Suppose that $G$ is a Lie group with Lie algebra $\mathfrak{g}$ and the center of $G$ is denoted by $Z(G)$ with its Lie algebra denoted $Z(\mathfrak{g})$. It's easy to show that $Z(\mathfrak{g})\subset\big\{X\in\mathfrak{g}\;|\;[X,Y]=0,\forall Y\in\mathfrak{g} \big\}$, my problem is that, when are they equal?
[Math] When is the Lie algebra of the center of Lie group the center of its Lie algebra
lie-algebraslie-groups
Best Answer
One can show the following:
Here, $\operatorname{Lie}$ denotes the functor from Lie groups to Lie algebras, i.e. $\operatorname{Lie}G$ is the Lie algebra of $G$.
There is a nice corollary:
To prove the statement, we'll need the following lemma:
Proof of the lemma:
"$\subset$": This follows immediately from the fact that conjugation by an element that lies in the center is the identity on $G$, in particular its differential is just $\operatorname{id}_{\operatorname{Lie}G}$.
"$\supset$": Let $g$ be in the kernel, i.e. $(\alpha_g)_* \equiv \operatorname{Ad}(g) = \operatorname{id}_{\operatorname{Lie}G}$, where $\alpha_g$ denotes conjugation by g. But the functor Lie is faithful on the subcategory of connected Lie groups (*), so since $(\alpha_g)_* = \operatorname{id}_{\operatorname{Lie}G} = (\operatorname{id}_G)_*$, it follows $\alpha_g = \operatorname{id}_G$, thus $g \in Z(G)$.
This proves the lemma. In particular, since the kernel is a Lie subgroup of $G$, this shows that $Z(G)$ is a Lie subgroup, as well, and thus a Lie group, so the notion of $\operatorname{Lie}Z(G)$ in the original claim is in fact well-defined.
Proof of the original claim:
Since $Z(G)$ is a Lie subgroup of $G$, we can consider $\operatorname{Lie}Z(G) \subset \operatorname{Lie}G$ to be a subalgebra in the canonical way and $\exp_{Z(G)} \equiv \exp_G|_{Z(G)}$ (which we'll just denote by $\exp$ in the following).
"$\subset$": Let $X \in Z(\operatorname{Lie}G)$, i.e. $\operatorname{ad}X = 0$. If we find a curve $\gamma$ in $Z(G)$ whose derivative is $X$ we're done. The somewhat obvious idea is $\gamma(t) := \exp(tX)$ which has derivative $X$. So it remains to show that indeed $\gamma(t) \in Z(G)$. But $\operatorname{Ad}(\gamma(t)) = \operatorname{Ad}(\exp(tX)) = \exp_{GL(\operatorname{Lie}G)}(\operatorname{ad}(tX))$ by a commutative diagram which, in turn, equals $\exp_{GL(\operatorname{Lie}G)}(0) = \operatorname{id}_{GL(\operatorname{Lie}G)}$ and the claim follows by the lemma.
"$\supset$": Let $X \in \operatorname{Lie}Z(G)$, in particular $\exp(tX) \in Z(G)\; \forall t$. We have to show that $\operatorname{ad}X = 0$. Now, using commutative diagrams one can show in a straightforward way that $\forall x, y \in \operatorname{Lie}G$:
$$ \exp x · \exp y · (\exp x)^{-1} = \exp(e^{\operatorname{ad}x}(y))$$
Here, $e \equiv \exp_{GL(\operatorname{Lie}G)}$ is the matrix exponential. We apply this identity in the following way: Let $Y \in \operatorname{Lie}G$ and let $U$ be a neighborhood of $0 \in \operatorname{Lie}G$ s.t. $\exp|U: U \rightarrow \exp U$ is a diffeomorphism. Pick $s, t \in \mathbb{K}, s,t \neq 0$ small enough s.t. $sX, tY, e^{\operatorname{ad}(sX)}(tY) \in U$. Then plugging in $x := sX, y := tY$ above and using $\exp(sX) \in Z(G)$ results in:
$$ \exp(tY) = \exp(e^{\operatorname{ad}(sX)}(tY))\,, $$
so by injectivity of $\exp$ on $U$ (cancel $t$ using linearity):
$$ Y = e^{\operatorname{ad}(sX)}(Y)$$
Since $Y$ was arbitrary, this implies $e^{\operatorname{ad}(sX)} = \operatorname{id}_{\operatorname{Lie}G}$ for all $s$ small enough. By choosing $s$ even smaller (if necessary), we achieve that $\operatorname{ad}(sX)$ is in a neighborhood of $0 \in \operatorname{gl}(\operatorname{Lie}G)$ where $e$ is injective as well and we obtain $\operatorname{ad}(sX) = 0$, implying the claim.
Proof of the corollary:
Let $G$ be Abelian. Then $Z(G) = G$, so by the above we have $Z(\operatorname{Lie}G) = \operatorname{Lie}G$, proving that $\operatorname{Lie}G$ is Abelian.
Let $\operatorname{Lie}G$ be Abelian, i.e. $\operatorname{Ad}_* = \operatorname{ad}: \operatorname{Lie}G \rightarrow \operatorname{gl}(\operatorname{Lie}G)$ is the zero map. But the map $G \rightarrow \operatorname{GL}(\operatorname{Lie}G), g \mapsto \operatorname{id}_{\operatorname{Lie}G}$ has differential $0$ as well, so once again by the faithfulness of the functor Lie we have: $\operatorname{Ad}(g) = \operatorname{id}_{\operatorname{Lie}G}$, i.e. $g \in Z(G)$, so $G$ is Abelian.
(*) The functor Lie is faithful on the category of connected Lie groups:
Sketch of proof: