Algebraic Matrix Groups

The basic fact we need is that on an algebraic matrix group S SL + t(/c) the functions xi bx, xh- x 1, and xi- x ibx for fixed b are continuous. This is clear, since they are given by polynomials, and polynomial maps are always continuous in the Zariski topology. It is worth mentioning only because multiplication is not jointly continuous (it is a continuous map S x S- S, but the topology on S x S is not the product topology). 

Condiary. Every affine algebraic group is isomorphic to an algebraic matrix group. 

Theorem. Let S be an algebraic matrix group. Let S° be the connected component containing the unit e. Then S° is a normal subgroup of finite index it is irreducible, and the other irreducible components are its cosets. 

We call G° the connected component of G. Unlike algebraic matrix groups, the G here need not have the other f( A isomorphic to A0 this fails in our introductory example of p3 over the reals. 

Theorem. Let g in an algebraic matrix group G(k) GL (fc) be separable. Then in any representation of G the element g acts as a separable transformation. 

Theorem. Let g be a unipotent element of an algebraic matrix group. Then g acts as a unipotent transformation in every linear representation. Homomar-phisms take unipotent elements to unipotent elements, and unipotence is an intrinsic property. 

If G comes from an algebraic matrix group, these are equivalent to ... 

Proof. Base-extending to k, we may assume we have a finite constant group scheme, say of order n. When we embed it as an algebraic matrix group, each g in it satisfies the separable equation X" — 1 = 0. If g is also unipotent, g — l. Thus the group is trivial. ... 

The argument at the start of the proof shows now that any homomorphism G-> H preserves Jordan decompositions. In particular, Jordan decomposition in an algebraic matrix group is intrinsic, independent of the choice of an embedding in GL . 

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. 

Let k be algebraically closed, G an algebraic matrix group. Show G is unipotent iff all elements of finite order have order divisible by char (k). [Use Kolchin s theorem to reduce to the abelian case, and look at diagonalizabie matrix groups.]... 

It is instructive to restate this proof geometrically for algebraic matrix groups. It first shows that the closure of the image of a product of connected sets is connected, then that the closure of the union of an increasing sequence of connected sets is connected. 

Corollary. Let S be a connected solvable algebraic matrix group. Then 2S is nilpotent. 

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. 

Let S be a connected solvable algebraic matrix group over a perfect field. If the separable elements form a subgroup, show that S is nilpotent. [S, is normal and S, n Su is trivial, so Ss and Su commute and S = S,x Su. Then S, is connected and hence abelian.]... 

Corollary. Let k be algebraically closed of characteristic zero. Then all algebraic affine group schemes come from algebraic matrix groups. 

The smoothness of algebraic matrix groups is a property not shared by all closed sets in /c". To see what it means, take fc = fc and let 5 fc" be an arbitrary irreducible closed set. Let s be a point in S corresponding to the maximal ideal J in k[S]. If S is smooth, n si k = O si /J us) has fc-dimension equal to the dimension of S. (This would in general be called smoothness at s.) If S is defined by equations fj = 0, the generators and relations for OUS] show that S is smooth at s iff the matrix of partial derivatives (dfj/dXi)(s) has rank n — dim V. Over the real or complex field this is the standard Jacobian criterion for the solutions of the system (f = 0) to form a C or analytic submanifold near s. For S to be smooth means then that it has no cusps or self-crossings or other singularities . 

Let A be a Hopf algebra. We are going to study the /(-linear operators T A - A which are translation-invariant. As in the previous part, we begin by seeing what this means when A is the ring of functions on an algebraic matrix group S. An operator T on functions there is left-invariant iff it commutes with all the left-translation operators Tg defined by (Tgf)(x) — f(9x)- Now on A the map / h- f(gx) is (g, x) ° A and since Tg makes... 

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