I’ve been struggling lately with Kervaire’s paper “A manifold which does not admit any differentiable structure.” The paper defines the Kervaire invariant of a 4-connected combinatorial 10-manifold and shows that it is automatically zero on smooth manifolds. He then constructs an example of a 4-connected combinatorial 10-manifold whose Kervaire invariant is ${1}$, concluding that it can admit no smooth structure.

The definition of the Kervaire invariant given in the paper is a little complicated, and I’d like to work through it carefully in this post.

1. Generalities on framed manifolds

Let ${M}$ be a framed ${k}$-dimensional manifold. One of the consequences of being framed is that ${M}$ admits a fundamental class in stable homotopy. Stated more explicitly, there is a stable map

$\displaystyle \phi_M: \Sigma^\infty S^{k} \rightarrow \Sigma^\infty_+ M$

which induces an isomorphism in ${H_{k}}$. One can construct such a map by imbedding ${M}$ inside a large euclidean space ${\mathbb{R}^{N+k}}$. The Thom-Pontryagin collapse map then runs

$\displaystyle S^{N+k} \rightarrow \mathrm{Th}( \nu),$

for ${\nu}$ the normal bundle of ${M}$ in ${\mathbb{R}^{N+k}}$. A choice of trivialization of this normal bundle (that is, a framing) allows us to identify ${\mathrm{Th}(\nu)}$ with ${S^{N} \wedge M_+}$, and this gives a map

$\displaystyle S^{N+ k}\rightarrow S^N \wedge M_+,$

which is an isomorphism in top homology. This gives the desired stable map.

A consequence of this observation is that, in a cell decomposition of ${M}$, the top cell stably splits off. That is, we can take the stable map ${\phi_M : S^{k} \rightarrow M }$ and compose it with the crushing map ${M \rightarrow S^{k}}$ (for instance, identifying ${S^{k}}$ with ${M/ M \setminus \mathrm{disk}}$) such that the composite

$\displaystyle S^{k} \stackrel{\phi_M}{\rightarrow} M \rightarrow S^{k}$

is an equivalence.

2. Example: oriented 3-manifolds

As an example, any oriented 3-manifold ${M}$ is parallelizable and thus stably framed by the following argument.

The Stiefel-Whitney class ${w_i(M)}$ can be computed via the Wu formula. Namely, consider the composite:

$\displaystyle H^*(M; \mathbb{Z}/2) \stackrel{\mathrm{Sq}}{\rightarrow} H^*(M; \mathbb{Z}/2) \stackrel{\int}{\rightarrow} \mathbb{Z}/2$

given by the composite of the total Steenrod square ${\mathrm{Sq}}$ and “integration:” pairing with the fundamental class. The operation is Poincaré dual to a class ${v \in H^*(M; \mathbb{Z}/2)}$, called the Wu class. The Wu formula states that

$\displaystyle w(M) = \mathrm{Sq} v.$

In the oriented 3-manifold case, we can see that the Stiefel-Whitney classes are zero. The top class ${w_3(M)}$ is the reduction of the Euler class, which is a multiple of the Euler characteristic and thus zero on an odd-dimensional manifold by Poincaré duality. The class ${w_1 (M) =0}$ by assumption of orientability. We just need to show that ${w_2(M) = 0}$.

For this, we observe first that the degree ${2}$-component of ${v}$ vanishes: applying Steenrod squares to elements in ${H^1}$ will never reach ${H^3}$. The degree ${1}$ component of ${v}$ must vanish because otherwise ${\mathrm{Sq} v}$ would contain a term in degree one, and we are given that ${w_1(M) =0}$. Thus ${v = 1}$, and ${w(TM) = \mathrm{Sq} v = 1}$.

The upshot of all this is that the tangent bundle of an oriented 3-manifold has trivial Stiefel-Whitney classes, and consequently admits a spin structure. That is, we can lift the classifying map ${M \rightarrow BSO(3)}$ to a map

$\displaystyle M \rightarrow B\mathrm{Spin}(3),$

which is necessarily zero: ${\mathrm{Spin}(3) \simeq S^3}$ is 2-connected, so that ${B \mathrm{Spin}(3)}$ is 3-connected.

As an example, we find that ${\mathbb{RP}^3}$ has a cell structure with one cell in degrees zero through three. In the suspension ${\Sigma^\infty \mathbb{RP}^3_+}$, the last cell splits off.

3. ${(k-1)}$-connected ${2k}$-manifolds

Now suppose ${M}$ is a manifold of dimension ${2k}$, which is ${k-1}$-connected (for ${k \geq 2}$). The study of such manifolds is closely related to the study of framed manifolds. Given a framed manifold of dimension ${2k}$, surgery theory allows us to modify it, while remaining in the same framed cobordism class, to make it ${k-1}$-connected, although there are obstructions to making it ${k}$-connected (and thus a homotopy sphere). Conversely, the preface to vol. III of Milnor’s collected works describe how the discovery of exotic spheres came out of trying to produce examples of ${k-1}$-connected ${2k}$-manifolds.

Anyway, the first thing we can say is that ${M}$ admits a decomposition up to homotopy

$\displaystyle M \simeq \bigvee S^k \cup_f e^{2k};$

that is, ${M}$ can be obtained by attaching a ${2k}$-cell to a wedge of ${k}$-spheres. In order to see this, use the Hurewicz theorem to produce a map

$\displaystyle \bigvee S^k \rightarrow M$

which is an isomorphism in homology for degrees ${\leq k}$ (and in homotopy). The map is actually an isomorphism in homology for degrees ${\leq 2k-1}$ by Poincaré duality, and we can choose a generator of

$\displaystyle \pi_{2k}(M, \bigvee S^k) \simeq \mathbb{Z}$

to attach a ${2k}$-cell to ${\bigvee S^k}$; this produces an equivalence of ${\bigvee S^k \cup e^{2k} }$ with ${M}$.

The benefit of this viewpoint is that only invariant up to homotopy of ${M}$ is given by the attaching map in ${\pi_{2k-1}(\bigvee S^k)}$. These homotopy groups can be understood in terms of the homotopy groups of spheres via the Hilton-Milnor theorem: one has a decomposition

$\displaystyle \pi_{2k-1}( \bigvee_{i=1}^r S^k) \simeq \bigoplus_{i=1}^r \pi_{2k-1}(S^k) \oplus \bigoplus_{i < j} \pi_{2k-1}(S^{2k-1}),$

where the copies of ${\pi_{2k-1}(S^{2k-1}) \simeq \mathbb{Z}}$ are imbedded in ${\pi_{2k-1}( \bigvee_{i=1}^r S^k )}$ via the Whitehead products of the canonical generators ${\iota_i, \iota_j \in \pi_k(\bigvee S^k)}$.

Moreover, the fact that the cohomology ring of ${M}$ is significantly restricted—that is, the cup square on ${H^{k}}$ has to be unimodular—restricts the possible choice of element in ${\pi_{2k-1}( \bigvee S^k)}$.

If we combine this with the previous section, we find:

Corollary 1 Let ${M}$ be a stably framed, ${k-1}$-connected ${2k}$-manifold. Then ${M}$ has the homotopy type of ${\bigvee S^k \cup e^{2k}}$, where the attaching map ${S^{2k-1} \rightarrow \bigvee S^k}$ is stably trivial.

4. The Kervaire invariant

The Kervaire invariant arises from a quadratic refinement of the intersection form on ${H_k}$ of a ${2k}$-dimensional framed manifold, where ${k}$ is odd. The Kervaire invariant is the (algebraically defined) Arf invariant of this quadratic refinement.

In order to define this, let’s suppose that ${M}$ is ${k-1}$-connected and stably framed. We saw in the previous section that the inclusion

$\displaystyle \bigvee S^k \rightarrow M$

inducing an isomorphism on homotopy, splits after applying ${\Sigma^\infty}$. In particular, given any class ${x \in H^k(M; \mathbb{Z})}$, we can define a map

$\displaystyle \Sigma^\infty_+ M \rightarrow \Sigma^\infty S^k$

realizing this cohomology class. Using the Freudenthal suspension theorem, we can define a map

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

which pulls back the fundamental class of ${\Sigma S^k}$ to ${x}$. In particular, we can find a factorization in the following diagram:

The Kervaire invariant comes from the obstruction to desuspending the map ${\Sigma M \rightarrow \Sigma S^k}$ to a map ${M \rightarrow S^k}$ realizing the cohomology class ${x}$.

As we’ve seen, given ${x \in H^k(M; \mathbb{Z})}$, we can realize ${x}$ by a map

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

which is generally not unique. The Kervaire invariant is supposed to be an obstruction to lifting this into ${S^k}$ under the map ${S^k \rightarrow \Omega \Sigma S^k}$. As such, it would be nice if ${S^k \rightarrow \Omega \Sigma S^k}$ were the homotopy fiber of a map. At the prime ${2}$, we can do this: there is a fibration sequence

$\displaystyle S^k \rightarrow \Omega \Sigma S^k \rightarrow \Omega \Sigma S^{2k}$

called the EHP sequence. In other words, a map ${M \rightarrow \Omega \Sigma S^k}$ lifts (2-locally) to ${S^k}$ if and only if the composite ${M \rightarrow \Omega \Sigma S^{2k}}$ is zero. We don’t need the fact that this is a 2-local fiber sequence, only that there is a map of spaces

$\displaystyle \Omega \Sigma S^k \rightarrow \Omega \Sigma S^{2k} ,$

called the James-Hopf map. It has the property that the push-forward in homology ${H_{2k}( \cdot; \mathbb{Z}/2)}$ is an isomorphism. Note that the homology of each side is given by a tensor algebra (on a generator in degree ${k}$ or ${2k}$, respectively), but the map in homology is not an algebra map.

Now, we have

$\displaystyle [M, \Omega \Sigma S^{2k}] = H^{2k}(M; \mathbb{Z}) \simeq \mathbb{Z},$

because ${\Omega \Sigma S^{2k}}$ is ${(2k-1)}$-connected. The identification can be made as follows: choose a generator of ${H^{2k}( \Omega \Sigma S^{2k}; \mathbb{Z}) \simeq \mathbb{Z}}$ and check where it pulls back under a map ${M \rightarrow \Omega \Sigma S^{2k}}$.

We will use this obstruction-theoretic machinery to define a quadratic refinement of the cup square on ${H^{k}}$.

Definition 2 Let ${x \in H^k(M; \mathbb{Z})}$. We define ${q(x) \in \mathbb{Z}/2}$ by choosing ${M \rightarrow \Omega \Sigma S^k}$ realizing ${x}$, and then taking the composite ${f: M \rightarrow \Omega \Sigma S^{2k}}$. We set ${q(x) = f^*(y) ( \mathrm{mod} 2)}$ for ${y \in H^{2k}( \Omega \Sigma S^{2k}; \mathbb{Z})}$ a generator.

We don’t actually need the James-Hopf map to define the form ${q}$; we could have looked at where the associated map ${M \rightarrow \Omega \Sigma S^{k}}$ pulls back the ${2k}$-dimensional class in ${H^{*}( \Omega \Sigma S^{k}; \mathbb{Z}/2)}$. However, we need some nontrivial homotopy theory to see that it is actually well-defined. We’ll state it as a proposition.

Proposition 3 The form ${q}$ is well-defined when ${k \neq 1, 3, 7}$.

Proof: We need to show that, although the map ${M \rightarrow \Omega \Sigma S^k}$ is not defined in terms of a cohomology class ${x \in H^k(M; \mathbb{Z})}$, the composite ${M \rightarrow \Omega \Sigma S^k \rightarrow \Omega \Sigma S^{2k}}$ is defined modulo 2. In fact, we note that ${M \simeq \bigvee S^k \cup e^{2k}}$, as before, so that a cohomology class ${x \in H^k(M; \mathbb{Z})}$ defines

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

on the five-skeleton ${\bigvee S^k}$ of ${M}$. Given a map ${\bigvee S^k \rightarrow \Omega \Sigma S^k}$, the extensions over ${M}$ form a torsor over maps ${S^{2k} \rightarrow \Omega S^6}$ by the Barratt-Puppe sequence.

In other words, the “indeterminacy” in the choice of the map ${M \rightarrow \Omega \Sigma S^k}$ (given a cohomology class) consists of the action by ${\pi_{2k}( \Omega \Sigma S^k)}$. So we need to show that given any

$\displaystyle S^{2k} \rightarrow \Omega \Sigma S^k ,$

the composite ${S^{2k} \rightarrow \Omega \Sigma S^k \rightarrow \Omega \Sigma S^{2k}}$ induces zero in homology mod ${2}$. But the whole point of the James-Hopf map is that it is a refinement of the Hopf invariant: that is, the map ${\pi_{2k}(\Omega \Sigma S^{k}) \rightarrow \pi_{2k}(\Omega \Sigma S^{2k}) \simeq \mathbb{Z}}$ can be identified with the Hopf invariant of an element of ${\pi_{2k+1}(S^{k+1})}$. The Hopf invariant is never odd for such an element, by Adams’s solution to the Hopf invariant one problem.

We conclude from this that any map ${S^{2k} \rightarrow \Omega \Sigma S^{k}}$ has the property that ${S^{2k} \rightarrow \Omega \Sigma S^k \rightarrow \Omega \Sigma S^{2k}}$ pulls back the generator in ${H^{2k}}$ to something divisible by ${2}$. This proves that the form is well-defined. $\Box$

The next thing to check is that the form is actually a quadratic refinement of the cup square. I’ll do this in the next post.