I am trying to understand how the introduction of a non-zero slow roll parameter ($\epsilon$) breaks the conformal invariance of the kinetic term in the scalar Lagrangian, particularly in the context of inflationary cosmology. Here is the background of my thought process:
Starting from de Sitter space, which is characterized by an exponential expansion, the metric in conformal time $\eta$ can be written as: $$ ds^2 = \frac{1}{H^2 \eta^2} \left( d\eta^2 - d\vec{x}^2 \right)$$
During inflation, the expansion is not exactly exponential but close to it. The slow roll parameter $\epsilon$ quantifies the deviation from the exact de Sitter case. The metric with a non-zero slow roll parameter is given by: $$ ds^2 = \frac{1}{H^2 \eta^2 (1 - \epsilon)^2} \left( d\eta^2 - d\vec{x}^2 \right)$$ In the context of scalar field theory, the Lagrangian for a scalar field $\phi$ in this spacetime is: $$ \mathcal{L} = \sqrt{-g} \left( \frac{1}{2} \partial_\mu \phi \partial^\mu \phi \right) $$
Under a conformal transformation, the scalar field transforms as: $$ \delta_\lambda \phi = \lambda (\phi' + \nabla \phi) $$ $$ \delta_b \phi = 2 \vec{b} \cdot \vec{x} (\phi' + \nabla \phi) + (\eta^2 - x^2) \vec{b} \cdot \nabla \phi $$
I have attempted to derive how the slow roll parameter affects the conformal invariance. Here is my derivation: \begin{equation} \begin{aligned} \delta \mathcal{L} &= \frac{\sqrt{-g}}{2} \delta (\partial_\mu \phi \partial^\mu \phi) = \sqrt{-g} \partial_\mu \phi \partial^\mu (\delta \phi) \\\\ &= \partial_\mu \left( \sqrt{-g} \partial^\mu \phi \delta \phi \right) - \partial_\mu \left( \sqrt{-g} \partial^\mu \phi \right) \delta \phi \end{aligned} \end{equation} The integral of first term would depend on the boundary where $\delta\phi$ vanishes and the second term is zero by the virtue of equation of motion of scalar field. I expected from this that kinetic term would be invariant regardless of the transformation. However, according to the paper by Baumann titled "High-Scale Inflation and the Tensor Tilt". The scalar field Lagrangian in the given metric is: $$ \mathcal{L} = \frac{1}{H^4 \eta^4 (1 - \epsilon)^4} \left( \frac{1}{2} (H^2 \eta^2 (1 - \epsilon)^2 \partial_\eta \phi \partial_\eta \phi - H^2 \eta^2 (1 - \epsilon)^2 \nabla \phi \nabla \phi) \right) $$ which simplifies to: $$ \mathcal{L} = \frac{1}{2} \left( (\partial_\eta \phi)^2 - (\nabla \phi)^2 \right) \frac{1}{H^2 \eta^2 (1 - \epsilon)^2} $$ Under conformal transformation, the variation of the Lagrangian is:
\begin{equation}\label{eq} \delta \mathcal{L} = \lambda \epsilon \left( (\partial_\eta \phi)^2 - (\nabla \phi)^2 \right)\tag{1} \end{equation}
It is easier to see that setting $\epsilon = 0$ restores conformal symmetry as it corresponds to the exact de Sitter spacetime. However, with $\epsilon \neq 0$, the conformal invariance is broken.
This result indicates that the kinetic term is no longer invariant under conformal transformations when $\epsilon$ is non-zero. The major issue I am facing in my understanding is how to derive this result.
My question is: How does the non-zero slow roll parameter $\epsilon$ specifically lead to the breaking of conformal invariance of the kinetic term in the scalar Lagrangian i.e. How to derive the equation \eqref{eq} and how can this be understood more intuitively in the context of inflationary cosmology?
