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!

X? (As for Wu orientations, I think you can safely ignore them). $\endgroup$ – Ilya Grigoriev Nov 17 '10 at 7:50X, 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