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What is the covariant derivative of the connection coefficients?

+ 3 like - 0 dislike
374 views

This question is inspired by this question on physics SE, but hasn't really received an answer.

The question: is it possible to take the covariant derivative of the connection coefficient? And if so, how are they defined?

Since the Riemann tensor is defined as:

\(R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z\)

which is equivalent to (the more well-known) equation:

\(R^\kappa{}_{\lambda \mu \nu} = \partial_\mu \Gamma^\kappa_{\nu \lambda} - \partial_\nu \Gamma^\kappa_{\mu \lambda} + \Gamma^\eta_{\nu \lambda} \Gamma^\kappa_{\mu \eta} - \Gamma^\eta_{\mu \lambda} \Gamma^\kappa_{\nu \eta} \)

it appears to me that the covariant derivative of the connection coefficients must exist. However, the connection coefficient are not honest tensors, and they are not really scalars, so how do we solve this problem (i.e. how do we define the covariant derivative of the connection coefficient)?

asked Apr 29, 2014 in Mathematics by Hunter (510 points) [ revision history ]
edited Apr 29, 2014 by Hunter

Geometrically, it doesn't really makes sense to talk about the covariant derivative of the connection coefficients. Given a fiber bundle $\pi:E\to B$ with typical fiber $F$ and a trivialization $\tau\times\phi:\pi^{-1}(U)\to U\times F$, the Christoffel symbols $\Gamma$ are elements of $\mathrm{Hom}(\tau^*(\mathrm TU),\phi^*(\mathrm TF))$. The most closely related 'nice' geometric object is the connection form (which is described locally via Christoffel symbols), and the covariant derivative of that is just the curvature.
 

@Christoph thanks for this comment. Unfortunately, I don't understand the language of fiber bundles, etc., but I'm currently self-studying differential geometry and I hope I will better understand your comment in the future.

@Hunter, you should look into the abstract definition of a connection due to Ehresmann - that's what helped me get some intuition about the geometry after I had previously only seen the more technical definitions in terms of differential operators (in case of linear connections) or as Lie-algebra valued forms (in case of principal connections).

2 Answers

+ 3 like - 0 dislike

The covariant derivative is initally defined on vector fields and then it is extended to all kinds of tensor fields by assuming that (a) this action is linear, (b) it works as a derivative  with respect to the tensor product:

$$\nabla_X (u \otimes v) = (\nabla_X u) \otimes v + u \otimes  \nabla_Xv \:,$$

(c) it acts as a standard vector field when acting on scalar fields: $\nabla_X f = X(f)$, and (d) it commutes with contractions:

$$ \nabla_X \langle Y, \omega \rangle = \langle \nabla_X Y, \omega \rangle + \langle Y, \nabla_X\omega \rangle\:.$$

With this definition, it is difficult to say if this question makes sense. Referring to a single connection coefficients, i.e., for fixed values of $\alpha,\beta,\gamma$:

$$\Gamma^\alpha_{\beta\gamma} = \langle \nabla_\beta \partial_{x^\gamma},  dx^\alpha\rangle $$

it makes sense, since it is a scalar field and thus the covariant derivative $\nabla_X$ coincides with the action of the vector field  $X$ on $\Gamma^\alpha_{\beta\gamma}$.

However, I do not think it was the intention of the OP. I guess that, instead, the OP thinks of the whole set of "components" $\Gamma^\alpha_{\beta\gamma}$ for all values of the indices. Unfortunately this set does not define a tensor so we cannot apply (a),(b),(c),(d) to this set of components. The only thing one can say is that the differences of connection coefficients referred to two different connections define a tensor field (the proof is evident) of components,

$$T^\alpha_{\beta \gamma}= \Gamma^\alpha_{\beta \gamma} - \overline{\Gamma}^\alpha_{\beta \gamma}\:.$$

Consequently,  the covariant derivative (w.r. to a third affine connection) of this difference is well defined.

What I am saying is that I do not think it make sense to assume that the Riemann tensor is constructed  out of something like covariant derivatives of connection coefficients...

answered Apr 29, 2014 by Valter Moretti (2,025 points) [ revision history ]
edited Apr 29, 2014 by Valter Moretti

Thanks for the answer and +1. Just for clarification: if I wanted to calculate the components of the Riemann tensor (in the coordinate basis):

\(R^\kappa{}_{\lambda \mu \nu} = \langle \mathrm{d} x^\kappa , R(e_\mu , e_\nu) e_\lambda \rangle\)

then I would encounter a term like:

\(R^\kappa{}_{\lambda \mu \nu} = \langle \mathrm{d} x^\kappa , \nabla_\mu( \Gamma^\eta_{\nu \lambda} e_\eta) \rangle + \cdots\)

For the above equation, can I treat the action of $\nabla_\mu$ on $\Gamma^\eta_{\nu \lambda}$ as a partial derivative, because the subscript $\mu$ is not contracted with any index on $\Gamma^\eta_{\nu \lambda}$ (and thus we can treat $\Gamma^\eta_{\nu \lambda}$ as scalar)? So, can I write:

\(R^\kappa{}_{\lambda \mu \nu} = \langle \mathrm{d} x^\kappa , (\partial_\mu \Gamma^\eta_{\nu \lambda}) e_\eta + \Gamma^\eta_{\nu \lambda} \nabla_\mu e_\eta \rangle + \cdots\)

or is this nonsense?

Yes,  $\Gamma^\eta_{\nu\lambda}$ is a scalar therein!

Great, thanks! It all makes sense now.

+ 2 like - 0 dislike

The meaningful way in which you can have a covariant derivative of the connection is the curvature. This can be made precise in the language of principal connections (the curvature form is the exterior covariant derivative of the connection form) or Ehresmann connections (curvature is given by the Frölicher–Nijenhuis bracket of the vertical projector with itself).

The Christoffel symbols describe the connection locally after choice of a trivialization of the fiber bundle, and are not tensors - honest or otherwise ;). Each index of the Christoffel symbols actually live in a different space (the bundle itself with possible non-linear dependence, the tangent space to the base manifold and the tangent space to the typical fiber - which we can identify with the fiber in case of vector bundles); it's just that the special case of linear connections on tangent bundles makes them look somewhat tensor-like.

answered Apr 30, 2014 by Christoph [ revision history ]
edited Apr 30, 2014

Thanks for your answer. As I said in the comment, I'm studying differential geometry in the hope I will understand fibre bundles, etc. (However, since I'm studying this on my own, I'm not sure how long it will take before I get it because I cannot neglect the other topics I'm studying.) Just for clarification: is what you have written in the second paragraph only valid for a torsion-free manifold (because you mention  the  Christoffel symbols which I always associate with a vanishing torsion tensor), or is it valid for any (pseudo-)Riemannian manifold?

@Hunter, do you differentiate between connection coefficients and Christoffel symbols? I use the terms synonymously, ie the 2nd paragraph applies to arbitrary (with torsion, non-metric or even non-linear) connections. Also, a quick reminder about torsion: $T^k{}_{ij}=\Gamma^k{}_{ij}-\Gamma^k{}_{ji}$

Yeah, maybe this is just semantics, but in General Relativity it seems to be conventional to use the following for Christoffel symbols:

\(\begin{Bmatrix} \kappa \\ \mu \nu \end{Bmatrix} = \frac{1}{2} g^{\kappa \lambda}( \partial_\mu g_{\nu \lambda} + \partial_\nu g_{\lambda \mu} - \partial_\lambda g_{\mu \nu} )\)

Then on a (pseudo-)Riemanniian manifold $(M,g)$, there exists a unique symmetric connection which is compatible with the metric $g$. This connenction is called the Levi-Civata connection. This is only valid if the torsion tensor vanish:

\(\Gamma^\lambda_{\mu \nu} = \Gamma^\lambda_{\nu \mu}\)

(Source: Mikio Nakahara's "Geometry, Topology and Physics")

The distinction between connection coefficients and Christoffel symbols is at least made in Sean Carroll's "Spacetime and Geometry" and Mikio Nakahara's "Geometry, Topology and Physics" (maybe other books don't make this distinction).

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