We continue our quest to understand when quotients of schemes by actions of group schemes exist. Last time we defined group schemes, group actions, geometric quotients, and gave some examples. In this post, I’ll define what it means to be a reductive group scheme. To give a full treatment of the theory of reductive group schemes here is impossible, so we’ll pick one way to define the notion of a reductive group scheme, state some equivalences, and give some examples. I apologize for the long time between the posts, I’m on vacation and have been have been spending more time outside than at a computer!

Fix a field (for simplicity assume that algebraically closed), and let the letter always stand for an affine group scheme over . The group is called a torus if . (See the comment below about non-split tori which arrise when working over non-algebraically closed fields). First, we’ll define the most general notion of a reductive group scheme.

**Definition:** An algebraic subgroup of affine group scheme is called normal if is a normal subgroup of for all k-algebras R. Recall that for an affine group scheme , denoted the functor of R-points, and is itself a group.

**Example:** Obviously the kernel of any algebraic group homomorphism is normal.

**Fact**: The Quotient of by a normal subgroup scheme always exists as a group scheme.

**Definition**: The derived group is the intersection of all normal subgroups of such that is commutative. Note that this is the smallest normal subgroup of such that the quotient is commutative. Denote by and similarly.

**Definition**: The group is called solvable if the derived series terminates with 1. Note that this is equivalent to the existence of a composition series of algebraic subgroups.

**Example**: Consider the group of upper triangular 2×2 matrices . Then is equal to the group of matrices of the form and the second derived group is trivial. Exercise: the group of upper triangular nxn matrices is also solvable.

**Remark 1**: In fact, over an algebraically closed field, the Lie-Kolchin Theorem states that if is a connected, smooth, and solvable algebraic sub-group of then there exists a basis of such that . Also, the condition is necessary since closed subgroups of solvable groups are solvable (quotients of solvable groups of quotients are as well).

**Remark 2**: One fact that we probably should have mentioned in the last post, is that actually every affine algebraic group can be realized as a subgroup of for some number n (though highly non-canonically). So the above theorem isn’t as restrictive as it might sound.

**Definition:** Suppose is a subgroup of consisting of unipotent endomorphisms, then there is a basis of so that is a subgroup of , the group of upper triangular matrices with ones on the diagonal. If is isomorphic to a subgroup of for some n then is called unipotent.

**Remark 3**: This is equivalent to: For every non-zero representation of there is a non-zero fixed vector.

**Remark 4:** If are algebraic subgroups of with normal, and and are solvable, then so is . This implies that contains a largest connected normal solvable (smooth) subgroup which is called the radical of .

**Definition:** Referring to Remark 4, if the algebraic group has trivial radical, it is called semisimple. It is called reductive if its radical is a torus. Being reductive is equivalent to the condition that the largest connected normal unipotent subgroup is trivial.

**Examples**: A torus is reductive. The groups are all semisimple (so reductive). The group is reductive but not semisimple. The group is neither.

We are now ready to give other versions of reductivity:

**Definition**: An algebraic group scheme is called linearly reductive if for any finite-dimensional representation of , and any , there is a -invariant linear function such that . If there is only a -invariant homogeneous polynomial such that then is called geometrically reductive.

**Lemma**: Linear Reductivity of is equivalent to the conditions:

i) Given a finite dimensional representation of and a surjective invariant linear form there is an invariant vector such that .

ii) For each surjection of representations the induced map on invariants is surjective.

ii’) For each surjection of finite dimensional representations as above, the induced map on invariants is surjective.

iii) For any finite-dimensional representation , if is -invariant modulo a proper sub-representation , then the coset contains a non-trivial -invariant vector.

**Proof**: Linear Reductivity is equivalent to condition i) by replacing with the dual vector space and noting that the space of invariant forms is where acts trivially on .

ii) implies ii’) is trivial. ii’) implies iii) is easy by looking at the quotient map . For iii) implies ii): if is a surjection of representations, suppose that . The vector is contained in a finite-dimensional subrepresentation (Exercise). This vector is invariant modulo the so by iii), there is a invariant vector such that . Then , so that the map on invariants is surjective.

ii) implies i) is clear by taking with trivial action.

i) implies ii’): If then decomposes as a representation of as (Exercise). Then by i), the composition is surjective.

Time for some **Examples**:

1) A finite group schemes with order prime to the characteristic of the field are linearly reductive:

Proof:

Suppose that is a finite dimensional representation and is invariant modulo a proper subrepresentation . Set . The vector is invariant by construction and is contained in (exercise) so that condition iii) above is verified.

2) Algebraic tori are linearly redutive.

Proof:

It is enough to show that is linearly reductive. Suppose as above that is a finite dimensional representation and there is a vector that is invariant modulo a proper subrepresentation . As discussed in the last post, to give a representation of is to give weight decompositions and . Now, to say that is invariant modulo means that for all . Thus is torus invariant and an element of the coset as required.

3) The group is not linearly reductive.

Proof:

Consider the 2-dimensional representation given by which corresponds to . The map is a surjection of representations but the map on invariants is not surjective.

4) In characteristic 0, is linearly reductive. In characteristic p, it is geometrically reductive.

Outline of proof when :

We have that . We’ll leave it as an exercise to show that for (the diagonal torus), then .

Now consider the set of rational functions . Set . Then we’ll leave it as another exercise to show that by sending .

Suppose that is a finite dimensional representation of and that is an invariant vector. By linear reductivity of , there is a invariant linear map such that . Define a map by the composition where the first arrow is the action and the second arrow is the map . Check that and that is a homomorphism of representations. Now, the image of is contained in for some n. A general element of looks like where . Consider the map which takes the determinant of the square (n + 1) by (n + 1) matrix given by the ; it is homogeneous and invariant. Check that .

Let be the linear coefficient in the expansion of (the formal differential of det at 1). Then is a invariant linear map and in characteristic 0, (exercise), and so the composition (the first map is , the second is ) is the required invariant linear form which doesn’t vanish at .

In characteristic p: in the above argument, choose for sufficiently large . Then check that and this is the homogeneous invariant form required.

Now unfortunately we can only list some non-trivial facts without going too deep into more theory.

**Facts**:

1) Clearly linear reductivity implies geometric reductivity. In characteristic 0, the converse is true. In characteristic p, as mentioned by Greg in the comments last time, linear reductivity is very restrictive; in fact, the only connected linear reductive groups are tori.

2) The property of reductivity implies geometric reductivity. This is a Theorem due to Haboush (1975). I see that there’s an outline of the proof on wikipedia!

It is actually the property of geometric reductivity of that can be used to prove that the invariant ring of an action on a affine scheme is finitely generated, and that closed orbits are separated by invariants. We’ll go more into what sorts of quotients exist under different hypotheses next time!

August 24, 2008 at 3:36 pm

Two things. First off, aren’t unipotent matrices the ones with 1’s along the diagonal, rather than 0’s? I’m pretty sure that if something is upper triangular with zeros along the diagonal, it’s nilpotent and so not invertible.

Second up, the first fact at the end is backwards, I think. Linear implies geometric reductivity, but should certainly be the stronger condition.

August 24, 2008 at 4:19 pm

Sorry about that. Possibly too much sun after all!

August 25, 2008 at 5:30 pm

Hi Matt! I like the post. Why is the quotient of an affine group scheme by a closed, normal subgroup scheme again a variety? What if the base is an affine scheme rather than a field?

August 26, 2008 at 7:44 am

Another good post.

I just wanted to point out a technical thing. At the beginning of your post you only define split tori, and as far as I recall if you want to work in the generality of non separably closed fields you need to deal with non-split tori, i.e. algebraic groups which are isomorphic to a finite product of multiplicative groups after base change to a finite separable extension field. I thought it was worth mentioning even if it isn’t needed above since it is one of the complications introduced when not working over an algebraically closed field.

August 27, 2008 at 8:50 am

Thanks for the comments.

Greg — I added a parenthetical remark about the existence of non-split tori above. Of course, I wanted to avoid this (interesting) topic by working over algebraically closed fields.

Jason — Quotients by closed normal subgroups do exist in our (restrictive) setup, that is when working over a field. I believe they exist when the subgroup scheme is flat over the base (I could be mistaken there). Is there an easy example over an affine scheme to see that the quotient need not exist? Thanks!

August 27, 2008 at 6:17 pm

Hi Matt,

I don’t know an easy example of a quotient of an affine group scheme by a normal, closed subgroup scheme which is not a scheme. I believe there are counterexamples if one drops the hypothesis that the subgroup scheme is normal . . .