Riemann tensor and covariant derivative in Sean Carroll's book

Question

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?

Answer 1

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.

Answer 2

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