knowledge, no precise conjectural expression for the mixing time of the <span style="font-variant:small-caps;">srw</span> has been put forward[^2].

Proof outline {#sec:outline}
=============

The proof of Theorem \[thm:main\] is divided into two (unequal) halves: for $$\begin{aligned}
\label{def:t}
t & = & t_\star+\lambda w_\star+o(w_\star),\end{aligned}$$ with $\lambda\in{\ensuremath{\mathbb{R}}}$ fixed, we will show that $$\begin{aligned}
\label{lowerbound}{\ensuremath{\mathbb{E}}}\left[{\ensuremath{\mathcal{D}}}\left(t\right)\right]& \geq & \Phi(\lambda) - o(1)\, ,\\
\label{upperbound}{\ensuremath{\mathcal{D}}}\left(t\right) & \leq & {\Phi}(\lambda)+o_{\ensuremath{\mathbb{P}}}(1).\end{aligned}$$ The lower bound (\[lowerbound\]) is proved in Section \[sec:lower-bound\]. The difficult part is the upper bound (\[upperbound\]), due to the maximization over all possible initial states. Starting from state $x\in{\ensuremath{\mathcal{X}}}$, the distance to stationarity is $$\begin{aligned}
{\ensuremath{\mathcal{D}}}_x(t) & = & \sum_{y\in{\ensuremath{\mathcal{X}}}} \left(\frac{1}{N}-P^t(x,y)\right)_+\\
& = &\sum_{y\in{\ensuremath{\mathcal{X}}}} \left(\frac{1}{N}-P^t(x,\pi(y))\right)_+,\end{aligned}$$ since $\pi$ is an involution. Using the symmetry (\[sym\]), we may re-write $P^t(x,\pi(y))$ as $$\begin{aligned}
\label{eq:reversible}
P^t(x,\pi(y)) & = & \sum_{(u,v)\in {\ensuremath{\mathcal{X}}}\times{\ensuremath{\mathcal{X}}}} P^{t/2}(x,u)P^{t/2}(y,v) \ind_{\{\pi(u)=v\}} \, .\end{aligned}$$ As a first approximation, let us assume that the balls of radius $t/2$ around $x$ and $y$ consist of disjoint trees, in agreement with Figure \[fig:tree\].

\[fig:tree\]

This is made rigorous by a particular exposure process described in Section \[sec:exposure-process\]. The *weight* ${\ensuremath{\mathbf{w}}}(u):=P^{t/2}(x,u)$ (resp. ${\ensuremath{\mathbf{w}}}(v):=P^{t/2}(y,v)$) can then be unambiguously written as the inverse product of degrees along the unique path from $x$ to $u$ (resp. $y$ to 