By Hanspeter Kraft

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A first result in this direction is formulated in the following exercise. 8 below. 2. Exercise. (1) The only algebraic group structure on the affine line C with identity element e = 0 is C+ . (Hint: If g ∗ h is such a multiplication, then g ∗ z = a(g)z + b(g) where a(g) ∈ C∗ and b(g) ∈ C. ) (2) The only algebraic group structure on C \ {0} with identity e = 1 is C∗ . ) (3) There is no algebraic group structure on C \ {z1 , z2 , . . , zr } for r > 1. (Hint: Use that Aut(C \ {z1 , z2 , . . 4. Connected component.

4 below). (2) Denote by nn ⊆ Mn the subspace of upper triangular nilpotent matrices. It ∼ follows from the above that exp induces an isomorphism nn → Un . 4. Exercise. Define the polynomials n En (x) := k=0 1 k x k! ex Ln (x) := k=1 (−1)k−1 (x − 1)k k and show that E(L(x)) = x mod x and L(E(x)) = x mod xn+1 . z (Hint: For all z ∈ C we have e = Ln (z) + z n+1 h(z) with a holomorphic function h, and for all y in a neighborhood U of 1 ∈ C we have ln(y) = Ln (y) + (y − 1)n+1 g(y) with g holomorphic in U .

Hence W = V and so ϕ = λ idV . lem Schur@Schur’s Lemma 36 CHAPTER II. ALGEBRAIC GROUPS An interesting application is given in the following lemma. 5). e. X := { λi xi | λi ∈ C, xi ∈ X}. i If G ⊆ GL(V ) is a subgroup, then G ⊆ End(V ) is a subalgebra which is stable under left- and right multiplication by G. 5. Lemma. A subgroup G ⊆ GL(V ) is irreducible if and only if G = End(V ). Proof. We can assume that V = Cn , and so G ⊆ GLn . (1) If A ∈ Mn is a matrix of rank 1, then GAG = Mn . In fact, A = uv t for some nonzero vectors u, v ∈ Cn , and so GAG ⊇ Gu · (Gt v)t .