GIT lecture 10

This lecture (GIT10-11) was given on 9/18/09

Excerpts

For a group G, G_0 denotes the conn. component of the identity.

  1. If G = G_s and G is abelian, then G = \Gamma \times G_0 where G_0 is a torus and \Gamma is finite.
  2. s \in G_s, then Lie(C_G(s)) = C_\mathfrak{g}(s).  Immediate corollary: C_\mathfrak{g}(s) = \mathfrak{g} and G connected imply s \in Z(G).
  3. If G is reductive then Z(G) = Z(G)_s.
  4. If G is reductive and char(k) > 0, then \mathfrak{g} = \mathfrak{g}_s.

Proposition 10.1 Suppose G is connected and \mathfrak{g} = \mathfrak{g}_s.  Then $lates G$ is abelian (and hence a torus by fact 1 at the start of lecture).

comments:Apply induction on \dim G and \dim V for G \to GL(V).

  1. Show a torus \exists H \subset G (using an element of infinite order); if G = H then done.
  2. Otherwise H abel (induct on \dim G) and can decompose V and \mathcal{g} via the (induced) action of H
  3. Either V decomposes into smaller peices (can apply induc. on \dim V), or can show H acts trivially on \mathcal{g} (otherwise would get nilpotent elements in \mathcal{g}.
  4. In the latter case can show H \subset Z(G) and [G \cap SL(V)]_0 has smaller dimesion, putting them together gives result for G.

Corollary 10.2 If G is reductive, connected and we’re in positive characteristic, then G is a torus.

Theorem 10.3 G reductive in positive characteristic p iff G_0 \subset G is a torus and p\not | |G/G_0|.

Proposition 10.4 G is redutive iff G_0 is reductive and G/G_0 is reductive.

10.5 There an aside on restricted lie algebras.

Reductive Groups in characteristic zero

\mathfrak{g} is simple if its not abelian and has no nontrivial ideals; semisimple = direct sum of simple.

Theorem 10.9 (Weyl) Every finite-dimensional representation over a semi-simple lie algebra is completely reducible.

Theorem 10.10 Any semi-simple algebraic group is redutive in characteristic zero.

comments: the strategy is just to reduce to verifying the reductive condition for lie algebra, i.e. reduce to Weyl’s thm.

If G reductive, then adjoint representation Ad_G is completely reducible, so can decompose into central and semisiple part \mathfrak{g} = \mathfrak{g}^{ss} \oplus Z(\mathfrak{g}).  Note, its clearly true [\mathfrak{g},\mathfrak{g}] \subset \mathfrak{g}^{ss}  but is there equality?

(exercise): if G conn. alg. grp. then G' = \{ghg^{-1}h^{-1}|g,h \in G\} is a closed conn. subgroup and G' \times Z(G)_0 \to G is surjective with Lie(G') = [\mathfrak{g},\mathfrak{g}]

If in addition G is reductive, the points 1 and 3 together give that Z(G)_0 is a torus, so we get

Theorem 10.12 In characteristic zero, G is reductive iff G is a quotient of G^{ss} \times T (semisimple and torus).

This theorem uses (exercise) if G_1 and G_2 are reducitve then so is G_1 \times G_2.

Advertisements

About this entry