GIT lecture 9

This lecture (GIT9a, GIT9b) was given on 9/16/09.


The goal of this class is to characterize reductive groups in characteristic p>0.

The notation is that G_s, G_u, \mathfrak{g}_s, \mathfrak{g}_u are the semisimple, unipotent elements in an algebraic group and its lie algebra.

Prop: If G_s = G and G is abelian, then G = \Gamma \times G_0 where \Gamma is a finite group such that p \not | |\Gamma|.

proof: Embedding G \subset GL(V), then G = G_s implies G lands in the set of diagonal matrices (which form a torus).  So it suffices to prove the thm considering G as a closed subgroup of a torus T.  Notation

R = k[T] = k[t_1^\pm, ..., t_n^\pm]

T^\vee = \mbox{Group of characters } \chi.

R = \oplus_\chi k\cdot t^\chi

t^\chi \mbox{ is monomial and } g.t^\chi = \chi(g)t^\chi

Define L = \{\chi|t^\chi(G) = 1\} \subset T^\vee.

|Aside For example if G = k^* \times 1 \subset k^* \times k^* = \mbox{Spec } k[t_1^\pm, t_2^\pm].  Then L = \{\chi|\chi(t_1, t_2) = (1, t_2^n), n \in \mathbb{Z}\}.  It can be checked that that way these t^\chi are defined they have the property that t^\chi(g) = \chi(g).  For example, if \chi(a,b) = 1*b^n then t^\chi = t_2^n.  This element has the property that g.t^\chi = t^\chi(gh) = t^\chi(g)t^\chi(g) = \chi(g) t^\chi(h). end Aside|

In any case, by definition there is an exact sequence 0 \to L \to T^\vee \to G^\vee \to 0. Hence G^\vee = T^\vee/L which is a finitely generated group, hence G^\vee = \Gamma^\vee \times \mathbb{Z}^m by the structure theorem for abelian groups.  Its a (nontrivial?) result about pontryagin duality that (G^\vee)^\vee \cong G and that taking duals is an exact functor (seems like maybe you don’t need taking duals is an exact functor).  The free part in G^\vee corresponds to a torus of rank m in G, and the finite group part corresponds to a dual finite group.  It follows that G = \Gamma \times G_0 where G_0 is a finite group of rank m.

It remains to show p \not | |\Gamma|.  First describe I_G. Notice the explicit example above the ideal of G is just (t_2 - 1).  That is, there is an exact sequence

0\to (t_2 -1) \to k[t_1^\pm, t_2^\pm] \to k[t_1^\pm] \to 0

For the general case set Q = k-span\{t^\chi - t^{\chi'}|\chi - \chi' \in L\}.  Let I^L = RQ be the ideal generated by Q.  The claim is that I^L = I_G.  Setting Q_\chi = k-span\{t^\chi - t^{\chi'}|\chi - \chi' \in L\} and R_\chi = \oplus_{\chi - \chi' \in L} k\cdot t^{\chi'} then

I^L = \oplus_{\chi \in T^\vee/L} Q_\chi

R = \oplus_{\chi \in T^\vee/L} R_\chi

\dim R_\chi/Q_\chi = 1

Now let J be any G invariant ideal.  The claim is that J = \oplus_{\chi \in T^\vee/L} (R_\chi \cap J).  Indeed, it suffices to show if t \in R_\chi and t' \in R_{\chi'} are different components ( meaning \chi - \chi' \not \in L) and t + t' \in J then t,t' \in J.  The assumptions guarantee that there is g \in G such that \chi(g) \ne \chi'(g).  Then \chi(g)t + \chi'(g)t' = g.(t + t') \in J, but so is -chi'(g)(t+t') hence (\chi(g) - \chi'(g)) t \in J, and the result follows.  Finally, if any R_chi \cap J = R_\chi,  then J = R because t^{\chi'} R_\chi = R_{\chi + \chi'} \subset J.  When veryfying this observe that (\chi' + \chi)(g) = \chi(g)\cdot \chi'(g).

It follows that I^L is a maximal invariant ideal so I_G = I^L.  To show p \not | |\Gamma| it suffices to show \Gamma doesn’t have elements of order p.  It it did, then \chi \not \in L but \chi^p \in L, hence t^{\chi^p} - 1 \in I_G, but since we are working with a variety we can factor and cancel so t^\chi - 1 \in I_G hence \chi \in I_GQED.

Here is the rest of the lecture all nice and stuff:GIT9r

lemma 9.2 Suppose G is a reductive group, then Z(G)_s = Z(G)

regarding proof: G invariance v \in \ker (u -1).  Then (u-1)gv = g(u-1)v =0.  By reductiveness this subspace must have a G invariant complement.  Clearly since 1 preserves everything and u \in G, then (u-1) should preserve everything.  But applying (u-1) will eventually takin elements in the complement into the kernel, contradition.

Proposition 9.3 If G is reductive than \mathfrak{g} = \mathfrak{g}_s.

Proposition 9.4 Suppose $lates G$ is connected and \mathfrak{g} = \mathfrak{g}_s.  Then G is abelian (and hence a torus).

comments: the proof says not to work over k = \overline{\mathbb{F}_p}; the point is if this is your field then just take a transcendental extension, because the result will be stable under base change.  The reason for doing this is that the proof utilizes an element of infinite order.  Now if \alpha \in k, then it satisfies a polynomial relation p(\alpha) = 0, so \alpha is in some finite extension \mathbb{F}_q.  So a general B \in GL(k^n) has entries in some finite field, hence it has finite order; so don’t work over this field!

Lemma 9.5 For s \in G_s, let C_G(s) = \{g \in G| sgs^{-1} = g\} and C_\mathfrak{g}(s) = \{x \in \mathfrak{g}|Ad_s(x) = x\}.  Then Lie(C_G(s)) = C_\mathfrak{g}(s).

comments: You have an embedding G \to GL(V), so you’d like to say something like Lie(C_G(s)) = \{X \in \mathfrak{gl}(V)| e^X \in G, se^Xs^{-1}= \exp(sXs^{-1}) = e^X \}.

the proof in the notes is rather eloborate, and it seems the proof that works for G = GL(n) should work in general but there are subtelties…


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