Quotients of uniform positroids

Flag matroids are a rich family of Coxeter matroids that can be characterized using pairs of matroids that form a quotient. We consider a class of matroids called positroids, introduced by Postnikov, and utilize their combinatorial representations to explore characterizations of flag positroids. Given a uniform positroid, we give a purely combinatorial characterization of a family of positroids that form quotients with it. We state this in terms of their associated decorated permutations. In proving our characterization we also fully describe the circuits of this family.


Introduction
If two matroids M 1 and M 2 on the same ground set of ranks r 1 , r 2 , respectively, are such that every circuit of M 2 is a union of circuits of M 1 , then we say that M 1 is a quotient of M 2 , or that M 1 and M 2 are concordant. Moreover, such pair of matroids are said to form a (two step) flag matroid. The main contribution of this paper is to utilize the combinatorics of a special family of matroids, called positroids, in order to characterize when certain pairs of positroids form a quotient, and thus form a flag positroid.
Introduced in [12], positroids have proven to be a combinatorially exciting family of matroids. One may describe positroids via several combinatorial objects such as Grassmann necklaces, decorated permutations, and Le-diagrams [8,12]. With such rich combinatorics, one can ask if certain matroidal properties may be better understood in the case of positroids through any of these objects.
Taking into account the fact that the uniform matroid U k,n is always a positroid, we combinatorially describe a family of positroids of rank k − 1 that are quotients of U k,n . Our characterization is a complete one for all possible quotients when n < 6. Our main result states that positroids of rank k − 1 that are a quotient of U k,n can be obtained from the decorated permutation of U k,n after performing a cyclic shift on some of its values. We conjecture that all positroid quotients can be described in a similar way.
Our results provide a partial answer to the problem stated in [8], namely, determine combinatorially when two positroids are concordant. Additionally, our work includes a concrete description of the circuits of this family of positroids concordant to U k,n .
A strong motivation for this work is the result of [2] which proves that positively oriented matroids are realizable. In this spirit, we believe our results provide a better understanding of certain flag positroids which can be useful in determining the realizability of positively oriented flag matroids.
In order to understand quotients of positroids, we introduce the poset of positroid quotients, whose elements are positroids on the same ground set and whose covering relation is given by N M if and only if N is a quotient of M and their ranks differ by one.
We conclude with a conjecture establishing a necessary condition for two arbitrary positroids to form a quotient. This conjecture is stated in terms of decorated permutations as well.
The paper is organized as follows. In Section 2 we provide the necessary background on positroids and quotients of matroids. In Section 3 we introduce the poset of quotients of positroids and explore some of its combinatorics. We also characterize families of positroids that are quotients of uniform positroids, and conjecture a general combinatorial rule for positroid quotients. In Section 4 we end with future work and some further questions.

Preliminaries
Matroids are combinatorial objects that generalize the notion of linear independence and can be defined through several equivalent ways. We suggest [9] for a wider view on matroid theory.
From here onwards we denote the set {1, 2, . . . , n} by [n] and the k-subsets of [n] by [n] k . Whenever there is no room for confusion, we will denote the set {a 1 , . . . , a n } by a 1 . . . a n .

Matroids
Definition 1. A matroid M is an ordered pair (E, B) that consists of a finite set E and a collection B of subsets of E that satisfies the following conditions: The set E is the ground set of M and the collection B := B(M ) is called the set of bases of M .
It can be shown that every element of B has the same cardinality, denoted r M , which is called the rank of M . We will say that a subset I of E is independent in the matroid M = (E, B) if there exists B ∈ B such that I ⊆ B. In particular, notice that ∅ is always independent. By I(M ) we denote the collection of independent sets of a matroid M . If I is not independent we call it dependent. In particular, a minimally dependent subset C of E is called a circuit of M . That is, C is dependent in M but every proper subset of C is independent. We denote by C(M ) the collection of circuits of the matroid M .
A classic example of a matroid, and in fact our main object of study, is the uniform matroid.
Definition 2. Let n be a positive integer and 0 k n. The uniform matroid of rank k on [n], denoted U k,n , is the ordered pair U k,n = [n], [n] k . That is, the bases of U k,n are all the k-subsets of [n]. The reader can see that the dual of a matroid of rank k is also a matroid and has rank n − k. More over, U * k,n = U n−k,n .

Quotients and flag matroids
In this paper we are concerned with quotients of a particular class of matroids that will be defined in Section 2.3. Thus we now recall the notion of quotients in matroid theory.
Definition 4. Given two matroids M and N on the same ground set E, we say that M is a quotient of N if every circuit of N can be expressed as the union of circuits of M .
Example 5. The matroid U 1,3 is a quotient of U 2,3 . This is clear since the only circuit of U 2,3 is the set {1, 2, 3}, which can be written as the union of {1, 2}, {2, 3}, {1, 3}. The latter are the circuits of U 1,3 . In general, U k,n is a quotient of U ,n as long as k .
Definition 4 has been studied in Chapter 8 of [16] where, given two matroids M and N on the same ground set, the property that M is a quotient of N is equivalent to the identity being a strong map. In fact, quotients can be defined in many equivalent ways. We present the following proposition whose proof we omit (see [16,Prop. 8.1.6]). Proposition 6. Let M and N be matroids on the same ground set E. The following statements are equivalent: (c) For any pair of subsets A and B of E, such that A ⊂ B, it follows that Notice that due to Proposition 6(c), if we take A = ∅ it follows that r N (B) r M (B) whenever M is a quotient of N . Moreover, equality holds in this case for all B at the same time if and only if M = N . In view of this, the following definition is in order.
Definition 7. Let M 1 , . . . , M k be a collection of distinct matroids on the ground set [n], of respective ranks 1 r 1 , r 2 , . . . , r k n. If for every 1 i < j k it holds that M i is a quotient of M j , we say that the collection {M 1 , . . . , M k } is a flag matroid. We denote this as M 1 ⊂ · · · ⊂ M k and refer to the matroids M i as the constituents of the flag matroid. When k = n = |E|, and thus r i = i for i ∈ [n], the collection {M 1 , . . . , M n } is called a full flag matroid.
If M 1 ⊂ · · · ⊂ M k is a flag matroid, we will say that its constituents are concordant. That is, a collection of matroids

Positroids
Consider a field F and let A be a k × n matrix with entries in F. Let I ⊂ [n] such that |I| = k. We think of the set [n] as indexing the columns of A and thus the set I is a k-subset of the columns. Let ∆ I (A) denote the maximal minor given by the determinant of the k × k submatrix of A whose columns are those indexed by I in the order they appear in A.
Let M = ([n], B) be a matroid of rank r M = k. We say that M is representable over F if there exists a full rank k × n matrix A with entries in F such that B ∈ B(M ) if and only if ∆ B (A) = 0. In this way, we say that the matrix A represents the matroid M over F and we denote the matrix A by A M . On the other hand, given a full rank k × n matrix A with entries in F, we construct the matroid M A with the set of bases Definition 9. A matroid P = ([n], B) is called a positroid if P is realizable via a matrix A P such that all the maximal minors of A are nonnegative.
Positroids are of particular interest as they have a strong connection to the positive Grassmannian. The Grassmannian Gr k,n (R) is the set of k-dimensional vector subspaces V in R n . Such a subspace V can be thought of as a k × n full dimensional matrix A by taking a basis of V as the rows of A. In this way, we can think of points in Gr k×n (R) as full dimensional k × n matrices over Then the matroidal decomposition of the Grassmannian is where the union is over all matroids M on [n] of rank k. Note this decomposition is not a stratification ( [4]). However, the restriction to Gr 0 k×n , the nonnegative part of Gr k×n (R), that is, does produce a matroidal decomposition. Postnikov proved that carrying the decomposition of the Grassmannian by matroids to the nonnegative Grassmannian with positroids provides a stratification for Gr 0 k×n [12,13]. That is, The reader can check that M A is a positroid as each of the 10 maximal minors of A is nonnegative. In fact, M A coincides with the matroid U 3,5 .
A non-example is the matroid M on [4] of rank 2 with bases {12, 14, 23, 34}. One can check that M is realizable but it is not a positroid.
Notice that unlike arbitrary representable matroids, positroids depend heavily on an ordering of the ground set. That is, changing the order of the columns of a matrix A can change the sign of its minors. However, being a positroid is closed under cyclic shift of the columns. That is, if the matrix As a last note, we provide a proof that uniform matroids are positroids.
Lemma 11. Let k, n be integers such that 0 k n. The matroid U k,n is a positroid.
Proof. Take a 1 , . . . , a n ∈ R such that 0 < a 1 < . . . < a n and consider the k × n matrix Since any maximal minor of A is a Vandermonde matrix, we have that for all I ∈ [n] As i 1 < i 2 implies that a i 1 < a i 2 , we have that a i 2 − a i 1 > 0 and this in turn shows that ∆ I (A) is a product of positive numbers. Therefore all maximal minors of A are positive and all collections of at most k columns of A form linearly independent sets. Thus, such a matrix A represents U k,n as a positroid.

Grassmann necklaces and decorated permutations
The following objects appear in [12] as part of the family of combinatorial objects parametrizing positroids. In order to define them we introduce < i , the i-order on [n], which is the total order given by that are totally ordered via < i . We say that S i T in the i-Gale order if and only if s j i t j for all j ∈ [k]. With this in hand, we can now define Grassmann necklaces.
A Grassmann necklace is associated to every matroid in the following way.
This is a shadow of a nice property of positroids. That is, if M is a positroid then its Grassmann necklace allows us to recover all of its bases in the following way. Making use of Grassmann necklaces we now define decorated permutations. These will be used as our main way to index positroids.
Definition 17. A decorated permutation is a bijection σ : [n] → [n] whose fixed points are decorated as σ( ) = or σ( ) = . We denote the set of all decorated permutations on [n] by D n .
Let D k,n be the set of all decorated permutations on [n] with k weak excedances. Then D n = n k=0 D k,n . The cardinality of D k,n has been computed in [12,Proposition 23.1] and corresponds to the sequence A046802 [10]. The following proposition gives us a bijective way to go between Grassmann necklaces and decorated permutations. If σ is a decorated permutation we will denote by P σ its corresponding positroid.
Remark 19. The bijection between Grassmann necklaces and decorated permutations outlined here, and used throughout the paper, is due to Postnikov in [12]. Our results can also be expressed using a different bijection given in [8]. It follows from Proposition 18 that σ(i) = i implies {i} is a loop (or 1-element circuit) of the associated positroid P σ , and thus is never contained in an element of I(P σ ). Similarly, σ(i) = i implies {i} is a coloop of P σ and is in every element of I(P σ ) and thus in every basis of P σ . As part of our notation, we denote by [a, b] any interval of [n], including cyclic ones. Namely, the sets of form {a, a + 1, b} if a b and {a, a + 1, . . . , n, 1, . . . , b} if b < a. This allows us to describe some details in a more compact way. For example, the positroid U k,n is such that its Grassmann necklace is given by and its decorated permutation is π k,n := (n − k + 1)(n − k + 2) · · · n12 · · · (n − k). That is, π(i) = n − k + i (mod n) for i ∈ [n].
As stated in the introduction, we are interested in a combinatorial characterization of quotients of positroids. Thus our main question is: Given two positroids P 1 and P 2 on the ground set [n], can we determine combinatorially whether P 1 is a quotient of P 2 , or vice-versa? We will make use of decorated permutations to give a partial answer to this question.

Poset of positroid quotients
For every n 1, we denote by P n the poset whose elements consist of the decorated permutations in D n (i.e. positroids on [n]) and whose order relation is the transitive closure of the following covering relation: τ π if and only if τ ∈ D k−1,n , π ∈ D k,n for some k ∈ {0, 1, . . . , n}, and P τ is a quotient of P π . We call P n the poset of positroid quotients on [n]. See Figure 1 for an illustration of P 3 . Recall that the decorated permutation corresponding to the uniform positroid U k,n is given by π k,n = (n − k + 1) · · · (n − 1) n 1 2 · · · (n − k). We state the following properties of the poset P n , whose proof we leave to the reader: 1. It is a poset with0 given by the decorated permutation π 0,n = 12 · · · n and1 given by the decorated permutation π n,n = 12 · · · n.
2. It is graded and the rank of each decorated permutation is its number of weak excedances. Thus, its rank polynomial is symmetric and unimodal. The number of elements in each rank is recorded in the sequence A046802.
On the other hand, if τ (i) = i, then σ(i) = i.
Similarly, one can construct the poset M n of matroid quotients whose elements are all matroids on the ground set [n] and whose order relation is M < N if and only if M = N and M is a quotient of N . This is implicitly done in [16]. Moreover, it is shown in [16,Prop. 8.2.5]  In other words, if M < N in M n , then there is a saturated chain M M 1 · · · N in M n . The existence of such saturated chains is made explicit via the Higgs lift (see [16,Prop. 8.2.5] for details). One may feel tempted to conclude the same in the poset P n . However, this is unclear as one needs to guarantee that the Higgs lift of a positroid is again a positroid.
Our main theorem identifies a set of positroids that are quotients of U k,n for any k ∈ [n]. We do this by defining the following sequence of moves on decorated permutations.
Definition 22. Given a decorated permutation π ∈ D n and a subset A of [n], we denote by ← − ρ A (π) the element of D n obtained from π by performing the following moves in order: Analogously, we denote by − → ρ A (π) the permutation obtained from π by performing the (F), (S), (D) moves, such that (S) shifts to the right instead of the left, and that (D) decorates any new fixed point i as i. We call this sequence of moves a FSD-shift of π.
Proof. Let A ⊆ [n], π := π k,n , and recall that π −1 = π n−k,n . With this in mind it is easy to see that freezing the elements π(i) for every π(i) ∈ A makes every element of B = {i ∈ [n] : i = π −1 (j) for j ∈ A} frozen in π n−k,n . Now consider the remaining values of π that are not frozen, in the order they appear. These values form a permutation ω over the set [n] \ A. The positions of ω are indexed by the ordered set [n] \ B. Thus a cyclic shift to the right of the values of ω is equivalent to a cyclic shift to the left of the indices of ω. Finally for i ∈ B, the index ω −1 (i) is precisely the entry π −1 (i) = π n−k,n (i). Thus, − → ρ A (π) −1 = ← − ρ B (π −1 ).

Uniform quotients
Now we provide a characterization using decorated permutations for a family of positroids of rank k − 1 that are quotients of the uniform positroid U k,n . For the remainder of the paper we will consider A ⊆ Theorem 26. Let k n and A ⊂ [n] be a union of disjoint cyclic intervals of [n] such that no interval in A has size greater than k−1. Then the positroid represented by ← − ρ A (π k,n ) has rank k − 1.
Let j = min{i ∈ [n] \ [k] : π(i) / ∈ A} if it exists. Since π(j) = n − k + j and π(j ) = j − k, we have that σ(j) = j − k. Recall that the intervals of A have length less than k. Thus j − j k, implying that j − k j and j is not a weak excedance in the inequality case. In the case of an equality, our construction tells us that fixed points are colored as σ(i) = i and we can assure that j ∈ W 1 (σ).
If such a j does not exist, then π(i) ∈ A for all i ∈ [k + 1, n]. This accounts for n − k frozen elements. Due to our construction of j, the k − j elements in [k] \ [j] are also frozen. Now notice that in this case, for j to be a weak excedance of σ, all the first j − (n − k) values of π would have to be frozen as well. This gives us a frozen interval of total size n − k + k − j + j − (n − k) = k which contradicts our assumption that A has no interval of size greater than k − 1. Thus we deduce that in both cases j / ∈ W 1 (σ).
Now let l ∈ [k + 1, n] and σ(l) = l for some l . If l < 1 l then π(l ) = l − k and σ(l) = l − k. Since l − l k, we have that l − k l and thus l / ∈ W 1 (σ). On the other hand, if l k then σ(l) = n − k + l . Again, the cyclic interval [l , l] contains at most k − 1 elements. Thus n + l − l k and n − k + l l and we conclude that l / ∈ W 1 (σ). Therefore W 1 (σ) = W 1 (π) \ j and P σ has rank k − 1 as desired.
We are now ready to state our main result.
The proof of Theorem 28 relies on showing that every circuit of U k,n is a union of circuits of the positroid represented by the decorated permutation ← − ρ A (π k,n ) for the set A ⊂ [n]. To this end, we state the following results.
We first show that the set of circuits of the positroid P σ represented by σ = ← − ρ A (π k,n ) has a simple description. Then we show how to obtain the circuits of π k,n as the union of circuits of P σ .
Theorem 29 (Circuit description of shifted uniform positroid). Let A = [a 1 , i 1 ] ∪ · · · ∪ [a m , i m ] be a subset of [n] composed of disjoint cyclic intervals of lengths l 1 , . . . , l m respectively. Then the circuits of the positroid P σ represented by σ = ← − ρ A (π k,n ) are given by the sets Moreover, the circuits of size less than k can be read from the decorated permutation σ.
Proof. Let π := π k,n . Assume that σ = ← − ρ A (π k,n ) for some A ∈ [n] and ∈ {0, . . . , k−1}. We begin by proving that each interval [i j +1, i j +k −l j ] satisfies σ(i j +k −l j ) = i j +1 and that for all r ∈ [n], the interval [σ(j r ), j r ] is a circuit of P σ . Let us suppose that within the frozen set A, there exist an interval [a, b] which has length b − a + 1. Notice that the interval [b + 1, a + k − 1] is the cyclic interval that would extend (clockwise) [a, b] into a interval of length precisely k. In fact [b + 1, a + k − 1] is the description of [i j + 1, i j + k − l j ] in terms of a and b. As we are freezing the values [a, b] and π k,n (x) = x − k (mod n), we have that we are freezing the positions [a + k, b + k] (mod n). Therefore we have that if σ(a + k − 1) = b + 1, and σ(i j + k − l j ) = i j + 1. For the rest of the proof we will denote such intervals as [σ(j r ), j r ] for some r ∈ [n].
We will now show the circuits described above and the k-subsets not containing these intervals are the only circuits of P σ .
If A = ∅, then C = [n] k , which are the circuits of the uniform positroid U k−1,n represented by σ. Now, let ∈ [k − 1] and A ∈ [n] . Recall that W 1 (π) = [k]. Moreover, for each r ∈ [n], it holds that W r (σ) = [r, r + k − 1] \ {j r } where j r = max{i ∈ [r, r+k−1] : π(i) / ∈ A}. That is, j r is the largest among the first k-positions in the < r order such that π(j r ) is not frozen. Recall that the Grassmann necklace I σ = (I 1 , . . . , I n ) corresponding to σ satisfies I r = W r (σ) and that for each r this means that I r is the minimal basis of P σ under the r-Gale order.
We now show that for each r ∈ [n], the interval [σ(j r ), j r ] is a circuit of P σ . We illustrate the proof with r = 1 as it is done analogously for each r. Set j := j 1 and notice that since j / ∈ W 1 (σ), then σ(j) < j and [σ(j), j] ⊆ [1, k]. If [σ(j), j] were independent, then [σ(j), j] would be contained in I σ(j) , but I σ(j) does not contain j since j σ(j) = j. Therefore, [σ(j), j] is dependent in P σ . To show it is a circuit we will show that each of the sets J x = [σ(j), j] \ {x}, for x ∈ [σ(j), j], is independent by constructing a basis B x of P σ such that J x ⊆ B x . If x = j, then B x := I 1 works. If x = j, we will prove the set B x := I 1 ∪ {j} \ {x} is a basis of P σ using Proposition 15. Since A quick comparison of the sets shows that B x r I 1 r I r for r ∈ [x]. For the other values of r, recall first that I r = [r, r + k − 1] \ j r . Arranging the elements of I r using the r-Gale order we get that I r = {r, . . . , j r − 1, j r + 1, . . . , r + k − 1} taking mod n where needed. Arranging B x we get B x = {r, . . . , k, 1, . . . , x − 1, x + 1, . . . , r − 1} if r ∈ [k] \ [x], whereas for r ∈ [n] \ [k] we get B x = {1, . . . , x − 1, x + 1, . . . , k}. In either case one can see that B x r I r . This allows us to conclude that [σ(j r ), j r ] is a minimally dependent set in P σ for every r ∈ [n] and therefore a circuit. The reader can verify that each of the sets [i j +1, i j +k−l j ] from equation (1) are of the form [σ(j r ), j r ] for some r.
We now proceed to show that any k-subset of [n] that does not contain a [σ(j r ), j r ] for some r is a circuit. Let D be a k-subset of [n] such that D does not contain any of the sets [σ(j r ), j r ] as given above. Since P σ has rank k − 1, D is automatically dependent. We only need now to show it is minimal. Consider a (k − 1)-subset F ⊂ D and let us see that F is a basis of P σ . If F were not a basis then there would be a r ∈ [n] such that F r I r . Thus if F = {c 1 < r · · · < r c k−1 } and I r = {b 1 < r · · · < r b k−1 } in r-Gale order, the fact that F r I r implies that for some l and p ∈ [l − 1], we have that c p b p and c l < b l . However, I r = [r, j r − 1] ∪ [j r + 1, r + k − 1] so l must be the position of j r , which implies that c l = j r . Therefore [r, j r ] ⊂ F and since σ(j r ) = r, we get that [σ(j r ), j r ] ⊂ D which is a contradiction to our assumptions on D. We thus conclude that F is a basis of P σ and D is a circuit.
So far we have shown that C A is contained in C σ , the set of circuits of the positroid P σ . To prove the reverse containment we will show that if S / ∈ C A then S / ∈ C σ . First, notice that we only need to consider sets S ⊂ [n] such that |S| < k and that do not contain any interval [σ(j r ), j r ]. We will prove that such S can be extended to a set D of cardinality k in a way such that D does not contain any of the [σ(j r ), j r ] and therefore, S will be independent. Suppose that the decomposition of the set A in cyclic intervals is A = J 1 ∪ · · · ∪ J s and let L i be the cyclic interval such that J i ∪ L i is an interval of size k for each i. Recall that all L i are circuits of P σ .
In order to prove that S can be extended to our desired D we will make use of J 1 ∪ L 1 . To this end, let D := J 1 ∪ L 1 ∪ S with cardinality at least k.
Case 1: Suppose D contains no L j except L 1 . If |S \ (J 1 ∪ L 1 )| 1 then S can be directly extended to a k-subset D of D \ {a} with an a ∈ L i \ S such that it does not contain any of the intervals in C A . On the other hand, if S ⊂ J 1 ∪ L 1 , let us take D := ((J 1 ∪ L 1 ) \ {a}) ∪ {b} as an interval where a ∈ L 1 \ S and b := c + 1 where c is the greatest element of L 1 . Notice that such an a exists since S does not contain L 1 by hypothesis. If b does not exist, it means that D = [n] and thus σ π = U n,n as any matroid is concordant to U n,n . Meaning there is nothing to prove. Otherwise if such b exists, then D has size k, contains S, and does not contain any interval of C A . Meaning S is independent.
Case 2: Suppose D contains L 1 and another different L j . Without loss of generality call it L 2 . If this is the case, then either L 1 ∪ L 2 or J 1 ∪ L 2 is a cyclic interval. In the former case, |D | k + 1 and D can be obtained by removing elements from D of the form a ∈ (L 1 ∪ L 2 ) \ S until it has size k. These elements exist as L 2 ⊂ S and L 1 ∩ L 2 = ∅. In the latter case, D can be obtained as a subset of D : S) and d := c + 1 where c is the clockwise greatest element of L 1 . As |D | > k and does not contain any L j but does contain S, we can extend S to a basis and make it an independent set. Case 2.1: If simultaneously, D contains L 2 and L 3 such that L 1 ∪ L 2 and J 1 ∪ L 3 are cyclic intervals, then either J 3 ∩ L 1 = ∅ or J 3 ∩ L 1 = ∅. In the former case, remove from D any pair of elements a ∈ (L 3 ∩ J 1 ) \ S and b ∈ (L 2 ∩ L 1 ) \ S. With this, D \ {a, b} will have still at least k elements as the cyclic components of S that intersect L 2 and L 3 have elements outside of D .
On the other hand, if J 3 ∩ L 1 = ∅, in order to maintain the cardinality of D \ {a, b} above or equal to k with a, b as above, we would need to guarantee that such elements can be substituted. This can be achieved directly if [n] \ (J 1 ∪ L 1 ) has at least 2 elements. If |[n] \ (J 1 ∪ L 1 )| = 0, we land again in the U n,n case. If instead |[n] \ (J 1 ∪ L 1 )| = 1, then J 3 ∩ J 1 = ∅ (as the reader can check) which is a contradiction since all J i are disjoint. This exhausts all the possibilities and the proof is complete.
Proposition 30. Every circuit of π k,n can be obtained as a union of elements in C A .
Proof. As π k,n corresponds to the uniform positroid U k,n , all of its circuits are all the (k + 1)-subsets of [n]. Let O be any such circuit and let k and there is one remaining element x in O we have yet to cover. Let us take the k-subset D := (C \ {y}) ∪ {x} for any y ∈ C. As D = C and |D| = k, then D / ∈ C O . Thus there is a cyclic interval L ∈ C A such that x ∈ L and L ⊂ D. This means that, O = L ∪ C and the claim is proved in this case.
Finally, if C O = ∅ then every k-subset C of [n] contained in O properly contains at least one of the cyclic intervals [i j + 1, i j + k − l j ] of A. Moreover, O contains at least two distinct intervals L 1 and L 2 . To see this, take any k-subset C 1 = O \ {x} where x ∈ O and let L 1 ⊂ C 1 be an interval in C A . Now let y ∈ L 1 and set C 2 = O \ {y}. Since C 2 ⊂ O, there is an interval L 2 ∈ C A such that L 2 ⊂ C 2 . As y ∈ L 1 ⊂ C 1 and x / ∈ C 1 , then L 1 = L 2 . Now assume L 1 , . . . , L m are the intervals in C A contained in O. Denote L = L 1 ∪ · · · ∪ L m . Clearly L ⊆ O. We will prove the reverse containment. Suppose that the intervals in L are not pairwise disjoint. Without loss of generality let L 1 ∩ L 2 = ∅. As L 1 , L 2 ∈ C A , there are two disjoint cyclic intervals J 1 , J 2 ⊆ A that give rise to L 1 , L 2 , respectively. That is, J r ∪ L r is a cyclic interval of length k for r = 1, 2. Now suppose that the least element in L 1 is smaller than the one in L 2 . Then J 2 ∪ L 2 ⊂ L 1 ∪ L 2 (otherwise J 1 overlaps J 2 which cannot happen) and this implies that k < |L 1 ∪ L 2 | k + 1. The first inequality follows from J 2 ∪ L 2 = k and the second since L 1 ∪ L 2 ⊆ L ⊆ O. Thus in this case, |L 1 ∪ L 2 | = k + 1 and we conclude that O = L 1 ∪ L 2 . Now suppose that the intervals in L are pairwise disjoint. Denote by J i ⊆ A the frozen interval that gives rise to L i . We know that for m 2. But since L ⊆ O and |O| = k + 1, we get that L = O. This finishes the proof.
Theorem 28 follows immediately as a consequence of Theorem 29 and Proposition 30.
Again, using SageMath we see that τ σ.
The dual version of Theorem 28 can be stated as follows.
Proof. Consider the decorated permutation π n−k,n , B ∈ [n] where ∈ {0, . . . , k−1} and τ = − → ρ B (π k,n ). Taking the corresponding positroids of π n−k,n and τ and using Proposition 6, we get that U n−k,n is a quotient of P τ if and only if P * τ is a quotient of U * n−k,n . As U * n−k,n = U k,n and P * τ = P τ −1 (see [7]), we know that U n−k,n is a quotient of P τ if and only if P τ −1 is a quotient of U k,n . Now τ = − → ρ B (π k,n ) implies that τ −1 = ← − ρ A (π n−k,n ) where A = τ −1 (B) because of Proposition 24. This with the fact that A ∈ [n] where ∈ {0, . . . , k − 1} and Theorem 28 gives us that τ −1 π k,n . Following back our trail of if and only ifs, this implies that U n−k,n is a quotient of P τ and π n−k,n τ as desired.
We point out that if N is a paving matroid then its simple truncation is a uniform matroid (see [11]). Thus, the positroids τ in Corollary 35 correspond to paving matroids. Hence, via our work we have characterized a family of paving positroids.
In view of Corollary 35, we see that π n,n covers r positroids if and only if π 0,n is covered by r positroids. However, since π 0,n and π n,n are the bottom and top elements of P n , respectively, then r = |D 1,n | = |D n−1,n | = 2 n − 1 and |D 1,n | = n−1 =0 n . This allows us to conclude that the converse of Theorem 28 also holds for π n,n .
We end by providing a conjecture that summarizes the findings detailed in this paper. This conjecture is based on evidence generated using SageMath for positroids on the ground set up to size 12.

Future work
A parallel approach to the matroid quotient problem can be taken via positroid polytopes. That is, one may study chains of positroids that form a flag via the inequalities of their corresponding flag polytope. Thus, we pose the problem of characterizing flag positroids via flag matroid polytopes. That is, what conditions must a polytope have to guarantee that it corresponds to a flag of positroids?
Finally, the path we took in this paper has unveiled the poset of positroid quotients which deserves to be explored further. Some interesting questions in this direction are: (a) What is the Möbius function of the poset P n ? Up to n = 4 the first values of µ(P n ) are 1, −1, 2, −9, 92.
(b) Is there an ER-labelling of P n ? A candidate is labelling the edge of the covering τ σ by the set that is frozen when passing from σ to τ . Answering this question may give a Whitney dual for this poset, in the sense of [6].