By Jürgen Müller

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N ] ν = µ, a contradiction. Let conversely µ be as asserted, and let ν = [ν1 , . . , νn ] n such that λ max ν µ. Hence for i ∈ {r, . . , s} we have νi = λi . Thus if s = r + 1 we conclude νr = λr +1 and νr+1 = λr+1 −1, thus ν = µ. If s > r+1 and hence λr = λs , then there are r ≤ r < s ≤ s such that νi = λi for i ∈ {r , s } as well as νr = λr + 1 and νs = λs − 1. Since λr = νr − 1 ≤ νr −1 − 1 = λr −1 − 1 < λr −1 , whenever r > 1, and λs = νs + 1 ≥ νs +1 + 1 = λs +1 + 1 > λs +1 this implies r = r and s = s, hence ν = µ in this case as well.

Hence on the associated Zariski tangential spaces we have A ∗ dEn (ϕ) : o2m+1 → sp2m : A = → A , implying dEn (ϕ)(o2m+1 ) = 0 0 o2m < sp2m , thus dEn (ϕ) is not an isomorphism of K-vector spaces. 1) Lemma. Let G be an algebraic group with Lie algebra g. a) Identifying T1 (G × G) := T[1,1] (G × G) ∼ = T1 (G) ⊕ T1 (G) ∼ = g ⊕ g, for the differential of the multiplication map we have d1 (µ) : g ⊕ g → g : [t, t ] → t + t . b) For the differential of the inversion map we have d1 (ι) = −idg : g → g. Proof.

Letting γ := γ(X)∂ ∈ DerK (K[X], K1 ), where γ(X) = γ(X − 1) := 1, for the right convolution γ associated to γ we have γ(X)(x) = γλ∗x−1 (X) = γ(xX) = x, for all x ∈ G, and thus γ(X) = X. Hence we have γ(X)∂ = X∂ ∈ L(Gm ) and thus L(Gm ) = X∂ K . 4) Example: General and special linear groups. a) Let G := GLn be the general linear group, hence K[GLn ] ∼ = K[X ]detn , where n X = {X11 , . . , Xnn } and detn = σ∈Sn (sgn(σ) · i=1 Xi,iσ ) ∈ K[X ] is the n-th determinant polynomial. Thus dimK (TEn (GLn )) = n2 , and TEn (GLn ) ∼ = DerK (K[X ]detn , KEn ) ∼ = DerK (K[X ], KEn ) ∼ = TEn (Kn×n ) ∼ = Kn×n , where δ = n n n×n .