In Einstein's Theory, A Rigorous Introduction for the Mathematically Untrained, by Grøn and Næss:

The change of the covariant components a vector by parallel transport around an indefinitely small closed curve enclosing a surface with area

$$dS^{\alpha\beta}$$ may now be written

(Equation 9.30) $${\Delta}A_{\mu}=\frac{1}{2}R^{\nu}_{\mu\alpha\beta}A_{\nu}\Delta{S}^{\alpha\beta}$$



is the Riemann Curvature Tensor, defined as:

(Equation 9.29) $$R^{\nu}_{\mu\alpha\beta}=\Gamma^{\tau}_{\mu\beta}\Gamma^{\nu}_{\tau\alpha}-\Gamma^{\tau}_{\mu\alpha}\Gamma^{\nu}_{\tau\beta}+\Gamma^{\nu}_{\mu\beta,\alpha}-\Gamma^{\nu}_{\mu\alpha,\beta}$$

But Equation 9.28 says


So where did the factor of one half come from?

  • $\begingroup$ Maybe I'm missing something, but it just looks like a mistake to me. This looks like what I would take as the definition of the Riemann tensor, which doesn't have the 1/2. $\endgroup$
    – user4552
    Jan 20, 2019 at 0:26
  • $\begingroup$ I don't think Equation 9.30 or 9.29 are mistaken. I think Eq. 9.28 might be the culprit. See the second part of here: scipp.ucsc.edu/~haber/ph171/parallel_transport15.pdf $\endgroup$
    – Romi
    Jan 20, 2019 at 0:47
  • $\begingroup$ Check how the infinitesimal surface area $dS^{\alpha\beta}$ is defined. Maybe for some reason it embeds a factor $2$. $\endgroup$ Jan 20, 2019 at 9:49

1 Answer 1


The explanation of the apparent contradiction is that the infinitesimal surface area embeds a factor $2$. In fact in your link it is stated:
$d\sigma^{\nu \alpha} = (a^\nu b^\alpha - a^\alpha b^\nu) d\xi d\eta$
$d\sigma^{\nu \alpha}$ infinitesimal surface area
$(\xi, \eta)$ parameterization coordinates of the surface

The expression for the infinitesimal change of a vector component vs. the infinitesimal surface area is:
$dv^\beta = \frac{1}{2} R^\beta_{\lambda \nu \alpha} v^\lambda (a^\nu b^\alpha - a^\alpha b^\nu)$
Being the Riemann tensor antisymmetric in the last two indices, after some relabeling of the indices we may write :
$dv^\beta = R^\beta_{\lambda \nu \alpha} v^\lambda a^\nu b^\alpha$

As you read, the standard expression of the infinitesimal change of a vector component vs. the sides of an infinitesimal parallelogram does not show any a half factor.

  • $\begingroup$ It's strange, because the "final" version of the equation in the book is the one with the factor of one half being shown. It doesn't make use of the antisymmetry to show how the 1/2 is eliminated nor does it show how the infinitesimal area is a parallelogram. Oh well. Thanks so much for your help! $\endgroup$
    – Romi
    Jan 20, 2019 at 21:25

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