4 1. Introduction matrix exponential exp(A) of A by the formula exp(A) := ∞ k=0 Ak k! = 1 + A + 1 2! A2 + 1 3! A3 + . . . which can easily be verified to be an absolutely convergent series. Exercise 1.0.1. Show that the map A → exp(A) is a real analytic (and even complex analytic) map from Mn(C) to Mn(C), and obeys the restricted homomorphism property (1.1) exp(sA) exp(tA) = exp((s + t)A) for all A ∈ Mn(C) and s, t ∈ C. Proposition 1.0.1 (Rigidity and structure of matrix homomorphisms). Let n be a natural number. Let GLn(C) be the group of invertible n × n complex matrices. Let Φ : R → GLn(C) be a map obeying two properties: (1) (Group-like object) Φ is a homomorphism, thus Φ(s)Φ(t) = Φ(s+t) for all s, t ∈ R. (2) (Weak regularity) The map t → Φ(t) is continuous. Then: (i) (Strong regularity) The map t → Φ(t) is smooth (i.e., infinitely differentiable). In fact it is even real analytic. (ii) (Lie-type structure) There exists a (unique) complex n × n matrix A such that Φ(t) = exp(tA) for all t ∈ R. Proof. Let Φ be as above. Let ε 0 be a small number (depending only on n). By the homomorphism property, Φ(0) = 1 (where we use 1 here to denote the identity element of GLn(C)), and so by continuity we may find a small t0 0 such that Φ(t) = 1 + O(ε) for all t ∈ [−t0,t0] (we use some arbitrary norm here on the space of n × n matrices, and allow implied constants in the O() notation to depend on n). The map A → exp(A) is real analytic and (by the inverse function theorem) is a diffeomorphism near 0. Thus, by the inverse function theorem, we can (if ε is small enough) find a matrix B of size B = O(ε) such that Φ(t0) = exp(B). By the homomorphism property and (1.1), we thus have Φ(t0/2)2 = Φ(t0) = exp(B) = exp(B/2)2. On the other hand, by another application of the inverse function theorem we see that the squaring map A → A2 is a diffeomorphism near 1 in GLn(C), and thus (if ε is small enough) Φ(t0/2) = exp(B/2).

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