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Let $G$ be a connected, simply-connected, complex, semisimple Lie group. Suppose that $H$ is a Zariski-closed subgroup of $G$ with reductive Lie algebra $\frak{h}$. Under what conditions may one conclude that $H$ is a linearly reductive complex Lie group? I would appreciate any and all references and suggestions.

Thanks!

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  • $\begingroup$ what is a reductive Lie algebra? $\endgroup$
    – Venkataramana
    Commented Mar 13, 2013 at 14:27
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    $\begingroup$ If $\mathfrak h$ is semisimple, not just reductive, then you're set. Otherwise you're going to have to handle the case of the unipotent group inside $SL(2)$, for starters. $\endgroup$
    – Allen Knutson
    Commented Mar 13, 2013 at 15:00
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    $\begingroup$ A reductive Lie algebra is one for which any ideal has a complementary ideal. I think that the condition that $\mathfrak{h}$ is reductive might be more naturally replaced with the condition that $\mathfrak{h}$ is a direct sum of a complex abelian subalgebra and a complex semisimple subalgebra, each acting completely reducibly on $\mathfrak{g}$. At least that is necessary to avoid $H$ being something like a parabolic subgroup of $SL(2,\mathfrak{C})$. $\endgroup$
    – Ben McKay
    Commented Mar 13, 2013 at 15:01
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    $\begingroup$ If $\mathfrak h$ acts completely reducibly on $\mathfrak g$, then the image of $H$ in the adjoint representation of $G$ has a faithful completely reducible representation and hence $H/(Z(G)\cap H$ is reductive. Since $H$ is algebraic, this means that $H$ is reductive. $\endgroup$
    – Venkataramana
    Commented Mar 13, 2013 at 15:30
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    $\begingroup$ If on the other hand, you drop the assumption of complete reducibility, then $h$ could be a semi-simple lie algebra plus the lie algebra of a unipotent group and hence $H$ will not be reductive (I think that is what @Ben McKay has remarked). $\endgroup$
    – Venkataramana
    Commented Mar 13, 2013 at 15:32

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As some of the comments suggest, "semisimple" correlates well for the groups and their Lie algebras in characteristic 0, but "reductive" doesn't work so well. For a Lie algebra in characterisic 0, Bourbaki Groupes et algebres de Lie (Chapter I, section 6.4) defines it to be reductive just when its adjoint representation is completely reducible; other convenient equivalent conditions follow. The groups don't show up until later chapters, where semisimple Lie groups are emphasized.

In your situation, $G$ can equally well be viewed as a semisimple algebraic group (from the viewpoint of Chevalley's classification over arbitrary algebraically closed fields). On the other hand, the Zariski-closed subgroup $H$ is also closed in the euclidean topology and thus is itself a Lie subgroup of $G$. But for example $H$ might well be commutative and unipotent (as an algebraic group), hence connected in either topology; then its Lie algebra is abelian and thus "reductive" even though $H$ itself isn't. (You might as well assume $H$ is connected in any case, since a finite component group doesn't affect linear reductivity in characteristic 0.)

Concerning the notion of "linearly reductive", it has been knwn since early work of Nagata and others that this is equivalent for algebraic groups to being reductive in charactristic 0: almost-direct product of central torus and semisimple derived group. But unlike the notion of "semisimple", there is no automatic equivalence between an algebraic group being reductive and its Lie algebra being reductive. (Most of the literature is old, including Benjamin lecture notes by John Fogarty. The 1975 Annals of Math. paper by Haboush has references, but worked in the prime characteristic setting to prove Mumford's conjecture that all reductive groups are "geometrically reductive".)

ADDED: In your situation you are mainly dealing with algebraic groups $G, H$ (the Lie algebra tells you even less for non-semisimple Lie groups). Since the Lie algebra is the same for $H$ and its identity component, the precise structure of $H$ is not revealed: your short exact sequence may or may not split. But one characterization of reductive Lie algebras, along with Chevalley's correspondence in characteristic 0 (highlights given in section 13 of my book), shows that $H$ (if connected) is the almost-direct product of its connected semisimple derived group and its center. There might be a finite intersection. But this center is just a commutative algebraic group, which isn't helpful for studying rational representations of $H$ as an algebraic group. [The scheme language used in section II.6 of Groupes algebriques by Demazure-Gabriel probably doesn't add anything here.]

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  • $\begingroup$ The Lie algebra of my subgroup $H$ is decidedly reductive but not necessarily semisimple. Is there some result I might invoke concerning the "closeness" of $H$ to being reductive? Is it a quotient/central extension, etc. of something reductive? $\endgroup$
    – Peter Crooks
    Commented Mar 13, 2013 at 17:38
  • $\begingroup$ (I am not assuming $H$ to be connected.) $\endgroup$
    – Peter Crooks
    Commented Mar 13, 2013 at 17:42
  • $\begingroup$ For instance, is there a splitting of the short-exact sequence $$1\rightarrow H_0\rightarrow H\rightarrow\pi_0(H)\rightarrow 1?$$ $\endgroup$
    – Peter Crooks
    Commented Mar 13, 2013 at 17:47
  • $\begingroup$ Your answer is excellent and much appreciated. The references are fantastic! $\endgroup$
    – Peter Crooks
    Commented Mar 14, 2013 at 14:45

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