This is the second post in a series on Kervaire’s paper “A manifold which does not admit any differentiable structure.” In the previous post, we described a form on the middle cohomology of a $k-1$-connected $2k$-dimensional manifold, for $k \neq 1, 3, 7$. In this post, we can define the Kervaire invariant of such a framed manifold, by showing that this defines a form. I’ll try to sketch the proof that there is no framed manifold of Kervaire invariant one in dimension 10.

1. The form $q$ is a quadratic refinement

Let’s next check that the form ${q: H^k(M; \mathbb{Z}) \rightarrow \mathbb{Z}/2}$ defined in the previous post (we’ll review the definition here) is actually a quadratic refinement of the cup product. Precisely, this means that for ${x, y \in H^k(M; \mathbb{Z})}$, we want

$\displaystyle q(x+y) - q(x) - q(y) = (x \cup y)[M].$

In particular, this implies that ${q}$ descends to a function on ${H^k(M; \mathbb{Z}/2)}$, as it shows that ${q}$ of an even class is zero in ${\mathbb{Z}/2}$. The associated quotient map ${q: H^k(M; \mathbb{Z}/2) \rightarrow \mathbb{Z}/2}$ is, strictly speaking, the quadratic refinement.

In order to do this, let’s fix ${x, y \in H^k(M; \mathbb{Z}/2)}$. As we saw last time, these can be obtained from maps

$\displaystyle M \rightarrow \Omega \Sigma S^k$

by pulling back the generator in degree ${k}$. Let ${f_x}$ be a map associated to ${x}$, and let ${f_y}$ be a map associated to ${y}$. We then have that

$\displaystyle q(x) = f_x^*(u_{2k}) [M], \quad q(y) = f_y^*(u_{2k})[M],$

for ${u_{2k}}$ the generator of ${H^{2k}(\Omega \Sigma S^k; \mathbb{Z}/2)}$. As we saw, this was equivalent to the definition given last time.

In order to compute ${q(x+y)}$, we need a map ${f_{x+y}: M \rightarrow \Omega \Sigma S^k}$ realizing ${x+y}$. In fact, we can use the composite

$\displaystyle M \stackrel{\Delta}{\rightarrow} M \times M \stackrel{f_x, f_y}{\rightarrow} \Omega \Sigma S^k \times \Omega \Sigma S^k \rightarrow \Omega \Sigma S^k;$

that is, we can use the monoidal product on ${\Omega \Sigma S^k}$. The claim that this map pulls back the generator of ${H^k(\Omega \Sigma S^k)}$ to ${x + y}$ is formal once one notes that ${H^*(\Omega \Sigma S^k)}$ is a Hopf algebra from this monoidal product, and the generator in degree ${k}$ is primitive because it is of minimal degree.

So the last thing we need to do is to understand ${f_{x+y}^* (u_{2k})}$, where ${f_{x+y}}$ is constructed as claimed. Equivalently, we need to understand where ${u_{2k}}$ pulls back to under

$\displaystyle \Omega \Sigma S^k \times \Omega \Sigma S^k \rightarrow \Omega \Sigma S^k.$

Let ${u_k \in H^*(\Omega \Sigma S^k; \mathbb{Z})}$ be the generator. Then the claim is that under the pull-back of the multiplication,

$\displaystyle u_{2k} \mapsto u_{2k} \otimes 1 + 1 \otimes u_{2k} + u_k \otimes u_k, \ \ \ \ \ (1)$

and pulling this back under ${(f_x, f_y)}$ would give

$\displaystyle f_{x+y}^*(u_{2k}) = f_x^*(u_{2k}) + f_y^*(u_{2k}) + f_x^*(u_k) \otimes f_y^*(u_k), \ \ \ \ \ (2)$

which is precisely the claim of the lemma.

In order to prove this, we need to understand better the Hopf algebra structure of ${H^*(\Omega \Sigma S^k)}$. Fortunately, this is equivalent to understanding the Pontryagin product structure on ${H_*(\Omega \Sigma S^k)}$, which by the theorem of Bott and Samelson is a tensor algebra ${T(w_k)}$ on a generator ${w_k}$ in degree ${k}$ (dual to ${u_k}$). The coproduct of ${u_{2k}}$ dualizes to understanding the different decompositions of ${w_k^2}$ as a product, and a little linear algebra gives (1).

What we’ve now done is to construct, on a ${(k-1)}$-connected ${2k}$-manifold which admits a framing, a quadratic refinement ${q}$ of the intersection pairing.

Definition 4 The Arf invariant of ${q}$ is the value (either ${0}$ or ${1}$) that ${q}$ assumes more often. The Kervaire invariant of a framed ${(k-1)}$-connected ${2k}$-manifold is the Arf invariant of the quadratic refinement ${q}$.

2. The main results

The main results of Kervaire’s paper are:

Theorem 5 (Kervaire) A smooth ${4}$-connected framed ${10}$-manifold has Kervaire invariant zero. However, there exists a combinatorial 4-connected ${10}$-manifold with Kervaire invariant one.

We note that this invariant is a homotopy invariant.

There is a bit of a wrinkle in what I’ve been saying, since so far I have not explained what the Kervaire invariant of a combinatorial 4-connected 10-manifold is. This is something Kervaire defines in his paper in exactly the same way. That is, he shows that any class in ${H^5(M; \mathbb{Z})}$ can be realized by a map ${M \rightarrow \Omega S^6}$. This is not obvious. We still find, by Poincaré duality, that

$\displaystyle M \simeq \bigvee S^5 \cup e^{10},$

and the obstruction to realizing a cohomology class from a map into ${\Omega S^6}$ is an element in ${\pi_9( \Omega S^6) = \pi_{10}(S^6) = 0}$; consequently any cohomology class can be realized this way. This additional fact is needed to prove existence of the above quadratic refinement. In the framed smooth case, we could prove that cohomology classes in ${H^5}$ were realized by maps into ${\Omega S^6}$ by using the stable splitting of ${M}$.

The major conclusion from this is:

Corollary 6 There exists a 4-connected combinatorial 10-manifold ${M}$ which is not homotopy equivalent to any smooth 10-manifold.

Proof: In fact, take for ${M}$ the combinatorial 10-manifold with Kervaire invariant one. Then if ${M}$ had the homotopy type of a smooth 10-manifold ${M'}$, the claim is that ${M'}$ would have Kervaire invariant zero, which is a contradiction. To see this, we need only see that ${M'}$ admits a framing: in fact, any 4-connected smooth 10-manifold ${M'}$ admits a framing.

This in turn is a consequence of the following argument. We note that ${M \simeq \bigvee S^5 \cup e^{10}}$. Since

$\displaystyle \pi_5(SO) = 0,$

by Bott periodicity, we conclude that the tangent bundle of ${M}$ is stably trivial (even trivial) over the 5-skeleton. Observe that ${M - \ast}$ is homotopy equivalent to the 5-skeleton, so that the tangent bundle of ${M}$ is stably framed over ${M - \ast}$. The obstruction to extending the framing over ${M}$ is an element of ${\pi_9(SO) \simeq \mathbb{Z}/2}$.

Now, the claim is that this obstruction is killed by the ${J}$-homomorphism

$\displaystyle \pi_9(SO ) \rightarrow \pi_9^{st} .$

Another way of saying this is that ${M - \ast}$ is framed, so the boundary ${S^9}$ of a small ball surrounding ${\ast}$ gets a framing, classified by the obstruction class in ${\pi_9(SO)}$. To say that this class is killed by ${J}$ is to say that this framed ${S^9}$ is a framed boundary: but that’s easy, take ${M - \mathrm{disk}}$. Here, one uses the geometric interpretation of the ${J}$-homomorphism, as the map that turns framed spheres—framed by an element of ${\pi_*(SO)}$—into framed manifolds.

However, by Adams’s results in “On the groups ${J(X)}$ IV,” the ${J}$-homomorphism is injective ${\pi_9(SO) \rightarrow \pi_9^{st}}$. It follows that the obstruction vanishes, so the manifold is actually (stably) framed. $\Box$

The above argument via the ${J}$-homomorphism seems a little ad hoc, but it’s one of the key observations that Kervaire and Milnor make in “Groups of homotopy spheres I” to argue that a homotopy sphere is stably parallelizable.

3. Sketch of a proof

Now let’s sketch the proof of Kervaire’s theorem. We need to show that any framed ${4}$-connected ${10}$-manifold has Kervaire invariant zero. Although we have not shown it, the Kervaire invariant is an actually an invariant of framed cobordism. That is, any framed ${10}$-manifold can be made framed cobordant (using surgery) to a framed ${4}$-connected one, to which we can apply Kervaire’s construction to get an invariant in ${\mathbb{Z}/2}$. In fact, the Kervaire invariant is defined as a map

$\displaystyle \pi_{10}^s \rightarrow \mathbb{Z}/2.$

Kervaire’s theorem states that this map is zero. How can this be proved? The Kervaire invariant of an exotic (i.e., homotopy) ${10}$-sphere is clearly zero, so it would suffice to prove:

Theorem 7 (Kervaire) Every framed ${10}$-manifold is framed cobordant to a homotopy sphere.

In fact, this is true with ${10}$ replaced by any dimension except a small finite set: this is a consequence of surgery theory and the solution to the general Kervaire invariant problem.

Proof: It is only of interest to consider ${\pi_{10}^s \rightarrow \mathbb{Z}/2}$ on the 2-component of ${\pi_10^s}$. It is known that on the 2-components, every element in ${\pi_{10}^s}$ is a product of ${\eta \in \pi_1^s}$ (the Hopf map) and an element of ${\pi_9^s}$. This is a nontrivial result, but it can be proved by computing the Adams spectral sequence out to that dimension and using the multiplicative structure: the differentials vanish in that range. In geometric manifold, this means that our framed ${10}$-manifold ${M}$ can be assumed to be of the form ${N \times S^1}$, for ${N}$ an appropriate framed ${9}$-manifold.

Anyway, Milnor and Kervaire prove (using surgery again) that any odd-dimensional framed manifold is framed cobordant to a homotopy sphere. Since I don’t yet understand surgery very well, and since I will be giving my next Kan seminar talk on the subject, I’ll defer a discussion of this fact to a later post. It tells us that ${N}$ can be assumed to be an exotic ${9}$-sphere.

In other words, we find that any framed 10-manifold (in the 2-component) is of the form ${N \times S^1}$ for ${N}$ an exotic ${9}$-sphere. We’d like to make this into an exotic ${10}$-sphere. But we can do framed surgery on the generator of ${\pi_1(N \times S^1)}$. Since surgery on ${S^p \times S^1}$ (for the generator of ${\pi_1}$) produces an ${S^{p+1}}$, this operation on ${N \times S^1}$ produces a homotopy ${10}$-sphere. $\Box$

Finally, Kervaire describes a construction of a topological 10-manifold with Kervaire invariant one. Here the strategy is to “plumb” two copies of the tangent bundle of $S^5$ together; the boundary is an exotic sphere. Coning off this exotic sphere gives a topological manifold which he checks is of Kervaire invariant one; consequently it is not smoothable. I’ll try to say a little more about this in a later post.