Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 37 (Interacting Particle Systems)

Orient the unique path from $s$ to $t$ in a uniform spanning tree $T$, and put $j_{xy}^T=1$ when it uses the edge from $x$ to $y$, $-1$ when it uses the reverse direction, and zero otherwise. Then the proposed current is $i_{xy}=\mathbb{E}{j}_{xy}^T$; set it to zero on nonedges. It is antisymmetric.

For each tree path, every intermediate graph vertex has one incoming and one outgoing path edge, while the source has one outgoing edge and the sink one incoming edge. HenceTaking expectations givesthe Kirchhoff node law for a unit flow.

To establish the Kirchhoff cycle law, it is not enough to check it separately for each tree path: those path currents need not satisfy the voltage law on the original graph. Instead, let $\mathcal{T}$ be all spanning trees and let ${\mathcal{F}}_{st}$ be all spanning forests with two components, one containing $s$ and the other $t$. For $F\in {\mathcal{F}}_{st}$ denote the source component by $C_s(F)$.

For an existing edge $xy$, deleting it from a tree whose $s$-$t$ path uses $x\to y$ produces precisely a forest $F\in {\mathcal{F}}_{st}$ with $x\in C_s(F)$ and $y\notin C_s(F)$. Conversely, adding $xy$ to any such forest produces exactly that tree and path orientation. This is a bijection. Applying it to both orientations givesDefine $h(x)=|\mathcal{T}{|}^{-1}{\sum }_F{1}_{x\in C_s(F)}$. Then $i_{xy}=h(x)-h(y)$ on every edge. Since the resistances are one, this is Ohm's law, and the voltage drops telescope around every oriented cycle:Together with the node law, this proves that the mean tree-path current is the electrical unit current. This argument is the mean spanning-tree path current identity.

For clarity, these laws determine the current uniquely. The difference of two solutions is a potential gradient $k(x)-k(y)$ with zero divergence. Multiplying its divergence by $k(x)$ and summing over graph vertices gives ${\sum }_{\{x,y\}}(k(x)-k(y){)}^2=0$. Thus every difference current is zero.

Let $e=\{a,b\}$ be an edge of $G$. Apply the preceding result with source $a$ and sink $b$. A tree contains $e$ exactly when its unique path from $a$ to $b$ is the single edge $e$, in the positive direction. Consequently the unit electrical current on $e$ is $\mathbb{P}(e\in T)$. By Ohm's law and the definition of effective resistance,since this edge has unit resistance. This is the unit-conductance case of the edge-inclusion formula for a uniform spanning tree.

The general principle needed is Rayleigh monotonicity principle: adding edges, or decreasing their resistances, cannot increase the effective resistance between fixed terminals. One precise variational formulation is the Thomson principle:with one orientation chosen for each unoriented edge. To see this, write any unit flow as $i+k$, where $i$ is the electrical unit current and $k$ has zero divergence. Since $r_e{i}_e$ is a potential difference, summation by parts gives ${\sum }_e{r}_e{i}_e{k}_e=0$. Thus its energy is ${\sum }_e{r}_e{i}_e^2+{\sum }_e{r}_e{k}_e^2$, minimized by $i$. The energy of $i$ equals its terminal voltage drop, hence the effective resistance.

Extend any unit flow on $G$ to $G^{\prime }$ by zero on all new edges. It is still a unit flow, including at new graph vertices, and its energy is unchanged. The variational minimum on $G^{\prime }$ is therefore no larger. Using the edge-inclusion formula in both graphs givesThe graphs are understood to be finite and connected, as required for their uniform spanning-tree laws. The comparison allows added graph vertices as well as added edges.

This is an asymmetrically almost-subadditive sequence. The limit must be allowed to take the value $-\infty$; the stated hypotheses alone do not imply a finite real limit. DefineFix $k$, and write $N=qk+r$ with $1\le r\le k$. Repeatedly apply the given inequality with the first summand equal to $k$. Induction givesFor fixed $k$, there are only finitely many possible residual terms $x_r$, so division by $N$ and passage to the upper limit yieldThis holds for every $k$, and therefore the upper limit is at most $I$. If $I$ is finite, its definition also gives $x_N/N\ge I-{\alpha }_N/N$, so the lower limit is at least $I$ because ${\alpha }_N/N\to 0$. If $I=-\infty$, the upper bound by every $(x_k+{\alpha }_k)/k$ makes the upper limit $-\infty$ directly. Thus in both casesIn particular,When $\lambda =-\infty$ this inequality is interpreted in the extended-real sense. For example, $x_n=-n^2$ and ${\alpha }_n=0$ satisfy the assumptions but have normalized limit $-\infty$. A finite-limit assertion would need an additional lower linear bound.

We use the Van den Berg-Kesten inequality: for increasing events $A,B$ under independent bond percolation, their disjoint occurrence $A\square B$, meaning that they have disjoint sets of open edges witnessing the two events, satisfies ${\mathbb{P}}_p(A\square B)\le {\mathbb{P}}_p(A){\mathbb{P}}_p(B)$. The finite-edge statement applies here; connection to a finite-box boundary can always be witnessed before the first exit from that box.

On the event $0\leftrightarrow \partial {\Lambda }_{m+n}$ choose a simple open path to that boundary, and let $y$ be its first point on $\partial {\Lambda }_m$. Its initial segment witnesses $0\leftrightarrow \partial {\Lambda }_m$ through $y$, inside ${\Lambda }_m$. The remaining segment eventually reaches maximum-norm distance at least $n$ from $y$, because its endpoint has norm $m+n$ while $\Vert y{\Vert }_{\infty }=m$. Stop that segment when it first reaches $y+\partial {\Lambda }_n$. These two segments use disjoint edges.

For fixed $y$, let $A_y$ be connection from zero to $y$ inside ${\Lambda }_m$, and $B_y$ be connection from $y$ to $y+\partial {\Lambda }_n$ inside $y+{\Lambda }_n$. Then ${\mathbb{P}}_p(A_y)\le {\beta }_m$ and, by translation invariance, ${\mathbb{P}}_p(B_y)={\beta }_n$. The union bound and the stated disjoint-occurrence inequality give

Set $x_n=\log {\beta }_n$ and ${\alpha }_n=\log |\partial {\Lambda }_n|$. The previous part applies, sinceMoreover a straight path of $n$ open bonds has probability $p^n$, so $p^n\le {\beta }_n\le 1$. Thus the limit is finite andEquivalently, for the nonnegative decay rate $\kappa =-\gamma$, the bound is ${\beta }_n\ge |\partial {\Lambda }_n{|}^{-1}{e}^{-n\kappa }$.

The printed final exponent has the opposite sign while retaining the definition of $\gamma$ as the limit of $n^{-1}\log {\beta }_n$. With that definition it cannot be correct in general. Indeed, if $(2d-1)p<1$, count self-avoiding paths of length $\ell \ge n$ to getHence $\gamma \le \log ((2d-1)p)<0$. A lower bound with $e^{-n\gamma }$ would eventually exceed one, because the boundary factor grows only polynomially. The valid bound is the boxed expression with $e^{n\gamma }$, or the equivalent expression using $\kappa =-\gamma$.

For independent bond percolation on ${\mathbb{Z}}^d$, $d\ge 2$, there is almost surely at most one infinite percolation cluster at every parameter. This is the Burton-Keane theorem. It does not require an assumption about the behavior at the critical probability. If the probability that the origin lies in an infinite percolation cluster is positive, translation ergodicity of Bernoulli percolation also gives almost-sure existence and hence exactly one.

The two structural ingredients are translation invariance and finite-energy property of Bernoulli percolation. For $0<p<1$, every prescribed configuration of finitely many bonds has positive conditional probability given all the other bonds. In particular, opening or closing a fixed finite set of bonds transforms events of positive probability into events of positive probability. The product measure is ergodic under translations. One way to see this is to approximate a translation-invariant event by a finite-cylinder event, translate the cylinder sufficiently far to make its coordinates disjoint, and use independence. Sending the approximation error to zero gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.

Let $K$ be the number of infinite percolation clusters. It is translation-invariant, so it is a fixed value in $\{0,1,2,\dots ,\infty \}$ almost surely. First exclude a finite value $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Force every internal bond in that box to be open. This joins those clusters and can only merge other clusters; it cannot split an infinite percolation cluster. It also cannot create a new infinite percolation cluster out of finite ones, since only finitely many clusters touch the modified region. Thus the number of infinite percolation clusters decreases. finite-energy property of Bernoulli percolation makes the modified event have positive probability, contradicting its almost-sure value $k$.

Suppose next that $K=\infty$. Some finite box meets three different infinite percolation clusters with positive probability. Each such cluster has an infinite component outside the box: deletion of a finite set in a locally finite graph leaves only finitely many components adjacent to that set, so an infinite percolation cluster cannot have all of them finite. Choose one infinite exterior component from each of the three clusters, and one bond connecting it to the box.

Inside the box, connect the three selected exterior endpoints by a finite tree, preserving just their three selected entrance bonds. Take the minimal subtree connecting these endpoints; they are leaves, since their only retained connection into the box is their chosen entrance bond. This tree has a branch graph vertex of degree three. Close every other internal or boundary bond of the box. Its three exterior components remain distinct outside, since they belonged to different original clusters. The branch graph vertex now has exactly three open incident bonds, and deleting it separates its cluster into three infinite components. Such a graph vertex is a trifurcation vertex in percolation. There are finitely many choices for the endpoints and tree within the chosen region, so at least one fixed finite modification of Bernoulli percolation has positive probability. finite-energy property of Bernoulli percolation consequently gives positive probability of a trifurcation at some fixed graph vertex. Translation invariance makes this probability the same number $q>0$ at every graph vertex.

The decisive fact is that a finite box cannot contain a positive-volume density of these branching points. Here is the boundary count, including its combinatorial justification. For a finite graph vertex set $W$, let ${\partial }^{+}W$ be the exterior graph vertices adjacent to it. In each infinite percolation cluster meeting $W$, contract each of its components outside $W$ that is adjacent to $W$ to a terminal graph vertex. The quotient graph consisting of the cluster's graph vertices in $W$ and these terminals is finite and connected. A trifurcation graph vertex in $W$ separates its terminals into at least three nonempty groups, since each of its infinite branches must leave $W$.

Take a minimal subtree of the quotient graph connecting all the terminals. Every such trifurcation graph vertex must belong to that subtree and have degree at least three: otherwise the terminals in one of its separated groups could not connect to the other groups. All leaves of the minimal subtree are terminals. A cluster with only one terminal has no counted trifurcation. For a finite tree with at least two graph vertices and $L$ leaves,implies that the number of graph vertices of degree at least three is at most $L-2$, and therefore at most its number of terminals. Distinct exterior components can be assigned distinct graph vertices of ${\partial }^{+}W$. Terminals from different clusters are disjoint as well. Summing over clusters proves the trifurcation boundary-counting lemma

Take $W=[-n,n{]}^d\cap {\mathbb{Z}}^d$. The expected left side is $q|W|$, of order $q{n}^d$, while the right side is $O_d(n^{d-1})$. Division by $|W|$ and passage to infinity force $q=0$, contradicting the positive probability obtained by finite modification of Bernoulli percolation. Thus infinitely many infinite percolation clusters are impossible. The surface-to-volume comparison is the geometric reason for uniqueness on the lattice; the same argument is tied to amenability on more general graphs.

At $p=0$ there is no infinite percolation cluster, and at $p=1$ the whole lattice is the unique infinite percolation cluster. This completes all parameter cases. Finally, if the origin's infinite-cluster probability is zero, the countable union over all graph vertices shows there are no infinite percolation clusters almost surely. If it is positive, the translation-invariant existence event has positive probability and hence probability one. Together with uniqueness this distinguishes absence from the case of a single infinite percolation cluster, without making an unsupported assertion that every graph vertex belongs to it.

Use recovery rate one and infection rate $\lambda$ along each directed nearest-neighbour edge. The graphical representation of the contact process places independent rate-one recovery marks at graph vertices and rate-$\lambda$ arrows on directed edges. Infection travels along time-directed paths following arrows and avoiding recovery marks. The same clocks couple all initial configurations and preserve coordinatewise order. Denote its semigroup by $P_t$.

The empty configuration is absorbing, so the lower invariant measure is $\underline{\nu }={\delta }_0$. To construct the upper one, put ${\mu }_t={\delta }_1{P}_t$, where $1$ is the all-infected configuration. Since ${\delta }_1{P}_s{\le }_{\mathrm{st}}{\delta }_1$, order preservation and the semigroup property implyThus these laws decrease in stochastic order. More concretely, use clocks on the entire time axis and start from all infected at time $-t$, looking at the configuration at time zero. These configurations decrease pointwise as $t$ increases: an infection arriving from an earlier start is also obtainable from the all-infected later start. Each coordinate therefore has a limit. Their joint law $\overline{\nu }$ satisfies ${\mu }_t\to \overline{\nu }$ in weak convergence of probability measures on the compact product space.

This limit is invariant. The graphical construction is Feller: for finitely many sites and a bounded time interval, the backward ancestor exploration is almost surely finite, since its size is dominated by a finite-rate branching process. Hence a cylinder function evolved for a fixed time is continuous in the initial configuration, by approximation with finite ancestor sets. Cylinder functions uniformly approximate continuous functions on the product space. We may therefore pass to the limit into get $\overline{\nu }{P}_s=\overline{\nu }$. These are indeed the extremal invariant measures of the contact process: if $\nu$ is any invariant probability law of a Markov process, apply $P_t$ to ${\delta }_0{\le }_{\mathrm{st}}\nu {\le }_{\mathrm{st}}{\delta }_1$ and take limits to obtain $\underline{\nu }{\le }_{\mathrm{st}}\nu {\le }_{\mathrm{st}}\overline{\nu }$.

For the relation with survival, reverse the graphical paths in the interval $[0,t]$, also reversing arrows. The law is unchanged because the rates are symmetric between neighbours. The set of ancestors of a site $x$ has the law of the contact process started from $\{x\}$. ConsequentlyThis is duality of the contact process. The events on the right decrease with time because the empty state is absorbing, and their limiting probability is $\theta (\lambda )$. It follows thatIf $\theta (\lambda )=0$, every coordinate is zero almost surely under $\overline{\nu }$; a countable union over the sites gives $\overline{\nu }={\delta }_0=\underline{\nu }$. Conversely, if the two invariant measures coincide, this one-site marginal is zero, so $\theta (\lambda )=0$. HenceA convention that divides the infection rate by the degree only rescales the parameter; the argument and equivalence are unchanged.

View the oriented square lattice as $(x,n)$ with $x+n$ even, and arrows $(x,n)\to (x\pm 1,n+1)$. Each graph vertex has two incoming bonds. Start with independent oriented bond percolation of parameter $p$, and declare a site open when at least one of its incoming bonds is open. Its probability of being open isIncoming bond sets for different heads are disjoint, so these site indicators are independent: they are precisely an oriented site percolation configuration with parameter $r$.

Every graph vertex after the first on an open oriented bond path is open as a site, because its incoming path bond is open. The root's two incoming bonds belong to earlier time layers and are independent of all bonds used by a forward path from the root. Thus the event that the root site is open, of probability $r$, is independent of bond survival from that root. On their intersection the entire infinite bond path is also an infinite open site path. Hence, writing ${\theta }_b,{\theta }_s$ for the two survival probabilities,This is the incoming-edge coupling of site and bond percolation. If site survival is defined conditional on an open root, the extra factor is omitted; the critical probability is the same. If the model is drawn only in a time half-plane, one can independently sample the root's incoming bonds as auxiliary variables.

For every $p>p_c(\mathrm{bond})$, bond survival is positive by monotonicity and the definition of the critical probability. Therefore $p_c(\mathrm{site})\le 1-(1-p{)}^2$. Let $p$ decrease to the bond critical probability and use continuity of this polynomial, obtainingIf the bond critical probability equals one, the inequality is trivial; no assertion about survival exactly at criticality is used.

The previous part and the permitted assumption $p_c(\mathrm{bond})<1$ imply $p_c(\mathrm{site})<1$. We build an independent oriented site percolation process whose open paths carry contact-process infections, and whose site probability tends to one as $\lambda$ increases.

Use the graphical representation of the contact process with recovery rate one and per-neighbour arrow rate $\lambda$. Fix $\delta >0$. On the parity lattice $(x,n)$ with $n\ge 0$ and $x+n$ even, declare $(x,n)$ good if there is no recovery mark at $x$ during $[(n-1)\delta ,(n+1)\delta ]$, and there is at least one arrow from $x$ to each of $x-1,x+1$ during $[n\delta ,(n+1)\delta ]$. Extend the Poisson clocks to negative times solely to define the root event; the process itself still starts at time zero. Goodness has probability

The independence here is important. At a fixed spatial site $x$, the allowed time indices differ by two. Its two-unit recovery intervals therefore have disjoint interiors, and its outgoing-arrow intervals are disjoint. At different spatial sites, recovery clocks are different and outgoing arrows have different tails, hence use different directed-edge clocks. Every defining random input belongs to just one parity site's event, up to deterministic interval endpoints at which a Poisson mark has probability zero. Thus the good-site indicators really are independent, with common probability $q$; no theorem about dependent percolation is needed.

Suppose $(x,n)$ and $(x\pm 1,n+1)$ are both good and $x$ is infected at time $n\delta$. The source has no recovery before $(n+1)\delta$, and it sends an arrow to the chosen neighbour during that interval. The target's good event excludes recovery throughout $[n\delta ,(n+2)\delta ]$, so the target is infected at time $(n+1)\delta$. Induction shows that any infinite oriented path of good sites starting at $(0,0)$ sustains infection at every time layer. Along each transmitting interval there is always an infected site, so the contact process never reaches its empty absorbing state.

Take $\delta ={\lambda }^{-1/2}$. ThenFor some finite $\lambda$ it exceeds $p_c(\mathrm{site})$, and the good-site process has positive survival probability. The contact process begun from the origin therefore also survives with positive probability. HenceThis is the independent oriented-percolation comparison for the contact process. If the infection convention assigns total rate $\lambda$ rather than rate $\lambda$ to each of the two neighbours, replace the arrow rate by $\lambda /2$; the same limit proves the same finiteness conclusion.

Order $\Omega =\{0,1{\}}^E$ coordinatewise. The statement ${\mu }_1{\ge }_{\mathrm{st}}{\mu }_2$ meansfor every real-valued order-preserving function $f$; equivalently, ${\mu }_1(A)\ge {\mu }_2(A)$ for every increasing event $A$. This is stochastic domination of probability measures. For strictly positive laws, the Holley condition is the sufficient conditionHere join and meet are coordinatewise maximum and minimum. The Holley inequality states that this condition implies the stated stochastic domination.

For a single positive law, the FKG lattice condition isWe prove that it gives positive association of random variables, namely ${\mathrm{Cov}}_{\mu }(f,g)\ge 0$ for all increasing $f,g$. Fix such a $g$, and for $t\ge 0$ form the strictly positive tilted lawThe lattice condition and $g(\omega \vee \eta )\ge g(\omega )$ implyThus the Holley condition applies to ${\mu }_t,\mu$, and ${\mathbb{E}}_{{\mu }_t}f\ge {\mathbb{E}}_{\mu }f$. Equality holds at $t=0$, so the right derivative there is nonnegative. Differentiating the finite sums givesHencewhich proves the requested positive association of random variables. Neither $f$ nor $g$ needs to be nonnegative; the finite space makes all differentiations justified. This is the exponential-tilt proof of positive association.

The law is a ferromagnetic Ising model with unit couplings and no field. Its masses are strictly positive. Put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, so $\pi (\sigma )=Z^{-1}{e}^{H(\sigma )}$. The order is coordinatewise with $-1<1$, which identifies the configuration space with a Boolean lattice.

Use the permitted reduction to pairs that disagree at no more than two graph vertices. If such a pair is comparable, its join and meet are the two original configurations, so the lattice inequality is equality. In the incomparable case, let the two differing graph vertices be $x,y$, with one configuration taking $(1,-1)$ there and the other $(-1,1)$. Their join takes $(1,1)$ and their meet $(-1,-1)$.

Every edge not joining $x$ to $y$ cancels ina fixed edge has the same contribution throughout, while an edge from a changed graph vertex to a fixed graph vertex has the same sum of its two endpoint-spin products before and after. An edge between $x$ and $y$ contributes $1+1-(-1)-(-1)=4$. Thus the difference is nonnegative, whether or not the two changed graph vertices are adjacent. Exponentiation, with the normalizing constants cancelling, proves the FKG lattice condition for these pairs, and hence for all pairs by the given reduction.

Equivalently one can see the full ferromagnetic Ising lattice inequality directly. Let $U=\{x:{\sigma }_x=1,{\tau }_x=-1\}$ and $W=\{x:{\sigma }_x=-1,{\tau }_x=1\}$. Exactly the edges between $U$ and $W$ contribute, each by four. ThereforeThe preceding part now givesso the measure is positively associated. The sign of the coupling is essential to this argument: positive couplings favor agreement and give the nonnegative lattice difference.

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 the given exchangeable law $\mu$, putwhich is well-defined by the stated assumption. Integrate the product self-duality of symmetric exclusion against $\mu$, and use boundedness to interchange expectation and integration:The finite-particle process conserves its cardinality. Hence the right side is ${\mathbb{E}}^A{m}_{|A_t|}=m_{|A|}$. All occupation-product moments are therefore unchanged by evolution.

These moments determine the law, not merely its one-site marginals. For disjoint finite sets $B,C$, the inclusion-exclusion principle givesEvery term is unchanged, so every finite cylinder probability is unchanged. Such cylinder events generate the product sigma-algebra on $\{0,1{\}}^{\mathbb{Z}}$, and probability laws agreeing on them agree everywhere. ConsequentlyThus every exchangeable probability law is invariant for the symmetric exclusion process. This proof of exchangeable invariant laws of symmetric exclusion needs only the given cardinality dependence and duality; no representation theorem for exchangeable measures is required.