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On a finite graph with symmetric edge-swap rates, the stirring representation of symmetric exclusion transports the occupations by a random permutation. An independent identically distributed Bernoulli distribution occupation field is invariant under every deterministic permutation, and hence under this random permutation when it is independent of the initial field. Therefore the displayed product measure is invariant, including densities zero and one. Equivalently, an edge swap preserves the number of particles and has the same reverse rate, giving detailed balance.

Let the common vertex degree be $r>0$. For each unordered edge place an independent Poisson process of rate $1/r$, and at a mark interchange the two endpoint occupations. If exactly one endpoint is occupied, this moves its particle across that edge at rate $1/r$; if both or neither are occupied, the occupied set stays unchanged. Thus each particle proposes jumps at total rate one, uniformly among its neighbours, with occupied-target proposals suppressed. This is the stated symmetric simple exclusion process. An edgeless graph has the identity dynamics and both conclusions below hold trivially.

In the stirring representation of symmetric exclusion, put a distinct label at every vertex and interchange labels at every edge mark, even when the occupations agree. There are finitely many marks on every finite time interval. Their chronological composition is a random permutation ${\pi }_t$ of the vertex set, and simultaneously for every initial set,Each edge swap is its own inverse. Reflecting the edge-mark times in $[0,t]$ preserves their joint Poisson process law and reverses the order of the swaps. Hence ${\pi }_t^{-1}$ and ${\pi }_t$ have the same distribution. Pathwise,Taking probabilities and then using the inverse-permutation distribution proves the product self-duality of symmetric exclusion in set form:No assumption about disjointness or equal cardinality of $A$ and $B$ is required; impossible inclusions have probability zero on both sides.

For invariance, independently choose the initial occupied set $A$ under the product measure ${\mu }_{\rho }$. The occupations are independent identically distributed Bernoulli distribution variables, so permuting their sites by any deterministic permutation preserves their joint law. Conditional on the independent stirring permutation ${\pi }_t$, the output therefore still has law ${\mu }_{\rho }$. Averaging proves the Bernoulli invariant laws of finite symmetric exclusion:There is also a direct check from the duality. For every $B\subseteq V$,because the exclusion process conserves particle number. These are exactly the joint occupation moments of ${\mu }_{\rho }$; on a finite set they determine the law by the inclusion-exclusion principle. The endpoint densities give the absorbing empty and fully occupied configurations, and the empty-set moment is one also at $\rho =0$.

The duality requested below uses the symmetric simple exclusion process, so we take that standard convention here. A configuration $\eta \in \{0,1{\}}^{\mathbb{Z}}$ records a particle at $x$ when $\eta (x)=1$. Particles attempt nearest-neighbour jumps with equal left and right rates, and any attempt into an occupied site is suppressed. Thus there is at most one particle at each site.

A convenient normalization puts independent rate-one Poisson clocks on the unoriented edges $\{x,x+1\}$. At an edge mark, interchange its two endpoint occupations. If both endpoints have the same occupation, nothing changes; if exactly one is occupied, that particle jumps to the vacant endpoint. The generator on cylinder functions isOnly finitely many terms can be nonzero for a cylinder function. A convention with total jump-attempt rate one per particle uses edge clocks of rate $1/2$, merely rescaling time.

For a pathwise construction, label all sites and swap their labels at each clock mark, carrying the initial occupations along with the labels. Each label path has jump rate two and is nonexplosive on a bounded time interval. The same is true of a backward path from any fixed space-time point. This stirring representation of symmetric exclusion therefore defines the occupations for every initial configuration, even with infinitely many particles. It also conserves the number of particles when finite.

A general exclusion process may have asymmetric jump rates, but then the plain occupation-product self-duality in the next part need not hold. The symmetry convention is therefore mathematically necessary here, rather than an irrelevant choice of time normalization.

For finite $A\subset \mathbb{Z}$, define the occupation-product functionwith empty product one. Use the stirring representation of symmetric exclusion on the time interval $[0,t]$. Trace the labels at the sites of $A$ backwards to time zero, and call their set of ancestors $B_t$. Labels remain distinct under swaps, so $|B_t|=|A|$. Pathwise,Reverse time in the Poisson clocks. Their joint law is unchanged, and each interchange is its own inverse. Thus the backward set $B_t$ has the same law as the finite-particle symmetric exclusion set $A_t$ started from $A$. Taking expectations gives the product self-duality of symmetric exclusionThe argument holds for every deterministic initial configuration $\eta$ and every finite set $A$, including the empty set, and makes no stationarity assumption.

The generator identity confirms the same mechanism. For any edge $xy$, let $A^{xy}$ interchange membership of $x,y$. ThenSumming over edges gives $L_{\eta }D=L_{A}D$. In the finite-particle generator only edges joining an occupied to a vacant site contribute; in the occupation generator the same transpositions act on the other argument. Symmetry supplies the same rate in both operations.

To see why this qualification matters, suppose right jumps have rate $a$ and left jumps rate $b\ne a$. Take $A=\{0\}$ and an initial configuration whose only particle is at $1$. The derivative at zero of the left-hand probability is $b$, since that particle must jump left into zero. The derivative of the right-hand probability for the same asymmetric finite-particle process is $a$, since its particle must jump right from zero to the initially occupied site $1$. Thus the printed same-process identity would be false in that asymmetric interpretation. Under the symmetric convention established in the preceding part, the proof is complete.

For finite $A\subset \mathbb{Z}$ put $D(\eta ,A)={\prod }_{x\in A}\eta (x)$, with empty product one. Interchanging occupations gives $D({\eta }^{xy},A)=D(\eta ,A^{xy})$, where $A^{xy}$ interchanges membership of the two sites. The edge-swap generators therefore agree on $D$. The stirring representation of symmetric exclusion proves the displayed identity at all times by tracing the ancestors of $A$ backwards. A general asymmetric jump kernel does not satisfy this product self-duality.

On $\mathbb{Z}$, put independent rate-$c$ Poisson processes on unoriented nearest-neighbour edges, and exchange their endpoint occupations at each mark. Its generator on cylinder functions is $Lf(\eta )=c{\sum }_x[f({\eta }^{x,x+1})-f(\eta )]$. Equal occupations do not change the state, and a particle next to a vacancy jumps across at rate $c$. The stirring representation of symmetric exclusion constructs it for arbitrary initial configurations. The simple occupation-product product self-duality of symmetric exclusion requires symmetry and is not the ordinary duality of an asymmetric process.