# Levi decomposition in disconnected linear algebraic group (characteristic 0)?

For algebraic groups or Lie groups, the subject of Levi decompositions tends to be surrounded by some mystery in the literature (and in an older question raised here). While I postpone further my intention to post a careful note on my homepage someday, I'm curious about the status of disconnected algebraic groups in this discussion. Take the ground field $K$ to be algebraically closed of characteristic 0 throughout.

If $G$ is a disconnected linear algebraic group over $K$, of positive dimension, does it always have a Levi decomposition; if so, what can be said about conjugacy of Levi subgroups?

First I have to outline briefly the history of the concept of Levi decomposition. This was first carried out for a finite dimensional Lie algebra $\mathfrak{g}$ and makes sense over an arbitrary field of characteristic 0: Levi's theorem says that the solvable radical of $\mathfrak{g}$ has as a semidirect complement some semisimple subalgebra, while Mal'cev showed the conjugacy of any two such "Levi subalgebras" under special automorphisms of $\mathfrak{g}$ (which can be defined algebraically). This is treated efficiently in modern language by Bourbaki in Chapter I, 6.8 of their treatise Lie Groups and Lie Algebras.

To carry this over to connected Lie groups (real or complex), one then has to use the nice correspondence between Lie subgroups and Lie subalgebras. (But this is developed using exponentials.) In the 1950s Chevalley experimented with linear algebraic groups in characteristic 0, using a sort of formal exponentiation process to imitate the Lie group framework.

Borel works some of this out in somewhat more modern language in II.7.9 of his Benjamin notes, reproduced in his GTM 126 second edition. For the Lie algebra $\mathfrak{g}$ of a connected algebraic group $G$, the basic observation is that any Lie subalgebra equal to its derived algebra is algebraic, i.e., is the Lie algebra of some connected closed subgroup of $G$. (And this subgroup is unique, since we are in characteristic 0.) In particular, a semisimple Levi subalgebra of $\mathfrak{g}$ is the Lie algebra of a connected semisimple subgroup $H$ of $G$ whose intersection with the solvable radical is finite. (In some sources the product of subgroups here is supposed to be semidirect, but this is artificial for cases like the general linear group.) Using Jordan decomposition, this becomes the more useful decomposition of $G$ into a semidirect product of a connected reductive subgroup and the unipotent radical.

The conjugacy theorem for $\mathfrak{g}$ then carries over to the group in a natural way. Unfortunately, there seems to be no complete account of this in a textbook, though Demazure-Gabriel translate some of it into scheme language in II, Section 6. In any event, my question above doesn't seem to be answered directly, since the characteristic 0 correspondence between closed subgroups and their Lie algebras requires connected subgroups.

Footnote: In prime charactertistic things tend to fall apart in general, as indicated in my 1967 note here. But a recent paper by George McNinch offers numerous useful refinements here.

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Results in model theory imply that many reasonable questions posed about an arbitrary algebraically closed field of characteristic zero can be answered by answering them specifically over $\mathbb C$. I assume that the $\mathbb C$ case is completely understood? Does model theory allow you to harness that understanding to answer the general case? –  Theo Johnson-Freyd Jun 10 at 0:44
@Theo: Model theory seems inappropriate to the situation. One needs only basic algebraic geometry at a level known to Zariski and Weil and Borel to show that an affirmative answer to Jim's question over $\mathbf{C}$ implies the same over any algebraically closed field of characteristic 0. It is the same exact algebraic geometry needed to prove that definitions in Borel's textbook over an algebraically closed field (e.g., reductive, or connected) are insensitive to extension to a bigger such field. –  user28172 Jun 10 at 0:59
@Theo: To clarify, I don't yet know how to handle disconnected groups over the complex field (or other fields of characteristic 0). My starting point was the good behavior of the Lie algebra, which by itself doesn't reflect more than the associated connected group. –  Jim Humphreys Jun 10 at 16:14

The only argument I know for the existence of a Levi decomposition over fields of characteristic 0 and its uniqueness up to $k$-rational conjugacy works without a connectedness hypothesis, as an application of Hochschild cohomology; it is probably discussed in Jantzen's book, which hopefully someone else can confirm. The main point is that a (possibly disconnected) reductive linear algebraic group over a field of characteristic 0 has completely reducible linear representation theory. I thought this argument is explained elsewhere on MO, but I cannot find it, so here is the argument.

Consider a field $k$ of characteristic 0 and a linear algebraic group $G$ over $k$, possibly disconnected. Let $U$ be the unipotent radical of $G^0$; it is a characteristic subgroup (and its formation commutes with any extension of the ground field), so it is equal to the unipotent radical of $G$ as well. We want to show that the exact sequence of $k$-groups $$1 \rightarrow U \rightarrow G \rightarrow \overline{G} \rightarrow 1$$ splits over $k$ as a semi-direct product, where $\overline{G}$ has reductive identity component, and that any two splittings are related by $U(k)$-conjugacy.

Using a filtration of $U$ by its (characteristic!) derived series, we reduce to the case where $U$ is commutative provided that we also show ${\rm{H}}^1(k,U)=0$ in the commutative case too (so that $k$-rational conjugators can be lifted through stages of the resulting composition series of $G$). Now $U$ is a vector group on which the $\overline{G}$-action is linear (since in characteristic 0 the automorphism functor of affine $n$-space as a $k$-group is represented by ${\rm{GL}}_n$), so the ${\rm{H}}^1(k,U)$-vanishing aspect is just additive Hilbert 90 (or whatever it should be called). Hence, we're trying to show that any extension of a possibly disconnected $\overline{G}$ over $k$ by a linear representation $V$ splits over $k$ as a semi-direct product when $\overline{G}^0$ is reductive, with any two splittings related by $V(k)$-conjugacy.

Observe that $G \rightarrow \overline{G}$ is a $V$-torsor for the etale topology. For any vector group $V$ over $k$, we claim that any $V$-torsor over an affine $k$-scheme $S = {\rm{Spec}}(A)$ is trivial (i.e., has a global cross-section). Indeed, we can use a composition series arising from a basis of $V$ to reduces to the case that $V = k$, which is to say $\mathbf{G}_{\rm{a}}$-torsors over $S$ (for the etale topology). It must be a classical fact that such torsors are trivial, or note that such
torsors "are" (necessarily quasi-coherent) $O_S$-linear extensions of $O_S$ by itself for the etale topology yet quasi-coherent sheaves for the etale and Zariski topologies coincides (by descent theory) and $A$-linear extensions of $A$ by itself clearly split.

Thus, the quotient map $G \rightarrow \overline{G}$ admits a section over $k$ as a map of affine scheme. Hence, as in the more classical setting of group cohomology, the obstruction to modifying a section to be a homomorphic section (over $k$) is a canonically associated class in the Hochschild cohomology group ${\rm{H}}^2(\overline{G},V)$ that imitates group cohomology by using algebraic cochains'', and if this class vanishes then the set of $V(k)$-conjugacy classes of such splittings is a torsor for ${\rm{H}}^1(\overline{G},V)$. It therefore suffices to show that all higher Hochschild cohomology of $\overline{G}$ with coefficients in a linear representation vanishes. The formation of such Hochschild cohomology groups commutates with extension of the ground field, so we can also now assume $k$ is algebraically closed (which ensures that all finite etale $k$-groups are constant and that all $k$-tori are split).

But in general Hochschild cohomology for an affine algebraic group scheme $H$ over a field (as a functor on the category of not necessarily finite-dimensional algebraic linear representations for the group) is the derived functor of the functor of $H$-invariants, so if $H$ has completely reducible finite-dimensional linear representation theory then the higher cohomology vanishes (as every linear representation is a direct limit of finite-dimensional ones, and the formation of Hochschild cohomology commutes with direct limits).

Thus, we are reduced to the assertion that over an algebraically closed field $k$ of characteristic 0, for any linear algebraic group $\overline{G}$ over $k$ with reductive identity component, the finite-dimensional linear representation theory of $\overline{G}$ is completely reducible. By the usual Hom trick, it is the same as showing the functor of $\overline{G}$-invariants is right-exact. For this formulation, it clearly suffices first to treat $\overline{G}^0$, and then to treat the finite constant component group. The case of finite constant groups goes via averaging since ${\rm{char}}(k)=0$, and the connected reductive case reduces separately to tori (also standard, valid in any characteristic) and then finally semisimple groups (for which it suffices to treat the analogue for the Lie algebra since ${\rm{char}}(k)=0$, so we win since ${\rm{char}}(k)=0$ and the Lie algebra is semisimple by various means, one of which is Levi's original theorem!).

QED

It is only at the final step, for treating the case of the finite etale component group and connected semisimple groups that we use ${\rm{char}}(k)=0$ rather than that $k$ is just a perfect field.

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Thanks for the detail. I'll look further into both Demazure-Gabriel and Jantzen (who based his foundations on D-G but then emphasized prime characteristic). The connected case struck me as fairly elementary, modulo Lie algebra theory in characteristic 0, but dealing with a finite component group seems to require a different level of prerequisites. (By the way, I assumed the field to be algebraically closed, to avoid getting into questions about splitting fields, etc. I'm not sure what the most natural set-up is.) –  Jim Humphreys Jun 10 at 16:10