Let $F_{n-1}$ be the free group of rank $n-1$ and $C^{*}(F_{n-1})$ be the universal group C*-algebra of $F_{n-1}$. And if $E_{n}$ is the $n$-dimensional operator space in $C^{*}(F_{n-1})$ spanned by the unit $U_{0}=1$ and the standard unitary generators $U_{1}, U_{2},..., U_{n-1}$ of $C^{*}(F_{n-1})$.
Then, can we show that $E_{n}$ is canonically isometric to $l_{n}^{1}$ (the $n$-dimensional $l^{1}$ space), or equivalently,
$$||\sum_{k=0}^{n-1}\alpha_{k}U_{k}||=\sum_{k=0}^{n-1}|\alpha_{k}|,$$
for all $(\alpha_{k})_{k=0}^{n-1}\in \mathbb{C}$.
Actually, the question is how to prove
$$||\sum_{k=0}^{n-1}\alpha_{k}U_{k}||\geq\sum_{k=0}^{n-1}|\alpha_{k}|?$$
because the "$\leq$" is obvious.
I try to answer the question by myself:
In order to prove the "$\geq$", from the definition of universal group C*-algebra $C^{*}(F_{n})$, we need only to find a special *-representation $\pi:\mathbb{C}(F_{n})\rightarrow B(H)$, such that $||\pi(\sum_{k=0}^{n-1}\alpha_{k} U_{k})||=\sum_{k=0}^{n-1}|\alpha_{k}|$. Well, It is natural to consider the *-representation $\pi(U_{k})=\frac{\overline{\alpha_{k}}}{|\alpha_{k}|}1_{H}$ (the "$1_{H}$" can be replaced by any fixed unitaries $u\in B(H)$).
The $\mathbb{C}(F_{n})$ is the group ring of $F_{n}$.