Theorem of Rokhlin

The goal of this post is to discuss Rokhlin’s Theorem and also the Rokhlin invariant which is for 3-manifolds. Theorem (Rokhlin): Let $M^4$ be a closed, smooth, spin 4-manifold. Then its signature $\sigma(M)$ is divisible by 16.

Remark: The Wu formulas allow us to define Stiefel-Whitney classes for a topological manifold even if it has no tangent bundle. Now, if $Q$ is the intersection form, then $Q(x,x)$ is not a morphism since $Q(x+y,x+y) = Q(x,x)+2Q(x,y)+Q(y,y)$. However, modulo 2, it is a morphism. And in fact, it is a functional $q:H_2(M,\mathbb{Z}/2) \to \mathbb{Z}/2$ and hence, is represented by pairing it against some vector $v$. That is, $q(x) := \langle x,v \rangle$ for some vector $v$; it is not unique in integral homology but is mod 2. The claim is that $w_2=v$ in general; a geometric argument can be made to prove this. An application is when $M$ is spin, then $w_2 = 0$ and so the intersection form is even since $Q(x,x) \equiv 0 \pmod{2}$. Moreover, it is a general fact independent of geometry/topology that even symmetric, bilinear, unimodular forms have signature divisible by 8. The $E_8$ lattice satisfies these conditions and is the intersection form of a closed, oriented, topological 4-manifold. Moreover, its signature is exactly 8. Hence, to get the extra divisibility by 16, one really needs the smoothness condition.

Rokhlin’s Proof

Rokhlin was the first to prove that $\Omega^{SO}_4 \cong \mathbb{Z}$; the cobordism group of closed, smooth, oriented 4-manifolds is isomorphic to the integers and the isomorphism is given by the signature $\sigma$. He also proved the formula $\sigma = p_1/3$.

Rokhlin then calculated by a geometric argument a la Pontryagin that if $M^4$ is almost parallelizable; i.e. parallelizable in the complement of a point, then the first Pontryagin number is actually divisible by 48. Since $\sigma = \frac{p_1}{3}$, this proves his theorem that $\sigma$ is divisible by 16.

To understand this, let’s first observe that being almost parallelizable for a oriented, closed, smooth 4-manifold means exactly that the intersection form is even which is implied by the manifold being spin. Here’s an argument. The bundle $TM$ is parallelizable if the classifying map for it $f:M \to BO(4)$ lifts to a map into $B1$ which is the same thing as saying that $f$ is nullhomotopic. We can model $B1 \simeq EO(4)$ so that the homotopy fiber of the map $B1 \to BO(4)$ is $O(4)$ itself. Then, the obstructions to this lift live in $H^*(M,\pi_{*-1}O(4))$. Now, we are talking about almost parallelizable; so we should remove a point from $M$, which I denote $M^\times$; this is homotopy equivalent to the 3-skeleton of $M$. Since $\pi_0 O(4) = \pi_1 O(4) = \mathbb{Z}/2$ and $\pi_2O(4) = 0$ (all Lie groups have vanishing $\pi_2$, the only obstuctions we need to consider are $w_1 \in H^1(M^\times,\mathbb{Z}/2)$ and $w_2 \in H^2(M^\times,\mathbb{Z}/2)$. Since $M$ is oriented, $w_1 = 0$. So we see that if $w_2(M)=0$, then $w_2(M^\times)=0$ as well and hence, $M^\times$ is parallelizable which means that $M$ is almost parallelizable. So spin $M$ is almost parallelizable. But the converse is also true; if $M$ is almost parallelizable, since $M^\times$ and $M$ have the same 1-skeleton and 2-skeleton, then $M$ is spin since $w_1=w_2=0$.

Nonsequitur: I think that if we want to state a slightly stronger statement, we could even say that a closed, smooth, 4-manifold $M$ with even intersection form has $\sigma(M)$ divisible by 16. If $M$ is simply-connected, then being spin and having even intersection form are equivalent but in the non-simply connected case, being spin implies even intersection form but the converse is not always true.

Anyways, I will be following the book Introductory Lectures on Manifold Topology: Signposts by Thomas Farrell and Yang Su, for the remainder of this section.

If $M$ is almost parallelizable, then $T(M \setminus B^4)$ is trivial where $B^4$ is a small ball. Of course, $TB^4$ is also trivial. So the fact that $TM$ is (possibly) nontrivial is dependent on the clutching map $S^3 \to SO(4)$ where we glue the two bundles together. This clutching map gives a way to form a bundle $\xi \to S^4$ in the following way: take $B^4_+ \times \mathbb{R}^4 \sqcup B^4_- \times \mathbb{R}^4$ and glue along the $S^3$ boundary of the balls with the clutching map.

We now define a map $f:M \to S^4 = B^4_+ \cup B^4_-$ which sends $B^4 \subset M \to B^4_+$, sends a collar neighborhood of the $S^3$ to $B^4_-$, and the rest of $M$ is collapsed to the south pole of $S^4$. Hence, $f^*\xi = TM$. So if $p_1(\xi)$ is divisible by 48, then so is $p_1(f^*\xi) = p_1(M)$ (and the converse is true as well). So let’s show that $p_1(\xi)$ is divisible by 48.

The next thing to observe is that $\xi$ has some nice properties. First, recall that the $J$-homomorphism is a map $J:\pi_iSO(n) \to \pi_{n+i}S^n$. Here, we view $\xi \in \pi_3 SO(4) = \pi_4 BSO(4)$. The claim is that the stablization $\xi \oplus \underline{\mathbb{R}}$ is mapped to 0: $J(\xi \oplus \underline{\mathbb{R}}) = 0$. But also, $\xi \oplus \underline{\mathbb{R}} \in \pi_3 SO(5)$ and this is mapped to $\pi_8 S^5$. $\pi_3 SO(5) \cong \mathbb{Z}$ by Bott periodicity while $\pi_8 S^5 = \pi^S_3$, the 3rd stable homotopy group of spheres. I’m not sure who was the first to compute this group, maybe Rokhlin himself or Serre. But it is $\mathbb{Z}/24$.

One way to try and compute this is to note Serre’s work shows that it is torsion. Also, the Lie group framing on $S^3$ is such that a disjoint union of 24 of them is bounded by a $K3$ surface with 24 points removed (the number of singular fibers for a generic elliptic fibration). This shows that $S^3$ has order which divides 24. Hence, $\pi^S_3$ contains a cyclic subgroup of order dividing 24. But it could have more cyclic pieces and therefore, it itself might not be cyclic. In order to prove $\pi^S_3 \cong \mathbb{Z}/24$, we need to know that $\Omega^{Spin}_3=0$. This will probably have to be saved for a different blog post.

Rokhlin has a very nice geometric argument (using Alexander duality) for showing why all closed, orientable 3-manifolds are bound some 4-manifold. The Wu formulas show that orientable 3-manifolds have $w_1=w_2=0$ and since 3-manifolds bound, $w_3 = 0$ as well (Thom showed that a manifold bounds if and only if all its Stiefel-Whitney numbers vanish; since $w_3$ is the top Stiefel-Whitney class, it is zero if and only if the 3rd Stiefel-Whitney number vanishes). We’ll discuss the Wu formulas and why closed, orientable 3-manifolds are parallelizable later. However, one needs to then prove that they bound spin 4-manifolds.

Anyways, assuming $\pi^S_3 = \mathbb{Z}/24$ and that $J(\xi \oplus \underline{\mathbb{R}}) = 0$ for the moment, we’ll complete the proof. The map $\pi_3 SO(5) \to \mathbb{Z}$ is given by sending an oriented rank 5 bundle $E \to S^4$ to $\langle p_1(E), [S^4] \rangle$. Let $E_0$ be the generator of $\pi_3 SO(5) = \mathbb{Z}$. Then $\xi \oplus \underline{\mathbb{R}}$ is some multiple of this, say $sE_0$. By the work of Adams, it is known that $J: \pi_3 SO(5) \to \pi_8 S^5 \cong \pi^S_3 =\mathbb{Z}/24$ is surjective. This means that $s=24t$ for some $t \in \mathbb{Z}$ because $J(\xi \oplus \underline{\mathbb{R}}) = 0$. Hence, we just need to show that $\langle p_1(E_0),[S^4] \rangle$ is divisible by 2.

Lemma: Let $S^4$ be viewed as $\mathbb{HP}^1$ which has a tautological quaternionic line bundle which we call $\eta$. The unit sphere bundle of this is the Hopf fibration $S^3 \to S^7 \to S^4$. The stable quaternionic bundle $\eta \oplus \underline{\mathbb{R}}$ generates $\widetilde{KO}^0(S^4) = \pi_3 SO(5)$.

Proof: First, $p_1(\eta) =-2$. This is because the tautological quaternionic line bundle has $p_1(\eta) = -c_2(\eta \otimes \mathbb{C}) = -c_2(\eta \oplus \bar{\eta})$ which can be computed by studying the degree 4 terms in $(1+c_1(\eta)+c_2(\eta)+…)(1-c_1(\eta)+c_2(\eta)-…) = 1+2c_2(\eta)-c^2_1(\eta)+…$ Since $c_1(\eta) = 0$ because $S^4$ doesn’t have 2nd cohomology, we find that $p_1(\eta) = -2c_2(\eta) = -2e(\eta)$. The Euler class of $\eta$ is 1 since it is the tautological quaternionic bundle $\mathcal{O}_\mathbb{H}(1)$.

Next, $SO(4) \cong SO(3) \times S^3$ as manifolds so $\pi_3 SO(4) \cong \pi_3 SO(3) \times \pi_3 S^3 = \mathbb{Z} \oplus \mathbb{Z}$. Let $\phi$ generate the $\pi_3 SO(3)$ component; $\eta$ generates the $\pi_3S^3$ part which is what classifies principal $SU(2)=S^3$-bundles over $S^4$. Next, let $\alpha \in \pi_3 SO(5)$ be a generator and consider the homotopy exact sequence for the fiber bundle $SO(3) \to SO(5) \to V_{2,5}$ where the last space is the space of orthonormal 2-frames in $\mathbb{R}^5$.

$\mathbb{Z}=\pi_3 SO(3) \to \pi_3 SO(5) \to \pi_3 V_{2,5} \to \pi_2 SO(3) = 0$.

Observe that $V_{2,5}$ is actually the unit sphere bundle of $TS^4$. Given an orthonormal frame ${v,w}$, $v$ can be viewed as a unit vector which picks out a point on $S^4$ and $w$, being orthogonal to $v$, lives in the tangent space $T_v S^4$. Since it’s unit length, we have the unit sphere bundle: $S^3 \to V_{2,5} \to S^4$. The Gysin sequence at the start is: $H^0(S^4) \to_{\cup e} H^4(S^4) \to H^4(V_{2,5}) \to H^1(S^4) = 0$. The Euler class of the tangent bundle for any even sphere is 2 so cupping with it is multiplication by 2. Hence $H^4(V_{2,5}) = H_3(V_{2,5}) = \mathbb{Z}/2$. Also, the LES of homotopy groups says that $\pi_1$ and $\pi_2$ of $V_{2,5}$ are zero. By Hurewicz, $\pi_3(V_{2,5}) = H_3(V_{2,5}) = \mathbb{Z}/2$ which we can apply to our sequence above which is now $\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2 \to 0$. $\phi$ is the generator of the first $\mathbb{Z}$ and because $\alpha \mapsto 1 \in \mathbb{Z}/2$, by exactness, $\phi$ mapsto twice a generator. So $\phi \oplus \underline{\mathbb{R}} = \pm 2 \alpha$. But since $\pi_3SO(4) \to \pi_3 SO(5)$ is a surjection, the images of $\eta$ and $\phi$ must generate $\pi_3 SO(5) = \mathbb{Z}$; with $\phi$ being mapped to an even multiple of $\alpha$, $\eta$ must map to an odd multiple for the sake of generation; i.e. $\eta \oplus \underline{\mathbb{R}} = s\alpha$ where $s$ is odd.

Now, $p_1: \widetilde{KO}^0(S^4) \to \mathbb{Z}$ which sends a bundle to its first Pontryagin number, is a morphism. Hence, $-2 = \langle p_1(\eta),[S^4] \rangle = s \langle p_1(\alpha),[S^4] \rangle$ which implies $s = \pm 1$. Therefore, $\eta \oplus \underline{\mathbb{R}}$ generates $\widetilde{KO}^0(S^4) = \pi_3 SO(5)$. $\square$

The immediate consequence is that $\langle p_1(E_0),[S^4] \rangle = \langle p_1(E_0 \oplus \underline{\mathbb{R}}),[S^4] \rangle = \langle p_1(t(\eta \oplus \underline{\mathbb{R}})),[S^4] \rangle = -2t$. We have proven that it is divisible by 2 and hence, $p_1(M)$ is divisible not only by 24 but by 48. Hence, the signature is divisible by 16. The last thing we need to prove is:

Proof that $J(\xi \oplus \underline{\mathbb{R}}) = 0$: We will consider the virtual bundle $-TM$ for the sake of KO-theory. To get a concrete bundle, embed $M$ into some large Euclidean space such as $\mathbb{R}^9$ and compactify to get a map $M \to S^9$ in which we take the normal bundle $\nu$ of the embedding. We find $TM \oplus \nu = \underline{\mathbb{R}}^9$ so $\nu$ is the inverse of $TM$. There is a natural degree 1 collapse map $\gamma:S^9 \to Th(\nu)$ where the latter is the Thom space of $\nu$. Since $TM$ is trivial on $M \setminus B^4$ and also $B^4$, then $\nu$ is also trivial on these pieces.

We therefore know that there is a clutching construction much like before which gives us a bundle map $\bar{f}:\nu \to \rho$ that covers $f:M \to S^4$. Then, $\bar{\gamma}:=\bar{f} \circ \gamma: S^9 \to Th(\rho)$ is a degree 1 map; we’ll use this to show $J(\rho) = 0$. Let $\sigma:D^5/S^4 \to Th(\rho)$ be the fiber inclusion map into the Thom space and take $\bar{\gamma} \vee \sigma: S^9 \vee S^5 \to Th(\rho)$. The Thom isomorphism says $\tilde{H}_{i+5}(Th(\rho)) = H_i(S^4)$. Hence, $\bar{\gamma}\vee \sigma$ induces an isomorphism on homology and is a homotopy equivalence by Whitehead’s theorem. But now, $Th(\rho) = S^5 \cup_{J(\rho)} D^9$ and also is $S^5 \vee S^9$ hence, $J(\rho)=0$ as it collapses everything on the boundary of $D^9$ to the wedge point. Since $\rho$ is the inverse to $\xi$, this shows that $J(\xi \oplus \underline{\mathbb{R}})=0$. $\square$

In fact, using the previous arguments, we can show that $p_1(\nu)$, a class for $\nu \to M$, is divisible by 48. But also, since $p_1(E \oplus F) = p_1(E)+p_1(F)$ (this only works for the 1st Pontryagin class), and $TM \oplus \nu = \underline{\mathbb{R}}^9$, then $p_1(TM) = -p_1(\nu)$ and hence, $p_1(M)$, the 1st Pontryagin number of $M$, is divisible by 48 as well.

$\widehat{A}$-Genus and Dirac Operators

The $\widehat{A}$-genus is a rational number defined for any manifold, but is in general not an integer. Borel and Hirzebruch showed that it is integral for spin manifolds and an even integer if in addition, the dimension $n \equiv 4 \pmod{8}$. Atiyah-Singer then proved that $\widehat{A}$ is in fact the index of the Dirac operator for spin manifolds of dimension $8k+4$. The extra factor of 2 in dimensions $8k+4$ comes from the fact that in this case the kernel and cokernel of the Dirac operator have quaternionic structure and so as complex vector spaces, they have even dimensions. So the index $\widehat{A}$ is even. In dimension 4 this result implies Rokhlin’s theorem. This is because in dimension 4, $\widehat{A} = -\frac{\sigma}{8}$.

Why is it that there is quaternionic structure? The explanation is rather simple: for $8k+4$, the Clifford algebra $Cl_{8k+4} \cong Cl_{8k}\otimes Cl_4 \cong Cl_8^{\otimes k} \otimes Cl_4$. Here, $Cl_8 \cong \mathbb{R}(16)$, that is, the algebra of $16 \times 16$ real matrices (which has dimension $2^8 = 16^2$. Also, $Cl_4$ is isomorphic to $\mathbb{H}(2)$, the algebra of $2 \times 2$ quaternionic matrices (which is dimension $2^4 = 16$). The representation theory of these are basically determined by each other and we say they are Morita equivalent. So we only need to consider $Cl_4$ and $Spin(4) = SU(2) \times SU(2)$. But also, since we need the spin representation for Dirac operators, we need to look at $Cl^04 \cong Cl_3 = \mathbb{H} \oplus \mathbb{H}$. So _this is where the quaternionic structure comes from. The kernel and cokernel of the Dirac operator of a spin manifold in dimensions $8k+4$ have quaternionic structure and hence, even complex dimension.

One nice thing to say about $\widehat{A}$ is the following. We can, in dimension $4k$, construct a plumbing using tangent bundle of spheres to obtain a manifold with boundary; the boundary is a topological sphere of dimension $4k-1$ and we can then cap off the boundary by coning off. This gives a PL manifold. If we plumb with the Dynkin diagram $E_8$, we obtain a PL manifold $M$ that has $E_8$ as its intersection form, by construction. One can then compute that the $\widehat{A}$-genus of this is $-1/28$. Since $\widehat{A}:\Omega^{SO}\otimes \mathbb{Q} \to \mathbb{Q}$ is a unital ring morphism, $\widehat{A}(M \sqcup M) = -1/14$ since disjoint union (or connected sum) is the additive structure of the cobordism ring.

It is no coincidence that $28$ appears here and is the order of the group of homotopy 7-spheres $\Theta_7 \cong \mathbb{Z}/28$. If we take the exotic 7-sphere $\Sigma = \partial M$, the $\widehat{A}$-genus shows us that it generates $\Theta_7$. So we see that $\widehat{A}$ is very closely tied to understanding smoothness and the gap between smooth and PL. We also see that if we want to define Dirac operators for PL manifolds, we’re basically doomed since $\widehat{A}$ will not be the index of the operator since it won’t be an integer.

I also want to make two remarks about plumbings.

Lemma: Let $M^{2n}$ be two copies of $TS^n$ plumbed together. Then $\partial M$ is homeomorphic to $S^{2n-1}$ if and only if $n$ is odd.

Proposition: Let $E^{2n}$ be the plumbinig of $TS^n$ via the $E_8$ Dynkin diagram. Then $\partial E$ is homeomorphic to $S^{2n-1}$ except when $n=2$, in which case, it is the Poincaré homology 3-sphere.

Wu Formulas and Closed, Oriented 3-Manifolds Bound

Without going into much details, there are cohomology classes with $\mathbb{Z}/2$ coefficients called Wu classes and they’re related to the Stiefel-Whitney classes of a smooth manifold. In fact, because of these formulas we can just define Stiefel-Whitney classes for topological manifolds using the Wu classes. The main thing is that the 1st Wu class equals the 1st Stiefel-Whitney class $\nu_1 = w_1$ and $w_2 = \nu_2 + \nu^2_1$. So if a 3-manifold is oriented, then the 1st SW class vanishes. But also, there are some degree formulas that force $\nu_2=0$ when the dimension is 3 and hence, the 2nd SW class vanishes. So closed, oriented 3-manifolds are automatically spin. Then, to show that the 3rd SW class vanishes, we just need to show that closed, oriented 3-manifolds bound. In fact, so do unoriented 3-manifolds. But anyways, we give a proof, due to Rokhlin himself.

Let $M$ be a closed 3-manifold; for simplicity, assume it is orientable. Then by Whitney’s immersion theorem, we can immerse $M$ into $\mathbb{R}^5$ which itself is immersed in $S^5$. We may perturb the immersion map so that the intersections are transverse. That is, the $\text{codim} (M \cap M) = \text{codim} M + \text{codim} M = 2+2 = 4$. So then, the intersections are circles. We may perform surgery to get rid of these circles and get some new 3-manifold $M’$ which is embedded.

But if two manifolds are surgery equivalent, then they are cobordant to each other. So if we find a 4-manifold with boundary $M’$, we can glue it to the cobordism to get a bounding 4-manifold for $M$.

Now, Alexander duality gives a relationship for reduced (co)homology: $\widetilde{H}_q(S^n \setminus M’) \cong \widetilde{H}^{n-q-1}(M’)$. Here, $n=5$ and so if $q=4$, we have $\widetilde{H}_4(S^5 \setminus M’) \cong \widetilde{H}^0(M’)$. But reduced (co)homology in degree 0 is just zero and is the same as singular (co)homology everywhere else. Hence, $H_4(S^5 \setminus M’) = 0$. This means that there is a 4-cycle which has $M’$ as boundary. Doing some surgery, we can realize the 4-cycle as a “Seifert” 4-manifold. Hence, $M’$ bounds and therefore, so does $M$. Note that this argument works just in these dimensions/degrees!

I believe the same proof works for $M$ unorientable but we need an improved version of the Whitney immersion theorem.

Also, it is possible to perform surgery on $M$ so that it becomes simply connected. Certainly, we know this in retrospect, now that we know that every 3-manifold bounds and hence, is cobordant to $S^3$. But if we did not know that fact, there should be a way to glue in 2-disks to kill off $\pi_1(M)$. Then, the manifold $M’ \cong S^3$, by Perelman. Since there is an obvious 4-manifold with $S^3$ with boundary, we have produced a 4-manifold with $M$ as boundary.

The Rokhlin Invariant and Triangulation

Recall that Rokhlin’s theorem states that the signature of a smooth spin 4-manifold is divisible by 16. Thus, one can start with a spin 3-manifold $(Y,s)$; it bounds a 4-manifold $M$ and it can be arranged that the 4-manifold is spin. If we take another bounding smooth spin 4-manifold $N$, then the signature of $M \cup_Y (-N)$ is divisible by 16. Also, we can still define the intersection form on $M$, despite its boundary though the form might not be unimodular. It turns out that the signature of $M$ will be divisible by 8. The argument is simply to take the double of $M$ which is $M \cup_Y (-M)$ and see that its signature is 16. On the other hand, one can arrange that basically all the action one sees from intersecting things can be kept in each copy of $M$ and what happens on the boundary $Y$ is cancelled out since $b_+(-M) = b_-(M), b_1(-M) = b_+(M)$.

Hence, given, any two bounding spin 4-manifolds $M,N$ of $Y$, after dividing their signatures by 8, their parities much agree (both odd or both even) since gluing them together gives us that the signature of $M \cup_Y (-N)$ is divisible by 16.

Thus, if we define the Rokhlin invariant $\mu(Y,s)$ in $\mathbb{Z}/16$ to be the signature of any smooth compact spin 4-manifold with spin boundary $(Y,s)$, we see that this value depends only on $Y$ and not on the choice of bounding $M$. Now, integral homology 3-spheres have a unique spin structure since $H_1(Y) = 0$. So we can define the Rokhlin invariant of a homology 3-sphere to be the element $\sigma(M)/8 \in \mathbb{Z} /2$ where $M$ is any spin 4-manifold bounding the homology sphere.

Why is this invariant important? Among other reasons, it plays a crucial role in the triangulation question: “Does every topological manifold admit a triangulation?” By triangulation, one means a homeomorphism to a simplicial complex. One can also ask the stronger question regarding whether every manifold admits a combinatorial triangulation which is one in which the links of the simplices are spheres. Such a triangulation is equivalent to a piecewise linear (PL) structure on the manifold. This is the content of the Triangulation Conjecture.

Amazingly, the question for all closed manifolds of dimension 5 and higher can be reduced to a question about the cobordism group $\Theta^3$ of homology 3-spheres and in particular, something about their Rokhlin invariants $\mu$. We have a short exact sequence $0 \to \ker \mu \to \Theta^3 \to \mathbb{Z}/2 \to 0$; this induces a LES where a particular class, the Kirby-Siebenmann obstruction to triangulating, is in the image of $\mu^*$. Galewski-Stern, Matumoto showed that studying these objects is not only necessary but also sufficient for determining whether a manifold admits a triangulation. There theorem can be rephrased:

Theorem: There exist non-triangulable manifolds of every dimension $n \geq 5$ if and only if the short exact sequence does not split.

Incredibly, using $\text{Pin}(2)$-equivariant Seiberg-Witten theory and stable homotopy theory, Ciprian Manolescu disproved the Triangulation Conjecture by proving:

Theorem (Manolescu): The short exact sequence does not split.

The history of the Triangulation Conjecture and these results is fascinating and deserves much more attention. But I think I’ll wrap it up here.