In Spacetime and Geometry, Sean Carroll defines the Riemann tensor in terms of the commutator of covariant derivatives:

$$R^\rho_{\sigma\mu\nu}V^\sigma = [\nabla_\mu, \nabla_\nu]V^\rho + T^\lambda_{\mu\nu}\nabla_\lambda V^\rho$$

At first I interpret this (the commutator part) as measuring how different the tensor would be after being parallel transported along two different paths ($\mu$ to $\nu$ vs $\nu$ to $\mu$) and this is the diagram he has in the book. However, he also says, on page 122,

the covariant derivative of a tensor in a certain direction measures how much the tensor changes relative to what it would have been if it were parallel transported.

This implies to me that we are moving the tensor through some means other than parallel transport. He also earlier says,

the covariant derivative quantifies the instantaneous rate of change of the tensor field in comparison to what the tensor would be if it were 'parallel transported.'

My questions are then,

  • Does the covariant derivative measure how a tensor field changes at different locations, or how a tensor changes when it is moved?
  • If the latter, how does the definition of the Riemann tensor still make sense?

As a note, is there a better way to format the indices so they have the proper placement?

  • $\begingroup$ I think the confusion is that you're mixing up tensors with tensor fields. When Carroll says "how much the tensor changes relative to parallel transport", he's talking about evaluating a tensor field at a different location. Nothing about the field is "moved". $\endgroup$
    – knzhou
    Jul 22 '16 at 18:32
  • $\begingroup$ I don't think so. In this context he only refers to a single vector and the act of moving it in an infinitesimal loop. There is no reference to a field. $\endgroup$ Jul 22 '16 at 18:35
  • $\begingroup$ You're right, this is a bit more subtle than I thought. I'll write a proper answer. $\endgroup$
    – knzhou
    Jul 22 '16 at 18:42
  • $\begingroup$ By the way, you can do indices like this: $R^{\mu}_{\; \nu}$. Right click on the math to see the LaTeX. Outside of this forum, use the tensor package. $\endgroup$
    – knzhou
    Jul 22 '16 at 18:55
  • $\begingroup$ Thanks! $R^{\mu}_{\; \nu}$ I'll make sure to do that in the future $\endgroup$ Jul 22 '16 at 19:04

The quote you give from Carroll about the covariant derivative is right: it quantifies the rate of change of a tensor field relative to parallel transport. The covariant derivative of a tensor at a point doesn't make sense. However, the commutator of covariant derivatives acting on a point does.

The situation is analogous to the vector field commutator. Earlier in Carroll, you read that given two vector fields $X$ and $Y$, the composition $XY$ is not a vector field, but the combination $$[X, Y] = XY-YX$$ is, because the nontensorial parts cancel out.

Now, heuristically, two covariant derivatives acting on a vector field gives $$\nabla_\mu \nabla_\nu V^\rho(x) = V^\rho(x + \epsilon(\hat{\mu} + \hat{\nu})) - \text{parallel transport of } V^\rho(x) \text{ along } \hat{\nu} \text{ first}$$ where I'm being a bit sloppy with $\epsilon$'s. This doesn't make sense if $V^\rho$ is a single vector, but compare this to the opposite ordering of covariant derivatives, $$\nabla_\nu \nabla_\mu V^\rho(x) = V^\rho(x + \epsilon(\hat{\mu} + \hat{\nu})) - \text{parallel transport of } V^\rho(x) \text{ along } \hat{\mu} \text{ first}.$$ If we subtract these two expressions, the dependence on $V^\rho(x + \epsilon(\hat{\mu} + \hat{\nu}))$ cancels out, giving a result that only depends on the single vector $V^\rho(x)$.

You can see this happening in Carroll too. In his computation of the components of the Riemann tensor in Eq. (3.111), the derivatives acting on $V^\rho$ itself drop out.


Consider a region in the manifold where a tensor field is everywhere well defined. Consider a point $x$ and a neighbouring point $x+dx$. The tensor field, say $V^\mu$ is given at both points as $V^\mu(x)$ and $V^\mu(x+dx)$. We can parallel transport the tensor(vector in this case) from $x$ to $x+dx$. This parallel transported vector $V_P^\mu(x+dx)$ is in general not the same as the value of the vector field at $x+dx$. Covariant derivative measures the difference between the two(I.e. $V_P^\mu(x+dx)$ and $V^\mu(x+dx)$) in the limit of $dx$ going to zero

  • $\begingroup$ So in this example, even though there's no reference to a field, there's a field implied? $\endgroup$ Jul 22 '16 at 18:37
  • $\begingroup$ Covariant derivative is only defined for a Tensor field. There is no way to define it for a single tensor. $\endgroup$ Jul 22 '16 at 18:39
  • $\begingroup$ One way to see it is that covariant derivative is a generalization of the partial derivative and that already involves a tensor field instead of a tensor $\endgroup$ Jul 22 '16 at 18:44
  • $\begingroup$ Thanks. I'd upvote your answer but I don't have the rep. Oh well. $\endgroup$ Jul 22 '16 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.