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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!

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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.

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    $\begingroup$ Nice, I totally overlooked that. Thank you! $\endgroup$ – Nukular Mar 23 '17 at 17:47

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