I am trying to understand the use of Cartan's formalism for calculating the Riemann and Ricci Tensors in differential geometry. My question is about the equation relating the curvature 2-forms with the Riemann Tensor:

$$\Theta^\mu _{\,\,\nu} = \frac{1}{2} R^{\mu}_{\,\,\nu\rho\sigma} \; e^\rho \wedge e^{\sigma}. $$

My understanding is that the indices of curvature 2-forms are lowered with the minkowskian metric, whereas the indices of the Riemann tensor are lowered with the general metric. This however, does not rhyme with the above equation. So how is this problem resolved?

  • $\begingroup$ Usually the Lorentz space indices are different to the spacetime ones, e.g., the two form is $R^{a}{}_{b} = R^a{}_{b \mu \nu} dx^{\mu} \wedge dx^{\nu}$. Then the metric $g_{\mu \nu}$ acts on spacetime indices and the flat Lorentz metric $\eta_{ab}$ on the local Minkowski tangent space indices. Using the frame fields you should be able to equate your definition with mine. $\endgroup$
    – Eletie
    Commented May 30, 2023 at 22:27
  • $\begingroup$ Thanks for your comment. I (think I) understand what you mean, but I am not sure that solves the issue regarding the fact that indices of curvature 2-forms are always lowered with $\eta_{\mu\nu}$ whereas this does not hold in general for the indices of the Riemann tensor. However, I believe I have found the solution - the tetrad basis is defined in such a way that the metric becomes minkowskian (just with a rather complicated basis). So in the tetrad basis, $g{\mu\nu} = \eta_{\mu\nu}$, resolving the issue. $\endgroup$
    – user206444
    Commented May 31, 2023 at 10:32
  • $\begingroup$ But you know you can easily convert between the two different types of indices by using the tetrad/frame fields $e_{a}{}^{\mu}$? And yes, the metric of the tangent space is Minkowski (if you're working in an orthonormal basis) $g_{\mu \nu} = e^a{}_{\mu} e^b{}_{\nu} \eta_{ab}$. $\endgroup$
    – Eletie
    Commented May 31, 2023 at 18:15

1 Answer 1


Always use the general metric tensor’s components in the given frame to do the index raising/lowering.

In more detail, say you have a pseudo-Riemannian manifold $(M,g)$, and you consider the Levi-Civita connection on it. Fix a local frame $\{e_1,\dots, e_n\}$ for the tangent bundle $TM$ over some open set $U\subset M$. Now, we have our metric $g$, so we can consider now the smooth functions $g_{ab}=g(e_a,e_b)$ defined on $U$. Relative to the frame, we can introduce the connection 1-forms and also the curvature 2-forms $\Theta^{\mu}_{\,\nu}$ as you call them.

We can then introduce a new collection of 2-forms, $\Theta_{ab}:=g_{ac}\Theta^{c}_{\,\,b}$. Notice how I use the components $g_{ac}$. There is no reason whatsoever for you to use $\eta_{ac}$ here (notice I didn’t even specify that $g$ is a Lorentzian metric).

Now, if you’re in the special case that the local frame $\{e_1,\dots, e_n\}$ you choose is “$g$-orthonormal”, in the sense that $g(e_a,e_b)$ is a constant function with value equal to either $0,1$ or $-1$ accordingly (so in the Riemannian case, $g_{ab}=\delta_{ab}$, and in the Lorentzian case, $g_{ab}=\eta_{ab}$), then you just count yourself fortunate that the $g_{ab}$’s are constant and diagonal.

But, in general there’s no reason to expect the frame you work with the be $g$-orthonormal. In fact, many coordinate-induced frames are not orthonormal. Of course, in many hands-on calculations, we would like to use a $g$-orthonormal frame, so in that case (assuming Lorentzian signature since GR is a tag here) we have the very special situation $g_{ab}=\eta_{ab}$, so some calculations become easy.

But again, the general thing to do is always define $g_{ab}=g(e_a,e_b)$, and use this (and the corresponding $g^{ab}$) to raise/lower the indices (unless you really mean to do something completely different, but I don’t think that’s the case for many of the “standard” calculations).


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