# About $G$-modules versus $Lie(G)$-modules for algebraic groups

Hello,

I would like to know clear references about the following facts:

Let $G$ be a connected algebraic group (over alg. closed field in char. 0), $Lie(G)$ its Lie algebra, $M$ a $G$-module. I don't assume that $G$ is affine, but if there is a nice simple reference with $G$ affine, then I'll like it too.

If $v \in M$ is a vector killed by $Lie(G)$, then it is fixed by $G$.

This will establish that $G-mod \to Lie(G)-mod$ is fully faithful (by the usual method of interpreting a morphism as an element of the inner $Hom$).

For tori, solvable, nilpotent, semi-simple groups, the expected characterizations of the essential image of $G-mod \to Lie(G)-mod$.

Thank you, Sasha

• You should perhaps have a look at the first couple of pages in Harish-Chandra "Discrete Series II". He describes the algebraic relations of the action of $Lie(G)$ and $G$ on the representation space. I would wonder, if the algebraic relation change, if $F$ is an arbitrary field of char $0$. – Marc Palm Feb 26 '12 at 15:35
• Thank you for the reference; But what I found there (if I understand correctly what are you referencing to) is nice general information about representations of locally compact groups or Lie groups, not what I wanted. – Sasha Feb 26 '12 at 15:50

## 3 Answers

I'm sure that there are a number of ways of answering this in the affine case; here's one. Let's assume $G$ affine and let's say that $k$ is our algebraically closed field. The following argument may not be the slickest one, but it has the benefit of working in arbitrary characteristic with the enveloping algebra replaced by the hyperalgebra. (In positive characteristic, instead of considering a Lie($G$)-stable vector we'd want to consider a hyperalgebra-stable vector. It is not true in positive characteristic that a Lie($G$)-stable vector is $G$-stable).

Now, a $G$-module structure on $M$ is given by a comodule morphism $$c : M \to k[G] \otimes M .$$ Let $U(G)$ denote the enveloping algebra of Lie($G$); equivalently, this is the so-called hyperalgebra of $G$ (hyperalgebra = enveloping algebra when char$(k) = 0$, but not in positive characteristic). We will view $U(G)$ as a subspace of the full linear dual of $k[G]$; namely, $U(G)$ is the subspace of $k[G]^*$ consisting of elements that vanish on some power of the ideal defining the identity. (You can look in, say, Jantzen's book Representations of Algebraic Groups for more details). Note that $v \in M$ is $G$-stable if and only if $c(v) = 1 \otimes v$.

Now, the action of $U(G)$ on $M$ also comes from the comorphism $c$. Namely, for $v \in M$, if $c(v) = \sum f_i \otimes v_i$ then for $X \in U(G)$ we have $$X.v = \sum X(f_i) \cdot v_i ,$$ where $X(f_i)$ is the dual action of $X$ on $f_i \in k[G]$. Let $U(G)^+$ denote the augmentation ideal; this is the two-sided ideal of $U(G)$ generated by the Lie algebra inside of $U(G)$. Then $v \in M$ is killed by Lie($G$) if and only if $X.v = 0$ for all $X \in U(G)^+$. (This part only works in characteristic 0; there are slight modifications to be made in positive characteristic).

So now let's assume that $v \in M$ is killed by Lie$(G)$. Let's write $$c(v) = \sum f_i \otimes v_i .$$ Without loss of generality we may assume that the $v_i$ are $k$-linearly independent. Since $X.v = 0$ for all $X \in U(G)^+$ this implies $$\sum X(f_i) \cdot v_i = 0$$ for all $X \in U(G)^+$. Since the $v_i$ are linearly independent, this means that $X(f_i) = 0$ for all $i$ and for all $X \in U(G)^+$. Now, any element $f \in k[G]$ that is killed by all elements of $U(G)^+$ must be constant. Hence we have $$c(v) = \sum 1 \otimes v_i .$$ But one of the properties of the comodule morphism is that $(\epsilon \otimes Id_M) \circ c = 1 \otimes Id_M$, where $\epsilon$ is the augmentation of $k[G]$. Hence $$\sum 1 \otimes v_i = 1 \otimes v$$ and $v$ is $G$-stable.

• As for a reference, see section I.7 (on algebras of distributions) in Jantzen's book "Representations of Algebraic Groups." – Chuck Hague Feb 26 '12 at 18:20

Maybe it is worthwhile indicating why it is suffices to prove the desired result for affine $G$.

Let $G$ be a connected algebraic group over $k$ (char. 0) and suppose that $V$ is a $G$-module (i.e. a linear representation of $G$). We want to know that $G$ acts trivially on $\operatorname{Lie}(G)$-fixed points in $V$; anyhow $G$ leaves invariant the subspace of $\operatorname{Lie}(G)$-fixed points in $V$, so we may and will suppose that $\operatorname{Lie}(G)$ acts trivially on $V$.

We now apply a theorem of Chevalley, a modern treatment of which can be found in [Conrad, Brian "A modern proof of Chevalley's theorem on algebraic groups", J. Ramanujan Math. Soc. 17 (2002), no. 1, 1–18] which you can also find on Brian's web page -- see the link in Jim Humphreys' answer. Chevalley's theorem says that $G$ has a normal subgroup $H$ for which $H$ is affine (linear) and the quotient $G/H$ is an Abelian variety.

We suppose known the desire result for the connected affine algebraic group $H^0$ (the connected component of the identity in $H$) -- namely, that $H^0$-fixed points in $V$ coincide with $\operatorname{Lie}(H)$-fixed points. We conclude that $H^0$ acts trivially on $V$, so we may and will suppose $H^0$ to be trivial. Now, $G$ is a connected group which is an extension of an Abelian variety by a finite group $H$, hence $G$ is itself an Abelian variety. [Indeed, the mapping $G \to G/H$ is finite, hence proper; thus $G$ is proper over $k$ -- i.e. complete -- and hence an Abelian variety].

The representation of $G$ on $V$ is defined by a homomorphism $G \to \operatorname{GL}(V)$; since the variety $G$ is complete and $\operatorname{GL}(V)$ is affine, this homomorphism must be constant. This shows that indeed $G$ acts trivially on $V$.

For affine (= linear) algebraic groups and their Lie algebras over an algebraically closed field of characteristic 0, there are fairly elementary discussions of the connection between their actions (including finite dimensional representations): see the three books titled Linear Algebraic Groups by Borel, Springer, and me. While the style of these treatments varies, the basic result is that the Lie group theory carries over well to the algebraic group setting in characteristic 0. For example, invariants of the group correspond naturally to invariants of the Lie algebra. In my book, this is all discussed in Sections 13-14. In prime characteristic everything gets more subtle, at which point Jantzen's more advanced book is helpful but requires more scheme language.

If an algebraic group in characteristic 0 isn't affine, older theorems of Barsotti and Chevalley show how to reduce most questions to the cases of affine groups and abelian varieties. (Brian Conrad provided a more modern proof of Chevalley's theorem. See his homepage here.) For the latter, Lie algebra ideas don't seem to contribute anything extra to the question asked here.

ADDED: Maybe it's helpful to comment further on the original question as well as the other answers.

1) If there is a definite reason to consider the notion of "Lie algebra of a non-affine algebraic group" over an algebraically closed field of characteristic 0, it's essential to define the notion explicitly or at least give a reference. In his 1950s work, Chevalley imitated successfully the correspondence between Lie groups and Lie algebras when the Lie group is replaced by a linear algebraic group in characteristic 0. This is essentially the material covered in the three textbooks I mentioned above, which provide clear answers to the questions raised in the affine/linear case. As noted, the Lie algebra by itself is too weak a tool in prime characteristic.

2) The paper by Brian Conrad updates the older language used for Chevalley's theorem on general algebraic groups. But note that Conrad always requires the groups to be connected, including the affine closed normal subgroup in the theorem. And there seems to be no point in working with Lie algebras in this general setting.