Determining the Lie algebra elements exponentiating to the center of a Lie group

Question

For a semi-simple compact Lie group $G$ with center $Z(G)$, one can characterize the preimage of $Z(G)$ in the Cartan subalgebra under the exponential map as the nodes of the Stiefel diagram (see for instance V.7.16 of "Representation of compact Lie groups", by Bröcker and tom Dieck).

Is there any generalization of this result, for instance for non-compact Lie groups, or for classes of infinite dimensional Lie groups?

Update: A look at $SL(2,\mathbb{R})$ shows that the preimage of the center consists in elements of the form

$ \left(\begin{array}{cc} a & b \\ -(k^2\pi^2 + a^2)/b & -a \end{array}\right) \;, \quad a \in \mathbb{R} \;, \quad b \in \mathbb{R}^\ast \;, \quad k \in \mathbb{N} $

One complication in this case is that the Cartan subalgebras are not all conjugate, and it looks indeed that the intersection with $\log Z(G)$ depends on the choice of Cartan subalgebra.

Answer 1

I have only some simple remarks here, valid also for non-compact Lie groups. Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. A first remark is that for $X\in Z(\mathfrak{g})$ and $Y\in \mathfrak{g}$, $\exp (X)$ and $\exp(Y)$ commute. Hence we have $\exp(Z(\mathfrak{g}))\subseteq Z(G)$. However, $\exp$ need not be injective, even if $G$ is simply connected. Also, $\exp$ need not be surjective (but there is a classification of all simple Lie groups with surjective exponential). There are some special cases, where we can say which elements do not exponentiate to the centre of $G$. As an example, the following is true:

Lemma: Let $G$ be a real or complex connected Lie group whose Lie algebra $\mathfrak{g}$ is linear and centerfree. If $X ∈\mathfrak{g}$ is nilpotent and $\exp(X) \in Z(G)$, then $X =0$.

Proof: By assumption $\exp({\rm ad} X)Y=Y$ for all $Y\in \mathfrak{g}$, so that $[X,Y]=0$ for all $Y\in \mathfrak{g}$. It follows $X=0$ because $Z(\mathfrak{g})=0$.