Closed subgroup

Let / be an ideal in a Hopf algebra A. Work out the conditions necessary for A/l to represent a closed subgroup which is normal. 

Theorem. Every algebraic affine group scheme over a field is isomorphic to a closed subgroup of some GL . 

Theorem. Let k be a field, G a closed subgroup of GL . Every finitedimensional representation of G can be constructed from its original representation on fc" by the processes of forming tensor products, direct sums, subrepresentations, quotients, and duals. 

Theorem. Every torus T has a largest split subtorus Td and a largest anisotropic subtorus Ta. The intersection Td n Ta is finite, and T equals Ta - Td in the sense that no proper closed subgroup contains them both. 

In any closed embedding of G in GL , some element of GL (fc) conjugates G to a closed subgroup of the strict upper triangular group U . 

Corollary., (a) If G is unipotent, so is any closed subgroup and any group scheme represented by a Hopf subalgebra. 

Theorem. Let k be perfect, S an abelian algebraic matrix group. Let Ss and Su be the sets of separable and unipotent elements in S. Then Ss and Su are closed subgroups, and S is their direct product. 

Eigenvectors for different characters are linearly independent, since p(v) = v Xv and we know by (2.2) that the different Xv are independent Hence there are only finitely many different jf > and the subgroup H = g xtv — Xv bas finite index in S. But for each n in the equality Xv n) = Xv(ff ifW) is a polynomial equation in g, and. thus H is closed. A connected S cannot have a proper closed subgroup of finite index, since by (S.2) the cosets would disconnect S. Thus H = S, and S acts on all gv by the same character. 

Theorem. Let Nbea connected nilpotent algebraic matrix group over a perfect field. Then the separable and unipotent elements form closed subgroups N, and Nu of which N is the direct product. 

Proof. The closure of N over k is still nilpotent, and by (9.2) the decomposition of elements takes place in k, so we may assume k is algebraically closed. The center of N is an abelian algebraic matrix group to which (9.3) applies. If the set Ns is contained in the center, it will then be a closed subgroup, and the rest is obvious from the last theorem. Thus we just need to show Nt is central. 

Theorem. Let G be an affine algebraic group scheme over a field. Assume G is smooth and connected, and let H be a proper closed subgroup. Then dim H < dim G. 

Consider connected closed subgroups H of G which are normal and solvable. If Ht and H2 are such, so is (the closure of) H H, since the dimensions cannot increase forever, there is actually a largest such subgroup. We denote it by R and call it the radical of G. By (10.3), the unipotent elements in R form a normal subgroup U, the unipotent radical. We call G semisimple if R is trivial, reductive if U is trivial. The theorem then (for char(fc) = 0) is that all representations are sums of irreducibles iff G is reductive. It is not hard to see this condition implies G reductive (cf. Ex. 20) the converse is the hard part. We of course know the result for R, since by (10.3) it is a torus we also know that this R is central (7.7), which implies that the R-eigenspaces in a representation are G-invariant. The heart of the result then is the semisimple case. This can for instance be deduced from the corresponding result on Lie algebras. 

Corollary. Let k = k. In a. homomorphism of algebraic matrix groups, the image is a closed subgroup. 

We showed back in (3.4) that an algebraic G can be embedded in the general linear group of some vector space we now must refine that so that we can pick out a specified subgroup as the stabilizer of a subspace. Recall from (12.4) that f W is any subspace of some V where G acts linearly, the stabilizer Hw(R)= ge G R) g(W R) W R does form a closed subgroup. 

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