John Stillwell - Naive Lie Theory.pdf - Index of

200 9 Simply connected Lie groups
Let ε be the minimum side length of the finitely many rectangular overlaps
of the squares S(P j ) covering the unit square. Then, if we divide the
unit square into equal subsquares of some width less than ε, each subsquare
lies in a square S(P j ). Therefore, for any two points P, Q in the subsquare,
we have d(P) −1 d(Q) ∈ N .
This means that we can deform p to q by “steps” (as described in the
previous section) within regions of G where the point d(P) inserted or removed
in each step is such that d(P) −1 d(Q) ∈ N for its neighbor vertices
d(Q) on the path, so Φ(d(P) −1 d(Q)) is defined. Consequently, Φ can be
defined along the path obtained at each step of the deformation, and we can
argue as in Stage 2 that the value of Φ does not change.
Stage 4. Verification that Φ is a homomorphism that induces ϕ.
Suppose that A,B ∈ G and that 1 = A 1 ,A 2 ,...,A m = A is a sequence of
points such that A −1
i A i+1 ∈ O for each i, so
Φ(A)=Φ(A 1 )Φ(A −1
1 A 2)···Φ(A −1
m−1 A m).
Similarly, let 1 = B 1 ,B 2 ,...,B n = B be a sequence of points such that
B i+1 ∈ O for each i, so
B −1
i
Φ(B)=Φ(B 1 )Φ(B −1
1 B 2)···Φ(B −1
n−1 B n).
Now notice that 1 = A 1 ,A 2 ,...,A m = AB 1 ,AB 2 ,...,AB n is a sequence of
points, leading from 1 to AB, such that any two adjacent points lie in a
neighborhood of the form CO. Indeed, if the points B i and B i+1 both lie in
CO then AB i and AB i+1 both lie in ACO. It follows that
Φ(AB)=Φ(A 1 )Φ(A −1
1 A 2)···Φ(A −1
m−1 A m)
× Φ((AB 1 ) −1 AB 2 )Φ((AB 2 ) −1 AB 3 )···Φ((AB n−1 ) −1 AB n )
= Φ(A 1 )Φ(A −1
1 A 2)···Φ(A −1
m−1 A m)
× Φ(B −1
1 B 2)Φ(B −1
2 B 3)···Φ(B −1
n−1 B n)
= Φ(A)Φ(B) because Φ(B 1 )=Φ(1)=1.
Thus Φ is a homomorphism.
To show that Φ induces ϕ it suffices to show this property on N ,because
we have shown that there is only one way to extend Φ beyond N .
On N , Φ(A)=e ϕ(log(A)) ,soforthepathe tX through 1 in G we have
d
dt ∣ Φ(e tX )= d t=0
dt ∣ e tϕ(X) = ϕ(X).
t=0