4 1. Introduction
matrix exponential exp(A) of A by the formula

= 1 + A +
+ . . .
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.
(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) = exp(B) =
On the other hand, by another application of the inverse function theorem
we see that the squaring map A
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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