# 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

## 1 Answer

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.

• Thank you for this detailed response. From what I understand, you suggest two reasons. Namely: 1) That the square-root matrix in interaction Lagrangian becomes more natural, perhaps aesthetically, in the vielbein formulations & 2) condition for symmetric vielbein. For the first one, I would say that yes it does make some sense. For the second one, I'm not so certain. My expectation for a root cause of the square-root was that it must arise to provide some extra-constraint to avoid a notorious ghost dof in the theory. Any thoughts on that? – topologically_astounded Jun 4 '18 at 16:48
• Well, yes the absence of the ghost is the underlying reason. I guess I was more motivating why you would even consider a square root, and it is because the theory really seems to want to be phrased in terms of vielbeins. Note that showing the absence of ghosts even in pure GR is a bit nontrivial, because the action involves second derivatives of the metric. Here too the vielbein formulation makes it more obvious that there are no second time derivatives, so the theory has a chance to be ghost free. It would then make sense that vielbeins should help in the massive case as well. – asperanz Jun 4 '18 at 17:47