What is Dirac's reasoning when saying parallel displacement creates vector field with vanishing covariant derivative?

Question

Section 12 of Dirac's book "General Theory of Relativity" is called "The condition for flat space", and he is proving that a space is flat if and only if the curvature tensor $R_{\mu\nu\rho\sigma}$ vanishes; he has just defined the curvature tensor (in the previous section) as the difference between taking the covariant derivative twice in two different orders.

He first shows that if space is flat then the metric tensor is constant, hence the curvature tensor vanishes. All fine.

For proving the other direction he first assumes the curvature tensor vanishes, then he states that in that case we can shift a "vector" $A_\mu$ by parallel displacement from point $x$ either to the point $x+dx+{\delta}x$ or to the point $x+{\delta}x+dx$ (ie two different paths) and get the same result -- so he says "we can shift the vector to a distant point and the result we get is independent of the path to the distant point". My first question, then, is: where does he get the path-independence from? In other words, why does the vanishing of the curvature tensor imply the path-independence of parallel displacement? I cannot see how it follows from what precedes it in this book.

He then says "Therefore, if we shift the original vector $A_\mu$ at $x$ to all points by parallel displacement, we get a vector field that satisfies $A_{\mu;\nu}=0$ " -- ie the covariant derivative of the parallel displaced vector is zero everywhere. My second question is: where does this conclusion come from? How does it follow from what he has just said?

I know that Dirac is famed for being terse and for assuming that his readers are able to make several jumps of logic all at once, so I cannot help but think that I am missing something obvious here in these two places. Please note that I am specifically looking for answers that fill in gaps in Dirac's reasoning.

I know that I am restricting my audience here to just those who gave access to Dirac's book, and I apologise for that.

Answer 1

The answer depends on how Dirac chooses to define curvature tensor. If he takes it by definition as that tensor which gives $$ R^a_{\lambda c d} v^\lambda = \nabla_c \nabla_d v^a - \nabla_d \nabla_c v^a \tag{1} $$ then you should note (i) this ignores torsion (which is ok for general relativity, but best to make it explicit), and (ii) one also needs a definition of covariant derivative $\nabla_c$. So the rest of the argument will depend on how he defined $\nabla_c$. We can postpone that definition and announce that the term 'parallel transport' refers to that way of defining a vector field $w^a$ which results in $$ \frac{d x^\lambda}{du} \nabla_\lambda w^a = 0 \tag{2} $$ where $u$ is an affine parameter along a curve $x^a(u)$. The idea is that we can use this idea to figure out that the operator $\nabla_a$ must respect $\nabla_\lambda g_{\mu \nu} = 0$ (zero covariant derivative of metric tensor), and this, together with Leibniz rule (product rule for differentiation) is sufficient to define $\nabla_a$. This is not the only way to set the theory up but I vaguely recall this is the way Dirac chose to do it.

So if $R^a_{bcd}$ is defined as (1), and parallel transport is defined as (2) (or has been shown to satisfy (2)) then the first statement in the argument in the question, that shift from $x$ to $x + dx + \delta x$ gives same outcome as $x + \delta x + dx$ when $R^a_{bcd}=0$, follows to first order in $dx$ and $\delta x$. By combining lots of such small shifts we then get the path-independence (but I would want to write down integrals and carefully check to be sure).

The previous paragraph concerned parallel transport, so for each path the vector in question (I am calling it $w^a$) went through a parallel transport and therefore satisfied (2). But the argument found that no matter what path was adopted, the same vector field results. It follows that our vector field must satisfy (2) for any path $x^\lambda(u)$. This implies that $\nabla_\lambda w^a = 0$. QED

Answer 2

I will try an intuitive approach valid for 2D.

If we have a perfect flat floor, we can fill it with square tiles. We start from a point in the bottom left and go to the bottom right, each tile perfectly parallel to its neighbour. From the bottom right we go up until the top right, also with each tile perfectly parallel to its neighbour. It is equivalent to parallel transport a vector.

The Riemann tensor be zero means that if we go first from the bottom left to the top left, and from there to the top right, the last tile is already there, from the first filling. (Probably it is necessary to cut some tiles, but only to change from square to rectangle. All the tile angles keep $90^{\circ}$). It is easy to see that all the floor can be filled that way.

On the other hand, if we can do that, the floor is flat by definition. One way to understand that is to observe that from a point out of a row (or column) there is a define row (or column) perfectly parallel to the first.

If the floor is on Earth and very big, limited by 2 Earth meridians, and 2 parallels (similar to the US state of Kansas for example), that is not possible. The tiles must be cut to angles different of $90^{\circ}$ to fill the space. The Riemann tensor is not zero and the surface is not flat.