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I was reading non-linear realisation of a group on Wikipedia. On this page, it states: Given a Lie algebra $\mathfrak{g}$ and its Cartan subalgebra $\mathfrak{h}$, $\mathfrak{g}$ splits into $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{f}$ where $\mathfrak{f}$ is the supplement of the Cartan subalgebra such that $[\mathfrak{f},\mathfrak{f}]\subset\mathfrak{h}$ and $[\mathfrak{h},\mathfrak{f}]\subset\mathfrak{f}$.

While the second statement $[\mathfrak{h},\mathfrak{f}]\subset\mathfrak{f}$ is clearly true by definition of Cartan subalgebra as $\mathfrak{h}=${$g\in\mathfrak{g}|\exists n\in\mathbb{N}_0, ad(h)^ng=0\forall h\in\mathfrak{h}$}. I am not so certain about the first statement $[\mathfrak{f},\mathfrak{f}]\subset\mathfrak{h}$.

For example, let X and Y be two vectors in two different root spaces of the Lie algebra. If these two roots add up to become another root then the commutator of X,Y will be in the root space of the added root, clearly not in the Cartan subalgebra.

Could someone offer an explanation?

Rescy_
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    Yeah as written that is just plain wrong. Perhaps they meant to talk about a Cartan decomposition instead where that would be true (or more generally any symmetric decomposition) – Callum Jun 08 '23 at 20:53
  • Well the question is if among all possible vector space complements of $\mathfrak h$,there might still be an $\mathfrak f$ as demanded. What you show is that such $\mathfrak f$ cannot be of a certain form. That does not exclude another one (that does not contain such weight spaces) exists. – Torsten Schoeneberg Jun 09 '23 at 05:53
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    @TorstenSchoeneberg This is a fair point but it certainly can't work for all Lie algebras. A counterexample being $\mathfrak{g} = \mathfrak{u}(n)$. If a Cartan subalgebra $\mathfrak{h}\leq\mathfrak{g}$ admits such a complement $\mathfrak{f}$ then the full real flag manifold $U(n)/T^n$ would be a symmetric space but it this is not the case unless $n=2$ (when it is just projective space). – Callum Jun 09 '23 at 10:18
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    @TorstenSchoeneberg in fact, thinking about this some more, a symmetric decomposition like this forces $\mathfrak{h}$ and $\mathfrak{f}$ to be orthocomplements so $\mathfrak{f}$ has to be the span of the root spaces (or its real part) which as noted is a problem. – Callum Jun 09 '23 at 12:16

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