# Riemann curvature tensor and parallel transport

I'm taking a course on general relativity and I'm confused about something in my course notes. First there is an explanation about a curved 2D plane where it is shown how curvature can be defined by using parallel transport along a closed circuit. This yields one scalar value that characterizes the curvature at a certain point.

Then, curvature in 4D spacetime is discussed and there is a sentence that reads (literally translated): "The difference with the Gaussian curvature of a 2D plane is that in 4 dimensions, we have 6 planes in which we can evaluate the convergence or divergence of geodesics or consider the parallel transport of vectors along a closed circuit. This gives rise to the rank 4 Riemann curvature tensor with 20 independent components."

The approach of introducing concepts from differential geometry intuitively by considering 2D spaces and then extending the concepts to 4D spacetime has been fairly helpful to me, but here I'm lost. I feel like there is a little more explanation necessary for the above sentence to make sense.

## 1 Answer

The six coordinate planes in 4 dimensions are $$xy$$, $$xz$$, $$yz$$, $$tx$$, $$ty$$, and $$tz$$. In $$n$$ dimensions there are

$$\binom n2=\frac{n(n-1)}{2}$$

such planes. You can parallel-transport a vector around a loop in any of these planes, and in general it will come back different from what it was before.

The fact that there are 20, not 256, independent components of the Riemann tensor in 4 dimensions is not obvious. It follows from that tensor’s multiple symmetries.

• Why 20 independent components of the Riemann tensor? The two pairs of indices are antisymmetric hence $6*6$ variables (cf. 6 planes). And Bianchi identity tells you that there are 4*4 relations between them. Therefore $6*6-4*4 = 36-16=20$. – MadMax Dec 11 '19 at 15:59
• The same excise for QED gives you $6*1 - 4*1 = 2$. – MadMax Dec 11 '19 at 16:07