of the map at the identity. For general Lie groupsG, something similar can
be done, showing that a representationπofGgives a representation of the Lie
algebra (by taking the derivative at the identity), and then trying to classify
Lie algebra representations.
Theorem.Ifπ:G→GL(n,C)is a group homomorphism, then
π′:X∈g→π′(X) =
d
dt
(π(etX))|t=0∈gl(n,C) =M(n,C)
satisfies
1.
π(etX) =etπ
′(X)
- Forg∈G
π′(gXg−^1 ) =π(g)π′(X)(π(g))−^1
3.π′is a Lie algebra homomorphism:
π′([X,Y]) = [π′(X),π′(Y)]
Proof. 1. We have
d
dt
π(etX) =
d
ds
π(e(t+s)X)|s=0
=
d
ds
π(etXesX)|s=0
=π(etX)
d
ds
π(esX)|s=0
=π(etX)π′(X)
Sof(t) =π(etX) satisfies the differential equationdtdf =fπ′(X) with
initial conditionf(0) = 1. This has the unique solutionf(t) =etπ
′(X)
- We have
etπ
′(gXg− (^1) )
=π(etgXg
− 1
)
=π(getXg−^1 )
=π(g)π(etX)π(g)−^1
=π(g)etπ
′(X)
π(g)−^1
Differentiating with respect totatt= 0 gives
π′(gXg−^1 ) =π(g)π′(X)(π(g))−^1