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Let $G$ be a compact connected Lie group and let $T$ be a maximal torus of $G$. Denote by $\mathcal{F}=G/T$ the flag manifold of $G$. Let $\theta : G \rightarrow G$ be an involution on $G$. This involution induces an involution on the Lie algebra of $G$, that we denote also by $\theta$.

Consider the natural action of $G$ on $\mathcal{F}$. Let $x=gT \in \mathcal{F}$, denote by $G_x$ the stabilizer of $x$ in $G$ and denote by $\mathfrak{g}_x$ its Lie algebra. (Note that $G_x =gTg^{-1}$).

Consider the set $M:= \lbrace x \in \mathcal{F}, \theta(\mathfrak{g}_x)=\mathfrak{g}_x \rbrace $.

$\textbf{Question}:$ Prove or disprove that $M$ is a submanifold of $\mathcal{F}$.

So far I didn't find any example for which $M$ is not a submanifold, and my thoughts are: Since the group $G$ acts transitively on the set $\tilde{\mathfrak{g}}:=\lbrace \mathfrak{g}_x , x \in \mathcal{F}\rbrace $ (this is because $\tilde{\mathfrak{g}}$ is the set of all cartan subalgebras of $\mathfrak{g}$) and since the set $M$ is a fixed point set of the involution $\theta$, then using this On the proof of that fixed point set of an involution is a submanifold), I conclude that $M$ is a submanifold. is this correct ?

$\textbf{Edit:}$ $T$ is assumed to be stable by $\theta$.

Asma
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  • By the argument in the link you've posted, this is already true, no? You have defined an involution on the flag manifold and then its fixed points form a submanifold. I don't think you need transitivity of the group action here unless you want to show that there is only one component (I am not sure that is automatic even then). I do suspect you want $\theta$ to be an involutive automorphism on the group/algebra in order to induce a proper involution on $\mathcal{F}$ as otherwise $\theta(\mathfrak{g}_x)$ may not be another point in $\mathcal{F}$ – Callum Sep 07 '22 at 20:00
  • Moreover, since you have defined the flag manifold using a compact group quotiented by a torus (rather than via parabolic subgroups) it has a natural $G$-invariant Riemannian metric and so you can presumably use the Theorem in Klingenberg your link refers to. – Callum Sep 07 '22 at 20:11
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    @Callum: As far as I can tell, $\theta$ need not induce a diffeomorphism of the flag manifold. For example, on $G = S^3$ thought of as unit quaternions with $T = S^1$ though of as unit complex numbers, and if $\theta$ is conjugation by $(i+j)/\sqrt{2}$, then $\theta$ does not seem to descend to a well defined map of $G/H\cong S^2$. – Jason DeVito - on hiatus Sep 08 '22 at 16:00
  • @JasonDeVito at which point does it fail? Any automorphism, inner or otherwise, should preserve the set of CSA's so is it just smoothness that breaks? – Callum Sep 09 '22 at 06:28
  • Upon reflection it occurs to me that a Cartan subalgebra does not stabilise a unique flag so while $M$ can be well defined $\theta$ is not well defined on $\mathcal{F}$ since $\theta(\mathfrak{g}_x)$ can stabilise multiple points. I had naively assumed that there was a bijective correspondence between $\mathcal{F}$ and $\tilde{\mathfrak{g}}$ as there is between $\mathcal{F}$ and the set of Borel subalgebras of some other Lie algebra but this is not the case. – Callum Sep 09 '22 at 09:17
  • So instead you have $\tilde{\mathfrak{g}}$ which is a quotient of $\mathcal{F}$ (by the action of the Weyl group?) and what we have is that $M$ is the preimage of a submanifold of $\tilde{\mathfrak{g}}$ by this quotient map. I don't believe that a submanifold of a quotient has to be a quotient of a submanifold in general but I'm not sure what can be done in this case. – Callum Sep 09 '22 at 09:29
  • I'm sorry, I now realize that I forgot to mention that we should assume that $T$ is $\theta$-stable (we can always find such $T$). In this case, I think that $\theta$ descends to $G/T$, right? – Asma Sep 09 '22 at 11:19
  • I have mentioned in my question that $G$ acts transitively on the set $\tilde{\mathfrak{g}}$, in order to prove that $\tilde{\mathfrak{g}}$ is a manifold, and to construct a diffemorphism between $M$ and the set $ \lbrace \alpha \in \tilde{\mathfrak{g}}, \theta (\alpha)= \alpha \rbrace$, which maps $x \in M$ to $\mathfrak{g}_x \in \lbrace \alpha \in \tilde{\mathfrak{g}}, \theta (\alpha)= \alpha \rbrace$ . – Asma Sep 09 '22 at 11:28
  • @Soumia Choosing $T$ to be $\theta$-stable doesn't change anything I'm afraid. In general the choice of a specific $H_x$ for a homogeneous space $G/H_x$ is arbitrary which is why we can denote it $G/H$ and not be too specific on which $H$. We need more restrictions on $\theta$ to make it descend to $G/T$ – Callum Sep 09 '22 at 18:03
  • @Callum, I had naively assumed that $\theta$ is an automorphism of Lie groups, that's why I said that $\theta$ descends to an involution on $G/T$. If we do assume that $\theta$ is an involutive autuomorphism on $G$, then my attempt to prove that $M$ is a submanifold of $\mathcal{F}$ is correct, or I'm missing something ? – Asma Sep 09 '22 at 19:15
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    I think that if $T$ is $\theta$-stable, this does change things. The natural way to define the $\theta$ action on $G/T$ is $\theta(gT) = \theta(g)T$. This is well-defined if $T$ is $\theta$-stable: for $t\in T$, we have $\theta(gtT) = \theta(gt)T = \theta(g)\theta(t) T = \theta(g)T$. In short, if $\theta$ stabilizes with $T$, then I agree with Soumia that we do have an action of $\theta$ on $G/T$. – Jason DeVito - on hiatus Sep 09 '22 at 19:23

1 Answers1

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$M$ is always a submanifold.

As you suggested, we can look at the space $\tilde{\mathfrak{g}}$. However, $\tilde{\mathfrak{g}}$ is not diffeomorphic to $\mathcal{F}$. Rather, it is a quotient of $\mathcal{F}$.

More specifically, we have:

Proposition: Let $N = N_G(T)$ be the normalizer of $T$ in $G$. Then $\tilde{\mathfrak{g}}$ is diffeomorphic to $G/N$.

Proof: As you noted, $G$ acts transitively on $\tilde{\mathfrak{g}}$, so we only need to determine the stabilizer at a point. Consider a point $\mathfrak{g}_x$ with $x\in T$, $g\in G$ stabilizes this set iff $Ad_g(\mathfrak{t}) = \mathfrak{t}$ (where $\mathfrak{t}$ denotes the Lie sub-algebra of $T$). Exponentiating out, we find that conjugation by $g$ stablizes $T$, so $g\in N$. Conversely, if $g\in N$, then $g$ stabilizes $T$, so $Ad_g$ stabilizes $\mathfrak{t}$. $\square$.

Simply because $G/N$ is easier to type, I'll stop writing $\tilde{\mathfrak{g}}$ and instead right $G/N$ for the duration of this post.

Note that since $T\subseteq N$, we have a natural projection $\pi:G/T\rightarrow G/N$. This projection is actually a covering with $|N/T|$ sheets. (More generally, the projection is a fiber bundle with fiber $N/T$, but in the case of compact Lie groups, it is well-known that the Weyl group $N/T$ is finite.)

Proposition: The $\theta$ action on $G$ induces a $\theta$ action on $G/N$. Under the diffeomorphism $G/N\rightarrow \tilde{\mathfrak{g}}$ this $\theta$ action is equivariantly diffeomorphic to the $\theta$ action on $\tilde{\mathfrak{g}}$ given by $\mathfrak{g}_x\mapsto \theta(\mathfrak{g}_x)$.

Proof: Define $\theta(gN) = \theta(g)N$. To see this is well-defined, let $n\in N$. Then $\theta(n) \in N_G(\theta(T)) = N_G(T) = N$, so $$\theta(gnN) = \theta(gn)N = \theta(g)\theta(n)N = \theta(g)N.$$

For the second statement, consider the diffeomorphism $\psi:G/N\rightarrow \tilde{\mathfrak{g}}$ given by $\psi(gN) = Ad_g \mathfrak{t}$. Then $$\psi(\theta(gN)) = \psi(\theta(g)N) = Ad_{\theta(g)} \mathfrak{t} = Ad_{\theta(g)} \theta(\mathfrak{t}) = \theta(Ad_g \mathfrak{t}) = \theta(\psi(gN)).$$

$\square$

The induced $\theta$ action on $G/N$ is obviously an involution, so, as you already know, this implies $\mathrm{Fix}(\theta)\subseteq G/N$ is an embedded submanifold of $G/N$.

But we also obviously have that $M = \pi^{-1}(\mathrm{Fix}(\theta))$. Thus, to finish off the proof, we need only note that the inverse image of a submanifold under a covering is itself a submanifold. Indeed, slice charts on $\mathrm{Fix}(\theta)$ pull back to to slice charts on $M$.

Callum
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Jason DeVito - on hiatus
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