# Which tensor describes curvature in 4D spacetime?

I heard these two statements which don't work together (in my mind):

1. In 4D spacetime the curvature is encoded within the Riemann tensor. He holds all the information about curvature in spacetime.

2. The metric describes the intrinsic geometry of a manifold/spacetime, including the curvature.

So, who encodes curvature in 4D spacetime?

I know that Riemann depends on the metric. The Riemann tensor is of rank 4. The metric is only a rank 2 tensor.

Edit: I just looked through my (old) lecture notes and found an explanation I was looking for. "The curvature tensor is a diagnostic tool which tells you if a given metric can be turned into the identity (or the Minkowski metric) for all points in spacetime with some transformation."

The metric can appear do describe some curved thing, but this may be due to the fact that the coordinates were choosen poorly. Given some metric it may be hard to find out if the described space is really curved or just expressed in bad coordinates. The Riemann tensor seems to allow us to figure out quickly if a given space is flat or curved.

• Something to ponder: which encodes the slope of a function, $f(x)$ or $f'(x)$? – Ryan Unger Dec 4 '15 at 13:23
• Or: which encodes the convexity of a function, $f(\vec x)$ or $\operatorname{Hess}f(\vec{x})$? – Ryan Unger Dec 4 '15 at 14:00

In general, it is the Riemann tensor that encodes curvature, not the metric. Although it is quite difficult to see why Riemann tensor describes curvature directly from its definition, due to its abstractness, it is fairly easy to see it geometrically from the equivalent notion of sectional curvature (https://en.wikipedia.org/wiki/Sectional_curvature).

Fortunately, in theories with Levi-Civita connection (torsionless and metric compatible), like General Relativity, the Christoffel symbols are given in terms of the metric (and its derivatives of course) and, in turn, the Riemann tensor is given as a function of the metric. Only in this case that Riemann tensor is a function of the metric.

You can derive the Riemann tensor from the metric tensor and vice versa so knowing either one is sufficient.

The Riemann tensor has a lot of symmetries that restrict its form. Not every possible rank 4 tensor is a valid Riemann tensor. That's why the two can be derived from each other even though the metric tensor appears to contain a lot less information.

• But Riemann still has twice as much components (information) that the metric. One must be more complete than the other. – Thomas Elliot Dec 4 '15 at 15:26
• @ThomasElliot The Riemann tensor is a function of the metric and its derivatives. You can solve the equation $R_{ijkl}=F(g,\partial g,\partial^2g)$ to get the metric back, in theory. – Ryan Unger Dec 4 '15 at 16:52
• @ThomasElliot Consider a function $f:\mathbb{R}^2\to\mathbb{R}$. In coordinates, this function has only one component. But its gradient $(f_x,f_y)$ has two components. Is the gradient "more complete"? – Ryan Unger Dec 4 '15 at 16:56
• @ThomasElliot: the Riemann tensor is obtained by differentiating the metric twice. The 20 independant components are basically the 20 degrees of freedom in the second derivatives of the metric (that can't be set to zero by fiddling with the coordinates). – John Rennie Dec 4 '15 at 17:01

The Riemann tensor can be related to the metric through the formula $$R_{\lambda\mu\nu\chi} = \frac{1}{2} \left( \frac{\partial^2 g_{\lambda\nu}}{\partial x^\chi \partial x^\mu} - \frac{\partial^2 g_{\mu\nu}}{\partial x^\chi \partial x^\lambda} - \frac{\partial^2 g_{\lambda\chi}}{\partial x^\nu \partial x^\mu} + \frac{\partial^2 g_{\mu\chi}}{\partial x^\nu \partial x^\lambda}\right) + g_{\eta\sigma} \left( \Gamma^\eta_{\nu\lambda} \Gamma^\sigma_{\mu\chi} - \Gamma^\eta_{\chi\lambda} \Gamma^\sigma_{\mu\nu}\right),$$ where $\Gamma_{\alpha\beta}^\gamma$ are the Christoffel symbols defined by $$\Gamma_{\mu\beta}^\chi = \frac{1}{2} g^{\chi\alpha} \left( \frac{\partial g_{\alpha\beta}}{\partial x^\mu} + \frac{\partial g_{\alpha \mu}}{\partial x^\beta} - \frac{\partial g_{\beta \mu}}{\partial x^\alpha} \right).$$ As you can see, the Riemann tensor is a rank 4 tensor completely defined by the metric $g_{\mu\nu}$ (a rank 2 tensor). Essentially, the four indices come from the fact that space-time is a four-dimensional vector space and that $\mu,\nu,\lambda,\chi$ take the values $1,2,3,4$.

After that, the two statement you posted before are perfectly coherent since $R_{\lambda\mu\nu\chi}$ is completely defined by the metric tensor and vice-versa. Indeed, the curvature of the space-time is encoded by the tensor solutions of Einstein's equation: $$R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R - \lambda g_{\mu\nu} = 8\pi G T_{\mu\nu}.$$ Here, $T_{\mu\nu}$ represents the energy-momentum tensor while $R_{\mu\nu},R$ are the Ricci tensor and Ricci scalar (two quantities directly related to $R_{\lambda\mu\nu\chi}$) . In simple words, $T_{\mu\nu}$ depends exclusively by the energy-momentum distribution inside space-time and, since mass can be traduced to energy from $E = mc^2$, we can interpret energy and momentum as a source of gravitational field.

All of the information you need to know about the intrinsic curvature of a manifold is stored in the metric, $g_{\mu\nu}$. While the Riemann tensor $R^\lambda_{\mu\sigma \nu}$ is the tensor which describes the intrinsic curvature, it is computed from $g_{\mu\nu}$ and its derivatives, so all the information about the curvature comes from $g_{\mu\nu}$, but you only actually see it, so to speak, when you compute the Riemann tensor.

Now, one should distinguish intrinsic curvature, described by the Riemann tensor and quantities computed from it, to the extrinsic curvature, which is due to the embedding of a manifold within another. In the case of the extrinsic curvature of a manifold, $K_{\mu\nu}$, it is not enough to know the metric of the manifold, you must also know the metric of the manifold it is embedded in.

Finally, note that if I give you some Riemann tensor $R^\lambda_{\mu\sigma\nu}$ which I have computed from some metric unbeknownst to you, then to determine the metric I would need to supplement the curvature by some initial conditions, as one would expect for any differential equation.

An example: For a Khaler manifold, there is something even simpler which holds just as much information as the metric: the Khaler potential. This is because for a complex Khaler manifold,

$$g_{ij} \sim \frac{\partial^2}{\partial z^i \partial \bar{z}^j} K(z,\bar{z}).$$

So, any tensor computed from $g_{ij}$ derives all its 'information' from the single scalar function $K(z,\bar z)$. In the case of the Ricci curvature,

$$R_{ij} \sim \frac{\partial^2}{\partial z^i \partial \bar{z}^j} \log (\mathrm{det} \, g).$$

Thus, because of the relatively simple form of the curvature tensor for a Khaler manifold, it is even more plain that there is no new information, but rather the metric determines the entire curvature.