An example Lie algebra $L$ with non-diagonalizable Killing form would have to be non-semisimple, and the Killing form complex. (Otherwise diagonalizability is obvious.) I tried with a few $L$, but (in a proper base) the result always was of the type $diag(0,\dots,0,1,\dots,1)$ however I chose $L$. Do you have an example $L$ or a proof of diagonalizability?
$\begingroup$
$\endgroup$
1
-
3$\begingroup$ Isn't every complex quadratic form diagonalizable? $\endgroup$– abxCommented May 7 at 10:08
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The Killing form is a complex quadratic-form and therefore given by complex symmetric (note - not Hermitian!) matrix (with respect to some basis). \begin{align} B(x,y) = x^T A y, \end{align} complex symmetric matrices may be diagonalised by a unitary and it's transpose (note - not conjugate-transpose!). I.e. for any complex symmetric $A$ there exists a unitary $U$ and diagonal matrix $D$, with real, non-negative entries such that \begin{align} D = U A U^T. \end{align} This is known as the Autonne–Takagi factorization (although it was also discovered by Hua and others).
I would like to emphasise that this is not the usual spectral theorem that applies to normal matrices as complex symmetric matrices do not have to be normal.
-
$\begingroup$ Does this mean that if my Killing form is A, I can choose the basis of the Lie algebra so that it is D? (Since this isn't the "usual" diagonalization.) $\endgroup$ Commented May 7 at 17:47
-
$\begingroup$ @HaukeReddmann yes - this diagonalization is exactly what you want in this case because your form is given by $x^T A y$, and not $x^\dagger A y$. $\endgroup$– orsCommented May 8 at 13:05