side inwards. You can make this move whenever you have a 2-simplex such that not all of its edges are connected to another 2-simplex; I leave it as an exercise to prove that this process doesn’t end in failure (that is, I can always push in 2-simplices until I’m down to the 1-complex) – a particularly careful proof will eventually invoke invariance of domain.
Paying more attention to the process we went through above, we actually see that $\Sigma'$ is homeomorphic to an arbitrarily small compact neighborhood of the 1-skeleton. Thus far, everything we have done would have worked for every compact surface. Now, invoke that $H_1(\Sigma) = 0$ to see that the 1-complex is a tree. To determine what the neighborhood of this is, first do so for a single edge (the neighborhood here must be homeomorphic to a disc, by invoking some variant of the tubular neighborhood theorem) and induct on the number of edges (every time you glue in a disc corresponding to a neighborhood of an edge you’re only doing so at a single vertex) to see that the result is homeomorphic to a disc. Hence $\Sigma' \cong D^2$; and gluing the disc we deleted back in, we see $\Sigma \cong S^2$ as desired.

It is worth noting that the previous theorem fails drastically for higher-dimensional manifolds: there are lots of 3-manifolds already with the same homology as the 3-sphere, but that are not homeomorphic. These are called homology spheres, and are a current object of interest in 3-dimensional topology. And once you pass to four dimensions or higher, you can have things with complicated (co)homology, even if they have trivial first homology or fundamental group.

Anyway, using the previous set of theorems, let’s make our second step towards classification.

Definition. A surface is prime if it admits no nontrivial connected sum decomposition $\Sigma \cong M \# N$ (that is, no such connected sum decomposition where neither $M$ nor $N$ are spheres).
Note that this is equivalent to saying that every embedded loop in the surface is either non-separating, or bounds a disc on one side. This motivates the following definition, which we’ll use later.
Definition. A surface is irreducible if every embedded loop in the surface bounds a disc on one side.

Lemma 4. Any surface has a prime decomposition.
Proof. If $\Sigma$ is not prime, decompose it. And so on, and so on. We only need some guarantee that this process will eventually stop. Now invoke lemmas 1, 3, and theorem 1; we see that every nontrivial connected summand will increase the rank of first homology, which is finite. So indeed the number of connected summands is bounded by the dimension of $H_1(\Sigma)$, as desired.

This lemma is true in dimension 3, but the proof no longer works, by the failure of theorem 1. The proof is significantly more difficult. I believe the lemma is no longer true in dimension 4.

So now it clearly suffices to identify what the prime surfaces are. First, on irreducible surfaces:
Lemma 5. The only irreducible surface is $S^2$.
Proof. First, note that every element in $\pi_1(\Sigma)$ can be represented by an immersed loop (one with transverse self-intersection); this can be proved by either a smooth or simplicial approximation theorem. Use irreducibility to show that every loop is contractible, as follows: start by tracing out the (immersed) loop; the first time you self-intersect, trace the loop backwards until you reach the self-intersection point. This gives a closed embedded loop, which bounds a disc by assumption, and hence can be contracted to the self-intersection point. Repeat until the self-intersections are gone and the loop itself is contracted. This proves that $\pi_1(\Sigma) = 0$; now invoke that $\pi_1(\Sigma)^{ab} = H_1(\Sigma; \Bbb Z)$, and the universal coefficient theorem to see that $H_1(\Sigma) = 0$ as desired.

Theorem 3. The prime, but not irreducible, surfaces are $T^2$ and $\Bbb{RP}^2$.
Proof. Take a non-separating loop in the surface. A small neighborhood of this, by the tubular neighborhood theorem, is homeomorphic to the total space of a line bundle over the loop; this is either a cylinder or a Mobius band.
Case 1: Mobius band. The boundary of the Mobius band is a separating loop in the surface, and hence bounds a disc on the other side, proving that we have obtained $\Bbb{RP}^2$.
Case 2: Cylinder. Take a loop from one side to the other (using the fact that it’s a non-separating loop) and take a neighborhood of this; we’ve glued on yet another cylinder (or Mobius band, but in that case we’ve already decided we’re either not prime or we’re a $\Bbb{RP}^2$). It is straightforward to see that the boundary of the resulting object is a circle, hence must bound a disc on the other side; and then by checking what the possible shapes we have are, we see we’ve got either a torus or a Klein bottle (which is not prime). This proves the theorem.

Thus, every surface is a connected sum of tori and projective planes. Once one proves that $T^2 \# \Bbb{RP}^2 = \#_3 \Bbb{RP}^2$, we see that every surface is either homeomorphic to $S^2, \#_g T^2$, or $\#_h \Bbb{RP}^2$.

Homology cobordisms and 3-manifold invariants

In this post, every manifold is smooth unless otherwise stated. Given a closed manifold $M$, one old and very natural question is what kind of manifolds it bounds. The simplest form of this question just asks: Is $M$ the boundary of some manifold $W$? This leads one to the theory of cobordisms: Two closed $M$ and $N$ are cobordant if $M \sqcup N$ is the boundary of some compact manifold $W$. (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 $n$ 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 $H^*(M;\Bbb Z/2)$).

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 $S^3$, that same link diagram provides an oriented cobordism from $S^3$ to the 3-manifold itself; filling in the $S^3$ 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 $S^3$; 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 $M_1, M_2$ to be the same if there is a manifold $W$ with boundary $M_1 \sqcup \overline{M_2}$ such that the inclusion maps $M_i \hookrightarrow W$ induce isomorphisms on every homology group. It is not hard to see that this forms a group under connected sum; this is called the $n$-dimensional homology cobordism group $\Theta^n$. Being trivial in this group is the same as bounding a homology ball, because the identity element is $S^n$; if you bound a homology ball, just delete a small ball from the interior to get a homology cobordism to $S^n$.

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 $\Sigma(p,q,r)$. 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 $\Theta^n$ 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 $\Theta^3$. 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) $M$, consider $V = H_2(M;\Bbb Z)/\text{tors}$. There is then a cup product pairing $V \otimes V \to \Bbb Z$, 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 $V$ called the intersection form; if $M$ is closed or has boundary with $H_1(\partial M;\Bbb Z) = 0$, this form is unimodular (has determinant 1). As an example, $\Bbb{CP}^2$ has $V = \Bbb Z$, intersection form just multiplication; $S^2 \times S^2$ has intersection form represented by $(x_1,y_1) \cdot (x_2, y_2) = x_1y_2 + x_2y_1$; $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ has intersection form given by $(x_1, y_1) \cdot (x_2,y_2) = x_1x_2 - y_1y_2$. 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 $V$ of maximal dimension such that the intersection form, restricted to the subspace, is positive (that is, has $x \cdot x > 0$ for all $x$ in the subspace). Call the dimension of this subspace $b_2^+$. You can find a complementary subspace on which the intersection form is negative; call the dimension of this $b_2^-$. Lastly, we say that the signature of our manifold is $b_2^+ - b_2^-$.

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 $\Bbb{CP}^2 \# \Bbb{CP}^2$ does not bound an orientable manifold, though it does bound a non-orientable manifold.

Given an intersection form, we call it even if $x \cdot x$ 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 $M$ with intersection form $Q$, if $Q$ is even, its signature is divisible by 16. A modern proof for the case where $H_1(M;\Bbb Z)$ has no 2-torsion (in particular, if it’s zero) might go as follows: By the assumption that $Q$ is even, some characteristic class wizardry shows that $M$ 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 $-\text{sign}(M)/8$. 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 $Y$, some fiddling with a surgery diagram for $Y$ shows that it bounds a smooth 4-manifold $X_1$ with even intersection form $Q_1$. Define $\mu(Y) = \text{sign}(Q_1)/8 \bmod 2$. The algebraic fact we invoked above means that this is indeed an integer mod 2; it’s well-defined because if $X_2$ is another such 4-manifold, $X_1 \cup_Y \overline{X_2}$ is a smooth closed 4-manifold, hence has signature divisible by 16; but because its intersection form is $Q_1 \oplus Q_2$, its signatures thus have $\text{sign}(Q_1) + \text{sign}(Q_2)$ 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 $W: Y \to Z$, if $X$ is a 4-manifold with boundary $Y$ and even intersection form, then $X \cup_Y W$ is a 4-manifold with the exact same even intersection form and boundary $Z$.

Second, it’s additive under connected sum. This is because if $Y_i$ bound smooth 4-manifold with even intersection form $X_i$ with intersection forms $Q_i$, then we can form the “boundary sum” of $X_1$ and $X_2$, a 4-manifold with intersection form $Q_1 \oplus Q_2$ whose boundary is $Y_1 \# Y_2$. These two facts combined tell us the Rokhlin invariant descends to a well-defined homomorphism $\mu: \Theta^3 \to \Bbb Z/2$.

Third, $\mu(\Sigma(2,3,5)) = 1$. This is because $\Sigma(2,3,5)$ can be described as surgery on a link whose linking form is the $E_8$ form, whose signature is 8. Thus, everything thus far gives us our first concrete result: $\Sigma(2,3,5)$ does not bound a homology ball! But the Rokhlin homomorphism doesn’t tell us anything about, for instance, $\Sigma(2,3,5) \# \Sigma(2,3,5)$.

Now, $\Theta^3$ 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 $Y$ such that $Y \# Y$ bounds a homology ball and $\mu(Y) = 1$? (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 $\nu: \Theta^3 \to \Bbb Z$ that reduces mod 2 to the Rokhlin homomorphism – then if $\mu(Y) \neq 0$, we have $\nu(Y \# Y) = 2\nu(Y) \neq 0$, so $Y \# Y$ 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 $\pi_1(Y) \to SU(2)$ mod conjugacy; if you’re lucky, there’s only finitely many of them. More precisely, define $\mathcal R(X)$ to be the space of representations $\pi_1(X) \to SU(2)$ mod conjugacy and denote the genus $g$ handlebody as $X_g$. Given a Heegaard decomposition of $Y$, we get two maps $\mathcal R(X_g) \to \mathcal R(\Sigma_g)$. The intersection of the image of these is $\mathcal R(Y)$. Inspired by this, we define the Casson invariant of $Y$ to be the intersection number of the two images of $\mathcal R(X_g)$. 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, $\lambda(Y)$, 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 $\mu$ doesn’t split, and it’s no good at showing that $\Sigma(2,3,5) \# \Sigma(2,3,5)$ 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 $Y$, we can define its instanton Floer homology $I_*(Y)$; this is a $\Bbb Z/8$-graded group, and it so turns out that its Euler characteristic (meaning $\sum_{n=1}^8 (-1)^n \text{rk } I_n(Y)$) 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 $\mu$, the topology of 4-manifolds it bounds; for $\lambda$, the representation theory of its fundamental group; and for $I_*$, 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 $h: \Theta^3 \to \Bbb Z$. It is unrelated to the Rokhlin invariant, but is a homomorphism. Because one calculates that $h(\Sigma(2,3,5)) = 1$, we’ve now proved one of our goals above: that $\Sigma(2,3,5) \# \Sigma(2,3,5)$ does not bound a homology ball.

Whether or not the Rokhlin homomorphism $\mu$ 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 $\mu$ lifts to an integer-valued homomorphism. Luckily, we don’t need quite that much. Manolescu, in 2013, defined an invariant $\beta: \Theta^3 \to \Bbb Z$ (motivated by much the same ideas as Froyshov’s invariant) such that $\beta(Y) = -\beta(\bar Y)$, that reduces mod 2 to the Rokhlin invariant. So if $Y \# Y$ bounds a homology ball, then $Y$ is homology cobordant to $\overline Y$, and hence $\beta(Y) = \beta(\overline Y) = - \beta(Y)$; hence $\mu(Y) = \beta(Y) \bmod 2 = 0$. 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 $Y$ bounds a 4-manifold with even, positive definite intersection form. (Positive definite means that $b_2^- = 0$.)  Then $Y$ cannot possibly bound a negative definite 4-manifold (in particular, a homology ball). If it could, then gluing our two manifolds bounding $Y$ would give us a smooth 4-manifold with positive-definite nonstandard intersection-form, contradicting Donaldson’s theorem. This proves, in particular, that neither $\Sigma(2,3,5)$ nor $\Sigma(2,3,5) \# \Sigma(2,3,5)$ bound homology balls. It says nothing about $\Sigma(2,3,5) \# \overline{\Sigma(2,3,5)}$, because the intersection form of the “obvious” manifold it bounds is $E_8 \oplus -E_8$, which is not positive definite (it has signature zero). Indeed, $\Sigma(2,3,5) \# \overline{\Sigma(2,3,5)}$ does bound a homology ball.

$\Theta^3$ is weird

To close out the post, some minor comments about $\Theta^3$. 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.

1. It’s countable, because there are only countably many compact manifolds up to homeomorphism.
2. It’s abelian, because the connected sum operation on the level of manifolds is abelian.
3. The Rokhlin homomorphism $\Theta^3 \to \Bbb Z/2$ is a nonzero homomorphism that does not split. Indeed, it has a lift to a non-homomorphism with $\beta(\overline Y) = -\beta(Y)$.
4. It’s infinitely generated! In fact, if you fix $p, q$ relatively prime, the Brieskorn spheres $\Sigma(p,q,pqk-1)$ for $k \geq 1$ are linearly independent in $\Theta^3$. So $\Theta^3$ has $\Bbb Z^\infty$ as a subgroup. This was first proved by Furuta using what’s known as Atiyah’s $\eta$ 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.
5. It has a nonzero homomorphism onto $\Bbb Z$ (the Froyshov invariant), and hence has a $\Bbb Z$ 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.

The dimension barrier

There is a curious phenomenon in geometric topology: some problems seem to get easier as you increase the dimension you’re looking at. The catchphrase is that “high dimensional topology is easy because you have more room to move around” – let’s see what this means.

The $n$-dimensional topological (resp. smooth) Poincaré conjecture says that if a closed $n$-manifold is homotopy equivalent to $S^n$, then it is homeomorphic (resp. diffeomorphic) to $S^n$.

The most well-known of this web of conjectures is the 3-dimensional topological case, which is equivalent to: “A simply connected closed 3-manifold must be homeomorphic to $S^3$.” This was famously resolved by Perelman in the early 2000s. The 4-dimensional topological Poincaré conjecture was proved by Freedman in 1982; and for $n \geq 5$, the n-dimensional topological Poincaré conjecture was proved in the mid-60s. We are, in some sense, going backwards in difficulty! (Rest assured, though, that the 2- and 1-dimensional cases are not so terrible that they won’t be solved in the millenium; they were known in the early 20th century. 3 is the first dimension which is complicated enough for these sorts of problems to be interesting.)

Because we know the topological Poincaré conjecture is true in all dimensions, we can reformulate the smooth Poincaré conjecture as follows: if a smooth manifold $M$ is homeomorphic to $S^n$, is $M$ necessarily diffeomorphic to $S^n$?

The question of determining the number of diffeomorphism classes of smooth manifolds that are homeomorphic to $S^n$ is known as the exotic spheres problem. (Such a smooth manifold – one homeomorphic, but not diffeomorphic, to $S^n$ – is called an exotic sphere.) The following table shows how many exotic spheres there are in each dimension $n$.

 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 ??? 1 1 28 2 8 6 992 1

While the 28 or 992 might seem mysterious, it’s less so than one might imagine – the exotic spheres problem, in dimensions above 5, has been essentially reduced to a fundamental (yet very difficult!) question in algebraic topology; more on this later. Worth noting is that dimension 3 was solved by Moise in 1952, and the 4-dimensional case is completely open; experts are not even entirely certain whether or not to believe it’s true.

So there is, fundamentally, some sort of difference between low-dimensional topology and high-dimensional topology. Why?
Everything is because of the h-cobordism theorem.

cobordism between two closed $n$manifolds $M$ and $N$ is a compact $(n+1)$-manifold $W$ whose boundary is the disjoint union of $M$ and $N$. Note that we need $W$ to be compact for this to be an interesting notion; otherwise a cobordism between any pair of closed manifolds $M, N$ is given by $(M \sqcup N) \times [0,\infty)$.
Some canonical examples are given by a pair of pants, a cobordism between $S^1$ and $S^1 \sqcup S^1$; the solid torus, a cobordism between the torus and the empty set; and for any manifold $M$, $M \times [0,1]$ is a cobordism between $M$ and itself.

An h-cobordism between two closed $n$-manifolds $M$ and $N$ is a cobordism $W: M \to N$ such that the inclusions $M \hookrightarrow W$, $N \hookrightarrow W$ are homotopy equivalences. An example is, as above, given by $M \times [0,1]$, an h-cobordism between $M$ and itself. Giving you more examples will be problematic, because:

h-cobordism theorem (Smale): Let $W: M \to N$ be a topological (resp. smooth) h-cobordism, and suppose $M, N$ and $W$ be simply connected. If, in addition, $n \geq 5$, then $W$ is homeomorphic (resp. diffeomorphic) to $M \times [0,1]$. In particular, $M$ is homeomorphic (resp. diffeomorphic) to $N$.

I still find this theorem absolutely astonishing. It is such an immensely powerful result from what don’t appear to be very strong conditions. (In fact, it’s still true topologically in low dimensions; Freedman’s proof of the 4-dimensional topological Poincaré conjecture starts by proving the topological 4-dimensional h-cobordism theorem; and the 3-dimensional h-cobordism theorem is now known to be true thanks to the 3-dimensional Poincaré conjecture. Perhaps part of what makes 4 dimensional topology so weird is that the smooth 4-dimensional h-cobordism theorem is absolutely, irrevocably, false. Not unknown, but false.)
Why does this fascinating, but seemingly unrelated statement help us? Well…

Corollary: If $M$ is a closed manifold, homotopy equivalent to $S^n$, $n \geq 6$, then $M$ is homeomorphic to $S^n$.
Proof: Embed two small closed balls $B_i: D^n \hookrightarrow M$ into $M$, whose images are disjoint, and delete their interiors. Call the resulting manifold $M'$. Some algebra shows that $M'$ is an h-cobordism between the two boundary copies of $S^{n-1}$; so that $M'$ is homeomorphic to $S^{n-1} \times [0,1]$. We want to extend this to a homeomorphism $M \cong S^n$; it suffices to show that given a homeomorphism $f: S^{n-1} \to S^{n-1}$, we can extend this to a homeomorphism $\tilde f: D^n \to D^n$. The process of doing so is known as Alexander’s trick; if $D^n$ is written in polar coordinates, simply write $\tilde f(t,\theta) = (t,f(\theta))$; the idea being that we’re repeating $f$ “levelwise” until we get to 0, which we fix. One can check that this is continuous. Even if $f$ was smooth, this $\tilde f$ is rarely smooth at the origin, which is why we haven’t simultaneously proved the smooth high-dimensional Poincaré conjecture (which, as above, is false). Indeed, there are many diffeomorphisms $S^{n-1} \to S^{n-1}$ which cannot be extended to a diffeomorphism of the disc.

(As a side note, the proof of the 5-dimensional Poincaré conjecture is slightly different; this was achieved, instead of deleting small balls, by showing that any closed 5-manifold, homotopy equivalent to $S^n$, must be h-cobordant to it.)

As for the exotic spheres problem, this helps us too. h-cobordisms are somehow more ‘algebraically assailable’ than diffeomorphisms; Milnor was able to write down a few exact sequences that help us with our problem.

In the following, $\theta_n$ is the group of h-cobordism classes of $n$-manifolds homeomorphic to $S^n$; the group operation is connected sum, though we won’t prove this actually gives you a group. $\theta_n^{bp}$ is the group of h-cobordism classes of $n$-manifolds homeomorphic to $S^n$ that bound parallelizable manifolds. While this sounds complicated, it’s not that hard (for Milnor) to understand this group; it’s cyclic, and we know its order in all dimensions. $\pi_n^s$ is the $n$th stable homotopy group of spheres, and $J_n$ is a certain (again, understood) subgroup; we know its order.

Milnor finds exact sequences, for $n \not\equiv 2 \mod 4$, $0 \to \theta_n^{bp} \to \theta_n \to \pi_n^s/J_n \to 0$; and for $n \equiv 2 \mod 4$, we get the slightly more ugly sequence $0 \to \theta_n \to \pi_n^s/J_n \xrightarrow{\Phi_n} \Bbb Z/2\Bbb Z \to \theta^{bp}_{n-1} \to 0.$ The map $\Phi_n$ is known as the Kervaire invariant; it was known in the 60s that it could only be nonzero for $n$ of the form $2^j-2$; recent work by Hill, Hopkins, and Ravenel shows that it is zero for all $n \geq 254$. It is known to be nonzero for $n \in \{6,14,30, 62\}$, and it’s unknown for $n = 126$. Dimensions close to these numbers are special; the order of $\theta_n$ is known in every dimension $n \leq 64$, and $\theta_n = 0$ in these dimensions iff $n \in \{1,2,3,4,5,6,12,61\}$. As a result of the above, except in dimensions 125 and 126, the order of $\theta_n$ is known as a function of the order of $\pi_n^s$. The calculation of this group, $\pi_n^s$, is one of the main motivating problems of homotopy theory, and is a major open problem. Nonetheless, it is expected that in dimension large enough, the smooth Poincaré conjecture is false.

How the h-cobordism theorem helps us here is obvious: the group $\theta_n$ is, as a result, isomorphic to the group of diffeomorphism classes of exotic spheres! So the above is entirely about exotic spheres, except in dimension $n =4$, where all exotic spheres are h-cobordant; but we have absolutely no idea whether or not they’re diffeomorphic.

Hopefully this convinces you that the h-cobordism theorem (or things related to it) is what causes this ‘high-dimensional simplicity’; but why is it true, and where does that catchphrase, “there’s more room to move around”, come from? Certainly the h-cobordism theorem doesn’t sound like this. So, next time: why the h-cobordism theorem is true, and how it relates to having extra wiggle room.