n!=
X
n=0
(−tX)n
n!=e−tX
.
Since
tX
and
−tX
commute, I get
etX (etX )∗=etX e−tX =etX−tX =e0= 1
. Thus,
etX ∈O(n)
. Of course,
e0X= 1
. Dierentiating termwise, as is allowed for entire functions on Banach algebras,
d
dt etX =
X
n=0
ntn−1Xn
n!=
X
n=1
(tX)n−1X
(n−1)! =X
X
n=0
(tX)n
n!=XetX
,
which at
t= 0
is
X1 = X
.
3
If
X∈o(n)
, then
Xi,j =−Xj,i
for indices
i, j
. If
i=j
, it follows that
Xi,j = 0
. If
i > j
, then
Xi,j
is de-
termined by
Xj,i
, where
j < i
. Thus,
X
is determined by
Xi,j
for
i < j
. Conversely, given any values for
Xi,j
for
i < j
, let
Xi,j
be
−Xj,i
when
i > j
and let
Xi,j
be
0
when
i=j
. Then
X∈o(n)
. Thus, a basis of
o(n)
is
{Ei,j
.
.
.
i < j}
, where
(Ei,j)i,j = 1
,
(Ei,j)j,i =−1
, and every other component of
Ei,j
is
0
. (That is,
(Ei,j)k ,l := δi,j
k,l
.) When
n= 3
, we have
{E1,2, E1,3, E2,3}
. Since
X1=−E2,3
,
X2=E1,3
, and
X3=−E1,2
,
another basis is
{X1, X2, X3}
.
4
If
v= (v1, v2, v3)
, then
(a·X)v=
0−a3a2
a30−a1
−a2a10
v1
v2
v3
=
a2v3−a3v2
a3v1−a1v3
a1v2−a2v1
=a×v
.
Thus, the
t
derivative of
eta·Xv
is
(a·X)eta·Xv=a×eta·Xv
.
5
Let
G
be any smooth function on the manifold
X
. Then
vF[G] = {F, G}=∂F
∂p1
∂G
∂q1
−∂F
∂q1
∂G
∂p1
+∂F
∂p2
∂G
∂q2
−∂F
∂q2
∂G
∂p2
+∂F
∂p3
∂G
∂q3
−∂F
∂q3
∂G
∂p3
= (a2q3−a3q2)∂G
∂q1
−(a3p2−a2p3)∂G
∂p1
+ (a3q1−a1q3)∂G
∂q2
−(a1p3−a3p1)∂G
∂p2
+ (a1q2−a2q1)∂G
∂q3
−(a2p1−a1p2)∂G
∂p3
= (a×q)· ∇qG+ (a×p)· ∇pG
.
Thus,
vF= (a×q)· ∇q+ (a×p)· ∇p
. Identifying the tangent spaces to
R3×R3
with
R3×R3
itself, this
becomes
vF= (a×q,a×p) = (a,a) (×,×) (q,p)
.
6
At
t= 0
, both sides of equation (1) reduce to
(q,p)
. The
t
derivative of the left side is
vF|exp (tvF)(q,p)=
(a,a) (×,×) exp (tvF)(q,p)
; the
t
derivative of the right side is
(a×eta·Xq,a×eta·Xp) = (a,a) (×,×)
eta·X(q,p)
. Thus, both sides satisfy the same initial value problem, so they are equal.
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