I wish to compute $$[\nabla_{\mu}, \nabla_{\nu}]e^{\lambda}_{~~a}. $$ To do so, I make use of $\nabla_{\nu}e^{\lambda}_{~~a} = \omega_{a~~~\nu}^{~~b}e^{\lambda}_{~~b}$, so that I may write $$\nabla_{\mu}\nabla_{\nu}e^{\lambda}_{~~~a} = \nabla_{\mu}(\omega_{a~~~\nu}^{~~~b})e^{\lambda}_{~~b}+\omega_{a~~\nu}^{~~b}~\omega_{b~~~\mu}^{~~c}e^{\lambda}_{~~c}$$In the end, I intend to anti-symmterize in $\mu, \nu$ to get the desired object. Therefore, I would like to know what is the covariant derivative of the spin-connection $\omega$ in order to finish my computation. Is $\omega$ a scalar, a vector or what? How do you decide? Can someone help?

  • $\begingroup$ The spin connection is a not a tensor. You cannot define its covariant derivative. $\endgroup$
    – Prahar
    Commented Apr 19, 2021 at 8:33
  • $\begingroup$ @Prahar Mitra: That can't be right: all connections are equivalently covariant derivatives. $\endgroup$ Commented Apr 19, 2021 at 9:03
  • $\begingroup$ @MoziburUllah connections define a covariant derivative for tensors (or tensor densities). You cannot take covariant derivative of the connection itself. $\endgroup$
    – Prahar
    Commented Apr 19, 2021 at 9:04
  • $\begingroup$ @Prahar Mitea: Ok, I thought he was asking for the spin connection to be expressed as a covariant derivative. $\endgroup$ Commented Apr 19, 2021 at 9:15
  • $\begingroup$ Could I ask where you see this problem? If it is your homework and you don't know the source, then would you tell me what book is used in your class? Thanks. $\endgroup$
    – ChoMedit
    Commented Apr 19, 2021 at 14:43

2 Answers 2


It's imporant o keep track of what is a vector, and and what are just numbers. The components of vectors, tensors etc are numbers, and the covariant derivative of a number-valued function is just the ordinary derivative. In particular the array of numbers ${\omega^a}_{b\mu}(x)$ are just number-valued functions, so $$ \nabla_\nu {\omega^a}_{b\mu} =\partial_\nu {\omega^a}_{b\mu}. $$ Let's use the definition $\nabla_\nu{{\bf e}_a} = {\bf e}_b {\omega^b}_{a\nu} $ together with Liebnitz' rule to work out $$ \nabla_\mu \nabla_\nu {{\bf e}_a} = \nabla_\mu ({\bf e}_b {\omega^b}_{a\nu})\\ = (\nabla_\mu {\bf e}_b) {\omega^b}_{a\nu}+ {\bf e}_b(\nabla_\mu {\omega^b}_{a\nu})\\ ={\bf e}_c {\omega^c}_{b\mu} {\omega^b}_{a\nu}+ {\bf e}_b\partial_\mu {\omega^b}_{a\nu} \\ ={\bf e}_c ({\omega^c}_{b\mu} {\omega^b}_{a\nu}+ \partial_\mu {\omega^c}_{a\nu}). $$ So $$ (\nabla_\mu \nabla_\nu {{\bf e}_a})^c = {\omega^c}_{b\mu} {\omega^b}_{a\nu}+ \partial_\mu {\omega^c}_{a\nu} $$ is the $c$-th compoent of $\nabla_\mu \nabla_\nu {{\bf e}_a}$.

Thus $$ [\nabla_\mu ,\nabla_\nu]{\bf e}_a = {\bf e}_c(\partial_\mu {\omega^c}_{a\nu}-\partial_\nu {\omega^c}_{a\mu}+ {\omega^c}_{b\mu} {\omega^b}_{a\nu}-{\omega^c}_{b\nu} {\omega^b}_{a\mu}). $$ or $$ ([\nabla_\mu ,\nabla_\nu]{\bf e}_a)^c = \partial_\mu {\omega^c}_{a\nu}-\partial_\nu {\omega^c}_{a\mu}+ {\omega^c}_{b\mu} {\omega^b}_{a\nu}-{\omega^c}_{b\nu} {\omega^b}_{a\mu}. $$ We can also write the components in the coordinate frame as ${\bf e}_a = {e_a}^\lambda {\boldsymbol \partial}_\lambda$ and then $$ ([\nabla_\mu ,\nabla_\nu]{\bf e}_a)^\lambda = (\partial_\mu {\omega^c}_{a\nu}-\partial_\nu {\omega^c}_{a\mu}+ {\omega^c}_{b\mu} {\omega^b}_{a\nu}-{\omega^c}_{b\nu} {\omega^b}_{a\mu}){e_c}^\lambda $$

It's a bad, but common, habit to write things like $\nabla_\mu X^\nu= \partial_\mu X^\nu+ X^\alpha {\Gamma^\nu}_{\alpha\mu}$ when you mean $$ (\nabla_\mu {\bf X})^\nu= \partial_\mu X^\nu+ X^\alpha {\Gamma^\nu}_{\alpha\mu}, \quad {\bf X}= X^\nu {\boldsymbol \partial}_\nu . $$

  • 3
    $\begingroup$ Just a note: what you call a bad habit is a common and well-founded convention known as the abstract index formalism formalized by Penrose and Rindler in 1984. $\endgroup$
    – Void
    Commented Apr 19, 2021 at 15:50
  • 4
    $\begingroup$ @Void: I agree that it is common and useful notation when you know what you are doing. I use it myself. It's just that there are endless questions here on Stack Exchange that show that it confuses beginners. Here is an example:physics.stackexchange.com/questions/628816/… $\endgroup$
    – mike stone
    Commented Apr 19, 2021 at 15:54
  • $\begingroup$ @Void: That's not the meaning of the abstract tensor formalism. It merely states that we can view the indexful tensor as a coordinate free tensor by considering the indices as labels for contractions. $\endgroup$ Commented Apr 19, 2021 at 18:43
  • $\begingroup$ @Void: What poster is pointing out is simply an ambiguity in the notation that he's being careful to keep clear. $\endgroup$ Commented Apr 19, 2021 at 18:45
  • $\begingroup$ @mikestone Thank you for providing clarity, I am most grateful. $\endgroup$ Commented Apr 21, 2021 at 7:03

Note that from the relation $e^\lambda{}_{a;\nu} = \omega_a{}^b{}_\nu e^\lambda{}_b$ you give you can deduce by contracting with $e_{\lambda c}$ $$e^\lambda{}_{a;\nu}e_{\lambda c} = \omega_{ac\nu}$$ Note, however, that I am using the definition of the covariant derivative that takes tetrad indices $a,b,c$ as mere labels and thus the covariant derivative of the tetrad leg vector field is $$e^\lambda{}_{a;\nu} = e^\lambda{}_{a,\nu} + \Gamma^\lambda{}_{\kappa\nu} e^\kappa{}_{a}\,.$$ That is, the covariant derivative is covariant with respect to coordinate transforms but not wrt vielbein transforms.

In other words, $\omega_{ab\nu}$ transforms as a tensor in the $\nu$ index and as such it will have the corresponding Christoffel connection term in the covariant derivative. This should help you derive the commutator (relation between the Riemann tensor and $\omega$) as desired.

  • $\begingroup$ So how does the action of the covariant derivative on $\omega$ look like? $\endgroup$ Commented Apr 19, 2021 at 10:59
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    $\begingroup$ @user44690 this answer is wrong or misleading. The spin connection is NOT a tensor. Covariant derivatives are defined only for tensors. $\endgroup$
    – Prahar
    Commented Apr 19, 2021 at 11:48
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    $\begingroup$ @Void - Yes, it is incorrect. The correct relation is $\nabla_\mu e^\lambda{}_a = 0$. The index $a$ is tensorial especially in the context being used in the question by OP. The covariant derivative $\nabla_\mu$ of course contains the Christoffel connection as well as the spin connection. $\endgroup$
    – Prahar
    Commented Apr 19, 2021 at 13:21
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    $\begingroup$ @Void - You may wish to define the covariant derivative to only act on spacetime indices $\mu,\nu,\dots$ but that is NOT the standard definition. If you want to use a non-standard definition, you'd have to make that clear in the beginning. If we do use your definition, then I actually agree with everything you have said. $\endgroup$
    – Prahar
    Commented Apr 19, 2021 at 13:23
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    $\begingroup$ I don't see how this is a theory dependent statement. The standard definition of the covariant derivative is $\nabla_\mu e_\nu{}^a \equiv \partial_\mu e_\nu{}^a - \Gamma^\lambda_{\mu\nu} e_\lambda{}^a + \omega_\mu{}^a{}_b e_\nu{}^b$ (with similar generalizations to other tensors). You may work with a different definition and that is perfectly OK! However, since it is NOT the standard definition, it should be mentioned beforehand. That's all I am saying. $\endgroup$
    – Prahar
    Commented Apr 19, 2021 at 13:31

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