Positroid Catalan numbers

Given a permutation $f$, we study the positroid Catalan number $C_f$ defined to be the torus-equivariant Euler characteristic of the associated open positroid variety. We introduce a class of repetition-free permutations and show that the corresponding positroid Catalan numbers count Dyck paths avoiding a convex subset of the rectangle. We show that any convex subset appears in this way. Conjecturally, the associated $q,t$-polynomials coincide with the generalized $q,t$-Catalan numbers that recently appeared in relation to the shuffle conjecture, flag Hilbert schemes, and Khovanov-Rozansky homology of Coxeter links.

f , and thus f andf determine each other. See Figure 1 for an example and Section 2.1 for further details. For a bounded affine permutation f , let Π • f ⊂ Gr(k, n) denote the corresponding open positroid variety of the Grassmannian. Let T ⊂ PGL(n) denote the natural torus of diagonal matrices acting on Gr(k, n).
Definition 1.1. For an n-cyclef ∈ Cyc(n), define the positroid Catalan number C f := χ T (Π • f ) to be the torus-equivariant Euler characteristic of Π • f . These numbers are positive integers which can be computed via an explicit combinatorial recurrence; see Section 3.3. Additionally, they have the following interpretations: (a) C f is equal to the number of maximal f -Deograms introduced in [GL20], which are in bijection with a class of distinguished subexpressions in the sense of Deodhar [Deo85]; see Section 7.4. (b) C f is equal to the q = 1 evaluation of the polynomialR f (q) := R f (q)/(q − 1) n−1 , where R f (q) is the Kazhdan-Lusztig R-polynomial [KL79,KL80]; see Section 3. By [GL20, Theorem 1.11],R f (q) may be obtained as a coefficient of the HOMFLY polynomial [FYH + 85, PT87] ofβ f . (c) C f is equal to the q = t = 1 evaluation of the mixed Hodge polynomial P(Π • f /T ; q, t). By [GL20], P(Π • f /T ; q, t) is equal to a coefficient of the Khovanov-Rozansky triplygraded link invariant ofβ f . We showed in [GL20], using results on torus knots that date back to Jones [Jon87], that for gcd(k, n) = 1 and f given by f (i) = i + k, the positroid Catalan number C f recovers the famous (rational) Catalan number C k,n−k := 1 n n k which counts Dyck paths above the diagonal inside a k × (n − k) rectangle. This explains the nomenclature "positroid Catalan number" and points towards a deeper investigation of positroid Catalan numbers from a combinatorial perspective. In this work, we make the first step in this direction.
The set Cyc(n) is in bijection with n−1 k=1 Θ k,n .  An inversion of f ∈ Θ k,n is a pair (i, j) of integers such that i < j, f (i) > f (j), and i ∈ [n]. The length (f ) is the number of inversions of f . For an inversion (i, j) of f , let f (i,j) : Z → Z be obtained by swapping the values f (i) and f (j) (and repeating this for f (i + rn) and f (j + rn) for all r ∈ Z). We say that f (i,j) is obtained from f by resolving the crossing (i, j); see Figure 2. We letf (i,j) ∈ S n denote the permutation obtained by reducing f (i,j) modulo n.
Definition 1.2. For f ∈ Θ k,n , the inversion multiset Γ(f ) contains a point γ(f (i,j) 1 ) for each inversion (i, j) of f . We say that f is repetition-free if Γ(f ) is actually a set, that is, if it contains exactly (f ) distinct points.
See Figure 3 for an example. When we draw the set Γ(f ) inside a k × (n − k) rectangle, we swap the horizontal and vertical coordinates; cf. Notation 4.1.
2. Bounded affine permutations 2.1. Affine permutations. An (n-periodic) affine permutation is a bijection f : Z → Z satisfying the periodicity condition f (i + n) = f (i) + n. We letS n denote the group (under composition) of n-periodic affine permutations. Inversions, and the length function (f ) (see Section 1.2) are defined for any f ∈S n . For k ∈ Z, letS (k) n ⊂S n be the subset of affine permutations satisfying the condition n is the Coxeter group of affine type A. The group S n is usually called the extended affine Weyl group.
A bounded affine permutation is an affine permutation f ∈S n that satisfies the additional condition i ≤ f (i) ≤ i + n for all i ∈ Z. Denote by B k,n the (finite) set of bounded affine permutations inS (k) n , called the set of (k, n)-bounded affine permutations. We see that if f ∈ Cyc(n) and k = k(f ) then the associated bounded affine permutation f (cf. Section 1.1) belongs to B k,n . In other words, we have Θ k,n ⊂ B k,n .  Figure 5. Moves for computingR f (q) and C f .
n ⊂S n be given by i → i + k for all i ∈ Z. Then {f k,n | k ∈ Z} is exactly the set of length 0 elements inS n , and for 0 ≤ k ≤ n we have f k,n ∈ B k,n .
For i ∈ Z, let s i ∈S n be the simple transposition given by i → i + 1, i + 1 → i, and j → j for all j ≡ i, i + 1 modulo n. For f ∈ B k,n and i ∈ Z, we have ( Given f ∈S n , define the cyclic shift σf ∈S n by In other words, we have σf = f 1,n f f −1 1,n . Note that σ preserves each of the subsetsS (k) n , B k,n , and Θ k,n .

Conjugation and double move reduction.
Definition 2.1. We say that f ∈ B k,n has a double crossing Equivalently, for a := f −1 (i + 1), b := f −1 (i), c := f (i + 1), d := f (i), f has a double crossing at i if and only if a < b < i < i + 1 < c < d. See Figure 5(right). In this case, we say that s i f s i is obtained from f by a double move.
Definition 2.2. Let f ∈ B k,n , i ∈ Z, and f := s i f s i . If (f ) = (s i f s i ) and s i f s i ∈ B k,n then we say that f and f are related by a length-preserving simple conjugation. We say that f, g ∈ B k,n are c-equivalent and write f c ∼ g if f and g can be related by a sequence of length-preserving simple conjugations. See Figure 5(left).
The following result describes the structure of Θ k,n under double moves and c-equivalence.

Proposition 2.3.
(i) The minimal length elements of Θ k,n are of length d := gcd(k, n) − 1 and all such elements are related by cyclic shift (2.1) and c-equivalence. (ii) Any f ∈ Θ k,n can be reduced to a minimal length element of Θ k,n by double moves and c-equivalence.
2.3. Proof of Proposition 2.3. We deduce these statements from the results of He and Nie [HN14] and He and Yang [HY12]. Following [HN14], we introduce the following notation. For f, f ∈S n , we write f → f if there is a sequence f = f 0 , f 1 , f 2 , . . . , f r = f such that f j = s i j f j−1 s i j for j = 1, 2, . . . , r, Lemma 2.4. Let f ∈ Θ k,n and f ∈S n be such that f → f . Then f ∈ Θ k,n .
Proof. Suppose f ∈ Θ k,n and f = s i f s i satisfies (f ) ≤ (f ). Since f ∈ Θ k,n , we have f (j) ∈ [j + 1, j + n − 1] for all j ∈ Z. It follows from this that f is also a bounded affine permutation, and thus f ∈ Θ k,n . Proof. For f ∈ Θ k,n , the imagef ∈ S n is an n-cycle. The n-cycles are elliptic elements in Proposition 2.7. The elements of Θ k,n all belong to a singleS n -conjugacy class O inS n .
We are ready to finish the proof of Proposition 2.3. By Proposition 2.7, there is anS nconjugacy class O ⊂S n containing Θ k,n . SinceS n =S It is easy to see that c(f k,n ) = gcd(k, n), where for f ∈S n , we denote by c(f ) := c(f ) the number of cycles of the permutationf . Now, for f ∈S n , we have c(s i f ) = c(f s i ) ∈ {c(f ) + 1, c(f ) − 1}. It follows that for f ∈ Θ k,n , we have (f ) ≥ gcd(k, n) − 1. On the other hand, it is easy to see that f k,n s 1 s 2 · · · s gcd(k,n)−1 ∈ Θ k,n . Thus the minimal length of f ∈ Θ k,n is d := gcd(k, n) − 1. Since c(·) is invariant under conjugation, we find that any f ∈ Θ k,n with (f ) = d has minimal length in itsS  n -conjugacy class containing f and g. Since (f ) = (g) = d is minimal, by Proposition 2.6, we get f c ≈ g. By Lemma 2.4, having f c ≈ g for f ∈ Θ k,n implies that f c ∼ g. This proves Proposition 2.3(i). As we showed above, the minimum length elements of Θ k,n are also minimum length elements in theirS in Π • f over a finite field F q with q elements (where q is a prime power). These R-polynomials are special cases of the R-polynomials of Kazhdan and Lusztig [KL79,KL80].
Proposition 3.1. The polynomials R f (q), f ∈ B k,n , may be computed from the following recurrence.
(a) If n = 1 then R f (q) = 1. (b) Iff has some fixed points then R f (q) = R f (q), where f is obtained from f by removing all fixed points off .
(e) If f has a double crossing at i ∈ Z then Proof. The results of [MS16] are formulated in the language of cluster algebras. For the convenience of the reader, we give an alternative proof of (a)-(e) not relying on cluster algebras, assuming familiarity with [KLS13]. We start by noting that the definition f under the natural projection map Gr(k, n) → Gr(k, n − 1) (resp., Gr(k, n) → Gr(k − 1, n − 1)) between Grassmannians that removes (resp., contracts) the i-th column. See e.g. [Lam16, Lemmas 7.8 and 7.9].
The Kazhdan-Lusztig R-polynomials R v,w (q) are indexed by pairs (v, w) of permutations. When v ≤ w (where ≤ denotes the Bruhat order on S n ), we have R v,w (q) = 0, and for v = w, we have R v,w (q) = 1. For v ≤ w ∈ S n , R v,w (q) can then be computed by a recurrence relation [KL79, Section 2]: Here, s = s i for some 1 ≤ i ≤ n − 1 is a simple transposition satisfying sw < w.
, where τ k,n ∈S n denotes a certain translation element; see [KLS13, Proposition 3.15]. From this, (3.2) implies (d)-(e) whenever we have a length-preserving simple conjugation or a double crossing at 1 ≤ i ≤ n − 1. Applying the cyclic shift, we see that properties (d)-(e) hold also for i = 0, which completes their proof.
Finally, a constructive algorithm to compute R f (q) from (a)-(e) is given in the proof of [MS16, Theorem 3.3].  Figure 6. Some examples ofR f (q) and C f .

Positroid Catalan numbers.
Recall that for a permutationf ∈ S n , we let c(f ) = c(f ) denote its number of cycles. For f ∈ B k,n , we let It is easy to see (for example using (3.5) below; see also [GL20, Proposition 4.5]) thatR f (q) is always a polynomial in q.
The definition of a positroid Catalan number C f (Definition 1.1) can be extended to all f ∈ B k,n by setting The relation to Definition 1.1 is given in Section 7.1. See Figure 6 for examples.
3.3. Recurrence for positroid Catalan numbers. If f has a double crossing at i ∈ Z then (3.1) implies Here, i, i + 1 are considered modulo n. The next result follows from Proposition 3.1 combined with (3.4)-(3.5).
Proposition 3.2. The positroid Catalan numbers C f , f ∈ B k,n , may be computed from the following recurrence.
Proposition 3.3. Let f ∈ B k,n . Then C f is a positive integer.
Proof. The proof of [MS16,Theorem 3.3] shows that C f may be expressed using (a )-(e ) in terms of C g for bounded affine permutations g satisfying either n(g) < n(f ) or n(g) = n(f ) and (g) > (f ). In particular, the recurrence in Proposition 3.2 is subtraction-free, which shows the result. See also [GL20, Remark 9.4 and Proposition 9.5].
Remark 3.4. It is not always true thatR f (q) has positive coefficients: see [GL20,Example 4.22]. This question is closely related to the odd cohomology vanishing phenomenon which appears for gcd(k, n) = 1 and f = f k,n (i.e., for torus knots) but not for all f ∈ Θ k,n . It is an important open problem to describe a wider class of positroids (or more generally, knots) for which this phenomenon occurs. We expect this class to contain all f ∈ Θ k,n which are repetition-free; see Conjecture 7.1. (1) nr ) is a product of r cycles. (The case r = 2 was considered in Section 1.2.) For each j ∈ [r], denote by f | S j ∈ B(k j , n j ) the restriction of f to the set S j of all integers congruent to one of a (j) 1 , . . . , a (j) n j modulo n. We deduce the following decoupling property from Proposition 3.2.
Corollary 3.5 (Decoupling). Let f ∈ B k,n and i ∈ Z. If i and i + 1 belong to different cycles off then s i f s i ∈ B k,n and In particular, in the above notation, for f ∈ B k,n we have Proof. Eq. (3.7) follows easily from Proposition 3.2. To deduce (3.8), we apply (3.7) repeatedly until each cycle off is supported on a cyclically consecutive interval [a, b] ⊂ [n] for some a, b ∈ [n].
After that, C f may be computed via Proposition 3.2 independently on each interval, which results in the product formula (3.8).
We will use a special case of (3.8) when r = 2.
Corollary 3.6. Suppose that f ∈ Θ k,n has a double crossing at i ∈ [n]. Then Our eventual goal will be to relate (3.9) to the recurrence for Dyck paths shown in Figure 12.
One other simple result we will need is the cyclic shift invariance of C f and Γ(f ).
Proposition 3.7. For any f ∈ B k,n , we have Proof. It is obvious that both the definition of Γ(f ) and the recurrence in Propositions 3.1 and 3.2 are invariant under the action of σ.

Big paths
The next few sections contain the main body of the proof of Theorem 1.3. From now on, we switch from working in the (k, n − k)-coordinates to working in the (k, n)-coordinates. For f ∈ Θ k,n , we let  f = (0, 3, 2, 5, 1, 4) in cycle notation P = P (f ) Figure 7. Computing the small path P (f ) . Its points are labeled according to Notation 5.4. and define the multiset Γ (f ) to be the image of Γ(f ) under the map (k 1 , n 1 − k 1 ) → (k 1 , n 1 ). We Our goal is to give a geometric interpretation of the multiset Γ (f ).
Notation 4.1. When referring to points in the plane, we swap their coordinates. For a point α = (a, b) ∈ Z 2 , we denote by n(α) := b (resp., k(α) := a) its horizontal (resp., vertical) coordinate. Figure 7. We usually drop the superscript and denote P ∞ := P (f ) ∞ . We refer to the points p f,r for r ∈ Z as the integer points of P ∞ .
Set δ := (k, n) and choose some α ∈ Z 2 . We will be interested in the intersection points of P ∞ with Q ∞ := P ∞ + α. First, observe that if α ∈ Zδ then P ∞ = Q ∞ . If α / ∈ Zδ then it is easy to see that no integer point of P ∞ belongs to Q ∞ , and that the set P ∞ ∩ Q ∞ is invariant under adding multiples of δ. We denote by |P ∞ ∩ Q ∞ | the size of this set when considered "modulo δ," that is, as a subset of the cylinder 1 Z 2 /Zδ. For l := |P ∞ ∩ Q ∞ |, we say that P ∞ and Q ∞ intersect l times. The number l is always finite and even.
Proposition 4.3. Let f ∈ Θ k,n . Then f is repetition-free if and only if for all α ∈ Z 2 \ Zδ, P ∞ and Q ∞ := P ∞ + α intersect at most two times. In this case, we have We will prove the more general statement that for all f ∈ Θ k,n , the multiplicity of α in the multiset Γ (f ) is given by 1 2 |P ∞ ∩ Q ∞ |. Indeed, suppose that P ∞ crosses Q ∞ from below at some non-integer point x. (That is, P ∞ is below Q ∞ when approaching x from the left and above Q ∞ when approaching x from the right.) Then x belongs to the segment of P ∞ connecting p f,r to p f,r+1 and to the segment of Q ∞ connecting α + p f,r−b to α + p f,r−b+1 , where α = (a, b). Let i := f r (0) and j := an + f r−b (0). Then we have i < j and f (i) > f (j), and thus (i , j ) := (i−tn, j −tn) form an inversion of f , where t ∈ Z is such that i−tn ∈ [n]. Moreover, it is easy to see that δ(f (i ,j ) 1 ) = α, where δ(·) was defined in (4.1). Conversely, given an inversion (i, j) of f with δ(f (i,j) 1 ) = α, we may find a (unique modulo n) index r ∈ Z such that f r (0) ≡ i modulo n, and we can also find a (unique modulo δ) shift α ∈ Z 2 such that Q ∞ passes through the point p f,r + ( j−i n , 0). This shows that the inversions We say that a multiset Γ is centrally symmetric if for each α ∈ [k − 1] × [n − 1], the multiplicities of α and of δ − α in Γ coincide.
Since p f,r+n = p f,r + δ for all r ∈ Z, we have δ ⊥ , P ∞ = j+n−1 r=j δ ⊥ , p f,r for all j ∈ Z, and similarly for δ ⊥ , Q ∞ . In particular, we have Observe that for each r ∈ Z, q f,r is above P ∞ if and only if it is above p f,r , which happens if and only if δ ⊥ , q f,r − p f,r > 0, since the vertical coordinate of δ ⊥ is positive. Thus (i)-(ii) follow, and (iii) follows by combining (i)-(ii) with (the proof of) Proposition 4.3, since if slope(α) = slope(δ) then Q ∞ contains integer points both below and above P ∞ , and therefore intersects P ∞ .

Convexity of the inversion multiset
Similarly to Section 1.3, we say that Γ (f ) is convex ifΓ (f ) contains all lattice points of its convex hull. (These sets were defined in the beginning of Section 4.) The goal of this section is to prove the following result.
Theorem 5.1. Let f ∈ Θ k,n be repetition-free. Then the set Γ (f ) is convex.
We start by stating some consequences of the results obtained in Section 2. Let | slope(α) = slope(δ)}. By Lemma 4.6(iii), we have Γ min k,n ⊆ Γ (f ) for all f ∈ Θ k,n . The next two statements follow directly from Proposition 2.3.
Corollary 5.2. Let f ∈ Θ k,n be repetition-free. Then at least one of the following holds: • Γ (f ) = Γ min k,n . • There exists g ∈ Θ k,n such that f c ∼ g and g has a double crossing at some i ∈ Z.
Notation 5.4. For each 0 ≤ r < n, let 0 ≤ j r < n be the unique index equal to f r (0) modulo n. Then we label p f,r by P [j r ] as in Figure 7. We extend this to all r ∈ Z using the convention that j r+n := j r + n, and we label p f,r by P [j r ] for r ∈ Z. Thus P [j + n] = P [j] + δ for all j ∈ Z. If P [i] appears to the left of P [j] for some i, j ∈ Z, we denote by P [i → j] the subpath of P ∞ connecting P [i] to P [j]. Thus P = P [0 → n] and we will be particularly interested in the subpaths P [0 → 1] and P [1 → n] of P (under the above assumption that f has a double crossing at 0).
Notation 5.7. Observe that the points P [1] and P [0] + δ 2 differ by (1/n, 0). Moreover, the two paths P ∞ and P ∞ + δ 2 form a double crossing at these two points, thus they form a small region as in Figure 8. Therefore no shift of P ∞ can contain an integer point in this region. In our analysis, we usually treat this region as a "single point" and write P [1] ≈ P [0] + δ 2 and P [n] ≈ P [1] + δ 1 . By an abuse of terminology, we will say that P ∞ is below P ∞ + δ 2 and above P ∞ + δ 1 .
Lemma 5.8. The bounded affine permutations f 1 and f 2 are repetition-free.
Proof. Let us compare the big path P (f 1 ) ∞ with (P [1 → n]) ∞ := t∈Z (P [1 → n] + tδ 1 ), where we identify the points P [n] + (t − 1)δ 1 ≈ P [1] + tδ 1 for all t ∈ Z. It is easy to see that these two paths are equivalent in the sense that for each α = (a, b) ∈ Z 2 , we have where the intersection points are counted modulo δ 1 . Thus we need to analyze the intersections of (P [1 → n]) ∞ with its shifts.
Note that P [1 → n] has low slope since it connects P [1] to P [n] ≈ P [1]+δ 1 while P [0 → 1] has high slope since it connects P [0] to P [1] ≈ P [0] + δ 2 ; cf. Lemma 5.5(ii).The next result states that a shifted segment of high slope cannot cross a segment of low slope from above.
Proof. Suppose otherwise that Q[0 → 1] crosses P [1 → n] from above. We consider the cases according to the positions of Q[0] and Q[1] relative to P ∞ . First, assume that Q[0] is below P ∞ and Q[1] is above P ∞ . Then Q[0 → 1] intersects P ∞ at least 3 times, a contradiction.
Definition 5.10. For an integer point q of Q, we say that q is vertically above P if there exists an integer point p of P with n(q) = n(p), and q is above p.
Since Q [0 → 1] intersects P [1 → n], it follows that Q [0] is vertically above P . Consider the path Q[1 → 1 + n]. It crosses P ∞ from above at a single point which belongs to Q[n → 1 + n] ∩ P [1 + n → 2n]. Moreover, it stays below Q which crosses P [1 → n] from above. Thus Q[1 → 1 + n] crosses P from above. Since Q[1 → 1 + n] cannot cross P [1 → 1 + n] from above, it must cross P from below. The remaining part of Q[1 → 1 + n] still has to cross P ∞ from above, however, it cannot cross P ∞ since it has already crossed P twice. Since P ∞ is below P ∞ , we get a contradiction. See Figure 9. Proof. Consider the two infinite unions R α := P [a → b] + Zα and R β := P [c → d] + Zβ. Observe that R α (resp., R β ) is a path-connected subset of R 2 . Thus it contains an infinite piecewise linear curve S α (resp., S β ) such that for each r ∈ Z, S α (resp., S β ) contains a unique point x α,r (resp., x β,r ) satisfying n(x α,r ) = n(x β,r ) = r. Here, we are additionally assuming that the vertical coordinates of x α,r and x β,r are increasing functions of r.
When r 0, x α,r is below x β,r , and when r 0, x α,r is above x β,r . Let r ∈ Z be the smallest integer such that x α,r is not below x β,r . Thus x α,r−1 is below x β,r−1 and either x α,r = x β,r or x α,r is above x β,r . In each case, it is straightforward to check that a shift P [a → b] + sα (passing through either x α,r or x α,r−1 or both) crosses a shift P [c → d] + tβ (passing through either x β,r or x β,r−1 or both) from below.
Given two paths Q, P , we say that Q is above P if whenever two integer points q ∈ Q and p ∈ P satisfy n(q) = n(p), we have that q is above p. Proof. Assume otherwise that they intersect twice. Our temporary goal is to show that We observe that Q satisfies the following properties: . In order to show (5.1), it suffices to prove that if Q satisfies (a)-(c) then either ). If (i) holds for Q then we proceed by induction, applying the same argument to Q − tδ 2 for t = 1, 2, . . . , until we find that (ii) holds for some Q − sδ 2 with s > 0. But then all integer points of P [0 → 1] are below s t=0 (Q − tδ 2 ), which proves (5.1). Assume that Q satisfies (a)-(c). Since Q[1 → n] and P [1 → n] intersect twice, Q[0 → 1] is above P ∞ and P [0 → 1] is below Q ∞ . Then Q [0 → 1] is above P ∞ , where P := P − δ 2 . We have the following situation: • apart from the double crossing, Q [1 − n → 1] is below Q ∞ ; • Q [1 − n] and Q [1] are above P ∞ ; is below Q ∞ . These statements imply that Q [1 − n → 1] crosses P [−n → 0] twice, first from above and then from below. Moreover, the second crossing (from below) must belong to P [1 − n → 0] since it has to come after both crossings of Q[1 − n → 0] with P [1 − n → 0]. In particular, Suppose that n(Q [1]) < n(P [1]). Then n(Q [0]) < n(P [0]). We have just shown that Q [0 → 1] is above P [0 → 1], so we arrive at case (ii).
Suppose now that n(Q [1]) ≥ n (P [1]). Then Q satisfies (a) and (c). Moreover, we also have n(Q [1 − n]) ≥ n(P [1 − n]) and n(Q[0]) ≥ n(P [0]). In view of the above statements, this implies that Q [1 − n → 0] intersects P [1 − n → 0] twice, i.e., Q also satisfies (b). We arrive at case (i). We are done with the proof of (5.1), and now we will use it to finish off the proof of the lemma.
The following straightforward result describes a natural transformation that swaps the notions of "above" and "below." We refer to it as the 180 • -rotation.
Proposition 5.15. For f ∈ Θ k,n , let g : Z → Z be given by Then g ∈ Θ k,n and the paths P (f ) and P (g) are related as For each point x ∈ R 2 , x is above (resp., below) P Given α, β ∈ R 2 , we say that α is weakly southwest of β and write α β if n(α) ≤ n(β) and k(α) ≤ k(β). We write α ≺ β if α β and α = β.
Case 2: t ≥ 0 and Q[0] is below P ∞ . Thus Q[0 → 1] intersects P ∞ once, and therefore so does Q[1 → n]. Thus Q[1 → n] must also intersect R[1 → n]. Since Q[1] is above P ∞ , In order for Q[0 → 1] to intersect L (<0) , it must intersect P . We claim that q belongs to P (t−1) ∞ . Indeed, suppose otherwise that q ∈ P (t) ∞ . Since Q[0 → 1] has to intersect L but it can no longer intersect P (t) ∞ , the first intersection point of Q[0 → 1] with L has to belong to L (s) for some s > t. In order for this to happen, Q[0 → 1] must intersect P (s) ∞ twice (with the second crossing at ), and therefore the remaining part of Q[0 → 1] will stay below P (s) ∞ , and thus below L. We see that Q[1] is below L, contradicting our assumption. Thus q ∈ P (t−1) ∞ . Specifically, we have q ∈ (P [1 → 1 + n] + (t − 1)δ 2 ), which is the lower boundary of the region bounded by P ∞ . Since Q[1 → n] stays below Q and intersects P (t) ∞ , we see that Q[1 → n] stays below P (t+1) ∞ , and the unique intersection point of Q[1 → n] with P (t) ∞ belongs to P [1 → 1 + n] + tδ 2 . Since t < 0, P [1 → n] stays above P Using Lemma 5.16, the result of Lemma 5.14 can be strengthened as follows.
Proof of Theorem 5.1. We proceed by induction on k and n. Suppose that the statement is known for all smaller k and n, and consider some lattice point x / ∈Γ (f ) which belongs to the convex hull ofΓ (f ). By Lemma 5.19, we have x / ∈ (∆ 1 ∪ ∆ 2 ) Z . By Corollary 5.23, δ 1 and δ 2 are vertices of the convex hull ofΓ (f ). By the induction hypothesis, we know that the sets Γ (f 1 ) and Γ (f 2 ) are convex. We obtain a contradiction with Corollary 5.21, so we must have x ∈Γ (f ).

Concave profiles and the counting formula
By Corollary 4.4 and Theorem 5.1, if f ∈ Θ k,n is repetition-free then Γ(f ) is convex and centrally symmetric. In this section, we show that each convex centrally symmetric set arises in this way, as stated in Theorem 1.3(ii). We will use this construction to prove the counting formula (1.2) in Section 6.2, completing the proof of Theorem 1.3. 6.1. Concave profiles.
Definition 6.1. A sequence H := (0 = H 0 , H 1 , . . . , H n = k) of real numbers is called a concave profile if Given a concave profile H, we let As before, we letΓ (H) := Γ (H) {(0, 0), (k, n)}. We also let P (H) be the path connecting the points (r, H r ) for r = 0, 1, . . . , n. ThusΓ (H) consists of all lattice points weakly below P (H) and weakly above the 180 • -rotation (k, n) − P (H) of P (H) . Proof. Choose a nonnegative strictly concave sequence = ( 0 , 1 , . . . , n ) whose values are sufficiently small, and let H be such that the difference H − records the maximal vertical coordinates of the intersection of the convex hull ofΓ with the vertical line n(x) = i for i = 0, 1, . . . , n. Then clearly H is a concave profile and we have Γ = Γ (H). See Figure 11 for an example.
The following construction uses H to find a bounded affine permutation f H ∈ Θ k,n satisfying the desired properties. Definition 6.3. Given a concave profile H, let f = f H ∈ Θ k,n be the unique bounded affine permutation such that for all 0 ≤ i, j < n, we havef i (0) <f j (0) if and only h i < h j , where h i , h j are defined in (6.1). In other words, writingf = (0, j 1 , j 2 , . . . , j n−1 ) in cycle notation, the indices (j 1 , j 2 , . . . , j n−1 ) have the same relative order as (h 1 , h 2 , . . . , h n−1 ). See Figure 11 for an example.
Proposition 6.4. Let H be a concave profile and f := f H . We have: Proof. (i): We prove the result by induction on r. The base case r = 0 is clear. Suppose that the result holds for 0 ≤ r < n. We have h r+1 = h r . If h r+1 > h r thenf r+1 (0) >f r (0), and thus f r+1 (0) and q ∈ P (H) + α have the same horizontal coordinate r ∈ Z then p is above q if and only if H r > a + H r−b . This condition is equivalent to having either H r > a + H r−b or H r = a+ H r−b and h r > h r−b . By (i), this is equivalent to having f r (0) n > a+ f r−b (0) n , which means that for the integer points p ∈ P (f ) and q ∈ P (f ) + α satisfying n(p ) = n(q ) = r, the point p is above q . Since the path P (H) is the plot of a concave function, it intersects P (H) + α at most once for each α ∈ Z 2 . Thus P ∞ + α if and only if α is below P (H) and (k, n) is below P (H) + α. This is equivalent to α ∈ Γ (H). Since |P (H) this is equivalent to α ∈ Γ (f ). 6.2. Counting formula for concave profiles. We prove (1.2) in two steps. We start by treating the case where f = f H arises from a concave profile. The case of arbitrary repetition-free f ∈ Θ k,n is considered in Section 6.3 below.
Proposition 6.5. Let H be a concave profile and let f := f H . Then Proof. Let us say that a slanted Dyck path is a lattice path connecting (0, 0) to (k, n) which stays above the main diagonal and consists of right steps (0, 1) and up-right steps (1, 1). Thus # Dyck(Γ(f )) counts the number of slanted Dyck paths which stay above P (H) (and do not share any points with P (H) except for the endpoints (0, 0) and (k, n)).
In order to keep track of the size of the rectangle in which H lives, let us refer to H as a (k, n)-concave profile. We proceed by induction on n using Proposition 3.2. The base case n = 1 is clear. Suppose now that n > 1 and that the claim has been shown for all n < n and also for all (k, n)-concave profiles H satisfying Γ (H ) Γ (H). Let 0 < r < n be the index such that 0 < h r < 1 is maximal among h 0 , h 1 , . . . , h n . Thusf r (0) = n − 1, and we let := 1 − h r . Assume first that r = 1. Let g ∈ Θ k−1,n−1 be given byḡ i (0) :=f i+1 (0) for 0 < i < n, and let H := (0 = H 0 , H 1 , . . . , H n−1 = k − 1) be given by H i := H i+1 + − 1 for 0 ≤ i < n − 1. It is easy to check that g = f H and that removing the first step (which must be up-right) of a slanted Dyck path above P (H) yields a slanted Dyck path above P (H ) and vice versa, thus # Dyck(Γ(f )) = # Dyck(Γ(g)). Applying parts (b )-(c ) of Proposition 3.2, we find C f = C g . Assume next that r = n − 1. Let g ∈ Θ k,n−1 be given byḡ i (0) :=f i (0) for 0 ≤ i < n − 1, and let H := (0 = H 0 , H 1 , . . . , H n−1 = k) be given by H i := H i + for 0 < i < n. Similarly to the above, we have # Dyck(Γ(f )) = # Dyck(Γ(g)) and C f = C g .
Let f := s i f s i . Our goal is to relate C f to C f 1 , C f 2 , and C f as shown in Figure 12. It follows that f ∈ Θ k,n and that f has a double crossing at i. Let f 1 , f 2 be obtained from f by resolving the crossing (i, i + 1). By Corollary 3.6, we have Let g = σf ∈ Θ k,n be the cyclic shift of f defined in (2.1). We have f r (0) ≡ n − 1 and g r (0) ≡ 1 modulo n, and for 1 ≤ s ≤ n such that s = r, we have g s (0) = f s (0) + 1. Choose > such that < 1 − h s for s = r and let H := (0 = H 0 , H 1 , . . . , H n = k) be given by H s := H s + for all 0 < s < n. One easily checks that g = f H . Since Γ (g) Γ (f ), by the induction hypothesis, we have C g = # Dyck(Γ(g)). By Proposition 3.7, we have C f = C g and Γ (f ) = Γ (g), thus C f = # Dyck(Γ(f )). It is straightforward to check that there exist concave profiles H (1) and H (2) such that f 1 = f H (1) and f 2 = f H (2) , thus by the induction hypothesis, (6.2) becomes (6.3) C f = # Dyck(Γ(f 1 )) · # Dyck(Γ(f 2 )) + # Dyck(Γ(f )).  Figure 15. Associating a knot K f inside T 2 × R to f ∈ Θ k,n . The dashed rectangle on the right represents the fundamental domain of T 2 .
corresponds to a Young diagram λ(w) that fits inside a k × (n − k) rectangle. An f -Deogram is obtained by placing a crossing or an elbow inside each box of λ(w) so that (i) the resulting strand permutation is v, and (ii) a certain distinguished condition is satisfied. An f -Deogram is maximal if it contains the maximal possible number of crossings, equivalently, assuming f ∈ Θ k,n , if it contains exactly n − 1 elbows. In view of Theorem 1.3(i), when f is repetition-free, C f = # Dyck(Γ(f )) also counts Dyck paths avoiding Γ(f ).
Problem 7.6. Let f ∈ Θ k,n be repetition-free. Find a bijection between Deo max f and Dyck(Γ(f )). 7.5. Fiedler invariant and knots in a thickened torus. Let T 2 := R 2 /Z 2 be a torus and K : S 1 → T 2 × R be a knot inside a thickened torus. To this data, Fiedler [Fie93] associates an isotopy invariant W K called the small state sum. Let us instead identify T 2 with R 2 / (0, n), (1, 0) . For f ∈ Θ k,n and P := P (f ) , letP be the image of P ⊂ R 2 under the quotient map R 2 → T 2 . The points whereP intersects itself correspond precisely to the inversions of f . Thus we may define a knot K f inside T 2 ×R whose projection to T 2 coincides withP , and for each inversion (i, j) of f , the line segment connecting P [i] to P [f (i)] lies above the line segment connecting P [j] to P [f (j)]. See Figure 15.
It is straightforward to check that the formal sum W K f contains essentially the same information as the inversion multiset Γ (f ). This leads to the following question: which parts of our story generalize to arbitrary repetition-free knots inside T 2 × R? Here we say that a knot K inside T 2 ×R is repetition-free if each nonzero coefficient of W K f is equal to ±1. For example, it would be interesting to determine which subsets of Z 2 /Zδ may appear with nonzero coefficients inside W K for a repetition-free K, and whether the HOMFLY polynomial of K (or its Khovanov-Rozansky homology) have nice properties when K is repetition-free.