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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 $[]$ 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}.

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    Yes, but they do not distinguish between different diffeomorphism classes. I guess I was less than clear by what I meant.2015-08-17

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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.

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    @wqr: see http://en.wikipedia.org/wiki/Intersection_theory#Topological_intersection_form .2012-06-16
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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.

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    You are right that the signature is not a homotopy invariant, but a orientation preserving-homotopy invariant. But wording it the way that you did might confuse @wqr. It is better to point out, that one has to restrict only to homotopies which preserve the orientation.2013-03-03