Algebraic Transformation Groups: An Introduction, Edition: by Hanspeter Kraft

By Hanspeter Kraft

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For the centers of GLn and SLn , an easy calculation shows that Z(GLn ) = {λE | λ ∈ C∗ } C∗ and Z(SLn ) = {λE | λn = 1} Z/nZ. 4 below). e. satisfies gW ⊆ W for all g ∈ G). 2. Example. If G ⊆ GL2 is a noncommutative finite subgroup, then G is irreducible. Otherwise, there is a one-dimensional G-stable subspace U = Cv ⊆ C2 , which admits a G-stable complement V = Cv ⊆ C2 , because G is finite. With respect to the new basis (u, v) the elements of G are diagonal matrices, contradicting the assumption.

1) For GLn the power map pm : g → g m is surjective for m = 0. (Hint: One can assume that g is in Jordan normal form, g = tu, where t is diagonal, u ∼ unipotent, and tu = ut. Then there is a subtorus T ⊆ Tn , T → C∗r , which commutes ∼ ∗r + with u and contains t. ) (2) Let G be an abstract group and Z ⊆ G its center. If pm is surjective for G, then so is for G/Z. If pm is surjective for Z and G/Z, then so is for G. (3) Study the power maps pm for SL2 . Are they surjective for m = 0? 3. 1. General and special linear groups.

It follows that hg belongs to the stabilizer of e1 , hg ∈ (SOn )e1 SOn−1 , and so g ∈ X := SO(U ) · (SOn )e1 which is, by induction, an irreducible closed subset of SOn containing g and En . 9 below). Now, similarly as in case (a), g ∈ X := SO(U1 ) · SO(U2 ) · (SOn )e1 , and the claim follows. 9. Exercise. Let V be a finite dimensional C-vector space with a nondegenerate quadratic form q. For any pair u, v ∈ V \ {0} there is a w ∈ V such that the subspaces u, w and v, w are nondegenerate, and one can even assume that q(w) = 1.

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