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Let $\mathfrak{g}$ be a simple Lie algebra with a compact subalgebra $\mathfrak{k}$ such that $(\mathfrak{g},\mathfrak{k})$ corresponds to an irreducible Riemann symmetric space. Denote by $\sigma$ be the involutive automorphism. Suppose that $\mathfrak{g}=\mathfrak{k}+\mathfrak{p}$ where $\mathfrak{p}$ is the eigenspace of $\sigma$ on $\mathfrak{g}$ with eigenvalue $-1$. Thus, $\mathfrak{p}$ must be an irreducible (CORRECT?) $\mathfrak{k}$-module.

Now suppose that $\mathfrak{g}$ is non-compact, and let $\tau$ be an arbitrary involutive automorphism of $\mathfrak{g}$. Then there exists a Cartan involution $\theta$ commuting with $\tau$, and one has $\mathfrak{g}=\mathfrak{g}^{\theta,\tau}+\mathfrak{g}^{\theta,-\tau}+\mathfrak{g}^{-\theta,\tau}+\mathfrak{g}^{-\theta,-\tau}$. Let $\mathfrak{u}:=\mathfrak{g}^{\theta,\tau}+\sqrt{-1}\mathfrak{g}^{\theta,-\tau}+\sqrt{-1}\mathfrak{g}^{-\theta,\tau}+\mathfrak{g}^{-\theta,-\tau}$, and then $\tau$ is a Cartan involution of $\mathfrak{u}$ with the maximal compact subalgebra $\mathfrak{k}':=\mathfrak{g}^{\theta,\tau}+\sqrt{-1}\mathfrak{g}^{-\theta,\tau}$. Thus $(\mathfrak{u},\mathfrak{k}')$ corresponds to an irreducible Riemann symmetric space. Write $\mathfrak{p}':=\sqrt{-1}\mathfrak{g}^{\theta,-\tau}+\mathfrak{g}^{-\theta,-\tau}$, and then $\mathfrak{p}'$ is irreducible as $\mathfrak{k}'$-module. (CORRECT?)

Moreover, is $\mathfrak{g}^{-\tau}=\mathfrak{g}^{\theta,-\tau}+\mathfrak{g}^{-\theta,-\tau}$ irreducible as $\mathfrak{g}^\tau=\mathfrak{g}^{\theta,\tau}+\mathfrak{g}^{-\theta,\tau}$-module?

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  • $\begingroup$ I assume you mean that ${\mathfrak g}$ is a real Lie algebra (and ${\mathfrak k}$ is a {\it maximal} compact subalgebra)? I think irreducibility can be checked case-by-case. Probably your second question can be checked in the same way. $\endgroup$ – Paul Levy Apr 30 '17 at 16:22
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It is true that $\mathfrak p$ is always an irreducible $\mathfrak k$-module. Be aware though that the complexification $\mathfrak p_{\mathbb C}$ might be reducible as an $\mathfrak k_{\mathbb C}$-module. This happens precisely in the case that $\mathfrak g$ is of Hermitian type.

Edit: If $\tau$ is not a Cartan involution then $\mathfrak g^{-\tau}$ may be a reducible $\mathfrak g^\tau$-module even over $\mathbb R$. The criterion is whether the center of $\mathfrak g^\tau$ is split or not. A typical example is $\mathfrak g=\mathfrak{sl}(2,\mathbb R)$ and $\tau$ is conjugation by $diag(1,-1)$.

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  • $\begingroup$ Thank you so much. I understand that if $\mathfrak{g}$ is of Heritian type with a Cartan decomposition $\mathfrak{g}=\mathfrak{k}+\mathfrak{p}$, then $\mathfrak{p}_\mathbb{C}=\mathfrak{p}_++\mathfrak{p}_-$ as $\mathfrak{k}_\mathbb{C}$-module. However, I am not sure that, if $\mathfrak{g}$ is a non-compact non-complex simple Lie algebra, and $\tau$ is an involutive automorphism (not a Cartan involution), is $\mathfrak{g}^{-\tau}$ irreducible as $\mathfrak{g}^\tau$-module in general? $\endgroup$ – Hebe May 1 '17 at 6:07
  • $\begingroup$ @Hebe See my edit. $\endgroup$ – Friedrich Knop May 1 '17 at 8:02

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