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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.

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  • $\begingroup$ Did you delete the question on M.SE? I tried to find it, to link between the two, but I couldn't. $\endgroup$
    – David Roberts
    Jun 4, 2020 at 6:49
  • $\begingroup$ @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. $\endgroup$
    – Hebe
    Jun 4, 2020 at 7:09
  • $\begingroup$ 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. $\endgroup$
    – David Roberts
    Jun 4, 2020 at 21:58
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    $\begingroup$ I guess $\sigma=id$ doesn't count? $\endgroup$ Jun 7, 2020 at 17:33
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    $\begingroup$ @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 $\mathfrak{g}$. $\endgroup$
    – Hebe
    Jun 8, 2020 at 8:52

1 Answer 1

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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 $\mathfrak g$.

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  • $\begingroup$ Thank you for the answer. So you just show that a maximal compact subalgebra in a simple Lie algebra is automatically a maximal subalgebra. Interesting. $\endgroup$
    – Hebe
    Sep 8, 2020 at 8:20

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