John Stillwell - Naive Lie Theory.pdf - Index of

5.2 The tangent space of SO(n) 97
involved is the same as the calculation for real numbers A and B, inwhich
case we know that e A+B = e A e B by elementary calculus. Therefore, the
formula is correct for any commuting variables A and B.
Now, the beauty of the matrices X and X T appearing in the condition
X + X T = 0 is that they commute! This is because, under this condition,
XX T = X(−X)=(−X)X = X T X.
Thus it follows from the above property of the exponential function that
e X e X T = e X+X T = e 0 = 1.
But also, e X T =(e X ) T because (X T ) m =(X m ) T and hence all terms in the
exponential series get transposed. Therefore
1 = e X e X T = e X (e X ) T .
In other words, if X + X T = 0 then e X is an orthogonal matrix.
Moreover, e X has determinant 1, as can be seen by considering the path
of matrices tX for 0 ≤ t ≤ 1. For t = 0, we have tX = 0, so
e tX = e 0 = 1, which has determinant 1.
And, as t varies from 0 to 1, e tX varies continuously from 1 to e X . This
implies that the continuous function det(e tX ) remains constant, because
det = ±1 for orthogonal matrices, and a continuous function cannot take
two (and only two) values. Thus we necessarily have det(e X )=1, and
therefore if X is an n × n real matrix with X + X T = 0 then e X ∈ SO(n).
This allows us to complete our search for all the tangent vectors to
SO(n) at 1.
Tangent space of SO(n). The tangent space of SO(n) consists of precisely
the n × n real vectors X such that X + X T = 0.
Proof. In the previous section we showed that all tangent vectors X to
SO(n) at 1 satisfy X + X T = 0. Conversely, we have just seen that, for any
vector X with X + X T = 0, the matrix e X is in SO(n).
Now notice that X is the tangent vector at 1 for the path A(t)=e tX in
SO(n). This holds because
d
dt etX = Xe tX ,