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:

  1. Show that {\alpha} is mapped to zero in the stable 3-stem (which is {\mathbb{Z}/24}) under the J-homomorphism.
  2. Deduce that {\alpha} is divisible by {24} in {\pi_3(\mathrm{SO})}.
  3. 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.

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