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 $R$.

  • 1
    $\begingroup$ 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'. $\endgroup$
    – LSpice
    Mar 10, 2020 at 18:33
  • $\begingroup$ (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 $T$.) $\endgroup$
    – LSpice
    Mar 10, 2020 at 20:21
  • $\begingroup$ (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 $\pm (x^2+y^2)$. $\endgroup$
    – YCor
    Mar 10, 2020 at 21:00
  • $\begingroup$ @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? $\endgroup$
    – LSpice
    Mar 10, 2020 at 21:32
  • 1
    $\begingroup$ @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. $\endgroup$
    – YCor
    Mar 10, 2020 at 21:43

1 Answer 1


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.

  • $\begingroup$ 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 $\operatorname{Lie} G^{\text{sc}} = \mathfrak g^{\text{ss}}$.) $\endgroup$
    – LSpice
    Mar 12, 2020 at 17:48
  • $\begingroup$ 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? $\endgroup$
    – LSpice
    Mar 12, 2020 at 17:52
  • $\begingroup$ @LSpice: In my answer ${\rm char}\,R=0$ ! $\endgroup$ Mar 12, 2020 at 18:18
  • $\begingroup$ 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.) $\endgroup$
    – LSpice
    Mar 12, 2020 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.