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Questions on connections and covariant differentiation

  1. Oct 20, 2011 #1
    The question is in the pdf file,thank you!:smile:

    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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  3. Oct 20, 2011 #2


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    g_jk is a function of U, so how could its derivative wrt to x_i be a vector field? It is a function still.

    Equally misterious to me is why you think the Christ. symbols are vectors. Like you wrote before, they are merely the coefficients(=smooth functions on U) of the vector field \partial_i \vdash \partial_j wrt to the basis (\partial_k)_k.

    Your second question kinda proves that there's some fundamental misconception you're having about the meaning of some of the symbols. Because indeed you get the vanishing of the Christ. symbols from the 2nd identity. In the euclidean riemannian metric, the metric coefficients g_ij (=smooth R-valued functions on R^n!) are just the constant function g_ii=1 and g_ij=0 (if i is not j). So their derivatives are indeed all 0.
  4. Oct 20, 2011 #3
    Oh,I get it! [tex]\partial_i g_{jk} [/tex] is only the derivative of g_{jk} in the direction of x_i,it is still a function. It was really a misconception I have.

    Thank you quasar987!:)
  5. Oct 21, 2011 #4


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    My pleasure. :)
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