# GIT Lecture 2

This lecture was given on august 28th 2009.

### Affine Geometric Quotient

is an affine algebraic group which can be equivalently be described as a hopf algebra. Let and

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

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

In addition there should be two commutative digrams corresponding to the properties that and . This latter property is the diagram:

where the diagonal map is . A subspace is invariant if

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 . Write (for some N)

where the are linearly independent. Set . By the commutative digram above:

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

where the left vertical arrows is just and the right vertical arrows is . The upshot is that this commutative diagram applied to gives

Comparing the first and last expression it follows (using the assumed independence of the ) that is contained in QED.

### Examples

Example 1. Let then and . For any integer m get a 1 dimensional representation .

Prop. Any representation of can be written as

where .

proof: For write (for some )

The using the diagram for to be an action get

it follows that so 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 . A character is a homomorphism . Let be the group of characters. The previous prop should suffice to prove

Cor. If is a representation of an algebraic torus then

where .

proof: basically the same argument as as the previous prop just write (for example in the case there are two copies of ) etc. QED

Example 3: with .

For a matrix A to be locally nilpotent means for all there is an integer such that .

Prop. Say char k = 0 and let be a representation of . Then there exists a locally nilpotent such that the representation is given by .

proof: In general where is a polynomial in . Reorder the sum and possibly change the definition of the so that

Then (as in previous arguments) ; in particular

Write the first expression as , then switch the order of the sum to conclude that

So then define a linear tranformation

Can check then that

QED.

Rmk: In the complex case the exponential map gives a homomorphism 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

- if is invariant then there is an invariant such that
- where the are irred. representations.

For any vector space set

Prop. TFAE

- is lin. reductive
- If is exact then exact.
- if then $\exists f \in (V^*)^G$ s.t.
- and .

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

Consider the surjection , pass to invariants to get a surjection , lift the identity to get a map and check that . Pretty easy to see that 1 implies 3. For 3 implies 4 try something like taking a basis of say then use the hypothesis of 3 get invariant functionals with . Then show with .

QED

Although its not in the statement, it can be shown that the decomposition is unique. For if then consider the natural projection . Then is also the zero map so and vice versa.

An algebraic torus over any characteristic is reducitve. If 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 is invariant then pick such that then

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 . In fact this is basically how you get all reductive groups.

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- August 30, 2009 / 12:43 pm

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