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Let $\mathfrak{g}$ be a simple finite-dimensional complex Lie algebra, and let $\theta$ be a complex linear involution on $\mathfrak{g}$. Let $\mathfrak{a}$ be a Cartan subspace, and choose a $\theta$-stable Cartan subalgebra $\mathfrak{h}$ containing $\mathfrak{a}$. Finally, make a choice of positive restricted roots in $\mathfrak{a}^*$, and extend this to a choice of positive roots in $\mathfrak{h}^*$, giving rise to a positive system for the roots of $\mathfrak{g}$.

Let $\omega$ denote the Chevalley involution of $\mathfrak{g}$ with respect to $\mathfrak{h}$ and the choice of positive system just described. To be explicit, $\omega$ is the unique automorphism of $\mathfrak{g}$ such that $\omega(h)=-h$ for $h\in\mathfrak{h}$, and $\omega(e_{\pm\alpha})=e_{\mp\alpha}$ for all simple roots $\alpha$.

Is it true that $\omega$ commutes with $\theta$?

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  • $\begingroup$ I think you have made a choice of pinning (choice of non-$0$ vector $e_\alpha$ in each root space for a simple root $\alpha$) that you are not mentioning; and then of course what you really mean is that $[e_\alpha, \omega(e_\alpha)]$ is the coroot $h_\alpha$, and $\omega^2(e_\alpha) = e_\alpha$. (A different choice of pinning gives a different, though conjugate, Cartan involution, and this can affect commuting with $\theta$.) This already forces $h_\alpha$ to be negated by $\omega$; so, since $\mathfrak g$ is simple, $\omega$ acts by negation on $\mathfrak h$—it's not a separate condition. $\endgroup$
    – LSpice
    Jan 2, 2021 at 16:09
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    $\begingroup$ Yes, you are right, I was being sloppy. My statement wasn't meant to depend on the choice of $\mathfrak{sl}_2$-triples, although as we now see it's false anyway :) $\endgroup$
    – freeRmodule
    Jan 4, 2021 at 9:29

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I've realized that the answer is no. I should have checked some examples! The case of $(\mathfrak{gl}(4),\mathfrak{gl}(3)\times\mathfrak{gl}(1))$ gives a counterexample. Indeed, this is induced by the involution $\theta$ given by conjugation by $$\begin{bmatrix}0 & 0 & 0 & i\\0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\-i &0 &0&0\end{bmatrix}.$$ Then $\theta$ preserves the Cartan subalgebra of diagonal matrices, which contains a Cartan subspace for this pair. One choice of positive system compatible with the Iwasawa decomposition in this case is the usual one, giving the Borel of upper triangular matrices. Thus a Chevalley involution $\omega$ is given by $X\mapsto-X^t$.

Now one can compute that $\omega\theta(e_{12})=-ie_{24}$ while $\theta\omega(e_{12})=ie_{24}$.

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  • $\begingroup$ Isn't the centraliser of your involution $\operatorname{GL}_2 \times \operatorname{GL}_1^2$, embedded as $(g, s, t) \mapsto \frac1 2\begin{pmatrix} s + t & & & s - t \\ & 2g \\ t - s & & & s + t \end{pmatrix}$, not $\operatorname{GL}_3 \times \operatorname{GL}_1$? Maybe I misunderstand the notation. I think if you choose your Cartan subalgebra to be the diagonal inside $\operatorname{GL}_2$ times these two $\operatorname{GL}_1^2$'s, then things go better. $\endgroup$
    – LSpice
    Jan 2, 2021 at 15:50
  • $\begingroup$ Namely, your involution is conjugate to $\operatorname{diag}(\zeta_8, 1, 1, -\zeta_8)$, where $\zeta_8$ is a primitive 8th root of unity, and it is clear that that automorphism behaves well with respect to the standard diagonal torus. $\endgroup$
    – LSpice
    Jan 2, 2021 at 16:04
  • $\begingroup$ More generally, everything works fine for an inner automorphism (necessarily semisimple) if you choose a Cartan subalgebra corresponding to a torus containing the conjugating element, so you're fine there if you only worry about existence. Then you need only worry about outer automorphisms. $\endgroup$
    – LSpice
    Jan 2, 2021 at 16:11
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    $\begingroup$ Ah I see, yes so a Cartan subspace is a maximal toral subalgebra of the (-1)-eigenspace, (not just abelian). Toral means all the elements are semisimple. And then yeah I think when it comes to symmetric pairs, the centralizer of interest is that of the Cartan subspace. So when I said centralizer I should have been more precise and said centralizer of the Cartan subspace. The fixed points of the involution are just the fixed points in the usual sense. $\endgroup$
    – freeRmodule
    Jan 4, 2021 at 9:27
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    $\begingroup$ Questions without upvoted answers get periodically bumped to the front page. $\endgroup$
    – LSpice
    Jun 1, 2021 at 20:10

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