We now suppose that $m_c$ is chosen to cancel $\tilde{k}_{cg}$. By further using $(\dot{m}_c^{-1}) = -m_c^{-1}(\dot{m}_c)m_c^{-1}$ and making use of eq. (E.21), one easily obtains that $\tilde{k}_{cc}$ is set equal to zero providing that we choose $n$ such that

$$ \dot{n}^T + k_{cg}n^T = m_{sol}^T r_{sol}, \qquad (E.26) $$

where $r_{sol}$ stands for the expression (E.25) where $m_c$ has been replaced by $m_{sol}$. The general solution is easily built from the solutions of eq. (E.21). It is given by

$$ n_{sol}^T(\tau) = m_{sol}^T(\tau) \left[ \int^\tau d\tau' r_{sol}(\tau') \right], \qquad (E.27) $$

where $m_{sol}^T$ is given by eq. (E.22). Because of the source matrix $r_{sol}$, the matrix $n_{sol}$ depends on $a$ and $m_p \in \text{Sp}(2, \mathbb{R})$ which have not been specified yet. However the above solution exists irrespectively of the choice of these two matrices. This shows that canceling the terms quadratic in the constraints is systematically feasible provided that the coupling between constraints and gauge degrees of freedom is canceled at the same time.

Fourth, we consider the couplings between the constraints and the physical degrees of freedom, which after a bit of algebra are given by

$$ \tilde{k}_{cp} = m_c^{-1} \dot{\Omega}_2 - a^T m_p^{-1} \Omega_2 (\dot{m}_p^{-1}) + k_{cg} a^T \Omega_2 - a^T m_p^{-1} k_{pp} m_p^{-1} + k_{cp} m_p^{-1} \Omega_2. \qquad (E.28) $$

This block can be set to zero if the matrix $a$ is solution of the following equation

$$ \dot{a}^T + k_{cg}a^T + a^T [m_p^{-1} \Omega_2 (\dot{m}_p^{-1}) + m_p^{-1} k_{pp} m_p^{-1}] \Omega_2 = k_{cp} m_p^{-1} \Omega_2. \qquad (E.29) $$

To solve the above, we write $a^T = m_{sol}^T a^T$. Since $m_{sol}^T$ is solution of eq. (E.21), it yields the following differential equation for $a$,

$$ \dot{a}^T + a^T \tilde{k}_{pp} \Omega_2 = m_{sol}^{-1} k_{cp} m_p^{-1} \Omega_2, \qquad (E.30) $$

where we further make use of eq. (E.19). The homogeneous equation, $\dot{a}^T + a^T \tilde{k}_{pp} \Omega_2 = 0$, describes standard Hamiltonian dynamics since $\tilde{k}_{pp}$ is symmetric. Hence it necessarily has solutions given by matrices of the symplectic group $\text{Sp}(2, \mathbb{R})$ (see e.g. ref. [4]). On denoting $a_{\text{hom}}(\tau)$ a solution of the homogeneous equation, the solutions of the inhomogeneous ones are given by

$$ a_{sol}^T(\tau) = \left[ \int^\tau d\tau' m_{sol}^{-1} k_{cp} m_p^{-1} \Omega_2 a_{hom}^{-1} \Omega_2(\tau') \right] a_{hom}^T(\tau). \qquad (E.31) $$

This finalises the proof that there exists a matrix $a$ ensuring the couplings between the constraints and the physical degrees of freedom to be set to zero. The cancellation of $\tilde{k}_{cp}$ is thus always possible. The final solution depends on $m_p \in \text{Sp}(2, \mathbb{R})$ but the solution always exists irrespectively of that choice. We note that despite of having used $m_{sol}$ to find a solution to $\tilde{k}_{cp} = 0$, it is not required to cancel the couplings between constraints and gauge degrees of freedom since the differential equation for $a$ does not depend on $m_c$. This is unlike canceling $\tilde{k}_{cc}$ for which we had to set $m_c$ to $m_{sol}$ in order to make the coupling between constraint and gauge degrees of freedom vanish. This means that canceling the block $\tilde{k}_{cp}$ can be done
independently of canceling the blocks $\tilde{k}_{cc}$ and $\tilde{k}_{cg}$. However, it is worth mentioning that to get the expression for $a$ that cancels the couplings between the physical degrees of freedom and the constraints, it is needed to solve the homogeneous part of eq. (E.30). Though we have argued that such an equation has solutions, we have not shown any concrete example of them (see the discussion below in section E.4).

We finally stress that the process of making $\tilde{k}_{cg}$, $\tilde{k}_{cc}$, and $\tilde{k}_{cp}$ equal to zero can be done irrespectively of the choice of the matrix $m_p$ which only generates a canonical transformation in the 2-dimensional sub-phase space of the physical degrees of freedom.

## E.3 Kinematics

The canonical transformation generated by $C$ allows one to define a new set of basis vectors on which solutions can be decomposed, and that we will call the dynamical basis. On denoting its vectors by $\{\vec{v}_a\}_{a\in\{0,1,2,3,4,5\}}$, the dynamical basis is related to the physical basis $\{\vec{e}_a\}$ by

$$ \vec{v}_a = \sum_{b=0}^{5} [C^{-1} {}^{\text{T}}]_{ab} \vec{e}_b. \qquad (\text{E.32}) $$

Although the physical basis is orthonormal, the matrix $C$ is not an orthogonal matrix. This implies that the dynamical basis is not an orthonormal basis. As a consequence, the different degrees of freedom in the dynamical basis cannot be extracted out of $\delta\vec{z}_k$ by projecting it onto the $\vec{v}_a$'s. Instead, its components are given by

$$ [\tilde{z}_k]_a = \sum_{b=0}^{5} [C]_{ab} (\vec{e}_b \cdot \delta\vec{z}_k). \qquad (\text{E.33}) $$

On denoting

$$ \vec{w}_a = \sum_{b=0}^{5} [C]_{ab} \vec{e}_b, \qquad (\text{E.34}) $$

any solution is decomposed on the dynamical basis as

$$ \delta \vec{z}_k = \sum_a (\vec{w}_a \cdot \delta \vec{z}_k) \vec{v}_a. \qquad (\text{E.35}) $$

Let us briefly discuss the new variables that are obtained by projecting $\delta\vec{z}_k$ onto the $\vec{w}_a$'s. Using eqs. (E.7) and (E.22), one can readily show that the new constraints are given by $\tilde{Q}_\mu = \vec{w}_\mu^C \cdot \delta\vec{z}_k$ where the two vectors $\vec{w}_\mu^C$ are

$$ \vec{w}_\mu^C = \beta_\mu \vec{S}_k + \left[ \alpha_\mu + \beta_\mu \left( \frac{\lambda_S}{\mu_S} - H_k \right) \right] \vec{D}_k. \qquad (\text{E.36}) $$

These two vectors belong to the plane of constraints. However, since $\alpha_\mu$ and $\beta_\mu$ must satisfy $\alpha_1\beta_2 - \alpha_2\beta_1 \neq 1$, it is not possible to find a choice of these four parameters such that e.g. $\vec{w}_1^C = \vec{D}_k$ and $\vec{w}_2^C = \vec{S}_k$. The new Lagrange multipliers associated to these constraints are

$$ (\alpha_1\beta_2 - \alpha_2\beta_1) \tilde{\Lambda}_\mu = \frac{\beta_\mu}{\lambda_D} k \delta N_1 - \frac{1}{\lambda_D} \left[ \alpha_\mu + \beta_\mu \left( H_k + \frac{\lambda_S}{\mu_S} \right) \right] \delta N. \qquad (\text{E.37}) $$
For the new gauge degrees of freedom, the vectors are

$$ \tilde{w}_{\mu}^{G} = \sum_{\mu'=1}^{2} \left\{ \left[ \mathbf{m}_{\text{sol}}^{-1} \mathbf{T} \right]_{\mu\mu'} \tilde{e}_{\mu'}^{G} + \left[ \mathbf{n}_{\text{sol}} \right]_{\mu\mu'} \tilde{e}_{\mu'}^{C} - \left[ \mathbf{m}_{\text{sol}}^{-1} \mathbf{T} \mathbf{a}_{\text{sol}}^{\text{T}} \mathbf{\Omega}_{2} \mathbf{m}_{\text{p}} \right]_{\mu\mu'} \tilde{e}_{\mu'}^{P} \right\}. \quad (\text{E.38}) $$

The first term guarantees that the new gauge degrees of freedom are canonically conjugated to the new constraints. The second and third terms guarantee the constraints and the physical degrees of freedom, respectively, to have vanishing contribution in the equations of motion of the gauge degrees of freedom.

Finally, the physical degrees of freedom are obtained by projecting onto the following vectors

$$ \tilde{w}_{\mu}^{P} = \sum_{\mu'=1}^{2} \left\{ \left[ \mathbf{m}_{\text{p}} \right]_{\mu\mu'} \tilde{e}_{\mu'}^{P} + \left[ \mathbf{a}_{\text{sol}} \right]_{\mu\mu'} \tilde{e}_{\mu'}^{C} \right\}. \quad (\text{E.39}) $$

The first term corresponds to a canonical transformation internal to the 2-dimensional phase space of physical degrees of freedom, while the second term ensures that constraints no longer contribute to the dynamics of the physical degrees of freedom.

## E.4 Equations of motion

The procedure detailed above leads to a Hamiltonian of the desired form (E.6), from which the equations of motion for the new set of variables are easily deduced. They read

$$ \dot{\tilde{Q}}_{\mu} = 0, \quad (\text{E.40}) $$

$$ \dot{\tilde{P}}_{\mu} = -\tilde{\Lambda}_{\mu}, \quad (\text{E.41}) $$

$$ \left( \begin{array}{c} \dot{\tilde{Z}}_1 \\ \dot{\tilde{Z}}_2 \end{array} \right) = \mathbf{\Omega}_2 \tilde{k}_{\text{pp}} \left( \begin{array}{c} \tilde{Z}_1 \\ \tilde{Z}_2 \end{array} \right), \quad (\text{E.42}) $$

where $\mu$ runs over 1, 2. The dynamical equations for the gauge degrees of freedom and the physical degrees of freedom now hold whether one solves the entire system on the surface of constraints or not. We further stress that the gauge degrees of freedom are no longer sourced by the physical ones. Their time-derivatives are only determined by the Lagrange multipliers, showing that these degrees of freedom are totally arbitrary and carry no physical information.

The unphysical sector is trivially solved and gives

$$ \tilde{Q}_{\mu}(\tau) = \text{constant}, \quad (\text{E.43}) $$

which is zero if initial conditions are selected to lie on the surface of constraints, and

$$ \tilde{P}_{\mu}(\tau) = \tilde{P}_{\mu}(\tau_{\text{in}}) + \int_{\tau_{\text{in}}}^{\tau} d\tau' \tilde{\Lambda}_{\mu}(\tau'), \quad (\text{E.44}) $$

where $\tilde{P}_{\mu}(\tau_{\text{in}})$ are arbitrary initial conditions for the gauge parameters. However, generic solutions cannot be found for the physical sector since this is where all the nontrivial dynamics enters.
We finally mention that if two linearly independent solutions of the dynamical equations of the physical degrees of freedom are found, one can build the fundamental matrix, $u_{\text{phys}}$, which is solution of

$$ \dot{u}_{\text{phys}} - \Omega_2 \tilde{k}_{\text{pp}} u_{\text{phys}} = 0. \qquad (\text{E.45}) $$

This is the same equation as the transpose of the homogeneous part of eq. (E.30), whose solutions are needed to cancel the couplings between the physical degrees of freedom and the constraints. As a consequence, finding the part of the canonical transformation that cancels $\tilde{k}_{\text{cp}}$ is equivalent to (and as difficult as) solving for the physical degrees of freedom ($\tilde{Z}_1, \tilde{Z}_2$).$^{30}$

## E.5 Discussion

Let us further compare the dynamical basis and the physical basis. Any solution can be decomposed on the dynamical basis using eq. (E.35), in which two sets of vectors need to be introduced, $\vec{v}_a$ and $\vec{w}_a$. This has to be contrasted with the decomposition on the physical basis,

$$ \delta \vec{z}_{\vec{k}} = \sum_{a} (\vec{e}_{a} \cdot \delta \vec{z}_{\vec{k}}) \vec{e}_{a}, \qquad (\text{E.46}) $$

which relies on one set of vectors only, thanks to the orthonormality of the physical basis. In that respect, the use of the physical basis is easier.

Obviously, the real advantage of the dynamical basis is at the level of the equations of motion. However, this supposes that one is able to find the matrix $a_{\text{sol}}$, which boils down to a dynamical problem equivalent to solving for the physical degrees of freedom. In practice then, the use of the dynamical basis does not really simplify the problem.

Nevertheless, it is still practically possible to set $a$ equal to the null matrix, making $\tilde{k}_{\text{cp}} \neq 0$ but still ensuring that $\tilde{k}_{\text{cc}}$ and $\tilde{k}_{\text{cg}}$ are vanishing. Fixing $a = 0$ leads to $b = 0$. In this case the matrix $C$ is the direct sum of two matrices, i.e. $C = M_{\text{unphys}} \oplus m_p$ where $m_p \in \text{Sp}(2, \mathbb{R})$ generates a canonical transformation in $\mathcal{P}$, and $M_{\text{unphys}} \in \text{Sp}(4, \mathbb{R})$ generates

$$ (m_p^{-1}) - \Omega_2 k_{\text{pp}} m_p^{-1} = 0. $$

We denote solutions of the above by $m_{p,\text{sol}}^{-1}$. The differential equation driving $a$, eq. (E.29), then simplifies to

$$ \dot{a}^{\text{T}} + k_{\text{cg}} a^{\text{T}} = k_{\text{cp}} m_{p,\text{sol}}^{-1} \Omega_2, $$

which is solved as

$$ a^{\text{T}}(\tau) = m_{\text{sol}}^{\text{T}}(\tau) \left[ \int^{\tau} d\tau' m_{\text{sol}}^{-1} k_{\text{cp}} m_{p,\text{sol}}^{-1} \Omega_2 \right]. $$

However, finding $m_{p,\text{sol}}$ is as complicated as finding $a_{\text{hom}}$ when setting $m_p = I_2$.

<footnote>$^{30}$One might be tempted to look for a specific choice of $m_p$ that simplifies the resolution of eq. (E.30). Up to our investigations, there is no such general choice for the matrix $m_p$. A first example consists in setting $m_p = I_2$. The equation of motion for $a$ is then given by eq. (E.30) where $\tilde{k}_{\text{pp}}$ is replaced by $k_{\text{pp}}$. Hence solving for $a$ is as complicated as solving for the dynamics of $(Z_1, Z_2)$. A second example consists in choosing $m_p$ such that</footnote>
a canonical transformation in $C \otimes G$. This leads to the following equations of motion

$$ \dot{\tilde{Q}}_{\mu} = 0, \qquad (E.47) $$

$$ \dot{\tilde{P}}_{\mu} = -\tilde{\Lambda}_{\mu} - [\tilde{k}_{\text{cp}}]_{\mu,1} \tilde{Z}_{1} - [\tilde{k}_{\text{cp}}]_{\mu,2} \tilde{Z}_{2}, \qquad (E.48) $$

$$ \left( \begin{array}{c} \dot{\tilde{Z}}_1 \\ \dot{\tilde{Z}}_2 \end{array} \right) \approx \Omega_2 \tilde{k}_{\text{pp}} \left( \begin{array}{c} \tilde{Z}_1 \\ \tilde{Z}_2 \end{array} \right). \qquad (E.49) $$

First, this has to be compared with the equations of motion in the dynamical basis, eqs. (E.40), (E.41) and (E.42). The dynamics of the constraints remains unchanged but the gauge degrees of freedom receive contributions from the physical ones. The dynamics of physical degrees of freedom remains unaffected at the condition that it is solved on the surface of constraints (which, we recall, is the meaning of the symbol “≈”). Second, the above has to be compared with the equations of motion (3.20) in the physical basis. Using the above compact notation, we remind that they are given by

$$ \dot{Q}_{\mu} = \sum_{\mu'=1}^{2} [\mathbf{k}_{\text{cg}}]_{\mu\mu'} Q_{\mu'}, \qquad (E.50) $$

$$ \dot{P}_{\mu} \approx -\Lambda_{\mu} - [\mathbf{k}_{\text{cp}}]_{\mu,1} Z_{1} - [\mathbf{k}_{\text{cp}}]_{\mu,2} Z_{2}, \qquad (E.51) $$

$$ \left( \begin{array}{c} \dot{Z}_1 \\ \dot{Z}_2 \end{array} \right) \approx \Omega_2 \mathbf{k}_{\text{pp}} \left( \begin{array}{c} Z_1 \\ Z_2 \end{array} \right). \qquad (E.52) $$

The dynamics of the constraints is now affected but it remains trivially solved. The gauge degrees of freedom have a similar dynamics provided that they are now solved on the surface of constraints. Finally, the physical degrees of freedom have an identical dynamics up to a canonical transformation internal to the 2-dimensional phase space of physical degrees of freedom. Indeed, it is possible to cast the equation of motion (E.52) in the form of eq. (E.49) without changing the structure of eqs. (E.50) and (E.51), by introducing the canonical transformation $\vec{z} \to (I_4 \oplus m_p)\vec{z}$, which leaves the space of constraints and of gauge degrees of freedom unchanged while the phase space of physical degrees of freedom is transformed along the internal canonical transformation generated by $m_p \in \text{Sp}(2, \mathbb{R})$.

## F Lagrange multipliers in the physical basis

Once the gauge is fixed, the Lagrange multipliers can be expressed as functions of the physical degrees of freedom only. To this end, we decompose the gauge conditions, $\vec{G}_i \cdot \delta\vec{z}_k^- = 0$, and eq. (4.26), on the physical basis. To lighten the calculation, we will directly work on the surface of constraints. Hence the two relevant planes are the plane of gauge degrees of freedom and the plane of physical degrees of freedom.

Let us first introduce the matrix made from the components of the gauge vectors in the plane of physical degrees of freedom, i.e.

$$ G_p = \begin{pmatrix} \vec{G}_1 \cdot \vec{e}_1^P & \vec{G}_1 \cdot \vec{e}_2^P \\ \vec{G}_2 \cdot \vec{e}_1^P & \vec{G}_2 \cdot \vec{e}_2^P \end{pmatrix}. \qquad (F.1) $$
The two gauge conditions can be decomposed on the physical basis and further expressed as an equality between vectors:

$$ G \begin{pmatrix} P_1 \\ P_2 \end{pmatrix} + G_p \begin{pmatrix} Z_1 \\ Z_2 \end{pmatrix} \approx 0, \qquad (F.2) $$

where **G** is defined in eq. (4.24) and is built from the components of the gauge vectors in the plane of gauge degrees of freedom. Since non-pathological gauges are considered here, the matrix **G** is invertible. This yields

$$ \begin{pmatrix} P_1 \\ P_2 \end{pmatrix} \approx -G^{-1}G_p \begin{pmatrix} Z_1 \\ Z_2 \end{pmatrix}, \qquad (F.3) $$

which can be used to replace the gauge degrees of freedom by the physical degrees of freedom in eq. (4.26).

Decomposing now the right-hand side of eq. (4.26) on the physical basis leads to

$$ \begin{pmatrix} \nabla_{\tau} \vec{G}_1 \cdot \delta \vec{z}_{\bar{k}} \\ \nabla_{\tau} \vec{G}_2 \cdot \delta \vec{z}_{\bar{k}} \end{pmatrix} \approx (\dot{G} + G K_1 + G_p K_2) \begin{pmatrix} P_1 \\ P_2 \end{pmatrix} + (\dot{G}_p + G K_3 + G_p K_4) \begin{pmatrix} Z_1 \\ Z_2 \end{pmatrix}, \qquad (F.4) $$

where the matrices $K_i$ read

$$ [K_1]_{\mu\nu} = \vec{e}_{\nu}^{G} \cdot (\nabla_{\tau} \vec{e}_{\mu}^{G}), \qquad (F.5) $$

$$ [K_2]_{\mu\nu} = \vec{e}_{\nu}^{G} \cdot (\nabla_{\tau} \vec{e}_{\mu}^{P}), \qquad (F.6) $$

$$ [K_3]_{\mu\nu} = \vec{e}_{\nu}^{P} \cdot (\nabla_{\tau} \vec{e}_{\mu}^{G}), \qquad (F.7) $$

$$ [K_4]_{\mu\nu} = \vec{e}_{\nu}^{P} \cdot (\nabla_{\tau} \vec{e}_{\mu}^{P}). \qquad (F.8) $$

These matrices are easily related to the elements of the Hamiltonian in the physical basis and we will use the notations introduced in section E. First, $\vec{e}_{\mu}^{G} = -\Omega_6 \vec{e}_{\mu}^{C}$, hence $K_1$ is related to the couplings between the constraints and the gauge degrees of freedom, i.e. $K_1 = -k_{cg}$. Second, $\vec{e}_{2}^{P} = -\Omega_6 \vec{e}_{1}^{P}$ (which also leads to $\vec{e}_{1}^{P} = \Omega_6 \vec{e}_{2}^{P}$), hence $K_2$ is given by the couplings between the gauge degrees of freedom and the physical degrees of freedom, which have been shown to vanish, i.e. $K_2 = 0$. Third, the matrix $K_3$ is shown to be given by the couplings between the constraints and the physical degrees of freedom, i.e. $K_3 = -k_{cg}$. Fourth, we obtain that the last matrix is related to the couplings internal to the physical degrees of freedom, i.e. $K_4 = \Omega_2 k_{pp}$.

Gathering the above results, it is straightforward to rewrite eq. (4.26) as

$$ G_{\text{LM}} \begin{pmatrix} \delta N \\ k \delta N_1 \end{pmatrix} \approx N \begin{pmatrix} Z_1 \\ Z_2 \end{pmatrix}, \qquad (F.9) $$

where

$$ N = (\dot{G} - Gk_{cg}) G^{-1} G_p - (\dot{G}_p + G_p \Omega_2 k_{pp}) + Gk_{cg}. \qquad (F.10) $$

We note that the above expression relating the Lagrange multipliers to the physical degrees of freedom holds providing that the gauge is fixed.
## G Non-pathological synchronous gauge

In this appendix, we show how to define a non-pathological gauge in which $\delta N$ and $\delta N_1$ are both vanishing. To this end, we start from eq. (4.27) and identify the conditions that $G$ and $G_p$ should satisfy to make $N = 0$, which ensures that the two Lagrange multipliers vanish on-shell. Since the matrices $G$ and $G_p$ are built from the components on the gauge vectors in the planes $G$ and $P$ respectively, they entirely define the two gauge vectors, hence the gauge choice. We note that in this identification, it is needed to impose the matrix $G$ to be non-singular for the gauge to be non-pathological.

Sufficient conditions for $N$ to equal the null matrix are

$$ \dot{G} - Gk_{cg} = 0, \qquad (G.1) $$

$$ \dot{G}_p + G_p \Omega_2 k_{pp} = G k_{cg}. \qquad (G.2) $$

The first equation is easily solved and general solutions read

$$ G_{\text{sol}} = m_{\text{sol}}^{\text{T}-1}, \qquad (G.3) $$

where $m_{\text{sol}}^{\text{T}}$ is non-singular (see appendix E).$^{31}$ We stress that $G_{\text{sol}}$ is non-singular, hence the gauge is guaranteed to be non-pathological. We also note that $G_{\text{sol}}^{-1}$ can be interpreted as the fundamental matrix generating the homogeneous evolution of the gauge degrees of freedom, i.e. eqs. (3.23) and (3.24). The second equation is solved by first solving the homogeneous problem which can be cast as

$$ \frac{d}{d\tau} G_p^{-1} - \Omega_2 k_{pp} G_p^{-1} = 0. \qquad (G.4) $$

This equation is a Hamilton equation generated by the Hamiltonian $k_{pp}$, hence it necessarily has a non-singular solution in the group $Sp(2, \mathbb{R})$. We note this solution $G_{p, \text{hom}}$ and it is no more than the fundamental matrix generating the dynamics of the physical degrees of freedom. Then, the general solutions of eq. (G.2) are

$$ G_{p, \text{sol}} = \left[ \int^{\tau} d\tau' G_{\text{sol}}(\tau') k_{cg}(\tau') \right] G_{p, \text{hom}}, \qquad (G.5) $$

which finalises the proof that one can set $N = 0$ with a non-singular $G$ matrix.

As a consequence, the gauge defined by setting $G$ and $G_p$ to be given by eqs. (G.3) and (G.5) is non-pathological and leads to $\delta N \approx 0 \approx \delta N_1$. In that sense, it can be dubbed as synchronous. However, it is not a priori synchronised with the homogeneous and isotropic FLRW space-time but rather a posteriori. Indeed, the Lagrange multipliers are now vanishing on-shell, which is unlike the standard (and pathological way) of defining the synchronous gauge by directly imposing $\delta N = 0 = \delta N_1$ off-shell. The foliation is synchronised with the FLRW background as a result of synchronising the gauge vectors with the dynamics of the gauge degrees of freedom and the physical degrees of freedom through $G_{\text{sol}}$ and $G_{p, \text{sol}}$, which

<footnote>$^{31}$We use the fact that if $M$ is solution of $dM/d\tau + A(\tau)M = 0$, then $M^{-1}$ is solution of $dM^{-1}/d\tau - M^{-1}A(\tau) = 0$. It is straightforwardly shown by noting that $dM^{-1}/d\tau = -M^{-1}(dM/d\tau)M^{-1}$. Similarly, if $M$ is solution of $dM/d\tau + MA(\tau) = 0$, then $M^{-1}$ is solution of $dM^{-1}/d\tau - A(\tau)M^{-1} = 0$.</footnote>
are given by solutions of the dynamical equations of the cosmological perturbations. Hence, a non-pathological implementation of the synchronous gauge is primarily synchronised with perturbative variables of the cosmological space-time, rather than with background variables.

It is worth noting that the above-defined gauge corresponds in fact to an entire class of non-pathological synchronous gauges since the matrices $G_{\text{sol}}$ and $G_{\text{p,sol}}$ are both defined up to two constants of integration. Moreover, we have only exhibited a set of solutions which makes $N$ vanish but other solutions may exist, hence other non-pathological synchronous gauges.

Finally, though conceptually interesting, it is fair to say that the class of non-pathological synchronous gauge defined above is likely to be of little practical use. Indeed, the full dynamics of the physical degrees of freedom needs to be solved in order to define such gauges. Because of this, it cannot be used to make the dynamics of cosmological perturbations easier to solve.

## H Pathology of the uniform-expansion gauge

We show in this appendix four different implementations of the uniform-expansion gauge, which are all pathological.

**Implementation 1.** Let us start with the common way of fixing the uniform-expansion gauge. It is based on the expression given in eq. (5.23), which implies that imposing $\delta\gamma_1 = 0$ and $\delta N_1 = 0$ as gauge conditions guarantees the perturbation of the integrated expansion rate to be zero. In our language, it corresponds to

$$ \vec{G}_1 = \vec{e}_1^\phi, \quad \vec{G}_2 = \vec{0} \quad \text{and} \quad \lambda = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. \qquad (H.1) $$

The second gauge condition is the same as the second condition in the Newtonian gauge. Thus it leads to a vector $\vec{J}_2$ aligned with $\vec{e}_2^\phi$. For time derivatives of the gauge degrees of freedom to be removed, the projection of $\vec{e}_2^\phi$ onto $\mathcal{G}$ has to be aligned with the projection of $\vec{e}_1^\phi$ onto $\mathcal{G}$. However, the study of the spatially-flat gauge in section 5.1 showed that it is not the case. Hence, the conditions $\delta\gamma_1 = 0$ and $\delta N_1 = 0$ lead to a pathological gauge.

**Implementation 2.** Alternatively, one can start from the Hamiltonian expression of the perturbations of the integrated expansion rate given in eq. (5.21). As a consequence, imposing $\delta N = 0 = \delta\Theta$ as gauge conditions guarantees a vanishing $\delta N_{\text{int}}$. Such a gauge is fixed by introducing

$$ \vec{G}_1 = -\frac{\theta}{4} \vec{e}_1^\phi + v^{1/3} \vec{e}_1^\pi, \quad \vec{G}_2 = \vec{0} \quad \text{and} \quad \lambda = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}. \qquad (H.2) $$

The rank of $\lambda$ equals 1 and we write the second gauge condition as $\vec{G}'_2 \cdot \delta\vec{z}_k^- + \vec{J}_2 \cdot \delta\dot{\vec{z}}_k^- = 0$, where

$$ \vec{G}'_2 = N \left( \frac{\sqrt{3}}{2} \frac{1}{v^{2/3}} \vec{e}_1^\phi - \frac{1}{\pi_\phi} \vec{e}_0^\pi \right), \qquad (H.3) $$

$$ \vec{J}_2 = \left( \frac{v}{\pi_\phi} \right) \vec{e}_0^\phi. \qquad (H.4) $$
A direct calculation shows that

$$ \vec{J}_2 \cdot (\Omega_6 \vec{D}_k) = 0 \text{ and } \vec{J}_2 \cdot (\Omega_6 \vec{S}_k) = 1, \qquad (H.5) $$

as well as

$$ \vec{G}_1 \cdot (\Omega_6 \vec{D}_k) = 0 \text{ and } \vec{G}_1 \cdot (\Omega_6 \vec{S}_k) = \frac{v^{2/3}}{\sqrt{3}} \left[ \frac{3}{2} \left( \frac{\pi_\phi}{v} \right)^2 + M_{\text{Pl}}^2 \left( \frac{k}{v^{1/3}} \right)^2 \right]. \qquad (H.6) $$

Consequently, the projection of $\vec{G}_1$ and $\vec{J}_2$ onto the plane of gauge degrees of freedom are aligned one with each other, and this implementation of the uniform-expansion gauge is free of time-derivatives of the gauge degrees of freedom. Let us now turn to the matrix $G_{\text{LM}}$ which has to be non-singular. Its first row is given by $\vec{G}_1 \cdot (\Omega_6 \vec{S}_k) \neq 0$ and $\vec{G}_1 \cdot (\Omega_6 \vec{D}_k) = 0$ and its second row by the projections of $\vec{G}'_2 - \vec{J}_2 + \dot{\alpha}\vec{G}_1$ onto the plane of gauge degrees of freedom. Thus, this matrix is non-singular providing that $(\vec{G}'_2 - \vec{J}_2 + \dot{\alpha}\vec{G}_1) \cdot (\Omega_6 \vec{D}_k)$ is not equal to zero. However, a direct calculation shows that $\vec{G}'_2 - \vec{J}_2 + \dot{\alpha}\vec{G}_1$ has a zero projection onto $\Omega_6 \vec{D}_k$ and the matrix $G_{\text{LM}}$ is singular. This shows that implementing the uniform-expansion gauge by imposing $\delta N = 0 = \delta \Theta$ leads to a pathological gauge.

**Implementation 3.** Another strategy consists in imposing $\Theta \delta N + N \delta \Theta = 0$ as a first gauge condition to be complemented with another gauge condition. This condition has to be of the form $\vec{G}_1 \cdot \delta \vec{z}_k^\phi = 0$ otherwise the gauge is pathological. As a result, the uniform-expansion gauge can be implemented introducing

$$ \vec{G}_2 = \frac{N\sqrt{3}}{v^{2/3}M_{\text{Pl}}^2} \left( -\frac{\theta}{4} \vec{e}_1^\phi + v^{1/3} \vec{e}_1^\pi \right) \quad \text{and} \quad \lambda = \begin{pmatrix} 0 & 0 \\ \Theta & 0 \end{pmatrix}, \qquad (H.7) $$

while the vector $\vec{G}_1$ is left unspecified. First, the time-derivative of the gauge degrees of freedom have to be removed for the gauge to be non-pathological. The second gauge condition leads to a derivative condition on the phase space where

$$ \vec{G}'_2 = \frac{N\sqrt{3}}{v^{2/3}M_{\text{Pl}}^2} \left( \frac{\theta}{2} \vec{e}_1^\phi - \frac{\sqrt{3}}{2} \frac{v^{2/3}\theta}{\pi_\phi} \vec{e}_0^\pi + v^{1/3} \vec{e}_1^\pi \right), \qquad (H.8) $$

$$ \vec{J}_2 = \frac{3\theta}{2M_{\text{Pl}}^2} \frac{v}{\pi_\phi} \vec{e}_0^\phi. \qquad (H.9) $$

This gives a first constraint on the vector $\vec{G}_1$ since its projection onto the plane of gauge degrees of freedom has to be aligned with the projection of $\vec{J}_2$ onto that same plane (note that without loss of generality, we can choose these two projections to be equal). Second, the matrix $G_{\text{LM}}$ has to be non-singular. Since $\vec{G}_1^G$ has to be aligned with $\vec{J}_2^G$, the first row of $G_{\text{LM}}$ is proportional to the projections of $\vec{e}_0^\phi$ onto the plane of gauge degrees of freedom. They read

$$ \vec{e}_0^\phi \cdot (\Omega_6 \vec{S}_k) = \frac{\pi_\phi}{v} \quad \text{and} \quad \vec{e}_0^\phi \cdot (\Omega_6 \vec{D}_k) = 0. \qquad (H.10) $$

Thus, the matrix $G_{\text{LM}}$ is not singular if $\vec{G}'_2 - \vec{J}_2 + \dot{\alpha}\vec{G}_1$ has a non-zero projection onto $\Omega_6 \vec{D}_k$. However, a direct computation shows that $\vec{G}'_2 \cdot (\Omega_6 \vec{D}_k) = 0$ and similarly for $\vec{J}_2$
and $\vec{G}_1$. Hence, this implementation of the uniform-expansion gauge is also pathological. Note that constraining the vector $\vec{G}_1$ to have a non-zero projection onto $\Omega_6 \vec{D}_k$ is necessary for $\det(G_{LM}) \neq 0$. However, this means that the projection of $\vec{G}_1$ onto the plane of gauge degrees of freedom is not aligned anymore with the projection of $\vec{J}_2$ onto $\mathcal{G}$. As a consequence, time-derivative of gauge degrees of freedom are not entirely removed and the gauge remains pathological.

**Implementation 4.** Finally, let us stress that the above implementation assumes gauge conditions such that $\mathrm{Rank}(\lambda) = 1$. Instead, one could start from a gauge with $\mathrm{Rank}(\lambda) = 0$ and design the gauge vectors such that a vanishing integrated-expansion appears as a dynamical consequences of the gauge conditions. For instance, we introduce the two following gauge vectors

$$ \vec{G}_1 = -\frac{\theta}{4} \vec{e}_1^{\phi} + v^{1/3} \vec{e}_1^{\pi}, \qquad (H.11) $$

$$ \vec{G}_2 = N^{-1} \nabla_{\tau} \vec{G}_1 + C_k(\tau) \vec{G}_1, \qquad (H.12) $$

where $C_k(\tau)$ can be any arbitrary function of time and scale. A direct calculation leads to

$$ N^{-1} \nabla_{\tau} \vec{G}_1 = -\frac{\sqrt{3}}{2} v^{2/3} V_{,\phi} \vec{e}_0^{\phi} - \frac{1}{6} \left[ 5 \left( \frac{\pi_{\phi}}{v} \right)^2 + V - M_{\mathrm{Pl}}^2 \left( \frac{k}{v^{1/3}} \right)^2 \right] \vec{e}_1^{\phi} - \frac{\sqrt{2} M_{\mathrm{Pl}}^2 k^2}{12 v^{2/3}} \vec{e}_2^{\phi} \\ + \frac{\sqrt{3}}{2} \frac{\pi_{\phi}}{v^{4/3}} \vec{e}_0^{\pi} + \frac{v^{1/3} \theta}{2 M_{\mathrm{Pl}}^2} \vec{e}_1^{\pi}. \qquad (H.13) $$

Let us first show that perturbations of the integrated expansion rate equal zero in that gauge. First, it is easily checked that the first gauge condition leads to $\delta\Theta = 0$. Second, the perturbed lapse function is obtained from eq. (4.26). Since $\vec{G}_1 \cdot (\Omega_6 \vec{D}_k)$ is vanishing, it boils down to

$$ \delta N = - \left[ \vec{G}_1 \cdot (\Omega_6 \vec{S}_k) \right]^{-1} \left( \nabla_{\tau} \vec{G}_1 \cdot \delta \vec{z}_k \right). \qquad (H.14) $$

The right-hand side of the above is proportional to $N [\vec{G}_2 - C_k(\tau)\vec{G}_1] \cdot \delta \vec{z}_k$, which is vanishing once the gauge conditions are imposed. As a consequence, the perturbed lapse function equals zero, which is here a dynamical consequence of the gauge choice rather than an imposed gauge condition. Combined with $\delta\Theta = 0$, this proves that $d\delta N_{\mathrm{int}}/d\tau = 0$ and so is the integrated-expansion rate. However, a direct calculation shows that $\vec{G}_2 \cdot (\Omega_6 \vec{D}_k) = 0$. Since $\vec{G}_1$ is also orthogonal to $(\Omega_6 \vec{D}_k)$, the matrix $G_{LM}$ is singular and this implementation of the uniform-expansion gauge is pathological too.

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