When the reduced $C^*$-algebra of $\Gamma$ admits character then $\Gamma$ is amenable [closed]

Question

Suppose that $C^*_r(\Gamma)$ admits some character (homomorphism into $\mathbb{C}$)-here $\Gamma$ is discrete group and $C^*_r(\Gamma)$ is the closure of the image of the group ring $\mathbb{C}\Gamma$ under the left regular representation $\lambda$ acting on the Hilbert space $\ell^2(\Gamma)$. Assuming that $C^*_r(\Gamma)$ admits a character $\tau$ one takes this $\tau$ and regard it as a state: as the state it can be extended to the whole of $B(\ell^2(\Gamma))$ (in particular its domain contains $\ell^{\infty}(\Gamma)$). Note that if $s.f$ denotes the action $s.f(t)=f(s^{-1}t)$ for $s \in \Gamma$ and $ f \in \ell^{\infty}(\Gamma)$ then we have (regarded both side as multiplication operators) $\lambda_sf\lambda_s^{*}=s.f$-this can be checked by straightforward computation. Therefore $$\tau(s.f)=\tau(\lambda_sf\lambda_s^*)=\tau(\lambda_s)\tau(f)\tau(\lambda_s^*)=|\tau(\lambda_s)|^2\tau(f)=\tau(f).$$ So our group has to be amenable! Also the converse is true: when $\Gamma$ is amenable then $C^*_r(\Gamma)$ admits character. So these conditions are equivalent-nevertheless I have a problem in understanding in which moment we really have used the assumption that $C^*_r(\Gamma)$ admits a character: more precise (since obviusly we use this assumption at the beginning of the argument) I'm asking

why this argument won't work for $C^*(\Gamma)$-the universal $C^*$-algebra of the group?

Well, yes, you can say that it is due to the presence of representation $\lambda$ or the fact that we can do the computations in $\ell^2(\Gamma)$, whence for $C^*(\Gamma)$ there is a lack of preferred Hilbert space on which $C^*(\Gamma)$ acts but still I have an impression that in the proof we haven't used our assumption heavily and that it can be somehow improved provided we assume some weaker (for the first sight-but in fact equivalent) condition. So in other words:

Is it possible to deduce from the above argument and the knowledge that we cannot improve our proof in order to work with $C^*(\gamma)$ something about the representation theory of $C^*(\Gamma)$?

EDIT: According to the comment below: I know that $C^*(\Gamma)$ always admits a character therefore it is clear that our proof cannot work in general since it would give that every group is amenable. My question was about what is essential in the above argument which prevents us to argue like this for $C^*(\Gamma)$.

Answer 1

It may be easier to understand what is going on if you interpret your character $\tau$ in a different way. Let us examine the argument for why amenability of $\Gamma$ implies existence of a character $\tau : C_r(\Gamma)\to \mathbb{C}$ (which I will call $\epsilon$ from now on to avoid confusion with the group trace).

Now, extension of the trivial representation $\epsilon : \mathbb{C}\Gamma$ to $C^*_r(\Gamma)$ exists iff $$|\epsilon(x)| \leq \Vert x\Vert_\min\qquad(*)$$ for any $x\in \mathbb{C}\Gamma$, where $\Vert x\Vert_\min = \Vert \lambda(x)\Vert_{B(\ell^2\Gamma)}=\Vert x\Vert_{C^*_r\Gamma}.$ Indeed, if the character does extend, the extension is unital and thus must have norm $1$; and if the inequality holds, then $\epsilon$ extends by continuity.

The inequality $(*)$ is equivalent to amenability (use weak containment of the trivial representation in the left regular representation for one direction, and the spectral radius characterization for the opposite direction).

If you consider $C^*\Gamma$ instead of $C^*_r\Gamma$, the question of whether $\epsilon$ extends to all of $C^*\Gamma$ once again has to do with a similar inequality: $|\epsilon(x) | \leq \Vert x\Vert_{C^*\Gamma}$. However, the latter inequality is always true (by definition of the full $C^*$ norm).