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M is a Riemannian manifold, $\vdash$ is a global connection on M compatible with the Riemannian metric.In terms of local coordinates $u^1,...,u^n$ defined on a coordinate neighborhood $U \subset M$, the connection $\vdash$ is determined by $\Gamma^k_{ij}$ on U,as follows. Let $\partial_k$ denote the vector field $\frac{\partial}{\partial u^k}$ on U.Then any vector field X on U can be expressed uniquely as $$X=\sum_{k=1} x^k \partial_k$$ where the $x^k$ are real valued functions on U.In particular the vector field $\partial_i \vdash \partial_j$ can be expressed as $$\partial_i \vdash \partial_j=\sum_k \Gamma^k_{ij}\partial_k$$

My 1st question is that there is an equation $$\partial_i g_{jk}=<\partial_i \vdash \partial_j,\partial_k>+<\partial_j,\partial_i\vdash \partial_k>$$,but how could it be? The left of the equation is a vector field but the right is a function.And there are the first Christoffel identity $$<\partial_i\vdash\partial_j,\partial_k>=\frac{1}{2}(\partial_ig_{jk}+\partial_jg_{ik}-\partial_kg_{ij})$$

and the second Christoffel identity $$\Gamma^l_{ij}=\sum_k\frac{1}{2}(\partial_ig_{jk}+\partial_jg_{ik}-\partial_kg_{ij})$$ for which I have the same question.How could a vector field equals a real function?

My 2nd question is why in Euclidean n-space,$R^n$,have $$\Gamma^k_{ij}=0?$$ I use the second Christoffel identity but obviously it don't equals zero.The metric is the usual Riemannian metric $dx_1^2+dx_2^2+...+dx_n^2$

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# Questions on connections and covariant differentiation

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