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I need to prove the statement below. Since my background on Lie theory is rather weak, I post it here.

Let $\frak{g}$ be a complex semi-simple Lie algebra. Fix a Cartan subalgebra $\frak{h}$ with roots $\Delta\subset\frak{h}^*$ and positive roots $\Delta^+\subset\Delta$, so that we have a decomposition $$ \frak{g}=\frak{h}\oplus\bigoplus_{\alpha\in\Delta}\frak{g}_\alpha. $$

Let $\eta\in\Delta^+$ be the longest root and $h_\eta\in\frak{h}$ be the corresponding coroot. Let $\rho: \frak{g}\rightarrow\frak{g}$ be a compact involution globally fixing $\frak{h}\oplus\frak{g}_{-\eta}\oplus\frak{g}_\eta$.

Problem. Prove that $\rho$ globally fixes $\mathbb{C}h_\eta\oplus\frak{g}_{-\eta}\oplus\frak{g}_\eta$.

For $\frak{sl}_n\mathbb{C}$, taking the usual $\frak{h}$ and $\Delta^+$, since $\rho$ corresponds to a hermitian metric on $\mathbb{C}^n$, I can prove the above statement by straightforward computations. Actually, $\rho$ globally fixes $\frak{h}\oplus\frak{g}_{-\eta}\oplus\frak{g}_\eta$ if and only if the corresponding hermitian metric is of the form $$ \begin{pmatrix} *&&&&&*\\ &*&&&&\\ &&*&&&\\ &&&*&&\\ &&&&*&\\ *&&&&&* \end{pmatrix}, $$ then computations show that $\rho$ globally fixes $\mathbb{C}h_\eta\oplus\frak{g}_{-\eta}\oplus\frak{g}_\eta$.

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  • $\begingroup$ It seems like everything you're asking transforms nicely under the Weyl group, so one could replace "highest root" by "any long root", even a simple one. $\endgroup$ Jun 14, 2014 at 2:37
  • $\begingroup$ What does it mean that a map $\rho$ globally fixes a subspace? I often get confused whether people mean that $\rho$ fixes the subspace (as in $\rho U \subseteq U$) or that $\rho$ fixes the points of the subspace (as in $\forall u \in U \rho u = u$). $\endgroup$ Jun 14, 2014 at 20:44
  • $\begingroup$ Globally fixing just means $\rho U\subset U$. This terminology is used in some literature in contrast to pointwise fixing. Some authors just say "fix" or "stabilize" though. $\endgroup$
    – Xin Nie
    Jun 15, 2014 at 5:29

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Actually the proof is just one line: because $$ [\frak{h}\oplus\frak{g}_\alpha\oplus\frak{g}_{-\alpha}, \frak{h}\oplus\frak{g}_\alpha\oplus\frak{g}_{-\alpha}]\subset \mathbb{C}h_\alpha\oplus\frak{g}_\alpha\oplus\frak{g}_{-\alpha}. $$

As noted by Allen, this works for any root and any map preserving the Lie bracket.

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