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The reference kernel printed in the hint has total mass $1/n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)/\left(\genfrac{}{}{0ex}{}{n}{2}\right)=1+1/n$, so it is not a Markov kernel. We explicitly replace it by the normalized random transposition shuffle: choose two positions independently and uniformly and swap them. Its probabilities areThese sum to one, and its ordinary spectral gap is $\tilde{\gamma }=2/n$, the value intended in the hint. This value can also be certified by the Aldous spectral gap theorem: the single-label generator has rate $2/n^2$ between distinct positions, so on centered functions it acts as multiplication by $-2/n$. Both shuffles have the same uniform distribution on a finite set as stationary distribution.

We use the Canonical paths comparison theorem in the following precise form. For two finite reversible Markov chains with the same positive stationary distribution $\pi$, route each directed transition $(x,y)$ of $\tilde{P}$ along a positive-capacity path ${\eta }_{xy}$ of $P$. Ifthen $\tilde{\mathcal{E}}(f,f)\le A\mathcal{E}(f,f)$ and $\gamma \ge \tilde{\gamma }/A$. Indeed, telescope each difference along its path, apply the Cauchy-Schwarz inequality, and sum with weights $\pi (x)\tilde{P}(x,y)/2$. The load definition gives the Dirichlet form of a Markov chain inequality; the common variance denominator then gives the spectral gap inequality. Identity transitions use empty paths.

Put $s_k=(k,k+1)$. A transposition $(i,j)$ is routed using the wordof length $L_{ij}=2(j-i)-1\le 2n$. Each generator occurs at most twice; write its occurrence count as $N_k(i,j)\le 2$. For a fixed directed adjacent edge $(\sigma ,\sigma s_k)$ and each occurrence of $s_k$ in a word, there is exactly one starting permutation whose translated word crosses this edge at that occurrence. This follows because right multiplication by the prefix is a bijection of the symmetric group. Hence the uniform stationary factors cancel in the comparison load, leavingThe crude bounds $L_{ij}\le 2n$, $N_k(i,j)\le 2$, and $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le n^2/2$ imply $A\le 4n^2$. ThereforeOnly pairs with $i\le k<j$ contribute, and there are $k(n-k)\le n^2/4$ of these. Keeping this restriction gives the stronger $A\le 2n^2$ and $\gamma \ge 1/n^3$. Neither comparison bound needs the exact adjacent-shuffle spectral gap; the normalization repair of the reference kernel is essential.

Let $P$ be a finite reversible Markov chain with stationary distribution $\pi$, and write $Q(e)=\pi (u)P(u,v)$ for an oriented transition edge $e=(u,v)$. For every ordered pair $(x,y)$ choose a directed path ${\gamma }_{xy}$ from $x$ to $y$ using positive-capacity edges, and define the congestionThen the Canonical paths comparison theorem gives the Poincaré inequalityfor every real function $f$, and hence the spectral gap is at least $1/\rho$.

Indeed,Write each difference as the sum of edge differences along ${\gamma }_{xy}$ and apply Cauchy-Schwarz inequality:Interchanging the pair and edge sums, then applying the definition of $\rho$, bounds the result by

Compare simple random walk $P$ with the independent sampler $\tilde{P}(x,y)=\pi (y)=1/n$, whose spectral gap is one. Both stationary distributions are uniform. Route each transition $(x,y)$ of the sampler along a uniformly selected shortest path. For a directed graph edge $e$,Part i therefore bounds the congestion byPart a, equivalently the Canonical paths comparison theorem, gives

For reversible chains $P,\tilde{P}$ with stationary laws $\pi ,\tilde{\pi }$, assign to every directed $\tilde{P}$-edge $(x,y)$ a path ${\Gamma }_{xy}$ of $P$-edges. Ifand $\tilde{\pi }(x)\le a\pi (x)$, the Canonical paths comparison theorem gives $\gamma (P)\ge \gamma (\tilde{P})/(aA)$, with equivalent conventions absorbing $a$ into the congestion.