rho invariant of manifolds [I thought that I had already posted this question, but I couldn't find it in a search, so I apologize if I'm posting twice.]
Let $G$ be a finite group. Then the rational oriented bordism ring $\Omega_{2k-1}^{STop}(BG)\otimes{\mathbb Q}$ is trivial. As a result, if $N^{2k-1}$is an oriented Top manifold with fundamental group $G$, then there is a Top manifold $(M^{2k}, \partial M)$ with fundamental group $G$ such that $\partial M$ is the disjoint union of a number (say $m$) of copies of $N$. 
Atiyah-Singer defines the rho invariant of $N$ to be $$\rho(N)=\frac{1}{m}\cdot\mathrm{sig}_G(\widetilde{M}) \in {\mathbb Q} R^\pm(G)/I_G,$$ where $\pm=(-1)^k$ and $I_G$ is the ideal generated by the regular representation. 
My question is about $I_G$. Certainly the manifold $M$ might not be uniquely chosen. How do we get a well-defined rho invariant independent of this choice if we mod out by $I_G$?
 A: The standard argument for this in the smooth case is to use Novikov additivity for the G-signature to express the difference between $\rho$ calculated using a manifold $M$ and a manifold $M'$ as the G-signature of a closed manifold. Then the G-signature theorem of Atiyah-Singer shows that the G-signature of a free action on a closed manifold lies in $I_G$. 
The proof by Gilmer (Pacific J. Math 97 (1981), 105--114) of the G-signature theorem for finite $G$ has an argument using bordism theory for this fact. He works in the smooth category, but you only need bordism tensored with Q, which is isomorphic to the topological version tensored with Q. By the way, you need some kind of bordism argument to show that the G-signature theorem used in the standard argument would also apply in the topological case as well.
A: The generalization from smooth to tame topological action is in Wall's book on surgery. A nice explanation which also answers your question is given here:
http://www.map.mpim-bonn.mpg.de/Rho-invariant
