Talk:Cartan decomposition

message: 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)
thread_title: Untitled

message: 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)
thread_title:  Definition of an involution 

message: 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)
replies: text: Right now, it says that the only involution over su(n) that is a Cartan involution is the identity one. --Svennik (talk) 08:38, 20 July 2020 (UTC)
thread_title:  puzzling examples 

message: Substituted at 01:51, 5 May 2016 (UTC)
thread_title: Assessment comment

message: Some authors, such as Fomin and Zelevinsky in Double Bruhat cells and total positivity, refer to the Cartan decomposition of \mathfrak{g} as \mathfrak{n}_{-} \oplus \mathfrak{h} \oplus \mathfrak{n}, which I think(?) is the root space decomposition (or is at least closely related to it). It would be very helpful to have a section explaining the relationship between these two meanings of Cartan decomposition, though I am incapable (right now) of writing it. 129.2.56.142 (talk) 13:39, 5 April 2017 (UTC)
replies: text: I know nothing about Bruhat cells, but guessing wildly, one of the subspaces is supposed to be the nilpotent part. And \mathfrak{h} would be the maximally compact part, see symmetric space which says stuff that this article doesn't. The nilpotent part is given by the Jordan–Chevalley decomposition aka just the Jordan decomposition for short; unfortunately, that article is about matrices in general and not about Lie algebras. All this should come together here. I've seen three-way decompositions like this before but was unclear on the details. 67.198.37.16 (talk) 05:31, 30 October 2020 (UTC)
thread_title:  <math>\mathfrak{n}_{-} \oplus \mathfrak{h} \oplus \mathfrak{n}</math> 

message: So, the Cartan involution is defined in terms of the Killing form. The comments "Untitled" above points out that this defintion won't work for GL(n,R).  Yet, later on in the article, it states that "For the general linear group,  X \mapsto (X^{-1})^T  is a Cartan involution." which clearly is incompatible with the Killing-form definition... but is apparently needed to resolve the complaint immediiately above ("Is the following line true?"). So there is something fishy going on here.  67.198.37.16 (talk) 06:52, 30 October 2020 (UTC)
thread_title: Inconsistency!

message: I'm reading texts which define the "generalized transpose" # and am tempted to add the following to the  article:
:The “generalized transpose” (\cdot)^\# is induced by the Cartan involution \theta as X^{\#}=-\theta\left(X\right) for any X\in\mathfrak{g}. It is always possible to find some matrix representation of the adjoint representation of \mathfrak{g}, such that this generalized transpose is the ordinary matrix transpose (this representation will in general be larger than the adjoint rep.). It is lifted to all of G  by defining \left(\exp X\right)^{\#}=\exp X^{\#} . If the subgroup of G  generated by the compact generators is orthogonal, then the fundamental rep can be written such that X^{\#}=X^{T}.  If the compact generators generate a unitary subgroup, then there is a fundamental rep such that X^{\#}=X^{\dagger}. After lifting \# to G, it is clear that the invariant subgroup of the Cartan involution is precisely the elements O\in G obeying O^\# O=1.  
But before I add this, is this just weird idiosyncratic notation, or is it more widely used? (FWIW, I cribbed the above from "The group theory of oxidation" Arjan Keurentjes and he ascribes the notation to an earlier paper E. Cremmer, B. Julia, H. Lu and C. N. Pope, “Dualisation of dualities. I,” Nucl. Phys. B 523 (1998) 73 [arXiv:hep-th/9710119]. which I have not looked at. I got to the oxidation paper via a third paper, so at least three different papers use this notation...) 67.198.37.16 (talk) 06:27, 30 October 2020 (UTC)
replies: text: I've added what I believe to be the correct definition of the "generalized transpose" to symmetric space. I found a nice, general definition in Jurgen Jost "Riemannian Geometry & Geometric Analysis" (Third edition section 5.3 page 256) I believe it's 100% identical to the above, but I still want to double-check that. 67.198.37.16 (talk) 06:25, 10 November 2020 (UTC)
thread_title:  Generalized transpose # 

message: Some authors define (in passing) the "Chevalley involution" as \omega:H_i\to -H_i and \omega:E_\alpha\to -E_{-\alpha} and it is then used as a synonym for "Cartan involution", in that it is used to perform a decomposition of real split-form algebras into into a maximally compact part plus the rest. Is this the same thing? The article semisimple Lie algebra that explains this just links back to here, leaving things unclear... 67.198.37.16 (talk) 17:03, 31 October 2020 (UTC)
replies: text: Yes, it is the same thing, at least in the case of sl(n) with all standard assumptions. The roots are then \alpha\in\{\varepsilon_i-\varepsilon_j, i\neq j\}. The element E_{\varepsilon_i-\varepsilon_j} is the matrix with a single 1 in the i^th row, j^th column. Negative transpose sends it to -E_{\varepsilon_j-\varepsilon_i}: minus the matrix with 1 in the j^th row, i^th column. Tition1 (talk) 12:01, 1 November 2021 (UTC)
replies: text:  Note that I added an indent to make your answer more readable. --Svennik (talk) 13:25, 2 November 2021 (UTC)
thread_title:  Chevalley involution