# The relation between an isotopy and a time-dependend flow (Exercise 9-21 Lee's Introduction to smooth manifolds)

The Exercise and most of the notation is from the book "Introduction to smooth manifolds" by Lee. A smooth isotopy of M is a smooth map $$H:M\times J\to M$$ where $$J\subset\mathbb{R}$$ and $$H_t:M\to M$$ defined by $$H_t(p)=H(p,t)$$ is a diffeomorphism. The time-dependend flow $$\Psi$$ of a time-dependend vectorfield $$V$$ is defined by $$\frac{d}{ds}|_{s=t}\Psi(s,t_0,p)=V(t,\Psi(t,t_0,p))$$.

Now in the exercise 9-21 the claim is $$V(t,p):=\frac{\partial}{\partial t}H(p,t)$$ is a time-dependend vectorfield and its flow is $$\Psi(t,t_0,p)=H_t\circ H_{t_0}^{-1}(p)$$.

Now my question towards this. We need to show that $$V(t,p)\in T_pM$$ so that $$V$$ is a time-dependend vectorfield. But for some fixed $$p$$, the curve $$s\mapsto H(p,s)$$ evaluated at $$t$$ dose not need to be a curve through $$p$$. I use the notation $$\frac{\partial }{\partial s}|_{s=t}f(s)=\frac{\partial }{\partial s}|_{s=t}(f(s))$$ for $$\frac{\partial}{\partial t}f(t)$$. Let $$f\in C^{\infty}(M)$$, then $$V(t,p)f=\frac{\partial}{\partial s}|_{s=t}(H(p,s))f=\frac{\partial}{\partial s}|_{s=t}(f\circ H(p,s))=\lim_{s \to t}\frac{f(H(p,s))-f(H(p,t))}{s-t}$$ is the differential quotient for $$f$$ at $$H(p,t)$$ and $$V(t,p)$$ would be a tangentvector in $$T_{H(t,p)}M$$.

Should it be $$V(t,p)=\frac{\partial}{\partial s}|_{s=t} (H(H_t^{-1}(p),s))$$ or equivalent $$V(t,H(p,t))=\frac{\partial}{\partial s}|_{s=t} (H(p,s))$$ That would fit to $$\Psi(t,t_0,p)=H_t\circ H_{t_0}^{-1}(p)$$. With $$V(t,\Psi(t,t_0,p))=\frac{\partial}{\partial s}|_{s=t}\Psi(s,t_0,p)=\frac{\partial}{\partial s}|_{s=t}H(H_{t_0}^{-1}(p),s)=V(t,H_t(H_{t_0}^{-1}(p))$$

Or do I have a missunderstanding?