Jacobson-Morozov on the algebraic group level

Question

Let $k$ be a field of characteristic 0 and $G$ a connected reductive algebraic group defined over $k$. Let $g \in G(k)$ be a unipotent element. Is it true that there is a homomorphism $\varphi: SL_2(k) \to G(k)$ mapping some unipotent element $u \in SL_2(k)$ to $g$?

Background: There is some nilpotent element $n \in Lie\, G$ which corresponds to $u$. If $Lie\, G$ is semisimple, the Jacobson-Morozov lemma states that $n$ is part of a $sl_2$-triple, so ''integrating'' would yield the desired homomorphism. (Apparently, Jacobson-Morozov holds more generally for completely reducible Lie subalgebras of $gl(V)$, though I don't have a precise reference.)

Answer 1

Much of the literature in characteristic 0 is older and may not immediately fit the exact format of your question. But here is a sample:

N. Jacobson's 1962 book Lie Algebras (later reprinted by Dover) discusses in II.5 (and II.11) the Lie algebra structure of a completely reducible linear Lie algebra in characteristic 0. His Theorem 8 shows that the Lie algebra decomposes into its center plus a semisimple ideal. (In the parallel algebraic group setting, the Lie algebra of a completely reducible linear algebraic group in characteristic 0 has this form.) The semisimple or just simple case is really the crucial one for Jacobson-Morozov theory.

In the 1968-69 IAS seminar volume published as Springer LN 131 in 1970, look at section III.4 in Conjugacy classes by T.A. Springer and R. Steinberg; here there are also adaptations to prime characteristic.

R.W. Carter's 1985 book Finite Groups of Lie Type (Chapter 5) has a nice treatment, though he usually works with simple algebraic groups.

There is also a Lie algebra treatment in section 11 of Bourbaki's Chap. 8 in the Lie groups series, supplemented by interesting exercises.

In characteristic 0 the exponential map works well to pass from the Lie algebra to the group, but in characteristic $p$ the Jacobson-Morozov argument only works for large enough $p$. For refinements involving the groups when $p$ is small, there are substantial papers by G. McNinch and D. Testerman in the past decade or so. At any rate, the case $g=1$ or $n=0$ of your question is trivial and can be left aside.

As BCnrd observes, you have to be careful to specify your maps to be algebraic group homomorphisms. In characteristic 0, working with an algebraically closed field isn't so important, but in general you have to treat points of the group over a field with care. (There is a careful study by Borel and Tits of the way abstract group homomorphisms relate to algebraic group morphisms, but you don't want to get into that here.)

Answer 2

$\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\SL{SL}$I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.

We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm $$ X\mathrel{:=} \log (g) \mathrel{:=} \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n}, $$ which is well defined in characteristic 0 since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel) \begin{align} \operatorname{G}_{a} & \to G \\ t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!} \end{align} which is an algebraic morphism in characteristic 0 since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $\Lie(U)$ consists of nilpotent elements and we can apply Jacobson–Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $\Lie(U)$. Following Borel, we define the closed algebraic subgroup $$ \mathcal{A}(M) := \bigcap_{\text{$H$ closed subgroup of $G$} \\ M \subset \Lie(H) \subset \Lie(G)} H $$ and take its commutator group $$ \SL_g \mathrel{:=} [\mathcal{A}(M),\mathcal{A}(M)] \mathrel{:=} \left< ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right>. $$ Using Prop 7.8 and Cor 7.9 in Borel, we get $$ \Lie(\SL_g) = \Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [\Lie(\mathcal{A}(M)),\Lie(\mathcal{A}(M))] = [M, M] = M, $$ where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-algebra of $\SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $\SL_g$ has to be isomorphic to $\SL_2$ (or contain a subgroup isomorphic to $\SL_2$).

In other words, an algebraic exponential map $ \operatorname{G}_a \to G$, $t \mapsto \exp(t\cdot \tilde{X})$ factors through $\SL_2$ via $$ t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right), $$ so indeed, the original element $g$ comes from a unipotent upper triangular element in $\SL_2$.