Let $\mathfrak g\subseteq(\mathbb C^{n\times n},\|\cdot\|)$ be a Lie algebra, $\|\cdot\|$ some matrix norm, and $G:=e^{\mathfrak g}$ the associated Lie group. Additionally, let us assume that $G$ is compact. I want to understand whether compactness guarantees that one can always find a norm-bounded subset of $\mathfrak g$ which generates $G$. More precisely:
If $G:=e^{\mathfrak g}$ is compact, does there exist $C>0$ such that every $X\in G$ can be written as $X=e^Y$ for some $Y\in\mathfrak g$ with $\|Y\|\leq C$?
Actually, I'm interested in this question for the case $\mathfrak g\subseteq\mathfrak u(n)$, but my intuition wants me to believe that the key property from the unitary group one needs here is compactness: after all, if $\mathfrak g$ generates a dense winding then this property does not hold (because closedness fails), but if $\mathfrak g$ is all of $\mathfrak u(n)$, then this is obviously true, with $C=\pi$.
Here are three approaches I tried so far:
- (Unbounded sequence of generators which converge after $\exp$)
The easiest thing to attempt is to just define $\mathfrak g_k:=\{Y\in\mathfrak g:\|Y\|<k\}$ and to assume by way of contradiction that $e^{\mathfrak g_k}\subsetneq e^{\mathfrak g}$ for all $k$. This yields a sequence $(X_k)_k$ in $G$ of elements which can only be written as $X_k=e^Z$ if $\|Z\|\geq k$. On the other hand, $G$ is compact so we can extract a subsequence $(X_{k_j})_j$ which converges to some $X'=e^{Z'}\in G$. The problem is that I don't quite see how the (necessarily) unboundedness of $(Z_{k_j})_j$ could contradict the limit point $X'$ having a valid geneator $Z'\in\mathfrak g$. I don't think there's an immediate way to "move" all the $Z_{k_j}$ into the same injective strip of the exponential to see that $(e^{Z_{k_j}})_j$ has to diverge, simply because $(Z_{k_j})_j$ does?
Next, I tried an approach which uses the covering space definition of compactness:
- (Construct open cover using that $\exp$ is a local diffeomorphism)
Because $\exp$ is a local diffeomorphism around $0\in\mathfrak g$, there exists $\delta>0$ such that $\exp$ maps $B_\delta(0)$ diffeomorphically to an open neighborhood $N$ of ${\bf1}\in G$. In particular there exists $\varepsilon$ such that $B_{\varepsilon}({\bf1})\cap G\subseteq N$. This yields the trivial open cover $\bigcup_{X\in G}B_{\varepsilon/M}(X)$ of $G$; here, $M$ is a constant such that $\|X\|\leq M$ for all $X\in G$ (exists because $G$ is bounded by assumption). But $G$ is compact so this open cover admits a finite subcover $G\subseteq B_{\varepsilon/M}(X_1)\cup\ldots\cup B_{\varepsilon/M}(X_m)$ for some $m\in\mathbb N$, $X_1,\ldots,X_m\in G$, and each of these can be written as $X_j=e^{Y_j}$, $Y_j\in\mathfrak g$. The idea now would be that the constant $C$ we are looking for can be somehow "built" from $\max_j\|Y_j\|$ and $\delta$: each $X\in G$ is in one of the balls $B_{\varepsilon/M}(X_j)$ so $X=(XX_j^{-1})X_j$ where $\|XX_j^{-1}-{\bf1}\|\leq\|X-X_j\|\|X_j^{-1}\|<\frac\varepsilon M\cdot M=\varepsilon$. Thus $XX_j^{-1}$ is in the neighborhood $N$ around ${\bf1}$ meaning there exists $Z\in B_\delta(0)\cap\mathfrak g$ such that $XX_j^{-1}=e^Z$. Combining things, what we showed is that $X=e^Ze^{Y_j}$. This of course screams Baker-Campbell-Hausdorff, but I didn't know how to guarantee convergence of the series (since $\|Y_j\|$ could be anything?). And I didn't see another way to construct a generator of $X$ which can be bounded using $\|Z\|<\delta$ and $\max_j\|Y_j\|$.
Finally, I had a similar but slightly modified version of this idea:
- (Construct open cover by exponentiating increasing bounded subsets of $\mathfrak g$)
Using the previously defined $\mathfrak g_k:=\{Y\in\mathfrak g:\|Y\|<k\}$ we could try to construct an open cover directly, e.g., via $G=\bigcup_{k\in\mathbb N}e^{\mathfrak g_k}$ or, using the interior ${}^\circ$ (relative to $G$), $G=\bigcup_{k\in\mathbb N}(e^{\mathfrak g_k})^\circ$. If either of these is an open cover of $G$, then by compactness there has to exist a maximal $k$, i.e., $G=e^{\mathfrak g_k}$ and we are done. For the first one, equality is trivial, but showing that $e^{\mathfrak g_k}$ is open (assuming that's even true) is tricky. For the second approach, each set is obviously open but showing that this is a cover is non-trivial to me. To be more precise, in this second case, we for all $X\in G$ would have to find $k$ and an open set $U$ such that $X=e^Y\in U\cap G\subseteq e^{\mathfrak h_k}$. If the exponential were locally diffeomorphic everywhere, then we could find a neighborhood of $Y$ which is mapped diffeomorphically to a neighborhood of $X$ and we should be done; however it seems that this property does not hold in general? Or could $\exp$ be locally diffeomorphic everywhere under additional assumptions on $G$ (e.g., $G\subseteq\mathsf U(n)$)? I also tried to apply the open function theorem to $d\exp_Y$ directly but the derivative $$d\exp_Y(Z)=Z+\sum_{n=1}^\infty\frac{\sum_{j=0}^{n-1}Y^jZY^{n-1-j}}{n!}$$ one ends up with seems rather messy, so I don't see immediately whether this is bijective or not.
Then again, maybe none of these proofs work because the statement I'm interested in isn't even true. Any insights or comments would be greatly appreciated!
