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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.

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    $\begingroup$ One big obstacle to getting a reasonable generalization is the fact that for non-compact (say semisimple) real or complex Lie groups, the exponential map may fail to be surjective. There's a lot of literature about such issues, touching also on the infinite dimensional case. Have you looked at small rank cases involving matrix groups? $\endgroup$
    – Jim Humphreys
    Commented Nov 16, 2013 at 14:47
  • $\begingroup$ @Jim Humphreys That's right... but one might still hope to characterize those elements which do exponentiate to the center. I updated my question to describe the case of $SL(2,\mathbb{R})$. $\endgroup$
    – Samuel Monnier
    Commented Nov 16, 2013 at 23:06
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    $\begingroup$ @JimHumphreys This is probably useless to the OP by now but just for the record (and because it is beautiful): There is this nice theorem of Hochschild asserting that for connected finite-dimensional Lie groups the center always lies in the image of the exponential map: "The center of an analytic group G is contained in an abelian analytic subgroup of G." (The Structure of Lie Groups, Holden-Day, 1965, page 189, Theorem 1.2) $\endgroup$
    – Yannick Voglaire
    Commented Mar 25, 2016 at 4:20
  • $\begingroup$ @Yannick: Thanks for the reference, though it doesn't help directly with the concrete question raised here. Hochschild studied carefully the general foundations for analytic groups and Lie groups, but without getting into the detailed study of special classes such as semisimple Lie groups (which probably need case-by-study at least initially).. $\endgroup$
    – Jim Humphreys
    Commented Mar 27, 2016 at 19:50
  • $\begingroup$ If $G$ is a {\it linear} connected semi-simple Lie group, then its centre lies in the centre of a maximal compact subgroup $K$ and therefore, you need only look at the case of a compact connected Lie group. $\endgroup$
    – Venkataramana
    Commented Nov 29, 2023 at 1:23

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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 $Z(\mathfrak{g})=0$.

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  • $\begingroup$ Thanks. But I expect that it should be possible to say more. Another simple fact is that if a Lie algebra element X exponentiates to the center, then it's whole adjoint orbit does. What I'm after is a characterization of these orbits. The lemma above is a first step... $\endgroup$
    – Samuel Monnier
    Commented Nov 19, 2013 at 20:06
  • $\begingroup$ "(but there is a classification of all simple Lie groups with surjective exponential". A pointer to where that can be found would be very helpful to me! $\endgroup$
    – Benjamin
    Commented Apr 14, 2016 at 21:00
  • $\begingroup$ @Benjamin: Michael Wuestner, The classification of all simple Lie groups with surjective exponential map, Journal of Lie Theory Vol. 15, No. 1, pp. 269–278 (2005) emis.de/journals/JLT/vol.15_no.1/19.html $\endgroup$
    – Yannick Voglaire
    Commented Nov 10, 2016 at 14:22

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