Talk:Cartan decomposition

There is a mistake in first given example. The Killing form, evaluated at (X,Y) is being defined as the trace of [ad(X),ad(Y)]. It contains in its kernel the center of gl(n,R). In particular, each non zero element in the center (assume here n is at least 1) will also be in the kernel of Tr[ad(X),ad(t(X))]. So this last form is not definite for n>0. It is only semidefinite.

Helgason (p. 185) defines Cartan involution (as is done here) only for semisimple algebras, which means not for gl(n,R), if n>0.

I'll replace gl(n,R) with sl(n,R) in this example. —Preceding unsigned comment added by 77.206.130.58 (talk) 17:10, 14 April 2009 (UTC)

Definition of an involution

There seems to be an inconsistency between the definition of an involution and the examples. In the defintion we have:

An involution on $\mathfrak{g}$ is a Lie algebra automorphism $\theta$ of $\mathfrak{g}$ which is not itself the identity...

and yet the examples give:

The identity map on $\mathfrak{g}$ is an involution, of course.

I'm not 100% sure which is correct, so I'm not going to change this.

130.88.123.49 (talk) 11:17, 29 October 2010 (UTC)

I'm not a Lie groups expert myself, but I think I understand what's going on, and you do need to allow the identity involution here. For example, the first source I found through Google (see http://www.jstor.org/stable/2160403 ) doesn't exclude the identity. I went ahead and changed it.

--Yzarc314 (talk) 17:57, 20 September 2011 (UTC)

puzzling examples

The last two examples seem puzzling:

1) the article states: Any real semisimple Lie algebra has a Cartan involution, and any two Cartan involutions are equivalent, and moreover the 2nd example clearly states Lie algebras of compact semisimple groups have a unique Cartan involution

2) but then we consider su(n) and show there are sometimes other inequivalent Cartan involutions

surely 2) contradicts 1)? — Preceding unsigned comment added by 78.239.179.184 (talk) 12:08, 14 February 2012 (UTC)