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the question is how the covariant derivative acts on the following?

$\nabla_\nu(\Gamma^\alpha_{\mu\lambda}R^{\beta\lambda})=?$ and $\nabla_\nu(\Gamma^\alpha_{\mu\lambda}R^{\beta\gamma\delta\lambda})=?$

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    $\begingroup$ the expression is meaningless as the Christoffel symbols do not form a tensor; however, if you use a more abstract way to define your connection (principal connection on the frame bundle, Ehresmann connections), there is a way to have something like the covariant derivative of it: its curvature $\endgroup$
    – Christoph
    Dec 14 '14 at 23:37
  • $\begingroup$ Hey Christopher I just edited the question. $\endgroup$
    – user67742
    Dec 14 '14 at 23:56
  • $\begingroup$ that makes more sense - I'm guessing this is supposed to be a step on the way towards evaluating the expression $\nabla_\nu\nabla_\mu R^{\beta\gamma}$ and $\nabla_\nu\nabla_\mu R^{\beta\gamma\delta\epsilon}$... $\endgroup$
    – Christoph
    Dec 15 '14 at 0:10
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It doesn't. The covariant derivative is a map from $(k,l)$ tensors to $(k,l+1)$ tensors that satisfies certain basic properties. As such it cannot act on anything except tensors. The collection of components $\Gamma^a_{bc}$ does not constitute a tensor.

If you got to this expression via something like $$ \nabla_d(\nabla_b A^a) = \nabla_d(\partial_b A^a) + \nabla_d(\Gamma^a_{bc} A^c), \tag{not recommended} $$ the problem is evaluating from the inside out. In order to express $\nabla_d$ in terms of partial derivatives and connection coefficients, you should imagine it acting on some arbitrary tensor with components $T_b{}^a$ first, and then later substitute $T_b{}^a = \nabla_b A^a$: $$ \nabla_d(\nabla_b A^a) = \partial_d(\nabla_b A^a) + \Gamma^a_{de} \nabla_b A^e - \Gamma^e_{db} \nabla_e A^a. $$

Now it turns out you'll get the same 6 or 8 terms fully expanded if you work the other way, treating $\Gamma^a_{bc} A^c$ as a (2,2) tensor and just applying the rules for covariant differentiation of such a thing, but I'm not sure this always works, and certainly the intermediate steps don't have any natural geometric interpretation.

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  • $\begingroup$ I got to it by acting d'alembert operator on Ricci and Riemann Tensor Chris it is one of the terms $\endgroup$
    – user67742
    Dec 15 '14 at 0:36
  • $\begingroup$ You can treat $\Gamma_{bc}^a A^c$ as a (1,1) tensor and you'll get the right answer always, you can check in my answer to see why :). $\endgroup$
    – Héctor
    Dec 15 '14 at 21:00
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There is no problem in treat Cristoffel symbols as tensors, because in some definitions they actually are tensors. If one defines abstractly a covariant derivative as an operator over tensors with the following properties:

  1. Linearity: $$ \nabla_c \left( \alpha A^{a_1,\dots,a_k}_{b_1,\dots,b_l} + \beta B^{a_1,\dots,a_k}_{b_1,\dots,b_l} \right)= \alpha \nabla_c A^{a_1,\dots,a_k}_{b_1,\dots,b_l} + \beta \nabla_c B^{a_1,\dots,a_k}_{b_1,\dots,b_l} $$

  2. Leibnitz rule: $$ \nabla_c \left( A^{a_1,\dots,a_k}_{b_1,\dots,b_l} B^{c_1,\dots,c_{k'}}_{d_1,\dots,d_{l'}}\right) = \nabla_c \left( A^{a_1,\dots,a_k}_{b_1,\dots,b_l}\right) B^{c_1,\dots,c_{k'}}_{d_1,\dots,d_{l'}} + A^{a_1,\dots,a_k}_{b_1,\dots,b_l} \nabla_c \left( B^{c_1,\dots,c_{k'}}_{d_1,\dots,d_{l'}} \right) $$

  3. Commutativity with contractions
  4. Consistency with the notion of tangent vectors as directional derivatives on scalar fields $$ t(f) = t^a \nabla_a f $$
  5. Torsion free $$ \nabla_a \nabla_b f = \nabla_b \nabla_a f $$ Then there are lots of differents covariant derivatives, in particular coordinate derivative $\partial_a$ is a covariant derivative. One can prove then that given two different covariant derivatives, their difference is a tensor, so $$ \left(\nabla_a - \widetilde{\nabla_a} \right) v^b = C_{ac}^b v^c $$ Now the covariant derivative used in general relativity is the levi-civita connection, is the only one who no changes the metric $$ \nabla_c g_{ab} = 0 $$ So given $\nabla_a$ the levi-civita covariant derivate and $\partial_a$ the coordinate derivative, there exist a tensor field $\Gamma_{ab}^c$ with the next property $$ \left(\nabla_a - \partial_a \right) v^c= \Gamma_{ab}^c v^b \implies \nabla_a v^c = \partial_a v^c + \Gamma_{ab}^c v^b $$ Now if you change of coordinate system, you have to change the tensor $\Gamma_{ab}^c$ because you change the reference connection $\partial_a$ you're using! so you use a different tensor for each coordinate system, this is the reason some treatments in general relativity say the cristoffel symbols are not tensors, but once is fixed a coordinate system, is valid to treat them like one (because they are the tensor that is the difference between the levi-civita connection and the coordinate derivative). This way of define things is given in the book "General relativity" of Wald, you can look there for reference.
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