Fundamental theorem on vector fields. Flow of a vector field

There is a better approach in @lee2013smooth page 314.
(I have to fix the notation)
What follows is from @warner prop 1.48:

Proposition
Let $X$ be a vector field on a manifold $M$ . For each $p \in M$ , there exist $a (p), b (p) \in R \cup {\pm \infty}$ and a differentiable curve

γ_{p} : (a (p), b (p)) ⟶ M

such that:

$0 \in (a (p), b (p))$ and $γ_{p} (0) = p$
$γ_{p}$ is an integral curve of $X$ (i.e. $d (γ_{p})_{s} (\frac{d}{d t}) = X (γ_{p} (s))$ )
Uniqueness: If $μ : (c, d) ⟶ M$ is another curve satisfying the above conditions, then $(c, d) \subseteq (a (p), b (p))$ and $γ_{p} |_{(c, d)} = μ$ .
For each $t \in R$ , we define a transformation $X_{t}$ with domain $D_{t} = {q \in M : t \in (a (q), b (q))}$ by the expression $X_{t} (q) = γ_{q} (t)$ , so that for each $p \in M$ , there exists an open neighborhood $V$ and an $ϵ > 0$ such that the map

(t, q) ⟶ X_{t} (q)

is defined on $(ϵ, ϵ) \times V$ . This transformation is called the flow of $X$ and is usually denoted by $ϕ_{X} (t, q) .$
5. $D_{t}$ is open for each $t$ .
6. $⋃_{t > 0} D_{t} = M$ .
7. $X_{t} : D_{t} ⟶ D_{- t}$ is a diffeomorphism with inverse $X_{- t}$ .
8. Let $s, t \in R$ . The domain of $X_{s} \circ X_{t}$ is contained in $D_{t + s}$ (but in general, it is not equal to $D_{t + s}$ ). It is equal to $D_{t + s}$ if $s$ and $t$ have the same sign. Moreover, in the domain of $X_{s} \circ X_{t}$ , we have $X_{s} \circ X_{t} = X_{t + s}$ . Therefore, we have a local group of transformations.

Important facts

Flow of the sum

See flow of the sum of two vector fields

Flow of the Lie bracket

See Lie bracket#Useful formula.

Re-scaling of the flow

Proposition
Let $X$ be a vector field, and $k$ a constant. Let's denote $ϕ_{X}$ and $ϕ_{k X}$ the flow of $X$ and $k X$ respectively. It is satisfied that:

ϕ_{k X} (t, q) = ϕ_{X} (k t, q) .

Proof
We only have to check that $\hat{ϕ} (t, q) = ϕ_{X} (k t, q)$ satisfies 1. and 2. for the vector field $k X$ , and then we will have the result. In fact:

$\hat{ϕ} (0, q) = ϕ_{X} (0, q) = q$ .
$\hat{ϕ} (t, q)$ are integral curves of $k X$ :

\frac{d}{d t} \hat{ϕ} (t, q) = \frac{d}{d t} ϕ_{X} (k t, q) = k X_{ϕ_{X} (k t, q)}

$◼$

Jacobian of the flow of a vector field

Proposition
Let $X$ be a vector field and $ϕ_{t} (p)$ its flow. For every $t$ the map $ϕ_{t} (-)$ is a diffeomorphism, so we can consider its Jacobian

J (t, x) := d e t (D_{x} ϕ_{t} (x)) .

It is satisfied that

\frac{d}{d t} |_{t = 0} J (t, x) = div X (x)

where $div$ is the divergence.

Moreover

\frac{d}{d t} J (t, x) = J (t, x) div X (ϕ_{t} (x)) .

Proof
(See this answer in Mathstackexchange)
It is satisfied that

\frac{d}{d t} ϕ_{t} (x) = X (ϕ_{t} (x))

and then, taking derivatives with respect to the coordinates $x_{1}, \dots, x_{n}$ we obtain the matrix equation

\frac{d}{d t} d (ϕ_{t})_{x} = d X_{ϕ_{t} (x)} d (ϕ_{t})_{x} .

We can consider the linear system given by

\dot{y} = d X_{ϕ_{t} (x)} y, y : R \to R^{n},

and then $d (ϕ_{t})_{x}$ is a matrix whose columns are solutions to it. We now apply Liouville's formula to this linear system, obtaining

J (t, x) = J (t_{0}, x) e^{\int_{t_{0}}^{t} tr d X_{ϕ_{s} (x)} d s}

and deriving with respect to $t$ :

\frac{d}{d t} J (t, x) = J (t, x) div X (ϕ_{t} (x)) .

$◼$
See xournal_155 page 5 for relation with Jacobi last multipliers.