Multisignature and homeomorphism type In classical surgery theory, there is a map
$$L_{n+1}(\pi_1M)\to S(M^n)$$
Element in $L_{n+1}(\pi_1M)$ is realized as surgery obstruction of a surgery problem to $M\times I$ with one boundary piece the identity map and the other a homotopy equivalence (Wall's realization). The map is defined by sending the element in $L$ to the boundary piece which is a homotopy equivalence (a structure on $M$).It is NOT clear to me if the domain manifold of the structure is homeomorphic to $M$.
In   Homology spheres and fundamental group 
(when $n+1$ is even) Danny Ruberman commented that "The effect of the action is to change the multisignature, and hence it changes the homeomorphism type of M"
What's the correct reference if i want to understand some details about the effect of Wall's realization on multisignature?
 A: There is perhaps some confusion over the terminology. Wall (chapter 13A) uses the term multisignature to denote a collection of invariants of certain Hermitian forms over group rings, giving rise to a function from $L_{2k}(\pi) \to \mathbb{Z}^n$. In that chapter, he interprets the multisignature in terms of equivariant signatures. Such signatures occur in many places in geometric topology, for instance as the Tristram-Levine signatures and Casson-Gordon invariants of knots.
One main use of the multisignature comes in the study of odd-dimensional manifolds. Given such a manifold $Y^{2k-1}$ and a finite regular covering with covering group $G$, classified by a homomorphism $\phi:\pi_(Y) \to G$, then one shows that some multiple (disjoint union) $N\cdot(Y,\phi)$ extends to a $2k$-manifold $(X,\Phi)$. Then (suitably normalized) the equivariant signature of $(X,\Phi)$ is an invariant $\rho(Y,\phi)$.
This can be packaged in various ways, see for instance chapter 14E in Wall, where for a manifold with finite fundamental group $G$, one obtains an invariant $\rho$. Technically one has to choose an identification of $\pi_1(Y)$ with G, and $\rho$ depends on this identification. A priori, the fact that one gets an invariant depends on the Atiyah-Singer index theorem, and so is really an invariant up to diffeomorphism. But a bordism argument (Wall 14B) shows that it is actually a homeomorphism invariant.
Putting these notions together, the way that the invariant $\rho$ is defined tells you that if one acts on (the identity map from $Y$ to itself) by an element $A \in L_{2k}(\pi_1(Y))$ then the invariant $\rho$ of the manifold $Y'$ at the other end of the resulting normal cobordism satisfies $\rho(Y') - \rho(Y) = $ multisignature of $A$. If this multisignature is non-trivial, then there is no homeomorphism of $Y$ and $Y'$ that respects the identification of their fundamental groups given by the normal cobordism. If you want to argue that they are simply not homeomorphic you need to look a little more closely.
