Conjugacy classes with elliptic limit points - MathOverflow most recent 30 from http://mathoverflow.net2013-05-20T11:31:58Zhttp://mathoverflow.net/feeds/question/41895http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/41895/conjugacy-classes-with-elliptic-limit-pointsConjugacy classes with elliptic limit pointsTobias Hartnick2010-10-12T10:36:55Z2010-10-14T19:09:55Z
<p>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. </p>
<p><strong>Question:</strong> Can one characterizes those conjugacy classes in $G$ which contain an elliptic conjugacy class in their closure?</p>
<p>(For $G = GL_n(\mathbb R)$ they are characterized by the fact that all eigenvalues are of modulus one, if I a not mistaken.)</p>
http://mathoverflow.net/questions/41895/conjugacy-classes-with-elliptic-limit-points/41941#41941Answer by Peter McNamara for Conjugacy classes with elliptic limit pointsPeter McNamara2010-10-12T20:14:53Z2010-10-14T19:09:55Z<p>For g in G, write g=g<sub>s</sub>g<sub>u</sub> 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 g<sub>s</sub> is elliptic.</p>
<p>Let us first suppose that g<sub>s</sub> is not elliptic. Choose an embedding of G into GL<sub>n</sub>(ℂ). Then by our assumption, g<sub>s</sub> 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 g<sub>s</sub> contained an elliptic element a in its closure. WLOG a is in the special unitary group SU<sub>n</sub>. Let h be in the conjugacy class of g<sub>s</sub>. 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.</p>
<p>Now suppose that g<sub>s</sub> is elliptic. We may replace G by the centraliser of g<sub>s</sub> is G, which is also reductive. So WLOG, g<sub>s</sub> is central in G. Now the Zariski closure of the group generated by g<sub>u</sub> 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 sl<sub>2</sub> triple E,F,H in Lie(G). Now consider conjugation by elements of the form exp(tH) with t real. This shows that g<sub>s</sub> is in the closure of the conjugacy class of g, and we're done.</p>