# Do rational points in a split reductive group act transitively on the orbits of the Cartan subalgebra (w.r.t. automorphism group of Lie algebra)?

Let $$(G,T,M)$$ be a split reductive group (over say, the integers), with Lie algebra $$(\mathfrak{g}, \mathfrak{t})$$, and let $$R$$ be a commutative ring. When $$R$$ is an algebraically closed field, it is a well-known theorem (of Cartan for $$R = \mathbb{C}$$, and Chevalley in characteristic zero, and I think Humphreys in general) that all Cartan $$R$$-subalgebras of $$\mathfrak{g}_R$$ are $$G(R)$$-conjugate, but this fails for arbitrary $$R$$. For example, when $$R$$ is the field of real numbers and $$G=\operatorname{SL}_2$$, the Lie algebra of a non-split maximal torus is not $$G(R)$$-conjugate to $$\mathfrak{t}_R$$). Instead of this general conjugacy, I'd like to know how well the action of $$G(R)$$ approximates the action of the group of $$R$$-Lie algebra automorphisms of $$\mathfrak{g}_R$$.

Question 1: Does $$G(R)$$ act transitively on the $$\operatorname{Aut}_{R\text{-Lie}} \mathfrak{g}_R$$-orbit of $$\mathfrak{t}_R$$?

There is a group scheme analogue of this:

Question 2: Does $$G(R)$$ act transitively on the $$\operatorname{Aut}_{R\text{-gp}.} G_R$$-orbit of $$T_R$$?

It seems pretty clear that a positive answer to the first question implies a positive answer to the second, but I don't know if the reverse holds. As far as I can tell, the answer to both questions is positive when the canonical isogeny $$Z(G) \times D(G) \to G$$ is an isomorphism (that is, when $$G$$ is a direct product of a torus and an adjoint type group). The problem is that, while the adjoint actions of $$G$$ on $$G_R$$ and $$\mathfrak{g}_R$$ factor through the quotient $$G/Z(G)$$ by the center, the scheme-theoretic surjection $$G \to G/Z(G)$$ may not yield a surjection on rational points! There is a cokernel that lies in a flat $$H^1$$ group. Even so, I have not been able to construct an example giving a negative answer to either question.

Naturally, I would appreciate any insights even in the case of fields, or with restricted characteristic.

Edit 9 hours later: L. Spice has helpfully pointed out in a comment that when $$R$$ is a field, then the answer to question 2 is "yes". I now see that this is theorem 20.9(ii) in Borel's Linear algebraic groups. I haven't gone through the details yet, but it looks like the proof in the text can be extended to conjugacy that is Zariski local in $$R$$.

• In $(G, T, M)$, I guess $G$ is the ambient reductive group and $T$ is a split maximal torus, but what is $M$? Anyway, over $R$ a field, all split tori are $G(R)$-conjugate, so the answer to Question 2 in that case is 'yes'. Mar 10, 2020 at 18:33
• (Based on my skimming of math.stanford.edu/~conrad/papers/luminysga3.pdf to try to answer your question, I guess that $M$ is the character lattice of $T$.) Mar 10, 2020 at 20:21
• (With $R$ field, $T$ nonsplit) For $G=\mathrm{SL}_2$, write $s=\mathrm{diag}(3,1)$ (so neither $s$ nor $-s$ is sum of two squares in $\mathbf{Q}$), $T$ the standard $\mathrm{SO}_2$. Then $sT_\mathbf{Q}s^{-1}$ is not in the $\mathrm{SL}_2(\mathbf{Q})$-orbit of $T_\mathbf{Q}$. Indeed, if it were equal to $s_1T_\mathbf{Q}s_1^{-1}$ with $s_1\in\mathrm{SL}_2(\mathbf{Q})$, then $t=s_1^{-1}s\in\mathrm{GL}_2(\mathbf{Q})$ normalizes $T_\mathbf{Q}$, so is in $\mathrm{O}(2)_\mathbf{Q}$ and has determinant $3$. But the determinant of a matrix in $\mathrm{O}(2)$ has determinant of the form $\pm (x^2+y^2)$.
– YCor
Mar 10, 2020 at 21:00
• @YCor, I thought the question was asking specifically about the case where $T$ is split? Also, what do you mean by asking whether a diagonal matrix is or is not a sum of two scalar squares? Mar 10, 2020 at 21:32
• @LSpice it's a typo: I just mean neither $3$ or $-3$ is a sum of 2 squares (in $\mathbf{Q}$. Yes I didn't guess whether OP's exclusively interested in split $T$ although it seems so; anyway I did the comment.
– YCor
Mar 10, 2020 at 21:43

When $$R$$ is a field, the answer to Question 1 is Yes (at least in char 0) with the same proof as for Question 2. For simplicity, we write $$G$$ for $$G_R$$, $$T$$ for $$T_R$$, etc.

Theorem. Let $$G$$ be a split (connected) reductive group over a field $$R$$ of characteristic 0, and let $$T\subset G$$ be a split maximal torus. Write $${\mathfrak g}={\rm Lie\,} G$$ and $${\mathfrak t}={\rm Lie\,} T$$. Let $$a\in{\rm Aut\,} {\mathfrak t}$$ be an automorphism of $${\mathfrak g}$$, and set $${\mathfrak t}'=a({\mathfrak t})\subset {\mathfrak g}$$. Then there exists $$g\in G(R)$$ such that $${\mathfrak t}'={\rm Ad}(g)({\mathfrak t})$$.

Proof. First let us canonically decompose $${\mathfrak g}$$ and $${\mathfrak t}$$. Write $${\mathfrak z}$$ for the center of $${\mathfrak g}$$, and write $${\mathfrak g}^{\rm ss}=[{\mathfrak g},{\mathfrak g}]$$, which is a semisimple Lie algebra. Then $${\mathfrak g}={\mathfrak z}\dotplus {\mathfrak g}^{\rm ss}$$ (direct sum) and hence, $${\rm Aut\,} {\mathfrak g}={\rm Aut\,} {\mathfrak z}\times {\rm Aut\,}{\mathfrak g}^{\rm ss}.$$ Set $${\mathfrak t}^{\rm ss}={\mathfrak t}\cap{\mathfrak g}^{\rm ss}$$; then $${\mathfrak t}={\mathfrak z}\dotplus{\mathfrak t}^{\rm ss}$$. Write $$a=(a^{\mathfrak z}, a^{\rm ss})\in {\rm Aut\,} {\mathfrak z}\times {\rm Aut\,}{\mathfrak g}^{\rm ss}={\rm Aut\,} {\mathfrak g}.$$ Then it is clear that $$a({\mathfrak t})=a^{\mathfrak z}({\mathfrak z})\dotplus a^{\rm ss}({\mathfrak t}^{\rm ss})={\mathfrak z}\dotplus a^{\rm ss}({\mathfrak t}^{\rm ss})=a^{\rm ss}({\mathfrak t}).$$ Therefore, we may and shall assume that $$a\in{\rm Aut\,}{\mathfrak g}^{\rm ss}$$.

Now let us canonically decompose $$G$$. Write $$Z^0=Z(G)^0$$ (the identity component of the center of $$G$$). Then $$Z^0$$ is a torus with Lie algebra $${\rm Lie\,} Z^0={\mathfrak z}$$. Write $$G^{\rm ss}=[G,G]$$ (the commutator subgroup of $$G$$), which is a connected semisimple group with Lie algebra $${\rm Lie\,} G^{\rm ss}={\mathfrak g}^{\rm ss}$$. Let $$G^{\rm sc}$$ denote the universal cover of $$G^{\rm ss}$$. Then $$G^{\rm sc}$$ is a simply connected semisimple group with Lie algebra $${\rm Lie\,} G^{\rm sc}={\mathfrak g}^{\rm ss}$$. We have a canonical homomorphism $$\rho\colon G^{\rm sc}\to G^{\rm ss}\hookrightarrow G$$ and a decomposition (in general not direct) $$G=Z^0\cdot\rho(G^{\rm sc}).$$

We can canonically decompose $$T$$. Set $$T^{\rm ss}=T\cap G^{\rm ss}$$, and let $$T^{\rm sc}$$ denote the preimage of $$T^{\rm ss}$$ in $$G^{\rm sc}$$. Then $$T^{\rm ss}$$ is a maximal torus in $$G^{\rm ss}$$, and $$T^{\rm sc}$$ is a maximal torus in $$G^{\rm sc}$$. We have $$T=Z^0\cdot T^{\rm ss}=Z^0\cdot \rho(T^{\rm sc}).$$

We may and shall identify $${\rm Aut\,}{\mathfrak g}^{\rm ss}$$ with $${\rm Aut\,} G^{\rm sc}$$; then $$a\in {\rm Aut\,} G^{\rm sc}$$. Set $$T'=Z^0\cdot \rho(a(T^{\rm sc}))\subset G$$. Then $$T'$$ is a maximal torus in $$G$$ with Lie algebra $${\rm Lie\,} T'={\mathfrak z}\dotplus{\rm Lie\,} a(T^{\rm sc})={\mathfrak z}\dotplus a({\rm Lie\,} T^{\rm sc})={\mathfrak z}\dotplus a({\mathfrak t}^{\rm ss})=a({\mathfrak t})={\mathfrak t}'.$$ Since $$T$$ is a split torus, the tori $$Z^0$$, $$T^{\rm ss}$$, $$T^{\rm sc}$$, and $$a(T^{\rm sc})$$ are all split. Thus $$T'$$ is a split maximal torus in $$G$$. By Theorem 20.9(ii) in Borel's book Linear Algebraic Groups, there exists $$g\in G(R)$$ such that $$T'=g\cdot T\cdot g^{-1}$$. Then $${\mathfrak t}'={\rm Ad}(g)({\mathfrak t})$$, as required.

• The decomposition $\mathfrak g = \mathfrak z \oplus \mathfrak g^{\text{ss}}$ can fail, for example, if $R = \mathbb F_p$ and $G = \operatorname{SL}_p$. (Also the equality $\operatorname{Lie} G^{\text{sc}} = \mathfrak g^{\text{ss}}$.) Mar 12, 2020 at 17:48
• Finally, why does $\operatorname{Aut} \mathfrak g^{\text{ss}}$ equal $\operatorname{Aut} G^{\text{sc}}$? For example, why can't there be some exotic Lie-algebra automorphisms in positive characteristic? Mar 12, 2020 at 17:52
• @LSpice: In my answer ${\rm char}\,R=0$ ! Mar 12, 2020 at 18:18
• Although you wrote that clearly, I skipped over it both in the parenthesis and in the statement of the theorem. Sorry! (Still, is it true algebraically, in characteristic 0, that automorphisms of a semisimple Lie algebra exponentiate to automorphisms of the simply connected group with that Lie algebra? I believe it, but don't know a reference.) Mar 12, 2020 at 18:31