By Armand Borel

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Extra info for Linear Algebraic Groups

Example text

Since G acts irreducibly on V , Z(G) acts by scalars by Schur’s lemma. As all elements of G have determinant 1, it follows that G is ﬁnite. But since it is connected, G = 1 and hence G is commutative, in contradiction to our assumption that d ≥ 2. 2 Let G be a connected, solvable subgroup of GLn . Then, G is conjugate to a subgroup of Tn . 1 to inductively construct a basis with respect to which G is triangular. 3 The assertion of the above corollary becomes false if G is no longer assumed to be connected.

V, with ﬁbers the cosets of H, which induces a bijection ϕ¯ : G/H → X. Using this bijection ϕ, ¯ we can endow G/H with the structure of a variety. v ⊆ P(V ) is a closed subset of P(V ), so it is a projective variety. v is open in its closure. v is what is called a quasi-projective variety. We call G/H, endowed with this structure of a variety, the quotient space of G by H. Then, by construction, the natural map π : G → G/H is a morphism of varieties. 4]. 6 Let H ≤ G be a closed subgroup of a linear algebraic group G.

Choose s ∈ T with distinct eigenvalues (which is clearly possible since k is inﬁnite). Then T1 ≤ CSO2n (T ) ≤ CGL2n (T ) ≤ CGL2n (s) = D2n . 2 that D2n ∩ GO2n = T , so T = T1 is a maximal torus and rk(SO2n ) = n. 19). A1 A2 ∈ SO2n (4) Let B := SO2n ∩T2n . A short calculation shows that 0 A3 tr tr if and only if Atr 1 Kn A3 = Kn , and A3 Kn A2 = −A2 Kn A3 , that is, A3 = −tr −1 tr Kn A1 Kn , and A3 Kn A2 = Kn A1 A2 is skew-symmetric. ) In particular, B is the product of In 0 Kn S In S ∈ k n×n skew-symmetric with {diag(A, Kn A−tr Kn ) | A ∈ Tn } ∼ = Tn , of dimension dim(B) = n n+1 2 + 2 n+1 = n2 , 2 = n2 , if n ≥ 2 if n = 1.