# Covariant derivative of the spin connection

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?

• The spin connection is a not a tensor. You cannot define its covariant derivative. Commented Apr 19, 2021 at 8:33
• @Prahar Mitra: That can't be right: all connections are equivalently covariant derivatives. Commented Apr 19, 2021 at 9:03
• @MoziburUllah connections define a covariant derivative for tensors (or tensor densities). You cannot take covariant derivative of the connection itself. Commented Apr 19, 2021 at 9:04
• @Prahar Mitea: Ok, I thought he was asking for the spin connection to be expressed as a covariant derivative. Commented Apr 19, 2021 at 9:15
• 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. Commented Apr 19, 2021 at 14:43

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 $$(\nabla_\mu {\bf X})^\nu= \partial_\mu X^\nu+ X^\alpha {\Gamma^\nu}_{\alpha\mu}, \quad {\bf X}= X^\nu {\boldsymbol \partial}_\nu .$$

• 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.
– Void
Commented Apr 19, 2021 at 15:50
• @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/… Commented Apr 19, 2021 at 15:54
• @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. Commented Apr 19, 2021 at 18:43
• @Void: What poster is pointing out is simply an ambiguity in the notation that he's being careful to keep clear. Commented Apr 19, 2021 at 18:45
• @mikestone Thank you for providing clarity, I am most grateful. 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.

• So how does the action of the covariant derivative on $\omega$ look like? Commented Apr 19, 2021 at 10:59
• @user44690 this answer is wrong or misleading. The spin connection is NOT a tensor. Covariant derivatives are defined only for tensors. Commented Apr 19, 2021 at 11:48
• @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. Commented Apr 19, 2021 at 13:21
• @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. Commented Apr 19, 2021 at 13:23
• 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. Commented Apr 19, 2021 at 13:31