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dt γ(t) =^ v(γ(t)),^ ∀t^ ∈^ R
We say a vector field v is integrable if ∀x ∈ X there exists an integral curve of v through x.
Example - X = (0, 1) and the vector field: (^) ∂x∂. If we try to get the integral curve through x ∈ (0, 1) we get γ(t) = x + t
but this is not in (0, 1) for t large! So, this vector field is not integrable.
Example - X = R. This is secretly the same, but anyway: let
v = x^2
∂x
Here our integral curve would satisfy: d dt
γ(t) = γ(t)^2
dy dt
= y^2 ∫ dy y^2
dt
y
= t + C
y = −
t + C
i.e.,
γ(t) = −
t + C The problem is that this solution is not defined for all t — it blows up at t = −C. So, this vector field is also not integrable.
Suppose v is an integrable vector field on a manifold X. Then:
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Theorem 1 for every x ∈ X the integral curve of v through x is unique.
This let’s us define a function: φ: R × X → X
by (t, x) 7 → φ(t, x) = φt(x)
such that φt(x) is the integral curve of v through x.
Theorem 2 φ: R × X → X is smooth.
Note also: φ 0 (x) = x
and φs(φt(x)) = φs+t(x)
Mathematicians summarize these equations by saying “φ is an action of the group (R, +, 0) on X”; note they imply: φ−t(x) = (φt)−^1 (x)
since φt ◦ φ−t = φ 0 = 1X
So: for any t ∈ R, φt: X → X
is smooth (by Theorem) with a smooth inverse, φ−t. A smooth map f : X → Y with smooth inverse is called a diffeomorphism.
Definition 3 If φ: R × X → X is a smooth map such that
- φ 0 (x) = x
- φs(φt(x)) = φs+t(x)
we call φ a flow.
We’ve seen that any integrable vector field v gives a flow φ: we call φ the flow generated by v. Conversely, any flow φ is generated by a unique (integrable) vector field v:
v(x) =
d dt φt(x)|t=0, x ∈ X
Now suppose X is a Poisson manifold. If H ∈ C∞(X) is any observable, thought of as the Hamil- tonian, we get a vector field {H, ·}: C∞(X) → C∞(X)
also called vH , the Hamiltonian vector field generated by H. If vH is integrable, it generates a flow φ: R × X → X
called time evolution or the flow generated by H. If our system is in the state x ∈ X initially, then at time t it will be at φt(x) ∈ X.
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