RSK tableaux and the weak order on fully commutative permutations

For each fully commutative permutation, we construct a"boolean core,"which is the maximal boolean permutation in its principal order ideal under the right weak order. We partition the set of fully commutative permutations into the recently defined crowded and uncrowded elements, distinguished by whether or not their RSK insertion tableaux satisfy a sparsity condition. We show that a fully commutative element is uncrowded exactly when it shares the RSK insertion tableau with its boolean core. We present the dynamics of the right weak order on fully commutative permutations, with particular interest in when they change from uncrowded to crowded. In particular, we use consecutive permutation patterns and descents to characterize the minimal crowded elements under the right weak order.


Introduction
First introduced in [Ste96a], the fully commutative elements of a Coxeter group have the property that every pair of reduced words are related by a sequence of commutation relations.This set of objects is combinatorially rich and has been studied extensively (see, for example, [MPPS20,Nad15,Ste98]).A permutation is fully commutative if and only if it avoids the pattern 321 [BJS93], and the fully commutative permutations are exactly those with fewer than three rows in their Robinson-Schensted-Knuth (RSK) tableaux [Sch61].In this paper, following up on recent work in [GPRT22], we examine the interplay between reduced words and RSK tableaux for fully commutative permutations and analyze the set of fully commutative permutations under the weak order.
Our previous work, which is a companion to this paper, proves that the RSK insertion tableaux for boolean permutations satisfy a certain sparsity condition that we call uncrowded [GPRT22].Boolean permutations are an important subset of fully commutative permutations, characterized by the fact that their principal order ideals in the Bruhat order are isomorphic to boolean algebras.Motivated by those results, we call a fully commutative permutation with an uncrowded insertion tableau an uncrowded permutation.In other words, an uncrowded fully commutative permutation shares its insertion tableau with some boolean element.A fully commutative permutation that is not uncrowded is called crowded.Central to this paper is the partition of the set of fully commutative permutations into crowded and uncrowded elements.
For each fully commutative element w, we identify a particular boolean element ŵ that is below w in the weak order and has the same support as w; we call this ŵ the boolean core of w (Theorem 3.2).We then view the fully commutative permutation w as an "elongation" of its boolean core, and we investigate the evolution of RSK insertion tableaux along chains of fully commutative elements in the right weak order.We prove that the second rows of insertion tableaux obey a containment property along covering relations in the right weak order (Theorem 3.4).
Applying this containment property, we show that if two fully commutative elements with the same support satisfy a covering relation in the right weak order and have different insertion tableaux then the larger one is necessarily crowded (Theorem 4.10).This has two important implications.First, a fully commutative element is uncrowded exactly when it has the same insertion tableau as its boolean core (Corollary 4.11).Second, within the set of fully commutative permutations under the right weak order, the uncrowded permutations form an order ideal and the crowded permutations form a dual order ideal (Lemma 5.1).Thus, knowing the minimal crowded elements in the poset is, in fact, enough information to identify each fully commutative element as being either crowded or uncrowded.Our final result, Theorem 5.15, proves a set of necessary and sufficient conditions for a fully commutative permutation to be minimal in the dual order ideal of crowded permutations.
This paper is organized as follows.Section 2 provides necessary background information and notation including several results from our companion paper on boolean RSK tableaux.Section 3 defines the boolean core of a fully commutative element and proves a containment property for RSK tableaux under the right weak order.Section 4 explores covering relations between fully commutative elements in the right weak order when the two permutations have the same support but different insertion tableaux.Finally, Section 5 characterizes the minimal elements of the dual order ideal of crowded fully commutative permutations in the right weak order, thus providing the key to classifying each fully commutative permutation as being either crowded or uncrowded.

Background and notation
Denote the symmetric group on n elements by S n .For a permutation w ∈ S n , we use the one-line notation w = w(1)w(2) • • • w(n) to represent w.For each i ∈ {1, . . ., n − 1}, we write s i ∈ S n to denote the simple reflection (or adjacent transposition) that swaps i and i + 1 and fixes all other letters.Every permutation can be expressed as a product of simple reflections.Given w ∈ S n , the minimum number of simple reflections among all such expressions for w is called the (Coxeter) length of w, and is denoted by (w).An inversion in the one-line notation for w is a pair of positions i < j such that w(i) > w(j).It is often convenient to recognize that (w) is the number of inversions in the one-line notation for w.A reduced decomposition of w is an expression w = s i 1 • • • s i (w) realizing the Coxeter length of w.To simplify notation, we refer to such a decomposition via its reduced word i 1 • • • i (w) .Let R(w) denote the set of reduced words for w.
The set of letters appearing in reduced words of a permutation w is the support supp(w) of w.For example, consider w = 51342 = s 4 s 2 s 3 s 2 s 4 s 1 ∈ S 5 .Because w has six inversions, we see that (w) = 6 and [423241] ∈ R(w).
The following technical lemma is related to the support of a permutation.It introduces a pair of values M and m which depend on the choice of v ∈ S n and i ∈ {1, . . ., n − 1}.These values play a central role in the arguments in Section 4.
Lemma 2.1.[Ten12, Lemma 2.8] Fix a permutation v ∈ S n and i ∈ {1, . . ., n − 1}, and let M := max{v(j) : j ≤ i} and m := min{v(j) : j ≥ i + 1}.Then the following statements are equivalent: The right weak order, denoted by ≤, is a partial order on S n obtained by taking the transitive closure of the cover relation w < ws i whenever (w) < (ws i ).We use w < w to denote when w ≤ w and w = w .The left weak order is defined analogously, with left multiplication by s i instead of right.In each order, the minimum element is the identity permutation and the maximum element is the long element n(n − 1) • • • 21.More details on the weak order can be found in, for example, [BB05, Section 3.1].
An order ideal of a poset is a subset C such that if y ∈ C and x ≤ y, then x ∈ C. A dual order ideal (or order filter, or upper order ideal) of a poset is a subset C such that if x ∈ C and x ≤ y, then y ∈ C.
2.1.Fully commutative permutations and boolean permutations.Let m ≤ n.The permutation w ∈ S n is said to contain the pattern σ ∈ S m if w has a (not necessarily contiguous) subsequence whose elements are in the same relative order as σ.In the case that w does not contain σ, we say w avoids σ.For instance, the permutation w = 314592687 contains the pattern 1423 because the subsequence 1927 (among others) has the same relative order as 1423.On the other hand, w avoids 3241 since it has no subsequences that follow the pattern 3241.We note also that the inversions of a permutation are exactly the instances of 21-patterns.
For |i − j| > 1, simple reflections satisfy commutation relations of the form s i s j = s j s i .An application of a commutation relation to a product of simple reflections is called a commutation move.In the context of reduced words, we will say adjacent letters i and j in a reduced word commute when |i − j| > 1.For a reduced word [u] of a permutation, the equivalence class of all words obtained from [u] by sequences of commutation moves is called the commutation class of [u].A permutation is called fully commutative if all of its reduced words form a single commutation class.As the following proposition shows, fully commutative permutations can be characterized in terms of pattern avoidance.

Proposition 2.2 ([BJS93]
).Let w be a permutation.The following are equivalent: • w is fully commutative, • w avoids the pattern 321, • no reduced word of w contains i(i + 1)i as a factor, for any i, and • no reduced word of w contains (i + 1)i(i + 1) as a factor, for any i.
Boolean permutations are an important subset of the set of fully commutative permutations.The following result gives a description of boolean permutations analogous to that of Proposition 2.2.

Proposition 2.3 ([Ten07]
).Let w be a permutation.The following are equivalent: • w is boolean, • w avoids the pattern 321 and 3412, • there is a reduced word of w that consists of all distinct letters, and • every reduced word of w consists of all distinct letters.

Heaps and commutation class.
In this section, we review the classical theory of heaps, which was used in [Ste96b] to study fully commutative elements of a Coxeter group.For a detailed list of attributions on the theory of heaps, see The following lemma follows directly from this definition.
Lemma 2.5.Let [u] be an arbitrary reduced word for a permutation, and let x < y be elements of the heap for [u].If y covers x, then the labels of x and y differ by exactly one.
Note that a heap is, in some sense, a partial ordering on the multiset of simple reflections occurring in a reduced word.For a fully commutative permutation, the heap structure on this multiset is, in fact, independent of the choice of reduced word (see Proposition 2.7).Throughout this paper, for a fully commutative permutation w, we will use H w to denote both the heap diagram for w and the poset of simple reflections of any reduced word [u] of w.The context should make it clear to which object H w refers.
From a linear extension of the heap, one can define a labeled linear extension essentially by replacing elements of the heap with their labels.
Definition 2.6.A labeled linear extension of the heap of a reduced word ) is a total order on {1, . . ., } that is consistent with the structure of the heap.That is, π(x) ≺ π(y) implies x < y.
As the next proposition illustrates, labeled linear extensions are related to reduced words and commutation classes.By definition, a fully commutative permutation has exactly one commutation class.Hence Proposition 2.7 implies that given any reduced word [u] for a fully commutative permutation w, the set of labeled linear extensions of the heap for [u] is exactly R(w), the set of reduced words of w.Proposition 2.3 states that a boolean permutation is a fully commutative permutation with no repeated letters in any of its reduced words.In the sense of heaps, this means that there are no two elements corresponding to the same simple reflection.For boolean-specific descriptions of heaps, see [GPRT22, Section 2.2].
2.3.Robinson-Schensted-Knuth tableaux.The well-known Robinson-Schensted-Knuth (RSK) insertion algorithm, as described in [Sch61], is a bijection w → (P(w), Q(w)) from S n onto pairs of standard tableaux of size n having identical shape.The tableau P(w) is called the insertion tableau of w, and the tableau Q(w) is the recording tableau of w.The shape of these tableaux is the RSK partition of w.We will also write P i (w) to denote the partial insertion tableau constructed by the first i letters in the one-line notation for w.For more details, see for example [Sta99,Section 7.11].
The following symmetry result is an important feature of the algorithm, and one that will simplify our own work.
Theorem 2.10.Given a permutation w, the length of the longest increasing (resp., decreasing) subsequence in the one-line notation of w is the size of the first row (resp., column) of P(w).
Due to Schensted's theorem, we can see that a permutation is fully commutative if and only if its RSK partition has at most two rows.We denote the set of values in the second row of the RSK insertion tableau of a permutation w by Row 2 (P(w)).More generally, we denote the set of values in the second row (resp., first row) of any tableau T by Row 2 (T ) (resp., Row 1 (T )).
Next we list some basic features of RSK insertion, which we may use without specific mention in the future.The following lemma is a consequence of the definition of RSK insertion.For permutation v ∈ S n and value q ∈ {1, . . ., n}, let c v (q) be the column of P(v) into which q is first inserted.Let LIS v (q) be the length of a longest increasing subsequence of v that ends with q.The following is a key result we will reference in our analysis.
Lemma 2.12 ([Sag01, Lemma 3.3.3]).For v ∈ S n and q ∈ {1, . . ., n}, we have c v (q) = LIS v (q).One consequence of Lemma 2.12 is that certain values must be part of every longest increasing subsequence of a permutation.
Corollary 2.13.For a permutation v, if q is the only value in v inserted into column c v (q) of P(v), then q is in every longest increasing subsequence in v.
The last result in this subsection highlights basic properties of RSK tableaux for fully commutative permutations.
Lemma 2.14.Let w be a fully commutative permutation with Row 2 (P(w For each i ∈ {1, . . ., t}, let b i be the value that bumps z i from the first row to the second row during the construction of P(w).Then we have the following.
(a) The sequence z 1 z 2 • • • z t is an (increasing) subsequence of w.In other words, the values z 1 , z 2 , . . ., z t appear from left to right in the one-line notation of w.(b) The sets {z 1 , . . ., z t } and {b 1 , . . ., b t } are disjoint.In other words, during RSK insertion, no value can both bump something and be bumped by something.(c) The sequence b 1 b 2 . . .b t is an increasing subsequence of w.(d) Let 1 ≤ i < j ≤ t.During RSK insertion, the value z i is bumped before z j . Proof.
(a) Suppose, to the contrary, that z i appears to the right of z i+1 for some i.Since b i bumps z i during the insertion algorithm, we know b i < z i , and the value b i occurs to the right of z i in the one-line notation of w.This means z i+1 z i b i is a 321-pattern in w, which is a contradiction.(b) By (a), we have that z 1 • • • z t is an increasing subsequence of w.Hence, there are no i and j such that z i bumps z j , and the sets {z 1 , . . ., z t } and {b 1 , . . ., b t } are therefore disjoint.(c) First, we show that b 1 < • • • < b t .Suppose, to the contrary, that b i > b i+1 for some i.Since z i appears to the left of b i in the one-line notation for w and z i > b i , the value b i+1 must appear to the left of b i in order to avoid a 321-pattern in w.We also know z i+1 appears to the left of b i+1 in the one-line notation for w and z i+1 > b i+1 .From (a), we know z i z i+1 is a subsequence of the one-line notation for w.Combining all of these observations, we conclude that 2.4.Characterization of boolean RSK tableaux.While Schensted's Theorem (Theorem 2.10) guarantees the insertion tableau of a boolean permutation has at most two rows, not every 2-row standard tableau is the insertion tableau of some boolean permutation.For example, the tableau T 1 below is the insertion tableau of the boolean permutation w = 315264 = [21435] ∈ S 6 , but T 2 cannot be obtained as the insertion tableau of any boolean permutation.
We review the characterization of these tableaux from [GPRT22].First we need to define when a set of integers is "uncrowded."Definition 2.15.Let L be a set of integers.If, for all integers x and y, with x > 0, we have then we will say that L is uncrowded.Otherwise, we say that L is crowded.
Let T be a standard tableau with at most two rows.When Row 2 (T ) is uncrowded, we also call the tableau T an uncrowded tableau, and T is a crowded tableau otherwise.In the example above, we can see that T 1 is an uncrowded tableau because its second row {3, 5, 6} is an uncrowded set, while T 2 is a crowded tableau because The following proposition, which is the combination of several results in [GPRT22], provides a characterization of RSK tableaux coming from boolean permutations.
Proposition 2.16.A standard tableau T with at most two rows is the insertion (or recording) tableau of a boolean permutation if and only if T is uncrowded.
We define an uncrowded (respectively, crowded) permutation to be a permutation with an uncrowded (respectively, crowded) insertion tableau.By Proposition 2.16, a permutation is uncrowded exactly when it shares an insertion tableau with some boolean permutation.

Fully commutative elements and the weak order
From Theorem 2.10, we know that the RSK partition for a permutation has at most two rows if and only if the permutation is 321-avoiding; that is, if and only if it is fully commutative.Boolean permutations, which avoid patterns 321 and 3412, are a special class of fully commutative permutations, and Proposition 2.16 fully characterized their RSK tableaux.In this section, we build upon Proposition 2.16 to study the insertion tableaux of fully commutative, but not necessarily boolean, permutations.
3.1.Boolean core.We set the stage using the following lemma, which is little more than a restatement of the definition of fully commutative.Lemma 3.1.Let w be a fully commutative permutation and [u] ∈ R(w).If j is a repeated letter in [u], then each pair of copies of j must be separated by both j + 1 and j − 1 in [u].Put another way, if x ≺ y are elements of the heap H w both with label j (i.e., u x = u y = j), then H w contains elements p and p with labels k + 1 and k − 1 such that x ≺ p ≺ y and x ≺ p ≺ y.
A key feature of boolean permutations is that their reduced words contain no repeated letters.This property fails to hold for arbitrary fully commutative permutations, but, as we will show in the next result, every fully commutative permutation can be thought of as having a "boolean core."More precisely, we can write any fully commutative permutation as the product of two permutations, one of which is boolean with the same support as the original permutation.As a result, every fully commutative permutation has a reduced word in which any repetition of letters occurs only after every letter in the support has appeared.

Proof. Fix a fully commutative permutation w and [
Because w is fully commutative, it has a unique heap H w .Elements with the same label are comparable in H w .Thus, for each i ∈ supp(w), we can take the smallest element x in H w such that u x = i.Let C denote the set of all such smallest elements, for i ∈ supp(w).
We claim that C is an order ideal of H w , and we will show that this is true using a proof by contradiction.Suppose x, y ∈ H w such that y ∈ C and x is covered by y.Let u x = j, and so by Lemma 2.5 u y = j ± 1. Suppose, for the purpose of obtaining a contradiction, that x ∈ C. Thus there exists x ≺ x with u x = j.Then, by Lemma 3.1, there exist p, p ∈ H w such that x ≺ p ≺ x, x ≺ p ≺ x, u p = j + 1, and u p = j − 1.But then we would have y ∈ C, which is a contradiction.
Because C is an order ideal of H w , we can choose a labeled linear extension of H w whose first |C| letters are precisely supp(w).This produces a reduced word for w whose leftmost |C| letters are precisely supp(w).
Finally we show that this ŵ is also unique.Recall that any prefix of a reduced word for w corresponds to an order ideal of H w .The condition supp( ŵ) = supp(w) requires that we pick an order ideal of H w having |supp(w)| elements of distinct labels.Elements with the same label are comparable in H w , meaning that we are forced to select the smallest one for each label.
We refer to the boolean permutation ŵ in Theorem 3.2 as the (right) boolean core of a fully commutative permutation, where "right" refers to the fact that ŵ is the maximal boolean permutation that is less than w in the right weak order.
Example 3.3.The heap of the permutation w = 345619278 in Example 2.8 is given in Figure 1.The boolean core of w is ŵ = 314569278, and its heap is given in Figure 2. Note that the reduced word [21873456] ∈ R( ŵ) appears as the left prefix of the reduced word [21873456234] ∈ R(w).
Theorem 3.2 can also be proved without the language of heaps, by inducting on the length of a permutation.
3.2.Containment under the weak order.Theorem 3.2 identifies the boolean core of a fully commutative permutation, which gives some sense of how fully commutative permutations can be viewed as "elongations" of boolean permutations.We can similarly consider lengthening a fully commutative permutation.This leads to an important property about insertion tableaux.
Theorem 3.4.Let v and w be fully commutative permutations such that w = vs i with (w) = (v)+1.
Proof.Let v and w be as in the statement of the result.So with v(i) < v(i + 1).The permutation w is fully commutative by assumption, so it is 321-avoiding.Therefore, in fact, we have v(j) < v(i + 1) for all 1 ≤ j ≤ i, and v(j) > v(i) for all i + 1 ≤ j ≤ n. (3.1) Set P i−1 := P i−1 (v) = P i−1 (w) to be the insertion tableau for the shared prefix v(1) • • • v(i − 1) in the two permutations.To compute P i+1 (v), we insert v(i) first and then v(i + 1); to compute P i+1 (w), we insert v(i + 1) first and then v(i).
Consider first what happens when we insert v(i) into P i−1 .There are two cases to consider: either v(i) bumps something out of the first row of P i−1 , or v(i) gets appended to the end of the first row of P i−1 .
Suppose first that v(i) bumps some z out of the first row of P i−1 .For v(i) to do this, the value z must have been the smallest number in that row larger than v(i).To create P i+1 (v) from P i (v), the value v(i + 1) must be appended to the first row of P i (v), because v(i + 1) > v(j) for all 1 ≤ j ≤ i, by (3.1).To construct P i (w), we again have that w(i) = v(i + 1) gets appended to the end of the first row of P i−1 .When w(i + 1) = v(i) is inserted into P i (w), it must bump the smallest value in Row 1 (P i−1 ) ∪ {v(i + 1)} that is larger than v(i); this value must be z, as above, because z < v(i + 1).Therefore, P i+1 (v) = P i+1 (w), with {v(i), v(i + 1)} in the top row and z in the second row.
Because the rest of the entries in the one-line notations of v and w are identical, we can conclude from here that P(v) = P(w).Now suppose, for the remainder of the proof, that when v(i) is inserted into P i−1 it is appended to the end of the first row of P i−1 .In other words, v(i) is larger than all values in Row 1 (P i−1 ).Then, when v(i + 1) is inserted into P i (v), this new value is also appended to the end of the first row because v(i + 1) > v(i).In other words, P i+1 (v) is created by appending both v(i) and v(i + 1) to the first row of P i−1 .
To construct P i (w), on the other hand, we first insert v(i + 1).This gets appended to the end of the first row of P i−1 because v(i + 1) is larger than all other values seen so far, by (3.1).In contrast, v(i) < v(i + 1), so v(i) will bump something out of the first row of P i (w) in order to form P i+1 (w).Everything in Row 1 (P i−1 ) is greater than v(i), so v(i) must bump v(i + 1) itself.Therefore, Row 1 (P i+1 (v)) = Row 1 (P i+1 (w)) ∪ {v(i + 1)}.And more to the point, Row 1 (P i+1 (v)) ⊃ Row 1 (P i+1 (w)).
Combining (3.1) with the fact that v(i) is larger than every letter in Row 1 (P i−1 ), we have that v(i + 1), . . ., v(n) must each be larger than every letter in Row 1 (P i−1 ) ∪ {v(i)}.Therefore, all future insertions performed during the computation of both P(v) and P(w) will not bump any letter of Row 1 (P i−1 ) ∪ {v(i)} out of the first row.That is, everything in the first row from v(i) leftward will remain unchanged during the remaining steps of the insertion algorithm.
We will prove that Row 1 (P(v)) contains all of Row 1 (P(w)), using an inductive argument with P k (v) and P k (w), for i + 1 ≤ k ≤ n.We have shown the base case: Row 1 (P i+1 (v)) ⊃ Row 1 (P i+1 (w)).Assume, inductively, that for some k i + 1, we have Row 1 (P k (v)) ⊇ Row 1 (P k (w)).There are two ways for v(k + 1) to be inserted into P k (v): either it gets appended to the end of the top row of the tableau, or it bumps some value z.
is larger than v(k + 1).We must now consider whether or not z was in Row 1 (P k (w)).
If not, then there is nothing to worry about and we are done.On the other hand, if z ∈ Row 1 (P k (w)), then, because Row 1 (P k (w)) ⊆ Row 1 (P k (v)), this z must also be the smallest number in Row 1 (P k (w)) that is larger than v(k + 1).Therefore, when we insert v(k + 1) into P k (w), we will also bump z.Thus the induction holds at all stages of the insertion algorithm, and hence Row 1 (P(v)) ⊇ Row 1 (P(w)).The tableaux have height at most 2, and so Row 2 (P(v)) ⊆ Row 2 (P(w)), as well.
We highlight several facts relevant to upcoming arguments in Section 4.
Remark 3.5.For v and w fully commutative permutations with w = vs i , (w) = (v) + 1, and P(v) = P(w), the following are established within the proof of Theorem 3.4: (a) v(k) < v(i + 1) for k < i, and v(k) > v(i) for k > i; (b) the value v(i) does not bump anything in P(v), and v(i) ∈ Row 1 (P(v)); (c) v(i) bumps v(i + 1) in P(w), and v(i) ∈ Row 1 (P(w)); Because the length of the first row of a permutation's shape is determined by the length of a longest increasing subsequence in the permutation, we can use Theorem 3.4 to characterize when the insertion tableaux of v and vs i are unequal.
Proof.Note that |Row 2 (P(w)) \ Row 2 (P(v))| ≤ 1, because the length of the longest increasing subsequence changes by at most one after swapping adjacent values in a position.Since Row 2 (P(v)) ⊆ Row 2 (P(w)) by Theorem 3.4, we have that Row 2 (P(v)) Row 2 (P(w)) if and only if the size of the first row of P(v) is one more than the size of the first row of P(w).By Schensted's theorem (Theorem 2.10), this holds if and only if the length of a longest increasing subsequence of v is one more than the length of a longest increasing subsequence of w.Swapping v(i) and v(i + 1) changes this length if and only if every longest increasing subsequence in v uses both v(i) and v(i + 1).It follows that when Row 2 (P(v)) Row 2 (P(w)), the set Row 2 (P(w)) contains exactly one more element than Row 2 (P(v)).
Theorem 3.4 has other implications for the weak order on fully commutative elements.
Corollary 3.7.Let v and w be fully commutative permutations.
(a) If v is less than w in the right weak order, then Row 2 (P(v)) ⊆ Row 2 (P(w)).
(b) If v is less than w in the left weak order, then Row 2 (Q(v)) ⊆ Row 2 (Q(w)).
There is another important implication of Theorem 3.4, in conjunction with Theorem 3.2.This allows us to show the relationship between the insertion tableaux of a fully commutative element and that of its boolean core.
The following example illustrates this result.

Insertion tableaux dynamics
Throughout this section, we will restrict our attention to certain important scenarios, and we will highlight our assumptions for the reader in centered boxed text.To begin, we will assume throughout this section that v and w are fully commutative permutations with w = vs i and (w) = (v) + 1.
Corollary 3.6 gave conditions that determine exactly when P(v) = P(w) in terms of the longest increasing subsequences of v.We next want to understand the entries of these tableaux when they are unequal.In particular, if P(v) = P(w), is it possible for P(w) to be uncrowded?Said another way, if the insertion tableau changes along a covering relation in the right weak order, can the covering permutation be uncrowded?If i ∈ supp(v), then this could certainly be the case.Consider, for example, when v is the identity.On the other hand, if i ∈ supp(v), then, as we shall see, the answer to the question is no.
Recall our assumptions in this section: v and w are fully commutative permutations (that is, they avoid 321) with w = vs i and (w) = (v) + 1.Let M and m be the values defined in Lemma 2.1: M := max{v(j) : j ≤ i} and m := min{v(j) : j ≥ i + 1}.
Our first lemma shows these values are part of a 3142-pattern in v whenever i ∈ supp(v).
Proof.Because i ∈ supp(v), it follows from Lemma 2.1 that m < M , and M ≥ v(i) and m ≤ v(i + 1) by definition.Because w = vs i and (w) > (v), we must have v(i) < v(i + 1).Next we argue that M > v(i).Suppose M = v(i).Then m < M = v(i) < v(i + 1), so v(i + 1) v(i) m will form a 321-pattern in w, which is a contradiction.Therefore M > v(i).Similarly we can show that m < v(i + 1).
Since w cannot have a 321-pattern, we also must have M < v(i + 1) and m > v(i).Therefore the subsequence M v(i) v(i + 1) m is a 3142-pattern in v.
In fact, 321-avoidance, the maximality of M , and the minimality of m force even more structure upon v.
Let us now further suppose, for the remainder of this section, that i ∈ supp(v) and P(v) = P(w).
Furthermore, we will maintain the notation established in Corollary 4.2.
Corollary 3.6 tells us that every longest increasing subsequence in v must use both v(i) and v(i + 1).In particular, this means that h ≥ 1 and j ≥ 1.
We will show that e occurs after v(i + 1) in v, that e > M , and, finally, that Row 2 (P (w)) is crowded as it contains too many integers in the interval {M, . . ., e}.
The next sequence of lemmas describe certain values in the rows of P(v) and P(w).Recall for a permutation v ∈ S n and value q ∈ {1, . . ., n}, we define c v (q) to be the column of P(v) into which q is first inserted.Lemma 4.3.In the construction of P(v) and P(w), the value M is bumped to the second row by one of a 1 , . . ., a h .Proof.By Remark 3.5(b), v(i) does not bump anything in P(v), so we have that c v (M ) < c v (v(i)).Since v(i) < M , this means some value that occurs between M and v(i) in v must bump M in P(v).Because the values prior to v(i) are unchanged in w, M will be bumped by that same value in P(w).
Just as we can track M in the RSK insertion algorithm, we can determine the role of m in the construction of P(v).
Lemma 4.4.The value v(i + 1) is bumped by m in P(v).
Proof.Corollary 4.2 and Remark 3.5(b) tell us that, just before m is inserted in the process of constructing P(v), the first row contains v(i) < v(i + 1) < e 1 < • • • < e j with no element between v(i) and v(i + 1).By Lemma 4.1, v(i) < m < v(i + 1), so m will bump v(i + 1) in P(v).
Next, we apply Remark 3.5 and Lemma 4.4 to determine the position of e in the one-line notation for v.
Lemma 4.5.The value e occurs after v(i + 1) in v.
Using Lemma 4.5, we can show that e does not bump anything during the construction of P(v).
Lemma 4.6.The value e does not bump anything in P(v).
Proof.Suppose, for the purpose of obtaining a contradiction, that e bumps something in P(v).Then there exists a value q such that e < q and q occurs before e in v.By Lemma 4.5, e occurs after v(i + 1) in v, so q occurs before e in w as well.However, e is bumped in w, so there is a value q with q < e and q occurring after e in w.This yields a 321-pattern in both v and w, which is not possible.Hence e does not bump anything in P(v).Define e 0 := v(i + 1).By Corollary 4.2, we see that For k > j, we can then define (if any) e k to be the first value in the one-line notation for v with c v (e k ) = c v (v(i + 1)) + k.Let r be maximal so that {e 0 , e 1 , . . ., e r } ⊆ Row 2 (P(v)).For all 0 ≤ k ≤ r, let t 0 := m, t 1 , . . ., t r be the values such that t k bumps e k in P(v).By Lemma 2.14(c) we have and these values appear from left to right in the one-line notation of v.
For a permutation v ∈ S n and a value q ∈ {1, . . ., n}, recall that we define LIS v (q) to be the length of a longest increasing subsequence of v that ends with q.The next lemma shows that the columns into which the values t k are first inserted are the same in P(v) and P(w), for 0 ≤ k ≤ r.Lemma 4.7.For all 0 ≤ k ≤ r, c v (t k ) = c w (t k ).
Proof.By construction, we have c v (t k+1 ) = c v (t k ) + 1 and c w (t k+1 ) > c w (t k ) for 0 ≤ k < r.Furthermore, since LIS w (t k ) ≤ LIS v (t k ), we know by Lemma 2.12 that c w (t k ) ≤ c v (t k ) for 0 ≤ k ≤ r.We prove the statement by induction on k.
First we show c v (t 0 ) = c w (t 0 ).By Corollary 4.2 and Remark 3.5(c), the first row of P i+j+1 (w) contains v(i) and e 1 , with no element between them.Because v(i) < m < e 1 , we know that m = t 0 bumps e 1 in P(w).Since c w (e 1 ) = c v (v(i + 1)) and t 0 bumps v(i + 1) in P(v), we have Next assume for some 0 ≤ k < r that c v (t k ) = c w (t k ).Then we have ), proving the statement.
Since e ∈ Row 1 (P(v)) occurs after v(i + 1) in v and does not bump anything in P(v), it follows that e = e k for some k > r.Therefore the value e r+1 exists, and by the definition of r, we have e r+1 ∈ Row 1 (P(v)) with e r+1 ≤ e.In fact, as a corollary to Lemma 4.7, we can show that e r+1 = e.
Proof.Since e r+1 is the only value in v inserted into column c v (e r+1 ) of P(v), we can apply Corollary 2.13 to conclude that e r+1 is in every longest increasing subsequence in v.By Corollary 3.6, this implies that every longest increasing subsequence in v ending with e r+1 must use both v(i) and v(i+1).As a result, LIS w (e r+1 ) = LIS v (e r+1 )−1.By Lemma 2.12, c w (e r+1 ) = c v (e r+1 )−1.We know c v (e r+1 ) − 1 = c v (t r ) by definition, and by Lemma 4.7, c v (t r ) = c w (t r ).Hence c w (e r+1 ) = c w (t r ).Since t r < e r+1 , we conclude that t r bumps e r+1 in P(w).Since e r+1 ∈ Row 1 (P(v)), it follows that e r+1 = e.
Next we show that e r and e are consecutive values.Lemma 4.9.With notation as above, e = e r + 1.
Proof.Since e occurs after e r and e r , e ∈ Row 2 (P (w)), Lemma 2.14(c) shows that e r < e. Suppose, for the purpose of obtaining a contradiction, that e = e r + 1, and so e > e r + 1.We analyze where e r + 1 could occur in the one-line notation of v. First we argue that e r + 1 cannot occur after e. Suppose it occurs after e.Before e r + 1 is inserted into P(v), e is in the first row and the element to the left of e is either e r or t r .Since t r < e r < e r + 1 < e, the value e r + 1 will bump e, which contradicts the fact that e ∈ Row 1 (P(v)).
Next we argue that e r + 1 cannot occur prior to e r .Suppose e r + 1 is to the left of e r .Before e r is inserted into P(v), if e r + 1 is in the first row, then e r will bump e r + 1, which contradicts Lemma 2.14(b).This forces e r + 1 ∈ Row 2 (P(v)), which, then, contradicts Lemma 2.14 (d).
Therefore e r + 1 must occur after e r and before e, which implies c v (e r ) < c v (e r + 1) < c v (e).However, this is impossible since c v (e r ) + 1 = c v (e).Hence e = e r + 1.
The maximality of M means that M + 1 appears to the right of v(i) in the one-line notation of v. Consider the set [M + 1, e r ] \ {v(i + 1) = e 0 , e 1 , . . ., e r }, which has e r − (M + 1) + 1 − (r + 1) elements.These elements occur after v(i) and are in Row 1 (P(v)), so they must bump (some of) the r elements {e 1 , . . ., e r } and nothing else, by definition of r.
This is a set of size e − M + 1, and we can use Lemma 4.9 and Equation (4.2) to get Moreover, the (r + 2)-element set {M, v(i + 1) = e 0 , e 1 , . . ., e r } is a subset of Row 2 (P(v)).
We are now able to state the main result.
Proof.As discussed above, there are r + 2 elements of the interval I in Row 2 (P(v)), and the interval I contains at most (2r + 3) elements.By Theorem 3.4, Corollary 4.8, and Lemma 4.9, there are r + 3 elements of the interval I in Row 2 (P(w)), which means that w must be crowded.
A corollary of this result is an alternate characterization of uncrowded permutations Corollary 4.11.Let w be a fully commutative permutation with boolean core ŵ.Then w is uncrowded if and only if P( ŵ) = P(w).

Minimal crowded permutations under the weak order
Consider the poset of fully commutative (that is, 321-avoiding) permutations in S n under the right weak order.The RSK partitions of such permutations have at most two rows, and we saw in Theorem 3.4 that the content of their second rows obeys a subset relation along covering relations in the weak order.We also saw, in Proposition 2.16, that a 2-row tableau is an insertion tableau for a boolean permutation if and only if it is an uncrowded tableau.
Recall that a fully commutative permutation w is called "uncrowded" if its insertion tableau is an uncrowded tableau.Otherwise a permutation is "crowded."Set Lemma 5.1.Consider the fully commutative elements of S n , partially ordered according to the right weak order.The uncrowded permutations form an order ideal of this poset, and the crowded permutations form a dual order ideal of this poset.
Proof.This follows from Theorem 3.4 and Proposition 2.16.
Thus we can identify this partition of the fully commutative permutations in S n by characterizing the maximal elements of the set uncrowded n or, equivalently, the minimal elements of the set crowded n .The minimal elements of this latter set satisfy a pattern containment condition.Before we state and prove that property, consider what it means for w to be a minimal element of crowded n : the fully commutative permutation w is crowded, while every fully commutative permutation ws i that it covers is uncrowded.
For the remainder of this section, we will assume that w is fully commutative; i.e., w is 321-avoiding.
We begin by recalling a standard definition: an integer d ∈ {1, . . ., n − 1} is a descent of w ∈ S n if w(d) > w(d + 1).
Lemma 5.2.Suppose that d is a descent of w, and that w(d + 1) does not bump w(d) during RSK insertion.Then P(w) = P(ws d ).In other words, if P(w) = P(ws d ), then either d is not a descent of w, or w(d + 1) bumps w(d) during RSK insertion.
Proof.Set v := ws d , and P := P d−1 (w) = P d−1 (v).Because w is 321-avoiding and d is a descent of w, the value w(d) must be larger than everything to its left in the one-line notation of w.Thus, in forming P d (w), this w(d) gets appended to the end of the first row of P , without bumping anything.
In forming P d+1 (w), the value w(d + 1), which is less than w(d) because d is a descent, will bump something.Let z be the value that it bumps; i.e., z is the smallest value in Row 1 (P d (w)) that is larger than w(d + 1).We know by assumption that z = w(d).In particular, z < w(d) and z appears to the left of w(d) in w.This last fact means that z ∈ Row 1 (P ).
In forming P d (v), the value v(d) = w(d + 1) bumps z from the first row of P to the second row.In forming P d+1 (v), the value v(d + 1) = w(d) is the largest value we have seen so far, so it gets appended to the end of the first row of P d (v), without bumping anything.Therefore P d+1 (w) = P d+1 (v), and hence P(w) = P(v).Somewhat akin to Lemma 5.2, we can make the following additional statement, which we phrase in terms of Knuth relations.Proof.The permutation w is 321-avoiding, so w(d)w(d + 1)w(d + 2) must be a 312-pattern.Knuth's theorem [Knu70] says that the insertion tableau is preserved under a Knuth relation, so P(w) = P(ws d ).
We now return to the characterization motivated by Lemma 5.1: identification of the minimal elements of crowded n in the poset of fully commutative permutations of S n .5.1.Consequences of minimality in crowded n .As it turns out, knowing that a permutation is minimal in the dual order ideal crowded n imposes substantial structure on the permutation.In this subsection, we will collect many of these consequences of minimality, with the ultimate goal of proving a characterization of minimality in Section 5.2.
Throughout this subsection we will consider permutations that are minimal elements of the dual order ideal crowded n , and we will identify features of the permutations that follow from that property.
We begin with an immediate corollary of Lemma 5.2.
Corollary 5.5.Let w be a minimal crowded permutation.Proof.Suppose, first, that some i < d is not fixed by w.Let i be minimal with this property, and let j be such that w(j) = i.Minimality of i means that j > i, and that j − 1 is a descent of w.By Corollary 5.5, the value i must bump w(j − 1) to the second row during RSK insertion.Moreover, this minimality means that w(d + 1) ≥ i.To avoid w(d)w(d + 1)i forming a 321-pattern in w, we must have that w(d + 1) = i.Minimality of i and the fact that d is the first descent mean that w(d + 1) = i < w(i) < w(i + 1) < • • • < w(d), and so i will actually bump w(i) during RSK insertion, contradicting the assumption that i < d and Corollary 5.5.Now suppose that some i > d + 1 is not fixed by w.Let i be maximal with this property, and let j be such that w(j) = i.Maximality of i means that j < i, and that j is a descent of w.And, by Corollary 5.5, this i must be bumped by w(j + 1) during RSK insertion.To avoid w(j)w(d )w(d + 1) forming a 321-pattern in w, we must have that j = d .Moreover, maximality of i > d means that w(j + 2) < i, and so Lemma 5.4 contradicts the minimality of w.
At this point, we have established several properties about the one-line representation of minimal elements of crowded n .In fact, we can go even further, showing that values in the interval [d, d + 1], in the language of Corollary 5.6 must be, in a sense, interwoven.
For the remainder of this subsection, define: Proof.First suppose that w is a minimal element of crowded n .Then Corollary 5.5 and Lemma 5.7 establish Properties (a) and (b).Property (c) is proved in Corollary 5.6, and Property (d) is a result of Corollaries 5.9 and 5.13.Finally, Property (e) follows from Lemma 5.14.Finally, we know from Property (a) and Corollary 5.5 that Row 2 (P(w)) = {w(d), w(d + 2), . . ., w(d + 2k)}.Now suppose that a permutation w has Properties (a)-(e) in the statement of the theorem.It follows from (a) and (c) and the fact that w is 321-avoiding that Row 2 (P(w)) = {w(d), w(d + 2), . . ., w(d + 2k)}.
This and Property (b) mean that w ∈ crowded n .It remains, now, to prove that w is minimal in that set.
Suppose, for the purpose of obtaining a contradiction, that w is not minimal in crowded n In fact, suppose that w is minimal with this property, meaning that anything covered by w is either not crowded, or minimal in crowded n .In particular, there must be at least one v = ws i in the latter category, by our assumption about w.Given Property (c), let us assume, without loss of generality, that d = 1 and d + 2k = n − 1.Because v < w, we have that w(i) > w(i + 1).Moreover, our assumptions about w mean that w(2j − 1) > w(2j) for all j, and each w(2j) bumps w(2j − 1) to Row 2 (P(w)).In particular, w(n − 1) = n and w(n − 3) = n − 1.On the other hand, Properties (a), (d), and (e) mean that in v, those bumping rules are no longer the case when 2j − 1 > i.Indeed, in v, it is w(2j + 2) that bumps w(2j − 1) to Row 2 (P(v)) when 2j − 1 ≥ i.Thus Row 2 (P(v)) = Row 2 (P(w)) \ {w(n − 1)}.
Since we have assumed that v ∈ crowded n is minimal, we know from previous results that Row 2 (P(v)) contains n − 1, n − 2 (which would have been w(n − 5), and either n − 3 or n − 4 (which would have been w(n − 7).Thus there can be no such v < w, and so w ∈ crowded n is minimal.

[ Sta12 ,
Solutions to Exercise 3.123(ab)].Given a reduced word [u] of a permutation, we can associate to [u] a heap, a poset whose elements are labeled by the simple reflections in [u].A heap diagram is the Hasse diagram for a heap in which poset elements are replaced by their labels.Definition 2.4.Given an arbitrary reduced word [u] = [u 1 • • • u ] of a permutation, consider the partial order on the set {1, . . ., } obtained via the transitive closure of the relations x ≺ y for x < y such that |u x − u y | ≤ 1.For each 1 ≤ x ≤ , the label of the poset element x is u x .This labeled poset is called the heap for [u].The Hasse diagram for this poset with elements {1, . . ., } replaced by their labels is called the heap diagram for [u].
Proposition 2.7 ([Ste96b, Proof of Proposition 2.2] and [Sta12, Solutions to Exercise 3.123(ab)]).Given a reduced word [u], the set of labeled linear extensions of the heap for [u] is the commutation class of [u].

Lemma 2. 11 .
Let w ∈ S n , and suppose b bumps z in the RSK insertion process for w.Then b < z and b appears to the right of z in the one-line notation of w.
is a subsequence of the one-line notation of w.So, since z i is bumped by b i , immediately before b i+1 is inserted, the value z i is still in the first row.This means that b i+1 must bump a number no larger than z i , which contradicts the assumption that b i+1 bumps z i+1 .Therefore b 1 < • • • < b t .Now say for some i that b i+1 occurs to the left of b i in the one-line notation for w.Since b i < b i+1 , we would have the 321-pattern z i+1 b i+1 b i in w, which is a contradiction.Hence b 1 , . . ., b t occur from left to right in w.(d) This follows from (c).
Therefore we get e r − M − (r + 1) ≤ r, and hence e r − M ≤ 2r + 1. (4.2) Now consider the interval uncrowded n := {w ∈ S n | w is fully commutative and uncrowded}, andcrowded n := {w ∈ S n | w is fully commutative and crowded}.The subset relation in Theorem 3.4 allows us to conveniently partition the fully commutative elements into two sets: uncrowded and crowded permutations.
Definition 5.3.Two permutations w and v differ by one Knuth relation if w is the result of replacing a consecutive 312-pattern in v by a consecutive 132-pattern (or vice versa), or replacing a consecutive 231-pattern in v by a consecutive 213-pattern (or vice versa).Lemma 5.4.Let d be a descent of w.If w(d + 2) < w(d), then P(w) = P(ws d ).
(a) Then d is a descent of w if and only if w(d + 1) bumps w(d) to the second row during RSK insertion.(b) Furthermore, every w(d) ∈ Row 2 (P(w)) is bumped by w(d + 1).Proof.It remains to prove Part (b).Suppose w(d) is an element of Row 2 (P(w)), bumped by w(j) with j > d + 1.Part (a) tells us that d is not a descent of w.Because w(d) > w(j), there exists a descent d ∈ [d + 1, j − 1].Lemma 5.2 implies that w(d + 1) bumps w(d ) during RSK insertion.The values w(j) and w(d + 1) cannot be equal, so d is in fact in [d + 1, j − 2].However, the fact that w(j)w(d + 1) is not a subsequence of w violates Lemma 2.14(c), and so in fact we must have j = d + 1. Lemma 5.4 and Corollary 5.5 impose rules on the values that are unaffected by bumping during RSK insertion.Corollary 5.6.Let w be a minimal crowded permutation, with first descent d and last descent d .Then the permutation w fixes all i ∈ [1, d − 1] ∪ [d + 2, n].