Signature of a manifold as an invariant

Question

Could you help me to see why signature is a HOMOTOPY invariant? Definition is below (from Stasheff) The \emph{signature (index)} $\sigma(M)$ of a compact and oriented $n$ manifold $M$ is defined as follows. If $n=4k$ for some $k$, we choose a basis $\{a_1,...,a_r\}$ for $H^{2k}(M^{4k}, \mathbb{Q})$ so that the \emph{symmetric} matrix $[<a_i \smile a_j, \mu>]$ is diagonal. Then $\sigma (M^{4k})$ is the number of positive diagonal entries minus the number of negative ones. Otherwise (if $n$ is not a multiple of 4) $\sigma(M)$ is defined to be zero \cite{char}.

Answer 1

You should be using a more invariant definition of the signature. First, cohomology and Poincaré duality are both homotopy invariant. It follows that the abstract vector space $H^{2k}$ equipped with the intersection pairing is a homotopy invariant. Now I further claim that the signature is an invariant of real vector spaces equipped with a nondegenerate bilinear pairing (this is just Sylvester's law of inertia). So after tensoring with $\mathbb{R}$ the conclusion follows.

Answer 2

I think that the "absolute value" of the signature is a homotopy invariant, not the signature itself! Indeed, $\sigma(-M) = -\sigma(M)$ ($-M$ is the manifold with the opposite orientation). And of course $M$ and $-M$ are diffeomorphic: the identity is a (reversing orientation) diffeomorphism from M to -M.