Why is the Riemann curvature tensor not zero?

Question

The Riemann curvature tensor for a torsion-free connection is given by:

$$R^d_{cab}V^c=(\nabla_a\nabla_b-\nabla_b\nabla_a)V^d$$

Where $\nabla_a$ and $\nabla_b$ are the covariant derivatives in the $a$ and $b$ directions. But a vector, $V^d$, parallel transported in a direction $a$ has $\nabla_aV^d=0$.

How can the Riemann curvature, which depends on the covariant derivative, measure the effect of parallel transporting a vector around a loop if the covariant derivative of a parallel transported vector is zero?

Edit: changed "vector field" to "vector"

Answer 1

I cannot see the problem. Fix $p\in M$, then $$(R^d_{cab}V^c)_p=((\nabla_a\nabla_b-\nabla_b\nabla_a)V^d)_p$$ implies that $(R^d_{cab}V)_p=0$ if the vector field $V$, defined in a neighborhood $U_p$ of $p$, is such that it is parallely transported along each coordinate curve $x^a$ -- that is $x^k= const$ for $k\neq a$ -- and each coordinste curve $x^b$, defined in $U_p$.

Indeed, in that case e.g. $\nabla_aV^d=0$ holds true in a neighborhood of $p$ so that also the second derivatives in $U_p$, especially $(\nabla_e \nabla_aV^d)_p$, vanish.

It is however by no means obvious that such vector field $V$ exists. That is because it has to satisfy the parallel transport equation along each coordinate curve $x^a$ in $U_p$ with the said fixed a (and b), not only the one exiting $p$!

This is the reason why your argument fails.

To conclude that $(R^d_{cab})_p=0$ (for every value of $a,b,c,d$) along this reasoning, one should prove that there are $n= dim(M)$ vector fields in a neighborhood $U_p$ of $p$ that define a basis of $T_pM$ and such that they are parallely transported along each coordinate curve in $U_p$. Notice that, as a consequence of the parallel transport, these vector fields form a basis at each point of $U_p$.

The point is that the existence of this basis is a quite difficult issue.

Actually, all that is a very known problem and the existence of those $n$ vector fields is not only sufficient for but even equivalent to $Riemann=0$ in a neighborhood of $p$ through the Frobenius theorem.

You can find this statements and its proof in several books (also in Sect. 9.2 Thm 9.18 of these lecture notes of mine which are still work in progress and thus affected by a number of typos).

The issue of the parallel transport along a loop is disentangled with the discussion above where one considers vector fields defined in a full neighbourhood of $p$ and not only on the loop.

In the loop argument one fixes a vector exactly at $p$ and next he/she transports that vector along the loop through $p$. There is no globally defined field in a neighbourhood of $p$ here.

Any attempt to extend the field from the loop to a neighbourhood $U_p$ of $p$ ( which is however impossible as we have two values of $V$ at $p$ in general) does not guarantee that $\nabla_aV^d=0$ in $U_p$ as instead requested above.

Answer 2

I think the OP is correct to call $V^d$ as a vector field rather than a vector. Say we define covariant derivative

$$ \nabla_X Y$$

Then, while evaluating the covariant derivative at a point $p$, it is fine for $X$ to be a vector at $p$, but we would need to know the behavior of $Y$ in the neighborhood of $p$ to perform the derivative operation. Hence, $Y$ needs to be a field (e.g., see this, where $Y$ is required to be a section of the tangent bundle).

However, I think $V^d$ (in the question) is not the vector that is being parallel transported as such $\nabla_a V^d \neq 0$, and therefore, the Riemann curvature is not zero as well.

Edit

I'm going to refer you to Wald (GR) chapter 3, Pg 37, for more info. But let me explain my above comment in more detail.

To quote Wald, "we first define the Riemann tensor in terms of the failure of successive operations of differentiation to commute when applied to a dual vector field", i.e., we define a tensor at a point $p$ as

$$ \nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c$$

As you can see, this is a (0,3) tensor. Moreover, $$ \nabla_a \nabla_b - \nabla_b \nabla_a$$ can be seen as a tensor that feeds on $\omega_c$ [ a (0,1) tensor] and returns a (0,3) tensor $\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c$.

Now, what tensor can consume a (0,1) tensor (i.e., a dual vector) and return a (0,3) tensor? The answer is that it is a (1,3) tensor, ${R_{abc}}^d$. In more detail $$ {R_{abc}}^d: \,\, {\rm dual \, vectors} \rightarrow (0,3) \,\, {\rm tensors}$$ $$ {R_{abc}}^d: \omega_d \rightarrow \nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c = {R_{abc}}^d \omega_d$$

So far, note that we have not yet talked about parallel transport at all.

Subsequently, Wald shows that ${R_{abc}}^d$ is directly related to the path-dependent nature of parallel transport, specifically, the failure of a vector to return to its original value when parallel transported around a small closed loop around $p$."

One can reduce ${R_{abc}}^d \omega_d$ to your form as ${R_{abc}}^d g^{ae} \omega_e = {R_{abc}}^d V^a$ but again $V^a$ is not parallel transported. It is some other vector that is transported along the loop while ${R_{abc}}^d \omega_d$ is evaluated at point $p$ only. It happens to be that if ${R_{abc}}^d = 0$, then the operation of parallel transport does not change the vector once it returns back to point $p$.