There seems to be some different conventions on the indexes of the tetrad. I am wondering which is the standard, which is correct, and which is an abuse of notation.

In Sean Carroll's notes and in Wikipedia I see the tetrad represented as $e^I_\mu$. This notation is safe to convey intentions of use for converting indexes from greek to latin and vice versa, but as soon as you start raising and lowering the tetrad's own indexes (as Wikipedia does) the representation becomes ambiguous. $e^I_\mu$ could represent either ${e_\mu}^I$ or could represent ${e^I}_\mu$, and these two values aren't equal.

Other sources like Einstein's vierbein field theory of curved space (Yepez 2008) make the distinction to write ${e_\mu}^I$ as the transform from $\eta_{IJ}$ to $g_{\mu\nu}$ and ${e^\mu}_I$ as the inverse. Other sources reverse the greek and latin indexes and use ${e^I}_\mu$ as the transform from $\eta_{IJ}$ to $g_{\mu\nu}$.

I'm going to use some matrix math to make my point. Let $G = ||g_{\mu\nu}||$ be the matrix representing the covariant metric tensor, $H = \eta_{IJ}$ be the matrix of the Lorentzian tensor, $E = ||{e_\mu}^I||$ be the tetrad transformation from $H$ to $G$, and $(E^T)^{-1} = ||{e^\mu}_I||$ be the transformation from $G$ to $H$ (transposed for consistency of order of indexes). The tetrad transformation rules and their matrix equivalents are as follows:

$$\begin{matrix} g_{\mu\nu} = {e_\mu}^I \eta_{IJ} {e_\nu}^J & & G = E H E^T \\ g^{\mu\nu} = {e^\mu}_I \eta^{IJ} {e^\nu}_J & & G^{-1} = (E^T)^{-1} H^{-1} E^{-1} \\ \eta_{IJ} = {e^\mu}_I g_{\mu\nu} {e^\nu}_J & & H = E^{-1} G (E^T)^{-1} \\ \eta^{IJ} = {e_\mu}^I g^{\mu\nu} {e_\nu}^J & & H^{-1} = E^T G^{-1} E \end{matrix}$$

These rules can be used to show that raising and lowering the forward transformation tetrad can produce the inverse transform tetrad, so index gymnastics works correctly on ${e_\mu}^I$ and ${e^\mu}_I$: $$\begin{matrix} {e^\mu}_I = g^{\mu\nu} {e_\nu}^J \eta_{IJ} & & (E^T)^{-1} = G^{-1} E H \end{matrix}$$

The ambiguity arises when we start with ${e_\mu}^I$ and use index gymnastics to get to ${e^I}_\mu$: $${e_\sigma}^I g^{\sigma\nu} {e_\nu}^J \eta_{JK} {e_\mu}^K = {e^I}_\mu $$ The matrix equivalent says: $$E^T G^{-1} E H E^T=E^*$$ This can be rearranged to show $$G^{-1} E H=(E^T)^{-1} E^* (E^T)^{-1}$$ combining this with the first of the identities above gives $$(E^T)^{-1} E^* (E^T)^{-1} = (E^T)^{-1}$$ rearrange: $$E^* = E^T$$ We only get $E^*=E$ in the case that $E=E^T$, which is not a constraint on the values of ${e_\mu}^I$. Therefore in general ${e_\mu}^I \neq {e^I}_\mu$. Therefore raising or lowering the ambiguous $e^I_\mu$ tensor by the metric tensor or the Lorentzian tensor could describe one of two different values.

What I've gathered overall from this is:

  • Using $e^I_\mu$ is communicable so long as you never attempt to simplify $g^{\mu\nu} e^I_\mu$ into either $e^{\nu I}$ or $e^{I \nu}$ since these values are different. Likewise for $e^I_\mu \eta_{IJ}$ into either $e_{\mu J}$ or $e_{J \mu}$. The Wikipedia entry I cited does commit this mistake.
  • Using either ${e_\mu}^I$ or ${e^I}_\mu$ as your tetrad transforming $\eta_{IJ}$ to $g_{\mu\nu}$ is more concise than $e^I_\mu$, though there is no standard as to which of these two options is correct.
  • Most sources will keep their greek first and latin second, or vice versa, and will never perform enough index gymnastics to rearrange this order. This is a safe bet to not run into the situation I'm describing above.

Okay, so all my work aside, what is the correct way of referencing the tetrad?

  • $\begingroup$ For those who want to look at Carroll's treatment, it's on p. 95 of the pdf at arxiv.org/abs/gr-qc/9712019 (p. 88 according to the page numbering of the pdf). Rather than the WP link given in the question, a more directly relevant article would be this one: en.wikipedia.org/wiki/Frame_fields_in_general_relativity $\endgroup$
    – user4552
    Oct 25, 2014 at 19:04
  • $\begingroup$ That Wikipedia article is a better example of my point. It goes back and forth between methods of representing the tetrad, rather than sticking to only one convention. $\endgroup$ Oct 25, 2014 at 19:27
  • $\begingroup$ My initial impression is that one simply has no right to expect an index like $I$ to behave like a tensorial index, nor is there any reason to expect that moving such an index around has any well-defined physical meaning. In this context, $I$ is simply an integer that labels the four basis vectors. Even if you were using completely coordinate-free notation ("mathematician notation" for differential geometry), you could still have these $I$ indices running around. If any of the usual index gymnastics stuff can be made to work and has physical meaning, that's gravy. $\endgroup$
    – user4552
    Oct 25, 2014 at 19:27
  • $\begingroup$ Comment to the question (v4): Please double-check formulas. E.g. fifth eq. has indices $IJ$ in wrong position. $\endgroup$
    – Qmechanic
    Oct 25, 2014 at 21:00
  • $\begingroup$ I am doing this on the presumption that the metric tensor coefficients are symmetric: $$\begin{matrix} g_{\mu\nu} = g_{\nu\mu} & G=G^T \\ g^{\mu\nu} = g^{\nu\mu} & G^{-1}=(G^{-1})^T \\ \eta_{IJ} = \eta_{JI} & H=H^T \\ \eta^{IJ} = \eta^{JI} & H^{-1}=(H^{-1})^T \\ \eta_{IJ} = \eta^{IJ} & H=H^{-1} \end{matrix} $$ That last rule comes from the fact that $\eta_{IJ} = \pm \delta_{IJ}$ I didn't state these above. Any false assumptions here? $\endgroup$ Oct 25, 2014 at 22:01

1 Answer 1


Comments to the question (v1):

  • As usual, be prepared that different authors use different conventions and notations. E.g. what some authors call a vielbein might be what other authors call a transposed vielbein.

  • A curved index (aka. as coordinate index) is raised and lowered vertically with the curved metric tensor, while a flat index (aka. as vielbein index) is raised and lowered vertically with the flat metric tensor.$^1$

  • On one hand, the curved indices $\mu,\nu,\lambda,\ldots,$ reflect covariance $e^{\prime I\nu} = e^{I\mu}\frac{\partial x^{\prime \nu}}{\partial x^{\mu}}$ under change of local coordinates $x^{\mu}\to x^{\prime \nu}=f^{\nu}(x)$ in the curved space time. On the other hand, the flat indices $I,J,K,\ldots,$ reflect covariance under local Lorentz transformations $\Lambda^I{}_{J}(x)$. In detail, a Lorentz transformation acts on a vielbein $e^I:=e^{I\mu} \frac{\partial}{\partial x^{\mu}}$ as $(\Lambda.e)^I:=\Lambda^I{}_{J}~e^{J}$.

  • If it is known which index is the the curved index and which index is the flat index on a vielbein/inverse vielbein, then the horizontal position of indices is not important.

  • In particular, the identity $e^{I}{}_{\mu}=e_{\mu}{}^{I}$ should not be interpreted as a condition for a symmetric matrix, but is merely the definition of the transposed tensor (that is given the same name $e$).

  • As OP already knows, when transcribing multiplications or rank 2 tensors into a matrix multiplication, repeated indices should be ordered horizontally next to each other. This often means that one might have to pass to a transposed tensor.

  • If the underlying spacetime manifold is a supermanifold, then care must be taken to implement Grassmann sign factors consistently. E.g. matrices are then replaced with supermatrices and transposition is then replaced with supertransposition, etc.


$^1$ It should be stressed that the curved index is a semantic name, which is related to a choice of local coordinates on a spacetime manifold, that generically is curved. Moreover, the flat index and the flat metric are also semantic names. They do not refer to the actual spacetime manifold in the vielbein formalism.

  • $\begingroup$ About point #4: My question is closely aligned with asking how important is the horizontal position of the tetrad. I would agree with you that it is not important ... so long as the authors don't attempt to perform gymnastics on the tetrad's indices (my point #1). My own confusion comes from trying to learn from Wikipedia sources that do perform gymnastics on the tetrad's indices. $\endgroup$ Oct 25, 2014 at 21:37
  • $\begingroup$ Thanks for the edit. I've removed my whiny comments and changed my -1 to a +1. $\endgroup$
    – user4552
    Oct 26, 2014 at 0:00

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