# Definition of flows on manifolds

I'm currently studying Differential Geometry with the book Introduction to Smooth Manifolds by Lee and got confused by the following definition of a flow:

A global flow $$\theta$$ on a smooth manifold $$M$$ is a continuous map $$\theta:\mathbb{R}\times M\rightarrow M$$ satisfying the following properties for all $$s,t\in\mathbb{R}$$ and $$p\in M$$:

(1) $$\theta(t,\theta(s,p))=\theta(t+s,p)$$

(2) $$\theta(0,p)=p$$

For specific subsets of $$\mathbb{R}\times M$$ we can also define a local flow with the same properties but some additional technicalities (usually if the time-domain cannot be the whole real axis). We know that every smooth vector field induces a unique maximal local flow (i.e. with a maximal time-domain).

Now to the problem: For $$M=\mathbb{R}$$ and the vector field $$V=x^2\frac{\partial}{\partial x}$$ we get the (local) flow $$\theta(t,p)=\frac{p}{1-t}$$ by solving the ODE $$\dot{x}=x^2$$ with initial value $$x(0)=p$$. However,

$$\theta(t,\theta(s,p))=\frac{p}{1-s}\frac{1}{1-t}\neq\theta(t+s,p)$$

so (2) is not fulfilled and $$\theta(t,p)$$ is not a flow - although it is the solution of the corresponding ODE... So apparently I made a mistake in interpreting above definition. I would be happy if someone could point it out. Thanks!

I think you made a slight mistake when solving the ODE. The solution should be $$\theta (t,p) = \frac p{1 - pt}.$$

Since $\theta(s,p) = p /(1-ps),$ we have $$\theta(t, \theta(s,p)) = \theta\left(t, \tfrac p{1-ps} \right) = \frac {\tfrac p{1-ps}}{1 - \tfrac p{1-ps}\times t} = \frac{p}{1-p(t+s)} = \theta(t + s, p),$$ as required.

• Nice, I totally overlooked that. Thank you! Mar 23 '17 at 17:47