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Is there a meaningful way to define the covariant derivative of the connection coefficients, $\Gamma^a_{bc}$? As in, does it make sense to define the object $\nabla_d\Gamma^a_{bc}$? Since the connection coefficients symbol doesn't transform as a tensor, it would seem like there should be some obstruction to defining this in the usual way, treating $a$ as a contravariant index and $b$ and $c$ a covariant indices.

Part of my motivation for thinking about this was for writing the Riemann tensor in terms of this symbol $\nabla_d\Gamma^a_{bc}$. If you work in a local Lorentz frame at a point where $\Gamma^a_{bc}$ all vanish, the expression for the Riemann tensor is just $$R^a_{\phantom{a}bcd}=\partial_c\Gamma^a_{bd}-\partial_b\Gamma^a_{cd}.$$ So then I'd like to "covariantize" this expression for a general coordinate system by writing \begin{equation} R^a_{\phantom{a}bcd}=\nabla_c\Gamma^a_{bd}-\nabla_b\Gamma^a_{cd}. \tag{*} \end{equation} If I pretend that $\Gamma^a_{bd}$ should have a covariant derivative defined by treating the indices as normal tensor indices, I get for this expression something pretty close to the right answer $$R^a_{\phantom{a}bcd}=\partial_c\Gamma^a_{bd}-\partial_b\Gamma^a_{cd} +2(\Gamma^a_{ce}\Gamma^e_{bd}-\Gamma^a_{eb}\Gamma^e_{cd})$$ and curiously, if I define $$\nabla_c\Gamma^a_{bd} \equiv \partial_c\Gamma^a_{bd} + \Gamma^a_{ce}\Gamma^e_{bd}-\Gamma^e_{cd}\Gamma^a_{eb} + \Gamma^e_{cb}\Gamma^a_{ed} $$ where the last term appears with the wrong sign from what you get with an ordinary $(1,2)$ tensor, the expression $(*)$ above for the Riemann tensor is correct. Is this just a coincidence, or is there some reason to define a covariant derivative of the connection symbol like that?

Update: The expression that gives the right form of the Riemann tensor for $(*)$ is actually $$\nabla_c\Gamma^a_{bd} \equiv \partial_c\Gamma^a_{bd} + \Gamma^a_{ce}\Gamma^e_{bd}-\Gamma^e_{cd}\Gamma^a_{eb}$$ so it is as if we are not treating $b$ as a tensor index, and we are just writing the covariant derivative of a $(1,1)$ tensor.

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For a fixed $b$, $\Gamma^a_{bd}$ is a $(1,1)$-tensor because $\nabla \vec{e}_b$ is a tensor, so you can it as a collection of tensors. But $\Gamma^a_{bd}$ is not a $(1,2)$-tensor. Those two facts at least gives some motivation why insisting on a 'covariant derivative' would (1) have the usual pattern in the upper and one of the lower indices but (2) break away from it in the other lower index. –  Stan Liou Jul 19 '13 at 18:55
Your last expression does not make sense : $\Gamma_{bd}^a$ is symmetric in $b$ and $d$, while the two last terms are antisymmetric in $b$ and $d$ –  Trimok Jul 19 '13 at 19:23
@Trimok, You're right, that seems to be a problem. I think I might have made a mistake in saying that that definition gives the right answer for the Riemann tensor. Actually, it seems to work if instead I just leave off that last term entirely. We might justify that since, as Stan noted, it's only the $a$ and $d$ indices that act like tensor indices. –  asperanz Jul 19 '13 at 19:32
@asperanz : Your "new" last expression is not correct, because the last term is not symmetric in $b$ and $d$ –  Trimok Jul 20 '13 at 7:55
Did you ever find an answer to this question? I would think that: $$\nabla_\nu R_{\mu \sigma}^\rho = \partial_\nu R_{\mu \sigma}^\rho$$ because in a way you can think of $R_{\mu \sigma}^\rho$ as a scalar (at least it is not a tensor). But I'm not sure, and am curious to find an answer. –  Hunter Apr 28 '14 at 11:32

2 Answers 2

up vote 3 down vote accepted

The formalism is explained very well in Landau-Lifshitz, Vol. II, par. 92 (properties of the curvature tensor). The Riemann curvature tensor can be called the covariant exterior derivative of the connection. The exterior derivative is a generalisation of the gradient and curl operators.

You might also consider looking at the geometry in differential forms language. The connection is seen as a 1-form (to be integrated along a line, the corresponding index is supressed), resulting in a (2-index) transformation matrix.* The Riemann curvature tensor is seen as a 2-form (to be integrated over a surface), again with values in a (2-index) transformation matrix. By doing so, you see Stokes' theorem appear, since integrating the connection (1-form) along a closed lines yields the same result as integrating the Riemann curvature (2-form) over the enclosed surface. That's why the Riemann curvature (2-form) needs to be the covariant exterior derivative of the connection (1-form).

Literature: Nakahara, Geometry, Topology and physics, chap. 5.4 and 7.

*Precisely: a Lie-algebra valued 1-form.

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I posted a question similiar to this one, link, however it concerns the representation of the covariant derivative. –  linuxfreebird Mar 5 '14 at 15:06

Your question:

Is there a meaningful way to define the covariant derivative of the connection coefficients...?

has a very simple answer: NO

It does not make sense (and it would be a very bad practice) to overload the operator "covariant derivative" and force it to somehow work on objects that are not tensors or scalars.

$\delta\Gamma$ (the variation of $\Gamma$ in an action ) is however a tensor with 3 indices, so you will find expressions like $$\delta\Gamma^a_{bc}$$ for the a,b,c component of this tensor and $$\delta\Gamma^a_{bc;d}$$ the a,b,c,d component of its covariant derivative, and even $$\delta R_{ab}=\delta\Gamma^l_{ab:l}-\delta\Gamma^l_{al:b}$$ for the variation of the Ricci tensor.

All this can be found in MTW page 492 and page 500

Pleasse note that: $$\delta\Gamma^a_{bc;d}$$ does not mean $$\delta (\Gamma^a_{bc;d})$$ which is meaningless, but rather $$(\delta\Gamma)^a_{bc;d}$$ since $\delta\Gamma$ is the tensor being differentiated

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Thanks for your answer. I have one more question though. The Riemann tensor is defined as: $$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z$$ and a term like $\nabla_X \nabla_Y Z $ will surely involve the covariant derivative of the connection coefficient, right? How do we interpret this? –  Hunter Apr 29 '14 at 15:23
No, it would not involve the covariant derivative of the connection coefficient if you know what you are doing with the notation. I suggest you ask a new question along these lines: in a double covariant differentiation, do we have terms involving the covariant derivative of the $\Gamma$? and I will answer it properly. –  magma Apr 29 '14 at 15:47
I have found an answer to the question somewhere else, so it is not really necessary anymore. But thanks for the offer. –  Hunter Apr 30 '14 at 8:33

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