Rigged Configurations and Kashiwara Operators

For types $A^{(1)}_n$ and $D^{(1)}_n$ we prove that the rigged configuration bijection intertwines the classical Kashiwara operators on tensor products of the arbitrary Kirillov-Reshetikhin crystals and the set of the rigged configurations.


Introduction
A search of a natural presentation of the basis for representations of infinite dimensional algebras sometimes reveals an unexpected and intriguing connection with other models of mathematical physics. Take the Feigin-Fuchs representation of the Virasoro algebra [7] as an example. In this case, a cerebrated integrable quantum many body problem called the Calogero-Sutherland model [35,36] plays the role. Indeed, it is conjectured in [30] and partially proved in [3] that the excited states of the Calogero-Sutherland model (usually called the Jack symmetric functions) behave particularly nicely as a basis for the Feigin-Fuchs module. For example, matrix elements of the Virasoro generators with respect to the Jack basis become factorized rational functions whose major part consisting of degree one polynomials of positive integer coefficients which admit a transparent combinatorial interpretation. This phenomenon seems not accidental since the theory includes Mimachi and Yamada's finding [22] which claims that the singular vectors of the Virasoro algebra coincide with special cases of the Jack symmetric functions. See, for example, [1,2,5,6,8] for recent developments on the subject.
The purpose of the present paper is to pursue a parallel investigation for the settings of the crystal bases [14] of an important class of finite dimensional representations (called the Kirillov-Reshetikhin modules) of the quantum affine algebras. We call the corresponding crystal the Kirillov-Reshetikhin (KR) crystal. In this case, the corresponding physical model is the box-ball system. The box-ball system is a prototypical example of the ultradiscrete (or tropical) soliton system whose simplest example was discovered by Takahashi and Satsuma [37]. For our purpose, an important aspect of the box-ball system is that the dynamics of the model is determined by the combinatorial R-matrices for the KR crystal [10,9]. This formalism allows us to define the box-ball system for any tensor products of the KR crystals for arbitrary quantum affine algebras (see, for example, [11,12]). See [29] for an introductory review on studies on the box-ball systems.
Recently, a complete set of the action and angle variables of the type A (1) n box-ball systems is found to be identical to the set of the rigged configurations [20]. The rigged configurations are combinatorial objects discovered by an insightful analysis of the Bethe ansatz for quantum integrable models [17,18]. Originally the rigged configurations are used to prove a combinatorial identity for the Kostka-Foulkas polynomials called the Fermionic formula as well as to give a description of the branching coefficients of finite dimensional representations of the quantum affine algebras with respect to the corresponding finite dimensional subalgebras. In such context the rigged configurations are labels of representations and thus naturally correspond to the classically highest weight tensor products of the KR crystals. On the other hand, since the box-ball system is defined for not necessarily highest weight tensor products, we need to consider generalized rigged configurations corresponding to any tensor products.
Therefore it is natural to introduce the Kashiwara operators on the set of the rigged configurations. Let I be the set of the Dynkin nodes of a quantum affine algebra g following Kac' convention [13] and set I 0 = I \ {0}. In [33] Schilling gave a definition of the classical Kashiwara operatorsẽ i andf i (i ∈ I 0 ) for simply laced algebras and proved that the crystal structure on the rigged configurations is actually isomorphic to the one on the tensor products of the KR crystals by using Stembridge's local characterization of crystals. In Section 3.2 of the present paper, we provide a direct proof of the fact that the classical Kashiwara operators on the set of the rigged configurations of types A (1) n and D (1) n indeed satisfy the axiom of the crystals.
The crux of the rigged configuration theory is the so called rigged configuration bijection which gives an one to one correspondence between the elements of tensor products of the KR crystals and the rigged configurations: Φ : rigged configurations −→ tensor products.
The algorithm of Φ is a rather complicated combinatorial procedure which is explained in Section 3.3. Nevertheless it is known that the map Φ has a lot of nice properties and seems to have a deep mathematical origin. For example, one of the most important properties of the rigged configuration bijection is that we can regard the algorithm Φ as a general form of the combinatorial R-matrices ( [19] for type A (1) n case and [31,34,25] for type D (1) n case). See Remark 23 for more precise meanings. This property is the basis for providing the action and angle variables of the box-ball systems [20,21]. Moreover, as an application of the relationship with the box-ball system we have a representation theoretical interpretation of the algorithm of Φ −1 for type A (1) n [28], though the full understanding of the rigged configurations themselves is yet to be seen.
The main result of the present paper is the proof of the following compatibility of the rigged configuration bijection and the classical Kashiwara operators for g = A (1) n and D (1) n (see Theorem 24). Here some remarks are in order.
• In Appendix C of the preprint version of [4], Deka and Schilling partially proved the relation (1) for type A (1) n . Given the fundamental importance of the problem, it seems appropriate to prove the result completely for type A (1) n also. Indeed, the proof for type A (1) n is a part of the proof for type D (1) n and easy to extract from it. 1 • The proof of (1) is direct and does not depend on other results.
• In [25], the relation (1) is effectively used to prove the above mentioned relationships between the rigged configuration bijection and the combinatorial R-matrices for general cases of type D (1) n .
Since our approach is direct, we expect that our result paves the way to deal with the rigged configuration bijection for arbitrary quantum affine algebras. Currently the following cases are established: • All cases of type A (1) n [17,18,19].
• Tensor products of the vector representations of arbitrary non-exceptional quantum affine algebras [27] as well as of type E (1) 6 [26].
In all cases the algorithms for the rigged configuration bijection share many common features. We expect that the rigged configuration bijection as well as all major properties will extend to the arbitrary quantum affine algebras.
We expect that such an extension of the rigged configuration theory will provide meaningful insights for both mathematical physics side and mathematical side which are not easily seen by other methods. One of the supporting evidences for our expectation is a bijection constructed in [23]. In that paper, we constructed a combinatorial bijection between the rigged configurations for arbitrary non-exceptional quantum affine algebras (under certain restrictions on ranks) and the set of pairs of a type A (1) n rigged configuration and a Littlewood-Richardson tableau. Remarkably, the construction is uniform for all types of algebras and the Littlewood-Richardson tableaux naturally appear as the recording tableaux of the algorithm. We expect that the bijection coincides with (and generalizes) a canonical Dynkin diagram automorphism (see Remark 3.2 of [23]).
Another supporting evidence is a numerical comparison of the rigged configuration bijection [27] for the vector representations of arbitrary non-exceptional quantum affine algebras and the corresponding combinatorial R-matrices (see Section 3 and subsequent sections of [20]). The conjecture provided there claims that seemingly complicated algorithms presented in [27] with many case by case definition possess uniform description in terms of the corresponding combinatorial R-matrices. In fact, we expect that the rigged configuration bijection gives a general form of the combinatorial R-matrices for general quantum affine algebras and thus provides action and angle variables for the corresponding box-ball systems. Finally, it should be mentioned that in [24] we see that the affine crystal structure of type D (1) n is essentially governed by the rigged configurations. Therefore we expect that the construction of the rigged configuration bijection for general cases should be an intriguing and meaningful future research topic. This paper is organized as follows. In Section 2, we collect necessary facts from the crystal bases and the tableau representations. In Section 3, we give the definition of the rigged configurations and the rigged configuration bijection and explain their basic properties (convexity relation of the vacancy numbers). We also introduce the classical Kashiwara operators on the rigged configurations and show that they satisfy the axiom of the classical crystals. In Section 4 we give the statement of the main result and clarify what are the essential points to be shown. Main proofs are given in Section 5, Section 6 and Section 7. During the proof we have tried to avoid unnecessarily long arguments. Nevertheless we have to deal with many subtle cases and some of them look very special. Thus some readers might wonder whether such case does exist and is worthy of careful analysis. In order to clarify that point, we include several examples for seemingly subtle cases.
2 Background on crystals and tableaux 2

.1 Crystal bases
Let us briefly recall some definitions about the crystal bases theory of Kashiwara [14]. For more detailed introduction, see, for example, [15]. Let g be an affine Kac-Moody algebra, g ′ be its derived subalgebra, g 0 be the corresponding finite dimensional simple Lie algebra obtained by removing the 0 node of the Dynkin diagram of g. Here we follow Kac' convention [13] for the labeling of the Dynkin nodes and denote by I the vertex set of the Dynkin diagram of g and I 0 := I \ {0}. Let U q (g), U ′ q (g) and U q (g 0 ) be the quantized universal enveloping algebra corresponding to g, g ′ and g 0 , respectively. In this paper we will mainly focus on the case g = D (1) n which has the following Dynkin diagram.
Definition 1. U ′ q (g)-crystal is a nonempty set B equipped with maps wt : B → P , ε i , ϕ i : B → Z ∪ {−∞} and the Kashiwara operatorsẽ i ,f i : B → B ∪ {0} for all i ∈ I such that In this paper, we only consider the case when the maps ε i , ϕ i are defined by for b ∈ B. In particular, we have ε i (b), ε i (b) < ∞ for all i ∈ I and b ∈ B. If we havẽ we write an arrow with color (or label) i from b to b ′ . In this way, the crystal B can be regarded as the colored oriented graph whose vertices are the elements of B. We call such graph the crystal graph.
One of the nice properties of the crystal bases is that it behaves nicely with respect to the tensor product. Let B 2 ⊗ B 1 be the tensor product of two crystals B 1 and B 2 . As the set, it coincides with the Cartesian product B 2 × B 1 . The action of the Kashiwara operatorsẽ i ,f i on an element b 2 ⊗ b 1 ∈ B 2 ⊗ B 1 is given explicitly as follows; where the result is declared to be 0 if either of its tensor factors are 0. We also have wt(b 2 ⊗ b 1 ) = wt(b 2 ) + wt(b 1 ).

Remark 2.
We use the opposite of the Kashiwara's tensor product convention.
The reduced i-signature of b is obtained by removing subsequence −+ of the i-signature of b repeatedly until it becomes the following form + · · · · · · · · · + ϕ i (b) times − · · · · · · · · · − ε i (b) times . By using this information the action ofẽ i andf i on b can be described as follows. If ε i (b) = 0, thenẽ i (b) = 0. Otherwisẽ where the leftmost − in the reduced i-signature of b comes from b j . Similarly, if ϕ i (b) = 0, thenf i (b) = 0. Otherwisẽ where the rightmost + in the reduced i-signature of b comes from b j . Finally the actions of e i andf i on each tensor factor b j are described by tableau representations of b j described in the sequel.

Kirillov-Reshetikhin tableaux
In order to perform explicit manipulations on the crystal bases, it is convenient to use an explicit realization of the elements of crystals. For the present purpose, it is more convenient to use a new kind of tableau representation which we call the Kirillov-Reshetikhin (KR) tableaux [24] than the usual Kashiwara-Nakashima (KN) tableaux [16].

Reviews on tableau representations
Let us briefly recall tableau representations of crystals. Let B(Λ) be the crystal associated to the highest weight representation of highest weight Λ of U q (g 0 ). Then in the present case g = D (1) n , the U ′ q (g) crystal B r,s has the following decomposition under the restriction to its finite dimensional subalgebra U q (g 0 ). If r < n − 1, we have and if r = n − 1, n, we have B r,s ≃ B(sΛ r ) as g 0 crystals.
Here in the first case r < n − 1, λ in the summation are determined as follows. For this we identify λ = i k iΛi with the Young diagram which has k i columns of height i for all i. Then one of λ in decomposition (4) is shape (s r ) rectangular diagram and the remaining λ are obtained by removing the vertical dominoes from (s r ) rectangular diagram in all possible ways. Then if b ∈ B r,s belongs to some B(λ) in the right hand side of the decomposition (4) or (5), then the KN tableau representation of b has shape λ.
KN tableaux for B r,s when r ≤ n − 2. For J ⊂ I, we say b is J-highest element if we haveẽ i (b) = 0 for all i ∈ J. Suppose that u λ ∈ B r,s (r ≤ n − 2) is the I 0 -highest element of B(λ) which appears in the decomposition (4). Then the corresponding KN tableau is obtained by filling the letter j to all the boxes at jth row of the Young diagram λ. In order to obtain the other tableaux, we define the action of the Kashiwara operators on tableaux as follows. The crystal graph of B 1,1 for g 0 = D n is as follows.
Weight of each vertex is wt i = ǫ i and wt ī = −ǫ i , respectively. For a given tableau t ∈ B r,s (r ≤ n − 2), we introduce the so-called Japanese reading word and regard t as the element of (B 1,1 ) ⊗N as follows: Via this identification we can introduce the Kashiwara operators on tableaux and obtain the KN tableau representation for the general case.
Thus we can represent each element ǫ i by half width tableau with entry i. We arrange the entries within a column according to the following order on the letters (smaller ones at higher places) Here we do not introduce the order between n and n.
For the general case, we start by the highest weight element u λ ∈ B n,2s+1 where λ = (2s + 1)Λ n . Then represent u λ by the spin column (+, . . . , +) and s columns of 1, 2, . . . , n such as In order to obtain the other tableaux, we embed u λ into B r,1 ⊗ (B 1,1 ) ⊗rs by using the Japanese reading word for the right s columns and apply the Kashiwara operatorsf i in all possible ways.
We can obtain the KN tableaux for the case B n,2s similarly if we ignore the leftmost spin column in B n,2s+1 case.
KN tableaux for B n−1,2s+1 . In the case B n−1,1 ≃ B(Λ n−1 ), we can represent the basis as (s 1 , s 2 , . . . , s n ) where s i = ± and s 1 s 2 · · · s n = −1. We can use the same formulae (6) and (7) for the Kashiwara operators and (8) for the weight. For the general case B n−1,2s+1 , we represent the highest weight element u (2s+1)Λ n−1 ∈ B n−1,2s+1 by Here we obtain the tableau for u (2s+1)Λ n−1 by replacing all n of the bottom row of u (2s+1)Λn byn. Then we can use the Kashiwara operators to obtain all the tableaux of B n−1,2s+1 . We can obtain tableaux for the case B n−1,2s if we ignore the leftmost spin column.

Filling map
The purpose to introduce the KR tableaux is to give a rectangular presentation of elements of crystals as opposed to the KN tableaux whose shapes depend on the classical decompositions given in (4) and (5). The definition of the KR tableaux is given explicitly for the highest weight element of B r,s and extended to the general case by using the Kashiwara operatorsf i as in the KN tableaux case.
To begin with we give a definition of the KR tableaux for the case B r,s (r ≤ n − 2). Let u λ be the I 0 -highest element of B r,s with weight λ = k rΛr + k r−2Λr−2 + · · · . We denote by fill(u λ ) the KR tableau representation of u λ . Now we describe the algorithm to obtain fill(u λ ) which we call the filling map. Let k c be the first odd integer in the sequence k r−2 , k r−4 , · · · . If there is no such k c , set k c = k −1 , that is, c = −1. Let t be the KN tableau representation of u λ . We place t at the top left corner of the r by s rectangle.
In order to fill all the empty places, we start from the leftmost column of t and move rightwards by the following procedure.
1. We do nothing for the height r columns. If c ≥ 0, remove one column from the height c columns and move the columns that are shorter than c to left by 1.
2. For the columns whose heights are equal to or larger than c, put the following stuff to the empty places as much as possible. If the height of the corresponding columns is h, put the transpose of If c = −1, we stop here since all empty places are filled by this procedure.
3. For the remaining columns except for the rightmost one, put the following stuff recursively from left to right with recursively redefining an integer x. As the initial condition, set x = c + 1.
If the corresponding column has height h, put the transpose of y · · · r − 1 r r · · · x + 1 x to the empty place. Since the number of the empty places is r − h, we have y = r − (x − h − 2). We redefine x = y and do the same procedure to the next column.
4. By using the final output of x from the last step, we put the transpose of the following to the rightmost column 1 2 · · · y y · · · x + 1 x .
Recall that the rightmost column is empty by Step 1 whenever c ≥ 0. Then in order to achieve the above pattern, we have y = (r + x − 1)/2.
Example 3. Let us consider the I 0 -highest element u λ of B 8,7 where λ =Λ 8 + 2Λ 6 +Λ 4 + 2Λ 2 . In this case, we have (k 6 , k 4 , k 2 , k 0 ) = (2, 1, 2, 1), thus c = 4. We put the KN tableau representation of u λ into the 8 by 7 rectangle. Then (3) Suppose that s L > 1. Let c i be the ith column of b L such that b L = c 1 c 2 · · · c s L . Then we define the operation called left-split by

Definition of the rigged configurations
Our strategy to define the rigged configurations of type g = D (1) n has three steps. First we define the highest weight rigged configurations. Next we define the classical Kashiwara operatorsẽ i andf i for i ∈ I 0 on the rigged configurations. Finally, we define the general rigged configurations by all possible applications of the operatorsf i on highest weight rigged configurations. Let us prepare several notation to describe the rigged configurations. The rigged configurations are combinatorial objects made up with partitions ν (a) and integers J (a) for a ∈ I 0 . ν (a) = (ν k ) is a sequence of integers called the rigging which are associated with rows of the corresponding configuration. We should regard that the pairs (ν (a) , J (a) ) are located on the vertices I 0 of the Dynkin diagram of D (1) n . We denote by m (a) l (ν) the number of length l rows of ν (a) . We will abbreviate the pair (ν (a) , J (a) ) by (ν, J) (a) and call the components of (ν, J) (a) = {(ν The rigged configurations also depend on the shape of the tensor product B = B r L ,s L ⊗ · · · ⊗ B r 1 ,s 1 . Let L (a) l be the number of the component B a,l within B. Then we define the partition µ (a) such that the number of length l rows of µ (a) is equal to L (a) l . Thus the rigged configuration (ν, J) is described by the sequence of configurations and the riggings together with additional data µ = (µ (1) , . . . , µ (n−1) , µ (n) ) under certain constraints to be described below. We denote by Q   l (µ, ν) (hereafter we will abbreviate as P (a) l (ν)) by the formula where A ab is the Cartan matrix of g 0 = D n and a ∼ b means that the vertices a and b are connected by a single edge on the Dynkin diagram. i ) and the corresponding vacancy numbers satisfy the following condition Remark 6. If ν (n−1) = ν (n) = ∅, then we can identify the rigged configuration as type A (1) n−2 . In this way we can recover proofs for type A (1) n from those for type D n .
Example 7. The following object is a highest weight rigged configuration corresponding to the tensor product Here we put the vacancy number (resp. rigging) on the left (resp. right) of the corresponding row of the configuration represented by a Young diagram. By the rigged configuration bijection Φ (see Section 3.3) the above rigged configuration (ν, J) corresponds to the following highest weight tensor product: The weight λ of the rigged configuration is defined by the relation (sometimes called the (L, λ)-configuration condition) If we expand λ by the basis ǫ i , we can rewrite as Then we can use the expressions (2) and (3) to obtain the explicit expressions for the weight λ i . We write the weight of the rigged configuration by wt(ν, J).
Following [33] we introduce the classical Kashiwara operators on the rigged configurations and use them to define the general rigged configurations. For the string (l, x) of (ν, J) (i) , we call the quantity P (1) Let ℓ be the minimal length of the strings of (ν, J) (i) with the rigging x ℓ . If x ℓ ≥ 0, defineẽ i (ν, J) = 0. Otherwiseẽ i (ν, J) is obtained by replacing the string (ℓ, x ℓ ) by (ℓ − 1, x ℓ + 1) while changing all other riggings to keep coriggings fixed.
(a) Let (k, x k ) be a string of (ν, J) (i) which satisfies k ≤ ℓ. Sincef i acts on the string with smallest rigging, we have x k ≥ x ℓ . Recall that we have P k (ν) sincẽ f i adds a box to the (ℓ + 1)-th column. Thus the string (k, x k ) remains as it is after f i and its rigging satisfies x k > x ℓ − 1.
(b) Let (k, x k ) be a string of (ν, J) (i) which satisfies ℓ < k. Sincef i acts on the longest string with rigging x ℓ , we have x ℓ < x k . Recall that we have P k (ν) − 2 in this case. Thus the string (k, x k ) becomes (k, x k − 2) afterf i . Then its rigging satisfies x k − 2 ≥ x ℓ − 1.
Finally, let us define the maps ε i , ϕ i : (ν, J) → Z by

Convexity relations of the vacancy numbers and their applications
In this subsection, we introduce a fundamental property of the vacancy numbers called the convexity relation. As an application, we provide a direct proof of the fact that the crystal structure on the set of the rigged configurations indeed satisfies the axiom of the U q (g 0 )-crystals. Finally, we prepare refined estimates for the vacancy numbers which are necessary in the later arguments. One of the basic properties of the vacancy number is the following convexity relation.
Proposition 10 (Convexity). Suppose that m (a) l (ν) = 0 for some l ≥ 1. Then the vacancy numbers satisfy the following convexity relation Proof. Recall that the functions Q k (ν) is a linear function between l − 1 ≤ k ≤ l + 1. Thus the combination in (10) gives the upper convex relation for the vacancy numbers.
By repeated use of the above convexity relation, we obtain several useful criteria.
(2) For some k satisfying l 1 < k < l 2 suppose that we have P (a) Then the corresponding rigged configuration is forbidden.
(3) For some k satisfying l 1 < k < l 2 suppose that we have P (a) Then the only admissible situation is P (a) Since we will use these properties so many times in the rest of the paper, we will sometimes use these relations without giving an explicit reference.
Let us give the first application of the convexity relations.
(b) Let (k, x k ) be an arbitrary string of (ν, J) (i) satisfying k < ℓ. Sinceẽ i acts on the shortest string with rigging x ℓ , we have x k > x ℓ . From k < ℓ, we see thatẽ i will not change corigging of (k, x k ), thus the string (k, x k ) remains as it is in (ν,J).
(c) Let (k, x k ) be an arbitrary string of (ν, J) (i) satisfying ℓ ≤ k. Since x ℓ is the minimal rigging of (ν, J) (i) , we have x ℓ ≤ x k . Recall that we have P (i) k (ν) = P (i) k (ν) + 2 sincẽ e i removes a box from ℓth column of ν (i) . Thusẽ i makes the string (k, x k ) into (k, x k + 2), in particular, its rigging satisfies x ℓ + 1 < x k + 2.
Step 1. Let us consider the case P (i) Suppose that x ℓ ≤ 0. Then we have ℓ > 0 and s = x ℓ , thus, P (i) ∞ (ν) = x ℓ . Let j be the maximal integer such that ℓ ≤ j and m (i) j (ν) > 0. Suppose if possible that ℓ < j. Let us consider the string (j, x j ). Sincef i acts on the longest string with rigging x ℓ , the assumption ℓ < j implies that Recall that the definition (10) implies that there exists a large integer L such that P (i) k (ν) satisfy the convexity relation between j ≤ k ≤ ∞. In particular, the relation P L+1 (ν) for j < L < L + 1 is a contradiction. Therefore we have ℓ = j and thus there is no string of (ν, J) (i) that is longer than ℓ. Then from the convexity relation of P (i) k (ν) between ℓ ≤ k ≤ ∞, the only possibility that is compatible with the requirement P Let us consider the string (ℓ + 1, x ℓ − 1) of (ν,J) (i) created byf i . Due to the extra box added byf i , we have P Thus the string (ℓ + 1, x ℓ − 1) of (ν,J) (i) has the rigging that is strictly larger than the corresponding vacancy number. Thus we havef i (ν, J) = 0 as requested.
Finally let us consider the case x ℓ > 0. Then we have P (i) ∞ (ν) = 0 by s = 0. Let j be the maximal integer such that m (i) j (ν) > 0 and consider the corresponding string (j, x j ). Since x ℓ is the minimal rigging, we have P Then for a sufficiently large integer L, we have P L+1 (ν) = · · · = 0. This is a contradiction since P (i) k (ν) must satisfy the convexity relation between j ≤ k ≤ ∞. Therefore we see that there is no string in (ν, J) (i) . This is a contradiction since we assume that x ℓ > 0. Hence this case cannot happen.
Step 2. Let us consider the case P      Let us show that If j = ℓ + 1 this relation is already confirmed. Thus suppose that ℓ + 1 < j. Suppose if possible that we have x ℓ ≥ P (i) ℓ+1 (ν). Then we have P . This is a contradiction since P (i) k (ν) must satisfy the convexity relation between ℓ ≤ k ≤ j. In conclusion, we have x ℓ < P (i) ℓ+1 (ν). Sincef i adds a box to the (ℓ + 1)-th column of ν (i) , we have P (15). Therefore the string (ℓ + 1, x ℓ − 1) of (ν,J) (i) created byf i has the rigging which is smaller than or equal to the corresponding vacancy number. Hence we havẽ f i (ν, J) = 0.
Let us determines.
(b) Let (k, x k ) be an arbitrary string of (ν, J) (i) satisfying ℓ < k. Sincef i acts on the longest string with rigging x ℓ , ℓ < k implies that x ℓ < x k . Recall that we have P k (ν) − 2 sincef i adds a box to the (ℓ + 1)th column of ν (i) . In order to keep the corigging, the string (k, x k ) becomes (k, x k − 2) afterf i . In particular, the new rigging satisfies (c) Let (k, x k ) be an arbitrary string of (ν, J) (i) satisfying k ≤ ℓ. Then we have x k ≥ x ℓ .
Sincef i does not change the corigging of the string, the string (k, x k ) remains as it is afterf i .
To summarize, we haves = x ℓ − 1 = s − 1. Since we have P (i) Finally consider the case x ℓ > 0. In this case we have s = 0. Thenf i adds the string (1, −1) to (ν, J) (i) . If we show thatf i (ν, J) = 0, then we haves = −1 = s−1 as requested. This can be done if we formally setting (ℓ, x ℓ ) = (0, 0) in the proof of (15) to deduce that P (i) Corollary 13. Let λ = wt(ν, J). Then we have Proof. From Theorem 12, we have . On the other hand, from (13) we have (see also definitions in Section 2.1) Example 14. Let us consider the following element of (B 1, We can see that ϕ 1 (b) = 3. Then we can applyf 1 as follows The leftmost rigged configuration corresponds to the path b. For (ν, J) = Φ −1 (b) we have P    For the later purposes, let us prepare some refined estimates related with the convexity relations. The first one is an intuitive estimate.
Proof. Let (ν ′ , J) be the fictitious rigged configuration obtained by replacing all length l strings of (ν, J) (i) by length l + 1 strings. From the similar argument of the proof of Proposition 10, we can show that P     For some purposes, we need a more refined estimate as follows.
Proof. Since Q Since L (a) l ≥ 0, we obtain the assertion. Similarly we can show the assertions for n − 2 ≤ a.
A nice property about these convexity relations is that they do not involve explicitly the information on L (a) l or µ. Indeed, we will not refer to L (a) l or µ explicitly during the arguments of Section 5, Section 6 and Section 7.

Rigged configuration bijection
Let us define the bijection between the rigged configurations and tensor products of crystals expressed by the KR tableaux which we call paths. For the map from the rigged configuration to paths, the basic operation is called δ where (ν, J) and (ν ′ , J ′ ) are rigged configurations and k ∈ {1, 2, . . . , n,n, . . . ,2,1}. In the description, we call a string (l, x l ) of (ν, J) (a) singular if we have x l = P  (1) For a < i ≤ n − 2, assume that ℓ (i−1) is already determined. Then we search for the shortest singular string in (ν, J) (i) that is longer than or equal to ℓ (i−1) .
(a) If there exists such a string, set ℓ (i) to be the length of the selected string and continue the process recursively. If there is more than one such string, choose any of them.
(b) If there is no such string, set ℓ (i) = ∞, k = i and stop.
(4) Once the process has stopped, remove the rightmost box of each selected row specified by ℓ (i) or ℓ (i) . The result gives the output ν ′ .
(5) Define the new riggings J ′ as follows. For the rows that are not selected by ℓ (i) or ℓ (i) , take the corresponding riggings from J. In order to define the remaining riggings, replace one B a,l in B by B a−1,1 ⊗ B a,l−1 (equivalently, replace one of the length l row of µ (a) by a length (l − 1) row and add a length 1 row to µ (a−1) ). Denote the result by B ′ . Use B ′ to compute all the vacancy numbers for ν ′ . Then the remaining riggings are defined so that all the corresponding rows become singular with respect to the new vacancy number.
We remark that the resulting rigged configuration (ν ′ , J ′ ) is associated with the tensor product B ′ . We write δ 2 δ 1 (ν, J) etc. for repeated applications of δ on the rigged configurations.
Definition 19. For a given rigged configuration (ν, J) corresponding to the tensor product of the form B = B r 1 ,s 1 ⊗ B r 2 ,s 2 ⊗ · · · ⊗ B r L ,s L , define the map Φ B (sometimes also just denoted Φ) as follows.
Remark 20. The inverse procedure Φ −1 is obtained by reversing all procedure in Φ step by step. As an example let us give a sketch of the algorithm for the operation (δ (1) 1 ) −1 . Suppose that we start from an unbarred letter k in a KR tableau. Then we look for the largest singular string of ν (k) . If there is no such string, add a length one row to the bottom of ν (k) . Otherwise add a box to the longest singular row of ν (k) . Suppose that we add the box to the ℓ (k) th column of ν (k) . Then we look for the singular string of ν (k−1) which is strictly shorter than ℓ (k) . We continue this process up to ν (1) . Finally change the riggings for the modified strings according to the new vacancy numbers. Here remind that we have to add a length one row to µ (1) for the computation of the vacancy numbers.
The following fundamental result is proved in [25].
Theorem 21. Consider a tensor product of crystals B = B r 1 ,s 1 ⊗ B r 2 ,s 2 ⊗ · · · ⊗ B r L ,s L . Then the map Φ B gives a well-defined bijection between the rigged configuration of type B and the elements of B expressed via the KR tableaux representation.
If there is no afraid of confusion, we use the abbreviation Φ in stead of Φ B .
Example 22. Let us consider the following rigged configuration corresponding to the tensor product B 2,2 ⊗ B 3,2 of type D We start by removing B 2,2 part. Then the first operation is δ 2 which start by searching the shortest singular string of ν (2) = (4, 3, 1) whose length is equal to or larger than 2. We work out all the procedure corresponding to B 2,2 in the following sequence of diagrams.
Here the boxes to be removed by δ (a) l indicated below the corresponding rigged configurations are marked by "×". The output of each δ (a) l is given on the right of the corresponding arrows. Note that after δ (2) 2 we recalculate all the vacancy numbers assuming that the resulting rigged configuration corresponds to the tensor product of type B 1,1 ⊗ B 2,1 ⊗ B 3,2 . We remark that the first δ (1) 1 cannot remove a box from ν (1) since there is no singular string and, δ (2) 1 starts by removing from ν (2) and ends at ν (1) . The final output for the given rigged configuration (ν, J) is as follows: Remark 23. In the above example, we can reverse the order of the tensor product. In this case, we obtain Φ B 3,2 ⊗B 2,2 (ν, J) = 1 4 52 31 Then the two tensor products in (17) and (18)  Therefore in this case we have R = Φ B 3,2 ⊗B 2,2 • Φ −1 B 2,2 ⊗B 3,2 . This relation is valid not only for two times tensor product but also for multiple applications of pairwise combinatorial R-matrices for arbitrary tensor products.

Rigged configurations and the Kashiwara operators 4.1 Statement of the main result
Let us state the main result of the present paper.
Theorem 24. For the rigged configuration bijection Φ and the Kashiwara operatorsẽ i andf i (i ∈ I 0 ) of types A (1) n and D (1) n , the following diagrams commute; Here for the tensor product of crystals B = B r L ,s L ⊗B r L−1 ,s L−1 ⊗· · ·⊗B r 1 ,s 1 , we denote by RC ( Therefore we have the compatibility of the rigged configuration bijection and the classical Kashiwara operators. We give a few words on the compatibility for the affine Kashiwara operatorsẽ 0 andf 0 of type D (1) n . Recall that in [32] the operatorsẽ 0 and f 0 are realized via the bijection between the set of J-highest element of B r,s and the combinatorial objects called the plus-minus diagrams. Here J = {2, 3, . . . , n}. In [24, Section 4], we introduce an analogue of the plus-minus diagram diagrams on the set of the rigged configurations and show that this indeed yields the affine isomorphism with the KR crystal B r,s .
The main points of the arguments in [24] are as follows. In [24, Theorem 5.9] we prove the combinatorial bijection Φ for the I 0 -highest element of B r,s directly. Then in [24,Proposition 4.3] we show that the same sequence of the classical Kashiwara operatorsf i gives the plus-minus diagram and the corresponding element in the set of rigged configuration. Therefore Theorem 24 asserts that the rigged configuration analogue of the plus-minus diagrams introduced in [24, Section 4] are indeed the image of the plusdiagrams under the rigged configuration bijection Φ. Hence, in view of the construction of [32], we have the compatibility of the affine Kashiwara operatorsẽ 0 ,f 0 and the rigged configuration bijection for the case B r,s . However, the compatibility of the affine Kashiwara operators and the combinatorial bijection Φ for more than two tensor factors is still an open problem.
The rest of the paper is devoted to the proof of Theorem 24.

Preliminary steps
Our main strategy of the proof is to reduce the essential combinatorial aspects of Theorem 24 to the fundamental operation δ = δ 1 . For this purpose we decompose the original bijection Φ into several steps according to the operations lh, lb and ls defined on the KR tableau representation of the crystals. In the following, we define the corresponding operations (summarized in the following table) on the set of the rigged configurations.
(1) Corresponding to B r L ,s L = B a,l , γ replaces a length l row of µ (a) by two rows of lengths l − 1 and 1 of µ (a) .
(2) Suppose that B r L ,s L = B r,1 where r > 1. If r < n − 1, then β removes a length one row from µ (r) , adds a length one row to each of µ (1) and µ (r−1) and adds a length one singular string to each of (ν, J) (a) for a < r.

Proposition 27.
We can decompose the map Φ into the following three steps.
(1) Let B = B r,s ⊗B with s > 1. Then we reduce the problem to the case of B r,1 by (2) Let B = B r,1 ⊗B with r > 1. Then we reduce the problem to the case of B 1,1 by Then we have to deal with the fundamental operation δ as Recall that γ replaces a length s row of µ (r) by two rows of length s − 1 and 1. Thus after application of γ there is no singular string of ν (r) which are strictly shorter than s. Then the length 1 row of µ (r) corresponds to the same column which appear as the leftmost column of the tableau b.
(2) If we apply δ 1 for RC(B ′ ), it automatically selects the length 1 singular strings of ν (a) (a < r) created by β. As for ν (r) , since β removes length 1 row from µ (r) and adds a length 1 row to ν (r−1) , all the coriggings for ν (r) do not change after β. Thus δ

Reduction steps
We show that the operations γ and β commutes with the Kashiwara operators.
Proposition 29. Let B = B r,1 ⊗B where r > 1. Then we have the following commutative diagrams Proof. Let us prove thef i part first. During the proof we denote thatf i (ν, J) = (ν,J) and β(ν, J) = (ν,J). Suppose that i > r. Recall that the operation β does nothing on (ν, J) (a) for a ≥ r. Thus β andf i commute since they do not have interaction.
Suppose that i = r. Since the operation β adds a length one string to (ν, J) (r−1) , the interaction between β andf r can occur iff r adds the string (1, −1) to (ν, J) (r) . In this case, since β does nothing on (ν, J) (r) ,f r adds the same string (1, −1) after β. Let us consider (ν, J) (r−1) . Then we have to check the coincidence of the string added by β; .
Here we use the fact thatf r changes the riggings so as to keep the coriggings.
Finally, suppose that i < r. Let x k be the smallest rigging of length k strings of (ν, J) (i) . Let x ℓ be the smallest rigging of (ν, J) (i) where ℓ is the length of the largest string with rigging x ℓ .
Suppose that ν (i) = ∅ and x ℓ ≤ 0. Recall that β adds the singular string (1, P then we see thatf i chooses the same string (ℓ, x ℓ ) even after β. Let j be the length of the smallest string of (ν, J) (i) . Suppose that j = 1. Then we have P by the definition of the rigged configurations and x 1 ≥ x ℓ by the minimality of x ℓ . Thus we have P (i) 1 (ν) ≥ x ℓ in this case. Next, suppose that j > 1. By the convexity relation of P then we have the following behaviors of the two strings of (ν, J) (i) ; and, on the other hand, .
• Suppose that j = 1. Since ν (i) = ∅ we have x ℓ > 0 by the assumption. On the other hand we have 0 ≥ P (i) 1 (ν) ≥ x 1 by the assumption and the definition of the rigged configurations. This is in contradiction to the relation x 1 ≥ x ℓ which follows from the minimality of x ℓ . Hence this case cannot happen.
• Suppose that j > 1. Again we have x ℓ > 0. By the convexity relation of P (i) k (ν) between 0 ≤ k ≤ j and the definition of the rigged configurations we have 0 = P Hence this case cannot occur.
Thus we have checked the commutativity off i and β in all possible cases.
Finally let us consider the proof forẽ i . Ifẽ i (ν, J) = 0, then the commutativity of e i and β follows from the result off i sinceẽ i andf i are mutually inverse in this case. Suppose thatẽ i (ν, J) = 0. Then all the riggings of (ν, J) (i) are non-negative. Recall that β acting on (ν, J) adds the string (1, P Let j be the length of the shortest string of (ν, J) (i) . If ν (i) = ∅, set j = ∞. In both cases we have P

Statements to be proved
Due to the arguments in the previous section, the essential points of the proof of Theorem 24 are reduced to the following two assertions on the case B = B 1,1 ⊗B. Up to Section 6 from here we will use the following notation when we consider the case B = B 1,1 ⊗B. Let (ν, J) ∈ RC(B). Denote the rigged configurations obtained by the actions off i or δ as in the following diagram: Here we denote as δ : (ν,J) → (ν,J) andf i : (ν,J) → (ν,J). We denote the smallest rigging associated with length k rows of ν (i) by x k and express such string as (k, x k ). For later purposes, let us prepare several notation for the specific strings as follows: • (ℓ, x ℓ ) is the string of (ν, J) that is selected byf i .
• (l, xl) is the string of (ν,J) that is selected byf i .
• ℓ (a) is the length of the singular string of (ν, J) (a) that is selected by δ.
•l (a) is the length of the singular string of (ν,J) (a) that is selected by δ.
Then our main claims to be proved are the following two assertions which are proved in Section 5, Section 6 and Section 7.
if allẽ i andf i in the above diagrams are defined.
Here from (1) to (4) we assume the commutativity off i with Φ forB and from (5) to (8) we assume the commutativity ofẽ i andf i with Φ forB.
Let us show how these properties lead to the proof of Theorem 24. In the proof, we will use the diagram of the following kind as in [19]: We regard this as a cube with front face given by the large square. Suppose that the square diagrams given by the faces of the cube except for the front face commute and i is the injective map. Then the front face also commutes since we have Proof of Theorem 24. First we prove thef i case. Then we can prove theẽ i case by a parallel argument. We follow the decomposition of Φ as described in Proposition 27. The simplest case B 1,1 can be shown directly. We give a list of the explicit rigged configurations in this case.
By using the crystal graph of B 1,1 presented in Section 2.2.1 we can verify the commutativity off i and Φ. Let us consider the case B = B 1,1 ⊗B. This case corresponds to an addition of a new tensor factor. For this we use the following diagram: Suppose that allf i which appear in the above diagram are defined. The top face and the bottom face commute by Proposition 27. The left face commutes by Theorem 30. The right face commutes by definition of lh on tensor products of crystals. The back face commutes by the assumption. Finally the map lh is injective if we consider the weight of the element of B 1,1 subtracted by the operation. Hence the front face commutes. This proves the commutativity of Φ andf i when bothf i before or after δ (or lh) are defined. On the other hand we know from Theorem 31 that Φ andf i also commute even if one of f i before or after δ (or lh) is undefined. By exclusion we have the commutativity for the case when bothf i before or after δ (or lh) are undefined.
Let us consider the the case B = B r,1 ⊗B with r > 1. Let B ′ = B 1,1 ⊗ B r−1,1 ⊗B. Consider the following diagram.
The top face and the bottom face commute by Proposition 27. The left face commutes by Proposition 29. The right face commutes by definition of lb on tensor products of crystals. The back face commutes by the assumption. Since lb is injective, the front face commutes.
Finally let us consider the the case B = B r,s ⊗B with s > 1. Let B ′ = B r,1 ⊗B r,s−1 ⊗B.
Consider the following diagram.

Spin cases
Let us consider the arguments which are needed to treat the spin cases. Here we will concentrate on the case B n,s . The other case B n−1,s can be obtained by replacing the Dynkin nodes n and n − 1 in the following description. To begin with, following [31], we introduce the three setsB n−1,1 ,B n,1 andB n,1 which are generated bŷ Here we regard them as the usual single columns and apply the Kashiwara operatorsf i (i ∈ I 0 ) as in the case for non-spin KN tableaux to obtain the whole elements ofB n−1,1 , B n,1 andB n,1 . On these sets, we define the operation left-box analogously to Definition 4.

See [31, Theorem 3.3].
Now we describe the algorithm for the rigged configuration bijection for the spin cases B n−1,l and B n,l following [31]. To begin with, we introduce the embedding of the rigged configurations emb : where the quantities with primes means that the everything is doubled as µ i . Similarly, if the rigged configuration {µ, (ν, J)} is composed of even integers, we can define emb −1 by dividing all its components by 2. Then the operation to obtain the leftmost column of an element of B n,l is Here γ is the same as Definition 26; γ replaces one of length l rows of µ (n) by two rows of length l − 1 and 1. The new operationsδ , k} is defined as follows. We look for the shortest singular string of (ν ′ , J ′ ) (n) which is longer than or equal to 2. If there is no such string, set ℓ (n) = ∞, k = n and stop. Otherwise we set ℓ (n−1) to be the length of the selected string of (ν ′ , J ′ ) (n) and do the same procedure as in the usual δ  except for the following point. We replace one of the length 2 rows of µ ′(n) by length 1 row and add a length one row to µ ′(n−1) .
, k} is defined as follows. Set ℓ (n−2) = 1. Then, as in the usual δ (a) l we begin to determine ℓ (n−1) and ℓ (n) and proceed similarly to the usual case. The new rigged configuration (ν ′′ , J ′′ ) is defined similarly as in the usual case except for the following point. We remove a length one row from each of µ ′(n−1) and µ ′(n) and add a length one row to µ ′(n−2) .
We regard the output of the sequence of δ's as the element ofB n,1 . Corresponding to the final emb −1 , we halve the width of the column with keeping all its contents. We regard the final output as the element of B n,1 .
Remark 33. In the above algorithm, some readers might feel that the change in µ is peculiar. To get a better understanding of this point, let us take a look at the following computations Let us compare with the special case ν (a) = ∅ for all a ∈ I 0 with weight 2Λ n (see (13)). In this case we should obtain a column filled by 1, 2, . . . , n from top to bottom. Note that each letter i in this column corresponds to ǫ i .
For the proof of Theorem 24, we have to modify β in Definition 26 as follows.β (n) replaces a length two row of µ (n) by a length one row, adds a length one row to each of µ (1) and µ (n−1) and adds a length one singular string to each of (ν, J) (a) for a ≤ n − 1. β (n−1) removes a length one row from each of µ (n−1) and µ (n) , adds a length one row to each of µ (1) and µ (n−2) and adds a length one singular string to each of (ν, J) (a) for a ≤ n − 2. Then we can use the same arguments of the proof of Proposition 29 to show thatβ (n−1) andβ (n) commute withẽ i andf i . Thus by using the same arguments we can show Theorem 24 in this case.

Outline
We concentrate on the proof forf i part since the other partẽ i follows from the former case. We divide the proof into the following five cases which exhaust all the possibilities: The actual proof for each case is based on the following behaviors of the vacancy numbers induced by δ.
Finally, let us prepare the following lemma which gives an useful criterion.
(1) Suppose that the string (ℓ + 1, x ℓ − 1) is singular. Then it satisfies P We can reverse the arguments to obtain the if part.
(2) If x ℓ ≥ x l , thenf i will act on (l, x l ) in stead of (ℓ, x ℓ ) since ℓ < l. Thus we have x ℓ < x l . Then we have P

Proof for Case A
The proof of this case depends on the following fundamental properties.
Case (I). We have to show P (i) which contradicts the minimality of x ℓ . Thus we have proved P Let j be the length of the longest string of ν (i) under the condition j < ℓ (i) . Since m (ii) The case j < ℓ (i) − 1. We can apply the convexity relation of P On the other hand, since ℓ < j, we have the condition x ℓ < x j ≤ P (i) j (ν). Combining this with the above result P . But then the string (j, x j ) is the singular string whose length satisfies ℓ (i−1) ≤ j < ℓ (i) , which is in contradiction to the definition of ℓ (i) . Hence we have proved j < ℓ (i−1) .
Then we can apply the convexity relation of P If ℓ < j, this contradicts the requirement x ℓ < x j . Hence we are left with the case j = ℓ. Since we are assuming ℓ < ℓ (i−1) − 1 we can refine the above relation as But this contradicts the requirement that the string (ℓ, x ℓ ) must satisfy x ℓ ≤ P (i) ℓ (ν). Thus we have proved P (i) Lemma 34. Since ℓ (i) = ℓ (i) we can use the same arguments of (I) to obtain the result.
Then by the convexity However this implies that x ℓ > x j or x ℓ > x ℓ (i) both of which are in contradiction to the minimality of x ℓ . Thus we conclude that P (i) . This is a contradiction since ℓ < j and ℓ < ℓ (i) .
Step 2. Based on the above observation, let us assume that ℓ (i−1) < ∞ in the sequel. Let us show that δ chooses the same strings before and afterf i . Again we havel (a) = ℓ (a) for all a ≤ i − 1 sincef i does not change coriggings of untouched strings. Recall that we have ℓ + 1 < ℓ (i−1) =l (i−1) by the assumption. Thus δ cannot choose the string (ℓ + 1, x ℓ − 1) created byf i . Thus we havel (i) = ℓ (i) which implies thatl (a) = ℓ (a) for all a ≥ i andl (a) = ℓ (a) for all a.
Step 3. Let us show thatf i chooses the same string before and after δ. Recall that δ creates the strings (ℓ (i) − 1, P Note that the string (ℓ, x ℓ ) remains as it is in (ν,J) (i) since δ acting on (ν, J) does not touch the string by the assumption ℓ < ℓ (i−1) . By Proposition 36 we have P Thereforef i acts on the same string (ℓ, x ℓ ) before and after the application of δ.
(2) It is enough to consider the strings (ℓ (a) , P (a) ℓ (a) (ν)) of (ν, J) (a) for a = i − 1, i, i + 1, sincef i adds a box to the same place before and after δ and since δ does not change riggings for untouched strings. For example, the string for a = i − 1 case behaves as follows; .
In all cases, the riggings are determined so thatf i and δ create the singular strings. Since we haveν =ν, we obtain P (i−1) The other cases are similar. Thus we conclude thatJ =J.

Proof for Case B
Let us show the commutativity off i and δ for the Case B, that is, We divide the proof into the two cases.
(2) We only need to consider J (i−1) and J (i) in this case. If the string is not touched by δ andf i , the riggings after δ •f i andf i • δ coincide since δ andf i choose the same strings in both cases respectively. The string (ℓ (i−1) , P Sinceν =ν we see the coincidence of the riggings. As for the string (ℓ, x ℓ ), we see that both δ •f i andf i • δ give the string (ℓ + 1, x ℓ − 1). Hence we haveJ =J.

Case 2:f i creates a singular string
This situation does not satisfy the condition of Theorem 30 due to the following fact.
Example 40. Consider the following rigged configuration (ν, J) of type ( The corresponding tensor product Φ(ν, J) is . f 2 acts on the string (4, −1) of (ν, J) (2) and makes it into (5, −2) of (ν,J) (2) . Since we have P 5 (ν) = −2, the latter string is singular. In particular, we see thatf 2 (ν, J) = 0. Now (ν,J) is Proposition 39). Indeed, we have P Recall that the defining condition of the Case C is We divide the proof into the three cases according to the following extra conditions.

A Common Property for Case C
In this case, we can show the following property. Note that the corresponding analysis for the case ℓ (i) = ℓ (i) is included in the analysis of P (i) ℓ (i) −1 (ν) which will be given later in this section.
Proof. We follow the classification in Lemma 34. Since ℓ (i) < ℓ (i) , we have to consider Case (V), (VI) and (VII). Let j be the largest integer satisfying j < ℓ (i) and m , we see that ℓ (i) ≤ j and in particular we have ℓ < j. In fact we only need the latter relation ℓ < j in the following proof. 3 Case (V). According to Lemma 34 we have to show P (i) By the assumption we see that ℓ < ℓ (i) . Then we have P Case (VI). According to Lemma 34 we have to show P (i) However this relation contradicts the requirement x j > x ℓ which follows from ℓ < j. Next, suppose that j < ℓ (i) − 1. Then, as in the Case (V), we have P . By the similar arguments in the previous case, we can deduce the contradiction. Therefore we obtain P (i) Case (VII). According to Lemma 34 we have to show P (i) Then the only possibility that is compatible with the assumption is P (i) j (ν) = x j = x ℓ + 1. In particular, the string (j, x j ) is singular whose length satisfies ℓ (i+1) ≤ j < ℓ (i) . However this is in contradiction to the definition of ℓ (i) .
Let us consider the case δ •f i . It is enough to analyze the strings (ℓ, x ℓ ) and (ℓ + 1, P .
Recall thatf i does not change coriggings other than the touched string. In particular, we . Thus the next δ can act on one of the length ℓ + 1 singular strings.
. Thereforef i will not act on the length ℓ (i) − 1 string of (ν,J) (i) that has been created by δ. On the other hand, if ℓ (i) = ℓ (i) we do not need to take care of this issue. Thus it is enough to analyze the behaviors of the lengths ℓ and ℓ + 1 strings; .
To summarize, we are left with lengths ℓ and ℓ + 1 strings in both δ •f i andf i • δ. Thus we haveν =ν.

Outline
As the second step, we treat the case when m (i) ℓ+1 (ν) = 0 andf i creates a singular string. In this case, we will show the following assertion by case by case analysis.
creates a singular string, then the following two conditions are satisfied: Note that we have ℓ To begin with, we explain how these properties provide the proof ofν =ν.
Proof. Let us analyzef i • δ and δ •f i respectively.
The remaining propertyJ =J will be checked from case by case analysis and it will be done along with the proof of Proposition 44.

Classification
We divide the proof into the two cases according to the following classification.

Proof for the case (27)
Under the present assumptions, let us show the following property.
Due to the existence of the length ℓ (i−1) = ℓ + 1 string at ν (i−1) and the assumption m ℓ+2 (ν). Let us show that this inequality is in fact equality. For this let j be the length of the smallest string of ν (i) that is longer than ℓ. Since m (i) ℓ+1 (ν) = 0 we have ℓ (i−1) = ℓ + 1 < j ≤ ℓ (i) . By Lemma 35 (2) we have Then by the convexity of P Combining this equality with (31) we obtain x j = P (i) j (ν), thus the string (j, x j ) is singular. By the relation ℓ (i−1) < j ≤ ℓ (i) we deduce that j = ℓ (i) .
Proof of Proposition 44 for the case (27). If m (i+1) k (ν) > 0 for some ℓ + 1 < k < ℓ (i) , the relation (29) must be strictly convex around such k, which is a contradiction. Therefore m This is in contradiction to the relation (30). Thus we have m This completes the proof of Proposition 44 for the case (27).
By using the factν =ν we showJ =J. For this purpose we note the following facts which are the consequences of the proof of Proposition 47.
Proof. From Corollary 48 and Proposition 45 we see that there are no strings of length from ℓ + 1 to ℓ (i) − 1 that are not touched by δ norf i during both δ •f i andf i • δ. Thus it is enough to analyze the strings (ℓ, x ℓ ) and (ℓ (i) , x ℓ (i) ) of (ν, J) (i) .
Let us analyze the behavior of the string (ℓ, x ℓ ). From the proof of Proposition 45 we have . By ℓ < ℓ (i−1) we have P Next let us analyze the behavior of the string (ℓ (i) , x ℓ (i) ). From the proof of Proposition 45 we have .
Recall that we have ℓ + 1 < ℓ (i) . Thus the firstf i of δ •f i decreases the rigging of the string (ℓ (i) , x ℓ (i) ) by 2.

Proof for the case (28)
Under the present assumptions, let us show the following property.
Proposition 51. Suppose that we have ℓ+1 (ν) = 0, the string (ℓ, x ℓ ) is non-singular andf i creates a singular string. Then we have Proof. Let j be the length of the shortest string of ν (i) that satisfies ℓ < j. By the assumption m In order to evaluate the minimum in (33), let us suppose if possible that P Combining the two inequalities we obtain P (i) ℓ+1 (ν) = P (i) j (ν). By the assumption, we conclude that P ≤ P (i) ℓ (ν) since the string (ℓ, x ℓ ) is non-singular.
Proof of Proposition 44 for the case (28). (32). From the result at the end of the proof of the previous lemma, we see that m (i) k (ν) = 0 for ℓ < k < ℓ (i) . Then the relation (32) implies that m (i+1) k (ν) = 0 for ℓ < k < ℓ (i) since the existence of such a string would imply that the relation (32) have to be strictly convex. Hence we complete the proof of Proposition 44.
By using the relationν =ν we showJ =J. For this purpose we note the following facts which are the consequences of the proof of Proposition 51.
Corollary 52. Under the same assumptions with the previous proposition, we have the following relations.
k (ν) becomes strictly convex around such the k, which is in contradiction to the relation (32). Thus we obtain the assertion.
Proof. By Proposition 45 and Corollary 52 (1) we see that it is enough to check the coincidence of riggings for the strings (ℓ, x ℓ ) and (ℓ (i) , x ℓ (i) ) that are touched by δ andf i .

Proof for Case C (3)
As the final step of the proof for Case C, let us treat the case where ℓ + 1 < ℓ (i) andf i creates a non-singular string.
Proposition 54. Assume that ℓ (i−1) ≤ ℓ + 1 < ℓ (i) < ∞. Iff i acting on (ν, J) creates a non-singular string, then the following relation is satisfied: Proof. In this case we have Throughout the proof of this proposition, let j be the largest integer such that j < ℓ (i) and m We divide the proof into four cases.
In particular we have P ℓ (i) (ν)) whose existence is assured by the definition of ℓ (i) . Then its rigging must satisfy P (i) ℓ (i) (ν) which is forbidden by the convexity of the vacancy numbers. Thus we have P (i) Case j = ℓ and (ℓ, x ℓ ) is non-singular. In this case we have P (i) ℓ (ν) > x ℓ since the string (ℓ, x ℓ ) is non-singular. Let us consider the string (ℓ (i) , P (i) ℓ (i) (ν)). Then its rigging must satisfy P (i) ℓ (i) (ν) > x ℓ since its length satisfy ℓ (i) > ℓ. Now we invoke the convexity of P Again by the convexity relation the only possibility is P (i) ℓ (ν) = · · · = P (i) ℓ (i) = x ℓ + 1. However this implies P (i) ℓ+1 (ν) = x ℓ + 1 and by Lemma 35 (1) this implies thatf i makes a singular string. This is in contradiction to the assumption thatf i makes a non-singular string. Thus we have P (i) Case j = ℓ and (ℓ, x ℓ ) is singular. In this case we have P (i) ℓ (ν) = x ℓ since the string (ℓ, x ℓ ) is singular. Also the rigging for the string (ℓ (i) , P (i) Then from the convexity relation of P Then the only possibility that is compatible with the convexity relation is In particular, we have P (i) ℓ+1 (ν) = x ℓ + 1. However, by Lemma 35 (1), this implies thatf i creates a singular string, which is the contradiction. Thus we conclude that P Proposition 55. Assume that ℓ (i−1) ≤ ℓ + 1 < ℓ (i) < ∞ andf i acting on (ν, J) creates a non-singular string. Then we have the following identities: (1)ν =ν, Proof. (1) We show thatf i acts on the same string (ℓ, x ℓ ) in both (ν, J) and (ν,J). For δ creates the strings (ℓ (i) − 1, P Proposition 41. Therefore the string (ℓ, x ℓ ) remains as the string with the smallest rigging of the largest length in (ν,J). Thusf i acts on the string (ℓ, x ℓ ) of (ν,J).
Also, sincef i creates the non-singular string and does not change other coriggings, δ chooses the same strings for both (ν, J) and (ν,J). Thus we haveν =ν.

Classification
Recall that the defining condition for Case D is ℓ (i) = ℓ. For the proof, it is convenient to further divide the case into the following four cases:

A Common Property for Case D
For the proof of Case D, we will need the following result.
Proposition 56. Assume that ℓ (i) = ℓ. Then the following relation holds: Proof. Throughout the proof of this proposition, let j be the largest integer such that j < ℓ and m If there is no such j, set j = 0. We divide the proof into the following four cases.

Case (a).
Denote the corresponding string of (ν, J) (i) as (ℓ − 1, x ℓ−1 ). Then we have x ℓ ≤ x ℓ−1 by definition of ℓ. Suppose that ℓ (i−1) < ℓ. Since we have ℓ (i−1) ≤ ℓ−1 < ℓ (i) the string (ℓ−1, x ℓ−1 ) cannot be singular. Thus we have P Next suppose that ℓ (i−1) = ℓ. In this case the string (ℓ − 1, x ℓ−1 ) can be singular. Therefore we have P Case (c). We can write the defining condition of case (c) as ℓ (i−1) ≤ j < ℓ − 1. Since ℓ (i−1) ≤ j < ℓ (i) the string (j, x j ) is non-singular. Thus we have P Case (d). We can write the defining condition of case (d) as j < ℓ (i−1) < ℓ. In this case the string (j, x j ) can be singular. Thus we have P We also have P However this is in contradiction to the fact that P (i) k (ν) is a strictly convex function between ℓ (i−1) − 1 ≤ k ≤ ℓ (i−1) + 1 due to the existence of the length ℓ (i−1) row at ν (i−1) . Thus we have P
(1) Let us analyze the action off i before and after the application of δ. In this case we can choose two distinct length ℓ strings (ℓ, x ℓ ) and (ℓ, P (i) ℓ (ν)) of (ν, J) (i) wheref i will act on the former one and δ will act on the latter one. Here note that if x ℓ = P (i) ℓ (ν) then all the riggings for the length ℓ strings are the same since the minimal value of the corresponding riggings is x ℓ and the maximal one is P (i) ℓ (ν). By Proposition 56 we have P (i) is shorter than (ℓ, x ℓ ) of (ν,J) (i) and its rigging is larger than or equal to x ℓ . Also, by ℓ < j, we can use the same arguments of Proposition 41 to show P (i) 1 (ν)) has the rigging that is strictly larger than x ℓ . Thereforef i will act on the same string (ℓ, x ℓ ) in both (ν, J) and (ν,J).
Let us analyze δ. Sincef i will not change the corigging of the string (ℓ (i) , P (i) ℓ (i) (ν)) of (ν, J) (i) , it remains as the shortest possible string in (ν,J) (i) starting froml (i−1) = ℓ (i−1) . So δ chooses this string after the application off i too. Let us show that the string (ℓ (i) , P (i) ℓ (i) (ν)) is chosen by δ in both before and after the application off i . Recall that we have two length ℓ strings (ℓ, x ℓ ) and (ℓ, P (i) ℓ (ν)) of (ν, J) (i) wheref i acts on the former one and δ acts on the latter one.
(i) Consider the case P (i) ℓ (ν) = x ℓ , that is, the string (ℓ, x ℓ ) is singular. Then the assumptions ℓ (i) < ℓ (i) and m (i) , as otherwise we would have ℓ (i) = ℓ (i) . Then a subtlety could occur only if ℓ (i) > ℓ (i+1) = ℓ + 1 andf i creates a singular string of length ℓ + 1. So suppose if possible that such a situation happens. In this case, we have P Also we have j ′ ≤ j by the assumption ℓ < j. Recall that we have P Since we have m Since ℓ < j ′ we have P (i) j ′ (ν) ≥ x j ′ > x ℓ . Then from the convexity relation of P (i) k (ν) between ℓ ≤ k ≤ j ′ , the only possibility that is compatible with P (i) j ′ (ν), in particular, the corresponding string is singular. Since j ′ ≥ ℓ+1 = ℓ (i+1) , we conclude that j ′ = ℓ (i) . However this relation contradicts the requirement j ′ ≤ j < ℓ (i) .

The case ℓ = j
We follow the classification of Lemma 34.
Proposition 58. Assume that ℓ = ℓ (i) < ℓ (i+1) = · · · = ℓ (i) , m (1)ν =ν, Proof. (1) In this case, we can use the same arguments of the proof of Case (V) of Proposition 41 to show P (i) ℓ (i) −1 (ν) > x ℓ . We also have P (i) ℓ (i) −1 (ν) ≥ x ℓ by Proposition 56. Thusf i acts on the same string before and after the application of δ. We notice that δ acts on the same string before and after the application off i even iff i creates a singular string since we are assuming that ℓ (i+1) = ℓ (i) . Here recall thatl (i) is determined as the length of the minimal possible string compared with the lengthl (i+1) = ℓ (i+1) and the length ℓ (i) singular string remains singular after the application off i .
Proof of (2) is the same as Proposition 57.

Case 2.
Let us consider the case ℓ < ℓ (i) −1. In this case, we can use the same arguments of Case (VI) of the proof of Proposition 41 to show P (i) Thus we see thatf i acts on the same string before and after the application of δ. Also, by the assumption ℓ (i+1) = ℓ (i) , we see that δ acts on the same string before and after the application off i . Thus we haveν =ν in this case.
Proof of (2) is the same as Proposition 57.
Step 1. Suppose that we have P (i) ℓ−1 (ν) ≥ x ℓ by Proposition 56, we see thatf i acts on the same string before and after δ. Let us consider the behavior of δ before and afterf i . Then a subtlety could occur only iff i creates a singular string.
Thus suppose if possible that P we see that all length ℓ strings of (ν, J) (i) are singular. This is a contradiction since we have m Let us write P Since we have m k (ν) between ℓ ≤ k ≤ ℓ (i) . Hence this case cannot happen. Therefore we haveν =ν andJ =J in this case.
Step 2. Suppose that we have ℓ = ℓ (i) − 1. By definition of the rigged configuration we have P Recall that we have the three strings (ℓ, x ℓ ), (ℓ, P (i) ℓ (ν)) and (ℓ + 1, P (i) ℓ+1 (ν)) of (ν, J) (i) . Then we have the remaining string (ℓ, x ℓ ) in (ν,J) (i) . This is a contradiction since we have P (i) ℓ (ν) < x ℓ . Thus suppose that P (i) ℓ (ν) ≥ x ℓ . As in Case 1 of the proof of the previous proposition we can show thatf i chooses the same string before and after δ. Let us consider the behavior of δ before and afterf i . In this case, there is no problem even iff i creates a singular string since we have ℓ + 1 = ℓ (i) . Hence we haveν =ν andJ =J.
Step 3. We assume that ℓ < ℓ (i) −1 and P (i) ℓ (i) −1 (ν) ≤ x ℓ in the following proof. According to Lemma 34 (VII) we have P (i) x ℓ by definition of the rigged configurations. Since we are assuming that ℓ = ℓ (i) = j < ℓ (i) − 1, the convexity relation of the vacancy numbers allows the following three possibilities: In the following we consider them case by case.
Let us analyzef i •δ first. From Lemma 34 (VII) we have P (i) ) has the smallest rigging of the largest length. Thus the nextf i acts on this string. To summarize, we haveν (a) =ν (a) for all a = i andν (i) is obtained by removing one box from the length ℓ (i) row.
Example 61. Here we give an example for Case (2) of Step 3 of the proof of Proposition 60. Consider the following rigged configuration (ν, J) of type ( The corresponding tensor product Φ −1 (ν, J) is .

Proof for Case D (2)
Proposition 62. Suppose that ℓ = ℓ (i) = ℓ (i) and m (1)ν =ν, Proof. By the assumption m (i) ℓ (ν) ≥ 3 we can choose the three distinct strings (ℓ (i) , P (i) ℓ (i) (ν)) = (ℓ (i) , P (i) ℓ (i) (ν)) and (ℓ, x ℓ ) of the i-th configuration. Here we may have P (i) Let us analyzef i • δ. After the application of δ the above three strings become (ℓ (i) − 1, P (i) 1 (ν)) and (ℓ, x ℓ ). By Proposition 56 we have P (i) Thus the string (ℓ, x ℓ ) remains as the string with smallest rigging of the largest length. Hencef i will act on it. To summarizeν (a) =ν (a) for all a = i andν (i) is obtained by removing a box from each of the two length ℓ (i) = ℓ (i) rows and adding a box to the length ℓ row.

Outline
Let us consider the third case. We prove the following properties in the latter part of this subsection by case by case analysis.
Proposition 63. Assume that ℓ (i) = ℓ and m (i) ℓ (ν) = 1. Then the following relations hold: Next let us show the following property: Proposition 64. Assume that ℓ = ℓ (i) < ℓ (i) and m (i) ℓ (ν) = 1. Then the following relation holds: Proof. We follow the classification of Lemma 34. Since we are assuming that ℓ (i) < ℓ (i) , we have to deal with Case (V), (VI) and (VII). Throughout the proof of this proposition, let j be the largest integer such that j < ℓ (i) and m (i) j (ν) > 0. Since we are assuming that ℓ (i) < ℓ (i) , we have ℓ (i) ≤ j.
Case (V). In this case, we have to show that P (i) ℓ (i) −1 (ν) ≥ x ℓ under the assumption ℓ (i) < ℓ (i+1) = · · · = ℓ (i) . We can use the same argument of the corresponding part of Proposition 41 to show the assertion.
Case (VI). In this case, we have to show that P (i) Let us consider the case j = ℓ (i) − 1. Note that by (42), which is shown independently to the present proposition, we have ℓ < ℓ (i+1) . Combining with the current assumption However this is in contradiction to the relation x ℓ (i) −1 > x ℓ which follows from the minimality of x ℓ under the condition ℓ (i) − 1 > ℓ.
Next let us consider the case j < ℓ . Then from the convexity relation of P This is in contradiction to the relation x j ≥ x ℓ that follows from the minimality of x ℓ .
We divide the proof of Case (VII) into three steps according to the position of j.
Step 1. First, let us consider the case j = ℓ (i) − 1. Then there is the corresponding ℓ (i) −1 (ν) by the minimality of x ℓ . Combining with the assumption P (i)

is singular and its length satisfies
. This contradicts the definition of ℓ (i) .
Step 2. Next, let us consider the case ℓ = ℓ (i) < j < ℓ (i) − 1. Since we are assuming that ℓ (i) < ℓ (i) we have P (i) Then, by the convexity of P (i) k (ν) between j ≤ k ≤ ℓ (i) , the only possibility that is compatible with the assumption P (i) 1. Consider the case ℓ (i+1) ≤ j. Then, combining with the relation P In particular, the string (j, x j ) is singular and its length satisfies that ℓ (i+1) ≤ j < ℓ (i) . This contradicts the definition of ℓ (i) .
Proofs ofJ =J will be given by case by case analysis assuming the relationν =ν along with the proof of Proposition 63.
To begin with, let us show the following property.
Step 1. Let us consider the case j = ℓ − 1. Then the string (ℓ − 1, x ℓ−1 ) is non-singular since its length satisfies ℓ (i−1) ≤ ℓ − 1 < ℓ (i) . Thus we have P Step 2. Let us consider the case j < ℓ − 1. Then we have x j ≥ x ℓ by definition of x ℓ and ℓ > 0. Thus we have P . Then by the convexity of P Then by the convexity of the vacancy numbers the only possibility that is compatible with P On the other hand, by the minimality of x ℓ , we have x j ≥ x ℓ . Thus . Then the length of the singular string (j, x j ) satisfies ℓ (i−1) ≤ j < ℓ (i) which is in contradiction to the definition of ℓ (i) . Thus we conclude that P (ii) Suppose that j < ℓ (i−1) . Then the relation of P (i) k (ν) between ℓ (i−1) − 1 ≤ k ≤ ℓ (i−1) + 1 must be strictly convex due to the existence of the length ℓ (i−1) string at ν (i−1) . This is in contradiction to the relation P Proof of Proposition 63 for the case ℓ (i−1) < ℓ. By (44) we have P (i) ℓ+1 (ν) = x ℓ + 1, that is, (40). By the assumption ℓ (i−1) < ℓ we see ℓ (i−1) ≤ ℓ − 1. Thus we obtain P that is, (41). Finally, from (45) we have m (i+1) ℓ (ν) = 0. Therefore we have ℓ < ℓ (i+1) , that is, (42).
In view of the proof ofν =ν, we can showJ =J for this case as follows.

Proof of Proposition 63 for the case ℓ (i−1) = ℓ
We begin by showing the following property.
Step 1. Let us consider the case j = ℓ − 1. Since the string (ℓ, x ℓ ) is singular, we have P (i) ℓ (ν) = x ℓ , and by the minimality of x ℓ we have Step 2. Let us consider the case j < ℓ − 1. Recall that the assumptions ℓ = ℓ (i) and m (i) ℓ (ν) = 1 imply that the string (ℓ, x ℓ ) is singular and thus P (i) ℓ (ν) = x ℓ . If j > 0, then by the minimality of ℓ with j < ℓ we have P In this case, we also have P which is the consequence of ℓ > 0. In both cases, by the convexity of P Proof of (2) is the same with the previous case.

Outline
We will prove the following proposition in the latter part of this subsection.
Step 1. Let us consider the casef i • δ. The first δ creates the two strings (ℓ − 1, P (i) ℓ−1 (ν)) of (ν,J) (i) . Since the operation δ does not change the riggings of unchanged strings, the relation (49) and m (i) ℓ (ν) = 2 imply that the strings (ℓ − 1, x ℓ ) are the longest strings among the strings with rigging x ℓ , hencef i will act on one of these strings.
Proofs ofJ =J will be given by case by case analysis along with the proof of Proposition 72 assuming the relationν =ν we have just proved.

Proof of Proposition 72 for the case ℓ (i−1) < ℓ
We divide the proof of Proposition 72 into two cases according to whether ℓ (i−1) < ℓ or ℓ (i−1) = ℓ. In this subsection, we consider the case ℓ (i−1) < ℓ. Throughout the proof of Proposition 72, let j be the largest integer such that j ≤ ℓ − 1 and m Moreover, we have For the proof of this proposition, we prepare the following lemma.
Assuming thatν =ν, we can showJ =J for this case as follows.
Proposition 75. Suppose that ℓ (i−1) < ℓ = ℓ (i) = ℓ (i) and m Proof. From the proof ofν =ν after Proposition 72, we see that it is enough to look at the string (ℓ, x ℓ ) of (ν, J) (i) which is changed by bothf i and δ in bothf i • δ and δ •f i . Then we have .

Recall that we havẽ
by the assumption ℓ (i−1) < ℓ = ℓ (i) = ℓ (i) and the arguments given in the proof ofν =ν. Then we have P Moreover, we have For the proof of this proposition, we prepare the following lemma.
Assuming thatν =ν, we can showJ =J for this case as follows. Proof. Recall that we can assume thatf i acts on the same string (ℓ, x ℓ ) of (ν, J) (i) in both f i • δ and δ •f i . Then we have .

Classification
Recall that the defining condition for Case E is ℓ (i) < ℓ. We further classify this case as follows.

A common property for Case E
The fundamental observation for this case is the following property.
Proposition 80. If ℓ (i) < ℓ, we have Proof. We follow the classification of Lemma 34.

Case (I).
Let us consider the case ℓ (i−1) = ℓ (i) . Then by Lemma 34, we have to show P (i) which is a contradiction since we have x ℓ ≤ 0 by 0 < ℓ. Therefore we assume ℓ (i) ≥ 2 in the sequel. From the assumption ℓ (i) < ℓ we have P (i) Let j be the length of the longest string of ν (i) that are strictly shorter than ℓ (i) . If j = ℓ (i) − 1, then we have x j ≤ P (i) j (ν) < x ℓ , which contradicts the minimality of x ℓ . Thus we assume that j < ℓ (i) − 1. We apply the convexity relation of P is a contradiction. Therefore we see that ν (i) does not contain strings that are strictly shorter than ℓ (i) . Again we can use the convexity relation of P (i) ℓ (i) (ν) and obtain the relation 0 = P This is a contradiction since we have x ℓ ≤ 0. Thus we have proved P (i) Case (II). Let us consider the case ℓ (i−1) < ℓ (i) . Then by Lemma 34, we have to show P (i) Let j be the length of the longest row of ν (i) that are strictly shorter than ℓ (i) . Suppose that j = ℓ (i) − 1. Then we have By the minimality of x ℓ this relation implies that x j = P (i) j (ν), that is, the string (j, x j ) is singular in (ν, J) (i) . This is in contradiction to the definition of ℓ (i) since we have Therefore we assume j < ℓ (i) −1 in the sequel. If P (i) ℓ (i) (ν) and by the convexity relation we have This is a contradiction. Therefore we must have P (i) by the similar argument we obtain x j < x ℓ , which is a contradiction. Thus we conclude that P (i) ℓ (i−1) (ν) > 0 the convexity relation for P (i) k (ν) between ℓ (i−1) − 1 ≤ k ≤ ℓ (i−1) + 1 must be strict. Thus we obtain x ℓ ≥ P (i) which is a contradiction. Therefore we conclude that (ν, J) (i) does not contain strings that are strictly shorter than ℓ (i) . Then from the convexity relation of P 0 (ν) = 0, which is in contradiction to the requirement x ℓ ≤ 0. Thus we have proved P (i) ℓ (i) −1 (ν) ≥ x ℓ for the case ℓ (i−1) < ℓ (i) .

Proof for Case E (1)
To begin with we prepare the following property.
Proposition 81. Suppose that ℓ (i) < ℓ and ℓ + 1 < ℓ (i+1) . Then we have Proof. We follow the classification of Lemma 34. From the assumption, we see that ℓ (i) < ℓ (i+1) . Then we only need to consider cases (V), (VI) and (VII). During the proof of the proposition, let j be the largest integer such that j < ℓ (i) and m (i) j (ν) > 0. Since ℓ (i) < ℓ < ℓ (i+1) ≤ ℓ (i) we have ℓ ≤ j. Proofs for cases (V) and (VII) are the same as the corresponding parts of the proof of Proposition 36.
Case (VI). In this case, we have to show P (i) First, let us consider the case j = ℓ (i) − 1. Then the corresponding string satisfies x ℓ (i) −1 ≤ P (i) ℓ (i) −1 (ν) ≤ x ℓ . Now recall that we have ℓ < ℓ (i) − 1 since we have ℓ + 1 < ℓ (i+1) ≤ ℓ (i) by the assumption. Thus the minimality of x ℓ requires that This is in contradiction to the requirements x ℓ < x ℓ (i) and x ℓ < x j derived from ℓ < ℓ (i) and ℓ < j, respectively.
Finally let us consider the case ℓ = j. Recall that we have P However these relations are in contradiction to the convexity relation of P (i) k (ν) between ℓ ≤ k ≤ ℓ (i) which we can use by ℓ < ℓ (i) − 1.
Step 2. Let us consider the casef i • δ. Since we are assuming that ℓ (i) < ℓ < ℓ (i) the string (ℓ, x ℓ ) remains as it is after δ. Recall that we have P (i) ℓ (i) −1 (ν) ≥ x ℓ by Proposition 80 and P (i) ℓ (i) −1 (ν) > x ℓ by the previous proposition. Thus the string (ℓ, x ℓ ) has the largest length among the strings of (ν,J) (i) that have the smallest rigging.
Step 3. To summarize, we haveν (a) =ν (a) =ν (a) for all a = i andν (i) =ν (i) is obtained by adding a box to the ℓ-th column ofν (i) . The statementJ =J is obtained by the equality P

Proof for Case E (2)
Let us consider the case ℓ (i) < ℓ and ℓ (i+1) ≤ ℓ + 1 < ℓ (i) = ∞. We divide the proof into two cases according to whetherf i creates a singular string or not.
Step 1. Let us consider the case δ •f i . Sincef i does not change coriggings of the untouched strings, and sincef i creates a non-singular string, we havel (a) = ℓ (a) and ℓ (a) = ℓ (a) for all a.
Step 2. Let us consider the casef i • δ. Since we are assuming that ℓ (i) < ℓ and ℓ (i) = ∞ the string (ℓ, x ℓ ) remains as it is after δ. Recall that we have P (i) ℓ (i) −1 (ν) ≥ x ℓ by Proposition 80. Thus the string (ℓ, x ℓ ) has the largest length among the strings of (ν,J) (i) that have the smallest rigging.
Proof. As in Case C we can reduce the proof to the previous proposition.

Proof for Case 3
We can use the same arguments in Section 5.7 if we replace ℓ (i) (resp. ℓ (i−1) ) there by ℓ (i) (resp. ℓ (i+1) ) and neglect the arguments concerning the string ℓ (i) there.
(1) Let us analyze the action off i before and after the application of δ. By the assumption m (i) ℓ (ν) > 1 we can choose two distinct length ℓ strings (ℓ, x ℓ ) and (ℓ, P (i) ℓ (ν)) of (ν, J) (i) wheref i will act on the former one and δ will act on the latter one. Recall that we have P ℓ (i) −1 (ν)) of (ν,J) (i) are shorter than (ℓ, x ℓ ) and their riggings are larger than or equal to x ℓ . Thereforef i will act on the same string (ℓ, x ℓ ) in both (ν, J) and (ν,J).
(2) Since eachf i and δ chooses the same strings in δ •f i andf i • δ, the coincidencē J =J is a consequence of P (i)

Proof for Case E (5)
In this case, we can show the following property.
As the consequence of this proposition, we have the following result.
Proof. Since we have P (i) ℓ (i) −1 (ν) ≥ x ℓ by Proposition 80 and P (i) ℓ (i) −1 (ν) ≥ x ℓ by Proposition 94, we see thatf i acts on the same string before and after δ. Also, since ℓ (i) ≤ ℓ (i) < ℓ and the fact thatf i does not change coriggings of untouched strings, we see that δ acts on the same strings before and afterf i . Thus we haveν =ν.J =J follows from the fact P For the proof, we start by preparing the following properties.
Proof. In Section 5, we analyze all possible cases such thatf i (ν, J) = 0. Then only cases such thatf i (ν, J) is defined andf i (ν,J) is undefined are the above two cases as described in Proposition 39 and Proposition 84.
(2) During the proof of Proposition 84 we show P (i) ℓ+1 (ν) = x ℓ . Then we can use the same arguments of (1) if we replace ℓ (i−1) , ℓ (i) and ℓ (i+1) etc. there by ℓ (i+1) , ℓ (i) and ℓ (i−1) etc., respectively. Note that in this case we have to work under the condition ℓ (i) < ℓ and ℓ (i) = ∞. Then we can use the fact that all strings of (ν, J) (i−1) that are strictly longer than ℓ (i−1) (< ℓ + 1) do not change after δ. In particular, we can show m Proof of Proposition 96. According to the previous observations, we have to consider two distinct cases.

Proof for (2)
To begin with, let us show the following property.
Proof. Note that if we show that the rigging for the shortened string in (ν,J) (i) is larger than or equal to s, then we have s ≤s. In particular, we can assume that ℓ (i) > 1, since otherwise the corresponding rigging will be erased by δ and will not contribute tos, thus s =s by ℓ (i) = ∞. Since ℓ (i) = ∞, it is enough to check Then we have the two possibilities; During the proof, let j be the largest integer such that j < ℓ (i) and m (i) j (ν) > 0. If there is no such j, we set j = 0.
Case (I). In this case we have to show that P (i) ℓ (i) −1 (ν) ≥ s under the assumption ℓ (i−1) = ℓ (i) . Let us consider the case j = ℓ (i) − 1. Then we are done since we have P (i) ℓ (i) −1 (ν) ≥ x ℓ (i) −1 ≥ s by the minimality of s. Next consider the case 0 < j < ℓ (i) − 1. Then by the convexity relation between j and ℓ (i) we have P (i) Suppose that j = 0. If s ≤ 0, we can use the same arguments of the previous case since P (i) 0 (ν) = 0.
Therefore we have to consider the case j = 0 and s > 0. Since s > 0,f i acting on (ν, J) adds the length 1 string to (ν, J) (i) . Recall thatf i does not change coriggings of untouched strings. Since we have ℓ (i−1) = ℓ (i) > 1, we see thatl (a) = ℓ (a) andl (a) = ℓ (a) for all a. However this is in contradiction to the assumptions ℓ (i) = ∞ andl (i) < ∞. Hence this case cannot happen.
Case (II). In this case we have to show that P (i) is singular. However this is a contradiction since its length satisfies Suppose that ℓ (i−1) ≤ j < ℓ (i) − 1. By definition of s we have P By the minimality of s and s ≥ P (i) ℓ (i) −1 (ν), we have P (i) j (ν) = x j , that is, the string (j, x j ) is singular. This contradicts the definition of ℓ (i) .
Suppose that j = 0. Then we have s ≥ P (i) this is a contradiction. Hence we assume that s > 0. Thenf i adds the length 1 string (1, −1) to (ν, J) (i) . If ℓ (i−1) > 1, we can use the same arguments in the last part of Case (I) to show that such case cannot happen. Thus assume that ℓ (i−1) = 1. If P (i) 1 (ν) > −1, then the added string (1, −1) is non-singular so that we can use the arguments in the last part of Case (I) to show that such case cannot happen.
To summarize, we have shown that we have s ≤s except for the case j = 0, s > 0, ℓ (i−1) = 1 and P Then by the convexity relation of P (i) k (ν) between 1 ≤ k ≤ ℓ (i) the only possibility that is compatible with 1 = P (i) To summarize, we have the following situation: Since j = 0 we have m We have m (ν) = 0. We also have m (i+1) k (ν) = 0 for 1 < k < ℓ (i) by the relation P (i) . Sincef i does not change the coriggings of (ν, J) (a) for a < i, we havel (i−1) = ℓ (i−1) = 1. Recall that the string (1, −1) of (ν,J) (i) is singular.
Proposition 100. Let b ∈ B 1,1 ⊗B and suppose that we have shown the commutativity off i and Φ for the elements ofB. Suppose that we havef i (b) = 0andf i (b ′ ) = 0 where b ′ ∈B is the corresponding part of b. Then we havef i (ν, J) = 0,f i (ν,J) = 0 and Proof. By the assumption, we have the following two possibilities; Case (a). In this case we have ℓ (i−1) < ∞ and ℓ (i) = ∞. Then we have P  ∞ (ν) =s wheres is the smallest rigging of (ν,J) (i) . Let s be the smallest rigging of (ν, J) (i) . Since ℓ (i) = ∞, we have s =s. Again by Theorem 12 we have , it is enough to show thatf i acting on (ν, J) creates a singular string and it satisfiesl (i−1) ≤ ℓ + 1 =l (i) <l (i+1) = ∞ under the assumptionsf i (ν, J) = 0 andf i (ν,J) = 0. Then this is the consequence of Proposition 97.

Proof for (3)
For the analysis of the present situation, let us prepare several properties of the rigged configurations. Let s (resp.s) be the smallest rigging of (ν, J) (i) (resp. (ν,J) (i) ).
Case (a-I). Suppose that j = ℓ (i) −1. Then we have x j ≤ P (i) j (ν) < s, which contradicts the minimality of s. Suppose that 0 < j < ℓ (i) − 1. Recall that by the minimality of s we have P (i) . Then by the convexity relation of P Since we are assuming that s > P (i) ℓ (i) −1 (ν), this contradicts the minimality of s. Finally suppose that j = 0. Then the previous inequalities become s > P (i) Note that this relation is always valid since we have ℓ (i) ≥ 1 by the definition. Therefore we have s > 0 which implies thatf i acting on (ν, J) adds the length 1 string (1, −1) to (ν, J) (i) . Since we are assuming thatf i (ν, J) = 0, the rigging of the string (1, −1) of (ν,J) (i) must be larger than the corresponding vacancy number. Thus we have P Therefore the rigging of the string (ℓ (i) , P (i) This is a contradiction since we have s > 0.
Case (a-II). Suppose that j = ℓ (i) − 1. Then we have x j ≤ P (i) j (ν) ≤ s. By the minimality of s we have x j = P (i) j (ν) = s, in particular, the string (j, x j ) of (ν, J) (i) is singular. However this is in contradiction to the definition of ℓ (i) since we have ℓ (i−1) ≤ j < ℓ (i) . Suppose that ℓ (i−1) ≤ j < ℓ (i) − 1. Then by the convexity relation of P Then by the minimality of s we have P (i) j (ν) = x j , in particular, the string (j, x j ) of (ν, J) (i) is singular. Again this is in contradiction to the definition of ℓ (i) since ℓ (i−1) ≤ j < ℓ (i) . Suppose that 0 < j < ℓ (i−1) . Since m (i−1) This contradicts the minimality of s.
Finally suppose that j = 0. Then the above relation becomes s ≥ P (i) 0 (ν) = 0, that is, s > 0. By the similar arguments of the latter part of Case (a-I), we obtain P . Hence the rigging of the string (ℓ (i) , P (i) ℓ (i) (ν)) of (ν, J) (i) satisfies that P (i) ℓ (i) (ν) < 0. This is a contradiction since we have s > 0.
Suppose that ℓ (i+1) ≤ j < ℓ (i) − 1. Since s is the minimal rigging of (ν, J) (i) , we have P By the minimality of s we have P (i) j (ν) = x j , in particular, the string (j, x j ) is singular. However this is in contradiction to the definition of ℓ (i) since we have ℓ (i+1) ≤ j < ℓ (i) .
Suppose that ℓ (i) ≤ j < ℓ (i+1) . As in the previous paragraph, we obtain s = P (i) However, this is a contradiction since we have m (i+1) ℓ (i+1) (ν) > 0, which implies that the vacancy numbers P Case (c-2). In this case we have s =s. This case is indeed possible as the example at the end of this subsection shows.
Case (d). By the assumption ℓ (i−1) < ∞, we have P We can use the same arguments of Case (c-1) to show that this case cannot happen.
Proof. (1) Step 1. Let us show s ≤s. It is enough to show that the new string of (ν,J) (i) has the rigging larger than or equal to s, that is, P Then we can use the same arguments of Case (a) of the proof of Lemma 101 to show that this case cannot happen. Thus s ≤s.
Step 2. Next, let us show s ≥s. Then it is enough to show P (i) ℓ (i) (ν) ≥s since the only rigging that is changed by δ −1 is the one associated with the lengthened string. By the assumption ℓ (i+1) = ∞ we have P (i) Thus the above relation is equivalent to P (i) Suppose if possible that there are strings of (ν,J) (i) that are longer than or equal to ℓ (i) . Let j be the length of the shortest such string. By the minimality ofs we have P (i) j (ν) ≥ x j by the convexity relation between ℓ (i) − 1 and j (if j = ℓ (i) we can directly show the inequality without the second term). By the minimality ofs, we haves ≤ x j . Therefore we obtains = x j which implies that the string (j, x j ) is singular with length larger than ℓ (i) − 1. However, we know that δ −1 will add a box to the length ℓ (i) − 1 string of (ν,J) (i) . This is a contradiction since ℓ (i+1) = ∞ implies that δ −1 will add a box to the longest possible string. Therefore we conclude that the longest string of (ν,J) (i) has length ℓ (i) − 1. Since we have P (i) , the convexity relation between ℓ (i) − 1 and ∞ gives P . However this is a contradiction since we already know that P  ℓ (i) (ν) >s is always satisfied. Hence we obtain s ≥s. By combining the two inequalities, we conclude that s =s.
(2) During the proof of the previous lemma, we have shown that the only possible case under the assumption is s =s.
Proof. According to Lemma 101, we see that there are only two possibilities. The first case is ℓ (i) < ∞ and ℓ (i+1) = ∞. In this case we have b = (i + 1) ⊗ b ′ and P Here we have used the relation s =s from Corollary 102 and the assumption ϕ(ν, J) = P (i) Then by the induction hypothesis we have ϕ i (b ′ ) = ϕ i (ν,J) = 1. On the other hand, recall that we have ε i (i + 1) = ε i (ī) = 1. Then from the tensor product rule at Section 2.1, we conclude thatf i (b) = 0. Hence we have Φ(f i (ν, J)) =f i (Φ(ν, J)).

Proof for (4)
Proposition 105. Let b ∈ B 1,1 ⊗B and suppose that we have shown the commutativity off i and Φ for the elements ofB. Suppose that we havef i (b) = 0 andf i (b ′ ) = 0 where b ′ ∈B is the corresponding part of b. Then we havef i (ν, J) = 0,f i (ν,J) = 0 and Proof. By the assumption, we have b = (i + 1) ⊗ b ′ or b = i ⊗ b ′ and ϕ i (b ′ ) = 1. In the first case we have ℓ (i) < ∞ and ℓ (i+1) = ∞ and in the second case we have ℓ (i) < ∞ and ℓ (i−1) = ∞. In both cases we have P (i) together with the induction hypothesis we have ϕ i (b ′ ) = ϕ i (ν,J) = 1, which implies P (i) ∞ (ν) =s + 1 by Theorem 12. Heres is the minimal rigging of (ν,J) (i) . Then we have If we show s ≥s we have ϕ i (ν, J) = 0 as desired. Note that this result leads to Let us show s ≥s. If ℓ (i) < ∞ and ℓ (i+1) = ∞ we can use the same arguments of Step 2 of the proof of Corollary 102 (1) since we do not use the assumption ϕ i (ν, J) = 0 there. If ℓ (i) < ∞ and ℓ (i−1) = ∞, again we can use the same arguments of Step 2 of the proof of Corollary 102 (1) by replacing ℓ (i) and ℓ (i+1) there by ℓ (i) and ℓ (i−1) respectively.

Proof of Theorem 31:ẽ i version
In this section we use a parallel notation that are introduced in Section 4.4 except for changing the role off i there byẽ i . For example,ẽ i acts on the string (ℓ, x ℓ ) of (ν, J) (i) . (5) Proposition 106. Let us consider the rigged configuration (ν, J) of type B 1,1 ⊗B. Suppose that we have the commutativity ofẽ i andf i with Φ forB. Suppose thatẽ i (ν, J) = 0 and e i (ν,J) = 0. Let b = Φ(ν, J) and b ′ = Φ(ν,J). Then we have one of the following two cases;

Proof for
For the proof, we show the following properties.
Then we can use a parallel arguments of the previous case (1) to deduce that the string (ℓ (i) − 1, P (i) ℓ (i) −1 (ν)) is the longest string of (ν,J) (i) .
Proof. By the assumption, we have the following two possibilities; In both cases we have ϕ i (b ′ ) = 0. Note that we haveẽ i (ν,J) = 0 by the assumption.
1. Let us consider the case ℓ (i) < ℓ (i+1) . Then we have P (i) ℓ (i) (ν) = P (i) ℓ (i) (ν) − 1. Therefore we have P (i) On the other hand, since we cannot have more than one strings of (ν,J) (i) which have negative riggings, we have m This contradicts the convexity relation of Lemma 17.
2. Next, let us consider the case ℓ (i) = ℓ (i+1) < ℓ (i+1) . Then we have P (i) On the other hand, we have m Again this contradicts the convexity relation of Lemma 17.
Let j be the largest integer such that j < ℓ (i) − 1 and m (i) j (ν) > 0. If there is no such string, set j = 0. Then we have P Since we have P (i) ℓ (i) −1 (ν) < 0 and ℓ (i) − 1 < ℓ (i) − 1, we see that there is the remaining negative rigging in (ν, J) (i) . This is a contradiction since we haveẽ i (ν, J) = 0. Thus we have m Since we have m ≥ 0. Combining this with the first relation of (79), we obtain 0 > n (i−1) . This is a contradiction since we have n  Case 2. Suppose that we have ℓ (i) −1 = ℓ (i+1) −1. In particular, we have m From n (i) ℓ (i) = n (i) ℓ (i) −1 − 1, the second relation of (80) gives n (i−1) ≥ −1. Then from the first relation of (80), we obtain 0 ≥ n (i−1) . Therefore we have n Proof. The proof of this assertion follows from the tensor product rule at Section 2.1.