# Center of Lie group and Lie algebra

Let $G$ be a Lie group and $\mathfrak{g} = T_eG$ its Lie algebra (where $e \in G$ is the neutral element). Denote $Z(\mathfrak{g})$ the center of $\mathfrak{g}$ and $Z(g)$ the center of $G$. I've read the following statement but don't see how to prove it: $$Z(\mathfrak{g}) = 0 \Longleftrightarrow Z(G) \text{ is zero dimensional.}$$ I'm mainly interested in the "$\Leftarrow$" direction, as this is used in a corollary of the Bonnet-Myers theorem in the text I'm reading.

Now, my knowledge of Lie groups is very basic. For instance, I guess that the dimension of $Z(G)$ means its dimension as a manifold. So I would think that $Z(G)$ is a submanifold of $G$. But how do I even see this in general?

According to this question When is the Lie algebra of the center of Lie group the center of its Lie algebra, it seems to be difficult to find conditions when $Z(\mathfrak{g}) = Z(G)$ holds in general, however the statement above looks more simple.

Would someone be able to prove the above equivalence?

• So there might be easier ways to do this, but do you know the fact that if two vector fields on a manifold have Lie bracket equal to zero, then their flows commute? Aug 5 '16 at 7:35
• Or if you only care about matrix groups, we could use $e^{A+B}=e^Ae^B$ whenever $A$ and $B$ are commuting matrices. Aug 5 '16 at 7:38
• Well, technically no, I don't know the fact that if two vector fields on a manifold have zero Lie bracket then their flows commute. Since you've mentioned it, I've quickly researched it now, although I can't say I fully understand the proof. If you have an easy way of showing the statement I'm interested in using this, please go ahead. I would also be interested in a proof that doesn't use it however. And, I'm interested in general Lie groups, not just matrix groups. Aug 5 '16 at 10:55
• I am probably generating more confusions here, but aren't the elements from each set different? One takes elements from $G$ and one select elements from the Lie Algebra $\frak{g}$ Aug 5 '16 at 12:19
• @TomBombadil, I have no idea what their definition of the center is in that question because I am pretty sure $Z(\mathfrak{g} ) = \{ X \in \mathfrak{g} : [X,Y] = 0, \forall Y \in \mathfrak{g} \}$ Aug 5 '16 at 23:34

This is an answer for $G$ connected.

Okay, for $Y\in \frak{g}$, you can show that the flow on $G$ given by the vector field, $Y$, is given by multiplication by elements of $e^{tY}, (t\in \mathbb{R})$. If $X\neq 0$, is in the center, then its flow commutes with the flow associated to all other members of $\frak{g}$ i.e. $e^{tX}$ commutes with $e^{tY}$ for all $Y\in \frak{g}$. The set $\{e^{tY}: t\in \mathbb{R}, Y\in \frak{g}\}$ contains an open neighborhood of the identity, so it generates $G$, since $G$ is connected. So $e^{tY}$ commutes with a subset that generates $G$, hence it commutes with all elements of $G$. Hence $e^{tY}$ is in the center of $G$. So $Z(G)$ contains a one-dimensional sub manifold, hence is not zero-dimensional.

I realize this probably uses more Lie-group theory than you had hoped, and maybe theres a more elementary way to do it, but i don't immediately see it.

Edit :Maybe there's an easier way to see that $X\in Z(\frak{g})$ implies $e^{tX}\in Z(G)$: Let $\gamma: \mathbb{R}\rightarrow \mathfrak{g}, t\mapsto L_{e^{tY}\ast} \circ R_{e^{-tY}\ast }(X_e)$. Then prove that $\gamma'(0)=[X,Y]$ (this requires the interpretation of the bracket as the Lie derivative). So if $X$ is in the center, then $\gamma'(s)= L_{e^{sY}_\ast }\circ R_{e^{-sY}_\ast }(0)=0$ (since $L_{e^{(s+t)Y}\ast}= L_{e^{sY} \ast} \circ L_{e^{tY} \ast}$ and likewise for $X$). But this implies $\gamma = X_e$ is constant. So $e^{tY}e^{uX}e^{-tY} = e^{uX}$ for all $t, u\in \mathbb{R}$, which says all $e^{uX}$ commute with all $e^{tY}$

• Thank you for your answer! I understand this proof modulo 2 things: The fact that if two vector fields have Lie bracket zero then their flows commute (as mentioned before), and the fact that an open neighbourhood of the identity generates G if G is connected. For the second statement however, I found a nice proof here: a neighbourhood of identity U generates G where G is a connected lie group Aug 7 '16 at 12:53
• So, I'm happy with your answer, as long as I know that the general statement is not immediate and somewhat difficult to prove:) Aug 7 '16 at 13:00
• You're welcome. The first statement is proved in John Lee's book on smooth manifolds. However I've edited the post to include a proof that doesn't use this fact. Aug 7 '16 at 17:31

I want to explain the fact in question with Lie theory which mostly appears very early in books of the topic, but yet isn't easy.

The center of $$G$$ is the set $$Z(G) = \{g \in G \ \vert \ gh = hg \ \forall h \in G\}$$. To tackle the equivalence, you were right with what $$0$$-dimensional in this case even means; it means $$Z(G)$$ is $$0$$-dimensional as an embedded submanifold of $$G$$. That is true since $$Z(G) = \ker(Ad)$$ is a closed subgroup and thus an embedded Lie-subgroup of $$G$$ (this is not easy to prove!).

Per definition, $$Z(\mathfrak{g}) = \{v \in \mathfrak{g} \ \vert \ [v,w] = 0 \ \forall w \in \mathfrak{g}\}$$. But one can also look at $$T_eZ(G)$$, and yes, they agree: $$T_eZ(G) = Z(\mathfrak{g})$$ (which is easy to see knowing that $$Z(G) = \ker(Ad)$$).

As a $$0$$-dimensional (embedded) submanifold of $$G$$, $$Z(G)$$ is a discrete subset of $$G$$. It is now clear that $$Z(G)$$ is $$0$$-dimensional iff $$Z(G) = T_eZ(G) = 0$$.

And by the way; one can also show that $$Z(G) = G \Longleftrightarrow Z(\mathfrak{g}) = \mathfrak{g}.$$