# 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.

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 ${ E }_{ i }$?

• Is there a preexisting notion of the covariant derivative of a vector field with respect to a vector field? I have never heard of such a thing, and it can't be expressed in abstract index notation. – Ben Crowell May 8 '17 at 0:19
• @BenCrowell: The covariant derivative of vector fields wrt vector fields is built in to the very definition of a connection in the usual formulation of DG. In index notation $\nabla_X Y$ becomes $X^i \nabla_i Y^j$. – Anthony Carapetis May 8 '17 at 2:07

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, 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 $\Gamma_{ij}^k = \theta^k(\nabla_{E_i} E_j)$.
• I think you mean that ${ \psi }_{ |M-W }=0$, so that $\psi$ is a function defined on all $M$. Also I think that in the 3rd "since" it should read $\psi { X }_{ 1 }=\psi { X }_{ 2 }$, right?. – Mr. X May 8 '17 at 16:59
• I was thinking that there may be not vector field extensions of the ${ E }_{ i }$ over $U$ at all!!!. Instead you pick $V\subseteq W\subseteq\ U$ and a bump function $\phi$ such that ${ \phi }_{ |V}=1$ and ${ \phi }_{ |U-W }=0$, take $\phi { E }_{ i }$ and the extend it by the zero section. This global vector field is smooth and coincides with ${ E }_{ i }$ on $V$, but not necesarilly on $U$. So how can we see that with respect to this kind of extensions the covariant derivative is independent?? I´m so sorry of asking again. Should I edit the original question? – Mr. X May 8 '17 at 17:59
• @Mr.X Thanks for the corrections, I've edited my answer. I'm not sure you're following the idea of the argument in your last objection - the whole bump function argument is just to show that the extension doesn't matter. We use a different $V,W,\psi$ for each point $p\in U$. – Anthony Carapetis May 9 '17 at 0:39