Are periodic points dense in the unitary group? In $U(1) = \{z \in \mathbb{C} : |z| = 1\}$, it is well known and easy to see that the set of $z$ so that $ z^n = 1 $ for some $n \in \mathbb{Z}_+$ are dense. Does this fact generalize to the group $U(d) = \{U \in \mathbb{C}^{d \times d} : UU^* = U^*U = I\}$?
What about $ U(\mathcal{H})$, the group of unitary operators on a separable, infinite-dimensional Hilbert space?  It seems reasonable to expect that the claim holds for $U(d)$, but not for $ U(\mathcal{H}) $, but I'm not sure.
 A: In fact it holds for $U(\mathcal{H})$ too. 
It is easy to see it visually if you know the spectral theorem for normal operators (a reference for it is Rudin's Functional Analysis): 
if $T\in U(\mathcal{H})$ then clearly $\sigma(T)\in\{z\in\mathbb{C}:|z|=1\}$ (i.e. its spectrum lies in the unit circle). 
Write $T$ in terms of its spectral decomposition: $T=\int_{\sigma(T)}\lambda \,dE(\lambda)$. 
Fix a large integer $N$ and cover the unit circle with $N$ half-closed arcs centered in the $N$-th roots of unity; so if we call $B_1,\dots,B_N$ these arcs we have $S^1=\sqcup_{i=1}^N B_i$. 
Let's now construct a unitary operator of order $N$ which is close to $T$: put $$T_N:=\sum_{i=1}^N \omega^iE(B_i)$$ (where $\omega$ is a primitive $N$-th root of unity and $\omega^i\in B_i$ for each $i$). 
Now functional calculus allows us to check that $T_N$ has the desired properties: 


*

*$T_NT_N^*=T_N^*T_N=\sum_i\sum_j \omega^i E(B_i)\overline{\omega^j} E(B_j)=\sum_i E(B_i)+\sum_{i\neq j}\omega^{i-j}E(B_i\cap B_j)=I$ since $B_i\cap B_j=\emptyset$ if $i\neq j$.

*$T_N^N=\left(\sum_i\omega^i E(B_i)\right)^N=\sum_i(\omega^i)^N E(B_i)=I$ (again all the mixed terms in the expansion are zero since the $B_i$'s are disjoint; moreover $E(B_i)^N=E(B_i)$ as $E(B_i)$ is an orthogonal projection, thus is idempotent).

*$T=\sum_i\int_{B_i}\lambda\,dE(\lambda)$, so $T-T_N=\sum_i\int_{B_i}\left(\lambda-\omega^i\right)\,dE(\lambda)=\int_{\sigma(T)}\left(\sum_i(\lambda-\omega^i)1_{B_i}(\lambda)\right)\,dE(\lambda)$ and finally
$\|T-T_N\|\le\|f\|_\infty$, where we define $f(\lambda):=\sum_i(\lambda-\omega^i)1_{B_i}(\lambda)$. 
But clearly $\|f\|_\infty\le \max_i \text{diam}(B_i)\le\frac{2\pi}{N}$, which is arbitrarily small when $N\to\infty$.

