In his classical 1956 paper Fully reducible subgroups of algebraic groups Mostow proves the following theorem:

Theorem 7.1. Let $G$ be an algebraic group over a field $K$ of characteristic 0, $\mathfrak{N}$ the set of nilpotent elements in the radical of its Lie algebra, and $N$ the connected algebraic group with Lie algebra $\mathfrak{N}$. Then $G=M\cdot N$ (semi-direct) with $M$ a maximal fully reducible group. Furthermore any two maximal fully reducible subgroups of $G$ are conjugate under an inner automorphism from $\mathfrak{N}$.

This paper was written long before the Three Books on linear algebraic groups were published, and I do not fully understand Mostow's terminology. Is "a fully reducible group" the same as "an algebraic group with reductive identity component" (in characteristic 0)? Does the following result follow from Mostow's theorem?

Theorem? Let $G$ be a linear algebraic group, not necessarily connected, over a field $K$ of characteristic 0. Let $G^0$ denote the identity component of $G$, and let $N$ denote the unipotent radical of $G^0$. Then the extension $$ 1\to N\to G\to G/N\to 1 $$ splits, i.e. $G=M\cdot N$ (a semidirect product), where $M\subset G$ is a $K$-subgroup of $G$ isomorphic to $G/N$.

Are there modern proofs of Mostow's theorem? References will be appreciated.

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    $\begingroup$ [I'm in char 0] According to modern terminology you can define $N$ as the unipotent radical of $G$ itself. Also, $G$ is reductive iff $G^0$ is reductive, and this is certainly what Mostow calls "fully reducible". I think your theorem indeed is contained in Mostow's theorem 7.1. $\endgroup$
    – YCor
    Sep 15, 2012 at 15:08
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    $\begingroup$ Let $U = R_u(G)$. To prove existence of a $K$-rational splitting and uniqueness up to $U(K)$-conjugacy, WLOG $U = V$ is a vector group. Conjugation of $G$ on $V$ defines an action of $G' = G/V$ on $V$ that is linear since char($K$)=0 (!). Also, $G \rightarrow G'$ is an etale $V$-torsor, so it admits a section by vanishing of degree-1 quasi-coherent etale cohomology of $G'$. Thus, the problem is vanishing of Hochschild cohomology H$^i(G',V)$ (algebraic cochains) for $i=1,2$. On the category of algebraic linear representations, Hochschild cohomology is the derived functor of "invariants". QED $\endgroup$
    – grp
    Sep 15, 2012 at 19:21
  • $\begingroup$ (cont'd) Of course, char. 0 is being used again at the very end, to see that the formation of $G'$-invariants is (right-)exact on the category of algebraic linear representations of the possibly disconnected reductive $G'$. Indeed, by the analogue of Hochschild-Serre (or bare hands) we reduce the cohomological vanishing (or equivalently, the exactness) to 3 cases: (i) connected semisimple, (ii) torus, and (ii) finite etale. Lie algebras over $K$ settle (i), and scalar extension to $\overline{K}$ reduces (ii) and (iii) to the easy cases of split tori and finite constant groups. $\endgroup$
    – grp
    Sep 15, 2012 at 19:33
  • $\begingroup$ @grp: Could you please write a detailed answer? I would be happy to accept it! $\endgroup$ Sep 16, 2012 at 13:56

2 Answers 2


I prefer the argument as I wrote it in the comments, since the brevity there conveys the structure of the proof most clearly. But since the OP requests it, below is a very detailed version (which might obscure the simplicity of the main idea due to its length).

Before explaining the "modern" cohomological proof more fully, note that the result is false in positive characteristic, even over algebraically closed fields. For example, following Tits, for any $n > 1$ and any algebraically closed field $K$ of characteristic $p > 0$, the $K$-group $G = {\rm{SL}}_n(W_2(K))$ (rigorous meaning is clear, I hope) has the evident maximal reductive quotient ${\rm{SL}}_n$ but admits no "Levi factor". This is proved by a root group analysis resting on the fact that the natural quotient map $W_2 \rightarrow {\mathbf{G}}_a$ admits no homomorphic section; the same argument works with ${\rm{SL}}_n$ replaced by any nontrivial connected semisimple $\mathbf{Z}/(p^2)$-group.

Now back to (possibly disconnected) linear algebraic groups $G$ over a field $K$ of characteristic 0, with unipotent radical $U = R_u(G) = R_u(G^0)$ (defined over $K$, since $K$ is perfect). We shall prove:

${\mathbf{Theorem}.}$ There exist closed $K$-subgroups $M \subset G$ such that $M \ltimes U \rightarrow G$ is an isomorphism, and all such $M$ are $U(K)$-conjugate to each other. (Obviously it is the same to assert $G(K)$-conjugacy rather than $U(K)$-conjugacy.) Equivalently, homomorphic sections to the quotient map $G \rightarrow G/U$ exist and are all $U(K)$-conjugate to each other.

In the argument below, we'll highlight exactly where the characteristic 0 hypothesis (rather than just perfectness) is used.

Step 1: The case $U = 1$ is trivial, so we may assume $U \ne 1$ and proceed by induction on $\dim U$. As for any smooth connected unipotent group over a perfect field, $U$ admits a "characteristic" composition series $1 = U_0 \subset U_1 \subset \dots \subset U_m = U$ with each $U_i/U_{i-1}$ a nonzero vector group (i.e., a power of $\mathbf{G}_a$). Here, by "characteristic" I mean that each $U_i$ is stable under the automorphism functor of $U$, and so each is normal in $G$. [In characteristic 0 we can use the derived series of $U$, since a commutative unipotent group in characteristic 0 is a vector group. Over perfect fields of positive characteristic we can use the derived series followed by $p$-power filtrations within those steps since a $p$-torsion commutative smooth connected unipotent group over a perfect field of characteristic $p > 0$ is a vector group.]

Step 2: Suppose for a moment that the case when $U$ is a vector group is settled, so in particular the cases with $m = 1$ are settled and we can suppose $m > 1$. We'll deduce the general case. The unipotent radical of $G' = G/U_{m-1}$ is the vector group $U/U_{m-1}$, so (by our temporary hypothesis that the case "$U$ is a vector group" is settled) the $K$-group $G'$ admits a Levi factor $M'$. The preimage $H$ of $M'$ under $G \twoheadrightarrow G'$ has unipotent radical $U_{m-1}$ of smaller dimension than $U$ and has maximal reductive quotient $M'$ (that maps isomorphically onto $G/U$). The induction on the dimension of the unipotent radical provides a Levi factor $M$ of $H$ unique up to $U_{m-1}(K)$-conjugacy, and this is clearly a Levi factor of $G$. So we get the existence for $G$, and uniqueness up to $U_{m-1}(K)$-conjugacy for those Levi factors which map isomorphically onto a common Levi factor $M'$ of $G'$ (as all such are Levi factors of the preimage $H$ in $G$ of such an $M'$).

If $M$ is any Levi factor of $G$ (so $M \cap U = 1$) then $M \cap U_{m-1} = 1$ and clearly $M$ maps isomorphically onto a Levi factor of $G'$. But any two Levi factors of $G'$ are related through conjugacy by $(U/U_{m-1})(K) = U(K)/U_{m-1}(K)$ (equality since ${\rm{H}}^1(K,U_{m-1}) = 1$, as $U_{m-1}$ is filtered by vector groups over $K$). Thus, to establish the desired $U(K)$-conjugacy result in $G$ we may apply a preliminary $U(K)$-conjugation to reduce to considering those Levi factors of $G$ which have the same image in $G'$, so these are Levi factors in an $H$ as considered above. We have seen that such $K$-subgroups of $G$ are $U_{m-1}(K)$-conjugate, so we'd be done.

Step 3: Now we may assume that $U = V = \mathbf{G}_a^n$ is a vector group over $K$. Let $G' = G/V$, so $(G')^0$ is reductive. Since $V$ is commutative, the conjugation action of $G$ on its normal subgroup $V$ factors through an action by $G'$ on $V$. We will reduce our problem to the vanishing of some higher Hochschild cohomology groups for $G'$ with coefficients in the commutative group functor $V$ equipped with its $G'$-action.

Consider the quotient map $G \twoheadrightarrow G'$. This is a $V$-torsor for the etale topology on $G'$ (since $V$ is smooth), so the isomorphism class of this torsor corresponds to a class in the etale cohomology group ${\rm{H}}^1(G',V)$. But since $V$ is a vector group, this is the etale cohomology of the quasi-coherent sheaf on $G'$ associated to a vector bundle. On any scheme, the etale cohomology of a quasi-coherent sheaf coincides with the Zariski cohomology (due to descent theory arguments). But on an affine scheme the higher Zariski cohomology of quasi-coherent sheaves vanishes, so the etale cohomology group ${\rm{H}}^1(G',V)$ vanishes. In other words, for the short exact sequence $$1 \rightarrow V \rightarrow G \stackrel{f}{\rightarrow} G' \rightarrow 1$$ there exists a section $s$ to $f$ (as a map of $K$-schemes, not necessarily a homomorphic section). We can arrange $s(1)=1$ via $G(K)$-translation, so $G$ as a pointed $K$-scheme equipped with an inclusion from $V$ and a surjection onto $G'$ can be identified with $V \times G'$ equipped with a $K$-group structure making it fit into an exact sequence as above and making the resulting $G'$-action on $V$ be the one that we introduced above.

Arguing exactly as for extensions in ordinary group theory (for which set-theoretic splittings always exist), the set of such $K$-group structures on $V \times G'$ is identified with the set ${\rm{Z}}^2(G',V)$ of "algebraic" 2-cocycles on $G'$ valued in $V$ (equipped with its $G'$-action!). This cocycle is a 2-coboundary (i.e., in the image ${\rm{B}}^2(G',V)$ of the group of "algebraic" 1-cochains ${\rm{C}}^1(G',V)$ under the usual differential) if and only if the resulting short exact sequence of $K$-groups splits as a semidirect product, which is to say that the obstruction to the existence of a Levi subgroup is a class in the Hochschild cohomology group ${\rm{H}}^2(G',V)$. Moreover, if the obstruction vanishes then the set of such splittings (i.e., the set of Levi $K$-subgroups) up to $V(K)$-conjugation is a principal homogeneous space under the Hochschild cohomology group ${\rm{H}}^1(G',V)$.

We have shown that the existence of a Levi $K$-subgroup is reduced to the vanishing of the Hochschild cohomology group ${\rm{H}}^2(G',V)$, and the uniqueness up to $V(K)$-conjugacy is reduced to the vanishing of the Hochschild cohomology group ${\rm{H}}^1(G',V)$. Note that we still haven't used that char($K$)=0, only that $K$ is perfect. (We also haven't yet done anything serious, just basic formalism.)

Step 4: Now it suffices to prove that the Hochschild cohomology groups ${\rm{H}}^i(G',V)$ vanish for all $i > 0$ and any action on the $K$-group $V = \mathbf{G}_a^n$ by a smooth linear algebraic group $G'$ whose identity component is reductive. For this we will finally use that $K$ has characteristic 0.

The first key point is to prove that the $G'$-action on $V$ is necessarily linear (which can certainly fail in positive characteristic). To see this, we just have to check that the endomorphism functor of $\mathbf{G}_a^n$ is represented by the functor ${\rm{Mat}}_n$ of $n \times n$ matrices (so its automorphism functor is represented by ${\rm{GL}}_n$, recovering also the uniqueness of the linear structure on vector groups in characteristic 0). This immediately reduces to the case $n = 1$, which is the obvious statement that an "additive polynomial" over a $\mathbf{Q}$-algebra is precisely a scalar multiple of $x$. Note in particular that $V^{G'}$ is a linear subspace of $V$.

Our problem is now one in the category of algebraic linear representations of a smooth linear algebraic group $G'$ with reductive identity component. Ignoring the reductivity property for a moment, let's consider the category of all algebraic linear representations of $G'$ (by which I mean algebraic linear actions on vector spaces of possibly infinite dimension, exhausted by $G'$-stable algebraic finite-dimensional subrepresentations). This category has enough injectives, and by a suitable "induction" construction inspired by the case of ordinary groups we see that the Hochschild cohomology $\delta$-functor ${\rm{H}}^{\bullet}(G',\cdot)$ on this category is erasable and thus is the derived functor of the functor $W \rightsquigarrow W^{G'}$ of $G'$-invariants. (This is valid without restriction on the characteristic.) Thus, our vanishing assertion when $(G')^0$ is reductive (and ${\rm{char}}(K)=0$) is equivalent to the (right-)exactness of the formation of $G'$-invariants, which in turn suffices to be checked for finite-dimensional representations.

The formation of the Hochschild cohomology groups commutes with extension of the ground field, so we can assume $K$ is algebraically closed. Thus, $G'$ has a finite composition series whose successive quotients are of 3 types: finite constant, split torus, or connected semisimple. To check that the functor $W \rightsquigarrow W^{G'}$ on finite-dimensional algebraic representations is right-exact, we thereby reduce to each of those 3 basic cases separately. The case of split tori is obvious by consideration of gradings, and the case of finite constant groups is obvious by averaging (since we're in characteristic 0). There remains the case of connected semisimple groups over a field $K$ of characteristic 0.

Since ${\rm{char}}(K)=0$, so surjective homomorphisms between linear algebraic $K$-groups induce surjections between Lie algebras, the connected smooth closed image of a $K$-homomorphism $G' \rightarrow {\rm{GL}}(W)$ is contained inside the smooth closed $K$-subgroup $P$ of automorphisms that restrict to the identity on a specific subspace of $W$ if and only if the induced map of Lie algebras ${\rm{Lie}}(G') \rightarrow {\rm{End}}(W)$ factors through the Lie subalgebra ${\rm{Lie}}(P)$ of endomorphisms that kill that specified subspace. Thus, the subspace of $G'$-invariants on a finite-dimensional algebraic representation $W$ of $G'$ is the same as the subspace of ${\rm{Lie}}(G')$-invariants under the corresponding representation of ${\rm{Lie}}(G')$ on $W$.

Step 5: We're reduced to proving the semisimplicity of the finite-dimensional representation theory of the Lie algebra $\mathfrak{h}$ of a connected semisimple group $H$ over a field $K$ of characteristic 0. By the theory of Lie algebras in characteristic 0, it suffices to show that $\mathfrak{h}$ is semisimple in the sense of Lie algebras. (This is not a tautology: it requires some input from the theory of algebraic groups, since solvable normal Lie subalgebras do not necessarily "integrate" to Zariski-closed subgroups. But I don't know a reference, so I give a proof below.)

It is harmless to extend the ground field so that $K$ is algebraically closed and hence $H$ admits a (split) maximal torus $T$ and the associated root system formalism. In particular, the weight-space decomposition of $\mathfrak{h}$ under $T$ has ${\rm{Lie}}(T)$ as the weight space for 0, and the nonzero weight spaces are 1-dimensional and come in pairs $\mathfrak{h}_a, \mathfrak{h}_{-a}$ generating the non-solvable $\mathfrak{sl}_2$.

Let $\mathfrak{s}$ be the solvable radical of $\mathfrak{h}$, so we want to show that $\mathfrak{s} = 0$. This radical is stable under all automorphisms of $\mathfrak{h}$, and so under the adjoint action of $H$. Thus, it admits a weight space decomposition under the $T$-action. If the $T$-action is nontrivial then $\mathfrak{s}$ would have to contain one of the lines $\mathfrak{h}_a$, and then applying the adjoint action of a suitable element of $N_H(T)(K)$ (representing the reflection $r_a$ that swaps $a$ and $-a$) we see that $\mathfrak{h}_{-a}$ is also contained in $\mathfrak{s}$. This would imply that the solvable $\mathfrak{s}$ contains the Lie subalgebra generated by $\mathfrak{h}_a$ and $\mathfrak{h}_{-a}$, a contradiction since this subalgebra is the non-solvable $\mathfrak{sl}_2$.

We conclude that the $T$-action on $\mathfrak{s}$ is trivial. But $T$ was arbitrary and (as for any connected reductive group over an algebraically closed field) $H$ is generated as an algebraic group by its maximal tori, so the adjoint action of $H$ on $\mathfrak{s}$ is trivial. The subspace of invariants in $\mathfrak{h}$ under the adjoint action of $H$ is the Lie algebra of the scheme-theoretic center $Z$ of $H$ (as in any characteristic), so since $Z$ is finite etale (as $H$ is connected semisimple and ${\rm{char}}(K) = 0$) we conclude that ${\rm{Lie}}(Z) = 0$, so $\mathfrak{s} = 0$. This proves that $\mathfrak{h}$ is semisimple.

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    $\begingroup$ You were asked for a detailed answer and you certainly complied! :-) $\endgroup$ Sep 17, 2012 at 6:14

Mostow's theorem actually has a relatively modern textbook treatment, by his early collaborator Gerhard Hochschild: see Theorem 4.3 in VIII.4 of his book Basic Theory of Algebraic Groups and Lie Algebras (GTM 125, Springer, 1981). Hochschild followed the original attempt by Chevalley (in volume 2 of his Theorie des groupes de Lie) to transfer to affine algebraic groups over arbitrary fields the main ideas of Lie group theory. Though this approach eventually becomes unsatisfactory in prime characteristic (especially over fields which are not algebraically closed), as Chevalley recognized, it does work fairly well in characteristic 0.

In particular, Hochschild is able to reformulate in somewhat more modern language some of Mostow's insights into Levi decomposition. For algebraic groups, the unipotent radical supplants the traditional solvable radical in Lie group theory due to the nicer description of solvable algebraic groups and the good behavior of tori. It's worth taking a look at Hochschild's proof of Mostow's theorem and related results, since he provides a fairly elementary unified framework that still makes sense to people working today on more sophisticated problems in algebraic geometry. Any way the proof is approached, it definitely has prerequisites, as the version here by grp illustrates. It's hard to say what version is "easiest", since that depends heavily on the reader's background.

In any case, Hochschild also points out why the restriction to characteristic 0 is essential. The notion of "reductive" algebraic group is meaningful in any characteristic, but "fully reducible" (or "linearly reductive" in Hochschild's more current language) applies only to tori among the connected algebraic groups in prime characteristic (Nagata). Indeed, there are serious problems with the notion of Levi decomposition (and conjugacy of Levi factors) in the latter case, which George McNinch has recently been studying. Fortunately Mumford's geometric invariant theory survived, due to the proof by Haboush of his conjecture that reductive groups are at least "geometrically reductive" in all characteristics. But Lie groups don't serve well enough as a model for linear algebraic groups in general.

  • $\begingroup$ Thank you, Jim, I will look in Hochschild's book later (I am at a conference now). $\endgroup$ Sep 23, 2012 at 3:20

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