# GIT Lecture 2

This lecture was given on august 28th 2009.

### Affine Geometric Quotient

$G$ is an affine algebraic group which can be equivalently be described as a hopf algebra.  Let $k[G] = \Gamma(G, O_G)$ and

$\mu \colon k[G] \to k[G] \otimes k[G]$

$\tau \colon k[G] \to k[G]$

$\epsilon \colon k[G] \to k$

be the ring homomorphisms corresponding to multiplication, inverse and the identity element of G.  For $V$ a finite dimensional vector space, a representation is a group homomorphism $G \to GL(V)$ which is also a morphisms of algebraic varieties.

Now consider an action of G on V.  The expected description as a morphism $G \times V \to V$ translates into a map of vector spaces

$\sigma \colon V \to k[G]\otimes V$

In addition there should be two commutative digrams corresponding to the properties that $g_1g_2.v =g_1.(g_2.v)$ and $1.v = v$.  This latter property is the diagram:

$\begin{array}{ccc} V & \xrightarrow{\sigma} & k[G]\otimes V \\ \downarrow & \swarrow & \empty \\ V & \empty & \empty\end{array}$

where the diagonal map is $\epsilon \otimes 1_V$.  A subspace $W \subset V$ is invariant if $\sigma (W) \subset k[G]\otimes W$

Prop. Any representation V of an affine group G is a union of finite dimensional representations.

proof: must show any vector of V lies in a finite dimensional invariant subspace.  Let $v \in V$.  Write (for some N)

$\sigma(v) = \sum_{i = 1}^N f_i\otimes v_i$

where the $f_i$ are linearly independent.  Set $M_v = span \{v_1, ..., v_N\}$.  By the commutative digram above:

$v = (\epsilon \otimes 1_V) \circ \sigma (v) = \sum_i \epsilon(f_i)v_i$

Recall that $\sigma$ should come equipped with the data of a group action so when you reverse all the arrows you get a commutative digram:

$\begin{array}{ccc} V & \xrightarrow{\sigma} & k[G]\otimes V \\ \downarrow & \empty & \downarrow \\ k[G] \otimes V & \xrightarrow{\mu\otimes 1_V} & k[G]\otimes k[G]\otimes V \end{array}$

where the left vertical arrows is just $\sigma$ and the right vertical arrows is $1_G \otimes \sigma$.  The upshot is that this commutative diagram applied to $v$ gives

$\sum_i f_i\otimes \sigma(v_i) = \sum_i \mu(f_i) \otimes v_i =: \sum_i (\sum_j g_{ij}) \otimes v_i$

Comparing the first and last expression it follows (using the assumed independence of the $f_i$)  that $\sigma(v_i)$ is contained in $k[G]\otimes M_v$ QED.

### Examples

Example 1.  Let $G = k^*$ then $k[G] = k[t,t^{-1}]$ and $\mu(t) = t\otimes t$.  For any integer m get a 1 dimensional representation $t.v = t^mv$.

Prop. Any representation $V$ of $k^*$ can be written as

$V = \oplus_{m \in \mathbb{Z}} V_m$

where $V_m = \{v \in V| \sigma(v) = t^mv\}$.

proof: For $v \in V$ write (for some $v_i$)

$\sigma(v) = \sum_i t^i \oplus v_i$

The using the diagram for $\sigma$ to be an action get

$\sum_i t^i\otimes \sigma(v_i) = \sum_i \mu(t_i) \otimes v_i = \sum_i t_i\otimes t_i \otimes v_i$

it follows that $\sigma(v_i) = t_i \otimes v_i$ so $V$ is the sum of the appropriate subspaces and its clear that these subspaces are only intersect at 0.QED

Example 2: An algebraic torus is a finite direct product of $k^*$.  A character is a homomorphism $\chi \colon G \to k^*$.  Let $G^\vee$ be the group of characters.  The previous prop should suffice to prove

Cor. If $V$ is a representation of an algebraic torus $G$ then

$V = \oplus_{\chi \in G^\vee} V_\chi$

where $V_\chi = \{ v| \sigma(v) = \chi(t) v\}$.

proof: basically the same argument as as the previous prop just write (for example in the case there are two copies of $k^*$) $\sigma(v) = \sum_{i,j} t_1^i \oplus t_2^j \oplus v_{ij}$ etc. QED

Example 3: $G = k$ with $\mu(t) = t\otimes 1 + 1 \otimes t$.

For a matrix A to be locally nilpotent means for all $v \in V$ there is an integer $n(v)$ such that $A^{n(v)} v = 0$.

Prop. Say char k = 0 and let $V$ be a representation of $G$.  Then there exists a locally nilpotent $A \in End_k(V)$ such that the representation is given by $t \mapsto \exp (At)$.

proof: In general $\sigma(v) = \sum_i g_i \otimes v_i$ where $g_i$ is a polynomial in $t$.  Reorder the sum and possibly change the definition of the $v_i$ so that

$\sigma(v) = \sum_{m = 0}^N t^m\otimes v_m$

Then (as in previous arguments) $\mu \otimes 1_V \circ \sigma(v) = 1_G \otimes \sigma \circ \sigma(v)$; in particular

$\sum_m (t\otimes 1 + 1 \otimes t)^m \otimes v_m = t^m \otimes \sigma(v_m)$

Write the first expression as $\sum_m \sum_k \binom{m}{k}t^k\otimes t^{m-k}\otimes v_m$, then switch the order of the sum to conclude that

$\sigma(v_m) = \sum_{n=m}^N \binom{n}{m} t^{n-m}\otimes v_n$

$\sigma(v_m) = \sum_{p = 0}^{N-m} \binom{m+p}{p} t^p \otimes v_{m+p}$

So then define a linear tranformation

$Av_m = \begin{cases} (m+1)v_{m+1} \mbox{ if m< N} \\ 0 \mbox{ if m = N} \end{cases}$

Can check then that $\sigma(v) = \sum_p t^p \frac{A^p v}{p!}$

QED.

Rmk: In the complex case the exponential map gives a homomorphism $\mathbb{C} \xrightarrow{exp} \mathbb{C}^*$ but this is not an algebraic map.

### (Linearly) reductive groups

An affine group G is (linearly) reductive if any representation of G is completely reducible, meaning

1. if $W \subset V$ is invariant then there is an invariant $W'$ such that $V = W oplus W'$
2. $V \cong oplus_i V_i$ where the $V_i$ are irred. representations.

For any vector space set $V^G = \{v \in V | gv = v\}$

Prop. TFAE

1. $G$ is lin. reductive
2. If $V \to W \to 0$ is exact then $V^G \to W^G \to 0$ exact.
3. if $\exists v \in V^G$ then $\exists f \in (V^*)^G$ s.t. $f(v) = 1$
4. $V = V^G\oplus W$ and $W^G = \{0\}$.

proof (sketch) 1 implies 4 implies 2 is clear.  Now 2 implies 1:

Consider the surjection $\hom_k(V,W) \to \hom_k(W,W) \to 0$, pass to invariants to get a surjection $\hom_G(V,W) \to \hom_G(W,W)$, lift the identity to get a map $\psi V \to W$ and check that $V = W \oplus \ker psi$.  Pretty easy to see that 1 implies 3.  For 3 implies 4 try something like taking a basis of $V^G$ say $e_1, ...., e_l$ then use the hypothesis of 3 get invariant functionals $f_1, ...., f_l$ with $f_i(e_i) = 1$.  Then show $V = V^G \oplus W$ with $W = \cap_i \ker f_i$.

QED

Although its not in the statement, it can be shown that the decomposition $V = V^G \oplus V'$ is unique.  For if $V = V^G \oplus V''$ then consider the natural projection $\pi \colon V \to V^G$.  Then $V''|_{\pi} \to V^G$ is also the zero map so $V'' \subset V'$ and vice versa.

An algebraic torus over any characteristic is reducitve. If $G$ is a finite group such that char k doesn’t divide the order of the group then G is reductive, this can be proved by appealing to property 3 above.  Namely if $v$ is invariant then pick $f \in V^*$ such that $f(v) = 1$ then

$\bar f (w) = \frac{1}{|G|} \sum_{g \in G} f(g(w))$

will do the trick (Mashcke’s thm).

Final rmk: Let G be a semisimple connected lie group, then its an algebraic group.  Then its a thm (Weyl) that every finite d dimensional representation of G is completely reducible (so G is reductive; maybe requires some transfinite induction to extend to infinite dimensional representations).

Also $GL(n, \mathbb{C}) = \frac{SL(n, \mathbb{C}) \times \mathbb{C}^*}{\mathbb{Z}_n}$.  In fact this is basically how you get all reductive groups.

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