John Stillwell - Naive Lie Theory.pdf - Index of

146 7 The matrix logarithm
If A(t) isasmoothpathinG with A(0)=1, then the sequence of points
A m = A(1/m) tends to 1 and
A ′ A m − 1
(0)= lim
m→∞ 1/m ,
so any ordinary tangent vector A ′ (0) is a sequential tangent vector. But
sometimes it is convenient to arrive at tangent vectors via sequences rather
than via smooth paths, so it would be nice to be sure that all sequential
tangent vectors are in fact ordinary tangent vectors. This is confirmed by
the following theorem.
Smoothness of sequential tangency. Suppose that 〈A m 〉 is a sequence in
a matrix Lie group G such that A m → 1 as m → ∞, and that 〈α m 〉 is a
sequence of real numbers such that (A m − 1)/α m → Xasm→ ∞.
Then e tX ∈ G for all real t (and therefore X is the tangent at 1 to the
smooth path e tX ).
A
Proof. Let X = lim m −1
m→∞ α m
. First we prove that e X ∈ G. Then we indicate
how the proof may be modified to show that e tX ∈ G.
Given that (A m − 1)/α m → X as m → ∞, it follows that α m → 0as
A m → 1, and hence 1/α m → ∞. Then if we set
a m = nearest integer to 1/α m ,
we also have a m (A m − 1) → X as m → ∞. Sincea m is an integer,
log(A a m m )=a m log(A m ) by the multiplicative property of log
[
Am − 1
= a m (A m − 1) − a m (A m − 1) − (A m − 1) 2 ]
+ ··· .
2 3
And since A m → 1 we can argue as in Section 7.2 that the series in square
brackets tends to zero. Then, since lim m→∞ a m (A m − 1)=X,wehave
X = lim log(A a m
m→∞
m ).
It follows, by the inverse property of log and the continuity of exp, that
e X = lim A a m
m→∞
m .
Since a m is an integer, A a m m ∈ G by the closure of G under products. And
then, by the closure of G under nonsingular limits,
e X = lim
m→∞
A a m m ∈ G.