- #1

Lactose

- 2

- 0

$$S := \int d^4 x e^\mu{}_ae^\nu{}_b R_{\mu \nu}{}^{ab}.$$

I managed to get the variation wrt. the connection to the form

$$ \varepsilon_{abcd}\varepsilon^{\mu \nu \rho \sigma} (D_\mu e^c{}_{\rho})e^d{}_\sigma = 0,$$

where latin indices denote frame (or flat) components and greek indices denote coordinate (or curved) components and ##D_\mu V^a= \partial_\mu V^a + \omega_{\mu}{}^a{}_c V^c## is the covariant derivative wrt. to the spin connection.

By the source (eq. 2.14) on page 14, the above is actually correct and then implies

$$T^a{}_{\mu\nu}:= 2D_{[\mu}e^a{}_{\nu]} = 0.$$

However, I do not see how this implication should work. I am pretty sure it has something to do with the contraction rules for epsilons and I already tried out:

- carrying out the summation in ##(d, \sigma)##, which gives contracted epsilons - did not work.
- multiplying the equation by epsilons corresponding to the free indices, giving contracted epsilons as well - did not work either.