the individual component functions of fare smooth. Thus fis smooth, and therefore its restriction
to the smooth submanifold R×so(n) of R×Rn2must be smooth. Whence, φ, whose two component
functions are both given by this restriction of f, is smooth as a map
φ:R×so(n)→so(n).
(The fact that the range is given as so(n), as opposed to Rn2, follows from exercise 1.)
3. For a skew adjoint matrix A, show that the observable
F(q, p) = 1
2Xaij(qipj−qjpi)
generates the flow from the previous exercise.
First note that from the skew-adjointness of Awe have
∂qkF=X
i
akipi= (Ap)kand ∂pkF=−X
i
akiqi=−(Aq)k,
from which it follows that the vector field generated by Fis
{F, ·} =X
k
(Ap)k∂pk+ (Aq)k∂qk,
and so the flow φt= (ψ(t), ϕ(t)) is determined by the following two systems of ODE’s:
ψ0(t) = Aψ(t)
ψ(0) = pand ϕ0(t) = Aϕ(t)
ϕ(0) = q.
But the solutions to these systems come easily as:
ψ(t) = etApand ϕ(t) = etA q.
This yields the desired flow.
4. Consider the observable
F(q, p) = q1p2−q2p1.
Determine the flow.
In light of the previous exercise, the skew-adjoint matrix associated with this observable is
A=0 1
−1 0 .
Whence the flow is given by:
φt(q, p) = etAq , etAp.
Now, notice that
Ax =−ix
where on the right the vector xis treated as the complex number x1+ix2,so that we have (continuing
to play loose with the identification between R2and C):
etAx=e−itx=Rtx
2
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