I'm not very familiar with differential geometry and am coming from a general relativity background, so would appreciate help with a question from that context. If this question could be posed in a more abstract sense, that would also be of tremendous help!

Suppose I have a curved space-time with metric $g_{\mu \nu}$ and with the connection coefficients $\Gamma^\alpha_{\mu \nu}$. Given some vector $v$, the expression for parallel transport of this vector along a curve $x^\mu(\sigma)$ is

$$ \frac{d v^\beta}{d\sigma} + \Gamma^\beta_{\mu \nu} v^\mu \frac{d x^\nu}{d \sigma} = 0. $$

Now, I want to consider parallel transport of this vector around an infinitesimal closed parallelogram spanned by the vectors $a$ and $b$ and study the change in $v$ upon performing this procedure. Suppose we start with vector $v$ at point $P$, specified by the initial condition

$$ v_P^\beta \equiv v^\beta(\sigma_P). $$

Then, the change in $v$ upon parallel transport along the closed infinitesimal loop is given by

$$ \Delta v = v^\beta_{|| P} - v_P^\beta = -\left(R^\beta_{\lambda \nu \mu} \right)_P v_P^\lambda a^\nu b^\mu $$

where the Riemann curvature tensor is $$ R^\beta_{\lambda \nu \mu} = \frac{d \Gamma^\beta_{\lambda \mu}}{d x^\nu} - \frac{d \Gamma^\beta_{\lambda \nu}}{d x^\mu} + \Gamma^\beta_{\alpha \nu} \Gamma^\alpha_{\lambda \mu} - \Gamma^\beta_{\alpha \mu} \Gamma^\alpha_{\lambda \nu} $$

My question is the following: suppose that the vector $v_P$ is orthogonal to the infinitesimal loop originally i.e., $$ v_P \cdot a =0,\quad v_P \cdot b = 0. $$ Then, is there a general class of metrics for which $\Delta v = 0$?

On playing around with this problem, I found that if the Riemann curvature tensor is maximally symmetric, then this is certainly true but I'm curious if there's a broader class of metrics for which this is true. For simplicity, we can even consider working in just 3 spatial dimensions and not worry about 3+1 space time dimensions to begin with. I'd love to know if there's some general statements made about when parallel transport of a vector along a loop orthogonal to the vector's initial orientation leaves the vector invariant.


In dimension $3$, the curvature tensor can always be written with respect to an orthonormal frame as something like $$ R_{ijkl} = R_{ik}\delta_{jl} + R_{jl}\delta_{ik} - R_{il}\delta_{jk} - R_{jk}\delta_{il} - \frac{1}{2}S(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) $$ If you let $v$, $a$, $b$ be the first 3 vectors of an orthonormal basis, then what you want is $$ R_{123\beta} = 0 $$ for $\beta = 1, 2$. It therefore suffices to consider $$ 0 = R_{1232} = R_{13} - \frac{1}{2}S\delta_{13}. $$ If you want this to hold for any orthonormal triplet, then this implies that all of the off-diagonal components of the Ricci tensor vanish with respect to any orthonormal frame and therefore the metric is Einstein. In dimension 3, this implies constant sectional curvature, so there are no new examples besides the ones you already found.

I believe that the same is true in higher dimensions using a similar argument. First, the argument above implies that $$ R_{ijkl} = 0 $$ for any 4 distinct indices $i,j,k,l$. This implies that the Weyl tensor vanishes. The curvature can then be written in the same form as above, and a similar argument shows that the Ricci is a scalar multiple of the metric. This also implies constant sectional curvature.

  • $\begingroup$ Thank you for that answer! Do you know of a good reference which shows the decomposition of the Reimann tensor as you've written it? I tried to show it using just the symmetries of the tensor but failed. $\endgroup$ – Aegon Feb 4 '18 at 21:30
  • $\begingroup$ Here's an explanation: en.wikipedia.org/wiki/Ricci_decomposition $\endgroup$ – Deane Yang Feb 4 '18 at 21:55
  • $\begingroup$ It's also explained here: projecteuclid.org/euclid.dmj/1077303336 $\endgroup$ – Deane Yang Feb 4 '18 at 21:56
  • $\begingroup$ Thank you again, that helps! I think I understand from your calculation that the same should hold true even for curved space-time, although the example was done for d=3 curved space. Is this correct? Thank you again, I appreciate your help with this. $\endgroup$ – Aegon Feb 5 '18 at 6:34
  • $\begingroup$ If the Ricci is a scalar multiple of the metric, does this imply that we also need torsion-free space? Since torsion is responsible for the anti-symmetric part of the connection, it would seem to me that is an additional requirement. Just confused about how torsion enters your calculation and if it would change anything in particular. $\endgroup$ – Aegon Feb 10 '18 at 21:55

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