I'm not very familiar with exterior derivatives. I've some trouble following argument (which is a part of a proof that if the Riemann tensor vanishes, $R^{\,\rho}_{\;\,\sigma \mu \nu}=0$, iff there exists a coordinate system where the components of the metric are constant.)

One finally get to the equation: $$\nabla_\mu \omega_\nu =0$$

Taking the anti-symmetrization of the last equation: $$\nabla_{[\mu} \omega_{\nu]} =\text{d}\omega=0 $$ Which means that $\omega$ is closed. However in general this does not mean that $\omega$ is exact, i.e. $\omega=\text{d}\alpha$, for some scalar function $\alpha$. Then 'since we have restricted the topology of the region in which we are working' the one-form must also be exact.

I don't see how the vanishing of the Riemann tensor directly leads to the bold-texted conclusion. More generally in what space is a closed form also always exact ?


I) When Ref. 1 writes

[...] since we have restricted the topology of the region in which we are working,

it is referring to a previous comment:

Technically, these statements should be restricted to a region of the manifold that is simply-connected (all loops in the region can be smoothly deformed to a point without leaving the region); we will implicitly assume this condition below.

II) Ref. 1 is apparently alluding to the Poincaré lemma in the text. The Poincaré lemma comes in various versions and generalizations, e.g.,

  1. In a contractible manifold, every closed form is exact (except for zero-forms).

  2. In a star-shaped neighborhood, every closed form is exact (except for zero-forms).

  3. In a sufficiently small neighborhood, every closed form is exact (except for zero-forms).

  4. In a connected and simply-connected manifold, every closed one-form is exact.

  5. In a manifold $M$ where the homotopy groups $\pi_0(M)=\pi_1(M)=\ldots=\pi_r(M)=0$ vanish, every closed $r$-form is exact.

Specifically, Ref. 1. is using version 4.

III) Example: Magnetic monopole. Let the manifold be $M=\mathbb{R}^3\backslash\{0\}$, which is connected and simply-connected, but not contractible. The two-form $$B~:=~\sum_{i,j,k\in\{1,2,3\}}\epsilon_{ijk}\frac{x^i}{r^3}\mathrm{d}x^j\wedge \mathrm{d}x^k, \qquad r~:=~\sqrt{\sum_{i\in\{1,2,3\}}(x^i)^2}~>~0, $$ of the dualized magnetic field is a closed two-form but not an exact two-form on $M$. (The magnetic potential one-form $A$ cannot be defined on a so-called Dirac-string.)


  1. S. Carroll, Spacetime and Geometry, Section 3.6, page 124-125.
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  • $\begingroup$ Can you explain how the fact that the Riemann tensor vanishes implies that one the Poincaré lemma is applicable, cfr 'contractible domain' and 'sufficiently small open neightbourhood' (while you need the complete manifold?)? $\endgroup$ – Anne O'Nyme Jan 3 '14 at 12:00
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 3 '14 at 20:12

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