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).