The answer to your question is affirmative in the following sense:

In the Riemann normal coordinates at $p$ the coefficients of the Taylor expansion of the metric $g_{ij}(x)$ are polynomials in the Riemann tensor at $p$ and its covariant derivatives at $p$. [Assuming the proof in this random thing I googled^{[a]} is correct, starting at (5.1)].

I think this is the correct formalization of your conjecture in the sense that if we are making a tensor out of $g$ the only thing we can use are $g$ and its expansion in normal coordinates. I'll maybe try to write out why I think this is case.

BTW, the local condition is very necessary, since otherwise we could define things like the length of the shortest loop containing $p$ that is in a certain homotopy class, which clearly "depends only on the metric" but is not made out of polynomials the curvature.

## Added after this answer was accepted

For those interested, I asked a question on Math SE that contains what I believe to be the correct formalization of the question: "What tensors can I produce from the metric tensor?" “Natural” constructions of tensor fields from tensor fields on a manifold

[a]: Guarrera, D.T., Johnson, N.G., Wolfe, H.F. (2002) *The Taylor Expansion of a Riemannian Metric*

$\mbox{ } $ $\mbox{ } $ $\mbox{ } $ $\mbox{ } $ $\mbox{ } $ $\mbox{ } $ $\mbox{ } $ http://www.rose-hulman.edu/mathjournal/archives/2002/vol3-n2/Wolfe/Rmn_Metric.pdf

If you choose a local inertial frame at a specific point of space-time, the metric tensor, around this point, is :

$g_{ij}= \delta_{ij} - \frac{1}{3} R_{ikjl} x^k x^l + O(x^3) \tag{1}$

And the space-time volume element (corresponding to the square root of the determinant of the metric) is :

$ d\mu_g = (1 - \frac{1}{6} R_{jk} x^j x^k + O(x^3)) ~d\mu_{Euclidean}\tag{2}$

The fact that the contraction of the Weyl tensor is zero, that is $C_{jk}=0$, looking at equation $(2)$, means that the $C_{jk}$ part of $R_{jk}$ is zero, so the Weyl tensor does not contribute to modifications of (infinitesimal) space-time volume. However, equation $(1)$ indicates you, that the Weyl tensor is participating to the modification of the metrics, because the $C_{ikjl}x^kx^l$ part of $R_{ikjl} x^k x^l$ is not zero.

So, finally, the Weyl tensor participates to a modification of the metrics, but without participating to the modification of a (infinitesimal) space-time volume, so the Weyl tensor is associated to modifications of the **shape** of(infinitesimal) space-time volume, but without modification of volume.

Typically, this involves tidal forces, gravitational waves. For instance, for a (basic) gravitational wave (here we suppose that the Ricci tensor is zero, and the Riemann tensor equals to the Weyl tensor) propagating along the $z$ axis, considering an infinitesimal space-time volume, the physical $\delta x$ could increase, and the physical $\delta y$ could decrease, so the shape is changing, however one variation is compensating the other, so that the infinitesimal space-time does not change. .

## Best Answer

I've always liked John Baez' coffee-grounds visualization of the Ricci and Weyl tensors.

Basically you look at a small ball of free falling coffee grounds and use them to test how the geodesics are moving with respect to each other. Ricci curvature measures the change in volume of this ball. Weyl curvature measures the stretching and squashing.