# GIT lecture 9

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

### Excerpts

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 $\chi \in I_G$QED.

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…