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.

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