By Jürgen Müller

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Similarly, from gWβ ⊆ Wαβ for g ∈ Wα we obtain gWβ ⊆ Wαβ , and thus Wα Wβ ⊆ Wαβ . Choose α ⊆ Λ such that dim(Wα ) ∈ N0 is maximal. For any β ⊆ Λ we since 1G ∈ Wβ have Wα = Wα · 1G ⊆ Wα Wβ ⊆ Wαβ , hence by the maximality of dim(Wα ) we conclude Wα = Wα Wβ = Wαβ , and Wβ = 1G · Wβ ⊆ Wα Wβ = −1 Wα . In particular, we have Wα Wα = Wα and Wα ⊆ Wα , implying that Wα ≤ G is a closed subgroup, such that Wβ ⊆ Wα for all β ⊆ Λ. Finally, there is ∅ = U ⊆ Wα open, and hence dense, such that U ⊆ Wα = im(ϕα ).

Then GLn acts morphically on the unipotent variety Gu = (GLn )u ⊆ G ⊆ GLn , which hence is a union of GLn -conjugacy II Algebraic groups 31 classes, and a union of G-conjugacy classes. Since G GLn = Zn · G, where Zn := K∗ · En = Z(GLn ) ≤ GLn , elements of G are G-conjugate if and only if they are GLn -conjugate. By the Jordan normal form theorem we conclude Gu = g∈G (Un )g = im(κ), where κ : Un × G → Gu : [u, g] → ug is the conjugation map. Now Un := {[aij ] ∈ GLn ; aij = 0 for i > j; aii = 1} ⊆ G ⊆ GLn ⊆ Kn×n is closed such that I(Un ) = Xij , Xii − 1; i, j ∈ {1, .

Let G be a connected algebraic group, and let Φ : G → G be a Frobenius homomorphism. Then the Lang map L : G → G : x → x−1 Φ(x) is surjective. Proof. We consider the Φ-conjugation action G×G → G : [z, x] → x−1 zΦ(x). For z ∈ G we have the orbit map Lz : G → G : x → x−1 zΦ(x) = x−1 zΦ(x)z −1 z, where in particular we have L1 = L: We have Lz = ρz µ(ι × (κz−1 Φ)), where κz : G → G : x → z −1 xz, implying d1 (Lz ) = d1 (ρz )(d1 (κz−1 Φ) − idT1 (G) ). Letting d ∈ N such that d1 (Φ)d = 0, we have (κz−1 Φ)d = κz−1 Φd , where z := z · Φ(z) · Φ2 (z) · · · · · Φd−1 (z) ∈ G.