Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
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 $(D^n,S^{n-1})$. –  Tom Goodwillie Nov 17 '10 at 4:19
    
@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). –  Ilya Grigoriev Nov 17 '10 at 7:50
3  
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 $X$. –  Oscar Randal-Williams Nov 17 '10 at 9:06
    
@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? –  Ilya Grigoriev Nov 17 '10 at 16:32

2 Answers 2

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!

share|improve this answer
    
Thanks, Elizabeth! I like this reference a lot, although I don't think it answers this specific question (Oscar did, though). –  Ilya Grigoriev Nov 17 '10 at 16:29

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.