Differential Geometry - Question about Limit

**slevvio** · November 10th 2011, 09:28 AM

Hello everyone, I had a question I was hoping I could get some help with. Let $M$ be a smooth manifold and let $X,Y$ be vector fields. Let $\phi_t$ be the flow of $X$ . So in my notation I have $Y_{p} \in T_p M$ and for a smooth function $g \in C^{\infty}(M)$ then $Y(g) \in C^{\infty}(M)$ . Let $h_t := \displaystyle\frac{g \circ \phi_t - g}{t}$ , so that $\displaystyle\lim_{t\rightarrow 0} h_t = X(g)$ pointwise.

In a proof in my course there is a statement
$\displaystyle\lim_{t\rightarrow 0} Y_{\phi_t(p)}\left(h_t\right) =^{?} Y_{\phi_0(t)}\left( \lim_{t\rightarrow 0}h_t\right)= Y_p( X(g))$ .

Why can we just take limits inside? It is not obvious to me that the map $t \mapsto Y_{\phi_t (p)}(h_t)$ is smooth.

Thanks for any help.

**Drexel28** · November 10th 2011, 10:12 PM

Originally Posted by slevvio

Hello everyone, I had a question I was hoping I could get some help with. Let $M$ be a smooth manifold and let $X,Y$ be vector fields. Let $\phi_t$ be the flow of $X$ . So in my notation I have $Y_{p} \in T_p M$ and for a smooth function $g \in C^{\infty}(M)$ then $Y(g) \in C^{\infty}(M)$ . Let $h_t := \displaystyle\frac{g \circ \phi_t - g}{t}$ , so that $\displaystyle\lim_{t\rightarrow 0} h_t = X(g)$ pointwise.

In a proof in my course there is a statement
$\displaystyle\lim_{t\rightarrow 0} Y_{\phi_t(p)}\left(h_t\right) =^{?} Y_{\phi_0(t)}\left( \lim_{t\rightarrow 0}h_t\right)= Y_p( X(g))$ .

Why can we just take limits inside? It is not obvious to me that the map $t \mapsto Y_{\phi_t (p)}(h_t)$ is smooth.

Thanks for any help.

No, but do you disagree that it's not continuous?

**slevvio** · November 11th 2011, 12:29 AM

I can't really see why it's continuous in t

**slevvio** · November 18th 2011, 03:52 AM

have you any ideas as to why this might be continuous in t?

Thanks very much

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