By Jürgen Müller

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**Extra resources for Algebraic Groups**

**Example text**

A) For x ∈ G let κx : G → G : y → x−1 yx and Ad(x) := d1 (κx−1 ) : g → g. Then Ad : G → AutLie (g) ⊆ GL(g) ∼ = GLdim(G) is a rational representation, called the adjoint representation; we have Z(G) ≤ ker(Ad). b) We have d1 (Ad) : g → EndK (g) : x → ad(x), where ad(x) : g → g : y → [x, y] is the left adjoint action. II Algebraic groups 40 Proof. a) Since κx is an isomorphism of algebraic groups, Ad(x) ∈ GL(g) is a Lie algebra automorphism. For x, y ∈ G we have Ad(xy) = d1 (κy−1 x−1 ) = d1 (κx−1 κy−1 ) = d1 (κx−1 )d1 (κy−1 ) = Ad(x)Ad(y), implying that Ad : G → GL(g) is a group homomorphism.

Xn }. For any x = [x1 , . . , xn ] ∈ K n let ∗ x : K[X ] → K : f → f (x) be the associated evaluation map, and for any f ∈ K[X ] let f • : K n → K : x → f (x) be the polynomial function afforded by f . a) Show that in general f is not necessarily uniquely determined by f • . b) Show that if K is infinite then f indeed is uniquely determined by f • . c) Show that an algebraically closed field is infinite. 2) Exercise: Unions of algebraic sets. Let K be an algebraically closed field, let X := {X1 , .

Thus there is n ∈ N such that H := Φn (G) = Φn+1 (G), implying that Φ|H : H → H is surjective. Since we have GΦ ≤ H, as far as fixed points are concerned we might restrict ourselves to surjective Frobenius endomorphisms. Moreover if G ≤ GLn is a Φq -invariant closed subgroup, and we have Φq |G = Φd , for some d Φ q d ∈ N, then we have GΦ ≤ GΦ = GΦq ≤ GLΦ n , hence G is finite. For the standard Frobenius endomorphism Φq on GLn we get the general q linear group GLn (Fq ) = GLΦ n , and since it is immediate that SLn ≤ GLn as well as S2m ≤ GLn and On ≤ GLn are Φq -invariant, we get the special Φq q linear group SLn (Fq ) = SLΦ n , the symplectic group Sp2m (Fq ) = S2m , for Φq char(K) = 2 the general orthogonal groups GO2m+1 (Fq ) = O2m+1 and Φq GO+ 2m (Fq ) = O2m as well as the special orthogonal groups SO2m+1 (Fq ) = Φq + GO2m+1 (Fq )∩SL2m+1 (Fq ) = SO2m+1 and SO+ 2m (Fq ) = GO2m (Fq )∩SL2m (Fq ) = Φq Φq SO2m , and for char(K) = 2 the general orthogonal group GO+ 2m (Fq ) = O2m ; since in the latter case SO2m = O◦2m is Φq -invariant we also get the special Φq orthogonal group SO+ 2m (Fq ) = SO2m .

### Algebraic Groups by Jürgen Müller

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