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Let $k$ be a field of characteristic 0 (not necessarily algebraically closed), let $G$ be a connected split reductive group over $k$ and let $\mathfrak{g}$ be the Lie algebra of $G$.

Let $X \in \mathfrak{g}$ be a nilpotent element. Does there exist a unipotent subgroup $U$ of $G$ such that $X$ is contained in the Lie algebra of $U$ ?

If $k$ is algebraically closed this is Theorem 5.1 of http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/applgr.pdf.

Here is a rough idea for a proof in the general case but I can't make the details work.

By the result over an algebraically closed field there exists a unipotent subgroup $U_{\overline{k}}$ of $G_{\overline{k}}$ such that $X$ is in the Lie algebra of $U_{\overline{k}}$. Via the expoential/logarithm there is an isomorphism between $U_{\overline{k}}$ and it's Lie algebra thus there exists $x \in U_{\overline{k}}(\overline{k})$ whose exponential is $X$.

Since $X$ is defined over $k$ I would expect it to be also true for $x$ (as an element of $G$). Then by Theorem 3.6 (very easy in characteristic 0) of the Conrad's notes, $x$ will be an element of a unipotent subgroup $U$ of $G$ and we would deduce that $X$ is contained in the Lie algebra of $U$.

Since I work in characteristic 0, I imagine there might be a much simpler way. Also I don't think the split reductive hypothesis is necessary (at least I don't use it in my "proof idea").

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    $\begingroup$ Could you say the definition of "nilpotent element" in the Lie algebra of a linear algebraic group? $\endgroup$
    – YCor
    Commented Jul 21, 2019 at 9:52
  • $\begingroup$ It means that for any representation $V$ of $G$ with induced representation $\rho : \mathfrak{g} \to \mathfrak{gl}(V)$, $\rho(X)$ is nilpotent is the usual sense. But I imagine it is equivalent to the fact there exists a faithful representation of $\rho$ of $\mathfrak{g}$ such that $\rho(X)$ is nilpotent (and in particular it does not depend on $G$) but I'm not 100% sure. $\endgroup$
    – Jdoe
    Commented Jul 21, 2019 at 9:55
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    $\begingroup$ As to the equivalence: suppose that $\rho(X)$ is nilpotent for some faithful representation, and let $X = X_{\text s} + X_{\text n}$ be the Jordan decomposition of $X$. Then $\rho(X) = \rho(X_{\text s}) + \rho(X_{\text n})$ is the Jordan decomposition of $X$ (Borel Corollary 4.4(2)), but $\rho(X)$ is nilpotent, so $\rho(X_{\text s}) = 0$, so $X_{\text s} = 0$ (by faithfulness), so $X = X_{\text n}$ is nilpotent. $\endgroup$
    – LSpice
    Commented Jul 21, 2019 at 11:53
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    $\begingroup$ @Jdoe no it's obviously not intrinsic to the Lie algebra. Just take the 1-dimensional Lie algebra and $X$ nonzero. As Lie algebra of the 1-dimensional additive group $X$ is nilpotent (not semisimple), and as Lie algebra of the 1-dimensional multiplicative group $X$ is semisimple (not nilpotent). For $G$ semisimple however, it's intrinsic, and $X$ is nilpotent/semisimple iff $\mathrm{ad}(X)$ is. $\endgroup$
    – YCor
    Commented Jul 21, 2019 at 12:02
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    $\begingroup$ To @YCor's point, I was assuming that your faithful representation of $\mathfrak g$ was the derivative of a representation of $G$. I agree that, without that assumption, it is not intrinsic. $\endgroup$
    – LSpice
    Commented Jul 21, 2019 at 12:04

3 Answers 3

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Indeed there's a much simpler way in characteristic zero, with $G$ an arbitrary $k$-defined linear algebraic group.

Let $X$ be nilpotent. Fix a faithful $k$-defined linear representation $\rho$ of $G$ and let $\rho'$ be the corresponding representation of $\mathfrak{g}$.

Now $\rho'(X)$ being nilpotent, it preserves a complete flag (defined over $k$) in the linear space $\mathfrak{g}$. Let $H$ be the subgroup of $G$ preserving this flag (for the representation $\rho$), and acting as identity on all successive 1-dimensional quotients. Then, since $\rho$ is faithful, $H$ is a $k$-defined unipotent subgroup of $G$, whose Lie algebra $\mathfrak{h}$ is the subalgebra of $\mathfrak{g}$ of those element preserving the flag (for $\rho'$), and acting as zero on all successive 1-dimensional quotients. Hence $X\in\mathfrak{h}$.

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    $\begingroup$ Why is $H$ the unipotent radical of a (rational) parabolic subgroup of $G$? Or maybe you mean to enlarge it to such a radical, in which case I guess that you would still need something like Lemma 8.3 of Borel and Tits - Groupes réductifs. $\endgroup$
    – LSpice
    Commented Jul 21, 2019 at 12:24
  • $\begingroup$ Thank you very much YCor ! Thank you also for the comment on the non-intrinsicness of nilpotency it corrected a big misconception of mine :) @LSpice I think I am happy with just having $X$ be in the Lie algebra of a unipotent subgroup. Would I gain something by knowing that it is in the Lie algebra of the unipotent radical of a Parabolic subgroup ? $\endgroup$
    – Jdoe
    Commented Jul 21, 2019 at 13:20
  • $\begingroup$ @LSpice I have no idea why you claim that I said that $H$ is unipotent radical of a parabolic subgroup. There's no parabolic subgroup in the question either. $\endgroup$
    – YCor
    Commented Jul 21, 2019 at 15:13
  • $\begingroup$ @YCor, you are right. I did not mean to claim that you said that, only that the problem asked for it; but I was wrong. I automatically read "unipotent subgroup" as "unipotent radical of a parabolic subgroup." $\endgroup$
    – LSpice
    Commented Jul 21, 2019 at 18:04
  • $\begingroup$ @Jdoe, the import of Lemma 8.3 which I cite is indeed that you gain nothing in a technical sense from this additional knowledge; whether it helps in your applications I don't know, but it might be convenient to have! $\endgroup$
    – LSpice
    Commented Jul 21, 2019 at 18:05
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This is too complicated, but it's the first approach that comes to mind. I take Conrad's ‘dynamical’ approach to parabolic subgroups, thinking instead of cocharacters defining them.

As you have observed, a nilpotent element lies in the Lie algebra of the unipotent radical of some parabolic subgroup over the algebraic closure, hence is strictly contracted by some cocharacter defined over the algebraic closure; so the closure of its orbit contains $0$. It follows from Corollary 4.3 of Kempf - Instability in invariant theory (MSN) that it is actually strictly contracted by some rational cocharacter, hence lies in the Lie algebra of the unipotent radical of the corresponding rational parabolic subgroup. (See also §2.5 of Adler and DeBacker - Some applications … (MSN).)

EDIT: The analogous result for unipotent elements is Lemma 8.3 of Borel and Tits - Groupes réductifs (MSN), and I suppose you could apply it to the group generated by your $x$.

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    $\begingroup$ Dear LSpice, thank you very much for your answer. Since I'm not very at ease with the dynamic method can you confirm the following argument ? The whole point is that $X \in \mathfrak{g}$ lies the unipotent radical of some parabolic of $G$ associated to a cocharacter $\lambda$ of $G$ if and only if $\lim\limits_{t\to 0} X^{\lambda(t)} = 0$. And then we use this fact in both directions by saying : 1) the result is true over $\overline{k}$ thus there exists a cocharater defined over $\overline{k}$ 2) by Kempf's result we can choose it to be defined over $k$ 3) the result is true over $k$. $\endgroup$
    – Jdoe
    Commented Jul 21, 2019 at 11:42
  • $\begingroup$ @Jdoe, yes, exactly. $\endgroup$
    – LSpice
    Commented Jul 21, 2019 at 11:48
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YCor's question about your definition of "nilpotent" is definitely in order, because Borel and Springer already defined this term in general for affine (=linear) algebraic groups over an arbitrary infinite field $k$ as an element lying in the Lie algebra of a unipotent subgroup of $G$ defined over $k$. This two-part paper came out of their collaboration at the 1965 AMS summer institute in Boulder, with the longer second part here. The first part is in the institute proceedings, published by AMS as PSPM 9.

In characteristic 0, there are some stronger results in the 1969 W.A. Benjamin notes from Borel's lectures written up by Bass (see $\S7$ of his second edition published by Springer as GTM 126) and in my Chapter V of Springer GTM 21. Some of his discussion of "replicas" goes back to Chevalley's treatment in vol. 2 of his later abandoned series of books on Lie groups, Lie algebras, and affine algebraic groups.

Note that your own assumptions on $G$ are too strict. For example, $G$ need not be reductive or semisimple.

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