Let $\Gamma$ be a discrete group. We can form two $C^*$-algebras: the universal (or full) and reduced, to be denoted by $C^*_u(\Gamma)$ and $C^*_r(\Gamma)$ (respectively). Both of them are completions of the group algebra $\mathbb{C}\Gamma$ but with respect to different norms: universal norm is defined as the supremum of $\| \pi(\cdot) \|$ over all $*$-representations of $\mathbb{C}\Gamma$: the reduced norm is the ordinary operator norm where $\Gamma$ acts on $\ell^2(\Gamma)$ by the left regular representation (and then we extend this action linearly). One can consider the identity mapping $id:(\mathbb{C}\Gamma,\| \cdot \|_u) \to (\mathbb{C}\Gamma,\| \cdot \|_r)$ and extend it to the whole $C^*_u(\Gamma)$ (call $\theta$ this extension). Since this is morphism beetween $C^*$-algebras its range is closed but it this dense. Therefore it is surjective: but there is no reason for this map to be injective (it is injective iff it is isomorphism iff the group is amenable).

What is the kernel of $\theta$?

My guess is that its kernel should be the set $\{x \in C^*_u(\Gamma): \tau(x^*x)=0\}$ where $\tau$ is the canonical trace on $C^*_u(\Gamma)$ defined on $\mathbb{C}\Gamma$ by $\tau(x)=\langle x \delta_e,\delta_e \rangle$ where $e$ denotes the neutral element.
EDIT: I corrected the typo, as pointed in the comment below.

  • 7
    $\begingroup$ Your guess is right (but I'm assuming you mean $x\in C^*_u(\Gamma)$ not $C^*_r(\Gamma)$ in your description of the kernel). Since the left regular representation is just the GNS representation of $C^*_u$ associated with the trace $\tau$ the kernel is as you describe (this is true for any C*-algebra) $\endgroup$ Commented Aug 27, 2015 at 0:24
  • $\begingroup$ Unfortunately, I don't see how this would imply my statement. Could you please explain why it is enough to know this? $\endgroup$
    – truebaran
    Commented Aug 27, 2015 at 12:23

1 Answer 1


(Caleb Eckhardt already answered, but this was too long for a comment)

This is indeed true for general $C^*$-algebra but use the fact that $\tau$ is a trace and not just a state.

An element of the maximal algebra is zero in the reduced algebra if it acts trivially on the regular representation, which is the GNS represenation assciated to the canonical trace $\tau$.

An element $h$ is in the kernel if for any $v$ in the regular representation one has $\Vert hv \Vert =0$. Up to an approximation, the element $v$ can be written $t \delta_0$ for $h$ in the $t$ in the maximal algebra (those are dense in the regular representation). Hence $h$ is in the kernel if and only if for all $t$: $0=\Vert ht \delta_0 \Vert^2 = \langle ht \delta_0| ht \delta_0 \rangle = \langle \delta_0 | t^* h^* h t \delta_0 \rangle = \tau( t^* h^* h t)$

In the general framework of a GNS representation this is the best you can obtain: $h$ is in the kernel if and only if for all $t$, $\tau( t^* h^* h t)=0$.

But because $\tau$ is not just a state but a trace one can improve:

$\tau( t^* h^* h t) = \tau(h^* h t t^*) = \tau(h tt^* h^*) \leqslant \Vert t \Vert^2 \tau(hh^*) = \Vert t \Vert^2 \tau(h^* h )$

(indeed $tt^* \leqslant \Vert t \Vert^2$ hence $htt^* h^* \leqslant \Vert t \Vert^2 hh^*$, because if $x$ is positive then $h x h^*$ is also positive)

Hence $h$ is in the kernel if and only if $\tau(h^* h ) =0$.


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