# Involutive automorphism of simple Lie algebra

I am sorry if this question is too elementary to be posted here, but no experts answer this question when I post it on Math Stackexchange.

Let $$\mathfrak{g}=\mathfrak{k}+\mathfrak{p}$$ be a Cartan decomposition for a noncompact real simple Lie algebra $$\mathfrak{g}$$ corresponding to a Cartan involution $$\theta$$, where $$\mathfrak{k}$$ is the maximal compact subalgebra of $$\mathfrak{g}$$. Suppose that $$\sigma$$ is another involutive automorphism of $$\mathfrak{g}$$ such that $$\sigma\theta=\theta\sigma$$. Then $$\sigma$$ preserves the Cartan decomposition, and $$\sigma|_\mathfrak{k}:\mathfrak{k}\rightarrow\mathfrak{k}$$. By the classification for symmetric pairs, it seems true that $$\sigma|_\mathfrak{k}$$ is never the identity map, but how to prove this fact theoretically (instead of case by case)?

In other words, there is no (non-Cartan) involutive automorphism $$\sigma$$ of a noncompact real simple Lie algebra $$\mathfrak{g}$$ such that the subalgebra $$\mathfrak{g}^\sigma$$ of the fixed points under the action of $$\sigma$$ on $$\mathfrak{g}$$ contains a maximal compact subalgebra. How to prove it? I shall be grateful if experts here may offer any hint.

• Did you delete the question on M.SE? I tried to find it, to link between the two, but I couldn't. – David Roberts Jun 4 '20 at 6:49
• @DavidRoberts Yes, I deleted the question on M.SE after I posted the question here. Do I need to undelete it? The link is math.stackexchange.com/questions/3703866/…, but I am not sure whether it appears. – Hebe Jun 4 '20 at 7:09
• No, it's ok. If it got no traction over there (like: no answers at all, no substantial comments), then it's better to not have a duplicate and zombie question floating around. The situation would be different if someone had put up a helpful answer but which still wasn't a full answer, or answered a slightly different question. – David Roberts Jun 4 '20 at 21:58
• I guess $\sigma=id$ doesn't count? – Torsten Schoeneberg Jun 7 '20 at 17:33
• @TorstenSchoeneberg I think that in general people do not regard the identity map as an involutive automrphism, but I am not sure. Anyway, here I just suppose that $\sigma$ is not identity on $\mathfrak{g}$. – Hebe Jun 8 '20 at 8:52

Let $$\mathfrak g$$ be a noncompact simple Lie algebra and let $$\mathfrak g=\mathfrak k+\mathfrak p$$ be a Cartan decomposition. The simplicity of $$\mathfrak g$$ implies that the adjoint representation of $$\mathfrak k$$ on $$\mathfrak p$$ is irreducible (indeed, if $$\mathfrak p_1$$ is an $$\mathrm{ad}_\mathfrak k$$-invariant subspace of $$\mathfrak p$$, one can show$$^\dagger$$ that $$\mathfrak g_1:=[\mathfrak p_1,\mathfrak p_1]+\mathfrak p_1$$ is an ideal of $$\mathfrak g$$). In particular, $$\mathfrak k$$ is a maximal (not only maximal compact) subalgebra of $$\mathfrak g$$. Indeed, if $$\mathfrak h$$ were a subalgebra containing $$\mathfrak k$$, then $$\mathfrak h =\mathfrak k +\mathfrak h\cap\mathfrak p$$ and $$\mathfrak h\cap\mathfrak p$$ would be an $$\mathrm{ad}_{\mathfrak k}$$-invariant subspace of $$\mathfrak p$$.
$$\dagger$$ Write $$\mathfrak p=\mathfrak p_1+\mathfrak p_2$$ $$\mathrm{ad}_{\mathfrak k}$$-invariant decomposition. Let $$X_i\in\mathfrak p_i$$ and $$Y=[X_1,X_2]\in\mathfrak k$$. Denote the Cartan-Killing form of $$\mathfrak g$$ by $$B$$; it is negative definite on $$\mathfrak k$$ and positive definite on $$\mathfrak p$$; in particular, we may assume $$B(\mathfrak p_1,\mathfrak p_2)=0$$. Now $$B(Y,Y)=B(Y,[X_1,X_2])=B([Y,X_1],X_2)\in B(\mathfrak p_1,\mathfrak p_2)=0$$, so $$Y=0$$. We have shown that $$[\mathfrak p_1,\mathfrak p_2]=0$$. Using Jacobi, one completes the check that $$\mathfrak g_1$$ is an ideal of $$\mathfrak g$$.