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.


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!}


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.


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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