free group factor- do the basis of generators strongly converge to each other? Consider $W=L(\mathbb{F}_{n})$, the group von Neumann algebra of free group with $n$ generators. $W$ is closed under strong operator topology(by bicommutant theorem). I am wondering for the algebra generated by the generator $\delta_{i}, i\in\{1,...n\}\backslash\{j\}$, will there exists an sequence in it such that it strongly converge to $\delta_{j}$?
I think the answer is no, since the generators are independent, but I don't see how it is related.
 A: I prefer to write $u_1,\dots,u_n$ for the generators as $\delta_j$ usually denotes elements of $\ell_2(G)$. Let $\mathcal A$ be the $\ast$-algebra generated by $\{u_k\mid k\neq j\}$.
Assume there is net in $\mathcal A$ that converges strongly to $u_j$. Then $\mathcal A$ must be strongly dense in $L(\mathbb F_n)$. By the Kaplansky density theorem this implies the existence of a norm bounded net $(x_\alpha)$ in $\mathcal A$ that converges strongly to $u_j$ (if you are only interested in sequences, you can skip this step and directly apply the uniform boundedness principle to see that the sequence is norm bounded).
Let
$$
\tau\colon L(\mathbb F_n)\to \mathbb C,\,x\mapsto \langle \delta_e,x\delta_e\rangle.
$$
This is a normal state on $L(\mathbb F_n)$. Hence if $(x_\alpha)$ is a bounded net in $L(\mathbb F_n)$ that converges strongly to $0$, then $y_\alpha^\ast y_\alpha\to 0$ weakly, which implies $\tau(y_\alpha^\ast y_\alpha)\to 0$.
However, for $y_\alpha=x_\alpha-u_j$, we have
$$
\tau((x_\alpha-u_j)^\ast(x_\alpha-u_j))=\tau(x_\alpha^\ast x_\alpha)-\tau(x_\alpha^\ast u_j)-\tau(u_j^\ast x_\alpha)+\tau(u_j^\ast u_j).
$$
Of course $\tau(u_j^\ast u_j)=1$. Moreover,
$$
\overline{\tau(u_j^\ast x_\alpha)}=\tau(x_\alpha^\ast u_j)=\langle x_\alpha \delta_e,\delta_{g_j}\rangle=0
$$
since $x_\alpha$ is a linear combination of elements $\lambda(g)$ with $g$ in the subgroup generated by $g_k$ for $k\neq j$ (here $g_1,\dots,g_n$ denote the generators of $\mathbb F_n$). This implies
$$
\tau((x_\alpha-u_j)^\ast(x_\alpha-u_j))\geq \tau(u_j^\ast u_j)=1
$$
in contradiction to our original claim.
As one might guess, this is not specific to free groups. Whenever $G$ is a countable discrete group and $H$ a proper subgroup, then $\{\lambda(h)\mid h\in H\}$ is properly contained in $L(G)$. As remarked in the comments, this boils down to the fact that the images of group elements in $L(G)$ are orthonormal with respect to the inner product induced by $\tau$ together with the fact that this inner product induces the strong toplogy on the unit ball of $L(G)$.
