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 , 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 be a framed -dimensional manifold. One of the consequences of being framed is that admits a fundamental class in stable homotopy. Stated more explicitly, there is a *stable* map

which induces an isomorphism in . One can construct such a map by imbedding inside a large euclidean space . The Thom-Pontryagin collapse map then runs

for the normal bundle of in . A choice of trivialization of this normal bundle (that is, a framing) allows us to identify with , and this gives a map

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 , the top cell stably splits off. That is, we can take the stable map and compose it with the crushing map (for instance, identifying with ) such that the composite

is an equivalence.

**2. Example: oriented 3-manifolds**

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

The Stiefel-Whitney class can be computed via the *Wu formula*. Namely, consider the composite:

given by the composite of the total Steenrod square and “integration:” pairing with the fundamental class. The operation is Poincaré dual to a class , called the **Wu class.** The Wu formula states that

In the oriented 3-manifold case, we can see that the Stiefel-Whitney classes are zero. The top class 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 by assumption of orientability. We just need to show that .

For this, we observe first that the degree -component of vanishes: applying Steenrod squares to elements in will never reach . The degree component of must vanish because otherwise would contain a term in degree one, and we are given that . Thus , and .

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 to a map

which is necessarily zero: is 2-connected, so that is 3-connected.

As an example, we find that has a cell structure with one cell in degrees zero through three. In the suspension , the last cell splits off.

**3. -connected -manifolds**

Now suppose is a manifold of dimension , which is -connected (for ). The study of such manifolds is closely related to the study of framed manifolds. Given a framed manifold of dimension , surgery theory allows us to modify it, while remaining in the same framed cobordism class, to make it -connected, although there are obstructions to making it -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 -connected -manifolds.

Anyway, the first thing we can say is that admits a decomposition *up to homotopy*

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

which is an isomorphism in homology for degrees (and in homotopy). The map is actually an isomorphism in homology for degrees by Poincaré duality, and we can choose a generator of

to attach a -cell to ; this produces an equivalence of with .

The benefit of this viewpoint is that only invariant up to homotopy of is given by the attaching map in . These homotopy groups can be understood in terms of the homotopy groups of spheres via the Hilton-Milnor theorem: one has a decomposition

where the copies of are imbedded in via the Whitehead products of the canonical generators .

Moreover, the fact that the cohomology ring of is significantly restricted—that is, the cup square on has to be unimodular—restricts the possible choice of element in .

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

Corollary 1Let be a stably framed, -connected -manifold. Then has the homotopy type of , where the attaching map is stably trivial.

**4. The Kervaire invariant**

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

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

inducing an isomorphism on homotopy, splits after applying . In particular, given any class , we can define a map

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

which pulls back the fundamental class of to . In particular, we can find a factorization in the following diagram:

The Kervaire invariant comes from the obstruction to desuspending the map to a map realizing the cohomology class .

As we’ve seen, given , we can realize by a map

which is generally not unique. The Kervaire invariant is supposed to be an obstruction to lifting this into under the map . As such, it would be nice if were the homotopy fiber of a map. At the prime , we can do this: there is a fibration sequence

called the EHP sequence. In other words, a map lifts (2-locally) to if and only if the composite is zero. We don’t need the fact that this is a 2-local fiber sequence, only that there is a map of spaces

called the **James-Hopf map.** It has the property that the push-forward in homology is an isomorphism. Note that the homology of each side is given by a tensor algebra (on a generator in degree or , respectively), but the map in homology is not an algebra map.

Now, we have

because is -connected. The identification can be made as follows: choose a generator of and check where it pulls back under a map .

We will use this obstruction-theoretic machinery to define a quadratic refinement of the cup square on .

Definition 2Let . We define by choosing realizing , and then taking the composite . We set for a generator.

We don’t actually need the James-Hopf map to define the form ; we could have looked at where the associated map pulls back the -dimensional class in . However, we need some nontrivial homotopy theory to see that it is actually well-defined. We’ll state it as a proposition.

Proposition 3The form is well-defined when .

*Proof:* We need to show that, although the map is not defined in terms of a cohomology class , the composite is defined **modulo 2.** In fact, we note that , as before, so that a cohomology class defines

on the five-skeleton of . Given a map , the extensions over form a torsor over maps by the Barratt-Puppe sequence.

In other words, the “indeterminacy” in the choice of the map (given a cohomology class) consists of the action by . So we need to show that given any

the composite induces zero in homology mod . But the whole point of the James-Hopf map is that it is a refinement of the Hopf invariant: that is, the map can be identified with the Hopf invariant of an element of . 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 has the property that pulls back the generator in to something divisible by . This proves that the form is well-defined.

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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