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). 



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> What is the kernel of $\theta$?
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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.



Kernel of the natural map between group $C^*$-algebras
(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 $\tau(h^\* h ) =0$.


