The complete $L^q$-spectrum and large deviations for return times for equilibrium states with summable potentials

Let $(X_k)_{k\geq 0}$ be a stationary and ergodic process with joint distribution $\mu$ where the random variables $X_k$ take values in a finite set $\mathcal{A}$. Let $R_n$ be the first time this process repeats its first $n$ symbols of output. It is well-known that $\frac{1}{n}\log R_n$ converges almost surely to the entropy of the process. Refined properties of $R_n$ (large deviations, multifractality, etc) are encoded in the return-time $L^q$-spectrum defined as $$\mathcal{R}(q)=\lim_n\frac{1}{n}\log\int R_n^q \,d\mu\quad (q\in\mathbb{R})$$ provided the limit exists. We consider the case where $(X_k)_{k\geq 0}$ is distributed according to the equilibrium state of a potential $\varphi:\mathcal{A}^{\mathbb{N}}\to\mathbb{R}$ with summable variation, and we prove that $$\mathcal{R}(q) = \begin{cases} P((1-q)\varphi) & \text{for}\;\; q\geq q_\varphi^*\\ \sup_\eta \int \varphi \, d\eta & \text{for}\;\; q<q_\varphi^{*} \end{cases}$$ where $P((1-q)\varphi)$ is the topological pressure of $(1-q)\varphi$, the supremum is taken over all shift-invariant measures, and $q_\varphi^*$ is the unique solution of $P((1-q)\varphi) =\sup_\eta \int \varphi \, d\eta$. Unexpectedly, this spectrum does not coincide with the $L^q$-spectrum of $\mu_\varphi$, which is $P((1-q)\varphi)$, and does not coincide with the waiting-time $L^q$-spectrum in general. In fact, the return-time $L^q$-spectrum coincides with the waiting-time $L^q$-spectrum if and only if the equilibrium state of $\varphi$ is the measure of maximal entropy. As a by-product, we also improve the large deviation asymptotics of $\frac{1}{n}\log R_n$.