John Stillwell - Naive Lie Theory.pdf - Index of

5.4 Algebraic properties of the tangent space 103
The tangent space of SU(2) should be the same as the space Ri + Rj + Rk
shown in Section 4.2 to be mapped onto SU(2) by the exponential function. This
is true, but it requires some checking.
5.3.6 Show that the skew-Hermitian matrices in the tangent space of SU(2) can
be written in the form bi + cj + dk, whereb,c,d ∈ R and i, j, andk are
matrices with the same multiplication table as the quaternions i, j,andk.
5.3.7 Also find the tangent space of Sp(1) (which should be the same).
Finally, it should be checked that Tr(XY)=Tr(YX), as required in the proof
that det(e A )=e Tr(A) . This can be seen almost immediately by meditating on the
sum
x 11 y 11 + x 12 y 21 + ···+ x 1n y n1
+x 21 y 12 + x 22 y 22 + ···+ x 2n y n2
.
+x n1 y 1n + x n2 y 2n + ···+ x nn y nn .
5.3.8 Interpret this sum as both Tr(XY) and Tr(YX).
5.4 Algebraic properties of the tangent space
If G is any matrix group, we can define its tangent space at the identity,
T 1 (G), to be the set of matrices of the form X = A ′ (0), whereA(t) is a
smooth path in G with A(0)=1.
Vector space properties. T 1 (G) is a vector space over R; that is, for any
X,Y ∈ T 1 (G) we have X +Y ∈ T 1 (G) and rX ∈ T 1 (G) for any real r.
Proof. Suppose X = A ′ (0) and Y = B ′ (0) for smooth paths A(t),B(t) ∈ G
with A(0)=B(0)=1, soX,Y ∈ T 1 (G). It follows that C(t)=A(t)B(t) is
also a smooth path in G with C(0)=1, and hence C ′ (0) is also a member
of T 1 (G).
We now compute C ′ (0) by the product rule and find
C ′ (0)= d dt ∣ A(t)B(t)=A ′ (0)B(0)+A(0)B ′ (0)
t=0
= X +Y because A(0)=B(0)=1.
Thus X,Y ∈ T 1 (G) implies X +Y ∈ T 1 (G).