Transform mixed vielbein expressions to $\sqrt{-g}$ times traces in massive gravity? In the Massive Gravity review by Claudia de Rham the massive gravity action is given by

with mass potential

in vielbein formulation.
Equivalently, the same action can then be described by

with

I would like to show that the vielbein formulation is equivalent to the metric formulation. For that one has to map the vielbein wedge products to the epsilon tensor contractions (traces essentially).
For the first term this is straightforward:
$$\epsilon_{abcd}e^a\wedge e^b \wedge e^c \wedge e^d = d^4x \epsilon_{abcd}\epsilon^{\mu\nu\rho\sigma}e^a_\mu e^b_\nu e^c_\rho  e^d_\sigma\\=d^4 x \det(e)4!=d^4 x \sqrt{-\det(g)}4!$$
while we also have $\epsilon_{\mu\nu\rho\sigma}\epsilon^{\mu\nu\rho\sigma}=4!$.
However, when not just the vielbein $e^a_\mu$ appears in the wedge product but also the vielbein $f^a_\mu$, I am not sure how to isolate a full $\sqrt{-g}$ (which is composed only of $e^a_\mu$ objects) from the expression.

How does one tackle the task for the mixed terms above?

 A: Let me work it out for $c_2$ only. The other ones probably can be done using similar computations. Let's first add two vielbeins with their inverses because at least 4 vielbeins need to show up. We will do it on the indices occupied by the $f$'s
$$
\begin{aligned}
\epsilon^{\mu\nu\rho\lambda} \epsilon_{abcd}\,e^a_\mu e^b_\nu f^c_\rho f^d_\lambda &=
\epsilon^{\mu\nu\rho\lambda} \epsilon_{abcd}\,e^{a}_{\mu} e^{b}_\nu  f^{c'}_{\rho'} \,\delta_{c'}^c\delta^{\rho'}_\rho\, f^{d'}_{\lambda'}\,\delta_{d'}^d\delta^{\lambda'}_\lambda\,
\\&
=\epsilon^{\mu\nu\rho\lambda} \epsilon_{abcd}\,e^{a}_{\mu}e^{b}_\nu  f^{c'}_{\rho'} e^c_\kappa e^\kappa_{c'}\,e^f_\rho e_f^{\rho'}\,e^d_\tau e^\tau_{d'}\,e^g_\lambda e_g^{\lambda'}\, f^{d'}_{\lambda'}\,.
\end{aligned}
$$
It's a bit messy, but it's clear what I'm doing. Now we extract the determinant of the $e$'s
$$
=\epsilon^{\mu\nu\rho\lambda} \det(e)\,\epsilon_{\mu\nu\kappa\tau}\, f^{c'}_{\rho'} e^\kappa_{c'}\,e^f_\rho e_f^{\rho'}\,e^\tau_{d'}\,e^g_\lambda e_g^{\lambda'}\, f^{d'}_{\lambda'}\,.
$$
Now we need some useful identities for the Levi-Civita tensor. We only need one of these, but for the other terms we will need all of them, so let me do a complete list.
$$
\begin{align}
\varepsilon^{\mu\nu\rho\lambda}\varepsilon_{\mu\nu\rho\lambda} &= -4!\;,\\
\varepsilon^{\mu\nu\rho\lambda}\varepsilon_{\kappa\nu\rho\lambda} &= -3! \delta^\mu_\kappa\;,\\
\varepsilon^{\mu\nu\rho\lambda}\varepsilon_{\kappa\tau\rho\lambda} &= -2 (\delta^\mu_\kappa \delta^\nu_\tau - \delta^\nu_\kappa\delta^\mu_\tau)\;,\\
\varepsilon^{\mu\nu\rho\lambda}\varepsilon_{\kappa\tau\omega\lambda} &= -\det( \delta_{a}^b)_{a = \{\kappa,\tau,\omega\}}^{b = \{\mu,\nu,\rho\}}\;,\\
\varepsilon^{\mu\nu\rho\lambda}\varepsilon_{\kappa\tau\omega\sigma} &=  -\det( \delta_{a}^b)_{a = \{\kappa,\tau,\omega,\sigma\}}^{b = \{\mu,\nu,\rho,\lambda\}}\;.
\end{align}
$$
In the conventions I am used to there are minus signs. Modify them accordingly to the correct choice of the signature. Now we get
$$
= -2\sqrt{-g}\,(f_{\rho'}^fe_f^{\rho'}\,e^{\lambda'}_gf^g_{\lambda'} - f_{\rho'}^fe_f^{\lambda'}\,e^{\rho'}_gf^g_{\lambda'})\,.
$$
This is of the form "inverse vielbein $e$"$\cdot$"vielbein $f$", so it looks promising. If we want to make it look like $\mathcal{L}_n[Q]$ we need to show that this is equivalent to traces of $\sqrt{g^{-1}f}$. But what is $g^{-1}f$ to begin with? Consider
$$
Q^\mu_\nu \equiv e^\mu_a f_\nu^a\,.
$$
We now need a symmetry property of the two vielbeins (See [1]: (C.12))
$$
e^\mu_a f_{\mu b} = e^\mu_b f_{\mu a}
$$
We can then compute
$$
\begin{align}
(Q^2)^\mu_\nu = Q^\mu_\rho Q^\rho_\nu &= e^\mu_a f_\rho^a\, \eta_{bc} e^{c\rho} f_\nu^b \\&=
e^\mu_a f_\rho^c\, \eta_{bc} e^{a\rho} f_\nu^b = \\&=
e^\mu_a e^{a\rho}\,f_{\rho}^c f_\nu^b = g^{\mu\rho}f_{\rho\nu} =(g^{-1}f)^\mu_\nu\,.
\end{align}
$$
So $Q = \sqrt{g^{-1}f}$. In the definition you gave is $Q_{\mathrm{yours}} = 1-Q_{\mathrm{mine}}$. But it's not hard to adjust things accordingly. Finally the expression above can be replaced with
$$
-2\sqrt{-g}\,(f_{\rho'}^fe_f^{\rho'}\,e^{\lambda'}_gf^g_{\lambda'} - f_{\rho'}^fe_f^{\lambda'}\,e^{\rho'}_gf^g_{\lambda'}) = -2 \sqrt{-g}\big((\mathrm{tr}\,Q)^2-\mathrm{tr}\, Q^2\big)\,,
$$
From the definitions of the $\mathcal{L}_n[Q]$ and the identities for the Levi-Civita that I showed before, it's clear that one will get traces of polynomials of $Q$. After all, the $\epsilon$'s get reduced to polynomials of Kronecker $\delta$'s.



*

*Interacting Spin-2 Fields - Kurt Hinterbichlera and Rachel A. Rosenb 1203.5783
A: One can use Mathematica to show the equivalence explicitly as follows.
First introduce a pair of $e$ vielbeins next to each $f$ vielbein like
$$\epsilon_{...a...}\epsilon^{...\mu...}f^a_\mu=\epsilon_{...a...}\epsilon^{...\mu...}e^a_\alpha e^\alpha_b f^b_\mu=\epsilon_{...a...}\epsilon^{...\mu...}e^a_\alpha Q^\alpha_\mu$$
where we used $e^a_\alpha e^\alpha_b=\delta^a_b$ and defined the quantity $Q^\alpha_\mu=e^\alpha_b f^b_\mu$.
These contractions can be calculated explicitly as
ep = LeviCivitaTensor[4];
epepEEEEQ = Sum[ep[[i, j, k, l]] e[i, m] e[j, n] e[k, o] Sum[ e[l, q] Q[q, p], {q, 1, 4}] ep[[m, n, o, p]], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}, {o, 1, 4}, {p, 1, 4}];
epepEEEQEQ = Sum[ep[[i, j, k, l]] e[i, m] e[j, n] Sum[ e[k, w] Q[w, o], {w, 1, 4}] Sum[e[l, q] Q[q, p], {q, 1, 4}] ep[[ m, n, o, p]], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}, {o, 1, 4}, {p, 1, 4}];
epepEEQEQEQ = Sum[ep[[i, j, k, l]] e[i, m] Sum[e[j, r] Q[r, n], {r, 1, 4}] Sum[ e[k, w] Q[w, o], {w, 1, 4}] Sum[e[l, q] Q[q, p], {q, 1, 4}] ep[[ m, n, o, p]], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}, {o, 1, 4}, {p, 1, 4}];
epepEQEQEQEQ = Sum[ep[[i, j, k, l]] Sum[e[i, t] Q[t, m], {t, 1, 4}] Sum[ e[j, r] Q[r, n], {r, 1, 4}] Sum[e[k, w] Q[w, o], {w, 1, 4}] Sum[ e[l, q] Q[q, p], {q, 1, 4}] ep[[m, n, o, p]], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}, {o, 1, 4}, {p, 1, 4}] // Simplify;

Then we need to compare the above to the determinant of $e^a_\mu$ times the corresponding contractions of epsilons with just $Q^\alpha_\mu$ quantities:
detE = Sum[ ep[[i, j, k, l]] e[i, m] e[j, n] e[k, o] e[l, p] ep[[m, n, o, p]], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}, {o, 1, 4}, {p, 1, 4}];
epepQ = Sum[ ep[[i, j, k, l]] ep[[i, j, k, m]] Q[l, m], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}];
epepQQ = Sum[ ep[[i, j, k, l]] ep[[i, j, n, m]] Q[k, n] Q[l, m], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}];
epepQQQ = Sum[ep[[i, j, k, l]] ep[[i, q, n, m]] Q[j, q] Q[k, n] Q[l, m], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}, {q, 1, 4}];
epepQQQQ = Sum[ep[[i, j, k, l]] ep[[w, q, n, m]] Q[i, w] Q[j, q] Q[k, n] Q[l, m], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}, {m, 1, 4}, {n, 1, 4}, {q, 1, 4}, {w, 1, 4}];

With this the comparisons work out to be:
detE epepQ/epepEEEEQ // Simplify
detE epepQQ/epepEEEQEQ // Simplify
detE epepQQQ/epepEEQEQEQ // Simplify
detE epepQQQQ/epepEQEQEQEQ // Simplify

in all four cases the answer is 24 which is equal to $4!$.
This way of doing it explicitly proves the proper factorization, but is a bit unsatisfying since it is brute force and does not introduce any fancy way to show it in symbols.
