States on the reduced group $C^\ast$-algebra of $\mathbb{F}_2$

Question

Consider the reduced group $C^\ast$-algebra $C_{\text{r}}^\ast (\mathbb{F}_2)$ of the free group in two generators $a$ and $b$. Further, let $\phi$ be a state on $C_{\text{r}}^\ast (\mathbb{F}_2)$. Let $(g_i)_i \subseteq \mathbb{F}_2$ be a sequence of group elements for which the word length (with respect to the generators $a$ and $b$) of $g_i$ is equal to $i$. Is it then true that $\phi(\lambda_{g_i}) \rightarrow 0$ for $i\rightarrow \infty $? Here $\lambda$ denotes the left-regular representation of $\mathbb{F}_2$. The statement looks kind of obvious, but I don't see how to prove it.

Answer 1

The answer is no.

Let $A$ be the closed *-subalgebra of $C_{\text{red}}^*(\mathbb F_2)$ generated by $\lambda _a$. It is easy to see that $A$ is isomorphic to $C(S^1)$, the continuous functions on the unit circle, via an isomorphism that maps $\lambda _a$ to the identity function "$f(z)=z$".

Consider the state $\varphi $ (actually a character) of $A$ defined by $$ \varphi :g\in A\mapsto g(1)\in {\mathbb C}. $$

By a well known result, $\varphi $ extends to a state, say $\tilde\varphi $, of $C_{\text{red}}^*(\mathbb F_2)$.

Setting $g_n = a^n$, one has that the word length of $g_n$ is $n$, but $$ \tilde \varphi (\lambda _{g_n}) = \varphi (\lambda _a^n) = z^n|_1 = 1, $$ which doesn't converge to zero.