# 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 $$M$$.

(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?

• I would look at Wall's book (search for "multisignature" in the index) and also Wall's paper "On the classification of hermitian forms VI. Group rings". – Igor Belegradek Jan 23 at 12:13

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.
• Is there an explicit example in the literature s.t. the action of $L(\pi_1 M)$ on $S(M)$ is nontrivial, but the homeomorphism type of the domain manifold is not changed? – student Jan 25 at 12:15