Let $G$ be a reductive algebraic group over $\mathbb R$ and $K$ a maximal compact subgroup. Then we refer to the conjugacy class in $G$ of some $k \in K$ as an elliptic conjugacy class.

Question: Can one characterizes those conjugacy classes in $G$ which contain an elliptic conjugacy class in their closure?

(For $G = GL_n(\mathbb R)$ they are characterized by the fact that all eigenvalues are of modulus one, if I a not mistaken.)


1 Answer 1


For g in G, write g=gsgu as its Jordan decomposition into semisimple and unipotent parts. I claim that the closure of the conjugacy class of g contains an elliptic element if and only if gs is elliptic.

Let us first suppose that gs is not elliptic. Choose an embedding of G into GLn(ℂ). Then by our assumption, gs has an eigenvalue of norm greater than one, let λ be the absolute value of such an eigenvalue. Suppose for want of contradiction that the conjugacy class of gs contained an elliptic element a in its closure. WLOG a is in the special unitary group SUn. Let h be in the conjugacy class of gs. Then h has an eigenvalue of absolute value λ. Letting v be an eigenvector, we see that |(h-a)v| is at least (λ-1)|v|, so |h-a|≥λ-1, a contradiction.

Now suppose that gs is elliptic. We may replace G by the centraliser of gs is G, which is also reductive. So WLOG, gs is central in G. Now the Zariski closure of the group generated by gu is a one-dimensional unipotent subgroup of G. Let E be a non-zero element in its lie algebra. This is a nilpotent element. Then by the Jacobson-Morozov theorem, we can extend E to a sl2 triple E,F,H in Lie(G). Now consider conjugation by elements of the form exp(tH) with t real. This shows that gs is in the closure of the conjugacy class of g, and we're done.


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