Let $k$ be a field. It might as well be algebraically closed, but I do not want to assume that it has characteristic $0$. I will write "group" for "affine group scheme over $k$", not assuming smoothness.

Two groups can have the same Lie algebras without being equal. For example, if $k$ has characteristic $2$, then every maximal torus in $\operatorname{SL}_2$ has the same Lie algebra as the centre $\mu_2$. Even two smooth groups can have the same Lie algebras without being equal: for example, all maximal tori in $\operatorname{SL}_2$ have the same Lie algebra. At least it is true that, if a smooth group $H$ is contained in a connected group $G$, and their Lie algebras are equal, then $H$ equals $G$; and so, if two connected subgroups $H_1$ and $H_2$ of $G$ have equal Lie algebras and smooth intersection, then they are equal.

I'm looking more for a result in line with Borel - Linear algebraic groups, Theorem 13.18(4)(d): given a maximal torus $T$ in a smooth, reductive group $G$, and a root $\alpha$ of $T$ in $G$, there is a unique smooth, connected subgroup of $G$ that is normalised by $T$ and whose Lie algebra is the $\alpha$-weight space of $T$ on $\operatorname{Lie}(G)$. The key ingredients here are reductivity and the torus action.

So I'm interested in any more general results of this sort that allow one to deduce equality of groups from equality of their Lie algebras. If that's too broad, I'll focus a bit: suppose that $G$ is a smooth, reductive group; $H_1$ and $H_2$ are smooth, connected, reductive subgroups; and $T$ is a torus in $H_1 \cap H_2$ that is not necessarily maximal in $G$, but is maximal in both $H_1$ and $H_2$. In this setting, if the Lie algebras of $H_1$ and $H_2$ are equal, then can we conclude that the groups are equal?

EDIT: I forgot to add, in case it helps, that, in my situation, $\operatorname C_G(T)^\circ$ (the connectedness automatic if $G$ itself is connected) is a torus.

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    $\begingroup$ I suggest the following question, that seems to be equivalent to yours: Let $U,\,U'$ be two 1-dimensional unipotent subgroups in a smooth connected reductive group $G$. Assume that ${\rm Lie\,}U={\rm Lie\,}U'$. Does it follow that $U=U'$? In char. 0 the answer is Yes...(I have no intuition in positive characteristic.) $\endgroup$ Aug 8, 2020 at 19:13
  • $\begingroup$ @MikhailBorovoi, I agree that question is natural. It is clear that it would suffice, but not clear to me that it is equivalent to mine. (I think that the torus action has got to play some role.) I originally hoped the torus action would be enough: if $T$ acts with "large orbits" on $U$ and $U'$, then it acts so on $U \cap U'$, so it must be smooth, right?—but it isn't. (The group often called $\alpha_p$, the $p$th-order neighbourhood of the identity, has such a torus action.) $\endgroup$
    – LSpice
    Aug 8, 2020 at 19:16
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    $\begingroup$ However, in line with @MikhailBorovoi's ideas, a reduction in the spirit of Borel's proof may be applied: by considering a fixed root $\alpha$ of $T$ in $\DeclareMathOperator\Lie{Lie}\Lie(H_1) = \Lie(H_2)$ and working inside $\operatorname C_G(\ker(\alpha)^\circ)$, we may assume that $H_1$ and (therefore) $H_2$ have semisimple rank $1$. $\endgroup$
    – LSpice
    Aug 8, 2020 at 19:21
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    $\begingroup$ The answer to my question is No, see this answer of Will Sawin. $\endgroup$ Aug 8, 2020 at 20:00
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    $\begingroup$ Indeed sorry I didn't read carefully. Possibly it would help writing "subgroups" instead of "groups" in the title. $\endgroup$
    – YCor
    Aug 8, 2020 at 20:18

1 Answer 1


$\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\Cent{C}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Lie{Lie}$The key point is not, as I expected, whether $\Cent_G(T)^\circ$ is a torus, but whether it equals $\Cent_G(\Lie(T))^\circ$. Certainly it is contained in the latter group, so this is the same as asking whether $T$ centralises $\Cent_G(\Lie(T))^\circ$.

If we do not require this, then we may adapt a construction by @WillSawin, pointed out by @MikhailBorovoi, to give a counterexample that is quite close to the one I attempted in the comments. Specifically, we give connected, reductive subgroups $H_1$ and $H_2$ of $G = \GL_4$ that contain a common maximal torus $T$ (for which $\Cent_G(T)^\circ$ is itself a maximal torus in $G$), and satisfy $\Lie(H_1) = \Lie(H_2)$, but $H_1 \ne H_2$. Namely, let $t$ be any non-scalar diagonal matrix in $\GL_2$, and put $H_1 = \left\{\begin{pmatrix} g & 0 \\ 0 & g^{[p]} \end{pmatrix} \mathrel\colon g \in \GL_2\right\}$ and $H_2 = \left\{\begin{pmatrix} g & 0 \\ 0 & t g^{[p]}t^{-1} \end{pmatrix} \mathrel\colon g \in \GL_2\right\}$, where $g^{[p]}$ is the matrix obtained by raising every entry of $g$ to the $p$th power.

Next we prove that, if $H_1$ and $H_2$ are connected, reductive subgroups of a common group $G$ that contain a common maximal torus $T$, and satisfy $\Lie(H_1) = \Lie(H_2)$, and if in addition $T$ centralises $\Cent_G(\Lie(T))^\circ$, then $H_1$ must equal $H_2$. As suggested by @MikhailBorovoi, it suffices to show that, for every root $b$ of $T$ in $\Lie(H_1) = \Lie(H_2)$, the corresponding root subgroups of $b$ in $H_1$ and $H_2$ are equal. Let $\mathfrak u$ be the common $b$-root subspace of $\Lie(H_1) = \Lie(H_2)$. Then we have $T$-equivariant isomorphisms $e_{i\,b} \colon \mathfrak u \to H_i$ such that $\Ad(e_{i\,b}(X))Y$ equals $Y - \mathrm db(Y)X$ for all $X \in \mathfrak u$ and all $Y \in \Lie(T)$. That is, $e_{1\,b}(X)e_{2\,b}(X)^{-1}$ lies in $\Cent_G(\Lie(T))$ for all $X \in \mathfrak u$, and hence, since $\mathfrak u$ is connected, in $\Cent_G(\Lie(T))^\circ$. Since this group is centralised by $T$, we see upon conjugating $e_{1\,b}(X)e_{2\,b}(X)^{-1}$ by $t$ that it equals $e_{1\,b}(b(t)X)e_{2\,b}(b(t)X)^{-1}$, for all $X \in \mathfrak u$ and all $t \in T$. In particular, $e_{1\,b}(X)e_{2\,b}(X)^{-1}$, as a function of $X$, is constant on $\mathfrak u \setminus \{0\}$, and hence, since it is continuous, is constant on $\mathfrak u$; but its value at $X = 0$ is the identity.

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    $\begingroup$ Could you please state explicitly, what assertions you prove or disprove in the second and third paragraphs of your answer? $\endgroup$ Aug 10, 2020 at 10:51
  • $\begingroup$ @MikhailBorovoi, thanks for the suggestion. I have tried to make it clearer. $\endgroup$
    – LSpice
    Aug 10, 2020 at 11:19
  • $\begingroup$ You write: " Then we have isomorphisms $e_{i\,b} \colon \mathfrak u \to H_i$ such that ${\rm Ad}(e_{i\,b}(X))Y$ equals $Y - \mathrm db(Y)X$ for all $X \in \mathfrak u$ and all $Y \in {\rm Lie}(T)$". What are these isomorphisms $e_{i\,b}$ ? $\endgroup$ Aug 10, 2020 at 13:24
  • $\begingroup$ @MikhailBorovoi, they are the automorphisms coming from a choice of Chevalley basis (see Carter - Simple groups of Lie type, p. 64). They are defined at first over $\mathbb Z$, so make sense over any field. A priori they exist only on the adjoint group, but the adjoint-quotient map is an isomorphism on unipotent subgroups, so they may be lifted uniquely to $H_i$. $\endgroup$
    – LSpice
    Aug 10, 2020 at 18:26
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    $\begingroup$ @MikhailBorovoi, $\operatorname C_G(T)^\circ = \operatorname C_G(T)$ in my counterexample is a maximal torus (the standard diagonal torus in $G = \operatorname{GL}_4$), whereas $\operatorname C_G(\operatorname{Lie}(T))^\circ = \operatorname C_G(\operatorname{Lie}(T))$ is $\operatorname C_G(T)^\circ\cdot(1 \times \operatorname{GL}_2)$. $\endgroup$
    – LSpice
    Aug 11, 2020 at 11:00

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