In his paper "Comparing analytic assemby maps", J. Roe considers a proper and cocompact action of a countable group $\Gamma$ on a metric space $X$. He constructs the Hilbert $C^*_r(\Gamma)$-module $L^2_\Gamma(X)$ as the simultaneous completion of the $\mathbb C\Gamma$-module $C_c(X)$ under the $\mathbb C\Gamma$-valued inner product given by the formula $$ \langle\psi\mid\psi'\rangle_\Gamma=\sum_{\gamma\in\Gamma}\langle\psi\cdot g\mid\psi'\rangle\,\gamma. $$ He shows (Lemmas 2.1-3) that the C$^*$-algebra of $C^*_r(\Gamma)$-compact operators $\mathbb K_{C^*_r(\Gamma)}(L^2_\Gamma(X))$ is isomorphic to $C^*_\Gamma(X)$, the $\Gamma$-invariant uniform Roe algebra.

All of this is fine, but there seems to be a problem with the following claims, and this is what my question is about.

Roe claims that $\mathbb K_{C^*(\Gamma)}(L^2_\Gamma(X))$ (and hence $C^*_\Gamma(X)$) is Morita equivalent to $C^*_r(\Gamma)$. I fail to see this and my question is: Is this true? If so, how to prove it? If not, how to disprove it?

Looking at Roe's arguments, he deduces the above statement from the more general claim that for any (countably generated) Hilbert $C^*_r(\Gamma)$-module $E$, $\mathbb K_{C^*(\Gamma)}(E)$ is Morita equivalent to $C^*_r(\Gamma)$. What is true is that it is Morita equivalent to $\langle E\mid E\rangle$ (with $E$ setting up the equivalence), that is immediate. The general statement that Roe makes however seems to be false unless $E$ is (right-) full. I find it hard to believe that it is true in general.

Concerning the particular case of $E=L^2_\Gamma(X)$, it would be great see that this is in fact a full HIlbert $C^*_r(\Gamma)$-module. I fail to see this, however. One computes in case of the above example: $$ \langle\delta_m\mid\delta_n\rangle=(m-n,+)+(m+n,-), $$ so the impression is that perhaps the closed ideal generator by the inner products is not the entire algebra.


1 Answer 1


To define the Roe algebra abstractly, we need a suitable covariant representation of the $\Gamma$-$C^\ast$-algebra $(C_0(X), \Gamma)$ on some Hilbert space $H$. In concrete situations, one usually takes some $L^2$-space on $X$. However, if the action of $\Gamma$ on $X$ is not free, then we need to be more careful about the representation of $\Gamma$. In particular, using $L^2(X, \mu)$ with an invariant measure $\mu$ is not quite appropriate if there are non-trivial stabilizers (see example of a trivial action below).

Moreover, I think in the mentioned paper the author had the "usual" Roe algebra in mind, that is not the uniform Roe algebra. This means, if $X$ is discrete, then we need to use e.g. $\bigoplus_{\mathbb{N}} \ell^2(X)$ rather than just $\ell^2(X)$ to define the Roe algebra. But—at least for the question about fullness of the Hilbert module $E = L^2_\Gamma(X)$—the relevant problem here seems to be about the $\Gamma$-representation, not about the representation of $C_0(X)$, so I'm going to ignore this aspect for now.

The free case

Suppose, in addition, that the action of $\Gamma$ on $X$ is free. Then the following argument should show that $E = L^2_\Gamma(X)$ is a full $C^*_r\Gamma$-module: Decompose $X = \bigsqcup_{g \in \Gamma} gF$, where $F$ is a fundamental domain. Now choose a $L^2$-unit vector $\psi \in C_c(X)$ with $\operatorname{supp}(\psi) \subseteq F$. Then $\langle \psi \mid \psi \rangle_\Gamma = \langle \psi \mid \psi\rangle\ 1 = 1$ because all non-trivial $\Gamma$-translates of $\psi$ are orthogonal to $\psi$. This shows that the ideal $\langle E \mid E \rangle$ contains the unit element, and hence is the entire algebra.

The case with non-trivial stabilizers

On the other hand, if the action is not assumed to be free, then first consider the following counterexample: Take $\Gamma$ to be a non-trivial finite group which acts trivially on the one-point space $X=\{*\}$. Then the Hilbert module you describe is not full because every element of $\langle E \mid E \rangle$ is a multiple of $N = \sum_{g \in \Gamma} g$. Also, the claimed Morita equivalence is not true because in this case the "Roe algebra" would just be the complex numbers if we defined it using $\ell^2(\ast)$ (or the compact operators if we defined it on $\bigoplus_{\mathbb{N}} \ell^2(\ast)$).

In the context of the usual Roe algebra, this problem is solved by insisting that $H$ contains sufficiently many copies of the left-regular representation of each stabilizer group. See for instance the concept of a locally free X-G-module, Definition 4.5.2 in [1]. Concretely, if $X$ is discrete, we could use $$H = \bigoplus_{\mathbb{N}} \ell^2(X) \otimes \ell^2(\Gamma),$$ where the $\Gamma$ acts by the left-regular representation on the $\ell^2(\Gamma)$ factor. Then Roe's construction yields a $C^*_r\Gamma$-module $H_\Gamma$ with $\mathbb{K}(H_\Gamma) \cong C^\ast_\Gamma(X, H)$ which is indeed full (for the latter e.g. use Fell's absorption principle), and hence a Morita equivalence $C^\ast_\Gamma(X, H) \sim C^\ast_r\Gamma$.

But for an infinite group this probably does not deserve to be called uniform Roe algebra even we use $\ell^2(X) \otimes \ell^2(\Gamma)$ without the infinite direct sum.

  1. R. Willett and G. Yu, Book draft "Higher index theory", available from https://math.hawaii.edu/~rufus/Papers.html
  • $\begingroup$ Thanks a lot, that clarifies the situation. I was under the impression that the stated equivalence cannot be true unless the action of $\Gamma$ is free, but I couldn't pin it down. Of course, under Roe's assumption, your counterexample only works for finite groups. But presumably, the original statement is still incorrect if the group is infinite but has non-trivial stabilisers. $\endgroup$ Mar 18, 2019 at 7:08
  • $\begingroup$ The free case that you discuss is also treated by Willett and Yu the book draft you cite, in their Proposition 5.2.7 on p. 171. $\endgroup$ Mar 18, 2019 at 10:04
  • $\begingroup$ The non-free case is then discussed in Proposition 5.2.9 on p. 175. A similar result holds: The equivariant Roe algebra is stably isomorphic to the reduced group C$^*$-algebra. Thanks again for the reference! $\endgroup$ Mar 18, 2019 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.