Killing form that is not diagonalizable


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?

Best Answer

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.

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