Timeline for Is the normalizer of a reductive subgroup reductive?

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Dec 7, 2012 at 15:51	comment	added	Venkataramana		I consulted Gopal Prasad (University of Michigan). He told me a proof which avoids Mostow's theorem but uses compact groups. Let $K_H$ be maximal compact in $H$ and $K$ a maximal compact in $G$ containing $K_H$, and $\theta $ the Cartan involution on $G$. \|It is enough to prove $Z_G(H)$ is reductive. Since $H$ is reductive, and $K_H$ is Zariski dense in $H$, it is the same as proving that the centraliser in $G$ of $K_H$ is reductive. But this centraliser s obviously closed under the Cartan involution.
Nov 24, 2012 at 1:06	comment	added	Venkataramana		I mean the matrix inverse of the transpose, not the map inverse to the transpose: the involution is the complex conjugate of $(^t A )^{-1}$.
Nov 23, 2012 at 20:32	comment	added	Robert K		Isn't the transpose its own inverse?
Nov 23, 2012 at 15:06	comment	added	Venkataramana		The Cartan involution on $GL_n({\mathbb C})$ is the complex conjugate of the inverse of the transpose.
Nov 23, 2012 at 15:02	history	answered	Venkataramana	CC BY-SA 3.0