Square-root matrix in Bimetric Gravity

The Hasan-Rosen formulation of bimetric gravity can schematically be written as[1]:

$$\mathcal{L}_{bi} = \mathcal{L}_g + \mathcal{L}_f + \mathcal{L}_{int}$$

where $\mathcal{L}_g$ is the Einstein-Hilbert action for metric $g_{\mu\nu}$ and similarly for metric $f_{\mu\nu}$. The interesting part really lies in the interaction term. This interaction term can schematically be written as:

$$\mathcal{L}_{int} = \Sigma_{i=0}^{4}\ c_i \ e_i(\mathbb{X})$$

Here: $e_i$ are the elementary symmetric polynomials. But the crux of this framework lies in the matrix $\mathbb{X}$, which is:

$$\mathbb{X} = \sqrt{g^{-1}f}$$

Question : Why the square-root?

I understand that the simplest choice for interaction between two metrics is to contract them completely. Fine. But why is it put under a square-root?

[1]-https://arxiv.org/abs/1109.3515 (and there are many more references, but they all start with this construction)

• Hmm, interesting question. Perhaps it's justifiable for similar reasons to why we scale everything with \sqrt{-g} (i.e. to ensure covariance). – astronat May 29 '18 at 21:24
• I believe not. I have only mentioned these terms schematically in my question. There is a factor of $\sqrt{-g}$ sitting out to do just what you've mentioned! – topologically_astounded May 29 '18 at 22:27

A partial answer to the question "why the square root" comes from realizing that the interaction terms are much simpler to write down in terms of vielbeins (aka tetrads/vierbeins/frame fields). These vielbeins are roughly like a square root of the metric, and hence it is less surprising that the square root of $g^{-1} f$ appears when writing interaction terms between vielbeins.
These statements can be made more precise. The vielbeins are a collection of one-forms $e_\mu^a$ indexed by the internal Lorentz index $a$, and $\mu$ is just the spacetime index. They are required to define an orthonormal frame, $$g^{\mu\nu}e_\mu^a e_\nu^b = \eta^{ab},$$ with $\eta^{ab} = \text{diag}(-1,+1,+1,+1)$. This equation also implies that $$e_\mu^a e_\nu^b \eta_{ab} = g_{\mu\nu},$$ so that $e_\mu^a$ acts like a square root of the metric, since a product of two of them gives back $g_{\mu\nu}$ (albeit with a necessary contraction with $\eta_{ab}$).
To write the bimetric interactions, we also introduce an independent set of vielbeins $h_\mu^a$ for the second metric, satisfying $$h_\mu^a h_\nu^b \eta_{ab} = f_{\mu\nu}$$ The bimetric interactions are then just given by \begin{align} L_1 &=e^a\wedge e^b\wedge e^c\wedge h^d \varepsilon_{abcd}\\ L_2 &=e^a\wedge e^b\wedge h^c\wedge h^d \varepsilon_{abcd}\\ L_3 &=e^a\wedge h^b\wedge h^c\wedge h^d \varepsilon_{abcd} \end{align} Here, $\varepsilon_{abcd}$ is simply the antisymmetric Levi-Civita symbol. The spacetime indices are suppressed, and the wedge product refers to these spacetime indices. We can convert these to Lagrangian densities by contracting with $\varepsilon^{\mu\nu\rho\sigma}$, the antisymmetric tensor density (i.e. not divided by $\sqrt{-g}$). Focusing just on $L_1$ for now, with a little work you can show that $$\frac1{4!}\varepsilon^{\mu\nu\rho\sigma}L^1_{\mu\nu\rho\sigma} = 3! \sqrt{-g} (e^{-1})_e^\alpha h_\alpha^e = 3! \sqrt{-g} \,\text{Tr}(e^{-1} h)$$ where we have introduced the inverse vielbein $(e^{-1})_e^\alpha = \eta_{ea}g^{\alpha\mu}e_\mu^a\equiv e^\alpha_e$, which acts like a square root of the inverse metric, $g^{\mu\nu} = e_a^\mu e_b^\nu \eta^{ab}$. The other interactions involve symmetric polynomials of the matrix $e^{-1}h$. So in particular, when written in terms of vielbeins, no square roots appear.
The last thing we should ask is if $e^{-1} h$ really is the square root of $g^{-1} f$. We can check this by squaring $e^{-1} h$, \begin{align} e^{-1} h e^{-1} h &=e^\mu_a h^a_\alpha e^\alpha_bh^b_\nu \\ &= g^{\mu\beta}e_\beta^c\eta_{ca}h^a_\alpha e^\alpha_b h^b_\nu \end{align} At this point, we would like to commute the $e^c_\beta$ past the $h^a_\alpha$, in order to contract it with another $e$ vielbein. It turns out that we can do this if we assume what is known as the symmetric vielbein condition, $$\eta_{ca} e^a_\beta h^c_\alpha = \eta_{ca}e^a_\alpha h^c_\beta.$$ Using this, we can complete the calculation, \begin{align} e^{-1} h e^{-1} h &= g^{\mu\beta} h^a_\beta \eta_{ac} e^c_\alpha e^\alpha_b h^b_\nu \\ &= g^{\mu\beta}h^a_\beta \eta_{ab} h^b_\nu\\ &= g^{\mu\beta} f_{\beta\nu} \;\checkmark \end{align}
Finally, you could ask whether imposing the symmetric vielbein condition is justified. This issue is discussed in this paper https://arxiv.org/abs/1208.4493, where the conclusion seems to be that the condition is justified precisely when $g^{-1} f$ has real square roots.