Definition of connection computed over local vector fields and Christoffel symbols

Question

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.

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

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

3. 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 ${ E }_{ i }$?

Answer 1

1. 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

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

2. Yes.
3. 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 $\Gamma_{ij}^k = \theta^k(\nabla_{E_i} E_j)$.