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

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

  • 3
    $\begingroup$ 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
    Commented Dec 14, 2014 at 23:37
  • $\begingroup$ Hey Christopher I just edited the question. $\endgroup$
    – user67742
    Commented Dec 14, 2014 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
    Commented Dec 15, 2014 at 0:10

2 Answers 2


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.

  • $\begingroup$ I got to it by acting d'alembert operator on Ricci and Riemann Tensor Chris it is one of the terms $\endgroup$
    – user67742
    Commented Dec 15, 2014 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
    Commented Dec 15, 2014 at 21:00

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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.