Given vector field $f: \mathbb{R}^2 \to \mathbb{R}^2$, with $f(0)=0$


ODE: $\dot{x}=f(x)$ generates a flow $\Phi^{t}$. so $\Phi^{t}(0)=0$ for all $t \in \mathbb{R}$


So time-one map $\Phi^1$ is diffeo. Assume $0$ is hyperbolic fixed point, 


let $W^s(0)$ be stable manifold for $\Phi^1$ through $0$. Then for any $x \in W^s(0)$, can we show:


$$f(x) \text{ is the tangent vector of } W^s(0) \text{ at } x?$$ 



flow, stable manifold and tangent
Yes. The stable manifold of $\Phi^1$ is the stable manifold of $(\Phi^t)\_{t\in \mathbb R}$. Since 0 is a fixed point, the stable manifold of 0 is time-invariant. So $\Phi^t(x)$ moves along the stable manifold. In particular, the derivative at $t=0$ is both $f(x)$ and a tangent vector to the stable manifold.


