The Pelletier-Ressayre hidden symmetry for Littlewood-Richardson coefficients

We prove an identity for Littlewood--Richardson coefficients conjectured by Pelletier and Ressayre (arXiv:2005.09877). The proof relies on a novel birational involution defined over any semifield.

One of the central concepts in the theory of symmetric functions are the Littlewood-Richardson coefficients c λ µ,ν : the coefficients when a product s µ s ν of two Schur functions is expanded back in the Schur basis (s λ ) λ∈Par . Equivalently, these coefficients are tensor product multiplicities of irreducible representations of GL n (noting that each partition λ having length n has a certain irreducible representation V (λ) of GL n corresponding to it, the so-called Weyl module for shape λ). Various properties of these coefficients have been found, among them combinatorial interpretations, vanishing results, bounds and symmetries (i.e., equalities between c λ µ,ν for different λ, µ, ν). A recent overview of the latter can be found in [2].
In [18], Pelletier and Ressayre conjectured a further symmetry of Littlewood-Richardson coefficients. Unless the classical ones, it is a partial symmetry (i.e., it does not cover every Littlewood-Richardson coefficient); it is furthermore much less simple to state, to the extent that Pelletier and Ressayre have conjectured its existence while leaving open the question which exact coefficients it matches up. In this paper, we answer this question and prove the conjecture thus concretized.
The conjecture, in its original form, can be stated as follows: Let n 2, and consider the set Par [n] of all partitions having length n. Let a and b be two nonnegative integers, and define the two partitions α = (a + b, a n−2 ) and β = (a + b, b n−2 ) (where c n−2 means c, c, . . . , c n−2 times , as usual in partition combinatorics). Fix another partition µ ∈ Par [n]. Then, the families c ω α,µ ω∈Par [n] and c ω β,µ ω∈Par [n] of Littlewood-Richardson coefficients seem to be identical up to permutation. We can restate this in terms of Schur polynomials in the n variables x 1 , x 2 , . . . , x n ; this then becomes the claim that the products s α (x 1 , x 2 , . . . , x n ) · s µ (x 1 , x 2 , . . . , x n ) and s β (x 1 , x 2 , . . . , x n ) · s µ (x 1 , x 2 , . . . , x n ), when expanded in the basis of Schur polynomials, have the same multiset of coefficients. Equivalently, this can be restated in terms of representations of GL n ; then, it becomes the claim that the tensor products V (α) ⊗ V (µ) and V (β) ⊗ V (µ) decompose into irreps (i.e., irreducible representations) with the same multiplicities (in the sense that there is a multiplicity-preserving bijection between the irreps in V (α)⊗V (µ) and the irreps in V (β) ⊗ V (µ)). 1 Pelletier and Ressayre have proved this conjecture for n = 3 (see [18,Corollary 2]) and in some further cases. We shall prove it in full generality, and construct what is essentially a bijection ϕ : Par [n] → Par [n] that makes it explicit (i.e., that satisfies c ω α,µ = c ϕ(ω) β,µ for each ω ∈ Par [n]). To be fully precise, ϕ will not be a bijection Par [n] → Par [n], but rather a bijection from Z n to Z n , and it will satisfy c ω α,µ = c ϕ(ω) β,µ with the understanding that c ω α,µ = c ω β,µ = 0 when ω / ∈ Par [n]. (Here, Par [n] is understood to be a subset of Z n by identifying each partition λ ∈ Par [n] with the n-tuple (λ 1 , λ 2 , . . . , λ n ).) We will define this bijection ϕ by explicit (if somewhat intricate) formulas that "mingle" the entries of the partition it is being applied to with those of µ (as well as a and b) using the min and + operators. These formulas are best understood in the birational picture, in which these min and + operators are generalized to the addition and the multiplication of an arbitrary semifield. (Our proof does not require this generality, but the birational picture has the advantage of greater familiarity and better notational support. It also reveals a connection with a known birational map known as a "birational R-matrix" (see Subsection 5.2), which could throw some light on the otherwise rather mysterious bijection.) Another ingredient of our proof is an explicit formula for s α (x 1 , x 2 , . . . , x n ) for the abovementioned partition α.

Remark on alternative versions
A number of proofs in this paper rely on long computations, inductions and laborious, if fairly straightforward, combinatorial arguments. In the present version of this paper, we only outline
• We fix a commutative ring k; we will use this k as the base ring in what follows.
• For any partition α and any positive integer i, we let α i denote the i-th entry of α (so that α = (α 1 , α 2 , α 3 , . . . )). More generally, we use this notation whenever α is an infinite sequence of any kind of objects.
• We let Par denote the set of all partitions.
• We will use the notation m k for "m, m, . . . , m k times " in partitions and tuples (whenever m ∈ N and k ∈ N). (For example, (2, 1 4 ) = (2, 1, 1, 1, 1).) • We let Λ denote the ring of symmetric functions in infinitely many variables x 1 , x 2 , x 3 , . . . We shall use the symmetric functions h n and s λ in Λ as defined in [10, Sections 2.1 and 2.2]. Let us briefly recall how they are defined: • For each n ∈ Z, we define the complete homogeneous symmetric function h n ∈ Λ by Thus, h 0 = 1 and h n = 0 for all n < 0.
• For each partition λ, we define the Schur function s λ ∈ Λ by where the sum ranges over all semistandard tableaux T of shape λ, and where x T denotes the monomial obtained by multiplying the x i for all entries i of T . We refer the reader to [ The family (s λ ) λ∈Par is a basis of the k-module Λ, and is known as the Schur basis. It is easy to see that each n ∈ N satisfies s (n) = h n .
(b) Let Par [n] be the set of all partitions having length n. In other words, (where we are using our convention that trailing zeroes can be omitted from partitions, so that a partition of length n can always be identified with an n-tuple).
(c) A family (u i ) i∈Z of objects (e.g., of numbers) is said to be n-periodic if each j ∈ Z satisfies u j = u j+n . Equivalently, a family (u i ) i∈Z of objects is n-periodic if and only if it has the property that (u j = u j whenever j and j are two integers satisfying j ≡ j mod n) .
Thus, an n-periodic family (u i ) i∈Z is uniquely determined by the n consecutive entries u 1 , u 2 , . . . , u n (because for any integer j, we have u j = u j , where j is the unique element of {1, 2, . . . , n} that is congruent to j modulo n).

Question 2.5.
Can the bijection ϕ in Theorem 2.3 be defined in a more "intuitive" way, similar to (e.g.) jeu-de-taquin or the RSK correspondence? (There is no tableau being transformed here, just a partition.)

A birational involution
The leading role in our proof of Theorem 2.3 will be played by a certain piecewise-linear involution (which is similar to the bijection ϕ in Theorem 2.3, but without the shifting by −a and b). For the sake of convenience, we prefer to study this involution in a more general setting, in which the operations min, + and − are replaced by the structure operations +, · and / of a semifield. This kind of generalization is called detropicalization (or birational lifting, or tropicalization in the older combinatorial literature); see, e.g., [12], [17], [5,Sections 5 and 9] or [19, §4.2] for examples of this procedure (although our use of it will be conceptually simpler).

Semifields
We recall the definition of a semifield (more precisely, the one we will be using, as there are many competing ones): 2 Definition 3.1. A semifield means a set K endowed with • two binary operations called "addition" and "multiplication", and denoted by + and ·, respectively, and both written infix (i.e., we write a + b and a · b instead of + (a, b) and · (a, b)), and • an element called "unity" and denoted by 1 such that (K, +) is an abelian semigroup and (K, ·, 1) is an abelian group, and such that the distributivity laws hold for all a ∈ K, b ∈ K and c ∈ K.
Thus, a semifield is similar to a field, except that it has no additive inverses and no zero element, but, on the other hand, has multiplicative inverses for all its elements (not just the nonzero ones).

Example 3.2.
Let Q + be the set of all positive rational numbers. Then, Q + (endowed with its standard addition and multiplication and the number 1) is a semifield. Example 3.3. Let (A, * , e) be any totally ordered abelian group (whose operation is * and whose neutral element is e). Then, A becomes a semifield if we endow it with the "addition" min (that is, we set a + b := min {a, b} for all a, b ∈ A), the "multiplication" * (that is, we set a · b := a * b for all a, b ∈ A), and the "unity" e. This semifield (A, min, * , e) is called the min tropical semifield of (A, * , e). Convention 3.4. All conventions that are typically used for fields will be used for semifields as well, to the extent they apply. Specifically: • If K is a semifield, and if a, b ∈ K, then a · b shall be abbreviated by ab.
• We shall use the standard "PEMDAS" convention that multiplication-like operations have higher precedence than addition-like operations; thus, e.g., the expression "ab + ac" must be understood as "(ab) + (ac)".
• If K is a semifield, then the inverse of any element b ∈ K in the abelian group (K, ·, 1) will be denoted by b −1 . Note that this inverse is always defined (unlike when K is a field).
• If K is a semifield, and if a, b ∈ K, then the product ab −1 will be denoted by a/b and by a b . Note that this is always defined (unlike when K is a field).
• Finite products i∈I a i of elements of a semifield are defined in the same way as in commutative rings. The same applies to finite sums i∈I a i as long as they are nonempty (i.e., as long as I = ∅). The empty sum is not defined in a semifield, since there is no zero element.

The birational involution
For the rest of Section 3, we agree to the following two conventions: We fix a positive integer n and a semifield K. We also fix an n-tuple u ∈ K n . Convention 3.6. If a ∈ K n is an n-tuple, and if i ∈ Z, then a i shall denote the i#-th entry of a, where i# is the unique element of {1, 2, . . . , n} satisfying i# ≡ i mod n. Thus, each n-tuple a ∈ K n satisfies a = (a 1 , a 2 , . . . , a n ) and a i = a i+n for each i ∈ Z. Therefore, if a ∈ K n is any n-tuple, then the family (a i ) i∈Z is n-periodic.
We shall soon use the letter x for an n-tuple in K n ; thus, x 1 , x 2 , . . . , x n will be the entries of this n-tuple. This has nothing to do with the indeterminates x 1 , x 2 , x 3 , . . . from Section 1 (that unfortunately use the same letters); we actually forget all conventions from Section 1 (apart from N = {0, 1, 2, . . . }) for the entire Section 3.
The following is obvious: If a ∈ K n is any n-tuple, then a k+1 a k+2 · · · a k+n = a 1 a 2 · · · a n for each k ∈ Z.
Definition 3.8. We define a map f u : K n → K n as follows: Let x ∈ K n be an n-tuple. For each j ∈ Z and r ∈ N, define an element t r,j ∈ K by Define y ∈ K n by setting Set f u (x) = y. Example 3.9. Set n = 4 for this example. Let x ∈ K n be an n-tuple; thus, x = (x 1 , x 2 , x 3 , x 4 ).
Let us see what the definition of f u (x) in Definition 3.8 boils down to in this case. Let us first compute the elements t n−1,j = t 3,j from Definition 3.8. The definition of t 3,0 yields Similarly, (since u 5 = u 1 and x 5 = x 1 and u 6 = u 2 and x 6 = x 2 ) .
We don't need to compute any further t 3,j 's, since we can easily see that t 3,j = t 3,j for any integers j and j satisfying j ≡ j mod 4. Thus, in particular, t 3,4 = t 3,0 and t 3,5 = t 3,1 . Now, let us compute the 4-tuple y ∈ K n = K 4 from Definition 3.8. By its definition, we have (by our formulas for t 3,0 and t 3,2 ). Similar computations lead to Of course, knowing one of these four equalities is enough; the expression for y i+1 is obtained from the expression for y i by shifting all indices (other than the "3"s that were originally "n−1"s) forward by 1.
Remark 3.10. Instead of assuming K to be a semifield, we could have assumed that K is an infinite field. In that case, the f u in Definition 3.8 would be a birational map instead of a map in the usual sense of this word, since the denominators x i+1 t n−1,i+1 in the definition of y can be zero. Everything we say below about f u would nevertheless still hold on the level of birational maps.
The map f u we just defined has the following properties: (b) Let x ∈ K n and y ∈ K n be such that y = f u (x). Then, (c) Let x ∈ K n and y ∈ K n be such that y = f u (x). Then, Theorem 3.11 will be crucial for us; but before we can prove it, we will need a few lemmas.
Lemma 3.12. Let x ∈ K n be an n-tuple. Let t r,j and y be as in Definition 3.8. Then: (a) We have t r,j = t r,j for any r ∈ N and any two integers j and j satisfying j ≡ j mod n.
In other words, for each r ∈ N, the family (t r,j ) j∈Z is n-periodic.

(b)
We have t 0,j = 1 for each j ∈ Z.
(c) For each r ∈ N and j ∈ Z, we have (d) For each r ∈ N and j ∈ Z, we have (e) For each a ∈ Z and b ∈ Z, we have (g) For each j ∈ Z and each positive integer q, we have Now, for each j ∈ Z and r ∈ N, let us define an element t r,j ∈ K by (This is precisely how t r,j was defined, except that we are using y in place of x now.) Then: (k) For each i ∈ Z, we have Proof of Lemma 3.12. The proof of this lemma is long but unsophisticated: Each part follows by rather straightforward computations (and, in the cases of parts (g) and (i), an induction on q) from the previously proven parts. 3 We therefore omit it.
Lemma 3.13. Let x ∈ K n be an n-tuple. For each j ∈ Z, let Let z ∈ K n be such that Proof of Lemma 3.13. Let t r,j and y be as in Definition 3.8. Then, t n−1,j = q j for each j ∈ Z (by comparing the definitions of t n−1,j and q j ). Hence, z i = y i for each i ∈ {1, 2, . . . , n} (by comparing the definitions of z i and y i ). Hence, z = y = f u (x) (since f u (x) was defined to be y).
For future convenience, let us restate Lemma 3.13 with different labels: Lemma 3.14. Let y ∈ K n be an n-tuple. For each j ∈ Z, let Let x ∈ K n be such that Proof of Lemma 3.14. Lemma 3.14 is just Lemma 3.13, with x, q j and z renamed as y, r j and x.
We are now ready for the proof of Theorem 3.11: Let t r,j and y be as in Definition 3.8. Then, f u (x) = y (by the definition of f u ). Let t r,j (for each r ∈ N and j ∈ Z) be as in Lemma 3.12. The definition of t n−1,j shows that t n−1,j = n−1 k=0 y j+1 y j+2 · · · y j+k · u j+k+1 u j+k+2 · · · u j+n−1 for each j ∈ Z. Lemma 3.12 (k) shows that Thus, Lemma 3.14 (applied to We have proved this for each x ∈ K n . In other words, f u • f u = id. This proves Theorem 3.11 (a).
(b) Let t r,j be as in Definition 3.8. Note that the y from Definition 3.8 is precisely the y in Theorem 3.11 (b) (because both y's satisfy f u (x) = y).
The definition of y in Definition 3.8 shows that This proves Theorem 3.11 (b). (c) Let t r,j be as in Definition 3.8. Note that the y from Definition 3.8 is precisely the y in Theorem 3.11 (c) (because both y's satisfy f u (x) = y).
Let i ∈ Z. Then, Lemma 3.12 (h) yields (by Lemma 3.12 (f)). Now, The same argument (applied to i + 1 instead of i) yields

Multiplying (3.2) with this equality, we obtain
This proves Theorem 3.11 (c). (d) Let t r,j be as in Definition 3.8. Every i ∈ Z satisfies (3.2) (as we have shown in the proof of Theorem 3.11 (c) above). Hence, taking the product of the equalities (3.2) over all i ∈ {1, 2, . . . , n}, we find Dividing both sides of this by which proves Theorem 3.11 (d).
Let us observe one more property of the involution f u (even though we will never use it): Remark 3.16. There is an alternative proof of Theorem 3.11 (a) that avoids the use of the (rather complicated) parts (g), (i), (j) and (k) of Lemma 3.12. Let us outline this proof: The claim of Theorem 3.11 (a) can be restated as the equality f u (f u (x)) = x for each x ∈ K n and each u ∈ K n (we are not regarding u as fixed here). This equality boils down to a set of identities between rational functions in the variables u 1 , u 2 , . . . , u n , x 1 , x 2 , . . . , x n (since each entry of f u (x) is a rational function in these variables, and each entry of f u (f u (x)) is a rational function in the former entries as well as u 1 , u 2 , . . . , u n ). These rational functions are subtraction-free (i.e., no subtraction signs appear in them), and thus are defined over any semifield. But there is a general principle saying that if we need to prove an identity between two subtraction-free rational functions, it is sufficient to prove that it holds over the semifield Q + from Example 3.2.
(Indeed, this principle follows from the fact that any subtraction-free rational function can be rewritten as a ratio of two polynomials with nonnegative integer coefficients, and thus an identity between two subtraction-free rational functions can be rewritten as an identity between two such polynomials; but the latter kind of identity will necessarily be true if it has been checked on all positive rational numbers.) As a consequence of this discussion, in order to prove Theorem 3.11 (a) in full generality, it suffices to prove Theorem 3.11 (a) in the case when K = Q + . So let us restrict ourselves to this case. Let x ∈ K n . We must show that f u (f u (x)) = x. Let y = f u (x), and let z = f u (y). We will show that z = x. Assume the contrary. Thus, z = x. Hence, there exists some i ∈ {1, 2, . . . , n} such that z i = x i . Consider this i. Hence, either z i > x i or z i < x i . We WLOG assume that z i > x i (since the proof in the case of z i < x i is identical, except that all inequality signs are reversed). But Theorem 3.11 (c) yields Likewise, Theorem 3.11 (c) (applied to y and z instead of x and y) yields (since z = f u (y)). Combining these two equalities, we find Cancelling the positive number u i +x i from this inequality, we obtain The same reasoning (but applied to i . Proceeding in the same way, we successively obtain z i+3 > x i+3 and z i+4 > x i+4 and z i+5 > x i+5 and so on. Hence, Also, Theorem 3.11 (b) (applied to y and z instead of x and y) yields z 1 z 2 · · · z n · y 1 y 2 · · · y n = (u 1 u 2 · · · u n ) 2 (since z = f u (y)). Comparing these two equalities, we find y 1 y 2 · · · y n ·x 1 x 2 · · · x n = z 1 z 2 · · · z n · y 1 y 2 · · · y n , so that In light of these two equalities, we can rewrite (3.3) as z 1 z 2 · · · z n > x 1 x 2 · · · x n . This, however, contradicts (3.4). This contradiction shows that our assumption was false, thus concluding our proof of z = x. Now, recall that f u (x) = y. Hence, f u (f u (x)) = f u (y) = z = x, as we wanted to prove. Hence, Theorem 3.11 (a) is proved again.
We shall return to the birational involution f u in Subsection 5.1, where we will pose several questions about its meaning and uniqueness properties.

Proof of the main theorem
We shall now slowly approach the proof of Theorem 2.3 through a litany of auxiliary results.

From the life of snakes
Recall the conventions introduced in Section 1 and in Convention 2.1. Let us next introduce some further notations.
n of Laurent polynomials in the n indeterminates x 1 , x 2 , . . . , x n over k. Clearly, the polynomial ring k [x 1 , x 2 , . . . , x n ] is a subring of L.
For any symmetric function f ∈ Λ, the evaluation is a polynomial in k [x 1 , x 2 , . . . , x n ] and thus a Laurent polynomial in L. Moreover, for any symmetric function f ∈ Λ, the evaluation is a Laurent polynomial in L as well.

(b)
A snake λ is said to be nonnegative if it belongs to N n (that is, if all its entries are nonnegative). Thus, a nonnegative snake is the same as a partition having length n. In other words, a nonnegative snake is the same as a partition λ ∈ Par [n].
(e) We regard Z n as a Z-module in the obvious way. Thus, if λ ∈ Z n and µ ∈ Z n are two n-tuples of integers, then   Note that what we call a "snake" here is called a "staircase of height n" in Stembridge's work [22], where he uses these snakes to index finite-dimensional polynomial representations of the group GL n (C). We avoid calling them "staircases", as that word has since been used for other things (in particular, ρ is often called "the n-staircase" in the jargon of combinatorialists).
The notations introduced in Definition 4.3 have the following properties: (a) If λ is a snake, and d is an integer, then λ + d and λ − d are snakes as well.
(b) If λ is a snake, then λ ∨ is a snake as well.
Proof of Proposition 4.5. Completely straightforward.
Let us now assign a Laurent polynomial a λ to each λ ∈ Z n : Definition 4.6. Let λ ∈ Z n be any n-tuple. Then, we define the Laurent polynomial where S n is the symmetric group of the set {1, 2, . . . , n} (and where sign w denotes the sign of a permutation w). This Laurent polynomial a λ is called the alternant corresponding to the n-tuple λ.
(The "a" in the notation "a λ " has nothing to do with the a in Theorem 2.3.) Example 4.7. We have The sum in Definition 4.6 is the same kind of sum that appears in the definition of a determinant. Therefore, we can rewrite the alternant as follows: Proposition 4.8. Let λ ∈ Z n be an n-tuple. Then, the alternant a λ ∈ L satisfies Thus, in particular, the alternant a ρ corresponding to the snake ρ = (n − 1, n − 2, . . . , 2, 1, 0) = (n − 1, n − 2, . . . , n − n) (by the classical formula for the Vandermonde determinant). We recall a standard concept from commutative algebra: An element a of a commutative ring A is said to be regular if it has the property that every x ∈ A satisfying ax = 0 must satisfy x = 0. (Thus, regular elements are the same as elements that are not zero-divisors, if one does not require zero-divisors to be nonzero.) is also a regular element of L (by the previous paragraph). Lemma 4.9 shows that fractions of the form u a ρ (where u ∈ L) are well-defined if u is a multiple of a ρ . (That is, there is never more than one b ∈ L that satisfies a ρ b = u.) We notice that the element x Π = x 1 x 2 · · · x n of L is invertible.  Proof of Lemma 4.11. Our proof will consist of two steps: Step 1: We will prove Lemma 4.11 in the particular case when λ is nonnegative.
Step 2: We will use Lemma 4.10 to derive the general case of Lemma 4.11 from this particular case.
We will use this strategy again further on; we shall refer to it as the right-shift strategy.
Here are the details of the two steps: Step 1: Let us prove that Lemma 4.11 holds in the particular case when λ is nonnegative. Indeed, let us assume that λ is nonnegative. We must show that a λ+ρ is a multiple of a ρ in L.
Step 2: Let us now prove Lemma 4.11 in the general case. The snake λ may or may not be nonnegative. However, there exists some integer d such that the snake λ + d is nonnegative (for example, we can take d = −λ n ). Consider this d.
The snake λ + d is nonnegative; thus, we can apply Lemma 4.11 to λ + d instead of λ (because in Step 1, we have proved that Lemma 4.11 holds in the particular case when λ is nonnegative). Thus we conclude that a (λ+d)+ρ is a multiple of a ρ in L. In other words, there exists some u ∈ L such that a (λ+d)+ρ = a ρ u. Consider this u. But Proposition 4.5 (c) yields (λ + ρ) + d = (λ + d) + ρ, and thus a (λ+ρ)+d = a (λ+d)+ρ = a ρ u.
Lemma 4.10 (applied to λ + ρ instead of λ) yields a (λ+ρ)+d = x d Π a λ+ρ . Since the element x Π of L is invertible, we thus obtain Hence, a λ+ρ is a multiple of a ρ . This completes the proof of Lemma 4.11.
Definition 4.12. Let λ be a snake. We define an element s λ ∈ L by s λ = a λ+ρ a ρ . (This is welldefined, because Lemma 4.11 shows that a λ+ρ is a multiple of a ρ in L, and because Lemma 4.9 shows that the fraction a λ+ρ a ρ is uniquely defined.) It makes sense to refer to the elements s λ just defined as "Schur Laurent polynomials". In fact, as the following lemma shows, they are identical with the Schur polynomials s λ (x 1 , x 2 , . . . , x n ) when the snake λ is nonnegative: Proof of Lemma 4.13. We know that λ is a partition of length n (since λ ∈ Par [n]). Hence, λ is a nonnegative snake. Furthermore, since λ is a partition of length n, we can apply [10, Corollary 2.6.7] and obtain s λ (x 1 , x 2 , . . . , x n ) = a λ+ρ a ρ = s λ (since s λ was defined to be a λ+ρ a ρ ).
The Schur Laurent polynomials s λ appear in Stembridge's [22], where they are named s λ . (The equivalence of our definition with his follows from [22,Theorem 7.1].) The following lemma is an analogue of Lemma 4.10 for Schur Laurent polynomials: Lemma 4.14. Let λ ∈ Z n be any snake, and let d ∈ Z. Then, s λ+d = x d Π s λ . Proof of Lemma 4.14. This follows easily by applying Lemma 4.10 to λ + ρ instead of λ. We have µ ∈ Par [n]. Hence, Lemma 4.13 (applied to λ = µ) yields the equality s µ = s µ (x 1 , x 2 , . . . , x n ). Likewise, s ν = s ν (x 1 , x 2 , . . . , x n ). Multiplying these two equalities, we obtain (where the last equality sign follows by substituting 0, 0, 0, . . . for x n+1 , x n+2 , x n+3 , . . . in (1.1)). But the sum on the right hand side of (4.3) can be split into two sums: one collecting all addends with λ ∈ Par [n], and one collecting all remaining addends. The second of these sums is 0, because if λ ∈ Par satisfies λ / ∈ Par [n], then λ has length > n and therefore satisfies s λ (x 1 , x 2 , . . . , x n ) = 0 (by (4.2)), so the corresponding addend vanishes. Thus, only the first sum survives. Hence,  Proof of Lemma 4.16. Let us define a strict snake to be an n-tuple α ∈ Z n of integers satisfying α 1 > α 2 > · · · > α n . It is easy to see that the map is a bijection. It is also easy to see that any two strict snakes α and β satisfy the coefficient of x β 1 1 x β 2 2 · · · x βn n in a α = δ α,β , where δ α,β is the Kronecker delta of α and β (that is, the integer 1 if α = β, or the integer 0 otherwise). From this, it is easily seen that the family (a α ) α∈{strict snakes} is k-linearly independent. But this family (a α ) α∈{strict snakes} is just a re-indexing of the family (a λ+ρ ) λ∈{snakes} (since the map (4.4) is a bijection). Hence, the latter family (a λ+ρ ) λ∈{snakes} must be k-linearly independent, too. Therefore, the family a λ+ρ a ρ λ∈{snakes} is also k-linearly independent (since any k-linear dependence relation between the a λ+ρ a ρ would yield a corresponding k-linear dependence relation between the a λ+ρ ). But this latter family is precisely the family (s λ ) λ∈{snakes} (by the definition of s λ ). Hence, the family (s λ ) λ∈{snakes} is k-linearly independent.
Lemma 4.16 is actually part of a stronger claim: The family (s λ ) λ∈{snakes} is a basis of the kmodule of symmetric Laurent polynomials in x 1 , x 2 , . . . , x n . We shall not need this, however, so we omit the proof (which follows easily from Lemma 4.14 and the analogous result for symmetric polynomials, which is well-known).

Recall Definition 4.3 (d).
Our next lemma connects the Laurent polynomials s λ and s λ ∨ for every snake λ; it is folklore (see [10, Exercise 2.9.15(d)] for an equivalent version), but we have not seen it stated in this exact form in the literature.

h +
k , h − k and the Pieri rule Definition 4.18. Let k ∈ Z. Then, we define two Laurent polynomials h + k ∈ L and h − k ∈ L by Note that if k ∈ Z is negative, then h + k = 0 (since h k = 0) and h − k = 0 (similarly). We begin by describing h + k as a Schur Laurent polynomial: Lemma 4.19. Let k ∈ N. Then, the partition (k) is a nonnegative snake (when regarded as the n-tuple (k, 0, 0, . . . , 0)), and satisfies Proof of Lemma 4.19. It is well-known that s (k) = h k . Substituting 0, 0, 0, . . . for the variables x n+1 , x n+2 , x n+3 , . . . on both sides of this equality, we obtain s (k) = h + k . This proves Lemma 4.19.
Next, we need to know what happens when a Schur Laurent polynomial s λ is multiplied by some h + k . We will answer this question using the classical first Pieri rule.

(b)
If λ ∈ Z n and d ∈ Z, then |λ + d| = |λ| + nd. We can now state the Pieri rule in the form we need: Proposition 4.24. Let λ be a snake. Let k ∈ Z. Then, Proof of Proposition 4.24. We follow the same right-shift strategy as we did in our proof of Lemma 4.11. Thus, our proof shall consist of two steps: Step 1: We will prove Proposition 4.24 in the particular case when λ is nonnegative.
Step 2: We will use Lemma 4.14 to derive the general case of Proposition 4.24 from this particular case.
Here are some details on the two steps (again, more can be found in [9]): Step 1: Let us prove that Proposition 4.24 holds in the particular case when λ is nonnegative. Indeed, let us assume that λ is nonnegative. We must prove the equality (4.7). If k < 0, then both sides of this equality are 0 (indeed, the sum on the right hand side is empty, since µ λ implies |µ| − |λ| 0). Thus, the equality (4.7) holds if k < 0. Therefore, for the rest of Step 1, we WLOG assume that k 0.
Note that λ is a partition of length n (since λ is a nonnegative snake). In other words, λ ∈ Par [n].
The following is easy to see:
[Proof of Claim 1: Unravel the definitions and recall that partitions in Par [n] are the same as nonnegative snakes. We leave the details to the reader.] From the first Pieri rule ([10, (2.7.1)] 5 , applied to k instead of n), we obtain s λ h k = λ + ∈Par; λ + /λ is a horizontal k-strip s λ + = µ∈Par; µ/λ is a horizontal k-strip s µ (here, we have renamed the summation index λ + as µ).
Evaluating both sides of this equality at x 1 , x 2 , . . . , x n , we find (where the last equality sign follows from (4.2) by a similar argument as in the proof of Lemma 4.15 above). Comparing this with where we used Claim 1 to rewrite the summation sign). This proves (4.7). Thus, Proposition 4.24 is proved under the assumption that λ is nonnegative. This completes Step 1.
Step 2: We now need to prove Proposition 4.24 in the general case. The idea is to find an integer d such that the snake λ + d is nonnegative (for example, d = −λ n ), and apply Proposition 4.24 to λ+d instead of λ (which we can do, since Step 1 has already covered this case). This yields h + k · s λ+d = µ is a snake; µ λ+d; |µ|−|λ+d|=k s µ = µ+d is a snake; µ+d λ+d; |µ+d|−|λ+d|=k s µ+d (here, we substituted µ + d for µ in the sum). The conditions under the summation sign on the right hand side can be simplified using Proposition 4.22 (b) and Proposition 4.23 (c), and the addends s µ+d can be rewritten as x d Π s µ using Lemma 4.14. Thus, the equality simplifies to Since Lemma 4.14 yields s λ+d = x d Π s λ , we can rewrite this as We can cancel x d Π from this equality (since x Π ∈ L is invertible), and thus obtain This proves Proposition 4.24.
Comparing this with µ is a snake; We have proved this equality for any snake λ. Thus, we can apply it to λ ∨ instead of λ. We obtain But because of (λ ∨ ) ∨ = λ, this equality is precisely (4.8). Thus, Proposition 4.25 is proved.

Computing s α
Convention 4.26. From now on, for the rest of Section 4, we assume that n 2.
Our next goal is to obtain a simple formula for the Schur polynomial s α , where α is as in Theorem 2.3. The first step is the following definition: Proof of Proposition 4.28. This is trivial when min {a, b} < 0, and otherwise follows easily from Proposition 4.25 (applied to λ = (b) and k = a). Details can be found in [9].
(Recall that every negative integer k satisfies h − k = 0 and h + k = 0.) Proof of Proposition 4.29. Apply Proposition 4.28 twice (once to a and b, and once to a − 1 and b − 1), and subtract. See [9] for the details.
Remark 4.30. The right hand side in Proposition 4.28 looks suspiciously like a determinant. This is no coincidence, and Proposition 4.28 can in fact be generalized to a determinantal formula for s λ where λ is any snake of the form (b 1 , b 2 , . . . , b q , 0 n−p−q , −a p , −a p−1 , . . . , −a 1 ). The latter formula can be obtained from an identity of Koike [13, Proposition 2.8] (see also [11, (6) and (10)]). See [9, §5.1] for some more details.
Corollary 4.31. Let a, b ∈ N. Define the partition α = (a + b, a n−2 ). Then, α is a nonnegative snake and satisfies Proof of Corollary 4.31. It is easy to see that α = (b a) + a (regarded as snakes). Hence, Lemma 4.14 (applied to λ = b a and d = a) yields This proves Corollary 4.31.

4.4.
The sets R µ,a,b (γ) and a formula for h − a h + b s µ We shall next aim for a formula for h − a h + b s µ (for a snake µ and integers a, b ∈ Z), which will be obtained in a straightforward way by applying Propositions 4.24 and 4.25. We will need the following definition: Lemma 4.33. Let µ, γ ∈ Z n and a, b ∈ Z. Assume that γ is not a snake. Then, |R µ,a,b (γ)| = 0.
Proof of Lemma 4.33. Let ν ∈ R µ,a,b (γ). We shall obtain a contradiction. Indeed, ν ∈ R µ,a,b (γ) means that ν is a snake satisfying the four conditions µ ν and |µ| − |ν| = a and γ ν and (by the definition of R µ,a,b (γ)). Thus, in particular, we have γ ν. Hence, Proposition 4.23 (a) (applied to γ and ν instead of µ and λ) yields that both ν and γ are snakes. Hence, γ is a snake. This contradicts the fact that γ is not a snake.
Proof of Corollary 4.35. Every snake γ satisfies s γ+a = x a Π s γ (4.15) (by Lemma 4.14, applied to γ and a instead of λ and d).
But α is a nonnegative snake; thus, α ∈ Par [n]. Hence, Lemma 4.15 (applied to α and µ instead of µ and ν) yields c λ α,µ s λ = λ is a snake; λ is nonnegative c λ α,µ s λ (since the partitions λ ∈ Par [n] are precisely the nonnegative snakes) = λ is a snake c λ α,µ s λ (where the last equality sign is owed to the fact that we understand c λ α,µ to mean 0 if λ is not a nonnegative snake). Hence, We can compare coefficients on both sides of this equality (since Lemma 4.16 shows that the family (s λ ) λ∈{snakes} of elements of L is k-linearly independent), and thus conclude that This proves (4.14) in the case when λ is a snake. However, it is easy to see that (4.14) also holds in the case when λ is not a snake 6 . Thus, (4.14) always holds. This proves Corollary 4.35.

Convention 4.36.
For the whole Subsection 4.5, we shall use Convention 3.6 (not only for ntuples a ∈ K n , but for any n-tuples a). This convention does not conflict with Convention 4.2, because both conventions define γ i in the same way when γ is an n-tuple and i ∈ {1, 2, . . . , n}.
• For any a, b ∈ Z, the quotient a b understood with respect to the semifield (Z, min, +, 0) is precisely the difference a − b understood with respect to the integer ring Z.
• For any a ∈ Z, the square a 2 understood with respect to the semifield (Z, min, +, 0) is the product 2a understood with respect to the integer ring Z.
We know that R µ,a,b (γ) is the set of all snakes ν satisfying the four conditions (by the definition of R µ,a,b (γ)).
The definition of ϕ then yields

The finale
Now, let us again use the convention (from Section 1) by which we identify partitions with finite tuples (and therefore identify partitions in Par [n] with nonnegative snakes). This is no longer problematic, since we are not using Convention 3.6 any more.
We are now ready to prove Theorem 2.3: Proof of Theorem 2. 3. The map f µ is an involution (by Theorem 4.38 (a)), thus a bijection. Let a − : Z n → Z n be the bijection that sends each ω ∈ Z n to ω − a. Let b + : Z n → Z n be the bijection that sends each ω ∈ Z n to ω + b. Now, Lemma 4.42 can be restated as follows: Hence, ϕ is a bijection (since b + , f µ and a − are bijections). This proves Theorem 2.

Questions on f u
We shall now pose several questions about the birational involution f u studied in Section 3. Convention 3.4, Convention 3.5 and Convention 3.6 will be used throughout Subsection 5.1.
Most of our questions are attempts at seeing the involution f u from different directions. The first one is inspired by what is now known as the "toggle approach" to dynamical combinatorics (see, e.g., [19]), but is really an application of the age-old "divide and conquer" paradigm to complicated maps: Question 5.1. Is there an equivalent definition of f u as a composition of toggles? (A toggle here means a birational map K n → K n that changes only one entry of the n-tuple. An example for a birational map that can be defined as a composition of toggles is birational rowmotion -see, e.g., [5]. Cluster mutations, as in the theory of cluster algebras, are another example of toggles.) Another set of questions concern the uniqueness of f u . While we defined the map f u explicitly, all we have then used are the properties listed in Theorem 3.11. Thus, it is a natural question to ask whether these properties characterize f u uniquely. A pointwise version of this question can be asked as well: Given x ∈ K n and y ∈ K n satisfying some of the equalities in parts (b), (c) and (d) of Theorem 3.11, does it follow that y = f u (x) ? (Keep in mind that u is fixed.) Of course, the answers depend on which equalities we require. Let us first ask what happens if we require the equalities from Theorem 3.11 (c) only: Question 5.2. Given x, y ∈ K n satisfying for all i ∈ Z. Does it follow that y = f u (x) or y = x ?
Note that the "or y = x" part is needed here, since y = x is obviously a solution to the equations (5.1).
The following example shows that the answer to Question 5.2 is "no" if K is the min tropical semifield (Z, min, +, 0) of the totally ordered abelian group Z. Example 5.3. Let k, g ∈ N with g k. Let K = (Z, min, +, 0) and n = 3 and u = (0, 0, g) and x = (1, 2, 0). Set y = (k + 1, 2, k) (where the "+" sign in "k + 1" stands for addition of integers, not addition in K). Then, the equations (5.1) hold in K for all i ∈ Z. (Restated in terms of standard operations on integers, this is saying that for all i ∈ Z.) This is straightforward to verify, and shows that for a given x there can be an arbitrarily high (finite) number of y ∈ K n satisfying the equations (5.1) for all i ∈ Z. (Incidentally, this number is always finite when K = (Z, min, +, 0); however, this does not generalize to arbitrary K.) However, the answer to Question 5.2 is "yes" if K = Q + and, more generally, if the semifield K embeds into an integral domain: Proposition 5.4. Assume that there is an integral domain L such that the semifield K is a subsemifield of L (in the sense that K ⊆ L and that the operations + and · of K are restrictions of those of L, whereas the unity of K is the unity of L). Let x ∈ K n . Then, the only n-tuples y ∈ K n satisfying the equations (5.1) for all i ∈ Z are y = f u (x) and y = x.
Another avatar of the uniqueness question is the following: Question 5.5. Given x, y ∈ K n satisfying both (5.1) for all i ∈ Z and y 1 y 2 · · · y n · x 1 x 2 · · · x n = (u 1 u 2 · · · u n ) 2 . (5.2) Does it follow that y = f u (x) ?
Another question concerns Lemma 3.12: Question 5.7. What is the "real meaning" of some of the more complicated parts of Lemma 3.12? In particular, Lemma 3.12 (g) reminds of the Plücker relation for minors of a 2 × m-matrix; can it be viewed that way (at least when K is a subsemifield of a field)?

The birational R-matrix connection
In this section, we shall connect the map f u from our Definition 3.8 with the birational R-matrix η defined in [14, §6] and studied further (e.g.) in [3]. We fix a positive integer n and a semifield K. We shall use Convention 3.4 and Convention 3.6. Let us recall the definition of the birational R-matrix η (no relation to the η in Theorem 2.3): Definition 5.8. We define a map η : K n × K n → K n × K n as follows: Let a ∈ K n and b ∈ K n be two n-tuples. For any i ∈ Z, define an element κ i (a, b) ∈ K by bp · a j+1 a j+2 · · · a i+n−1 = i+n−1 p=j+1 ap .