On the sandpile model of modified wheels II

Definition 1.1

The family of graph denoted by W ˆ n is constructed from an n-wheel W n by adding one vertex on each of the cycle edge of the wheel. In particular, let W n be the wheel with vertices q , v 1 , v 2 , … , v n , where we fixed the central vertex as q. Then adding one extra vertex in between vertices v i for all i = 1 , 2 , … , n , we obtained the family W ˆ n . This family has 2 n + 1 vertices and 3 n edges. The set of vertices are given by V ( W ˆ n ) = { q , v 1 , v 2 , … , v 2 n } .

Definition 1.2

A configuration s is said to be stable if it satisfies the following inequality:

Note that a configuration with 2 dollars at each vertex is stable for W n , see [1], but not for W ˆ n .

Example 1.3

Consider the graph W ˆ 2 . A configuration on this graph is determined by a vector ( s ( v 1 ) , s ( v 2 ) , s ( v 3 ) , s ( v 4 ) ) denoting the number of dollars at the vertices v 1 , v 2 , v 3 , v 4 . Instead of referring to the vertices as v 1 , v 2 , v 3 , and v 4 , we number the vertices as 0, 1, 2, 3 with the number 0 assigned to a vertex with degree 3. Also note that the vertices 0, 1, 2, 3 are non-central vertices. A configuration is stable if and only if 0 ≤ s ( v ) ≤ 2 for v = 0 , 2 and 0 ≤ s ( v ) ≤ 1 for v = 1 , 3 . There are 12 critical configurations. The zero element is (2, 0, 2, 0) and the sandpile group S P ( W ˆ 2 , q ) is the direct sum of two cyclic groups of orders 2 and 6 generated by configurations a and b, respectively.

Lemma 2.1

If n is even number, i.e., n = 2 k , then b is a critical configuration and

Proof

Clearly b is stable. It is easy to check that the sequence of vertices q , 2 , 1 , 0 , 2 n − 1 , 2 n − 2 , 2 n − 3 , … , 4 , 3 is q -legal for b, and since each vertex is fired once the resulting configuration is b − Q u = b , which shows that b is critical. The proof of second statement is obtained by taking induction on n. The statement is true for n = 4 and n = 6 . The inductive hypothesis states that for all integers l less than or equal to n, where l = 2 k ′ . We have a l + 1 ⋅ b = c ′ when k ′ ≡ 0 mod 2 , and a l + 3 ⋅ b = c when k ′ ≡ 1 mod 2 . We shall show that a l + 5 ⋅ b = c ′ . Since a l + 5 = 4 a l + 3 − a l + 1 and by inductive hypothesis, we have a l + 5 ⋅ b = γ ( 4 c − c ′ ) = γ ( s ) , where

It can be easily verified that the following sequence is a q -legal sequence for this configuration and results in c ′

Now, we shall prove that a l + 7 ⋅ b = c . Since a l + 7 = 4 a l + 5 − a l + 3 and by induction hypothesis, we have a l + 7 ⋅ b = γ ( 4 c ′ − c ) = γ ( s ′ ) , where

It can be easily verified that the following sequence is a q -legal sequence for this configuration and results in c

Lemma 2.2

The configurations c and c ′ are critical configurations of order 3 in the abelian sandpile group S P ( W ˆ n , q ) .

Proof

Clearly c ′ is stable and the following sequence of vertices is q -legal for c ' q , 2 , 3 , 4 , … , 2 n − 1 , 0 , 1 , since each vertex is fired once, c ′ is a recurrent configuration and hence critical. The following sequence of firings is a q -legal sequence and leads to identity for 3 ⋅ c ′ .

By definition c is stable and the given sequence of vertices is q -legal for c 0 , 1 , 2 , 3 , … , 2 n − 1 , which shows that c is a recurrent configuration and hence critical. Similarly, it can be checked that the following sequence is a q -legal sequence and leads to identity for 3 ⋅ c .

Lemma 2.3

Let t be a critical configuration on W ˆ n such that

1. α ⋅ t = t + t + ⋯ + t makes all the vertices unstable (except for those which have zero dollars) for any positive integer α ,

2. all the even numbered vertices have the same number of dollars,

3. there is only one odd numbered vertex with zero dollars.

Then the only odd numbered vertex with zero dollars will always have zero dollars at the same position after stabilizing.

Proof

Let us consider a configuration t that satisfies the aforementioned conditions and let an odd numbered vertex say v i has zero dollars where index i is odd. In the very first round of firings, the vertex v i does not fire but receives 2 dollars from its neighboring vertices (as they were destabilized initially). In other rounds of firings, the number of dollars at odd numbered vertices remains constant because the odd numbered vertices are not directly connected to the central vertex so whatever they loose they gain it back through their neighbors.□

Example 2.4

Let p = ( 2 , 0 , 2 , 1 , 2 , 1 , 2 , 1 ) be a configuration on W ˆ 4 that satisfies the conditions stated in the aforementioned lemma and α = 3 , 22 , 105 , then

Example 2.5

It is easy to see that the following configuration m = ( 2 , 0 , 1 , 1 , 1 , 1 , 1 , 1 ) does not satisfy the conditions stated in the aforementioned lemma and α = 13, 16, 18 positive integer then,

Example 2.6

Let r = ( 2 , 1 , 2 , 0 , 2 , 1 , 2 , 0 , 2 , 1 ) be a configuration on W ˆ 5 , which satisfies the conditions of the aforementioned lemma, but has zero dollars at more than one vertex and α = 3 , 5 is the positive integer then, we obtained the following critical configuration after the stabilization of vertices, 3 ⋅ r = ( 2 , 1 , 0 , 1 , 2 , 1 , 2 , 1 , 0 , 1 ) , 5 ⋅ r = ( 1 , 1 , 1 , 1 , 2 , 0 , 2 , 1 , 1 , 1 ) , which shows that the conditions in the lemma are not sufficient.

Note.

For defining permutations of the rim vertices, we denote the vertices on the rim of W ˆ n by the residue classes − ( n − 1 ) , − ( n − 2 ) , … , − 1 , 0 , + 1 , + 2 , … , + ( n − 1 ) ( mod n ) as shown in Figure 1.

Theorem 2.7

If n is even, the abelian sandpile group S P ( W ˆ n , q ) is isomorphic to the direct product of two cyclic groups of orders a n and 3 a n . In particular,

where the cyclic groups are generated by configurations a and b, respectively.

Proof

Let π − 1 , n − 1 and π − 2 , − 3 denote the permutations of the rim vertices defined as follows:

Every multiple of a is symmetrical with respect to π − 1 , n − 1 , that is, ( α ⋅ a ) ( v ) = ( α ⋅ a ) ( π − 1 , n − 1 ( v ) ) . On the other hand, ( β ⋅ b ) ( v ) = ( β ⋅ b ) ( π − 2 , − 3 ( v ) ) . Since both the defined permutations generate a group which acts transitively on the rim, the only configurations that are symmetrical with respect to both permutations are those in which every rim vertex has the same number of dollars or all the even numbered vertices have the same number of dollars and all the odd numbered vertices have the same number of dollars. So, if ( α ⋅ a ) = ( β ⋅ b ) , both configurations are in fact the critical configurations in which all the even numbered vertices have the same number of dollars and all the odd numbered vertices have the same number of dollars.

The required configuration cannot be a configuration in which every rim vertex has the same number of dollars because such a configuration is not critical for this particular family of graphs. Therefore, after deleting the non-critical configurations we are left with the configurations ( 2 , 0 , 2 , 0 , … ,   2 , 0 ) and ( 2 , 1 , 2 , 1 , … ,   2 , 1 ) . By using Lemma 2.3, the choice ( 2 , 1 , 2 , 1 , … ,   2 , 1 ) is ruled out and hence ( 2 , 0 , 2 , 0 , … ,   2 , 0 ) is the required critical configuration, which is also the zero element of the sandpile group S P ( W ˆ n , q ) . It follows that the subgroup generated by a and b is the direct sum of the cyclic groups, which completes the proof.

Case II: If n = 2 k + 1 is an odd number and define some configurations on W ˆ n .

Lemma 2.8

If n is an odd number, then g is a critical configuration and a n + 1 ⋅ g = f .

Proof

Clearly g is stable. It easy to check that the following sequence of vertices

is q-legal for g ⇒ g − Q u = g . So g is recurrent, and therefore critical. The proof of the second statement is obtained by taking induction on n. It is easy to see that n = 5 , i.e., a 6 ⋅ b = 19 ⋅ b = γ ( 38 , 19 , 38 , 0 , 38 , 0 , 38 , 19 , 38 , 19 ) = ( 2 , 1 , 2 , 1 , 2 , 1 , 2 , 1 , 2 , 1 ) = f . Suppose that the statement is true ∀ k ≤ n , where n , k are both odd, i.e., a k + 1 ⋅ g = f . We shall prove that a k + 3 ⋅ g = f . Since a k + 3 = 4 a k + 1 − a k − 1 and by induction hypothesis we get, a k + 3 ⋅ g = γ ( 4 f − f ) = γ ( 3 f ) , where

It can be verified that the following sequence is a q-legal sequence for this configuration and results in f.

Lemma 2.9

Prove that f is a critical configuration and its order is 2 in the abelian sandpile group S P ( W ˆ n , q ) .

Proof

By definition, f is a stable configuration and since the following sequence of vertices 0 , 1 , 2 , 3 , … , 2 n − 1 is q-legal for f, therefore, f is a recurrent configuration and hence critical. The following sequence is a q-legal sequence for 2 ⋅ f = f + f 0 , 1 , 2 , … , 2 n − 1 , 0 , 1 , 2 , … , 2 n − 1 , 1 , 2 , 3 , … , 2 n − 1 , 0 , 1 , 3 , 5 , 7 , … , 2 n − 1 , 0 , 2 , 4 , 6 , 8 , … , 2 n − 2 , 2 n − 1 , 1 , 3 , 5 , 7 , … , 2 n − 3 and stabilization results in the identity element of the group S P ( W ˆ n , q ) .□

Lemma 2.10

Let t be a critical configuration of the abelian sandpile group S P ( W ˆ n , q ) such that

1. every even numbered vertex has the same number of dollars,

2. the number of dollars at v 1 and v 2 n − 1 is the same,

3. the number of dollars at remaining odd vertices is zero,

4. multiplying by α destabilizes all the vertices,

then after stabilizing v n will always have zero dollars.

Proof

It is easy to identify that the only configurations that satisfy the aforementioned conditions are d and e given as:

Theorem 2.11

If n is an odd number, then the abelian sandpile group S P ( W ˆ n , q ) is isomorphic to the direct product of two cyclic groups of order a n and 2 a n . In particular,

where the cyclic groups are generated by the configurations d and g, respectively.

Proof

Let π 0 , n and π ′ denote the permutations of the rim vertices defined as follows:

Every multiple of d is symmetrical with respect to π 0 , n , that is, ( α ⋅ d ) ( v ) = ( α ⋅ d ) ( π 0 , n ( v ) ) . On the other hand, every multiple of g is symmetrical with respect to π ′ , that is, ( β ⋅ g ) ( v ) = ( β ⋅ g ) ( π ′ ( v ) ) . Since π 0 , n and π ′ generate a group which acts transitively on the rim, the only configurations which are symmetrical with respect to both permutations are those in which all the even numbered vertices have the same degree and all the odd numbered vertices have the same degree or every rim vertex has the same degree. So, if α ⋅ d = β ⋅ g , then both configurations are in fact equal to the critical configuration e = ( 2 , 0 , 2 , 0 , … , 2 , 0 ) because Lemma 13 ensures that the required critical configuration cannot be ( 2 , 1 , 2 , 1 , … , 2 , 1 ) = f , which completes the proof.□