Quantum Mechanics for Mathematicians

a more subtle example of this:SU(2) andSO(3) are different groups with the
same Lie algebra.
We haveG⊂GL(n,C), andX∈M(n,C), the space ofnbyncomplex
matrices. For allt∈R, the exponentialetXis an invertible matrix (with inverse
e−tX), so inGL(n,C). For eachX, we thus have a path of elements ofGL(n,C)
going through the identity matrix att= 0, with velocity vector

d

dt

etX=XetX

which takes the valueXatt= 0:

d

dt

(etX)|t=0=X

To calculate this derivative, use the power series expansion for the exponential,
and differentiate term-by-term.
For the caseG=GL(n,C), we havegl(n,C) =M(n,C), which is a linear
space of the right dimension to be the tangent space toGat the identity, so
this definition is consistent with our general motivation. For subgroupsG⊂
GL(n,C) given by some condition (for example that of preserving an inner
product), we will need to identify the corresponding condition onX∈M(n,C)
and check that this defines a linear space.
The existence of such a linear spaceg⊂M(n,C) will provide us with a
distinguished representation on a real vector space, called the “adjoint repre-
sentation”:

Definition(Adjoint representation).The adjoint representation(Ad,g)is given
by the homomorphism

Ad:g∈G→Ad(g)∈GL(g)

whereAd(g)acts onX∈gby

(Ad(g))(X) =gXg−^1

To show that this is well-defined, one needs to check thatgXg−^1 ∈gwhen
X∈g, but this can be shown using the identity

etgXg

− 1

=getXg−^1

which implies thatetgXg
− 1
∈GifetX∈G. To check this identity, expand the
exponential and use

(gXg−^1 )k= (gXg−^1 )(gXg−^1 )···(gXg−^1 ) =gXkg−^1

It is also easy to check that this is a homomorphism, with

Ad(g 1 )Ad(g 2 ) =Ad(g 1 g 2 )

A Lie algebragis not just a real vector space, but comes with an extra
structure on the vector space: