A Stationary Planar Random Graph with
Singular Stationary Dual: Dyadic Lattice Graphs

The graph $\Gamma_{\circ}$ has an automorphism $\phi$ that takes a binary string $w$ of length $n$ to the binary string $2^{n}-w$ of equal length. This may be seen as a reflection in Figure 2.3, where it fixes each string of the form $000\dots 0$ and $100\dots 0$, and interchanges the two horizontal intervals between these fixed points. Equivalently, moves to the left and moves to the right are swapped. This is also a reflection in the real axis in Figure 2.4.

This notion of interchanging the notions of right and left is evident in the action of $\phi$ at any fixed depth. The depth $n$ portion of $\Gamma_{\circ}$ may be seen as the Cayley graph of the additive group of integers modulo $2^{n}$ with generators $\pm 1$. There is a unique nontrivial automorphism $\phi_{n}$ of this graph that preserves the identity, defined by writing an arbitrary element $w$ as a sum of generators and reversing the sign of each of these generators, producing $\phi_{n}(w)=-w$. This operation of reversing the sign exchanges the roles of the generators $+1$ and $-1$, or equivalently of right and left.

Any automorphism of $\Gamma_{\circ}$ must fix $\varnothing$, which is the sole vertex with a double loop, must fix $0$, which is the only vertex connected to $\varnothing$, and must fix $1$, which is the only vertex connected to $0$ by two edges. Finally, it must fix the only other neighbours of $0$ and $1$, which are $00$ and $10$. There are only two vertices connected to both $00$ and $10$, namely, $01$ and $11$. Therefore any automorphism of $\Gamma_{\circ}$ either fixes these two vertices or interchanges them.

To see that $\Gamma_{\circ}$ has no nontrivial automorphisms other than $\phi$, it suffices to show that an automorphism of $\Gamma_{\circ}$ that fixes the vertices $01$ and $10$ must fix the entire graph, as automorphisms that swap $01$ and $11$ may be composed with $\phi$.

If an automorphism of $\Gamma_{\circ}$ fixes each $k$-bit string for some $k\geq 2$, then it fixes each $(k+1)$-bit string that ends with a zero, and then also fixes each $(k+1)$-bit string that ends with a one, because each string ending with a one is adjacent to different pairs of strings ending with zeros, as long as $k\geq 2$. This completes the proof. ∎

Any horizontal edge is between a vertex of degree $3$ and a vertex of degree $4$, so an edge with two degree-$4$ endpoints must be vertical.

Consider an edge $e$ between a vertex $x$ of degree $3$ and a vertex $y$ of degree $4$. If $y$ is adjacent to two vertices of degree $4$, then $e$ is a horizontal edge. Otherwise, let $z$ be the unique vertex of degree $4$ adjacent to $y$. If a shortest path between $x$ and $z$ that does not go through $y$ has length $3$, then $e$ is horizontal. Otherwise such a path has length $7$ and $e$ is vertical. Examples of these paths are shown in red and green, respectively, in Figure 2.6. ∎

Let $e$ be a vertical edge between two vertices $x$ and $y$. There are exactly two paths between these two vertices that take one horizontal step, then one vertical step, then two horizontal steps. The initial vertex of these paths is above the final vertex. ∎

If there is an automorphism that takes $(m,b)$ to $(n,c)$ with $m\neq n$ or $b\neq c$, then by Proposition 2.7, either $b=c$ or $b=-c$ (as dyadic integers). Since $(m,b)$ and $(n,c)$ belong to the same connected component of $\mathbf{\Gamma}$, it follows from Remark 2.2 that $b$ is eventually periodic, whence so is $a$.

Conversely, if $a$ is eventually periodic, then $\Gamma_{a}$ has a vertex $(m,b)$ where $b$ is periodic. Thus, without loss of generality, assume that $a$ itself is periodic. Choose $m<0$ so that $(a_{m+k})_{k\leq 0}=a$. Then there is an orientation-preserving automorphism that takes $(0,a)$ to $(m,a)$, namely, $(n,b)\mapsto(n+m,b)$. ∎

By considering all possible sequences of two steps from the vertex $X_{2t}$, we calculate that

Thus, $\operatorname{\mathbf{E}\mathopen{}}[D_{t+1}-D_{t}\mid X_{2t}]\geq 1/6$ in all cases. Since the random variables $Z_{t}:=D_{t+1}-D_{t}-\operatorname{\mathbf{E}\mathopen{}}[D_{t+1}-D_{t}\mid X_{2t}]$ are uncorrelated and take values in $[-2,2]$ for $t\geq 0$, the strong law of large numbers yields that their average tends to 0 a.s.; see, e.g., [14, Theorem 13.1]. This gives the first result. Moreover, $Z_{t}$ are martingale differences, whence the Hoeffding–Azuma inequality yields

Note that each level is reached for the first time at a vertex of degree 4, except possibly for the starting level. The portion $\Gamma_{+}$ of the graph at the levels equal to or greater than that of a given vertex and rooted at that vertex does not depend on that vertex or on which graph in $\mathbf{\Gamma}$ the walk takes place, in the sense that it is the same up to rooted isomorphism, so $T_{k+1}-T_{k}$ are IID in $k\geq D_{0}$ and have distribution independent of $D_{0}$. Likewise, for the random walk on $\Gamma_{\circ}$, the random variables $T_{k+1}-T_{k}$ are IID in $k\geq D_{0}$; in addition, a walk on $\Gamma_{+}$ projects to a walk on $\Gamma_{\circ}$ in a way that preserves changes in levels, whence the distribution of $T_{k+1}-T_{k}$ is the same on $\mathbf{\Gamma}$ as on $\Gamma_{\circ}$. Finally, let $F_{n}$ be the first time the random walk is at level $n$. For $u\geq 1$,

and

whence exponential tail bounds are provided by Proposition 3.3. ∎

Because $\Gamma_{\circ}$ is a quotient of $\Gamma_{+}$, it suffices to prove that $\nu_{+}$ is well defined in order to prove that $\nu_{\mathrm{h}}$ is well defined. For $n\geq 0$, let $Z_{n}$ be the number of times that the random walk is at level $n$ and $H_{n}$ be the total (signed) change in horizontal position while at level $n$. At most $Z_{n}$ sideways steps are taken on level $n$, and each of these changes the horizontal position by $1/2^{n}$, so $|H_{n}|\leq Z_{n}/2^{n}$. By Proposition 3.2, $s:=\operatorname{\mathbf{E}\mathopen{}}[T_{n+1}-T_{n}]\vee\operatorname{\mathbf{E}\mathopen{}}[T_{0}]<\infty$, so $\operatorname{\mathbf{E}\mathopen{}}[T_{n}]\leq s(n+1)$. Because $Z_{n}\leq T_{n}$, it follows that $\operatorname{\mathbf{E}\mathopen{}}\mkern-1.5mu\bigl{[}\sum_{n}Z_{n}/2^{n}\bigr{]}\leq\sum_{n}s(n+1)/2^{n}<\infty$. Therefore $\sum_{n}H_{n}$ converges a.s., which is the result. Because the tails of $\sum_{n}H_{n}$ are arbitrarily small, the rest of the proposition follows. ∎

It suffices to prove the second statement. Fix $x\in\mathbb{R}$. Define $S_{n}$ to be the first time after $T_{n}$ that the walk makes a horizontal step $s_{n}=\pm 1$, and let $L_{n}$ be the level at which the walk makes that step. Because each $T_{n}$ is a.s. finite, we may choose an increasing sequence $(n_{k})_{k\geq 1}$ so that for each $k$, we have $\operatorname{\mathbf{P}\mathopen{}}[S_{n_{k}}<T_{n_{k+1}}]\geq\nu_{+}(x)/2$. Let $A$ be the set of paths that tend to $x$, so that $\operatorname{\mathbf{P}\mathopen{}}(A)=\nu_{+}(x)$. Let $A_{k}$ be the set of paths in $A$ for which $S_{n_{k}}<T_{n_{k+1}}$, so that $\operatorname{\mathbf{P}\mathopen{}}(A_{k})\geq\operatorname{\mathbf{P}\mathopen{}}(A)/2$. Write $A^{\prime}_{k}$ for the sequences obtained from $A_{k}$ by changing the step $s_{n_{k}}$ to $-s_{n_{k}}$; all sequences in $A^{\prime}_{k}$ still correspond to paths in $\Gamma_{+}$, and $\operatorname{\mathbf{P}\mathopen{}}(A^{\prime}_{k})=\operatorname{\mathbf{P}\mathopen{}}(A_{k})$. However, paths in $A^{\prime}_{k}$ tend to $x-s_{n_{k}}/2^{L_{n_{k}}-1}$. Note that this limit depends on $L_{n_{k}}$, which is not the same for all paths in $A^{\prime}_{k}$. However, by the definition of the sequence $(n_{k})$, paths in $A^{\prime}_{j}$ and $A^{\prime}_{l}$ cannot have the same limit for $j\neq l$. Therefore the sets $A^{\prime}_{k}$ ($k\geq 1$) have disjoint sets of limits in $\mathbb{R}$, yet each of these sets has probability at least $\nu_{+}(x)/2$. Therefore, $\nu_{+}(x)=0$, as desired. ∎

The key idea is to consider a random walk on $\Gamma_{a}$ as developing a longer and longer past history, which converges to the stationary measure. Every time the random walk leaves a level for the last time, the future moves of the random walk are independent of the graph above that level, and therefore have the same distribution no matter what $a$ is. The existence of such regeneration times shows that there is at most one stationary measure. Furthermore, we can construct a stationary measure from them as follows. We break up the walk into segments between the last times it leaves successive levels. These segments are IID, and have finite expected length by Proposition 3.2. The usual length-biasing and uniform choice then yields a stationary process of random walk moves. These random walk moves must still be converted to dyadic integers. Whether the walk is presently at a vertex of degree $3$ or degree $4$ depends only on the most recent (partial) segment. If it is at a vertex of degree $4$, then the degree of the vertex immediately above depends only on the most recent two segments, and so on. Considering increasing numbers of these segments will allow us to obtain the stationary measure as a limit. Lastly, the IID segments of random walk moves have distribution that is invariant under switching left and right moves; this leads to continuity of $\nu_{\mathrm{s}}$.

We now begin the proof. Consider simple random walk $(X_{t})_{t\geq 0}$ on $\Gamma_{a}$ starting at $(0,a)$. Each transition is a move on $\Gamma_{a}$ that is either left, right, up, or down from the current position; code this as a symbol $M_{t}\in\{\mathrm{L},\mathrm{R},\mathrm{U},\mathrm{D}\}$, where the walk moves from $X_{t-1}$ to $X_{t}$ via $M_{t}$. The sequences $(M_{T_{n}+t})_{1\leq t\leq T_{n+1}-T_{n}}$ are IID for $n\geq 0$. (Here, $M_{T_{n}+1}=\mathrm{D}$ for all $n$.) Let $\nu$ denote their common law. Because $\operatorname{\mathbf{E}\mathopen{}}[T_{n+1}-T_{n}]<\infty$, we may define $\mu_{1}$ to be the distribution on $\mathbb{Z}^{+}$ defined by

and then $\mu_{2}$ to be the law of $V$ when $V$ is uniformly chosen from $\{0,1,2,\ldots,W-1\}$ and $W$ has law $\mu_{1}$. Let $\nu_{0}$ denote the law of $(M_{T_{0}+t})_{1\leq t\leq V}$, where $V$ is independent of the walk and has law $\mu_{2}$. Consider now sequences of $\{\mathrm{L},\mathrm{R},\mathrm{U},\mathrm{D}\}$ indexed by the nonpositive integers. Then renewal theory shows that the finite-dimensional distributions of $(M_{t+s})_{-t\leq s\leq 0}$ tend as $t\to\infty$ to those of the concatenation of $S_{n}$ ($n\leq 0$), where $S_{n}$ are independent for $n\leq 0$ with law $\nu$ when $n<0$ and with law $\nu_{0}$ when $n=0$. In particular, this is independent of $a$. Indeed, the graph $\Gamma_{a}$ has cycles of length $5$, whence the distribution of $T_{1}-T_{0}$ is nonlattice. Therefore, the discrete key renewal theorem shows that for the delayed renewal process $(T_{n})_{n\geq 0}$ with renewals at $T_{n}$, the age distribution tends to $\nu_{0}$. This gives convergence to the equilibrium renewal process; e.g., [13, Theorem 3.1]. Usually one looks into the future, but one can just as well look into the past, as we do here. Given this, the fact that the sequences $(M_{T_{n}+t})_{1\leq t\leq T_{n+1}-T_{n}}$ are IID yields our claim.

We next convert these random walk moves to dyadic integers. Each finite sequence $\sigma$ from $\{\mathrm{L},\mathrm{R},\mathrm{U},\mathrm{D}\}$ corresponds to a function $f_{\sigma}\colon\mathbb{Z}_{2}\to\mathbb{Z}_{2}$ by composing the individual-symbol functions $f_{\mathrm{L}}(b):=b-1$, $f_{\mathrm{R}}(b):=b+1$, $f_{\mathrm{U}}(b):=b/2$, $f_{\mathrm{D}}(b):=2b$, provided $f_{\mathrm{U}}$ is applied only to even dyadic integers. Here, for a sequence $\sigma=(m_{1},m_{2},\ldots,m_{k})$, we compose in the order $f_{\sigma}:=f_{m_{k}}\circ\cdots\circ f_{m_{2}}\circ f_{m_{1}}$. Say that a finite sequence $\sigma$ is ‘definable’ if $f_{\sigma}$ satisfies that restriction when applied to every dyadic integer and is ‘permissible’ if for every nonempty initial segment of $\sigma$, the number of symbols $\mathrm{D}$ is strictly larger than the number of symbols $\mathrm{U}$; in particular, the first symbol of each permissible $\sigma$ is $\mathrm{D}$. The sequences $S_{n}$ above are guaranteed to be both definable and permissible. Furthermore, the value of $f_{\sigma}(b)$ mod 2 does not depend on $b$ for definable and permissible $\sigma$. More generally, $f_{\sigma_{1}}\circ f_{\sigma_{2}}\circ\cdots\circ f_{\sigma_{k}}(b)$ mod $2^{k}$ does not depend on $b$ for any definable and permissible sequences $\sigma_{1},\ldots,\sigma_{k}$. Therefore, $f_{S_{0}}\circ f_{S_{-1}}\circ\cdots\circ f_{S_{n}}(b)$ has a limit a.s. as $n\to-\infty$ and does not depend on $b$; its law is $\nu_{\mathrm{s}}$.

Finally, for $n<0$, the moves in $S_{n}$ result in a net change to the depth of $1$ and in a net horizontal change compared to the location of $M_{T_{n}+1}$. That is, we may write $\bar{X}_{T_{n+1}}=2\bar{X}_{T_{n}}+H_{n}$ for some IID $\mathbb{Z}$-valued random variables $H_{n}$, where $X_{t}=\bigl{(}0pt(X_{t}),\bar{X}_{t}\bigr{)}$ with $0pt(X_{t})$ the depth of $X_{t}$. The law of $\nu$ is invariant under switching $\mathrm{L}$ and $\mathrm{R}$, whence $-H_{n}$ has the same law as $H_{n}$. The law $\nu^{*}$ of the limit of $f_{S_{-1}}\circ f_{S_{-2}}\circ\cdots\circ f_{S_{n}}(b)$ is that of $\sum_{n<0}2^{1-n}H_{n}$, a series that converges in the $2$-adic metric by the estimates in the proof of Proposition 3.7. Now an argument very similar to that proving Lemma 3.9 shows that the law of $\nu^{*}$ is continuous, whence so is $\nu_{\mathrm{s}}$. ∎

There are only countably many finite binary strings, whence there are only countably many eventually periodic strings. Thus, the result is immediate from Theorem 3.10. ∎

Corollary 3.11 says that almost no $a$ are eventually periodic, so for $\nu_{\mathrm{s}}$-a.e. $a\in\mathbb{Z}_{2}$, the map $0pt_{a}(b):=m$ for vertices $(m,b)$ of $\Gamma_{a}$ is well defined in view of Proposition 2.10. Consider the stationary simple random walk $(X_{t})_{t\geq 0}$ on $\mathbf{\Gamma}_{{\bullet}}$. If $X_{0}=a$, then for each $t\geq 0$, there is some $m\in\mathbb{Z}$ such that $(m,X_{t})$ is a vertex of $\Gamma_{a}$; with the preceding notation, we may write $m=0pt_{a}(X_{t})=0pt_{X_{0}}(X_{t})$. Given $X_{0}=a$, we may ($\nu_{\mathrm{s}}$-a.s.) identify the walk $(X_{t})_{t\geq 0}$ with simple random walk on $\Gamma_{a}$ starting at its root, $(0,a)$. Now consider the standard two-sided stationary extension $(X_{t})_{t\in\mathbb{Z}}$ of the stationary simple random walk on $\mathbf{\Gamma}_{{\bullet}}$. Thus, $(X_{t+t_{0}})_{t\geq 0}$ has the same law as $(X_{t})_{t\geq 0}$ for every $t_{0}\in\mathbb{Z}$. Because the walk on $\Gamma_{X_{0}}$ drifts downward, this two-sided extension almost surely has the property that for every $m\in\mathbb{Z}$, the number of $t\in\mathbb{Z}$ with $0pt_{X_{0}}(X_{t})=m$ is finite. Let $(Z_{1},Z_{2},\ldots,Z_{N})$ be the list of times $t\in\mathbb{Z}$ with $0pt_{X_{0}}(X_{t})=0pt_{X_{0}}(X_{0})=0$, listed in increasing order. Thus, $Z_{N}=T_{0}$, with $T_{n}$ the leaving times of Definition 3.1.

Write $B_{k}(a):=(a_{-k-\ell+1},a_{-k-\ell+2},\ldots,a_{-k-1},a_{-k})$. Lemmas 3.8 and 3.9, together with stationarity, imply that

Because one of the times $Z_{1},\ldots,Z_{N}$ is 0, we obtain

Let $A_{k,n}$ be the event that $B_{k}(X_{T_{n}})=\sigma$; using the definition of $A_{k,0}$ and the fact given earlier that $Z_{N}=T_{0}$, the preceding equation can be written as

Let $L$ be the set of random walk trajectories where the level at time $0$ is never visited again. The law of $(X_{t})_{t\in\mathbb{Z}}$ is invariant and ergodic under the left shift because $\nu_{\mathrm{s}}$ is invariant and ergodic, and the leaving times $T_{n}$ correspond to shifts that bring the trajectory to $L$. Because the return map to $L$ is also measure-preserving and ergodic, it follows that the sequence $(\mathbf{1}_{A_{k,n}})_{n\in\mathbb{Z}}$ is stationary and ergodic. Therefore, for each $k\geq 0$, the ergodic theorem yields

On the other hand, Proposition 3.7 shows that $\mathbf{1}_{A_{n-\ell,n}}$ converges a.s. as $n\to\infty$ with $\lim_{n\to\infty}\operatorname{\mathbf{P}\mathopen{}}(A_{n-\ell,n})=\nu_{\mathrm{h}}(\sigma)$; since by stationarity, $\operatorname{\mathbf{P}\mathopen{}}(A_{k,n})$ is the same for all $n$, we obtain

Since $(\mathbf{1}_{A_{n-\ell,n}})_{n\geq 0}$ is a Cauchy sequence a.s., we have

By stationarity, $\operatorname{\mathbf{P}\mathopen{}}(A_{n-\ell,n+r}\mathbin{\triangle}A_{m-\ell,m+r})$ does not depend on $r$. Therefore,

Choosing $n=k+\ell$, $\>m=k+\ell+s$, and $r=-k-\ell-s$ yields

Combining (3.14), (3.15), and (3.16), we obtain

which is the same as

In view of (3.13), we may write this as

This is the desired result. ∎

Consider a simple random walk on $\mathbf{\Gamma}$ or $\Gamma_{\circ}$ that starts at a vertex at depth $n$. This random walk may be broken up into the segments between the leaving times $T_{k}$ and $T_{k+1}$ for $k\geq n$, in addition to the initial segment up to time $T_{n}$. By Proposition 3.2, the expected time in each such interval spent at vertices of degree $3$ is bounded, and the times spent at vertices of degree $3$ in each of these intervals are IID, except for the initial segment. Therefore the limit $p_{3}$ exists and is equal to the ratio of the average time spent at vertices at degree $3$ between times $T_{k}$ and $T_{k+1}$ to the average interval length $T_{k+1}-T_{k}$. ∎