How does covariant derivative act on Christoffel Symbols?

Question

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

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

Answer 1

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.

Answer 2

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.