# GIT lecture 8

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

In this lecture the lie algebra 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 for some finite dimensional vector space .

From gives . I didn’t mention this the first time around, but picking a basis for , then say . From this, define by sending the coordinate function , and you get a map . Anton’s way of writing this

iff

In any case, the action should be given be described as is sent to the endomorphism . Recall is a left invariant derivation, and composing with the canonical projection clarifies the meaning of . There should be a good way of justifying this…

Defining and we get

Exercise . proof: according to Anton’s description, means in particular so the are constant function on . Looking at the action of any , we get , since kills constants.

Prop. If is connected and , then .

proof: just need the opposite inclusion of the previous argument, so let , and consider the map given by . Say . Then the map can be described as

By assumption, for every we have . There is map of differentials defined by , and dualizing we get the differential map, looking at the identity, we get . Under this map, maps to a vector . A vector is equivalent to a map sending . The backwards map is , so the vector can be recovered from . In our case, maps to

so maps to the vector corresponding to the vector . Now by equivariance there is a commutative diagram

Where there are vertical maps and . Note 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 to be constant. It suffices to show that implies , i.e. a constant. The appropriate lemma is that if there is a section , then the kernel of is just . In our case, , hence the are constant. QED.

Now if , then setting , and , then always have . The converse is true in characteristic zero (don’t need connected). Apparently , perhpas this is supposed to be (i.e. tangent space at the identity is the lie algebra…).

On the matter of adjoint stuff, takes to the automorphism . The differential at the identity is taking , and .

Results:

If connected and , then . If is connected and is abelian, then is abelian. Both of these claims fail in positive characteristic, and example is given in the notes.

### Chevalley-Jordan Decomposition

For algebraically closed of arbitrary characteristic, we have

Thm: If , then there exists a unique decomposition with and there exists polynomials such that

part of proof: The first part is the usual Jordan decomposition. For the second part, use algebraically closed to write the char poly . Now use the chinese reminader theorem to write

to find a single polynomial such that mod . Then mod . In particular, this is true on each generazlied eigenspace, but on each generalized eigenspace, is the zero matrix. So is the zero matrix on each generalized eigenspace hence is the zero matrix. The case there there aren’t at least two distint eigenvalues needs to be handeled seperately. :).

Lemma: If are affine groups and is the ideal of , then and can be recovered as follows

proof: acts on itself from the left and the ring maps correspnds to taking to . By definition, consists of the points where all elements of vanish. Then if and , then . Now assuming , then surely because the identity is in . It follows that . The case of lie algebra is similar. QED.

Anton said . Hmm…

Thm: Let be a closed algebraic subgroup. Then , implies , and .

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

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- September 19, 2009 / 4:47 pm

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