Can you generate all unitary operators from the unit ball of Hermitian operators? I know that the exponential map $\mathcal H\rightarrow \mathcal U$ is surjective where $\mathcal H$ is the set of all skew hermitian matrices and $\mathcal U$ is the set of all unitary matrices of some fixed rank $n$, however what if we restrict to the unit ball of the skew hermitians? Can every unitary be written as $e^{tH}$ for some $H$ with $\|H\|\leq 1$ and $t\in[0,2\pi)$?
 A: The unit ball of skew hermitians has the sup norm of the spectrum equal to $1,$ the spectrum being purely imaginary. So the answer is "yes".
A: I believe that I have at least a partial solution and I invite others to suggest edits to finish off the proof, or provide your own answer filling in the remaining details.
Let $U$ be a unitary operator, then $U=\exp(iH)$ for some hermitian operator $H$. Note that all skew hermitians can be expressed as $iH$ for some hermitian $H$, so I can write my skew hermitians in this form without loss of generality. Now taking the complex logarithm of $U$, it follows that $$\log(U) = iH+2\pi ik I$$ for all integers $k\in\mathbb{Z}$. Thus find some $m\in\mathbb{Z}$ such that $\|iH +2\pi imI\|\leq 1$ and define $X:=iH +2\pi imI$.
Now recall that for any operators $A$ and $B$, $\exp(A+B)=\exp(A)\exp(B)$ if and only if $AB=BA$. Since $iH$ and $2\pi imI$ commute, it follows that $\exp(X)=\exp(iH)\exp(2\pi imI) $. So let's examine $\exp(2\pi im I)$. Notice that
$$ \exp(2\pi imI)=\Big(\exp(2\pi iI)\Big)^m = \Big(\sum_{k=0}^\infty\frac{(2\pi iI)^k}{k!} \Big)^m = \Big(I\sum_{k=0}^\infty \frac{(2\pi i)^k}{k!}\Big)^m=e^{2\pi im}I=I$$
Therefore $\exp(X)=\exp(iH)$. The only part that remains proving is that there actually exists such an $m$, I feel like such an $m$ should exist but I'm not sure how to prove it. If anybody wants to fill in the gap here, feel free.
