Super RSK correspondence with symmetry

Super RSK correspondence is a bijective correspondence between superbiwords and pairs of semistandard supertableaux. Such a bijection was given by Bonetti, Senato and Venezia, via an insertion algorithm closely related to Schensted insertion. Notably, the symmetry property satisfied by the classical RSK bijection holds only in special cases under this bijection. We present a new super RSK bijection, based on the mixed insertion process defined by Haiman, where the symmetry property holds in complete generality.


Introduction
The work of Robinson [R] in 1938, and Schensted [S] in 1961, describes a bijection between permutations and pairs of same-shape standard tableaux, now known as the Robinson-Schensted (RS) correspondence. A key ingredient in the bijection is an algorithm called Schensted insertion. In 1970, Knuth showed that Schensted insertion could be adapted to a more general setting to achieve a bijection between two-line arrays of letters called 'biwords' (which are in natural bijection with matrices of non-negative integers) and pairs of same-shape semistandard tableaux [K]. This bijection is known as the Robinson-Schensted-Knuth (RSK) correspondence.
A celebrated feature of the RSK correspondence is a certain symmetry property; namely, exchanging the rows of a biword (or, transposing the matrix from the matrix perspective) translates via the RSK correspondence to exchanging the positions of the associated pair of semistandard tableaux. This property, proven for the RS correspondence by Viennot [V] in 1977 and extended to the full RSK correspondence by Fulton [F] in 1997, is far from obvious from the workings of the RSK algorithm itself.
The RSK correspondence has applications in a variety of settings; of particular relevance for this paper is its application in representation theory and invariant theory, where it describes a bijection between various important bases for associative algebras and Lie algebras. We consider here the generalization of RSK correspondence to combinatorial objects associated with the representation theory and invariant theory of superalgebras. These 'super' combinatorial objects are restricted superbiwords and semistandard supertableaux. In contrast with the classical situation, letters in restricted superbiwords can have even or odd parity, with repetition of mixed-parity biletters disallowed. Semistandard supertableaux are nondecreasing tableaux in which letters of even parity strictly increase down columns, and letters of odd parity strictly increase along rows. See for example [CPT,DR,GRS,LNS,MZ] for a few instances of these combinatorial objects arising in the study of bases of superalgebras and their representations.
We prove that an adaptation of the mixed insertion algorithm defined by Haiman [H] can be used to define a 'super-RSK' correspondence between restricted superbiwords and same-shape pairs of semistandard supertableaux. This correspondence fully generalizes the classical RSK correspondence, in the sense that classical RSK can be viewed as a specialization of super-RSK to the case of even-parity superbiwords, and the classical symmetry property described above holds for super-RSK in full generality.
Numerous variants of super-RSK correspondence exist in the literature. Most notably, Bonetti, Senato, and Venezia [BSV] presented a different correspondence between the same sets of combinatorial objects considered in this paper. At the heart of their algorithm are dual insertion processes which are very much like the classical Schensted insertion process, in that insertion progresses linearly from one row to the next (or one column to the next), and the number of 'bumps' in a given insertion is bounded by the number of rows (or the number of columns) in the Young diagram. By contrast, the Haiman insertion process utilized in this paper progresses in a less direct fashion, where the number of 'bumps' in an insertion is bounded only by the number of nodes in the Young diagram. A more crucial difference between the two algorithms is the fact that the super-RSK of [BSV] does not have the symmetry property in general (see Example 6.7). La Scala, Nardozza and Senato describe [LNS,Proposition 4.7] a subset of superbiwords where symmetry is known to hold for the [BSV] correspondence, but a complete description of such biwords is still an open problem.
Another variant of super-RSK correspondence appears in work by Shimozono and White [SW]. Their algorithm is based around the same Haiman insertion algorithm as used here, but adapted to work with a different class of combinatorial objects: unrestricted superbiwords (repetitions of mixed biletters allowed), and supertableaux which are row-weak and column-strict with respect to both parities. They demonstrate a bijection between these objects, and prove that their correspondence generalizes the classical symmetry property as well. While the [SW] algorithm generally yields different supertableaux from ours (see again Example 6.7), we note that they agree, crucially, in the special case of 'standard' superbiwords-those with no repeated letters. This is a key ingredient in the proof of the symmetry of our super-RSK correspondence. Now for a description of the structure of this paper. In §2, we set up basic notation and definitions of the relevant combinatorial objects. In §3, we describe the ε-insertion process which drives the super-RSK algorithm, and prove some useful lemmas about the process. In §4, we prove some bounds on the distribution of bumped nodes during the insertion process, which are necessary for the results in the subsequent section, and perhaps of independent interest in the study of tableau growth. In §5, we define the super-RSK map 'sRSK' and prove the first main theorem of the paper, which appears as Theorem 5.2 in the text: Theorem 1 (Super-RSK correspondence). The map sRSK defines a bijection between restricted superbiwords and same-shape pairs of semistandard supertableaux.
In §6, we prove some lemmas related to standardizing superbiwords, and prove the other main theorem of the paper, which appears as Corollary 6.6 in the text: Theorem 2 (Super-RSK symmetry). Under super-RSK correspondence, exchanging rows in the superbiword w is equivalent to exchanging the positions of the pair of supertableaux sRSK(w).
1.1. Acknowledgements. The author would like to thank Alexander Kleshchev, who originally suggested the topic of the paper as an approach to a representation theoretic problem, and provided helpful suggestions. The author would also like to thank Scott Cook for numerous fruitful discussions.

Preliminaries
Since all combinatorial objects considered in this paper are Z 2 -colored, we will henceforth suppress the prefix 'super' from most of our terminology.

2.1.
Alphabets. An alphabet X is a set equipped with a parity function X → Z 2 , x → x, and a total order < X . Elements of alphabets are called letters. We call x ∈ X even if x = 0 and odd if x = 1. Let ≺ X be the total order on X defined by Note then that The dual alphabet X * of an alphabet X has underlying set {x * | x ∈ X }, parity function defined by x * = x + 1, and total order < X * defined so that The standardizing alphabet X • of an alphabet X has underlying set {x (i) | x ∈ X , i ∈ Z >0 }, parity function defined by x (i) = x, and total order < X • defined so that Define the 'forget superscripts' function• : Going forward, we will suppress the subscripts and write < or ≺ when the underlying alphabet is clear from context. 2.2. Tableaux. We set N := Z >0 × Z >0 and refer to the elements of N as nodes. Define a partial order ≤ on N as follows: (r, s) ≤ (r ′ , s ′ ) if and only if r ≤ r ′ and s ≤ s ′ . For u = (r, s) ∈ N we will write u ′ := (s, r) ∈ N.
I.e., N(ε, i) is the ith row of nodes if ε = 0, and the ith column of nodes if ε = 1.

Insertion and Extraction
It will be convenient in practice to formally extend the domain and range of an (X , λ)-tableau T to a function T : N → X ∪{∞} by setting T(u) = ∞ for all u / ∈ [λ]. We extend the order < on X to X ∪ {∞} by setting x < ∞ for all x ∈ X . We define the symbols 0 < :=< and 1 < :=≤. 3.1. Insertion. Let λ ⊢ n, and assume T is a semistandard (X , λ)-tableau. Let ε ∈ Z 2 and x ∈ X . From this data we construct an (X , µ)-tableau (T ε ← − x), where µ ⊢ n + 1, via the method of ε-insertion.
(2) Set b j to be the smallest node in N(ε + x j , i) such that x j ε < T(b j ).
Assuming the process terminates when j = m, we call b 1 , . . . , b m the bumped node sequence, and call A(T, ε, x) := b m the added node. We call x 1 , . . . , x m the bumped letter sequence.
Remark 3.1. Informally speaking, under0-insertion, bumped even letters are inserted in the next row down and bumped odd letters are inserted in the next column to the right. In1-insertion, this is reversed. The fuss over the differing comparisons 0 < and 1 < is needed to assure that semistandardness is maintained under ε-insertion, as will be shown in Lemma 3.9.
Remark 3.2. If X = N, where < is the usual order on integers and every element is of even parity, then 0-insertion is Schensted insertion [S]. For general X and standard tableaux, 0-insertion is mixed insertion as defined by Haiman (where odd letters are referred to as circled) [H].
Assuming the process terminates when j = m, we call y 1 , . . . , y m the unbumped letter sequence., and define R(U, ε, u) := y m to be the extracted letter. We call c 1 , . . . , c m the unbumped node sequence.
Example 3.5. With X as in Example 2.3 and T as shown below, we have The unbumped node sequence for (T 0 − → (3, 1)) is only the node (3, 1), and the unbumped letter sequence is3.
3.3. Some results on insertion and extraction. As noted in Remark 3.2, εinsertion is an adaptation of 'mixed insertion' defined by Haiman [H]. Although Haiman works with standard tableaux, he remarks that his results may be extended to the semistandard case in a straightforward manner-this is outlined in [H, §1] and we take some pains to make the idea explicit in Lemma 3.11. Though some of the results in this subsection would follow from those in [H] and Lemma 3.11, we nevertheless include full proofs working in the general semistandard case, for the sake of clarity and self-containment, and since our notation and approach differ substantially from that of [H].
The following two lemmas follow directly from definitions of the algorithms.
Lemma 3.6. Let T be a standard (X , λ)-tableau, and assume x ∈ X \T. Then , and both insertions have the same bumped node sequence.
Lemma 3.8. Let b 1 , . . . , b m be the bumped node sequence, and let x 1 , . . . , x m be the bumped letter sequence for the ε-insertion (T ε ← − x). Assume i < j. Then: In the latter case, semistandardness implies that T(b i ) = 0, hence by the insertion algorithm, Proof. (i) Let b 1 , . . . , b m be the bumped node sequence, and let x 1 , . . . , x m be the bumped letter sequence for the insertion (T Let l, r be the nodes to the immediate left and right of b k , respectively. Then Let u, d be the nodes directly above and below b k , respectively. Then Thus T k is semistandard if ε + x k = 0. Assume on the other hand that ε + x k = 0. Let U = T ′ * . Then, applying the above argument to the insertion (U ε ← − x * ), we have that U k is semistandard. But by Lemma 3.6, U k = T ′ * k . Thus by Lemma 2.2, T k is semistandard. This completes the proof of (i).
(ii) Let c 1 , . . . , c m be the unbumped node sequence, and y 1 , . . . , y m be the unbumped letter sequence for the extraction (T otherwise. We have that T 1 is semistandard by assumption. We show by induction that T k is semistandard for all k.
Assume ε + y k−1 = 0. Let l, r be the nodes to the immediate left and right of c k . Then , yet both are odd, a contradiction since T k−1 is semistandard by assumption. Thus T k is row-strict with respect to odd letters.
Let u, d be the nodes directly above and below c k , respectively. Then c k−1 ր c k and c k is in the row above c k−1 . So c k−1 → d or c k−1 = d. In either case we have , and thus T k is column-strict with respect to even letters.
Therefore T k is semistandard if ε+y k−1 = 0. On the other hand assume ε+y k−1 = 1. Let U = T ′ * . Then, applying the above argument to the extraction (U ε − → u ′ ), we have that U k is semistandard. But U k = T ′ * k by Lemma 3.6. Thus by Lemma 2.2, T k is semistandard.
The following lemma is a key tool in generalizing some results proved for standard tableaux to the more general case of semistandard tableaux.
). We will also write x 1 , . . . , x k be the bumped letter sequence for the insertion (T ε ← −•(y)), and x • 1 , . . . , x • m for the bumped letter sequence for the insertion (T • ε ← − y). We will first prove by induction that But this cannot happen in0-insertion for distinct i and j. This leaves the cases where exactly one of u, v is a bumped node. We consider the two cases separately: (a) Assume that v = b i is a bumped node, and u is not.
Note that if y = 1, then•(y) = 1, and b 1 is in the first column. Then by Lemma 3.8,•(y) = T• (y) (b 1 ) < T• (y) (w) for every node w directly below b 1 . Then, since T• (y) is semistandard, T• (y) (b 1 ) < T• (y) (w) for every node w such that b 1 ց w. Thus b 1 ր w for every node w such that•(y) = T• (y) (b 1 ) = T• (y) (w). Thus if i = 1, then y = 0 and by the assumption on y, This completes the proof of the lemma when ε = 0. Now assume ε = 1. Then x > y for all x ∈ T • . This proof proceeds along the same lines as the first part, but because there is an inherent discrepancy in the comparisons 0 < =< and 1 < =≤ we will provide the details in full. We'll write T • y for (T • 1 ← − y) and T• (y) for (T 1 ← −•(y)). We will prove by induction that If y = 0 (resp. if y = 1), let u be the node directly above (resp. directly to the left of) b • 1 . Then T • (u) < y, so T(u) ≤•(y). Moreover, by the assumption on y, we have that•(y) = T(u), so•(y) > T(u).
, let u be the node directly to the left of (resp. directly above) b • i+1 , and note that u On the other hand if both are bumped nodes, say u = , as required. This leaves the cases where exactly one of u, v is a bumped node. We consider the two cases separately: (a) Assume that v = b i is a bumped node, and u is not.
Note that if y = 0, then b 1 ր w for every node w such that•(y) = T• (y) (w). Thus if i = 1, then y = 1 and T . Assume x i = 1. Since T• (y) (u) = T• (y) (v) = x i and T• (y) is semistandard, it cannot be that v is directly to the right of u. If it is not the case that b i−1 ր v, then it must be that v ց b i−1 . Moreover since T(v) = T(b i−1 ) and T is semistandard, it cannot be that b i−1 is directly to the right of v, so , as required. This completes the proof of the lemma in the case ε = 1.

Behavior of bumped nodes
4.1. Bumped node distribution. In this section we prove some technical results on the distribution of bumped nodes in ε-insertion, which will be of repeated use in §4.2.
Lemma 4.2. Let T be a semistandard (X , λ)-tableau, ε ∈ Z 2 , and x ∈ X . Let b 1 , . . . , b m be the bumped node sequence for the insertion (T ε ← − x). Let i, j, k be such that Then there exists some l > k such that Proof. We will call a triple (i, k, j) which satisfies (i)-(iii) a stair triple. For compactness we'll write r a for (b a ) 0 and c a for (b a ) 1 . Define: Take n i,j = 4, the least possible value for n i,j . Then m i,j,k = 2. Then b k = (r i − 1, c j − 1), so b k ∈ [λ] and thus cannot be the last bumped node. By Lemma 3.8, either Taking l = k + 1, this completes the base case.
We argue by induction. Assume that i, j, k satisfy (i)-(iii), and further assume that the claim holds for all i ′ , j ′ , k ′ such that n i ′ ,j ′ < n i,j , or n i ′ ,j ′ = n i,j and m i ′ ,j ′ ,k ′ < m i,j,k .
Assume ε + T(b k ) = 0 (the argument in the other case is exactly dual to what follows). Then r k+1 = r k + 1. If r k+1 = r j , then, taking l = k + 1, we are in case (i). Assume r k+1 < r i . If c k+1 = c k , we may apply the induction assumption to the stair triple (i, k + 1, j). Thus assume c k+1 < c k . Since k + 1 > j, it must be that c k+1 > c j . Now, apply the induction assumption to the stair triple (i, k + 1, k). This either gives a node b l which satisfies (i), or b l is such that c l = c k and r k < r l < r i . In the former case we are done, so assume the latter. Now apply the induction assumption to the stair triple (i, l, j), and we are done.

Bumped nodes in successive insertions.
In this section we prove a key result which bounds the distribution of bumped nodes appearing in successive insertions. Theorem 4.4, together with Corollary 4.6 can be viewed as a generalization of [K, Theorem 1] to the realm of superalphabets and ε-insertion. We begin by defining a certain set partition of the nodes of a Young diagram that naturally results from the ε-insertion process.
Let y ∈ X , ε ∈ Z 2 , and let T be a semistandard (X , λ)-tableau. Let b y 1 , . . . , b y m be the bumped nodes for the insertion (T is the nearest node directly to the left of v which was bumped in the y insertion. If no such element exists then l(v) = 0. Similarly, b y u(v) is the nearest node directly above v which was bumped in the y insertion. If no such element exists then u(v) = 0. Let [µ] A be the set of all nodes of [µ] together with all addable nodes of [µ]. Now define the sets Remark 4.3. Informally, NE(T, ε, y) represents the set of nodes 'northeast' of a rough perimeter delineated by tracing the path of the bumped nodes in sequence, and SW(T, ε, y) represents the nodes to the 'southwest' of that perimeter. For example, if ε + y = 1, and the bumped nodes are those labeled in the diagram of [µ] below, then the red-colored nodes represent the set NE(T, ε, y), and the blue-colored nodes represent the set SW(T, ε, y). Though we will not need this fact, it follows from the definition that b y i ∈ NE(T, ε, y) if and only if (T ε ← − y)(b y i ) + ε = 1, as can be verified in the example above.
Proof. Since NW(T, ε, y)⊔SE(T, ε, y) = [µ] A , we may prove the equivalent statement: y ≺ z and ε = 0, or y ≻ z and ε = 1, or y = z and y = 0, (4.5) for all 1 ≤ i ≤ m 2 . First we prove that the lemma holds when T is a standard tableau, y, z / ∈ T, y = z, and y = 0. Note that in this situation we need not consider the third case in the right side of (4.5). We will go by induction on 1 ≤ i ≤ m 2 . We will write NE for NE(T, ε, y) where the context is clear.
Base case i = 1. Assume y ≺ z. Since y = 0, we have y < z and z = 0. If ε = 0, then b y 1 is in the first row, and T y (b y 1 ) = y < z, so b y 1 ⇒ b z 1 . Then l 1 > 0 = u 1 , so b z 1 ∈ NE. If ε = 1, then b y 1 is in the first column, and b z 1 ⇒ b y 1 so l 1 = 0 < u 1 , so b z i / ∈ NE. Now assume y ≻ z and ε = 0. Then b y 1 is in the first row. If z = 0, then z < y, so b z 1 ⇒ b y 1 . Then l 1 = 0, so b z 1 / ∈ NE. If z = 1, then b z 1 is in the first column, so l 1 = 0, and again b z 1 / ∈ NE. Now assume y ≻ z and ε = 1. Then b y 1 is in the first column. If z = 0, then z < y, so b y 1 ⇒ b z 1 . Then u 1 = 0, so b z 1 ∈ NE. If z = 1, then b z 1 is in the first row, so again u 1 = 0 and b z 1 ∈ NE. Induction step. So (4.5) holds when i = 1. Now we show that Then l i > u i , or l i = u i = 0 and ε = 1. We assume the former, and will later address the latter case.
There are two cases to consider: (a) Assume ε + T y (b z i ) = 0. Then (b z i+1 ) 0 = (b z i ) 0 + 1. First we show that b z i = b y r for any r. Indeed, if we did have that b z i = b y r , then by Lemma 4.1 applied to (b y l i , b y r ), there is some b y s in the column to the left of b y r , with l i ≤ s < i and ε + T(b y s ) = 1. Then by the above paragraph, b y s ր b y i . But then b y s+1 is in the same column as b y i , but cannot be above or below b y i since u i < s < s + 1 ≤ i. Then the only option is s + 1 = r. But this cannot be, since by assumption . First we prove that u i+1 < l i . Assume this is not the case. Then by Lemma 4.1, there exists a sequence t 0 , . .
, a contradiction of the ε-insertion algorithm. Thus u i+1 < l i .
, it must be that b y ta ⇒ b z i+1 . Then l i+1 ≥ t a ≥ u i+1 , and thus l i+1 > u i+1 , so b z i+1 ∈ NE. Now assume that u i = l i = 0 and ε = 1. We will show that u i+1 = 0. By way of contradiction assume u i+1 > 0. There are two cases to consider: (a) Assume T y (b z i ) = 0. Then, since b y 1 is in the first column, by Lemma 4.1 there exists a sequence t 0 , . . . , t k , where k = (b y u i+1 ) 1 − 2, such that 1 ≤ t 0 < · · · < t k < u i+1 , (b y t j ) 1 = 1 + j and ε + T(b y t j ) = 1 for all j.
Then, applying Lemma 4.1, we have u i+1 = 0 unless b z i ⇓ b z i+1 and b z i = b y r for some r. Then T(b y r−1 ) = T y (b y r ), and T y (b y r ) = 1, so b y r−1 is in the row above b y r , and b y r ⇒ b y r−1 . Then, since b y 1 is in the first column, by Lemma 4.1 there exists a sequence t 0 , . . . , t k , where k = (b y r−1 ) 1 − 2, such that 1 ≤ t 0 < · · · < t k < r − 1, (b y t j ) 1 = 1 + j and T(b y t j ) = 0 for all j. Then there exist some t j such that b y t j is in the same column as b y r . Then, since t j < r, we have b y r ⇑ b y t j , a contradiction, since u i = 0. This completes the proof that ε, y). Let c y 1 , . . . , c y m 1 and c z 1 , . . . , c z m 2 be the bumped node sequences for the insertions (T ′ ε+1 ←− − y) and ((T ′ ε+1 ←− − y) ε+1 ←− − z), respectively. Then by Lemma 2.2, c y j = (b y j ) ′ and c z j = (b z j ) ′ for all j. But then u(b z j ) = l(c z j ) and l(b z j ) = u(c z j ) for all j, so c z i ∈ NE(T ′ , ε + 1, y). Then, applying the 'only if' direction of the claim proved above, we have c z i+1 ∈ NE(T ′ , ε + 1, y). Then and ε + y = 0, so b z i+1 / ∈ NE(T, ε, y), as required. This completes the proof of the lemma when T is a standard tableau, y, z / ∈ T, y = z, and y = 0. Now we maintain the above assumptions but consider the case y = 1. Let c y * 1 , . . . , c y * m 1 and c z * 1 , . . . , c z * m 2 be the bumped node sequences for the ⇐⇒ y * ≺ z * and ε + 1 = 0, or y * ≻ z * and ε + 1 = 1 ⇐⇒ y ≻ z and ε = 1, or y ≺ z and ε = 0.
This completes the proof of the lemma when T is a standard tableau, y, z / ∈ T, y = z. Now, let T be an arbitrary semistandard tableau, with arbitrary y, z ∈ X . We may choose elements z • , y • ∈ X • , and a •-standardization T • of T, such that Then by this choice we have y ≺ z and ε = 0, or y ≻ z and ε = 1, or y = z and y = 0, by application of Lemma 3.11.
Proof. Let b y 1 , . . . , b y m 1 , b z 1 , . . . , b z m 2 be as in Lemma 4.4. By that lemma, b z m 2 ∈ NE(T, ε, y) if and only if the right side holds.
( ⇐= ) Assume by way of contradiction that b z m 2 ∈ NE and b z m 2 ր b y m 1 . Then b y m 1 cannot be in the same column as b z m 2 , else u m 2 > l m 2 . First assume l m 2 > 0. Then by Lemma 4.1, there exists a sequence t 0 , . . . , t k , where k = (b y m 1 ) 1 − (b y lm 2 ) 1 − 1, such that l m 2 ≤ t 0 < · · · < t k < m 1 , (b y t j ) 1 = (b y lm 2 ) 1 + j and ε + T(b y t j ) = 1 for all j. Then there is some t j such that b y t j is in the same column as b z m 2 . Moreover, we have b z m 2 ⇑ b y t j , hence u m 2 ≥ t j > l m 2 , a contradiction. Now assume l m 2 = 0. Then u m 2 = 0 and ε + y = 1. Then b y 1 is in the first column, and by Lemma 4.1, there exists a sequence t 0 , . . . , t k , where k = (b y m 1 ) 1 − 2, such that 1 ≤ t 0 < · · · < t k < m 1 , (b y t j ) 1 = 1 + j and ε + T(b y t j ) = 1 for all j. Then there is some t j such that b y t j is in the same column as b z m 2 . Moreover, we have b z m 2 ⇑ b y t j , hence u m 2 ≥ t j > 0, a contradiction.
( =⇒ ) Applying the 'if' statement proved above to the conjugate situation (as in the proof of claim Lemma 4.4), we have that b m 2 / ∈ NE implies that b z m 2 ր b y m 1 , completing the proof.

Super RSK correspondence
5.1. Biwords. Given alphabets X and Y , we call an element of X ×Y an (X , Y )biletter. We call a biletter (x, y) mixed if x + y = 1. We define a total order ⊳ on (X , Y )-biletters by setting (x 1 , y 1 ) ⊳ (x 2 , y 2 ) if For k ∈ Z >0 , we call an element w = ((x 1 , y 1 ), . . . , (x k , y k )) ∈ (X × Y ) k an (X , Y )-biword of length k. We say that w is restricted if it is multiplicity free with respect to mixed biletters; i.e. (x i , y i ) = (x j , y j ) for i = j only if x i + y i = 0. We say that w is ordered if (x i , y i ) (x j , y j ) for all i ≤ j. The left content lcon(w) of w is the multiset {x 1 , . . . , x k } and the right content rcon(w) of w is the multiset {y 1 , . . . , y k }.
If L is a multiset of elements of X and R is a multiset of elements of Y , with |L| = |R| = k, we say (L, R) is an (X , Y )-content pair of length k. For an (X , Y )-content pair, define RBiw(L, R) to be the set of restricted (X , Y )-biwords w with lcon(w) = L and rcon(w) = R. Let RBiw(L, R) = {w ∈ RBiw(L, R) | w is ordered}. Finally, define Tab(L, R) to be the set of pairs (L, R) of tableaux such that sh(L) = sh (R), con(L) = L and con(R) = R. Let SStd(L, R) ⊆ Tab(L, R) be the subset of semistandard tableau pairs. 5.2. Super RSK algorithm. Let (L, R) be an (X , Y )-content pair of length k. Let w = ((x 1 , y 1 ), . . . , (x k , y k )) ∈ RBiw(L, R) . We define T 0 w := ∅, then for and define a i w to be the added node of this insertion. We say T w := T k w is the insertion tableau of w. The recording tableau of w is the (Y , sh(T w ))-tableau T w defined by T w (a i w ) := y i . We then define sRSK(w) := (T w , T w ). Proof. Let |L| = |R| = k, and w = ((x 1 , y 1 ), . . . , (x k , y k )) ∈ RBiw(L, R) . We have that T w is semistandard by inductive application of Lemma 3.9 . Define T w i by T w i (a j w ) := y j for all 1 ≤ j ≤ i. By induction, assume: (i) T w i is semistandard (ii) If r < s ≤ i, y r = y s , and y r = 0, then a r w ր a s w (iii) If r < s ≤ i, y r = y s , and y r = 1, then a s w ր a r w . If y i+1 > y i , then T w i+1 automatically satisfies (i)-(iii). Assume y i+1 = y i . Then either x i ≺ x i+1 or x i = x i+1 and y i+1 + x i+1 = 0. Note that T w i+1 is non-decreasing since the upper row of w is non-decreasing. There are two cases: (a) Assume y i+1 = 0. Then (iii) holds, and T w i is column-strict with respect to odd letters. Moreover by Corollary 4.6, a i w ր a i+1 w , so (ii) holds, and T w i is row-strict with respect to even letters. (b) Assume y i+1 = 1. Then (ii) holds, and T w i is row-strict with respect to even letters. Moreover by Corollary 4.6, a i+1 w ր a i w , so (iii) holds, and T w i is column-strict with respect to odd letters. Thus, by induction T w k = T w is semistandard. Thus sRSK(w) ∈ SStd(L, R). Now let (L, R) ∈ SStd(L, R). We define L k = L, R k = R, and then for 1 ≤ i ≤ k inductively define R i−1 and L i−1 in the following manner. Define y i ∈ Y to be the <-maximal element of R i . If y i = 0 (resp. y i = 1), let u i be the rightmost (resp. bottommost) node in R i such that and let x i be the extracted letter. Then define sRSK * (L, R) = ((x 1 , y 1 ), . . . , (x k , y k )) Let w := sRSK * (L, R). By construction and Lemma 3.8 we have that T j w = L j , T w j = R j , and u j = a j w for all 1 ≤ j ≤ k. We argue by induction on i that w i := ((x 1 , y 1 ), . . . , (x i , y i )) is an ordered restricted (X , Y )-biword. By construction, y i+1 ≥ y i , so w i+1 is an ordered restricted biword if y i+1 = y i . Assume y i = y i+1 . Then there are two cases.
(a) Assume y i+1 = 0. If x i ≻ x i+1 or x i = x i+1 and x i = 1, then by Corollary 4.6, u i+1 = a i+1 w ր a i w = u i , which by the choice of u i+1 implies that u i ⇓ u i+1 , R(u i ) = R(u i+1 ), and R(u i ) = 0, a contradiction, since R is semistandard. (b) Assume x i+1 = 1. If x i ≻ x i+1 or x i = x i+1 and x i = 0, then by Corollary 4.6, u i = a i w ր a i+1 w = u i+1 , which by the choice of u i+1 implies that u i ⇒ u i+1 , R(u i ) = R(u i+1 ), and R(u i ) = 1, a contradiction, since R is semistandard. Thus w i+1 is an ordered restricted biword. Thus by Lemma 3.8, sRSK and sRSK * are mutual inverses on RBiw(L, R) and SStd(L, R).
Remark 5.3. When X = Y = N, where < is the usual order on integers and every element is of even parity, the super RSK correspondence of Theorem 5.2 reduces to the classical RSK correspondence [K].
Remark 5.4. As noted in §1, the existence of a bijection between the sets in Theorem 5.2 was proved by Bonetti, Senato, and Venezia [BSV], using a more straightforward insertion algorithm which yields a different bijection than the one in Theorem 5.2. Our motivation in presenting this new bijection is in the direction of fully generalizing the symmetry property of classical RSK, which we do in §6.
Remark 5.5. In [SW,§3], Shimozono and White present a close relative to our super RSK correspondence-the algorithm they use to construct the upper and lower tableaux of an (X , Y )-biword (called in their paper a doubly-colored biword) is very similar in spirit to the super RSK algorithm presented here (see Remark 3.3).
However, they work with the set of all (not just restricted) biwords, and their semistandard tableaux are defined to be row-weak and column-strict for both parities. Consequently, the fact that a bijective correspondence exists between these objects (as noted in [SW,Theorem 22]) can be deduced from classical RSK correspondence, while this is not true of the correspondence in Theorem 5.2, which involves distinct (and distinctly-sized) sets of combinatorial objects.
Note that RBiw(L, R) is a set of orbit representatives for RBiw(L, R) under the action of the symmetric group S k . For w ∈ RBiw(L, R), write w for the unique element of RBiw(L, R) which belongs to the S k -orbit of w. By precomposing with the function w → w , we may extend sRSK to a function sRSK : RBiw(L, R) → SStd(L, R) which is constant on S k -orbits.

Symmetry
In this section we prove that the super RSK algorithm defined in §5.2 satisfies the symmetry property that holds for the classical RSK algorithm. In this section we assume that (L, R) is an (X , Y )-content pair of length k.
6.2. Standardizing biwords. We will say a biword is standard if no letter occurring in the biword has multiplicity greater than one. For a multiset L of letters in X , we define where mult L (x) is the multiplicity of x in L.
Definition 6.1. Let (L, R) be an (X , Y )-content pair of length k, and let w ∈ RBiw(L, R) . We construct a related biword in RBiw(L • , R • ) as follows.
We call w • the •-standardization of w.
Let T i w (resp. T i w • ) be the ith insertion tableaux in the Super RSK algorithm applied to w (resp. w • ), and let a i w (resp. a i w • ) be the added node of this insertion, using notation in §5.2. By induction, assume that•(T i w • ) is a •-standardization of T i w , and a i w • = a i w , for all i < n. If (x (c) i , y j ′ ) is the nth biletter in w • , then (x i , y j ′ ) is the nth biletter in w. If y j ′ + x i =0, then, by the above claim, we have x (c) i ≻ z for every z ∈ T n−1 w • such that•(z) = x i . On the other hand if y j ′ + x i =1, then, by the above claim, we have x (c) i ≺ z for every z ∈ T n−1 w • such that•(z) = x i . Therefore by Lemma 3.11, it follows that T n w • = (T n−1 i ) is a •-standardization of (T n−1 w y j ′ ← − − x i ) = T n w , and a n w • = a n w , as desired. Thus•(T w • ) = T w and•(T w • ) = T w , so•(sRSK(w • )) = sRSK(w), proving the lemma. 6.3. Symmetry. As noted in Remark 5.5, Shimizono and White [SW] define a super-analogue of the RSK algorithm which is identical to the super RSK algorithm presented here when restricted to standard biwords. Thus their symmetry result proves a special case of the symmetry of the sRSK map: Lemma 6.4. If w ∈ RBiw(L, R) is a standard biword, then we have sRSK(w inv ) = (sRSK(w)) inv .
Proof. This follows from [SW,Theorem 21(3), (6)]. Now we extend this result to the general case.
Theorem 6.5. The following is a commuting diagram: Proof. The top face commutes by Lemma 6.2. The bottom face commutes since 'forgetting superscripts' then swapping tableaux clearly yields the same result as swapping tableaux and then 'forgetting superscripts'. The front and back faces commute by Lemma 6.3. The right face commutes by Lemma 6.4. Thus we have so the left face commutes, proving the theorem.