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.


3 Answers 3


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.

  • $\begingroup$ I posted a question similiar to this one, link, however it concerns the representation of the covariant derivative. $\endgroup$ Mar 5, 2014 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

  • $\begingroup$ 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? $\endgroup$
    – Hunter
    Apr 29, 2014 at 15:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – magma
    Apr 29, 2014 at 15:47
  • $\begingroup$ I have found an answer to the question somewhere else, so it is not really necessary anymore. But thanks for the offer. $\endgroup$
    – Hunter
    Apr 30, 2014 at 8:33

Whether or not this makes sense rigorously$^\ast$ is beyond me right now (would need to brush up a bit on (gauge-)natural bundle functors and jet prolongations), but the covariant derivative of connections can be defined.

At first let us work with a gauge connection instead of a tangent space connection.

Suppose that $\rho:G\rightarrow\mathrm{GL}(V)$ is a representation of the structure group $G$, and $\psi$ is a field that transforms under the representation $\rho$ as $$ \psi^\prime=\rho(g^{-1})\psi. $$ Then infinitesimally ($g\approx1+\epsilon X$, where $X\in\mathfrak g$ is a Lie algebra element) we have $$ \delta\psi=-\rho(X)\psi, $$ where I have used the same $\rho$ symbol for the induced Lie algebra representation.

The heuristic idea behind gauge connections is that (infinitesimal) parallel transport should have the same effect on the field components as a gauge transformation, so the parallel transport of $\psi$ from $x$ to $x+\mathrm dx$ is $$ \psi_{x+dx}=\rho(U^{-1}(x,dx))\psi_x\approx(1-\mathrm dx^\mu \rho(A_\mu(x)))\psi_x, $$ where $A_\mu$ is $\mathfrak g$-valued.

Then $$ \delta\psi=-\rho(A_\mu)\psi\mathrm dx^\mu $$ is the infinitesimal change in field components during parallel transport. The covariant differential is then $$ D\psi(x)=\psi(x+\mathrm dx)-(\psi(x)-\rho(A_\mu(x))\psi(x))=\mathrm d\psi(x)+\rho(A_\mu(x))\psi(x) \\ =(\partial_\mu\psi+\rho(A_\mu)\psi)\mathrm dx^\mu, $$ so the covariant derivative is $$ D_\mu\psi=\partial_\mu\psi+\rho(A_\mu)\psi. $$

For clarity, let us assume we are given a different gauge connection $B_\mu$. Under a gauge transformation, it transforms as $$ B^\prime_\mu=g^{-1}B_\mu g+g^{-1}\partial_\mu g\approx(1-\epsilon X)B_\mu(1+\epsilon X)+(1-\epsilon X)\partial_\mu(\epsilon X) \\ \approx B_\mu -\epsilon XB_\mu+\epsilon B_\mu X+\epsilon\partial_\mu X=B_\mu +\epsilon(\partial_\mu X+[B_\mu,X]), $$ thus $$ \delta B_\mu=\partial_\mu X+[B_\mu,X]. $$

The field $B_\mu$ does not transform under a linear representation of $G$, but under a nonlinear action of the group's first jet prolongation, but nontheless we can use the formal $X\mapsto A_\mu\mathrm dx^\mu$ rule to determine $B_\mu$'s variation under parallel transport by $A_\mu$ to get $$ \delta B_\mu=\partial_\mu A_\nu\mathrm dx^\nu+[B_\mu, A_\nu]\mathrm dx^\nu. $$

Then the derivative of $B_\mu$ is $$ DB_\mu=\mathrm dB_\mu-\partial_\mu A_\nu\mathrm dx^\nu-[B_\mu,A_\nu]\mathrm dx^\nu \\ =(\partial_\nu B_\mu-\partial_\mu A_\nu +[A_\nu,B_\mu])\mathrm dx^\nu, $$ and so the covariant derivative is $$ D_\mu B_\nu=\partial_\mu B_\nu-\partial_\nu A_\mu +[A_\mu,B_\nu]. $$

If we now substitute $A$ instead of $B$, we obtain $$ D_\mu A_\nu=D_{[\mu}A_{\nu]}=\partial_\mu A_\nu -\partial_\nu A_\mu +[A_\mu,A_\nu]=F_{\mu\nu}, $$ the gauge curvature.

For the tangent space connection, the fundamental representation is $$ V^{\mu^\prime}=V^\mu\frac{\partial x^{\mu\prime}}{\partial x^\mu},$$ so $g=\frac{\partial x}{\partial x^\prime}\equiv J$ (I'll use the notation $J$ here). The infinitesimal transformation is then a vector field's ordinary derivative $\partial_\nu \xi^\mu$.

For the connection we have (I'll use index notation only in the vector index) $$ \Gamma^\prime_\mu=J^{-1}\Gamma_\alpha J J^\alpha_\mu+J^{-1}\partial_\alpha J J^\alpha_\mu, $$ so infinitesimally $$ \delta\Gamma_\mu=-\partial\xi \Gamma_\mu + \Gamma_\mu\partial\xi+\Gamma_\alpha\partial_\mu\xi^\alpha+\partial_\mu\partial\xi. $$

The covariant derivative is then ($\partial\xi\mapsto \Gamma_\mu$) $$ \nabla_\nu\Gamma_\mu=\partial_\nu\Gamma_\mu+\Gamma_\nu\Gamma_\mu-\Gamma_\mu\Gamma_\nu-\partial_\mu\Gamma_\nu-\Gamma_\alpha\Gamma_{\nu\mu}^{\alpha}. $$

Skew-symmetrizing this gives $$ \nabla_{[\mu}\Gamma_{\nu]}=\partial_\mu\Gamma_\nu-\partial_\nu\Gamma_\mu+[\Gamma_\mu,\Gamma_\nu]=R_{\mu\nu}, $$ where $R_{\mu\nu}$ is the Riemann tensor, just the two other indices had been suppressed, and the last term drops out ($\Gamma$ is symmetric).

In case $\Gamma$ is not symmetric, we probably should have ignored the vector index, since we essentially want to take a funky covariant exterior derivative.

*: The curvature can be interpreted rigorously as the covariant derivative of the connection without using heuristic "$X\mapsto A_\mu$" rules or jet prolongations if we lift everything to the overlying principal fibre bundle and use the interpretation of the exterior covariant derivative as $$ \Omega=\mathrm D\omega\equiv \mathrm d\omega\circ \mathrm h, $$ where $\omega$ is a connection form, $\mathrm h$ is the horizontalization and $\Omega$ is the curvature form.


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