# GIT lecture 8

This lecture ( GIT8a , GIT8b) was given on 9/14/09.

In this lecture the lie algebra $\mathfrak{g}$ was changed from right invariant derivations to left invariant derivations.  Starting at the bottom of the first page, firs recall from the lecture 3 (lemma after Hilbert’s thm), that we can embed $G \to GL(V)$ for some finite dimensional vector space $V$.

From $G \to GL(V)$ gives $V \xrightarrow{\sigma} \to k[G] \otimes V$.  I didn’t mention this the first time around, but picking a basis for $V$, then say $\sigma(e_i) \mapsto \sum_j f_{ij}\otimes e_j$.  From this, define $k[GL(V)] \to k[G]$ by sending the coordinate function $x_{ij} \to f_{ij}$, and you get a map $G \to GL(V)$.  Anton’s way of writing this

$\sigma(v) = \sum_i f_i \otimes e_i$ iff $g\cdot v = \sum_i f_i(g) v_i$

In any case, the action $\mathfrak{g} \to \mathfrak{gl}(V)$ should be given be described as $X \in \mathfrak{g}$ is sent to the endomorphism $V \xrightarrow{\sigma} k[G] \otimes V \xrightarrow{X \otimes id} V$.  Recall $X\colon k[G] \to k[G]$ is a left invariant derivation, and composing with the canonical projection $k[G] \to k[e] \cong k$ clarifies the meaning of $X \otimes id$.  There should be a good way of justifying this…

Defining $V^G = \{v | g.v = v\}$ and $V^\mathfrak{g} = \{v| Xv = 0\}$ we get

Exercise $V^G \subset V^\mathfrak{g}$.  proof: according to Anton’s description, $v \in V^G$ means in particular $\sigma(v) = 1 \otimes v = \sum_i f_i \otimes e_i$ so the $f_i$ are constant function on $G$.   Looking at the action of any $X$, we get $Xv = \sum_i X(f_i)e_i = 0$, since $X$ kills constants.

Prop. If $G$ is connected and $char(k) = 0$, then $V^G = V^\mathfrak{g}$.

proof: just need the opposite inclusion of the previous argument, so let $v \in V^\mathfrak{g}$, and consider the map $G \to V$ given by $g \mapsto g.v$.  Say $\sigma(v) = \sum_i f_i \otimes e_i$.  Then the map $G \to V \cong \mathbb{A}^n$ can be described as

$k[x_1, ..., x_n] \to k[G]$

$x_i \mapsto f_i$

By assumption, for every $X \in \mathfrak{g}$ we have $\sum_i X(f_i)e_i = 0$.  There is map of differentials $\Omega_V \to \Omega_G$ defined by $dx_i \mapsto df_i$, and dualizing we get the differential map, looking at the identity, we get $D\phi_e \colon T_e G \to T_\phi(e) V \cong V$.  Under this map, $X$ maps to a vector $w = (a_1, ..., a_n) \in V$.  A vector $v$ is equivalent to a map $k \to \mathbb{A}^n$ sending $1 \to v$.  The backwards map is $x_i \mapsto a_i$, so the vector can be recovered from $k[x_1,..., x_n] \to k$.  In our case, $X \colon k[G] \to k$ maps to

$k[x_1,..., x_n] \to k[G] \to k$

$x_i \mapsto X(f_i)$

so $X$ maps to the vector corresponding to the vector $\sum_i X(f_i) e_i = 0$.  Now by $G$ equivariance there is a commutative diagram

$T_g G \to T_V$

$T_e G \to T_V$

Where there are vertical maps $T_g G \xrightarrow{DL_{g^{-1}}} T_e G$ and $T_V \to D_g T_V$.  Note $g\colon V \to V$ is an automorphism hence its differential is everywhere nonzero.  It follows since the bottom map is zero, the top map is zero.  This shows the entire differential map is zero.  Argue that this forces $\phi$ to be constant.  It suffices to show that $df_i = 0$ implies $f_i \in k + I_G$, i.e. a constant.  The appropriate lemma is that if there is a section $A \to B \xrightarrow{s} A$, then the kernel of $d \colon B \to \Gamma_{B/A}$ is just $A$.  In our case, $f_i \in \ker (d \colon k[G] \to \Omega_{k[G]/k}$, hence the $f_i$ are constant.  QED.

Now if $G \subset GL(V)$, then setting $G_v = stab(v)$, and $\mathfrak{g}_v = \{X| Xv = 0\}$, then always have $Lie(G_v) \subset \mathfrak{g}_v$.  The converse is true in characteristic zero (don’t need connected).   Apparently $T_v(G \cdot v) = \mathfrak{g} \cdot v$, perhpas this is supposed to be $\mathfrak{g}_v$ (i.e. tangent space at the identity is the lie algebra…).

On the matter of adjoint stuff, $Ad\colon G \to GL(\mathfrak{g})$ takes $g$ to the automorphism $X \mapsto gXg^{-1}$.  The differential at the identity is $ad \colon \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ taking $X \mapsto ad_X$, and $ad_X(Y) = [X,Y]$.

Results:

If $G$ connected and $char(k) = 0$, then $\ker Ad = Z(G)$.  If $G$ is connected and $\mathfrak{g}$ is abelian, then $G$ is abelian.  Both of these claims fail in positive characteristic, and example is given in the notes.

### Chevalley-Jordan Decomposition

For $k$ algebraically closed of arbitrary characteristic, we have

Thm: If $X \in End_k(V)$, then there exists a unique decomposition $X = X_s + X_n$ with $[X_s, X_n] = 0$ and there exists polynomials $p(t), q(t)$ such that $X_s = p(X), X_n = q(X)$

part of proof: The first part is the usual Jordan decomposition.  For the second part, use algebraically closed to write the char poly $p_X(t) = \prod_i (t-\lambda_i)^{n_i}$.  Now use the chinese reminader theorem to write

$k[t] \cong k[t]/(t - \lambda_1)^{n_1} \times \cdots \times k[t]/(t - \lambda_k)^{n_k}$

to find a single polynomial $p(t)$ such that $p(t) \equiv \lambda_i$ mod $(t - \lambda_i)^{n_i}$. Then $p(X) = \lambda_i I$ mod $(X - \lambda_i)^{n_i}$.  In particular, this is true on each generazlied eigenspace, but on each generalized eigenspace, $(X - \lambda_i)^{n_i}$ is the zero matrix.  So $p(X) - X_s$ is the zero matrix on each generalized eigenspace hence $p(X) - X_s$ is the zero matrix.  The case there there aren’t at least two distint eigenvalues needs to be handeled seperately. :).

Lemma: If $H \subset G$ are affine groups and $I_H$ is the ideal of $H$, then $H$ and $Lie(H) = \mathfrak{h}$ can be recovered as follows

$H = \{g \in G| g(I_H) \subset I_H\}$

$\mathfrak{h} = \{ X \in \mathfrak{g}| X(I_H) \subset I_H\}$

proof: $G$ acts on itself from the left and the ring maps correspnds to taking $f \in k[G]$ to $f(g\cdot -)$.  By definition, $H$ consists of the points where all elements of $I_H$ vanish.  Then if $g \in G$ and $g \in I_H$, then $f(g) = g.f(e)$.  Now assuming $g.f \in I_H$, then surely $g.f(e) = 0$ because the identity is in $H$.  It follows that $g \in H$.  The case of lie algebra is similar. QED.

Anton said $g.f = f(g^{-1} (-)$Hmm…

Thm: Let $G \subset GL(V)$ be a closed algebraic subgroup.  Then $g \in G$, $X \in \mathfrak{g}$ implies $g_s, g_u \in G$, and $X_s, X_u \in \mathfrak{g}$.

rmk: This is an immediate consequence of the lemma and the Jordan decomposition thm, and is false for general lie algebras.