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.