Before we prove this, let’s draw a picture to explain it!
picture of U(1) in the complex plane with tangent space at identity before and after left multiplication
A left-invariant vector field on U(1) is one of constant length, always pointing the same way. Given
x∈T1Gwe get a left-invariant vector field Von Gby
v(g) = (Lg)∗x
Now let’s prove the theorem:
Proof - We construct an inverse map:
g→ {left-invariant vector fields}
x7→ vx
where
vx(h) = (Lh)∗x∈ThG.
Let’s check that vxis left-invariant:
(Lg)∗vx(h) = (Lg)∗(Lh)∗x
= (LgLh)∗x
= (Lgh)∗x
=vx(gh).
Next check it is an inverse map. First: start with a left-invariant w, turn it into w(1) ∈g, then turn
that back into a left-invariant vector field vw(1). Check: w=vw(1).
vw(1)(h) = (Lh)∗w(1)
=w(h).
Second: start with x∈g, turn it into a left-invariant vector field vx, then turn that back into
vx(1) ∈g. Check: x=vx(1).
vx(1) = (L1)∗x
= (1G)∗x
=x.
picture of Lie-algebra element going to left-invariant vector field on the circle and vice versa
We henceforth use this isomorphism to freely think of geither as T1Gor as the space of all left-
invariant vector fields on G. We use this to define a bracket operation on g, using the fact that
V ect(G) is a Lie algebra. Using g⊆V ect(G) to make ginto a Lie algebra we just need:
Lemma 3 If v , w ∈V ect(G)are left-invariant, so is [v, w].
Proof - For this we will use a general fact: if φ:M→Nis a diffeomorphism, then given v∈V ect(M)
picture of push-forward of vector
If φis a diffeomorphism there is a unique x∈Mmapping to any y∈N(namely φ−1(y)), so we can
define a vector field φ∗v∈V ect(N) by:
(φ∗v)(y) = φ∗v(φ−1(y))
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