Compactness of unitary group

Question

How to see the unitary group is not compact in norm topology of $B(H)$ for infinite dimensional Hilbert space? What if norm topology is replaced by SOT?

Answer 1

Say $(e_n)_{n\in\Bbb Z}$ is an orthonormal basis. For each $k$ there is a unitary operator $T_k$ with $$T_ke_0=e_{k},$$ $$T_ke_k=e_0,$$ and $$T_ke_j=e_j\quad(j\ne0,k).$$Hence $\|T_k-T_j\|=\|T_0-T_1\|>0$ for $j\ne k$, so the group is not compact in the norm topology.

And $T_ne_1$ has no norm-convergent subsequence, so the group is also not compact in the SOT.

(Not that you asked , but the same example shows that the unitary group is not even closed in the WOT, since $T_k\to0$ in the WOT...)