Rectification of Vector fields

I am trying to understand how to rectify vector fields on manifolds. Namely if there is a manifold $M$ and a given vector field $X_p$ where for a $p_0$ we have $X(p_0)\neq0$ then there exists a neighborhood of $p_0$ and some other coordinate systme $y$ where $X=\frac{\partial}{\partial y_1}$. I would like some refference to study this phenomenon, mainly excersices or examples. I have read the proof in Spivak's book but I have a hard time solving actual excersises. Also I found Arnold's book on Ordinary Differential Equations lacking of any worked examples (although I may be wrong here).

For example, I would appreciate if someone could give me some on direction on rectifying the following $\mathbb{R^2}$ vector field, given on cartesian coordinates, around the point $p=(1,2)$:

$$V=x_1\frac{\partial}{\partial x_1} -2x_2\frac{\partial}{\partial x_2}.$$

I would be perticularly grateful if you could illuminate the procedure using flows.

2 Answers

I) Generally stratification/rectification of a non-vanishing vector field is guaranteed locally by the local existence theorem for ODEs.

II) In OP's example, it is easiest to stratify/rectify the vector field $V$ in steps:

1. Choose first $$(y_1,y_2)~=~(\ln |x_1|,\ln|x_2| ).$$ Then the vector field becomes $$V ~=~ \frac{\partial}{\partial y_1} - 2\frac{\partial}{\partial y_2}$$ by the chain rule.

2. Next choose $$(y_1,y_2)~=~(z_1+f(z_2),-2z_1+g(z_2)),$$ so that the Jacobian is non-vanishing. Then the vector field $$V ~=~ \frac{\partial}{\partial z_1}$$ is stratified.

• Thank you ! I would really appreciate it if you could also describe how to solve this example using flows. – Nick A. Dec 29 '17 at 14:29

After a few days I (the O.P.) finaly understood how to solve such problems using flows (usually this is reffered to as "streightening out", and can be found for example in Lee's book on differentiable manifolds). I will work on my example:

First you need to find the flow of the vector field. To do so you start by finding the integral curves, namely solving the diffeq system:

$\frac{dγ_1(t)}{dt}=γ_1(t)$

$\frac{dγ_2(t)}{dt}=-2γ_2(t)$

Setting initial conditions to be $γ(0)=(x,y)$ we get the flow $φ_t(x,y)=(xe^t,ye^{-2t})$.

Since at the point (1,2) we have $X_{(1,2)}=\frac{\partial}{\partial x}-2\frac{\partial}{\partial y}\neq0$ let's choose ${x}$ to play the role of time.

So to get the coordinate system we have $U(t,u)=φ_{t}(0,u)+(1,-2)=(e^t,e^{-2t}u-2)=(x,y)$ and $U^{-1}=(lnx,x^{2}(y+2))=(t,u)$. Using the chain rule to write:

$\frac{\partial}{\partial x}=\frac{\partial t}{\partial x}\frac{\partial}{\partial t}+\frac{\partial u}{\partial x}\frac{\partial}{\partial u}$

$\frac{\partial}{\partial y}=\frac{\partial t}{\partial y}\frac{\partial}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial}{\partial u}$

We arrive to the desired answer.