Soliton cellular automata for the affine general linear Lie superalgebra

The box-ball system (BBS) is a cellular automaton that is an ultradiscrete analogue of the Korteweg--de Vries equation, a non-linear PDE used to model water waves. In 2001, Hikami and Inoue generalised the BBS to the general linear Lie superalgebra $\mathfrak{gl}(m|n)$. We further generalise the Hikami--Inoue BBS to column tableaux using the Kirillov--Reshetikhin crystals for $\hat{\mathfrak{gl}}{(m|n)}$ devised by Kwon and Okado (arXiv:1804.05456), where we find similar solitonic behaviour under certain conditions.


Introduction
The box-ball system (BBS) is an integrable nonlinear dynamical system that has connections to both classical and quantum integrable systems. The Korteweg-De Vries (KdV) equation is one such classical system which describes shallow water waves in a one-dimensional channel [KdV95]. It was shown by Kruskal and Zabusky that solutions to the KdV equation separate into solitonic waves that move with speed proportional to their amplitude and maintain their shape under collision with other solitons [KZ64]. It was later discovered that the ultradiscretisation of these soliton solutions produces the BBS [TTMS96].
In addition to its connection to classical systems, the BBS also emerges from quantum integrable systems such as the six-vertex lattice model from statistical mechanics [Bax89]. The symmetries of this particular system are governed by the quantum group U ′ q (sl 2 ). Under crystallisation (the q → 0 limit) the system is frozen to the ground state and produces the BBS [HIK99]. The position of BBS within the realm of classical and quantum integrable systems opens its analysis to a variety of methods. Moreover, the discrete nature of the system provides an important connection to combinatorics.
An important development in the analysis of the BBS comes from the construction of crystal bases by Kashiwara [Kas91,Kas90]. This allowed for the reformulation of crystallisation in terms of the crystal theory of quantum affine algebras [HKT00], leading to a crystal theoretic formulation of the BBS. This formulation utilises the 'classical' crystal B 1,s , which is the crystal base of an s-fold symmetric tensor representation of U q (sl n ) promoted to the Kirillov-Reshetikhin (KR) crystal of U ′ q ( sl n ) [KKM + 92] by adding additional crystal operators. States of the system are then defined as elements of (B 1,1 ) ⊗∞ . The time evolution of these states utilises the existence of the combinatorial R-matrix, a unique isomorphism between the tensor product of KR crystals, R : B ⊗ B ′ → B ′ ⊗ B [KKM + 92]. The time evolution is given by repeated applications of this R-matrix together with a carrier. Below is an example of a BBS constructed from the KR crystal B 1,1 in A (1) 1 .
Example 1.1. For B 1,1 in A (1) 1 with carrier B 1,3 , the time evolution of the state 3 ⊗ 3 ⊗ 2 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ · · · is the state where each crossing represents the application of the R-matrix, R : B 1,3 ⊗ B 1,1 → B 1,1 ⊗ B 1,3 . The top row represents the initial state, the bottom row represents the state after one time evolution, and the middle represents how the carrier changes during the evolution.
Within the BBS there exist states exhibiting solitonic behaviour; that is, states containing elements of (B 1,1 ) ⊗d within (B 1,1 ) ⊗∞ that move with speed corresponding to their length and are stable under collisions (this stability is called scattering). Such elements are called solitons. These solitons are the ultradiscrete analogue of the KdV solitons. In Example 1.1, 3 ⊗ 3 ⊗ 2 is a soliton. For more detail we refer the reader to [FOY00, HHI + 01].
In 2001, Hikami and Inoue generalised the BBS using crystals for the general linear Lie superalgebra gl(m|n) and showed that similar solitonic behaviour existed in this supersymmetric system [HI00]. Work by Yamada [Yam04] generalised the system in a different manner by considering the crystal B r,s of U ′ q ( sl n ), producing a system with r rows. This paper uses the KR crystals for gl(m|n) devised by Kwon and Okado [KO21] to generalise the Hikami-Inoue BBS analogously to Yamada's generalisation of the U ′ q ( sl n ) BBS. Each gl(m|n) KR crystal is parameterised by a Young diagram Y , and the crystal is identified with the set of semistandard Young tableaux (SSYT) of shape Y . We are primarily interested in B r,s ; the crystal of rectangular SSYT of height r and width s. In our generalised BBS, we define states as elements of B r,1 ⊗∞ . We similarly have an R-matrix giving a bijection of tensor products of crystals B r1,s1 ⊗ B r2,s2 → B r2,s2 ⊗ B r1,s1 . The R-matrix can be explicitly calculated with the RSK algorithm, using the modified Schensted's bumping algorithm outlined in Section 2.3. This allows us to define the time evolution of the system analogously to the classical case. Taking r = 1 reduces our system to the Hikami-Inoue BBS [HI00]. The following example gives a two-soliton state within our generalised system. Example 1.2. Consider the U q ( gl(3|1)) crystal B 2,1 . In the following diagram is a state (in (B 2,1 ) ⊗∞ ) evolved over four time steps starting at time t = 0. The maximal weight element ( 3 2 ) is represented as a dot.
states have a form that is not analogous to any of the highest weight states in the non-super symmetric system of [Yam04].
The paper is organised as follows. In Section 2 we quickly review the crystal base theory required for our purposes, including the computation of crystal operators, the combinatorial R-matrix and the energy function. In Section 3 we present the explicit structure of our generalised system and outline the process of time evolution. In Section 4, we present our two main theorems and our conjectures. The proofs of the theorems can be found in Appendices A and B, respectively. Appendices C and D contain the proofs of some technical lemmas used in Appendix B.
. If we instead consider the fundamental representation of the finite type U q (gl(m|n)), the crystal graph the same as above but without the 0 arrow. We can interpret the finite type crystal graph (with arrow labels removed) as a total ordering; explicitly, m < · · · < 1 < 1 < · · · < n. For a more detailed explanation of crystals for U q ( gl(m|n)), see [KO21].
2.2. Finite type crystals and tableaux. Now we restrict our attention to the finite type crystal. Let V ⊗N be the N -th tensor power of the fundamental representation of U q (gl(m|n)). It can be shown that all tensor powers with N ≥ 1 are completely reducible. Moreover, the irreducible subrepresentations (up to isomorphism) are in bijection with Young diagrams of (m|n)-hook shape [BR87,BKK00]. This bijection is derived using a map from crystal base elements to semistandard Young tableaux. In this context, a tableau is called semistandard if the rows are weakly (resp. strictly) increasing for letters in B − (resp. B + ) and the columns are weakly (resp. strictly) increasing for letters in B + (resp. B − ).
We map crystal base elements to Young diagrams using a modified version of Schensted's bumping algorithm. For inserting i ∈ B into a tableau T , which we will denote i → T , the bumping algorithm is as follows: (1) For i ∈ B + , (resp. i ∈ B − ): if none of the boxes in the first column of T are strictly larger than i (resp. larger than or equal to i) then add a box containing i at the bottom of the column.
(2) Otherwise, for the topmost j with j > i (resp. j ≥ i) in the first column, replace j with i . Then, insert j into the second column following analogous steps 1 and 2.
(3) Repeat until the bumped number can be put in a new box.
Example 2.1. The following is an example computation of the bumping algorithm: Let v b1 ⊗ v b2 ⊗ · · · ⊗ v bN ∈ V ⊗N be a crystal base element with b 1 , . . . , b N ∈ B. The SSYT associated with this crystal base element is the insertion which, for brevity, we will denote b 2 · · · b N → b 1 . Note that the map from crystal base elements to SSYT is not injective (for example, v 3 ⊗v 2 ⊗v 5 ⊗v is mapped to the same tableau in Example 2.2). However, this map sends crystal base elements of isomorphic irreducible subrepresentations to the same set of SSYT (in particular, this map gives a bijection between irreducible subrepresentations (up to isomorphism) and Young diagrams). Note also that it is possible to construct a bijection between crystal base elements and ordered pairs of tableaux using the RSK algorithm (which makes use of Schensted's bumping algorithm) [BR87].
Using SSYT allows us to give combinatorial descriptions of U q (gl(m|n)) crystals. We will now restrict our attention to rectangular tableaux, but much of the discussion in this section applies more generally.
Let B r,s be the set of rectangular SSYT with height r and width s. Take, an arbitrary tableau, We define a function, col by reading the tableau from top-to-bottom, right-to-left; explicitly, Moreover, for T 1 , T 2 ∈ B r,s , we define col(T 1 ⊗ T 2 ) = col(T 1 ) col(T 2 ). For i ∈ I even , the action of the crystal operators e i and f i can be computed by a signature rule similar to that for U ′ q ( sl n )-crystals [Yam04].
Definition 2.3. For some positive integer d, let T ∈ (B r,s ) ⊗d and let i ∈ I even . If i = k ∈ I − , we denote i + 1 = k + 1. We define the i-signature, denoted sg i (T ), to be the sequence of + and − obtained by deleting all letters in col(T ) that are not i or i + 1, and then replacing all i with a − symbol and replacing all i + 1 with a + symbol.
We define the reduced i-signature, denoted rsg i (T ), to be equal to the i-signature, except with +− pairs (in that order) successively deleted, so that rsg i (T ) is of the form (where a or b can be zero).
For a tableau T ∈ B r,s and for i ∈ I even where i ∈ I + (resp. i ∈ I − ): • To evaluate f i (T ) (resp. e i (T )), find the rightmost − symbol in rsg i (T ) and change the corresponding i in T to i + 1 . If there are no − symbols, then f i (T ) = 0 (resp. e i (T ) = 0).
• To evaluate e i (T ) (resp. f i (T )), find the leftmost + symbol in rsg i (T ) and change the corresponding i + 1 in T to i . If there are no + symbols, then e i (T ) = 0 (resp. f i (T ) = 0).
The f 0 and e 0 operators have a different algorithm: • If the first occurrence of 1 in col(T ) is before the first occurrence of 1, then e 0 (T ) = 0 and f 0 (T ) replaces the corresponding 1 in T with 1 .
• If the first occurrence of 1 in col(T ) is before the first occurrence of 1, then f 0 (T ) = 0 and e 0 (T ) replaces the corresponding 1 in T with 1 .
Example 2.4. We will compute e 3 (T ) for .
We can also use SSYT to describe the weights (in the representation theoretic sense) of the crystal elements. Weights are linear combinations in the set b∈B Zε b (for our purposes, ε b can be treated as formal symbols). In the weight of a SSYT T , the coefficient corresponding to ε b is equal to the number Example 2.5. Let T be as in Example 2.4. Then, the weight of T is ε 4 + 3ε 3 + 2ε 1 + ε 2 + 2ε 3 .
Note that the operators e i (i ∈ I \ {0}) raise the weight and the operators f i (i ∈ I \ {0}) lower the weight. We say that T is a highest weight element if e i T = 0 for all i ∈ I \ {0}.
Definition 2.6. A crystal element T with weight λ is a genuine highest weight element if (i) given any other crystal element with some weight µ, the expression λ − µ has only positive coefficients; and (ii) no other crystal element has weight λ.
Every genuine highest weight element is a highest weight element, but not every highest weight element is a genuine highest weight element [BKK00].
For crystals whose elements are the SSYT of the same shape, the genuine highest weight element exists and is unique [BKK00]. Each connected component of B r1,s1 ⊗ B r2,s2 is isomorphic to such a crystal (this isomorphism is given by (T 1 ⊗ T 2 ) → (col(T 2 ) → T 1 ) for T 1 ⊗ T 2 in the connected component of interest). Thus, each connected component of B r1,s1 ⊗ B r1,s1 has a unique genuine highest weight element. This property is of great utility in the proofs of the main theorems.
2.3. Combinatorial R-matrix. Consider two U q ( gl(m|n))-crystals B r1,s1 and B r2,s2 . Then there exists [KO21] a unique isomorphism that commutes with e i and f i (for all i ∈ I) called the combinatorial R-matrix : We can describe the action of the combinatorial R-matrix using Schensted's bumping algorithm.
Theorem 2.7 ([KO21, Theorem 7.9]). The combinatorial R-matrix maps T 1 ⊗ T 2 to T 2 ⊗ T 1 if and only if col(T 2 ) → T 1 = col( T 1 ) → T 2 Example 2.8. Set We similarly find that There is a more explicit method of determining the R-matrix of two tableaux. This involves an inversion of the modified bumping algorithm outlined previously. Let T 1 ∈ B r1,s1 and T 2 ∈ B r2,s2 . Then R(T 1 ⊗ T 2 ) is determined by the following process. Begin with P = col(T 2 ) → T 1 . Let Q be a rectangular reverse semi-standard tableau of height r 1 and width s 1 . We construct Q using the weight vector given by µ = ( d i − d i , · · · , d s − d s ) where d i and d i are the heights of the ith column of P and T 2 respectively, and s is the width of P . Then Q is given by the unique reverse conjugate semi-standard tableau of shape T 1 and weight µ.
We then perform a reverse insertion on P by reading Q bottom-to-top, left-to-right. Each element in the reading of Q gives the next column on which we perform the bumping algorithm in reverse. The elements removed from P are placed bottom-to-top and left-to-right into a rectangular tableau of height r 1 and width s 1 . This tableau is T 1 . Continuing until Q is empty, we obtain the resultant tableaux as P = T 2 and T 1 .
Example 2.9. This example will demonstrate the explicit R-matrix computation for T 1 and T 2 defined as follows The weight vector associated with Q is then given by (2, 3, 3, 1). There exists a unique reverse conjugate semi-standard tableau of shape T 1 and weight (2, 3, 3, 1) given as follows, . Now reading Q from bottom-to-top, left-to-right. We first read the element 3, beginning in column 3 of P we pop the final row element 3, then performing reverse insertion from column 2, this 3 switches with the 2 which then subsequently switches with 1 in column 1. The 1 left over then begins filling a tableau T 1 bottom-to-top, left-to-right. We are then left with the following P and Q, and T 1 : Continuing this process until Q is empty we obtain the following tableaux The energy function exists and is unique up to additive constant [KO21]. Moreover, we can compute the energy function using the bumping algorithm.
Proposition 2.11 ([KO21, Theorem 7.9]). Up to additive constant, H(x ⊗ y) is given by the number of boxes in col(y) → x that are strictly to the right of the max(s 1 , s 2 )-th column.
By convention, we will choose the additive constant so that the maximum value of H is zero. Explicitly, if H(x ⊗ y) is given by the number of boxes as in Proposition 2.11, with additive constant equal to 0, then we define H(x ⊗ y) = H(x ⊗ y) − min(r 1 , r 2 ) min(s 1 , s 2 ).

Super Box-Ball System
3.1. Box-ball system definition. A box-ball system possesses a vacuum element representing the absence of a ball. We require that the combinatorial R-matrix acts as an identity on the vacuum element; that is, if u is the vacuum element then R(u ⊗ u) = u ⊗ u.
We define the vacuum element to be the genuine highest weight element of B r,1 as a finite-type U q (gl(m|n))-crystal where r ≤ m (see [BKK00]). Such a vacuum element will have the desired property. More generally, the genuine highest weight element for B r,s has the form The vacuum element is then denoted by u 1 . We can think of the elements of B r,1 \ {u 1 } as representing different balls in the system. Within the super box-ball system, a state consists of B r,1 elements in a one dimensional lattice with only finitely many non-vacuum elements. More precisely, a state is of the form where b j ∈ B r,1 can be any element (including u 1 ). The state evolves in time via a carrier element which 'picks up' and 'puts down' B r,1 elements. The carrier is an element of B r,ℓ which changes based on its location in the state. It is initialised as the genuine highest weight element u ℓ . The action of moving the carrier through the state is performed by the combinatorial R-matrix. In particular, this is performed by functions R a where We can then define the time evolution operator , T ℓ , by for any state b. This is well-defined because there are finitely many non-vacuum elements in the state, so we eventually have R(u ℓ ⊗ u 1 ) = u 1 ⊗ u ℓ . The time evolution operator computes the state for the next time step. For convenience, we will write T ∞ = lim ℓ→∞ T ℓ .
Proposition 3.1. T ∞ is well-defined Proof sketch. A simple insertion argument shows that the action of the R-matrices in the definition of T K and T ℓ are the same (under the inclusion B r,K ֒→ B r,ℓ , v → u ℓ−K v ). This shows that T ℓ = T ℓ ′ for ℓ, ℓ ′ ≥ K and hence T ∞ is well-defined.
Pictorially, we can represent the computation of the time evolution That is, Remark 3.3. We only consider BBSs with r ≤ m. This is because we encounter difficulties if we consider r > m. For instance, the empty carrier u ℓ will contain fermionic boxes which will increase in value horizontally. So, we can no longer think of u ℓ as containing ℓ vacuum elements. More concerning, the time evolution operator may no longer be well-defined. Consider a BBS defined from a U q ( gl(1|3))crystal with r = 2 and p = 1 Then, which is not of the form p ⊗ u 2 = p ⊗ 1 1 1 2 that is required for our above definition of T 2 (p). Additionally, we do not want to define T 2 (p) to be because the boxes inside T 2 (p) are different from p; intuitively, the 'mass' is no longer a conserved quantity. This indicates that other conserved quantities may not be present, which impacts the integrability of the system. Nevertheless, such a system with r > m may exhibit interesting behaviour or applications, but is beyond the scope of this article.
Remark 3.4. Many authors will define a state so that the tensor product extends infinitely in both directions (with only finitely many non-vacuum states). If we define a state in this way then the system is time reversible, since the uniqueness of the combinatorial R-matrix implies R −1 a = R a .
3.2. Properties of the time evolution operator.
Proposition 3.5. Time evolution operators commute: The proof of this fact is identical to [FOY00, Theorem 3.1], and relies on the Yang-Baxter equation: The time evolution operator also respects the crystal structure, i.e., T ℓ commutes with some of the crystal operators, as outlined in the following Lemma: Lemma 3.6. For all i ∈ I \ {0, m − r}, and for a state p, we have that T ℓ (e i (p)) = e i (T ℓ (p)) and T ℓ (f i (p)) = f i (T ℓ (p)).
The proof is similar to [Yam04, Lemma 2.8]. Let B 0,m−r be the U q ( gl(m|n))-crystal of BBS states where the operators f 0 , e 0 , f m−r and e m−r have been removed. Note that B 0,m−r is isomorphic to a U q (gl(r)) ⊗ U q (gl(m − r|n))-crystal. Lemma 3.6 allows us to prove results by only considering a single element from each connected component of B 0,m−r . In practice, this means that it is sufficient to consider genuine highest weight elements of B 0,m−r .

4.1.
Properties of solitons and coupled solitons. In this paper, a soliton is an element of (B r,1 \ {u 1 }) ⊗d that moves with constant speed (not necessarily equal to its length).
for some positive integer c. We call c the speed of the soliton.
This definition is very broad, and these solitons do not satisfy many of the properties of we want.
However, this broad definition is convenient for our purposes. Many of the important properties will be satisfied by a specific type of soliton (uncoupled solitons) which is defined by conserved quantities N ℓ .
Let p = p 1 ⊗ p 2 ⊗ p 3 ⊗ · · · be a state. Let u (j) ℓ be the carrier after applying the R-matrix j times; that is Define a function E ℓ by where H is the energy function (note that the above sum is finite because we chose H such that H(u ℓ ⊗ u 1 ) = 0).
The proof of this proposition is the same as [FOY00, Theorem 3.1]. Define N ℓ by with E 0 = 0. We can now use N ℓ to define uncoupled solitons. Note that s ≤ d. Indeed, the proof of Proposition 3.1 shows that T ℓ = T ℓ ′ for ℓ, ℓ ′ ≥ d. In fact, the same insertion argument shows that E ℓ = E ℓ ′ for ℓ, ℓ ′ ≥ d and hence N j = 0 for j > d. Moreover, we believe that s = d for every uncoupled soliton (see the discussion after Conjecture 4.11) so we can interpret s as the length of the uncoupled soliton.
Intuitively, a coupled soliton contains overlapping uncoupled solitons that are not interacting with one another. With this intuition in mind, we can interpret N ℓ as the total number of uncoupled solitons of length ℓ in a state. With this interpretation, we should expect that a coupled soliton has N s = 0 for exactly one positive integer s. Indeed, were this not the case, then a coupled soliton would contain overlapping uncoupled solitons of different speeds, and we would expect these uncoupled solitons to separate given enough time (contradicting the fact that the coupled soliton is a soliton -Definition 4.1).
We formalise this intuitive interpretation of N ℓ in the following conjecture, which claims that a coupled soliton v can be split into uncoupled solitons after collision.
, and for some integers c 1 , c 2 , . . . , c B+A . Note that each v j has speed s and that each w j has speed d j .
Example 4.5. Consider the U q ( gl(3|1)) coupled soliton v = 1 1 . Note that N 1 (v ⊗ u ⊗∞ 1 ) = 2 and N j (v ⊗ u ⊗∞ 1 ) = 0 for j = 1. So we expect that v contains two overlapping uncoupled solitons of length one. In the following diagram, we pass the uncoupled soliton 3 3 3 1 1 1 through a state containing v. The maximal weight element 3 2 is represented as a dot. We find that v splits into two copies of the uncoupled soliton 3 1 after collision.
So we expect that v contains three overlapping uncoupled solitons of length one. In the following diagram, we pass two copies of the uncoupled soliton We know that N ℓ is a conserved quantity (since E ℓ is). We can interpret this fact as a form of stability under collision, which is one of the important properties of solitons.
In the height 1 BBS, every state separates into solitons given enough time. We conjecture that this is also true in our system (though the solitons might be coupled).
Conjecture 4.7. Let p be any state. There exists some positive integer t, some solitons v 1 , v 2 , . . . , v D of speeds d 1 , d 2 , . . . , d D (respectively) and some positive integers c 1 , c 2 , . . . , c D , such that for any t > t, We have verified this conjecture for all U q ( gl(2|2)) and U q ( gl(3|3)) states with only the first five factors being non-vacuum elements.
There are two important properties that the solitons of the KdV equation and of height 1 BBSs satisfy: they move with speed corresponding to their length and are stable under collision. In general, the solitons of Definition 4.1 do not satisfy these two properties (even though N ℓ is conserved). However, we will show that uncoupled solitons do satisfy these two properties.

4.2.
Solitons with speed equal to their length. One of the properties of the height 1 BBS is that the speed of the solitons is equal to their length. In general, this is not true of the solitons in our system.
In this section, we provide a large class of solitons which move with speed corresponding to their length. We conjecture that this is the largest such class and that it contains all of the uncoupled solitons.
x rs · · · x r2 x r1 form a SSYT and that there is a row number k (1 ≤ k ≤ r) such that x ij < m − r for all j and for i < k x ij ≥ m − r for all j and for i ≥ k.
for all positive integers t.
We first use the R-matrix insertion algorithm to prove the theorem for a genuine highest weight state, and then generalise using Lemma 3.6. The proof is given in Appendix A.
Conjecture 4.9. The subset of ∞ d=1 (B r,1 ) ⊗d defined by Theorem 4.8 is the largest such subset of solitons which move with speed equal to their length.
The value of k relates the structure of these solitons to the value of N s .
Applying the bumping algorithm while moving the carrier through the soliton (c.f. Appendix A), we find that H(u In particular, note that if k = r then x is an uncoupled soliton. We propose the following generalisation of the above proposition. (4.1) Intuitively, we interpret r + 1 − k j as the number of overlapping solitons at the j-th position of the soliton.
Proposition 4.12. If we assume Conjecture 4.11, then the solitons of Theorem 4.8 with k = r are the only uncoupled solitons.
Proof. Since r + 1 − k j ≥ 1, the left-hand side of (4.1) is at least d, and for an uncoupled soliton, the right-hand side of (4.1) is s. But we already know s ≤ d. We deduce that s = d and hence k j = r for all j.

4.3.
Scattering of two solitons. One of the main properties of the height 1 BBS is that solitons are stable under collision. This behaviour is also called scattering.
Definition 4.13. Let v, w be uncoupled solitons of lengths d 1 , d 2 respectively, with d 1 > d 2 . We say that v and w scatter if there exist non-negative integers c 2 , t such that for any t > t and c 1 ∈ Z ≥0 , for some non-negative integers c 3 , c 4 (dependent on t) and some uncoupled solitons w, v of lengths d 2 , d 1 respectively.
We can interpret this definition as saying that the longer soliton, v, eventually collides and interacts with the shorter soliton, w, after which the states separate into two solitons again. However, it is important to note that w and v are generally different from w and v, respectively. We have already seen an example of scattering for gl(m|n) in Example 1.2.
Using the same notation in Definition 4.13, let Let V be a SSYT. Let V ↓ denote the bottom row of V , and V ↑ denote the other rows of V . We will just consider the case where V ↓ only has entries greater than or equal to m − r, and V ↑ only has entries strictly less than m − r (where r is the height of V ). In the notation from Theorem 4.8, we are only considering the case where k = r.
Theorem 4.14. Consider uncoupled solitons composed of elements of U q ( gl(m|n))-crystals with height r ≤ m. Then any two uncoupled solitons of the form given in Theorem 4.8 (i.e. with k = r) scatter.
Moreover, let v, w be uncoupled solitons and let w, v be obtained from v, w as in Definition 4.13. The elements v and w are related to w and v via their semistandard Young tableaux (SSYT). Let V, W, W , V be the SSYT corresponding to v, w, w, v, respectively. Then, and The phase shift is given by The proof of the above theorem is given in Appendix B Note that the assumption of uncoupled is not a necessary condition, and some other coupled solitons also scatter as in Defintion 4.13.
Example 4.15. Consider the following time evolution of a BBS composed of elements from the U q ( gl(4|1))crystal with r = 2.
However, in general, the collisions of coupled solitons are more complicated. Calculating N ℓ , we find that N 1 = 2, N 2 = 2 and N ℓ = 0 for ℓ > 2. Acknowledgements. The authors would like to thank Masato Okado for helping them understand coupled solitons through the use of numbers N ℓ (Section 4.1). The authors are especially grateful to Travis Scrimshaw for all of his invaluable support and guidance. The authors would also like to thank the referee for taking their time to read the paper and providing comments.
Appendix A. Proof of Theorem 4.8 Fix a positive integer k ≤ r. Let h = min{r − k + 1, m − r} and j ∈ B with 1 ≤ j ≤ n. For convenience, we use the following notation: Note that if k = 1 then X is an empty tableau. Similarly, if r − k + 1 − h = 0 then j is empty.
Since any state can be reached by successively applying crystal operators to a genuine highest weight state, by Lemma 3.6 it is sufficient to prove the theorem for genuine highest weight states (with respect to the crystal B 0,m−r ).
The genuine highest weight states are of the form Proof. First we compute the bumping algorithm for the left-hand side. That is, we want to compute: To insert i, observe that all of the columns of the tableau contain i except the column after S s (possibly empty). Hence, i will replace itself in all of these columns, leaving them unchanged. Because the cells inserted before i are smaller than i, the column after S s can contain only cells smaller than i, Hence, i gets inserted at the end of this column. Thus, The remaining items to be inserted are larger than the last cell of the first column (m − r + 1), and are in ascending order, so they all get appended on to the end of the first column: Similarly, we compute the bumping algorithm for the right-hand side: Hence, as required to prove the lemma.
Proof. Observe that Proof. In the process of performing the bumping algorithm for the right-hand side, we find Using Lemmas A.1, A.2 and A.3, we can easily show that Using Lemma 3.6, this is sufficient to prove Theorem 4.8 for general states.
Appendix B. Proof of Theorem 4.14 B.1. Reductions. Consider gl(m|n) with height r tableaux, and assume m ≥ r + 2. Set, We employ the following notation: To prove Theorem 4.14 in the case where m ≥ r + 2, it suffices to prove the following proposition: Proposition B.1. Let p be a genuine highest weight state in the U 1 (gl(r)) ⊕ U q (gl(m − r|m))-crystal B 0,m−r . That is, Indeed, by Lemma 3.6, certain crystal operators commute with the time evolution operator T ℓ .
Since any state can be reached by successively applying crystal operators to a genuine highest weight state [BKK00], it is sufficient to prove Theorem 4.14 for genuine highest weight states of B 0,m−r .
All genuine highest weight states are given by B.1. Indeed, the vacuum elements are invariant under the crystal operators of B 0,m−r , so can be safely be ignored. Additionally, given the assumption k = r in Theorem 4.14, we can treat the bottom row and the other rows as two seperate crystals, (which are a U q (gl(r))-crystal and a U q (gl(m − r|n))-crystal respectively). Finally, due to the semistandard assumption of each soliton, the bottom row is isomorphic to the U q (gl(r))-crystal B 1,d1 ⊗ B 1,d2 and the other rows are isomorphic to the U q (gl(m − r|n))-crystal B (r−1),d1 ⊗ B (r−1),d2 (the isomorphisms are derived from Schensted's bumping algorithm). We know the general form for genuine highest weight elements for these crystals [KO21, Proposition 7.1].
for the genuine highest weight states (using the notation from Theorem 4.14) then it will also hold for a general state, since the crystal operators commute with the R-matrix. Note that we can explicitly calculate R(V ↑ ⊗W ↑ ) and R(V ↓ ⊗W ↓ ) for the genuine highest weight states, and we do so in Lemma B.2. Then, by inspection, we see that The above two insertions are identical, proving the first part of the lemma.
Proof. From the proof of Lemma B.2, we know There are d 2 (r − 1) − L boxes to the right of the max{d 1 , d 2 } = d 1 column. By definition, we have From the proof of Lemma B.2, we know There are d 2 − M boxes to the right of the max{d 1 , d 2 } = d 1 column. By definition, we have Finally, we have that (T ℓ ) t = (T d2+1 ) −t ′ (T ℓ ) t (T d2+1 ) t ′ since time evolution operators commute by Proposition 3.5, (note that (T d2+1 ) −1 exists; see Remark 3.4). Therefore, if we prove the theorem for T d2+1 and choose t ′ sufficiently large (so that the solitons have finished colliding after applying (T d2+1 ) t ′ ) and choose t > t ′ (so that (T d2+1 ) −t ′ (T ℓ ) t does not undo the collision; this follows from Theorem 4.8), then we can prove the theorem in the general case.
To prove Proposition B.1, we could use direct (and tedious) computation. However, in the rest of this section we give an alternate proof, in which we reduce the behaviour of the general system to the behaviour of a modified height 1 system. B.2. Behaviour of the R-matrix. It is first useful to classify how the R-matrix acts on common tableaux.
The proof of the lemma follows by a direct application of Schensted's bumping algorithm. See Appendix C for details.
Remark B.5. The above lemma only lists the cases required to prove the main theorem and does not completely classify the action of the R-matrix.
Remark B.6. In Lemma B.4, X does not affect the action of R-matrix on genuine highest weight states. Therefore, it is sufficient to only consider height 2 states.
Note that many cases of the lemma can be informally described using the height 1 R-matrix: Example B.7. Assuming b 1 > 1, R( 2 b2 1 b1 0 b0 ⊗ 0 ) = 1 ⊗ 2 b2 1 b1−1 0 b0+1 . Compare this with the bottom row of (2a): 1 ⊗ 2 b2+1 1 b1−2 0 b0+1 . Additionally, note that the second last row has an extra 1. One interpretation of this is that when the carrier picks up the 2 in the second last row, it gets swapped with a 1 in the bottom row.
With these examples in mind, we can see when the carrier picks up a 2 in the second last row, it generally gets swapped with a 1 in the bottom row; and when the carrier 'should' put down a 2 in the bottom row, it generally gets swapped with a 1 from the second last row. Note that sometimes this happens simultaneously. Note also that there are exceptions ((2d) for example).
This observation tells us that when computing time evolution by moving the carrier through a genuine highest weight state, the bottom row of the super BBS is similar to the single row case, except for some swapping of 2 and 1. This swapping can occur simultaneously and is hard to keep track of. Our strategy is to swap all of the 2 from the second last row with 1 from the bottom row before passing the carrier through, so that no swapping occurs when passing the carrier through. After employing this strategy to create a 'swapped state', the bottom row would theoretically behave identically to the single row case. This would reduce the problem of computing the time evolution to the single row case. This is advantageous because the behaviour of the single row BBS is simpler and also easier to compute since we can use an alternate algorithm for the R-matrix detailed in [HI00, Section 2]. Unfortunately, the first soliton moves slower in the 'swapped state' than in the original state. We can fix this by 'speeding up' the first soliton.
B.3. L-sped box-ball system. With this goal of 'speeding up' the first soliton, we define a new time evolution operator: Definition B.9. Consider a general state where y d2 · · · y 2 y 1 is semistandard and y j = u 1 for any j = 1, . . . , n. (Note that d or c 2 can be 0.) For some non-negative integer L, set Let T ℓ (P L ) = P 1 ⊗ P 2 ⊗ P 3 ⊗ · · · and assume that P c1+1 is the first non-vacuum element in T ℓ (P L ).
Define the L-sped time evolution operator , denoted T L ℓ , by Note that in the case where d ≤ ℓ − L, we have T L ℓ (P ) = T ℓ (P L ).
Example B.10. Let L = 2 and ℓ = 5, and consider T L ℓ acting on the state We have that Note that we have only defined the L-sped time evolution operator for some two-soliton states. This is sufficient for our purposes, but the definition can be extended using crystal operators if so desired.
Proposition B.11. Let Proof. We follow a similar strategy as [HI00] and [FOY00]. Consider applying the time evolution operator to P . For c 2 suficiently large, the first soliton initially moves with speed d 2 + 1 and the second soliton with speed d 2 . Eventually, we arrive at the state: By direct computation (using either the R-matrix algorithm in Section 2.3 or in [HI00, Section 2]), we find that after another t ′ time units (with t ′ ≤ d 1 − d 2 ), this state becomes: When t ′ > d 1 − d 2 the solitons do not interact, and the first soliton moves with speed d 2 and the second soliton moves with speed d 2 + 1.
Let δ 3 be the phase shift of the speed d 2 + 1 soliton, and δ 2 be the phase shift of the speed d 2 soliton. After time t ′ = d 1 − d 2 , the phase shifts will remain the same, so computing both phase shifts at time Thus, the total phase shift of the system is δ = δ 3 = −δ 2 = 2d 2 − M − L.
Note that Proposition B.11 essentially shows us that two-soliton states behave solitonically with respect to L-sped time evolution. Our aim is to reduce the behaviour of the general super BBS to the behaviour of a height 1 L-sped BBS.
B.4. Reducing to height 1. As alluded to in Section B.2, we can define a map from a genuine highest weight super BBS state to a L-sped state by swapping all of the 1 in the bottom row with 2 in the second last row, and then removing all but the bottom row. More precisely, assuming d 1 − L > 0, the super BBS state (recall that we are omitting X as per Remark B.6), where y 1 y 2 · · · y d2 = 0 M 1 d2−M and (Q > 0 only if Call this map F .
Example B.12. Let The map F is invertible. Indeed, F −1 maps the L-sped state where d > 0, to the super BBS state Remark B.13. In defining the domain of F , we assumed that d 1 − L > 0. This assumption ensures that F −1 is well-defined. We sometimes encounter states where d 1 = L and we need to deal with these edge cases separately.
Recalling that we can alternatively compute F −1 by the following process: (1) Add a row above the L-sped state, and fill it entirely with 3 (if needed, add rows filled with X above this added row).
(2) In the second soliton, replace the first L positions of this added row with 2.
(3) Replace the first L occurrences of 2 that occur immediately after the first soliton (regardless of the row they appear in) with 1.
The only states encountered when evolving a genuine highest weight super BBS are the form P (or of the form discussed in Remark B.13); this will be a consequence of the proof of Proposition B.14.
Therefore, Proposition B.14 allows us to compute the time evolution of a genuine highest weight super BBS state using almost entirely height 1 L-sped states. Since L-sped states behave solitonically by Proposition B.11, this will complete the proof of Proposition B.1 and hence Theorem 4.14 (except in the edge cases of Remark B.13).
To prove Proposition B.14, first observe that Since R(u ℓ ⊗ u 1 ) = u 1 ⊗ u ℓ , (for u 1 , u ℓ of any height) and F −1 does not change u ⊗C1 1 , the result holds for the first C 1 tableaux of the states. We can thus assume, without loss of generality, that C 1 = 0. Additionally, if C 2 ≥ min(d 1 , ℓ) then the solitons do not interact in either system at this time step, and the result is obvious. So, assume C 2 < min(d 1 , ℓ).
Let P = P 1 ⊗ P 2 ⊗ P 3 ⊗ · · · and let u(j) denote the carrier after applying the R-matrix to P j-times. That is, for some P 1 , . . . , P j . Note that u(0) = u ℓ . Let where A 3 , A 2 , A 1 , B 2 , B 1 , B 0 are functions on non-negative integers. We will often write A 3 for A 3 (j), with j being clear from context. Similarly for the other functions. From Lemma B.4 (3c), with the assumption C 1 = 0, we find that P 1 = · · · = P min(d1−Q,ℓ) = u 1 and u(min(d 1 − Q, ℓ)) = 3 ℓ 2 B2 1 B1 with B 1 = min(d 1 − Q, ℓ) and B 2 = ℓ − B 1 . If Q = 0, then we clearly have P 1 = · · · = P min(d1,ℓ) = u 1 , and if Q > 0 (so that C 2 = 0) then, by (1a), these equalities also hold. That is, the first C = min(d 1 , ℓ) tableaux in T ℓ (P ) are u 1 . This matches the first C tableaux in F −1 T L ℓ F (P ) (cf. (B.2) and (B.3)). Note that In the case where d 1 ≤ ℓ and C 2 > 0 or in the case where d 1 − Q > ℓ, we have Therefore, the ( C + 1)-th tableaux in T ℓ (P ) is the same as the ( C + 1)-th tableaux in F −1 T M ℓ F (P ). In particular, the first soliton starts at the same point in both of these states.
Remark B.15. Note that the only way to leave (2d) is to have B 2 = A 3 = 0. Since A 3 is always decremented by one in (2a), (2b), (2c), and (2d), to have A 3 = 0 we would need ( However, this inequality does not hold, because we assumed L < ℓ. We have thus shown that (2d) is the last possible case, and consequently that these four cases are the only four which we can encounter. However, note that some of these cases may be skipped, and that we might not reach the last case before we reach u((d 1 − Q) + C 2 + L).
Remark B.16. Under the assumptions of Lemma B.4, Cases (3a), (3b), (3c) and (3d) are the only possible cases. However, note that some of these cases may be skipped.
Observation B.17. The cases which put down a 1 in the bottom row occur before the cases which put down a 2 in the bottom row. Moreover, the cases following a 2 in the bottom row only put down a 2 (except in the final case, (3d)). Only 3's are put down in the second last row, except in (3d).
Let N (j) be the number of 1's in P 1 ⊗ · · · ⊗ P j .
Lemma B.18. For every case except (3c) and (3d), we have Proof. Observe that We proceed by induction.
Proof. We consider three cases.
From Observation B.17, we can conclude that the tableaux in T ℓ (P ) before we reach Case (3d) are of the form .
for some non-negative integers d 1 and C 2 . Observe that (3d) puts down the columns of the carrier in reverse order, so we simply need to examine the carrier u( C + d 1 + C 2 ).
Observation B.20 tells us that A 2 ( C + d 1 + C 2 ) + A 1 ( C + d 1 + C 2 ) = L. Therefore, the number of 1 and 2 in the second last row of T ℓ (P ) is L, so F T ℓ (P ) is a L-sped state. Now, we determine the value of d 1 by examining u( C + d 1 ). Observe that (3c) and (3d) do not apply before u( C + d 1 ), so Lemma B.18 applies. That is, at If C 2 > 0, then (2d) or (3c) applies to u( C + d 1 ); in both of these cases we must have A 1 = 0, which implies d 1 = d 1 − max(B 1 − A 2 , 0). We find that the first C + d 1 factors in F T ℓ (P ) are which matches (B.2) and (B.3). If C 2 = 0 then (3d) applies to u( C + d 1 ), and the columns are put down in reverse order. Thus, in T ℓ (P ), there are A 1 ( C + d 1 ) factors with 1 in the second last row, and d 1 factors 1 in the bottom row immediately before them. Therefore, the first C + d 1 factors in A 2 , 0), this is the same as (B .2) and (B.3).
appearances of 1 in the state or carrier at u( C + d 1 + C 2 ). Therefore, of the 1 have come from the second soliton's bottom row in P .
• In the case B 2 = 0 we have B 1 + B 0 = ℓ. If y = 2 then the previous case applies, so assume y = 2. If y = 1, then the carrier has picked up all of the 0, so B 0 = M . If y = 0, then we must have B 1 = 0 for (3d) to apply, and so ℓ = B 0 ≤ M , contradicting M < ℓ. We deduce that, in this case, y = 1 and B 0 = M . Since B 1 − D = (ℓ − M ) − D of the 1's have come from the second soliton (using (B.6)), we can deduce that C + d 1 We conclude that Comparing with (B.2) and (B.3), this is what we desire. That is, the second soliton in F T ℓ (P ) starts at the same place as the second soliton in T L ℓ F (P ). Since the first solitons in both of these are the same length, we have that the second solitons are also the same length. Since (3d) puts the columns down in reverse order, we can thus deduce that This proves Proposition B.14.
Observe that T ℓ (P ) satisfies the assumptions of Proposition B.14, except we may have B.5. Edge cases. As previously noted, Propositions B.14 and B.11 together prove the main theorem except in edge cases. It is not difficult to see that these edge cases are exactly the genuine highest weight states in which L = d 2 .
Let p be a genuine highest weight state (as in (B.1)) in which L = d 2 , and let t 0 = ((c 2 + L) − (d 2 − M )) + (d 1 − d 2 ) = c 2 + M + d 1 − d 2 . We can use repeated applications of Proposition B.14 to compute At this point, we can no longer apply Proposition B.14. So, we manually compute the time steps after this point using Lemma B.4.
Therefore, the case where L = d 2 behaves solitonically. Moreover, we can easily calculate that the phase shift is δ = d 2 − M = 2d 2 − L − M . This takes care of the edge cases and completes the proof of Theorem 4.14.
Remark B.21. We could possibly reformulate L-sped solitons to allow size 0 (i.e. speed L) solitons. This would have the benefit of removing the edge cases, but would also make the definitions and proofs more technical.
After these modifications, the proof of Lemma B.4 remains similar to Appendix C. Then, with some care, the arguments for the m ≥ r + 2 case can be adapted to prove Theorem 4.14 in the m = r + 1 case. B.7. The m = r case. The m = r case is slightly different than the other cases, owing to the fact that the genuine highest weight states have a different form: for some non-negative integers L, M ≤ d 2 . Note that, for m = r, we have 3 = 2 and 2 = 1. Note also that solitons can not be longer than n in this case and the value of n may limit which of these highest weight states are possible. When evolving the time of this genuine highest weight state, we encounter cases that would not be covered by Lemma B.4. So, the proof presented in Section B.4 does not work for m = r (at least, not without major modifications).
However, Theorem 4.14 holds for the m = r case.
Example B.22. Consider a state composed of elements from the U q ( gl(2|4)) crystal B 2,1 (so that m = r = 2). Note that this state satisfies the assumptions of Theorem 4.14 with m = r. We observe that this state behaves solitonically. Moreover R(V ↑ ⊗ W ↑ ) = R( 2 2 2 ⊗ 2 1 ) = 2 2 ⊗ 2 2 1 and R(V ↓ ⊗ W ↓ ) = R( 1 3 4 ⊗ 1 2 ) = 1 4 ⊗ 1 2 3 . The phase shift is 2 = 4 − 1 − 1 = We believe that it would be possible to prove this theorem by reducing it to a modified version of the [HI00] system by using something similar to an L-sped BBS. However, we would likely need to modify the speed of both solitons (as opposed to only of them in the L-sped BBS), which would make the proof much more techinical. So, we instead prove the theorem by direct computation.
We first compute general results about the carrier, similar to Lemma B.4.
We apply (1e) until A 1 = 0 or B 2 = A 3 = 0. Next, we apply (1f) until B 2 = A 3 = 0. From then onwards, (1g) applies. Note that we might not reach all of the above cases before reaching u(d 1 + L − Q) and we might skip over cases.
After u(d 1 − Q + d 2 ), (2d) puts down the columns of the carrier in reverse order.
. That is, not apply (but the state immediately before them does). Thus, x d1−M is still in u(d 1 + L − Q) (and, incidentally, will remain in the carrier until we apply (2d)). After u(d 1 + L − Q), (2a) and (2b) put down members from (x d1−j ) M−1 j=0 in order (with 2 in the row above) until B 2 = 0. Observe that (using also the fact that A 3 = ℓ − L). Applying (2d) proves (ii).
so modified versions of (B.7) and (B.8) do not apply (but the state immediately before them does). Thus, x d1−M is still in u(d 1 + C − M − 1) = u(d 1 + C 2 + L).