5
$\begingroup$

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.)

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.