Proof. This lemma follows immediately from the definition of $\operatorname {Div}^0(G)$ and the fact that G is connected.

Proof. We will prove the lemma by contradiction. The second equality follows from Definition 2.10. For the first equality, suppose that for some vertex $v \in V(G)$ , the rotor at v moves past $T^{\prime }_s{\langle } v{\rangle }$ on $(G,\widetilde \chi )$ (otherwise the proof works analogously after switching the roles of $(G,\chi )$ and $(G,\widetilde \chi )$ ). Consider the first instance where such an event occurs. At this moment, the chip must have reached v more times than it did during rotor-routing on $(G,\chi )$ . Because no edge is crossed more than once in each direction (by Proposition 3.3), the chip must enter v along an edge that it did not cross during rotor-routing on $(G,\chi )$ . It follows that the rotor on some vertex y that is adjacent to v must have already passed $T^{\prime }_s {\langle } y {\rangle }$ . This contradicts our minimality assumption.

Proof. Let x be a cut vertex. We say x separates two edges or vertices g and h if all paths between g and h must contain x. We claim that there is some $D \in \operatorname {Div}^0_x(G)$ such that $[D] = [c-s]$ and $D = \sum _{v \in V} (v - x)$ , where V is a multiset of vertices which x does not separate from f. Assume that the claim holds. Then, by Definition 2.10, it follows that

$$\begin{align*}r_{(G,\chi)}([c-s],T) = \widehat r_{(G,\chi,x)}(D,T).\end{align*}$$

When rotor-routing with sink x, all chips halt upon reaching x. As a result, any rotor that x separates from f will not rotate during the implementation of Algorithm 1. It follows that for any edge e that is separated from f, we have

$$\begin{align*}e \in T \Longleftrightarrow e \in \widehat r_{(G,\chi,x)}(D,T) \Longleftrightarrow e \in r_{(G,\chi)}([c-s],T).\end{align*}$$

To prove the claim, we use a modified version of the proof of Lemma 2.3. If $s = x$ , we can take $D = c-s$ . Otherwise, let $V_1 \subseteq V(G \setminus x)$ be the set of vertices in the same component as s in $G \setminus x$ , and let $V_2 = V(G \setminus x) \setminus V_1$ . Let $\delta _x \in \operatorname {Div}_x^0(G)$ be the divisor which places $\deg (v)$ chips on every $v \in V_1$ , 0 chips on every $w \in V_2$ , and $-\sum _{v \in V_1} \deg (v)$ chips on x. Define $\delta _x^{\circ }$ to be the divisor we obtain after repeatedly firing any vertex v with at least $\deg (v)$ many chips. Notice that $\delta _x - \delta _x^{\circ }$ has a positive value on each $v \in V_1$ and is equal to 0 on each $w \in V_2$ . Furthermore, $[\delta _x - \delta _x^{\circ }]= 0$ . We prove the claim by letting $D = c - s + \delta _x - \delta _x^{\circ }$ .

Question 4.5. Is Condition 3 of Definition 4.3 implied by Conditions 1 and 2?

Proof. We begin with a paragraph giving the outline of our argument. We reduce the result to six specific cases. If an edge e is in both T and $T' := r_{(G,\chi )}([c-s],T)$ , then the rotor at e may be untouched during rotor-routing, it may complete a full rotation or it could be oriented in opposite directions in $T_s$ and $T^{\prime }_s$ (note that by Proposition 3.3, the rotor will never move more than a full rotation). When the edge e is in neither tree, the traveling vertex may never have crossed e, it may have crossed e in both directions or it may have only crossed e in one direction. For some of these cases, the path of the traveling vertex will change after deleting or contracting e, but we use Lemma 3.6 to show that it still crosses the same edges, just in a different order. The planarity condition comes into play for the final case (where the edge is crossed in one direction). We apply Lemma 3.5 to show that this case is never realized.

Throughout this proof, we let ${\mathfrak i}(e) = \{x,y\}$ and $T' = r_{(G,\chi )}([c-s],T)$ . Let $\mathcal R$ be the ordered list of directed edges crossed when rotor-routing on $(G,\chi )$ with input $c-s$ and T.

We will consider three cases to prove that the first condition of Definition 4.3 holds, and then three cases to prove that the second condition holds. The theorem then follows from Proposition 4.4. For the first three cases, assume that $e\in T \cap T'$ . We further assume without loss of generality that $T_s\langle x\rangle = e$ . Let $T^e = r_{(G/ {e},\chi /e)}([c-s],T \setminus e)$ and let $ {\mathcal R_e}$ be the ordered list of directed edges crossed when rotor-routing on $(G/e,\chi /e)$ with input $c-s$ and $T\setminus e$ . Finally, let z be the vertex formed by contracting e. Notice that $(T\setminus e)_s\langle z\rangle = T_s\langle y\rangle $ if $y \neq s$ and z does not have a rotor in $(T\setminus e)$ when $y = s$ .

Case 1: The chip never enters the vertex x while rotor-routing.

For this case, we claim that $\mathcal R$ and ${\mathcal R_e}$ are identical.

Assume for the sake of induction that this statement holds for the first k edges. After crossing these rotors, the chip must be on the same vertex of both graphs (or the chip is on y in G and z in $G/e$ ). By assumption, this vertex has been entered the same number of times for each graph, so the rotor here must be in the same position. It follows that the chip will also exit along the same rotor for both graphs.

This proves our claim that $\mathcal R = {\mathcal R_e}$ . For every $v \in V(G) \setminus \{s,x,y\}$ , it follows immediately from the claim that $T^{\prime }_s{\langle } v{\rangle } = T^e_s{\langle } v{\rangle }$ . Furthermore, by construction, we also have $T^{\prime }_s{\langle } y{\rangle } = T^e_s{\langle } z{\rangle }$ and $T^{\prime }_s{\langle } x{\rangle } = e$ . This implies that $T^e = T' \setminus e$ as desired.

Case 2: The rotor at x spins around completely. In other words, $T_s\langle x\rangle = T^{\prime }_s\langle x\rangle $ , but $T_s\langle x\rangle \not = \rho \langle x\rangle $ for some intermediate rotor configuration $\rho $ .

We begin with a brief, high-level explanation of the proof of this case. We start by showing that $x \neq c$ and $y\neq s$ to avoid any edge cases that may arise. We then carefully define an alternative (possibly nonplanar) ribbon structure $\widetilde {\chi /e}$ for the contracted graph $G/e$ for which the steps of the rotor-routing algorithm used to compute $\widehat {r}_{(G/e,\widetilde {\chi /e},s)}(c-s,T\setminus e)$ and $\widehat {r}_{(G,\chi ,s)}(c-s,T)$ are nearly identical. It is then easily argued that $\widehat {r}_{(G/e,\widetilde {\chi /e},s)}(c-s,T\setminus e) = T'\setminus e$ . We then apply Lemma 3.6 to show that

$$\begin{align*}T^e = \widehat{r}_{(G/e,\chi/e,s)}(c-s,T\setminus e) = \widehat{r}_{(G/e,\widetilde{\chi/e},s)}(c-s,T\setminus e) = T'\setminus e.\end{align*}$$

Now, we give a complete proof. First, we note that by Proposition 3.3, it is impossible for x to complete more than one full rotation. Let $\chi (x) = (e,e_1,\dots ,e_p)$ and $\chi (y) = (e,\widehat e_1,\dots ,\widehat e_l)$ after removing edges parallel to e, so that $(\chi /e)(z) = (e_1,\dots ,e_p,\widehat e_1,\dots ,\widehat e_l)$ . Recall that we require $e \not = f$ for the first condition of Definition 4.3. Suppose that $x=c$ . By assumption, the rotor must cross f from c to s at some point during rotor-routing. As soon as this happens, the algorithm terminates, and thus $T^{\prime }_s{\langle } x {\rangle } = f$ . This contradicts the assumption that $T^{\prime }_s{\langle } x {\rangle } = e$ , so we must have $x \not = c$ .

Likewise, $y \neq s$ . Otherwise, since $x \neq c$ , for the rotor at x to make a full rotation, the chip must enter x precisely $\deg (x)$ many times. By Proposition 3.3, it must enter x along every edge, including the edge e from y to x, before the rotor makes a full rotation. However, this is clearly a contradiction as the rotor-routing algorithm halts as soon as the chip reaches $y=s$ .

Let $T_s {\langle } y{\rangle } = \widehat e_i$ and $T^{\prime }_s{\langle } y{\rangle } = \widehat e_j$ . By Proposition 3.3, each edge may only be crossed once in each direction. By assumption, the chip exits x along each incident edge. Because $x \not = c$ , this means that x must be entered its degree many times. Thus, the chip must traverse every edge incident to x in both directions. In particular, the rotor at y must cross e, which implies that $i\ge j$ . When rotor-routing on $(G,\chi )$ , keep an ordered list of every edge other than e that the chip crosses when moving out of either x or y and call this list $\mathcal E$ . This list must consist precisely of the edges

$$\begin{align*}\{e_1,\dots,e_p\} \cup \{\widehat e_1,\dots,\widehat e_j\} \cup \{\widehat e_{i+1},\dots,\widehat e_l\}.\end{align*}$$

Furthermore, the last edge in $\mathcal E$ must be $\widehat e_j$ . This is because after exiting x for the final time, the rotor enters y across e, and $T^{\prime }_s{\langle } y{\rangle } = \widehat e_j$ .

We now define an alternate ribbon structure on $G/e$ , which we will call $\widetilde {\chi /e}$ . For each vertex $v \not = z$ , let $(\widetilde {\chi /e})(v) = (\chi /e)(v)$ . Let $(\widetilde {\chi /e})(z)$ begin with $\mathcal E$ and then be ordered arbitrarily. Let $\mathcal R - {\overline e}$ be the sublist of $\mathcal R$ that is obtained after removing any directed edges between x and y. Finally, let $\widetilde {\mathcal R_e}$ be the ordered list of directed edges crossed when rotor-routing on $(G/e,\widetilde {\chi /e})$ with input $c-s$ and $T\setminus e$ . We claim that $\mathcal R - {\overline e}$ and $\widetilde {\mathcal R_e}$ are identical.

Suppose that the claim holds and let $\widetilde T = \widehat r_{(G/e,\widetilde {\chi /e},s)}(c-s,T\setminus e)$ . For all $v \in V(G) \setminus \{x,y,s\}$ , we know that $T^{\prime }_s{\langle } v {\rangle } = \widetilde T_s{\langle } v {\rangle }$ . Furthermore, by construction, we also have $T^{\prime }_s{\langle } y {\rangle } = \widetilde T_s{\langle } z {\rangle }$ and $T^{\prime }_s{\langle } x {\rangle }= e$ . The fact that $T^{\prime }_s{\langle } y {\rangle } = \widetilde T_s{\langle } z {\rangle }$ , along with our construction of $\widetilde {\chi /e}$ , implies that $\widetilde {\chi /e}$ satisfies the requirements of Lemma 3.6 (when comparing to $\chi /e$ on $G/e$ ). Thus, we know that $\widetilde T = T^e$ . The result follows.

Now we just need to prove the claim that $\mathcal R - {\overline e} = \widetilde {\mathcal R_e}$ . Assume for the sake of induction that the first k rotors match. After crossing these rotors, the chip must be on the same vertex of both graphs (or the chip is on z for $G/e$ and on either x or y for G). If the chip is on some $v \not =z$ , then the argument is analogous to the argument from Case 1. Thus, it suffices to consider the case where the chip is on z.

Out of the first k entries of $\mathcal R - {\overline e}$ , let a be the number of edges that exit x or y. Then, by assumption, for the first k entries of $\widetilde {\mathcal R_e}$ , there must be a edges that exit z. By construction, the next edge crossed in either case must be the $(a+1)$ th edge of $\mathcal E$ . It follows by induction that $\mathcal R - {\overline e} = \widetilde {\mathcal R_e}$ .

Case 3: The rotor e is directed differently for $T_s$ and $T^{\prime }_s$ . In other words, $T^{\prime }_s{\langle } y {\rangle } = e$ .

This case is similar to the previous case, and we will use the same notation for $\chi (x)$ and $\chi (y)$ . Suppose that $T_s {\langle } y{\rangle } = \widehat e_i$ and $T^{\prime }_s{\langle } x{\rangle } = e_j$ . As before, let $\mathcal E$ be the ordered list of edges other than e that the chip crosses when moving out of x or y. This list must consist precisely of the edges

$$\begin{align*}\{\widehat e_{i+1},\dots,\widehat e_l\} \cup \{ e_1,\dots, e_j\}.\end{align*}$$

Furthermore, after exiting y for the final time across e, the chip enters x. This means that the last edge in $\mathcal E$ must be $e_j$ . Similarly to the previous case, we define a ribbon structure $\widetilde {\chi /e}$ on $G/e$ . For $v \not = z$ , let $(\widetilde {\chi /e})(v) = (\chi /e)(v)$ , and let $(\widetilde {\chi /e})(z)$ begin with $\mathcal E$ and then be ordered arbitrarily.

The rest of the argument is analogous to the argument used for Case 2. In particular, we define $\mathcal R - {\overline e}$ and $\widetilde {\mathcal R_e}$ and then show that they are identical. We also use this idea to show that $\widetilde {\chi /e}$ satisfied the requirements of Lemma 3.6. The only slight difference from the previous case is that the final positions of the rotors at x and y swap.

For the last three cases, $e \not \in T\cup T'$ . For these cases, we define ${}^eT = r_{(G\setminus e,\chi \setminus e)}([c-s],T)$ . We need to show that ${}^eT = T'$ . Also, let ${}_e\mathcal R$ be the ordered list of directed edges crossed when rotor-routing on $(G\setminus e,\chi \setminus e)$ with input $c-s$ and T.

Case 4: The chip never crosses e.

Since the chip never crosses e, removing this edge has no effect on the output of the rotor-routing algorithm. We can use an analogous argument to what we used for Case 1. In particular, we show that $\mathcal R = {}_e\mathcal R$ , and the result follows.

Case 5: The edge e is crossed in both directions, but $T^{\prime }_s{\langle } x {\rangle } \not = e \not = T^{\prime }_s{\langle } y {\rangle }$ .

First, we observe that ${\mathfrak i}(e) \not = \{c,s\}$ ; otherwise, it is impossible for the chip to cross e in both directions. Thus, c and s must remain incident on $G\setminus e$ . This allows us to apply Proposition 3.3 to say that no edge is crossed more than once in each direction.

Define $\chi (x)$ and $\chi (y)$ as in Case 2. Without loss of generality, suppose that the first time the chip crosses e, it moves from y to x. Let $\rho $ be the first rotor configuration reached while rotor-routing on $(G,\chi )$ (with input $c-s$ and T) such that $\rho {\langle } y {\rangle } = e$ . Further, suppose that $\rho {\langle } x {\rangle } = e_q$ . We introduce a new ribbon structure $\widetilde {\chi }$ on G. For $v \in V(G) \setminus x$ , let $\widetilde {\chi }(v) = \chi (v)$ , and let

$$\begin{align*}\widetilde{\chi}(x) = (e,e_{q+1},e_{q+2},\dots,e_p,e_1,e_2,\dots,e_q).\end{align*}$$

Let ${\mathrm E}_1$ be the set of edges after $T_s {\langle } x {\rangle }$ and before $T^{\prime }_s {\langle } x {\rangle }$ with respect to $\chi $ and ${\mathrm E}_2$ be the set of edges after $T_s {\langle } x {\rangle }$ and before $T^{\prime }_s {\langle } x {\rangle }$ with respect to $\widetilde \chi $ . Because we only rearranged the position of e, we must have ${\mathrm E}_1\setminus e={\mathrm E}_2\setminus e$ . Furthermore, $e \in {\mathrm E}_1$ by assumption, and $e \in {\mathrm E}_2$ by construction (because the rotor-routing algorithm is equivalent on $(G,\chi )$ and $(G,\widetilde \chi )$ until the chip crosses e from x to y). It follows that ${\mathrm E}_1 = {\mathrm E}_2$ .

By the reasoning in the previous paragraph, $\widetilde {\chi }$ satisfies the conditions for Lemma 3.6, which means that $\widehat r_{(G,\widetilde {\chi },s)}(c-s,T) = T'$ .

Let $\widetilde {\mathcal R}$ be the ordered list of directed edges crossed when rotor-routing on $(G,\widetilde \chi )$ with input $c-s$ and T. Let $\widetilde {\mathcal R}-e$ be the remaining edges of $\widetilde {\mathcal R}$ after removing e (but not edges parallel to e). We claim that $\widetilde {\mathcal R}-e = {}_e\mathcal R$ . This follows from construction, because when the chip crosses e from y to x in $\widetilde {\mathcal R}$ , we know by construction that $e_q$ is the rotor at x. Therefore, the next step of the algorithm is to cross back along e in the other direction. Thus, it is as though this edge did not exist.

Given our claim, the result follows from the same reasoning we used for the previous cases.

Case 6: The edge e is only crossed in one direction.

We will use Lemma 3.5 to show that this case is never realized. This is the one case where the planarity condition is vital.

Without loss of generality, suppose that the chip crosses e from x to y. Then, by assumption, it must reach x again without crossing e. Consider the next occurrence when the chip returns to x. Let the rotor configuration at this moment be $\rho $ .

Consider the sequence $(x,y,v_1,v_2,\dots )$ of vertices where $\rho {\langle } y{\rangle } = v_1$ and each $v_i$ satisfies ${\mathfrak i}(\rho {\langle } v_i{\rangle } ) = \{v_i,v_{i+1}\}$ . Since there are a finite number of vertices, and the chip has just returned to x, there must be some j such that $x \in {\mathfrak i}(\rho {\langle } v_j {\rangle })$ . Thus, the rotors incident to these vertices form a directed cycle, which does not include the sink vertex. By Lemma 3.5, this cycle must reverse before the rotor-routing algorithm terminates. In order for the cycle to reverse, the chip must cross e in the other direction. This gives a contradiction.

Proof. It is straightforward to show that the defining properties of sandpile torsor actions, as well as consistency, are preserved if we replace $\alpha $ with any of the other three possibilities.

Let $E_3$ be the triple edge with a planar ribbon structure. Reversing the ribbon structure on $E_3$ is a ribbon graph automorphism that is the identity on $V(E_3)$ and $\operatorname {Pic}^0(E_3)$ but is not the identity on $E(E_3)$ or $\mathcal {T}(E_3)$ . It follows that $\alpha _{(E_3)} \not = \overline \alpha _{(E_3)}$ and $\alpha ^{-1}_{(E_3)} \not = \overline \alpha ^{-1}_{(E_3)}$ .

Let $C_3$ be the circuit with three edges and its unique ribbon structure. Then, $\operatorname {Pic}^0(C_3) \cong \mathbb {Z}/3\mathbb {Z}$ , which means that there are elements of $\operatorname {Pic}^0(G)$ that are not equal to their own inverse. It follows that $\alpha _{(C_3)} \not = \alpha ^{-1}_{(C_3)}$ and $\overline \alpha _{(C_3)} \not = \overline \alpha ^{-1}_{(C_3)}$ .

By combining the arguments from the previous two paragraphs, we know that for any two of these algorithms, there exist plane graphs for which the sandpile torsor actions are distinct.

Proof. Suppose for the sake of contradiction that there is some plane graph $(G,\chi )$ such that $\alpha _{(G,\chi )} \not = \beta _{(G,\chi )}$ . By the contrapositive of Corollary 3.2, this means that for some adjacent vertices c and s, and some $T\in \mathcal {T}(G)$ , we have

$$\begin{align*}\alpha_{(G,\chi)}([c-s],T) \not= \beta_{(G,\chi)}([c-s],T).\end{align*}$$

We can assume $(G,\chi )$ is not 2-connected, or else the contradiction is immediate. Let f be an edge such that ${\mathfrak i}(f) = \{c,s\}$ . By Condition 3 of consistency (Definition 4.3), any edges separated from f after removing cut vertices must be unchanged when acting on T by $[c-s]$ . For all such edges, we contract the edges that are in T and delete the edges that are not. We are left with a 2-connected plane graph $(G',\chi ')$ and a spanning tree $T' \in \mathcal {T}(G')$ . By Conditions 1 and 2 of consistency, the inequality is preserved. In particular,

$$\begin{align*}\alpha_{(G',\chi')}([c-s],T') \not= \beta_{(G',\chi')}([c-s],T').\end{align*}$$

This is a contradiction.

Proof. If c is a leaf vertex, then the unique edge of T incident to c must be equal to $T_s{\langle } c {\rangle }$ . The result follows from the definition of rotor-routing.

Proof. This lemma follows immediately from Definitions 2.16 and 5.9.

Proof. For the if direction, Lemma 5.10 says that $T_s {\langle } x {\rangle } = g$ . Thus, Algorithm 1 terminates after a single step and the output tree is $T \setminus g \cup f$ . For the only if direction, $c \not \prec _{T_s} x$ implies that $T_s {\langle } x {\rangle } \not = g$ . If the rotor-routing algorithm terminates after a single step, the resulting tree must include the edge g. Thus, it is not a single-step pair from g to f.

Proof. Suppose that $T_{\mathfrak {r}}$ is rotatable at c from g to f. By Definition 5.12, this implies that $g = T_{\mathfrak {r}}{\langle } c {\rangle }$ , $f = \chi (c,g)$ and $r_c(T_{\mathfrak {r}})$ is acyclic. Let $s = {\mathfrak i}(f) \setminus c$ and $x = {\mathfrak i}(g) \setminus c$ . Let P be the unique path from c to s in T. If $g \not \in P$ , then $T \setminus g \cup f$ must contain a cycle, which contradicts the condition that $r_c(T_{\mathfrak {r}})$ is acyclic. Thus, $g \in P$ and $c \prec _{T_{s}} x$ . By Lemma 5.11, $(c-s,T)$ is a single-step pair from g to f.

Alternatively, suppose that there exists an $s \in V(G)$ such that $(c-s,T)$ is a single-step pair from g to f and $T_{\mathfrak {r}} {\langle } c {\rangle } = g$ . A single step through the rotor-routing algorithm produces the rotor configuration $r_c(T_{s})$ . By Lemma 5.11, we know that $c \prec _{T_s} x$ , which implies $T_s {\langle } c {\rangle } = g$ by Lemma 5.10. Because $T_{\mathfrak {r}} {\langle } c {\rangle } = T_{s} {\langle } c {\rangle }$ , it follows that $r_c(T_{\mathfrak {r}}){\langle } c {\rangle } = r_c(T_s){\langle } c {\rangle } = f$ . This implies that $r_c(T_{\mathfrak {r}}) = (T \setminus g \cup f)_{\mathfrak {r}}$ , which is acyclic by Lemma 2.17.

The proof is analogous when working with source-rotatability and source-turn pairs noting that the c will remain a source vertex after rotating the rotor at c from g to f.

Proof. Both of these results follow directly from Definition 5.12. All of the defining properties of (source) rotatability still hold after contracting an edge in T or deleting an edge not in T (as long as this edge is not f or g).

Proof. Because F has an edge incident to $\mathfrak {r}$ , each vertex of $(\widetilde {G},\widetilde {\chi })$ other than $\widetilde {\mathfrak {r}}$ has a single preimage in $(G,\chi )$ . Furthermore, for any $\widetilde T \in \mathcal {T}(\widetilde G)$ , and any $c \not = \widetilde {\mathfrak {r}} \in V(\widetilde G)$ , it is immediate that $\widetilde T \cup F\in \mathcal {T}(G)$ and $\widetilde T_{\mathfrak {r}'}{\langle } c {\rangle } = (\widetilde T \cup F)_{\mathfrak {r}}{\langle } c {\rangle }$ .

Fix some tree $T" \in \mathcal {T}(G)$ and suppose there exists a vertex $c \in V(\widetilde {G})\setminus \widetilde {\mathfrak {r}}$ and edges $f,g \in E(\widetilde {G})$ such that $(T"\setminus F)_{\widetilde {\mathfrak {r}}}$ is source-rotatable at c from g to f in $(\widetilde {G},\widetilde {\chi })$ . Then by Definition 5.12, $g = (T"\setminus F)_{\widetilde {\mathfrak {r}}}{\langle } c{\rangle } = T^{\prime \prime }_{\mathfrak {r}}{\langle } c{\rangle }$ , and $f = \widetilde {\chi }(c,g) = \chi (c,g)$ . Lastly,

$$\begin{align*}r_c((T"\setminus F)_{\widetilde{\mathfrak{r}}}) = ((T"\setminus F)\cup f\setminus g)_{\widetilde{\mathfrak{r}}} = (T"\cup f\setminus g)_{\mathfrak{r}}\setminus F = r_c(T^{\prime\prime}_{\mathfrak{r}})\setminus F.\end{align*}$$

Thus, $r_c(T^{\prime \prime }_{\mathfrak {r}}) = r_c((T"\setminus F)_{\widetilde {\mathfrak {r}}})\cup F$ . Given that $r_c((T"\setminus F)_{\widetilde {\mathfrak {r}}})$ is acyclic (by Definition 5.12) and that F contracts to a single vertex, $r_c(T^{\prime \prime }_{\mathfrak {r}})$ can only contain a cycle if F contains a cycle. However, $F \subseteq T$ , where T is a spanning tree, so this is impossible. Therefore, $r_c(T^{\prime \prime }_{\mathfrak {r}})$ is acyclic. Then, by Definition 5.12, $T^{\prime \prime }_{\mathfrak {r}}$ is rotatable at c from g to f in $(G,\chi )$ . Lastly, $T^{\prime \prime }_{\mathfrak {r}}$ is source-rotatable at c because if $c\neq \widetilde {\mathfrak {r}}$ is a leaf of $T"\setminus F$ , then it is a leaf in $T"$ .

Given that for any $\widetilde {T} \in \mathcal {T}(\widetilde {G})$ , $\widetilde {\mathfrak {r}}$ has no rotors in $\widetilde {T}_{\widetilde {\mathfrak {r}}}$ , this implies that any sequence of source-rotations that take $(T\setminus F)_{\widetilde {\mathfrak {r}}}$ to $(T'\setminus F)_{\widetilde {\mathfrak {r}}}$ on $(\widetilde {G},\widetilde {\chi })$ must also take $T_{\mathfrak {r}}$ to $T^{\prime }_{\mathfrak {r}}$ on $(G,\chi )$ . Furthermore, the rotors corresponding to edges in F are not present in $ (\widetilde {G},\widetilde {\chi })$ , so they are not rotated by any of the source rotations taking $T_{\mathfrak {r}}$ to $T^{\prime }_{\mathfrak {r}}$ on $(G,\chi )$ .

Proof of Theorem 5.8

First, choose an arbitrary root vertex $\mathfrak {r} \in V(G)$ which we will fix throughout this proof. Our goal will be to show that there is a series of source-rotations whose composition maps from $T^{start}_{\mathfrak {r}}$ to $T^{goal}_{\mathfrak {r}}$ . The theorem then follows from repeated applications of Lemma 5.13.

Next, we set up an induction argument. When G consists of a single edge, the result is trivial. Consider the following inductive assumption.

Inductive Assumption: Suppose $(G',\chi ')$ is a 2-connected proper minor of $(G,\chi )$ . For any pair $\widehat {T}^{start}, \widehat {T}^{goal} \in \mathcal {T}(G')$ , and any vertex $\mathfrak {r}' \in V(G')$ , there is a series of source-rotations whose composition maps from $\widehat {T}^{start}_{\mathfrak {r}'}$ to $\widehat {T}^{goal}_{\mathfrak {r}'}$ .

Let $\xi $ be the map from $\mathcal T(G) \times (V(G)\setminus \mathfrak {r}) \to \{0,1\}$ defined as follows:

$$\begin{align*}\xi(T,x) = \begin{cases} 0 & \text{if }T_{\mathfrak{r}} {\langle} x {\rangle} = T^{goal}_{\mathfrak{r}} {\langle} x {\rangle}, \\ 1 & \text{otherwise.}\end{cases} \end{align*}$$

Notice that $\xi (T,x) = 0$ for all x if and only if $T = T^{goal}$ . We can use $\xi $ and the vertex partial order from Definition 5.9 to define a partial order on spanning trees, which we will write as $\prec _{\xi }$ (or $\preceq _{\xi }$ for the corresponding weak partial order).

$$ \begin{align*} T \preceq_{\xi} T' \Longleftrightarrow \text{ For all } &x \in (V(G)\setminus \mathfrak{r}) \text{ such that }\xi(T,x)> \xi(T',x) \text{, there exists }\\ & y \in (V(G)\setminus \mathfrak{r})\text{ such that }\xi(T,y) < \xi(T',y) \text{ and } x \prec_{T^{goal}_{\mathfrak{r}}} y. \end{align*} $$

We now show that $\prec _{\xi }$ is a partial order. Irreflexivity and asymmetry are immediate; we just need to show transitivity. Let T, $T'$ and $T"$ be spanning trees such that $T \prec _{\xi } T'$ and $T' \prec _{\xi } T"$ . Let x be an arbitrary vertex such that $\xi (T,x)> \xi (T",x)$ (if no such vertex exists, then $T \prec _{\xi } T"$ is immediate).

First, suppose that $\xi (T',y) = \xi (T,y)$ for all y where $x \prec _{T^{goal}_{\mathfrak {r}}} y$ . Then, $\xi (T',x) \geq \xi (T,x)> \xi (T",x)$ . Because $T' \prec _{\xi } T"$ , there is some z such that $x \prec _{T^{goal}_{\mathfrak {r}}} z$ and $\xi (T',z) < \xi (T",z)$ . Furthermore, because $\xi (T',z) = \xi (T,z)$ , we also have $\xi (T,z) < \xi (T",z)$ .

Otherwise, there must be some y such that $x \prec _{T^{goal}_{\mathfrak {r}}} y$ and $\xi (T,y) < \xi (T',y)$ . We can take y to be a maximal such vertex (with respect to $\prec _{T^{goal}_{\mathfrak {r}}}$ ). If $\xi (T',y) \le \xi (T",y)$ , then $\xi (T,y) < \xi (T",y)$ . Otherwise, because $T' \prec _{\xi } T"$ , there is some z such that $y \prec _{T^{goal}_{\mathfrak {r}}} z$ and $\xi (T',z) < \xi (T",z)$ . By maximality of y and the definition of $\prec _{\xi }$ , we know that $\xi (T,z) \le \xi (T',z)$ . It follows that $\xi (T,z) < \xi (T",z)$ . Since x was arbitrary, $x \prec _{T_{\mathfrak {r}}^{goal}} y$ , and $x \prec _{T_{\mathfrak {r}}^{goal}} z$ , we know that $T \prec _{\xi } T"$ .

The partial ordering $\prec _{\xi }$ is used to describe deviations from $T^{goal}_{\mathfrak {r}}$ , prioritizing rotors closest to the root. Because $|\mathcal {T}(G)|$ is finite and $T^{goal}$ is the unique minimal spanning tree, it suffices to show that given any $T \in \mathcal {T}(G)\setminus T^{goal}$ , there is a sequence of source-rotations from $T_{\mathfrak {r}}$ to some $T_{\mathfrak {r}}"$ such that $T" \prec _{\xi } T$ .

Suppose that there exists some $x \in V(G)$ such that $\xi (T,x) = 1$ and x is a leaf vertex of T. Then, we can freely rotate the rotor at x using source-rotations until we get a configuration $T_{\mathfrak {r}}"$ such that $\xi (T",x) = 0$ . Because we only rotated the rotor at x, it follows that $T" \prec _{\xi } T$ and we are done.

Otherwise, there must exist some $x \in V(G)\setminus \mathfrak {r}$ such that $\xi (T,x)=1$ and for all $v \prec _{T_{\mathfrak {r}}} x$ , we have $\xi (T,v) = 0$ . Let $g = T_{\mathfrak {r}}{\langle } x {\rangle }$ , $f = \chi (x,g)$ and $\{x,y\} = {\mathfrak i}(f)$ . Consider the sets

$$\begin{align*}V^x = \{v \mid v \prec_{T_{\mathfrak{r}}} x\} \text { and } E^x = \{T_{\mathfrak{r}}{\langle} v {\rangle} \mid v \in V^x\}.\end{align*}$$

By assumption, all of the rotors associated with vertices in $V^x$ are already in their final position and are directed toward x. This implies that

$$\begin{align*}V^x \subseteq \{v \mid v \prec_{T^{goal}_{\mathfrak{r}}} x\}.\end{align*}$$

We now introduce the following claim:

Claim: There exists some $T'\in \mathcal {T}(G)$ such that x is a leaf of $T'$ , and $T_{\mathfrak {r}}'$ can be reached from $T_{\mathfrak {r}}$ after a sequence of source-rotations at vertices in $V^x$ .

If the claim is true, then we can rotate the rotor at x until it reaches $T^{goal}{\langle } x {\rangle }$ and call the new tree $T"$ . By construction, $\xi (T",x) < \xi (T',x) = \xi (T,x)$ . Furthermore, since all source-rotations were at vertices in $V_x\subseteq \{v \mid v \prec _{T^{goal}_{\mathfrak {r}}} x\}$ , it follows that $\xi (T",v) = \xi (T,v)$ for all $v \not \prec _{T^{goal}_{\mathfrak {r}}} x$ . Thus, $T" \prec _{\xi } T$ .

All that is left is to prove the claim. By Lemma 5.10, the set $T \setminus E^x$ has only one edge incident to x. Because $(G,\chi )$ is 2-connected, x is not a cut vertex. Thus, there exists a spanning tree $T' \in \mathcal {T}(G)$ such that $(T \setminus E^x) \subseteq T'$ and x is a leaf edge of $T'$ . Let F be the set of edges in $T\setminus E^x$ . Notice that F consists of all of the edges of T that are on the same component of T as $\mathfrak {r}$ when the tree is severed at the point x. In particular, F is connected and contains at least one edge incident to $\mathfrak {r}$ .

If we contract all of the edges in F, we obtain a new ribbon graph $(\widetilde G,\widetilde \chi )$ such that $V(\widetilde G) = V^x\cup x$ . Furthermore, $T \cap E(\widetilde {G})$ and $T' \cap E(\widetilde {G})$ are both in $\mathcal {T}(\widetilde G)$ . Let $\widetilde {\mathfrak {r}}$ be the vertex formed by contraction. The vertices other than $\widetilde {\mathfrak {r}}$ cannot be cut vertices of $(\widetilde G,\widetilde \chi )$ because they are not cut vertices of $(G,\chi )$ . If $\widetilde {\mathfrak {r}}$ is a cut vertex, we can consider each 2-connected component separately. In either case, we can apply the induction hypothesis.

By the induction hypothesis, there is a sequence of source-rotations whose composition maps from $T \cap E(\widetilde {G})$ to $T' \cap E(\widetilde {G})$ on $(\widetilde G,\widetilde \chi )$ . Thus, by Lemma 5.15, there is a sequence of source-rotations that take $T_{\mathfrak {r}}$ to $T^{\prime }_{\mathfrak {r}}$ on $(G,\chi )$ without turning any rotors corresponding to edges in F. The claim follows from the fact that the edges in $T \setminus F$ correspond to the rotors at vertices in $V^x$ .

Proof. $(1) \to (2)$ is immediate. $(2) \to (3)$ follows from Lemma 5.7. $(3) \to (1)$ follows from Theorem 5.8 and Lemma 5.5.

Proof.

$$ \begin{gather*}r_{(G,\chi)}([s-c],T) = r_{(G,\chi)}\left([s-c], r_{(G,\chi)}([c-s],T\setminus f \cup g)\right) \\ = r_{(G,\chi)}([0],T\setminus f \cup g)) = T\setminus f \cup g.\\[-36pt] \end{gather*} $$

Proof of Theorem 5.4

Let $\alpha $ be an arbitrary consistent sandpile torsor algorithm. We will prove the theorem by induction, but we first consider a few special plane graphs.

Let $C_3$ be the 3-cycle along with its unique ribbon structure, and let $E_3$ be the triple edge with a planar ribbon structure. Each of these plane graphs has two sandpile torsor actions. On $C_3$ , one of these actions is equivalent to both $r_{C_3}$ and $\overline r_{C_3}$ , whereas the other is equivalent to both $r^{-1}_{C_3}$ and $\overline r^{-1}_{C_3}$ . On $E_3$ , one of the actions is equivalent to both $r_{E_3}$ and $\overline r^{-1}_{E_3}$ , whereas the other is equivalent to both $\overline r_{E_3}$ and $r^{-1}_{E_3}$ . This means that $\alpha _{C_3}$ and $\alpha _{E_3}$ will simultaneously match precisely one of the four sandpile torsor algorithms with the same structure as rotor-routing. For this proof, we assume that $\alpha _{C_3} = r_{C_3}$ and $\alpha _{E_3} = r_{E_3}$ . By Proposition 5.2, we can make this assumption without loss of generality.

Let $(G,\chi )$ be an arbitrary 2-connected plane graph. For the induction hypothesis, assume that $\alpha _{(G',\chi ')} = r_{(G',\chi ')}$ for all proper minors $(G',\chi ')$ of $(G,\chi )$ . By Corollary 5.18, it suffices to show that $\alpha _{(G,\chi )}([c-s],T) = r_{(G,\chi )}([c-s],T)$ for every source-turn pair $(c-s,T)$ . Throughout this proof, $(c-s,T)$ is an arbitrary source-turn pair from g to f and

$$\begin{align*}\widehat T := \alpha_{(G,\chi)}([c-s],T).\end{align*}$$

To prove the theorem, we need to show that $\widehat T = r_{(G,\chi )}([c-s],T) = T \setminus g\cup f $ .

First, suppose that there exists some $e\not = g$ such that $e \in T\cap \widehat T$ . This implies that ${\mathfrak i}(e) \not = \{c,s\}$ because $T_s{\langle } c {\rangle } = g$ (if ${\mathfrak i}(e) = \{c,s\}$ and $e \in T$ , then $T_s{\langle } c {\rangle } = e \neq g$ , which presents a contradiction). By the condition that $\alpha $ is consistent, we must have $\alpha _{(G/e,\chi /e)}([c-s],T \setminus e) = \widehat T \setminus e$ . By the induction hypothesis, $\alpha _{(G/e,\chi /e)}([c-s],T \setminus e) = r_{(G/e,\chi /e)}([c-s],T \setminus e)$ . Furthermore, it follows from Lemmas 5.14 and 5.13 that $(c-s,T)$ is a source-turn pair for $(G/e,\chi /e)$ . This means that $r_{(G/e,\chi /e)}([c-s],T) = (T \setminus g \cup f)\setminus e$ . It follows that $\widehat T = T \setminus g \cup f = r_{(G,\chi )}([c-s],T)$ . Similarly, suppose that there exists some $e \not = f$ such that $e \not \in T \cup \widehat T$ . Then, by an analogous argument, we have $\alpha _{(G\setminus e,\chi \setminus e)}([c-s],T) = \widehat T$ and $\widehat T =T \setminus g \cup f= r_{(G,\chi )}([c-s],T)$ . In both cases, the theorem holds by induction. Therefore, we may assume that the only shared edges or non-edges of T and $\widehat T$ are in $\{f,g\}$ . In particular, we have the relation

(2) $$ \begin{align} E(G)\setminus (T\Delta \widehat T) \subseteq \{f,g\}, \end{align} $$

where $\Delta $ denotes the symmetric difference operator. Note that this argument still holds when $(c-s,T)$ is a single-step pair or a reverse single-step pair. This leaves us with four cases to consider corresponding to the four subsets of $\{f,g\}$ .

Case 1: We have $f \in \widehat T$ and $g \in \widehat T$ .

In this case, the number of edges of G must be one less than double the size of each spanning tree.

Let ${\mathfrak i}(g) = \{x,c\}$ . Because $g \in T \cap \alpha _{(G,\chi )}([c-s],T)$ , we can apply Definition 4.3 and the induction hypothesis to show that

$$\begin{align*}\alpha_{(G,\chi)}([c-s],T)\setminus g = \alpha_{(G/g,\chi/g)}([c-s],T\setminus g) = r_{(G/g,\chi/g)}([c-s],T\setminus g). \end{align*}$$

Because $(c-s,T)$ is a source-turn pair (and thus also a single-step pair) from g to f, we must have $T_s{\langle } c {\rangle } = g$ . By Definition 5.9 and Lemma 5.10, this implies that the path of edges $(e_1,e_2,\dots ,e_k)$ on T from x to s does not contain g or any edges parallel to g. Furthermore, these edges must also form a path from z to s on $G/g$ (where z is the vertex formed by contracting g). By Lemma 3.4, any edges to the right of the directed cycle formed by $(e_1,e_2,\dots ,e_k,f)$ are in $r_{(G/g,\chi /g)}([z-s],T\setminus g) = \widehat T\setminus g$ if and only if they are in $T \setminus g$ . This implies that for any such edge e, we have $e \notin \{f,g\}$ and $e \notin T\Delta \widehat {T}$ , which contradicts (2). Thus, there can be no edges to the right of the cycle. In particular, this means that $(\chi /g)(z,f) = e_1$ and $\chi (c,f) = g$ .

By Definition 5.6, we know that $\chi (c,g) = f$ . Together, the facts that $\chi (c,g) = f$ and $\chi (c,f) = g$ imply that c is a vertex of degree 2. This means that a single firing of vertex c sends $2c -s - x$ to $0$ . Thus, $[2c - s - x] = [0]$ and $[c-s] = [x-c]$ .

Let $T' = T\setminus g \cup f$ . It follows from Definition 5.6 that $(c-x,T')$ is a source-turn pair from f to g. Suppose that $\alpha _{(G,\chi )}([c-x],T') = T$ . Then, this would mean that $\alpha _{(G,\chi )}([x-c],T) = T'$ . However, we showed that

$$\begin{align*}\alpha_{(G,\chi)}([x-c],T) = \alpha_{(G,\chi)}([c-s],T) = \widehat T\neq T'.\end{align*}$$

Because $\alpha _{(G,\chi )}([c-x],T') \not = T$ , this action must swap every edge of $T'$ other than f and g. Furthermore, since the number of edges in G is one less than double the size of each spanning tree, both f and g must be in $\alpha _{(G,\chi )}([c-x],T')$ . It follows that $\alpha _{(G,\chi )}([c-x],T') = \widehat T$ .

Notice that the edges $(g,e_k,e_{k-1},\dots ,e_1)$ form a directed cycle on the graph $G/f$ . Using our earlier argument, we can conclude that there are no edges to the right of this cycle. Furthermore, the rotors of this cycle are oriented in the opposite direction as they were on $G/g$ . Together with the fact that c is a degree 2 vertex, this implies that $\{e_1,e_2,\dots ,e_k,f,g\}$ forms a cycle in G with no edges on either side. In particular, this means that G is a cycle.

It follows that $\widehat T = \{f,g\}$ . The only cycle with a 2-edge spanning tree is the 3-cycle $C_3$ . Thus, $\alpha _{(G,\chi )} = r_{(G,\chi )}$ by our assumption that $\alpha _{C_3} = r_{C_3}$ .

Case 2: We have $f \notin \widehat T$ and $g \in \widehat T$ .

Let ${\mathfrak i}(g) = \{x,c\}$ . By the same argument that we used in the previous proof, c has degree $2$ and $[c-s] = [x-c]$ . This means that

$$\begin{align*}\alpha_{(G,\chi)}([x-c],T) = \alpha_{(G,\chi)}([c-s],T) = \widehat T. \end{align*}$$

Because $f\not \in T \cup \widehat T$ , we can use consistency to conclude that $\alpha _{(G\setminus f,\chi \setminus f)}([x-c],T) = \widehat T$ . However, g is a cut edge of $G \setminus f$ . Thus, $[x-c] = 0$ on $G\setminus f$ . This is a contradiction by (2), since $T \not = \widehat T$ unless $E(G) = \{f,g\}$ . If $E(G) = \{f,g\}$ , we still have a contradiction because $[c-s]$ is a nontrivial element of $\operatorname {Pic}^0(G)$ , so it cannot fix any spanning trees. Thus, Case 2 is impossible.

Case 3: We have $f \notin \widehat T$ and $g \notin \widehat T$ .

This case is a bit more complicated than the previous two, so we begin with a brief overview of our argument.

• Step 1: First, we show that for any source turn pair $(c"-s",T")$ , we either have $\alpha _{(G,\chi )}([c" - s"],T") = r_{(G,\chi )}([c" - s"],T")$ or $\alpha _{(G,\chi )}([2(c" - s")],T") = r_{(G,\chi )}([c" - s"],T")$ (see (3)).

• Step 2: This observation allows us to define a sequence of sandpile elements such that the action by $\alpha _{(G,\chi )}$ on $T"$ rotates the rotor at $c"$ all the way around until returning to $T"$ . In particular, these sandpile elements must add to the identity (see (4)).

• Step 3: Using properties of the sandpile group, we prove that all the sandpile elements defined in the previous step have a particular form.

• Step 4: Next, we use the fact that $\alpha _{(G,\chi )}$ must always output a spanning tree in order to deduce that there must be exactly three edges incident to c.

• Step 5: After this, we show that when Case 3 occurs and c is incident to exactly three edges, these must be the only edges of G. In particular, it follows that $(G,\chi ) = E_3$ (a planar embedding of the triple edge).

• Step 6: Finally, we know that $r_{E_3} = \alpha _{E_3}$ by a base case of our induction.

Let $T' = T \setminus g$ . By assumption, $\alpha _{(G,\chi )}([c-s],T) \not = T'\cup f$ . Thus, $\alpha _{(G,\chi )}([s-c],T'\cup f) \not = T$ . However, note that $(s-c,T'\cup f)$ is a reverse single-step pair. Because (2) holds for reverse single-step pairs, every edge other than f and g must swap. In addition, because the total number of edges in each spanning tree is consistent, neither f nor g can be in $\alpha _{(G,\chi )}([s-c],T'\cup f)$ . This means that $\alpha _{(G,\chi )}([s-c],T'\cup f) = \widehat {T}$ and $\alpha _{(G,\chi )}([c-s],\widehat {T}) = T'\cup f$ . We obtain the following chain of equalities:

$$\begin{align*}\alpha_{(G,\chi)}([2c-2s],T) = \alpha_{(G,\chi)}([c-s],\widehat T) = T' \cup f = r_{(G,\chi)}([c-s], T). \end{align*}$$

Note that the reasoning in the previous paragraph holds for any source-turn pair. In particular, if $(c"-s",T")$ is a source-turn pair, then

(3) $$ \begin{align} \alpha_{(G,\chi)}([a(c" - s")],T") = r_{(G,\chi)}([c" - s"],T") \text{ for some } a \in \{1,2\}. \end{align} $$

Let $\chi (c) = (e_1,e_2,\dots ,e_k)$ , where $e_1 = g$ , and set ${\mathfrak i}(e_i) = \{c,s_i\}$ . Note that $e_2 = f$ , $s_2 = s$ , and the $s_i$ are allowed to coincide. Because $(c-s,T)$ is a source-turn pair, it follows that $(c-s_{i+1},T'\cup e_i)$ is also a source-turn pair for any $i\in [1,k]$ . Thus, by (3), there exists a function $\phi : [1,k] \to \{1,2\}$ such that for all $i \in [1,k]$ ,

$$\begin{align*}\alpha_{(G,\chi)}([\phi(i)(c-s_{i+1})],T' \cup e_i) = r_{(G,\chi)}([c-s_{i+1}],T' \cup e_i),\end{align*}$$

where $i+1$ is replaced with $1$ for $i = k$ . It follows that

(4) $$ \begin{align} \alpha_{(G,\chi)}\left(\left[\sum_{i=1}^k\phi(i)(c-s_i)\right],T\right) = r_{(G,\chi)}\left(\left[\sum_{i=1}^k(c-s_i)\right],T\right) = r_{(G,\chi)}([0],T) = T, \end{align} $$

where we use the fact that

$$\begin{align*}\left[\sum_{i=1}^k(c-s_i)\right] = [0], \end{align*}$$

because after firing c, we return to the identity. Since $\alpha $ is free, this implies that

$$\begin{align*}\left[\sum_{i=1}^k\phi(i)(c-s_i)\right] = \left[\sum_{i=1}^k(\phi(i)-1)(c-s_i)\right]= [0]. \end{align*}$$

The divisor $\sum _{i=1}^k(\phi (i)-1)(c-s_i)$ has $-1$ chips on each $s_i$ for which $\phi (i) = 2$ , chips on c equal to the number of such i, and no chips elsewhere. By [Reference McDonoughMcD21a, Lemma 17], and the assumption that c is not a cut vertex, this divisor is only equivalent to the identity if $\phi (i) = 2$ for all i or $\phi (i) = 1$ for all i. By the assumption that we are in Case 3, $\phi (1) \not = 1$ , which means that $\phi (i) = 2$ for all $i \in [1,k]$ .

Next, we consider the possible values of k (the number of edges incident to c). Recall that $\widehat T = \alpha _{(G,\chi )}([c-s],T)$ . Because $(c-s,T)$ is a source-turn pair, $e_j\not \in T$ for $j> 1$ . Furthermore, since we are in Case 3, this implies that $e_j \in \widehat T$ if and only if $j>2$ . If $k \le 2$ , then $\widehat T$ has no edges connecting to c, which is impossible. Since $e_1 \not \in \widehat T$ , there must be a path from $s_1$ to some other $s_i\neq s_1$ along $\widehat T$ that does not pass through the vertex c. If $k>4$ , then by pigeonhole principle, there must be a pair $s_j$ and $s_{j+1}$ such that neither vertex is $s_1$ or $s_i$ .

We know that $\phi (j) = 2$ , which means that $\alpha _{(G,\chi )}([c-s_{j+1}],T' \cup e_j) \not = T' \cup e_{j+1}$ . The only other possibility is that

$$\begin{align*}\alpha_{(G,\chi)}([c-s_{j+1}],T' \cup e_j) = \widehat T \cup \{e_1,e_2\} \setminus \{e_j, e_{j+1}\}. \end{align*}$$

However, $\widehat T \cup \{e_1,e_2\} \setminus \{e_j, e_{j+1}\}$ contains $e_1$ , $e_i$ and a second path from $s_1$ to $s_i$ . This means that it must contain a cycle, which is a contradiction.

When $k=4$ , the situation is similar. The argument that we used for $k>4$ still works if we can show that there is a path along $\widehat T$ from $s_1$ to $s_2$ or $s_4$ that does not pass through c. In particular, we let $j = 3$ in the first case and $j = 2$ in the second. We will show that such a path always exists (where throughout this paragraph, all of our paths are along $\widehat T$ ). First, we note that if every path from $s_1$ to $s_4$ goes through c, then there must be a path $P_1$ from $s_1$ to $s_3$ that does not pass through c. If there is a path from $s_3$ to $s_2$ that does not pass through c, then by combining that path with $P_1$ , we get a path from $s_1$ to $s_2$ that does not pass through c as desired. Otherwise, there must be a path $P_2$ from $s_2$ to $s_4$ that does not pass through c. Furthermore, note that $s_4$ and $s_2$ are on opposite sides of the cycle $P_1 \cup \{e_1,e_3\}$ . By planarity, this implies that $P_1$ and $P_2$ must intersect at some point v. If we combine the path from $s_1$ to v along $P_1$ and the path from v to $s_2$ along $P_2$ , we get a path from $s_1$ to $s_2$ that does not pass through c.

We have now shown that we must have $k=3$ for this case to be possible. Recall that $\widehat T := \alpha _{(G,\chi )}([c-s_2],T) = (E(G) \setminus T)\setminus e_2$ . Because $k=3$ , $\widehat T$ contains exactly one edge incident to c, namely $e_3$ . In particular, $([c-s_1],\widehat T)$ is a source-turn pair. Let $\widetilde T = \alpha _{(G,\chi )}([c-s_1],\widehat T)$ . By (2), there are only 2 possibilities for $\widetilde T$ : either $\widetilde T = \widehat T \setminus e_3 \cup e_1$ or $\widetilde T = T' \cup e_2$ .

In either case, we have the following:

$$\begin{align*}\alpha_{(G,\chi)}([c-s_3],\widetilde T) = \alpha_{(G,\chi)}([(c-s_3) + (c -s_1) + (c - s_2)],T) = \alpha_{(G,\chi)}([0],T) = T.\end{align*}$$

If $\widetilde T = \widehat T \setminus e_3 \cup e_1$ , then $e_2 \not \in \widetilde T \cup T$ . Thus, by consistency, we have

$$\begin{align*}\alpha_{(G\setminus e_2,\chi\setminus e_2)}([c-s_3],\widetilde T) = T.\end{align*}$$

By our induction hypothesis,

$$\begin{align*}\alpha_{(G\setminus e_2,\chi\setminus e_2)}([c-s_3],\widetilde T) = r_{(G\setminus e_2,\chi\setminus e_2)}([c-s_3],\widetilde T) = \widetilde T \setminus e_1 \cup e_3.\end{align*}$$

This is a contradiction because $e_3 \not \in T$ . It follows that we must have $\widetilde T = T' \cup e_2$ . Notice that $(c-s_3,T' \cup e_2)$ is a source-turn pair and $r_{(G,\chi )}([c-s_3],T' \cup e_2) = T' \cup e_3 \not = T$ . By (2), this means that $\alpha _{(G,\chi )}([c-s_3],T' \cup e_2)$ cannot share any edges with $T' \cup e_2$ , and the only shared non-edge is $e_3$ . However, we showed above that

$$\begin{align*}\alpha_{(G,\chi)}([c-s_3],T' \cup e_2) = \alpha_{(G,\chi)}([c-s_3],\widetilde T) = T = T' \cup e_1. \end{align*}$$

It follows that $E(G) = \{e_1,e_2,e_3\}$ and, by the 2-connectedness condition, $(G,\chi )$ must be $E_3$ , a planar embedding of the triple edge. Thus, $\alpha _{(G,\chi )} = r_{(G,\chi )}$ by our assumption that $\alpha _{E_3} = r_{E_3}$ .

Case 4: We have $f \in \widehat T$ and $g \notin \widehat T$ .

Most of the work for this case is given in Appendix B, and there is an overview of our argument just before Lemma B.1. If $(G,\chi )$ is a telescope graph (see Definition B.8), then $\alpha _{(G,\chi )}([c-s],T) = r_{(G,\chi )}([c-s],T)$ by Lemma B.13. If $(G,\chi )$ is not a telescope graph, then $\alpha _{(G,\chi )}([c-s],T) = r_{(G,\chi )}([c-s],T)$ by Lemma B.12.

Proof. We showed in Theorem 5.4 that every consistent sandpile torsor action must be equivalent to $r_{(G,\chi )}$ , $\overline {r}_{(G,\chi )}$ , $r^{-1}_{(G,\chi )}$ or $\overline {r}^{-1}_{(G,\chi )}$ .

If $(G,\chi )$ is 2-connected, but not equal to $C_k$ for some k, then it must have a vertex c with degree at least $3$ . By 2-connectedness, there exists some $T \in \mathcal {T}(G)$ such that c is a leaf vertex of T. Let g be the unique edge of T incident to x. Furthermore, let $f = \chi (c,g)$ , $h = \chi (c,f)$ , and $\{c,s\} = {\mathfrak i}(f)$ . It follows from the definition of rotor-routing that

$$\begin{align*}r_{(G,\chi)}([c-s],T) = T \setminus g \cup f = \overline{r}_{(G,\chi)}([c-s],T\setminus g \cup h).\end{align*}$$

Because $T \not = T\setminus g \cup h$ , we can conclude that $r_{(G,\chi )} \not = \overline {r}_{(G,\chi )}$ and $r^{-1}_{(G,\chi )} \not = \overline {r}_{(G,\chi )}$ for any 2-connected $(G,\chi )$ not equal to $C_k$ for some k.

If $(G,\chi )$ is 2-connected, but not equal to $E_k$ for some k, then it must have at least three vertices. If we remove all but one edge out of every set of parallel edges, the resulting graph must contain a cycle or there would be a cut edge. After adding in the parallel edges (taking the innermost edge of each), the cycle remains and forms a face of G that is bounded by at least three edges. Let C be the set of edges which bound this face and choose an arbitrary $f \in C$ . Since $C \setminus f$ is acyclic, there must exist some $T \in \mathcal {T}(G)$ such that $C \setminus f \in T$ . Suppose that when considering the edges of C in counterclockwise order, the edge h comes just before f, and the edge g comes right after f. Furthermore, let $c = {\mathfrak i}(f) \cap {\mathfrak i}(g)$ and $s = {\mathfrak i}(f) \cap {\mathfrak i}(h)$ . It follows from the definition of rotor-routing that

$$\begin{align*}r_{(G,\chi)}([c-s],T) = T \setminus g \cup f \text{ and } \overline{r}^{-1}_{(G,\chi)}([c-s],T) =\overline{r}_{(G,\chi)}([s-c],T) = T \setminus h \cup f.\end{align*}$$

Because $T\setminus g \cup f \not = T\setminus h \cup f$ , we can conclude that $r_{(G,\chi )} \not = \overline {r}^{-1}_{(G,\chi )}$ and $r^{-1}_{(G,\chi )} \not = \overline {r}_{(G,\chi )}$ for any 2-connected $(G,\chi )$ not equal to $E_k$ for any k.

When $(G,\chi ) = E_k$ for some k, the actions $r_{(G,\chi )}$ and $\overline {r}^{-1}_{(G,\chi )}$ as well as $\overline {r}_{(G,\chi )}$ and $r^{-1}_{(G,\chi )}$ are equivalent since both vertices are incident to the same edges, but in the opposite order. When $(G,\chi ) = C_k$ for some k, the actions $r_{(G,\chi )}$ and $\overline {r}_{(G,\chi )}$ as well as $r^{-1}_{(G,\chi )}$ and $\overline {r}^{-1}_{(G,\chi )}$ are equivalent since reversing the ribbon structure has no effect on $C_k$ .

The result follows if we show that $r_{(G,\chi )} \not = r^{-1}_{(G,\chi )}$ when $(G,\chi )$ is a 2-connected ribbon graph other than $E_1=C_1$ or $E_2=C_2$ . In particular, we need to show that for some $D \in \operatorname {Div}^0(G)$ , we have $[2D] \not = [0]$ . There are a few ways to show this, but we will use the chip-firing perspective.

First, suppose that every vertex of G has degree at most $2$ . By our previous reasoning, $(G,\chi )$ must be equal to $C_k$ for some k. It is well-known and straightforward to check that $\operatorname {Pic}^0(C_k) = \mathbb {Z}/k\mathbb {Z}$ ([Reference Corry and PerkinsonCP18, Problem 2.7]). Thus, for $k>2$ , there are elements not equal to their own inverse. This means that we can assume that there is some $s \in V(G)$ such that $\deg (s)> 2$ . Let $c\in V(G)$ be any other vertex. We will show that $[2c-2s] \not = [0]$ and the result follows. First, recall that $[2c-2s] = [0]$ if and only if there is a sequence of firing moves from $2c-2s$ to $0$ . Suppose that such a sequence of firings exist. Since firing every vertex is equivalent to firing no vertices, we can assume that not every vertex is fired. The only way to reduce the number of chips on a vertex is by firing it, which means that c must fire. After c fires, any adjacent vertices other than s have a positive number of chips, so they must also fire. By recursion, it follows that a vertex v must fire if there is a path from c to v that does not pass through s. By 2-connectedness, this means that every vertex other than s must fire at least once. After these firings, s has $\deg (s)-2$ chips on it, which is positive by assumption. Thus, s must fire as well. However, this is a contradiction because we assumed that not every vertex is fired. Thus, $[2c-2s] \not = [0]$ and $r_{(G,\chi )} \not = r^{-1}_{(G,\chi )}$ .