Rohlin’s theorem on the signature of four-manifolds

The signature ${\sigma(M)}$ of a ${4k}$ -dimensional compact, oriented manifold ${M}$ is a classical cobordism invariant of ${M}$ ; the so-called Hirzebruch signature formula states that ${\sigma(M)}$ can be computed as a complicated polynomial in the Pontryagin classes of the tangent bundle ${TM}$ (evaluated on the fundamental class of ${M}$ ). When ${M}$ is four-dimensional, for instance, we have

$\displaystyle \sigma(M) = \frac{p_1}{3}.$

This implies that the Pontryagin number ${p_1}$ must be divisible by three.

There are various further divisibility conditions that hold in special cases. Here is an important early example:

Theorem 1 (Rohlin) If ${M}$ is a four-dimensional spin-manifold, then ${\sigma(M)}$ is divisible by ${16}$ (and so ${p_1}$ by ${48}$ ).

I’d like to describe the original proof of Rohlin’s theorem, which relies on a number of tools from the 1950s era of topology. At least, I have not gotten a copy of Rohlin’s paper; the proof is sketched, though, in Kervaire-Milnor’s ICM address, which I’ll follow.

1. Obstruction theory

The first observation is that:

Proposition 2 If ${M}$ is a (connected) four-dimensional spin manifold, then ${TM}$ restricted to ${M \setminus \left\{p\right\}}$ is trivial for any ${p \in M}$ . (This is equivalent to the spin condition.)

To see this, note that the tangent bundle ${TM}$ comes from a principal ${\mathrm{Spin}(4)}$ -bundle ${P \rightarrow M}$ (which is not unique). The claim is that there is a section over ${M \setminus \left\{p\right\}}$ for any ${p}$ , which will imply that ${M \setminus \left\{p\right\}}$ is parallelizable.

There are a number of obstructions to finding a section of ${P}$ over ${M \setminus \left\{p\right\}}$ . They live in ${H^2(M \setminus \left\{p\right\}, \pi_1( \mathrm{Spin}(4)), H^3(M \setminus \left\{p\right\}, \pi_2(\mathrm{Spin}(4)) }$ , and ${H^4( M \setminus \left\{p\right\}, \pi_3(\mathrm{Spin}(4))}$ . These groups are all zero; the spin groups are simply connected and have zero ${\pi_2}$ (like all Lie groups). Also, ${M \setminus \left\{p\right\}}$ has no ${H^4}$ .

What does this mean? If we take a small neighborhood ${D}$ of ${p}$ which is isomorphic to the four-disk, then ${TM}$ is trivial on ${M \setminus D}$ and ${D}$ , and is uniquely determined by an element of ${\pi_3(\mathrm{SO}(4))}$ . The stable normal bundle is, similarly, determined by an element ${\alpha}$ of ${\pi_3(\mathrm{SO}) = \mathbb{Z}}$ .

The strategy of Rohlin’s theorem is the following:

Show that ${\alpha}$ is mapped to zero in the stable 3-stem (which is ${\mathbb{Z}/24}$ ) under the J-homomorphism.
Deduce that ${\alpha}$ is divisible by ${24}$ in ${\pi_3(\mathrm{SO})}$ .
Deduce that ${p_1}$ of the stable normal bundle is divisible by ${48}$ , which is equivalent to Rohlin’s theorem.

The first observation is actually mostly formal in retrospect, but it is a hugely important idea, because the image of J-homomorphism has been completely determined. (Incidentally, Rohlin’s theorem and its generalizations can be used to bound below the image of the J-homomorphism, by reversing the argument.) Since the ${J}$ -homomorphism in dimension three is a surjection

$\displaystyle \mathbb{Z} \rightarrow \mathbb{Z}/24,$

the second step is then clear. (This fact is not obvious, but it follows from Adams’s and Mahowald’s work; apparently Rohlin had to prove it himself, though.) Computing the Pontryagin classes as a function of an element of ${\pi_3(\mathrm{SO})}$ is classical.

2. The J-homomorphism and the Pontryagin-Thom construction

The classical J-homomorphism is the map

$\displaystyle \pi_r(SO(q)) \rightarrow \pi_{r+q}(S^q)$

which can be obtained as follows: there is a natural map ${\pi_r(SO(q)) \rightarrow \pi_r(\Omega^q S^q)}$ since every element of ${SO(q)}$ determines an element of ${\Omega^q S^q }$ by the action of ${SO(q)}$ on the sphere ${S^q}$ (as the one-point compactification of ${\mathbb{R}^q}$ ). Some (rather crude) notes on the J-homomorphism, describing a paper of Adams, can be found here.

Another way to think of it, which is equivalent (up to a sign), is the following. Suppose ${q > r}$ . There is a canonical trivialization on the normal bundle ${\nu_q}$ of ${S^r}$ (the trivialization that extends to ${D^{r+1}}$ ) in a high-dimensional imbedding ${S^r \hookrightarrow \mathbb{R}^{r+1} \hookrightarrow \mathbb{R}^{r + q}}$ . Any element ${\beta: \pi_r(SO(q))}$ determines a “twist” of this trivialization, a different isomorphism

$\displaystyle \beta: \mathbb{R}^q \simeq \nu_q.$

By the Pontryagin-Thom construction, this determines an element of the stable homotopy group ${\pi_{r}^s}$ .

Proposition 3 The element determined by the Pontryagin-Thom construction as above is precisely ${J}$ applied to ${\beta}$ .

I won’t prove this; if you unwind the definition of the Pontryagin-Thom collapse map and the J-homomorphism, it is visually clear.

As a result, we can give a criterion for when an element in ${\pi_r(SO(q))}$ goes to zero under the J-homomorphism: this happens precisely when there exists a manifold ${M}$ (of dimension ${r+1}$ ) such that ${\partial M = S^r}$ together with a framing of its stable normal bundle which restricts to the framing of ${S^r}$ . This is the Pontryagin-Thom theorem, essentially.

Consequently, we have:

Proposition 4 Let ${X}$ be any manifold with boundary ${S^r}$ . Choose a trivialization of the stable normal bundle of ${X}$ which restricts to an element ${\alpha \in \pi_r(SO)}$ (regarded as a framing of ${S^r}$ ). Then this is annihilated by the J-homomorphism.

In fact, this is now essentially tautological: the manifold ${S^r}$ , with the framing given by the element ${\alpha \in \pi_r(SO)}$ , is the (framed) boundary of ${X}$ . Consequently, ${\alpha}$ goes to zero under ${J}$ .

3. Rohlin’s theorem

We now have essentially all the ingredients to prove Rohlin’s theorem, at least if we believe a certain fact about the J-homomorphism.

Theorem 5 The ${J}$ -homomorphism ${\pi_3(\mathrm{SO}) \rightarrow \pi_3^s}$ is a surjection ${\mathbb{Z} \rightarrow \mathbb{Z}/24}$ .

In general, the order of the image of the J-homomorphism in dimensions ${4k-1}$ (the cases when ${\pi_\bullet(\mathrm{SO}) = \mathbb{Z}}$ , by Bott periodicity) is given by the denominator of ${\frac{B_{2k}}{4k}}$ ; this is a theorem of Mahowald and Adams.

Proof: Let ${M}$ be a four-dimensional spin-manifold. If ${p \in M }$ , we have seen that we can describe the stable normal bundle ${\nu}$ of ${M}$ by an element ${\alpha \in \pi_3(\mathrm{SO})}$ by trivializing ${\nu}$ away from ${p}$ . Alternatively, we choose a trivialization of ${\nu}$ on ${M}$ minus a small disk around ${p}$ , and restrict the trivialization to the boundary ${S^3}$ to get an element of ${\alpha \in \pi_3(\mathrm{SO}(3))}$ . By the discussion in the previous section, ${\alpha}$ must map to zero under ${J}$ , so it is divisible by 24.

Now the Pontryagin class ${p_1}$ of a vector bundle given by a clutching function ${\alpha}$ in this way are linear in ${\alpha}$ . In fact, I claim that up to sign, ${p_1}$ is given by ${2 \alpha}$ . If we prove this, it will follow that

$\displaystyle 48 \mid p_1(\nu) = - p_1(TM),$

which is Rohlin’s theorem.

To see this, note that we can collapse ${M \setminus D}$ to a point (for ${D}$ a small neighborhood of ${p}$ ) and using the trivialization ${\alpha}$ , collapse ${\nu}$ to a vector bundle on ${S^4}$ . This (stable) vector bundle is given precisely by the clutching function ${\alpha: S^3 \rightarrow \mathrm{SO}}$ .

So, to complete the argument, we may as well prove:

Lemma 6 The Pontryagin class ${p_1}$ of a stable vector bundle on ${S^4}$ as a function of the clutching function ${\alpha \in \pi_3(\mathrm{SO}) = \mathbb{Z}}$ is ${\pm 2 \alpha}$ .

To see this lemma, we note that ${p_1}$ is linear in ${\alpha}$ , and the image of ${p_1}$ is exactly ${2\mathbb{Z}}$ . To see this, note that the image of the second Chern class ${c_2}$ for complex vector bundles on ${S^4}$ is precisely ${\mathbb{Z}}$ (this is a special case of a theorem of Bott) and complexification

$\displaystyle \mathbb{Z} = \widetilde{KO}(S^4) \rightarrow \widetilde{K}(S^4) = \mathbb{Z}$

has image of index two. Since ${p_1}$ of a real vector bundle is ${-c_2}$ of the complexification, this completes the argument. $\Box$

The Kervaire-Milnor talk is not actually about deducing Rohlin’s theorem; it is about reversing the argument. Rohlin’s theorem was later generalized by the statement that the $\hat{A}$ –genus of a spin manifold of dimension $\equiv 4 \mod 8$ is an even integer; Atiyah and Singer, for instance, deduced this by interpreting it as an index. As a result, Kervaire and Milnor showed that if an element $\alpha \in \pi_r(\mathrm{SO})$ mapped to zero under $J$ , one could use the same Pontryagin-Thom argument to construct a manifold with only its top Pontryagin class nontrivial and determined by $\alpha$ ; the integrality of the $\hat{A}$ -genus now gives a strong divisibility condition on the top Pontryagin class and thus on $\alpha$ .