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We can describe a vector on a manifold $M$ of dimension $n$ as follows: Let $p:I\rightarrow M$ with $I$ open interval in $\mathbb{R}$ be a curve in $M$. Now look to $p_0=p(0)$. Locally we can find a coordinate system $(U,x_U)$ around $p_0$ such that we can describe $p(t)$ by $(x_U^1(t),\ldots,x_U^n(t))$. Now we can desribe $p'(0)$ be the $n$-tuple $(\frac{dx_U^1(0)}{dt},\ldots,\frac{dx_U^n(0)}{dt})$ (i hope that i write down this correctly. My question is now the following:

Suppose $p_0$ lies also in the coordinate system $(V,x_V)$, then we have a diffeomorphism on the overlap, namely $x_{UV}:x_U^{-1}\circ x_V:x_U(U\cap V)\rightarrow x_V(U\cap V)$. Moreover we have that we can describe $p'(0)$ by the $n$-tuple $(\frac{dx_V^1(0)}{dt},\ldots,\frac{dx_V^n(0)}{dt})$ (locally). My qeustion is know how to get the relation between $\frac{dx_V(0)}{dt}$ and $\frac{dx_U(t)}{dt}$ on $U\cap V$?.

My idea was the following: Since we have the overlap map $x_{UV}$ we can say the following:

$$(x_V^1(0),\ldots,x_V^n(0))=x_{UV}(x_U^1(0),\ldots,x_U^n(0))$$

Now it should be possible to differentiate with repsct to $t$ to get an expression for $\frac{dx_v^i(0)}{dt}$ for all $1\leq i\leq n$ depending op $x_U^i$. I think you have to use the chain rule, but i don't get on the right way. Can someone help me with this. I get the inspirations from "The Geometry of Physics - An Introduction".

Thanks a lot.

Berk89
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1 Answers1

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Yes, it's the chain rule, as you surmised. Your definition of $x_{UV}$, however, is backwards. To get a map on open subsets of Euclidean space, you want something like $$x_{UV} = x_U\circ x_V^{-1}\colon x_V(U\cap V)\to x_U(U\cap V).$$ Now, $x_U(p(t)) = x_{UV}\big(x_V(p(t)\big)$, so what you're writing becomes $$\big(x_U^1(t),\dots,x_U^n(t)\big) = x_{UV}\big(x_V^1(t),\dots,x_V^n(t)\big),$$ from which the chain rule gives us $$\frac{dx_U(t)}{dt} = Dx_{UV}\big(\frac{dx_V(t)}{dt}\big),$$ where $Dx_{UV}$ is the Jacobian matrix of $x_{UV}$ at the appropriate point $x_U(t)$.

Ted Shifrin
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