Browder's book "Surgery on simply-connected manifolds" defines the Kervaire invariant in a very general setting. My question is: how does one get the more usual definition of the invariant for a framed (or Wu-orinented) manifold from Browder's general definition?

Browder's setting: For Poincare pairs $(X,Y)$ and $(A,B)$ with specified Spivak normal bundles, a normal map is a map $f: (X,Y) \to (A,B)$ that has degree 1 and is covered by a bundle map $b$ from one Spivak normal bundle to the other. Browder manages to define the $\mathbb Z /2$-valued Kervaire invariant $\sigma$ for any such map (no additional information needed), at least as long as $f^*$ takes the Wu class of $A$ to the Wu class of $X$. I find it hard to understand his definition.

Some simplification: For the sake of understanding, it should be safe to replace all the Poincare pairs with closed manifolds, and Spivak normal bundles with normal bundles. Then, a normal map becomes a degree-1 map of manifolds embedded in a high-dimensional Euclidean space that is covered by a bundle isomorphism of normal bundles. If the manifolds can be framed (or Wu-oriented), the Wu classes will be zero, so there's no reason to worry about them.

Question: How does this situation relate to the more usual situation of having a manifold with a framing or Wu orientation (which are, of course, necessary to define the Kervaire invariant in the usual way)? I'd imagine that there is some standard thing I could fix $(A,B)$ to be so that Browder's version of Kervaire invariant computes the usual Kervaire invariant of $(X,Y)$. I couldn't figure out what it should be. This choice of $(A,B)$ should somehow encode the Wu orientation on $(X,Y)$.

Please tell me if there are any mistakes in the above. Thank you!

  • $\begingroup$ I'm not familiar with the term "Wu orientation" (or "Wu oriented"). If $X$ is framed then surely $(A,B)$ can be taken to be $(D^n,S^{n-1})$. $\endgroup$ – Tom Goodwillie Nov 17 '10 at 4:19
  • $\begingroup$ @Tom: What would the map of degree 1 be? Is there some obvious way to construct one map for every framing of X? (As for Wu orientations, I think you can safely ignore them). $\endgroup$ – Ilya Grigoriev Nov 17 '10 at 7:50
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    $\begingroup$ I would have thought that you take $(A,B) = (D^n, \partial D^n)$. Then for any closed framed manifold $X$ take $f$ to be the map that collapses the complement of a ball, with the isomorphism $TX \cong f^* TD^n$ determined by the framing of $X$. $\endgroup$ – Oscar Randal-Williams Nov 17 '10 at 9:06
  • $\begingroup$ @Oscar: Yeah, that's probably it! I was confused about whether we can get any framing in this manner, but this seems clear to me now (as $f^∗ TD^n$ is a trivial bundle on X, the bundle isomorphism is precisely the same as a framing); thanks for straightening me out! Also, is there some deep meaning to the fact that this doesn't seem to work for manifolds with boundary, or am I missing something simple again? $\endgroup$ – Ilya Grigoriev Nov 17 '10 at 16:32

My algebraic theory of surgery gives the following approach to the definition of the Kervaire invariant. Let $Q_n(C,\gamma)$ be the Weiss twisted quadratic $Q$-group defined for any chain bundle $(C,\gamma)$. A spherical fibration $\nu:X \to BG(k)$ determines a chain bundle $(C(X),\gamma(\nu))$ with a Hurewicz-style group morphism $$h~:~\pi^S_{n+k}(T(\nu)) \to Q_n(C(X),\gamma(\nu))$$ from the stable homotopy groups of the Thom space $T(\nu)$. The image $h(\rho)$ of a stable homotopy class $\rho$ relates the evaluations on the Hurewicz-Thom image fundamental homology class $$[X]~=~[\rho] \in H_{n+k}(T(\nu))~=~H_n(X)$$ of the Steenrod squares of $X$ and the cup products with the Wu classes $v_r(\nu) \in H^r(X)$, verifying on the chain level the formula of Wu and Thom $$\langle v_r(\nu) \cup y,[X] \rangle = \langle Sq^r(y),[X] \rangle~(y\in H^{n-r}(X))~.$$ An $n$-dimensional geometric Poincare complex $X$ (e.g. an $n$-dimensional manifold) has a canonical class of pairs $(\nu_X:X \to BG(k),\rho_X:S^{n+k} \to T(\nu_X))$ with $\nu_X$ the Spivak normal fibration (= sphere bundle of the normal bundle of an embedding $X \subset S^{n+k}$ for a manifold $X$). A fibre homotopy trivialization $b:\nu_X \simeq *:X \to BG(k)$ (e.g. one determined by a framing of a manifold) determines a morphism $Q_n(C(X),\gamma(\nu_X)) \to {\mathbb Z}_2$ such that the image of $h(\rho_X)$ is the Kervaire invariant $K(X,b)\in {\mathbb Z}_2$. More generally, a Wu-orientation $b$ of $X$ (for which Browder's 1969 Annals paper The Kervaire invariant of framed manifolds and its generalization is a good reference) determines a morphism $Q_n(C(X),\gamma(\nu_X)) \to {\mathbb Z}_8$ such that the image of $h(\rho_X)$ is the Brown generalized Kervaire invariant $K(X,b)\in {\mathbb Z}_8$. Most of this is already explained in my paper Algebraic Poincare cobordism.


I found a useful reference on Doug Ravenel's website that first explains Browder's definition and then relates it to Kervaire's. Here it is (pp. 142-143; the file's a whopping 5MB, but just because it's scanned in).

Also, hi, and looking forward to your presentation!

  • $\begingroup$ Thanks, Elizabeth! I like this reference a lot, although I don't think it answers this specific question (Oscar did, though). $\endgroup$ – Ilya Grigoriev Nov 17 '10 at 16:29

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