Definition of connection computed over local vector fields and Christoffel symbols Let $M$ be a manifold and $\nabla$ a connection on $M$. If $X$ and $Y$ are smooth vector fields defined on a open set $U$ of $M$ then I define ${ \nabla  }_{ X }Y ={ \nabla  }_{ \widetilde{X} }\widetilde{Y} $ where $\widetilde{X}$ and $\widetilde{Y}$ are global vector fields which extend $X$ and $Y$, respectively. 


*

*Is this well-defined?, that is, is this definition independent of the vector field extensions? 

*Let ${ E }_{ i }$ form a local frame on $U$. Is this the way that ${ \nabla  }_{ { { E }_{ i } } }{ E }_{ j }$ is defined?

*If so, then we have that ${ \nabla  }_{ { { E }_{ i } } }{ E }_{ j }={ \nabla  }_{ { \widetilde { E } _{ i } } }{ \widetilde { E }  }_{ j }=\sum _{ k }^{  }{ { \Gamma  }_{ ij }^{ k } } { E }_{ k }$ on $U$, where the ${ \Gamma  }_{ ij }$ are functions defined on $U$ (a.k.a Christoffel symbols). Are the Christoffel symbols independent of the extensions of the the ${ E }_{ i }$?
 A: *

*To show that this is well-defined, we must take extensions $X_1, X_2$ and $Y_1, Y_2$ of $X$ and $Y$ and show that $\nabla_{X_1} Y_1 |_p = \nabla_{X_2} Y_2 |_p$ for any $p \in U$. Let $V,W$ be open sets such that $p \in V \subset W \subset U$ and construct a smooth bump function $\psi$ such that $\psi|_V = 1$ and $\psi|_{M\setminus W} = 0$. Then $\psi(p)=1$ and $d\psi|_p = 0$, so


$$
\begin{align*}
\nabla_{X_1} Y_1|_p &= \psi(p) \nabla_{X_1}Y_1|_p \\
&=\nabla_{X_1}(\psi Y_1)|_p-(X_1|_p \psi) Y_1|_p \tag{by the Leibniz rule} \\
&= \nabla_{X_1}(\psi Y_1)|_p \tag{since $d\psi|_p=0$} \\
&= \nabla_{X_1}(\psi Y_2)|_p \tag{since $\psi Y_1 = \psi Y_2$} \\
&= \nabla_{\psi X_1}(\psi Y_2)|_p \tag{by $C^\infty$-linearity in $X$} \\
&= \nabla_{\psi X_2}(\psi Y_2)|_p \tag{since $\psi X_1 = \psi X_2$} \\
&= \nabla_{X_2}(\psi Y_2)|_p \\&= \psi(p)\nabla_{X_2} Y_2|_p + (X_2|_p \psi)Y_2|_p \\
&=\nabla_{X_2}Y_2|_p.
\end{align*}$$


*Yes.

*Yes, since $\nabla_{E_i} E_j$ is independent of the extension and the formula $\nabla_{E_i} E_j = \sum_k \Gamma_{ij}^k E_k$ uniquely determines the $\Gamma_{ij}^k$. You can see this by letting $\theta^i$ denote the dual frame to $E_i$ (i.e. $\theta^j(E_i) = \delta^j_i$) and showing $\Gamma_{ij}^k = \theta^k(\nabla_{E_i} E_j)$.

