I'm struggling with Anthony Zee's chapter on differential forms in Einstein Gravity in a Nutshell, page 600. He asks us to prove that $$\omega= \Lambda \omega' \Lambda^{-1} - (d\Lambda) \Lambda^{-1}$$ using $$e=\Lambda e'$$ and $$ de+ \omega e= 0.$$ I'm not sure what $ \Lambda $ is exactly. He describes it as not being a coordinate transform but merely a local Lorentz transform (or in euclidean space a rotation) of the orthonormal frame. I originally thought it was Infinitismal transform but now I'm thinking I can put in any Lorentz Transform. Everytime I start the calculation I get lost somewhere in the notation and I'm not sure if I'm even starting from the correct transform anyway.

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    $\begingroup$ Make $\Lambda$ a function of $x$, use the Leibniz rule for differentiation after substituting for $e$ in the parallel transport equation and you should be able to get there. $\endgroup$ – Void Jun 19 '15 at 9:32
  • $\begingroup$ How would I make it a function of x? Maybe I should just do it for the rotation case first then move to lorentz. $\endgroup$ – mathmath12 Jun 19 '15 at 9:41
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    $\begingroup$ Simply: $e^\mu(x) = \Lambda^\mu_\nu(x) e'^\nu(x) $. You do not really have to think of $\Lambda$ as anything else than a non-degenerate matrix which varies from point to point, that's the beauty of diffeomorphism invariance of general relativity. $\endgroup$ – Void Jun 19 '15 at 9:52
  • $\begingroup$ Ok. I will try thst $\endgroup$ – mathmath12 Jun 19 '15 at 10:01
  • $\begingroup$ Related: physics.stackexchange.com/q/225413/2451 $\endgroup$ – Qmechanic Nov 8 '15 at 19:34

Well, this is linked to what the cotetrad $e_\mu^I$ is.

It is customary to present the cotetrad as a diagonalization of the metric and indeed, we have: $$g_{\mu\nu} = e_\mu^I e_\nu^J \eta_{IJ}$$ Note here that the cotetrad has two kind of indices. The greek type corresponds to spacetime coordinates but the latin type indices are indices in the tangent space. So, in a way, the cotetrad is a natural basis for the metric.

The physical idea behind this is actually quite simple: the metric is always locally minkowskian. It takes this form (locally) in a free falling frame of reference. So given a metric, you can assign to each point a free-falling frame (described by the cotetrad) in which the metric is diagonal (and even minkowskian).

To be a bit more precise, the cotetrad is the inverse of the tetrad, usually noted $e_I^\mu$ (note the position of the indices - this is kind of a loose notation but it is standard). The tetrad is the collection of four vectors labelled by $I$ with coordinates labelled by $\mu$. These vectors are orthogonal and normed. One of them is timelike and corresponds to the normed tangent vector along the free-falling trajectory.

But when you do this, you have an arbitrary choice: there are an infinite amount of free-falling frames at a given point and they are linked together by Lorentz transformations. So at each point, you have possible Lorentz freedom. This is your local Lorentz transform: it relates the free-falling frames of reference at each point.

Back to the math now, your Lorentz transform will not act on the coordinates $\mu$, these are fixed here. It will act on the tangent space. So, writing down the indices, you have: $${e'}_\mu^I(x) = \Lambda^I_{~J}(x) e_\mu^J(x)$$ With that you can conclude quite easily since the second equation defines $\omega_{\mu~J}^{~I}(x)$ as a connection.


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