Noncrossing Partition Lattices from Planar Configurations

The lattice of noncrossing partitions is well-known for its wide variety of combinatorial appearances and properties. For example, the lattice is rank-symmetric and enumerated by the Catalan numbers. In this article, we introduce a large family of new noncrossing partition lattices with both of these properties, each parametrized by a configuration of n points in the plane.


Introduction
For each subset P of the complex plane, a noncrossing partition of P is a way of dividing P into subsets with pairwise disjoint convex hulls.The collection of all noncrossing partitions of P , denoted NC(P ), is a partially ordered set under refinement.When P is the vertex set for a convex n-gon, NC(P ) is the classical noncrossing partition lattice NC n introduced by Kreweras [3].Among other things, Kreweras showed that the size of NC n is counted by the combinatorially ubiquitous Catalan numbers C n = 1 n+1 2n n and, more specifically, the number of lattice elements with rank k is the Narayana number N n,k = 1 n n k n k−1 .Since N n,k = N n,n−k , this further says that NC n is a rank-symmetric lattice.In the fifty years since its definition, the noncrossing partition lattice has made countless appearances in algebraic and geometric combinatorics -see the survey articles [4] and [1] for more information.
Returning to the more general case prompts a natural question: for which subsets P ⊂ C does the poset NC(P ) have similar properties to NC n ?While some existing work studies the size of NC(P ) in asymptotic and extremal cases (e.g.[10] [6]), similarities to NC n seem uncommon in the literature.In our first main theorem, we introduce a convexity condition on P which guarantees that NC(P ) has the same size as NC n .
Let P ⊂ C be a set of n points in general position.We say that a subset A ⊆ P is convex if no point in A lies in the convex hull of the others, and we say that P satisfies Property ∆ k if, for all convex A ⊆ P , the convex hull of A contains at most |A| + k − 3 elements of P in its interior.
Theorem A (Theorem 5.4).Let P ⊂ C be a set of n points with Property ∆ 1 .Then NC(P ) is a rank-symmetric graded lattice, and the number of elements with rank k is the Narayana number N n,k .In particular, |NC(P )| = C n = |NC n |.It is worth noting that if P contains a point which lies in the convex hull of the others, then NC(P ) is not isomorphic to NC n .Thus, Theorem A introduces a large new class of lattices with the same number of elements in each rank as the noncrossing partition lattice.
If the conditions on P are weakened to only require Property ∆ 2 , then the number of noncrossing partitions may increase compared to those in Theorem A. Nevertheless, some symmetry is preserved.
Theorem B (Theorem 6.4).Let P ⊂ C be a set of n points with Property ∆ 2 .Then NC(P ) is a rank-symmetric graded lattice.
See Figures 1 and 2 for examples of configurations which satisfy Properties ∆ 1 and ∆ 2 respectively.One can quickly find examples of P without Property ∆ 1 such that NC(P ) fails to be rank-enumerated by the Narayana numbers and examples of P without Property ∆ 2 such that NC(P ) fails to be rank-symmetric.In fact, NC(P ) can even fail to be graded if P does not have Property ∆ 2 (see Proposition 4.4 and Figure 13).The extent to which these properties are necessary conditions for Theorems A and B is not immediately clear; this is an interesting direction for future research.
Some of the techniques used in proving Theorems A and B can be interpreted in a stronger topological context.Recall that the (unordered) configuration space of n points in C is the topological space of all n-tuples in C n with distinct entries, considered up to permutations of the coordinates.Also, if P barely fails to be in general position (i.e.there is a single triple of collinear points in P ) but otherwise satisfies Property ∆ k , we say that P satisfies the weak Property ∆ k .We are unaware of any prior appearances of the space described in Theorem C. It would be interesting to know the homology of this space for each k and, in particular, whether it is a classifying space for the n-strand braid group.
The article is structured as follows.In Section 2, we introduce some background on posets and partitions, along with basic properties of NC(P ).Section 3 concerns the transformation of configurations with Property ∆ k and includes the proof of Theorem C. We provide some technical details on "skewers" in Section 4, then give the proofs of Theorem A in Section 5 and Theorem B in Section 6.

Noncrossing Partitions
To start, we establish some basic definitions and properties for partitions, posets, and configurations -see [7] for a standard reference.Recall that a partition expresses a set S as the union of a collection of pairwise disjoint subsets of S (called blocks).The set of all partitions for a fixed set S forms a partially ordered set under refinement: one partition lies "below" another in the partial order if each block in the latter partition can be obtained as a union of blocks in the former.This partially ordered set is a lattice in the sense that each pair of elements has a unique meet (greatest lower bound) and a unique join (least upper bound).Let Π(S) denote the lattice of partitions for S; in the standard case where S = {1, . . ., n}, the associated partition lattice is denoted Π n .
The partition lattice Π(S) is bounded in the sense that it contains a unique minimum element 0 (in which each block is a singleton) and a unique maximum element 1 (in which all of S belongs to the same block).The partition lattice is also graded : if |S| = n and we let bl(π) denote the number of blocks in a partition π ∈ Π(S), then the map ρ : Π(S) → N given by ρ(π) = n − bl(π) is a rank function for this lattice.Note that the minimum 0 and maximum 1 have ranks 0 and n − 1 respectively.The atoms and coatoms of this lattice are defined to be the elements of rank 1 and n − 2 respectively.
Our main object of study in this article is a subposet of the partition lattice for a finite set of points in the complex plane.Definition 2.1.Fix P ⊂ C finite.For any A ⊆ P , the convex hull of A, denoted Conv(A), is the smallest convex subset of C which contains A. Note that Conv(A) is a convex polygon with up to |A| vertices.A partition of P is noncrossing if the convex hulls of its blocks are pairwise disjoint.The set of all noncrossing partitions for P forms a subposet of the partition lattice Π(P ), and we refer to this subposet as NC(P ).
Example 2.2.Let P = {z 1 , z 2 , z 3 , z 4 } be a set of points in C such that z 1 , z 2 , and z 3 form the vertices of a triangle which contains z 4 in its interior.Then every partition of P is noncrossing except for {{z 1 , z 2 , z 3 }, {z 4 }}, so the noncrossing partition lattice NC(P ) has 14 elements, arranged according to the diagram in Figure 3.Note that NC(P ) has the same size as the classical noncrossing partition lattice NC 4 , but the two are not isomorphic since NC(P ) has 15 maximal chains, while NC 4 has 16.
As a poset, the noncrossing partitions of P inherit some useful properties from the larger partition lattice Π(P ).
Proposition 2.3.If P ⊂ C is finite, then NC(P ) is a bounded lattice.
Proof.Since the minimum and maximum elements of Π(P ) are noncrossing, we know that NC(P ) is bounded.To show that NC(P ) is a lattice, we need only prove that NC(P ) is a meet-semilattice (i.e. that each pair of elements in NC(P ) has a unique meet) by a standard property of finite bounded posets [7,Prop 3.3.1].First, suppose that π ≤ π ′ in Π(P ).If π ′ is noncrossing but π has a pair of crossing blocks A, B ∈ P , then π ′ must have a block which contains the union A ∪ B. It follows that if π 1 , π 2 ∈ NC(P ), then the meet π 1 ∧ π 2 must be noncrossing; if it had a pair of crossing blocks A and B, then π 1 and π 2 would each have blocks containing A∪B by the argument above and thus π 1 ∧π 2 would have a block containing A∪B, a contradiction.Therefore, NC(P ) is a meet-semilattice and thus a lattice.□ Although Π(P ) is graded for every configuration P , it is possible to construct P so that the poset NC(P ) is not graded.In this article, however, we are interested in configurations satisfying particular conditions (Properties ∆ 1 and ∆ 2 , defined in Section 3), and we will see in Section 4 that NC(P ) is graded in these cases.
We close the section with a few important examples and some observations.For more information on the combinatorial significance of these connections, see [8].
Noting that N n,k = N n,n−k for all 1 ≤ k ≤ n, one can see that NC n is a ranksymmetric lattice.In fact, the classical noncrossing partition lattice is self-dual in the sense that it admits a bijection α : NC n → NC n with the property that π 1 ≤ π 2 if and only if α(π 2 ) ≤ α(π 1 ) [9].However, this stronger condition is rarely held by NC(P ) more generally.
Remark 2.5.Nica and Speicher showed in 1997 that intervals in the noncrossing partition lattice NC n are isomorphic to products of smaller noncrossing partition lattices [5].With this in mind, we note that if P is a set of n points of C in general position and if some point of P lies in the convex hull of the others, then NC(P ) has an interval which is isomorphic to the lattice described in Example 2.2, which cannot be expressed as a product of noncrossing partition lattices.Therefore, NC(P ) is only isomorphic to NC n if the elements of P form the vertices of a convex n-gon.
Example 2.6.If P ⊂ C with |P | = n such that all points in P are collinear, then NC(P ) is isomorphic to the Boolean lattice Bool n−1 , which is defined as the set of all subsets of a set with n − 1 elements, partially ordered under inclusion.To see this, observe that each partition in NC(P ) is determined precisely by choosing a subset of the n − 1 gaps between the n points; see Figure 5.
When |P | = 4, there are only four possibilities for NC(P ) (up to isomorphism), and three of them are depicted in the preceding figures.All three (indeed, all four) possess several useful lattice properties, including rank-symmetry, self-duality, and a simple counting formula.However, these properties do not always hold for larger sizes of P .
Example 2.7.If P consists of five points in general position with three points on the boundary of the convex hull and two points in the interior, then |NC(P )| = 43 (whereas |NC n | = 42), although NC(P ) remains rank-symmetric.Furthermore, if P consists of six points in general position, arranged so that the three extremal points form an equilateral triangle and the three interior points form a shrunken equilateral triangle with the same center, then NC(P ) is not rank-symmetric: it has 15 atoms (rank 1) and coatoms (rank 4), but 55 elements at rank 2 and 57 elements at rank 3.

Configurations
Before moving on to the main theorems, we introduce some tools for studying the geometry of planar configurations, by which we mean finite unordered sets of points in the Euclidean plane.We begin with some helpful terminology, partially inspired by [2].Throughout this section, let P denote a configuration of n points in C in general position (i.e.no three points in P are collinear), unless otherwise specified.Definition 3.1.Let A ⊆ P and recall that Conv(A) denotes the convex hull of the points in A. Define the closure A by Conv(A) ∩ P and the interior int(A) to be int(Conv(A)) ∩ P .We say that A is convex if int(A) ∩ A is empty.Also, a point p ∈ P is internal if p ∈ int(P ) and extremal otherwise.Definition 3.2.A configuration P in general position satisfies Property ∆ k if, for every convex subset A in P , the interior int(A) contains at most (|A|−3)+k points.Equivalently, P has Property ∆ k if P is in general position and each subset B ⊆ P (not necessarily convex) has at most ⌊ |B|−3+k 2 ⌋ internal points.Finally, we say that P instead has the weak Property ∆ k if it satisfies the same convexity criteria, but has at most one instance of three collinear points.
It is worth noting that Property ∆ 1 is equivalent to a simpler condition which is easier to check: for any A ⊆ P with |A| = 3, we have |int(A)| ≤ 1.To see this, consider that each convex subset of m points in P forms the vertices of a convex mgon, and any triangulation of this polygon consists of m − 2 triangles; Property ∆ 1 is equivalent to the requirement that each of those m − 2 triangles has at most one point of P in its interior.Unfortunately, this does not generalize to Property ∆ k when k > 1.
If P satisfies Property ∆ k , then any small perturbation of P will also satisfy Property ∆ k since P is in general position.However, deformations which move a point in P across the line between two other points in P might not preserve this property.The main goal of this section is to provide some tools for moving points z Figure 6.From left to right: an arrangement A, the subarrangement A z with the region R z highlighted, and the subarrangement A ex of lines between pairs of extremal points.In each image, only the core of each line has been drawn.in P while preserving Property ∆ k .To start, we establish some terminology for the lines connecting points in P .Definition 3.3.Each pair of distinct points in P determines a line in C; let A denote the arrangement of the n 2 lines obtained in this way.If ℓ is the line obtained from the points z and w in P , then we say that z and w are the endpoints of ℓ and write V (ℓ) = {z, w}.We also refer to the line segment between z and w as the core c(ℓ) = Conv(V (ℓ)) of ℓ.More generally, we write V (ℓ 1 , . . ., ℓ k ) to mean the 2kelement set of endpoints belonging to the lines ℓ 1 , . . ., ℓ k and we write c(ℓ 1 , . . ., ℓ k ) to mean the convex hull Conv(V (ℓ 1 , . . ., ℓ k )).Definition 3.4.For each z in int(P ), let A z denote the subarrangement of A obtained by deleting the lines which pass through z.Also, define A ex to be the subarrangement of A which consists of all lines with two extremal endpoints, i.e. the intersection of all A z for z ∈ int(P ).We associate two regions to each z ∈ int(P ): the connected component of C − A z containing z (which we denote R z ) and the connected component of C−A ex containing z (denoted R ex z ).Note that each region is a convex polygon since it is a bounded subset of the plane determined by removing some number of half-planes.Finally, we say that a line in A is separating if it has points from int(P ) on either side of it, and a boundary line is one which contains a boundary edge for the convex hull Conv(P ).
For the sake of clarity, we will typically illustrate the line arrangement A by its intersection with the convex hull of P -see Figure 6 for an example.Definition 3.5.A move is a bijection m : P → m(P ) such that m fixes all of P except some element z, which is instead sent to a point m(z) in the interior of a region adjacent to R z .If ℓ is a line in the arrangement A which separates the regions R z and R m(z) , then we say that m moves z across ℓ.If both P and m(P ) satisfy Property ∆ k , we say that m is a ∆ k -move.Finally, note that m induces an isomorphism m * : Π(P ) → Π(m(P )) by replacing z with m(z) in each partition.
It is worth noting that for any z ∈ P , we can replace z with any other point in the region R z without changing the isomorphism type of Π(P ), so moves on P can be described solely by the regions involved.Proof.Let m : P → m(P ) be a move which takes z across a line ℓ in A ex , and let w 1 and w 2 be the endpoints of ℓ.Suppose for the sake of contradiction that m(P ) does not satisfy Property ∆ k ; then there is a subset First, note that the supposed bound on |int(A)| precludes m(z) from being an internal point of A. If it were internal, then A would necessarily contain w 1 and w 2 , and A would therefore be a subset of H + m(z),ℓ since z ̸ ∈ A. However, m(z) is the unique internal point of m(P ) in H + m(z),ℓ since we assumed ℓ was non-separating in the initial configuration P , so the inequality would not hold.
Thus, m(z) is an extremal point of since z is an internal point for B but not A), and we can combine these to find the following chain of inequalities: Since B is a subset of P , this contradicts our assumption that P has Property ∆ ksee Figure 7 for an example when k = 1.Therefore, m(P ) must satisfy Property ∆ k and we are done.□ Proof.Let ℓ be a line in A z adjacent to z with endpoints w 1 and w 2 , suppose that w 2 is an internal point of P , and let m : P → m(P ) be the move which takes z across ℓ.As above, we suppose for the sake of contradiction that m(P ) does not satisfy Property ∆ k and can thus find a subset A ⊆ m(P Consider the three lines in A for which one endpoint is w 1 and the other belongs to the set {z, w 2 , m(z)}.Since z, w 2 , and m(z) are internal points of P and both z and m(z) are adjacent to ℓ, all three of these lines must pass through the same side of the polygon Conv(P ).Let a 1 , a 2 ∈ P be the extremal points which determine this side, where a 1 is on the same side of ℓ as z -see Figure 8 for an illustration.
, then A must contain w 1 and w 2 , and the fact that z / ∈ A tells us that A does not contain any points in the half-plane H + z,ℓ .Thus, in this case we have that |B| ≤ |A| + 2 and |int(B)| ≥ |int(A)| + 1 (since w 2 is internal for B but not A), which provides the same chain of inequalities as described in the proof of Lemma 3.7.Therefore, P does not satisfy Property ∆ k , which is a contradiction.
If m(z) is instead an extremal point of A, then we see that |B| ≤ |A| + 4 and |int(B)| ≥ |int(A)| + 2, so we have a similar sequence of inequalities: This also contradicts our assumption that P satisfies Property ∆ k , so we conclude that m(P ) must satisfy Property ∆ k .□ Note that the following lemma supposes only that P has Property ∆ 2 , so in particular it holds when P satisfies Property ∆ 1 as well.Lemma 3.9.If P has Property ∆ 2 , then there is a point z ∈ int(P ) such that at least one side of the region R ex z belongs to a non-separating line in A ex .
Proof.Let P be a configuration satisfying Property ∆ 2 and let D be the convex hull of the regions R ex z , where z is an interior point of P .Let ℓ 1 , . . ., ℓ k denote the lines (not necessarily in A) which contain the k sides of D, arranged so that they appear in counter-clockwise order.If at least one ℓ i contains a side of some region R ex z , then ℓ i belongs to the arrangement A ex , and it follows that ℓ i must be non-separating since all internal points lie on one side of it.
Suppose for the sake of contradiction that this is not the case.Then we can fix points z 1 , . . ., z k ∈ int(P ) such that for each i, the region R zi intersects the boundary of D at the point where the lines ℓ i and ℓ i+1 (evaluated mod k) intersect.Since neither ℓ i nor ℓ i+1 are in A ex , it follows that the half-planes H − zi,ℓi and H − zi,ℓi+1 intersect in an unbounded region which contains an extremal point of Psee Figure 9 for an illustration.
If E is the set of k extremal points obtained in the manner above, then one can show that the convex hull Conv(E) contains D. However, this means that z 1 , . . ., z k lie in the interior of E, which contradicts our assumption that P satisfies Property ∆ 2 .Therefore, at least one side of D must belong to a non-separating line in A ex and we are done.□ Putting all of these tools together, we obtain a useful connectivity property.
Theorem 3.10.Suppose P satisfies Property ∆ k , where k ∈ {1, 2}.Then there is a sequence of ∆ k -moves which transforms P into a convex configuration.
Proof.We proceed by induction on the number of internal points for P .If P has no internal points, then P is already convex and we are done.Now, suppose the theorem holds for configurations with fewer internal points than P and define nsnb(P ) to be the set of non-separating non-boundary lines in A ex ; we will prove that the theorem holds for P as well by a second induction on |nsnb(P )|.If |nsnb(P )| = 0, then each internal point of P is adjacent to a boundary line, and by Lemma 3.7, we can bring one of these internal points across a boundary line by a ∆ k -move, which reduces the number of internal points by one and allows us to apply the first inductive hypothesis.Next, suppose the claim holds for all configurations Q with |int(Q)| = |int(P )| and |nsnb(Q)| < |nsnb(P )|.By Lemma 3.9, we know that there is an internal point z in P such that one side of the region R ex z is contained in a non-separating line in A ex .This means that while z may not be adjacent to a non-separating line, there is a finite sequence of moves across lines with at least one internal endpoint which takes z to a region which is adjacent to a non-separating line in A ex .Each move across lines with an internal endpoint preserves Property ∆ k by Lemma 3.8, after which we can perform a ∆ k -move across the non-separating line by Lemma 3.7.This new configuration has fewer non-separating non-boundary lines, so by the second inductive hypothesis, it can be further transformed via ∆ k -moves into a convex configuration and the proof is complete.□ From a topological perspective, Theorem 3.10 can be interpreted as a statement about the configuration space of n points in the plane.To do so, recall that the weak Property ∆ k is a slight weakening of Property ∆ k which allows for one instance of three collinear points.Applying a move to a configuration P will certainly affect the noncrossing partition lattice NC(P ), and possibly even its isomorphism type.To close this section, we introduce a natural map between the larger partition lattices.Definition 3.12.Let m : P → m(P ) be a move which brings the point z ∈ P across the line ℓ in A z and let w 1 and w 2 be the two points of P on ℓ.The blockswitching map BS m : Π(P ) → Π(m(P )) is defined for each π ∈ Π(P ) as follows: if w 1 and w 2 share a block in π and z belongs to a different block, then define BS m (π) to be the result of removing {w 1 , w 2 } and {m(z)} from their respective blocks in m * (π) and swapping them; otherwise, define BS m (π) = m * (π).See Figure 10 for an illustration.Note that BS m is a rank-preserving bijection, but not an isomorphism.A skewering tree for a configuration of 14 points which satisfies Property ∆ 2 , but not Property ∆ 1 .The tree consists of two skewering sequences: ℓ 0 ⊣ α 1 ⊣ α 2 and ℓ 0 ⊣ β 1 .For the sake of visual clarity, only the cores of the four lines have been drawn.

Skewers
In this section, we introduce the notion of "skewering" in the line arrangement A and prove some technical results which come from Property ∆ 2 .These tools are used in this section to prove that NC(P ) is graded when P satisfies Property ∆ 2 (Proposition 4.4) and in Section 6 to prove Theorem B. Throughout the rest of this section, let P be a configuration of n points in C and let A be the corresponding line arrangement.Definition 4.1.For each pair of distinct lines ℓ 1 , ℓ 2 ∈ A, the intersection ℓ 1 ∩ ℓ 2 can be classified into one of four different types (without loss of generality): As a useful shorthand, we write ℓ 1 ⊣ ℓ 2 to mean that ℓ 1 skewers ℓ 2 .When this is the case, note that the convex hull c(ℓ 1 , ℓ 2 ) contains one of the endpoints of ℓ 1 as an internal point -we refer to this as the link vertex for the skewer.Finally, a skewering sequence is a collection of lines ℓ Definition 4.2.Suppose that P satisfies Property ∆ 2 .A skewering tree is a subset T ⊂ A, together with two additional pieces of data -a special element ℓ 0 ∈ T called the initial line and a chosen closed half-plane bounded by ℓ 0 , which we call the positive side of ℓ 0 -with the following properties: • no two elements of T intersect internally; • for all ℓ ∈ T with ℓ ̸ = ℓ 0 , there is a unique line ℓ ′ ∈ T which skewers ℓ; • no element of P is the link vertex for more than one skewer.We can also build a skewering tree inductively as follows: begin with an initial line ℓ 0 ∈ A, select one of the two half-planes bounded by ℓ 0 to be the positive side, and define T = {ℓ 0 }.Next, either stop here or add a line from A to T which is skewered by another element of T such that the requirements above remain satisfied.Repeat this process and stop at any point; the resulting set T is a skewering tree.See Figure 11 for an example.In either construction, we refer to the non-initial lines in T which do not skewer any other lines as leaves.Finally, we say that a skewering tree is maximal if it is not properly contained in any other skewering tree in A.
The following lemma places fairly strong restrictions on the planar structure of skewering trees in configurations which satisfy Property ∆ 2 .Lemma 4.3.Suppose P satisfies Property ∆ 2 , let T ⊂ A be a skewering tree, and let ℓ, ℓ 1 , and ℓ 2 be distinct lines in T such that ℓ is a leaf and ℓ 1 ⊣ ℓ 2 .Then ℓ does not intersect the convex hull c(ℓ 1 , ℓ 2 ).
Proof.First, suppose that ℓ is the last element in a skewering sequence which includes ℓ 1 and ℓ 2 , i.e. that there is a skewering sequence ℓ 0 ⊣ α 1 ⊣ • • • α k such that ℓ = α k , and α i = ℓ 1 and α i+1 = ℓ 2 for some i.The set V ({ℓ 0 , α 1 , . . ., α k }) then consists of 2(k + 1) points, of which at least k are internal: one for each non-leaf element in the skewering sequence.If ℓ were to intersect the convex hull c(ℓ 1 , ℓ 2 ), then one of the endpoints of ℓ would lie in the interior of the convex hull c(ℓ 1 , ℓ 2 , ℓ), but that would mean that the 2(k+1)-element set described above has k+1 internal points, which contradicts our assumption that P satisfies Property ∆ 2 .
On the other hand, suppose that ℓ is in a different skewering sequence than ℓ 1 and ℓ 2 .That is, suppose that T contains the sequences where α k = ℓ, and β i = ℓ 1 and β i+1 = ℓ 2 for some i.Then V ({ℓ 0 , α 1 , . . ., α k , β 1 , . . ., β m }) is a set of 2(k + m + 1) points in P , of which at least k + m are internal.Once again, if ℓ intersected the convex hull c(ℓ 1 , ℓ 2 ), this would imply that our set of 2(k + m + 1) points has k + m + 1 internal points, which Figure 13.No two blocks of this noncrossing partition can be combined to make a coarser noncrossing partition without also including the third, so this is a partition with three blocks which is covered by the maximum element 1, which has one block.again would violate Property ∆ 2 .In both cases, we see that ℓ does not intersect c(ℓ 1 , ℓ 2 ).□ Note that one may "prune" a skewering tree by iteratively removing leaves, and the result remains a skewering tree at each step.Thus, Lemma 4.3 further shows that the non-leaf elements of a skewering tree are similarly constrained.
As mentioned in Section 2, the full partition lattice Π(P ) is graded by the rank function ρ(π) = n − bl(π) for any choice of configuration P , but it is possible to construct P such that NC(P ) is not graded.For example, Figure 13 depicts a noncrossing partition with three blocks which is covered by the maximum element 1 in NC(P ), but not in Π(P ).In the following proposition, we use Lemma 4.3 to show that configurations with Property ∆ 2 avoid this sort of behavior.Proposition 4.4.If P satisfies Property ∆ 2 , then NC(P ) is graded.
Proof.Suppose P has Property ∆ 2 and let π < π ′ in NC(P ) be a covering relation, i.e. there is no π ′′ ∈ NC(P ) with π < π ′′ < π ′ .Then π ′ is obtained from π in the following manner: there is a collection B of k ≥ 2 blocks in π which are removed and replaced by the union U = ∪B to create π ′ .If we can show that k = 2, then we have that ρ(π) + 1 = ρ(π ′ ), and therefore ρ is a rank function for NC(P ).
To this end, let A π denote the lines in A which contain a side of Conv(A) for some block A ∈ π and let A 1 ∈ B. Since k ≥ 2, there is a side of Conv(A 1 ) which does not lie in the boundary of Conv(U ); let ℓ 1 ∈ A π be the line which contains this side.If ℓ 1 does not skewer any other line in A π , then each block in π lies on one side of ℓ 1 or the other, and therefore one may define a new partition π ′′ by replacing B with two blocks, each of which is the union of all blocks in B with interior on a particular side of ℓ 1 .In this case, we would have π ≤ π ′′ < π ′ , which implies π = π ′′ by our initial assumptions, thus k = 2 and we are done.Suppose ℓ 1 does skewer another line ℓ 2 ∈ A π .Without loss of generality, we may choose ℓ 2 to be one of the "closest" to ℓ 1 in the sense that the segment of ℓ 1 between the intersection ℓ 1 ∩ ℓ 2 and the core c(ℓ 1 ) does not intersect any elements of A π .Then ℓ 2 does not contain a side of Conv(U ), so we may repeat the analysis in the previous paragraph: if ℓ 2 does not skewer any other line in A π , then we are done.Otherwise, we can continue this process to obtain a skewering sequence ℓ 1 ⊣ ℓ 2 ⊣ • • • ⊣ ℓ m such that each ℓ i is an element of A π which does not contain a side of Conv(U ), and by Lemma 4.3, we can guarantee that this sequence terminates in a line ℓ m which does not skewer any line in A π .Applying the argument above to ℓ m , the proof is complete.□ Next, we investigate how skewering trees decompose configurations into regions.
Definition 4.5.The union of all lines in a skewering tree T forms a subset of the plane with the structure of an unbounded graph, consisting of vertices, line segments, rays, and lines (one may equivalently view this as a graph embedding on the 2-dimensional sphere, viewed as the stereographic projection of the plane); let Γ T denote the unbounded graph obtained from this by removing any ray which does not contain the core of its corresponding line in T .We refer to Γ T as the planar realization of T , observing that Γ T is an acyclic graph, i.e. a tree.Note that while a pair of rays in Γ T might overlap, Lemma 4.3 implies that no ray in Γ T has a transverse intersection with a line segment.
Definition 4.6.For each non-initial line ℓ in T , the core c(ℓ) is contained in the boundary for exactly one region of the complement C − Γ T (since the other side of ℓ contains the core of the line which skewers it).The cell associated to ℓ, denoted C ℓ , is the union of the interior of this region together with the line segment or ray which contains c(ℓ).Note that C ℓ is a convex (possibly unbounded) subset of the plane which contains exactly one side of its boundary.The initial line ℓ 0 bounds two regions and thus corresponds to two cells: C + ℓ0 , which contains the core c(ℓ 0 ) and belongs to the positive side of ℓ 0 , and C − ℓ0 , which does not.See Figure 14 for an illustration.For any cell C ± ℓ , we write V (C ± ℓ ) to mean the intersection of P with C ± ℓ .Lemma 4.7.The cells of a skewering tree are pairwise disjoint.
Proof.The interior of a cell for a skewering tree is a connected component of the complement C − Γ T , so it follows that the interiors of two cells overlap if and only if the interiors are identical.What remains to be shown is that no connected component of C − Γ T belongs to the cores of two different lines in T .
First, we consider lines α i and α j which come from the same skewering sequence is a connected subset of the plane and by Lemma 4.3, it cannot intersect α j .Thus, this subset lies on one side of α j (the side which contains the cell C αi ), while the cell C αj lies on the other.Thus, the cells associated to α i and α j are disjoint.Now, we consider lines α i and β j which come from distinct skewering sequences Figure 14.Cells associated to the skewering tree in Figure 11.In this case, the positive side of the initial line ℓ 0 is chosen to be the upper-right side; the corresponding cell is shaded orange.
Similar to the previous case, we observe that is a connected subset of the plane and by Lemma 4.3, it must be wholly contained in one of the four components of the complement C − (α i ∪ β j ).In particular, it must belong to the unique component which contains the core of α i and the core of β j in its boundary.Thus, this region of C − (α i ∪ β j ) does not contain the cells for either α i or β j , which means that the two cells are disjoint -see Figure 15 for an illustration.□ Lemma 4.8.Let P be a configuration of n points which satisfies Property ∆ 2 .
Then the cells of a skewering tree cover the elements of P .
In the complement of the lines α i and β j , the region which has the cores of α i and β j in its boundary (shaded blue here) must contain the cores of lines which skewer α i and β j , which means that the cells C αi and C βj lie outside the shaded region.
Proof.Let z ∈ P , let T be a skewering tree and let Ω be the connected component of C − Γ T which contains z.Suppose for the sake of contradiction that Ω does not belong to a cell of T .This immediately rules out the possibility that Ω touches only a single line in T , since that line would necessarily be a leaf and thus Ω would belong to the cell associated to that line.The remaining case to consider is that that there are distinct lines ℓ 1 , ℓ 2 ∈ T such that (1) ℓ 1 and ℓ 2 bound adjacent sides of Ω, and (2) Ω lies in the unique component of C − (ℓ 1 ∪ ℓ 2 ) which has neither c(ℓ 1 ) nor c(ℓ 2 ) in its boundary.Notice that this implies that the endpoints of ℓ 1 , the endpoints of ℓ 2 , and z form a 5-element subset of P where ℓ i and ℓ j each have one (non-link) endpoint in the interior of the convex hull.Similar to the proof of Lemma 4.7, we consider two cases according to whether ℓ 1 and ℓ 2 belong to the same skewering sequence or not.
First, suppose T contains a skewering sequence α 0 ⊣ α 1 ⊣ • • • ⊣ α k such that ℓ 1 = α i and ℓ 2 = α j , where 0 ≤ i < j ≤ k and let A be the set containing both z and the endpoints of α i , α i+1 , . . ., α j .Then |A| = 2(j − i + 1) + 1 and we know that j − i points in A are link vertices for skewers involving other lines in A. Together with the two points mentioned above, we have that A is a set of 2(j − i + 1) + 1 points where j − i + 2 of them are internal, which contradicts our assumption that P satisfies Property ∆ 2 .
Next, suppose that T contains skewering sequences and ℓ 2 = β j for some i, j ≥ 1, and let B be the set containing z together with the endpoints of ℓ 0 , α 1 , . . ., α i , β 1 , . . ., β j .Then |B| = 2(i + j + 1) + 1, and at least i + j elements of B are internal due to being link vertices of skewers in this set.By the same reasoning as in the previous case, we know that i + j + 2 of the 2i + 2j + 3 points in B are internal, which again contradicts our assumption that P satisfies Property ∆ 2 .

0T 1T
Figure 16.The minimum and maximum elements for the interval associated to the skewering tree drawn in Figure 11 Since both cases lead to a contradiction of Property ∆ 2 , we conclude that z must in fact belong to a cell of the skewering tree T , and this completes the proof. □ We now conclude this section by defining a subposet of NC(P ) associated to each skewering tree.Definition 4.9.Let P be a configuration of n points satisfying Property ∆ 2 and let T be a skewering tree for P .We define the skewering interval NC(P, T ) to be the subposet of NC(P ) consisting of all partitions π satisfying the following conditions: (1) for each ℓ ∈ T , the endpoints of ℓ share a block in π; (2) endpoints of distinct lines in T belong to distinct blocks in π; (3) each block in π has a convex hull which lies in a cell of T .
As the name suggests, NC(P, T ) is an interval in NC(P ).Let 0T be the partition in NC(P ) for which the only non-singleton blocks are V (ℓ) for each ℓ ∈ T .Let 1T be the partition where two points in P belong to the same block if and only if they belong to the same cell of T (note that this requires us to choose a positive side for ℓ 0 before discussing the skewering interval).Then NC(P, T ) is the interval [ 0T , 1T ].See Figure 16.
We close the section with some features of skewering intervals which will be useful in the proof of Theorem B. To begin, we provide a product decomposition for skewering intervals.Definition 4.10.Let ℓ be a boundary line in the arrangement A corresponding to P .Define π ℓ to be the partition of P in which the two endpoints of ℓ share a block, while every other block is a singleton.Further, let NC ℓ (P ) denote the interval [π ℓ , 1] in NC(P ).Lemma 4.11.Let T be a skewering tree for P .Then the skewering interval NC(P, T ) is isomorphic to the product Proof.This follows immediately from Definitions 4.9 and 4.10.□ Next, we note that each skewering interval is "centered" in the sense that the rank of its minimum element is equal to the corank of its maximum.Definition 4.12.Let π 0 and π 1 be elements of NC(P ) with π 0 ≤ π 1 .We say that the interval [π Proof.Let T and T ′ be distinct maximal skewering trees and suppose that the partition π is contained in both NC(P, T ) and NC(P, T ′ ).In other words, both 0T ≤ π ≤ 1T and 0T ′ ≤ π ≤ 1T ′ .Since T and T ′ are distinct and maximal, there must be lines α, ℓ, and ℓ ′ such that α, ℓ ∈ T , α, ℓ ′ ∈ T ′ , and α skewers both ℓ and ℓ ′ .Note that by Lemma 4.3, ℓ and ℓ ′ cannot skewer one another, and by definition of a skewering tree, ℓ and ℓ ′ do not intersect internally.Therefore the two lines are either parallel or have an external intersection.Let H − α,ℓ and H − α,ℓ ′ denote the closed half-planes bounded by ℓ and ℓ ′ respectively which do not include the core of α.Since α skewers ℓ, we know that the cell C ℓ (and therefore the core c(ℓ)) must belong to H − α,ℓ ; the analogous statement holds for ℓ ′ .By combining the inequalities above, we see that 0T ≤ 1T ′ , which means that the core c(ℓ) must be contained in the cell C ℓ ′ , which implies that c(ℓ) belongs to H − α,ℓ ′ .Similarly, the fact that 0T ′ ≤ 1T tells us that c(ℓ ′ ) is contained in H − α,ℓ .Combining all of the above, we know that the intersection H − α,ℓ ∩ H − α,ℓ ′ includes both c(ℓ) and c(ℓ ′ ), but not c(α).However, by the same reasoning used in the proof of Lemma 4.7, this is precisely the region in which the core of α must be placed for it to skewer both ℓ and ℓ ′ .Therefore, we have a contradiction, so the skewering intervals for T and T ′ must be disjoint.□

Property ∆ 1 and Catalan Numbers
Our strategy for proving Theorem A is to show that if P satisfies Property ∆ 1 , then applying a ∆ 1 -move to P does not change the size of NC(P ).From here, applying Theorem 3.10 completes the proof.We begin with a useful lemma and some terminology, then give the proof of Theorem A. Proof.We know by definition that R z and R ex z are convex polygons with R z ⊆ R ex z , so we just need to show that each of its sides is a subset of a line in A ex .Suppose that one of the sides for R z is a subset of a line ℓ with endpoints u and v, where v (and possibly u as well) is internal.Then ℓ must not contain any other points in P (since P is assumed to be in general position), so it must eventually intersect the boundary of Conv(P ) in some edge between extremal vertices w 1 and w 2 , and thus v lies within the triangle with vertex set {u, w 1 , w 2 }.Since ℓ was assumed to contain a side of R z , we know that z must also be contained in the same triangle -see Figure 17 for an illustration.But this implies that P does not satisfy Property ∆ 1 , which is a contradiction.□ Remark 5.2.Let P be a configuration satisfying Property ∆ 1 , suppose that z ∈ P is adjacent to a line ℓ in A ex , and let π be a partition of P .Since P satisfies Property ∆ 1 , we know by Lemma 5.1 that ℓ contains an edge of the region R z .In other words, there are no lines in the arrangement A z which lie between ℓ and z.This implies that if B z and B ℓ are the blocks in π containing z and the endpoints of ℓ respectively, then Conv(B z ) and Conv(B ℓ ) are disjoint if and only if B z ∩H − z,ℓ = ∅ and B ℓ ∩ H + z,ℓ = ∅.Definition 5.3.Let m : P → m(P ) be a move.We say that π ∈ Π(P ) is pre-mnoncrossing if π is noncrossing, but its image m * (π) ∈ Π(m(P )) is not.Similarly, we say that µ ∈ Π(m(P )) is post-m-noncrossing if µ is noncrossing, but its preimage m −1 * (µ) ∈ Π(P ) is not.Then NC(m(P )) can be obtained from NC(P ) by the following procedure: remove partitions which are pre-m-noncrossing, apply m * to all remaining elements, then add in the post-m-noncrossing partitions.
We are now ready to prove the main theorem of this section.Proof.By Theorem 3.10, we know there is a sequence of ∆ 1 -moves which transforms P into a convex configuration, for which we know the lattice of noncrossing partitions is isomorphic to NC n .Therefore, we need only show that if m : P → m(P ) is a ∆ 1 -move, then the lattices NC(P ) and NC(m(P )) have the same number of elements in each rank.
Our strategy is to show that the block-switching map BS m restricts to a rankpreserving bijection NC(P ) → NC(m(P )).Suppose that m : P → m(P ) is a ∆ 1 -move which brings the point z across the line ℓ in A ex , where the endpoints of ℓ are w 1 and w 2 , and let m(z) = y.
Let π be a partition of P .Applying Remark 5.
Thus, the block-switching map BS m induces a rank-preserving bijection between the pre-m-noncrossing partitions of P and the post-m-noncrossing partitions of m(P ), and we are done.□

Property ∆ 2 and Rank Symmetry
We now turn our attention to Theorem B, in which we demonstrate that if P satisfies Property ∆ 2 , then NC(P ) is rank-symmetric.As a first step, one could write an explicit bijection between the atoms and coatoms of NC(P ), where P is an arbitrary configuration of n points.This was previously demonstrated by Razen and Welzl in the special case where P is in general position [6] and is straightforward to generalize.In short: each atom is determined by a single line in A, and by slightly rotating this line counterclockwise about the midpoint of its core, we obtain a line which divides P into two pieces, thus producing a coatom in NC(P ) -see Figure 18 for an illustration.
One could hypothetically prove Theorem B by extending the map above to a rank-reversing bijection from NC(P ) to itself, but this seems intractable in general.Instead, our proof technique is similar to that of Theorem A, although the weakening of our hypotheses from Property ∆ 1 to Property ∆ 2 means that Lemma 5.1 no longer holds.That is, it is possible that while some points in P are adjacent to non-separating lines in A, none of these lines are in A ex .As a result, we must account for moves which take an interior point across a line with an interior endpoint, which in turn means we need to understand situations where a partition block overlaps with the line being moved across.To do so, we use the results on skewers developed in Section 4.
First, we require a technical lemma regarding the interval NC ℓ (P ) introduced in Definition 4.10.For the remainder of this section, let P denote a configuration of n points in C which satisfies Property ∆ 2 .Definition 6.1.Let ℓ be a boundary line in the arrangement A corresponding to P .We say that P satisfies Property ∆ 1 relative to ℓ if each B ⊆ P with V (ℓ) ⊆ B has at most ⌊ |B|−2 2 ⌋ internal points.The motivation for the preceding definition comes from its appearance in certain skewering trees for configurations which satisfy Property ∆ 2 .Lemma 6.2.Let T be a skewering tree for P with initial line ℓ 0 .Suppose that ℓ 1 is a line in T with ℓ 0 ⊣ ℓ 1 , and that y is a point in P which lies in the convex hull c(ℓ 0 , ℓ 1 ) on the non-positive side of ℓ 0 .Then V (C + ℓ0 ) satisfies Property ∆ 1 relative to ℓ 0 , and for each ℓ ∈ T with ℓ ̸ = ℓ 0 , V (C ℓ ) satisfies Property ∆ 1 relative to ℓ.
Proof.We prove both claims by contradiction.To start, suppose that V (C + ℓ0 ) does not satisfy Property ∆ 1 relative to ℓ 0 .Then there is a subset B ⊆ V (C + ℓ0 ) which contains the endpoints of ℓ 0 such that B has more than ⌊ |B|−2 2 ⌋ internal points.If we define B ′ = B ∪ V (ℓ 1 ) ∪ {y}, then the internal points of B, together with y and one endpoint of ℓ 0 , are all internal points of B ′ .Therefore, B ′ has more than points, which violates the assumption that P satisfies Property ∆ 2 .Thus V (C + ℓ0 ) satisfies Property ∆ 1 relative to ℓ 0 .Similarly, let ℓ ∈ T with ℓ ̸ = ℓ 0 and suppose that there is a subset B ⊆ V (C ℓ ) which contains the endpoints of ℓ such that B has more than ⌊ |B|−2 2 ⌋ internal points.If ℓ = ℓ 1 , then we can define B ′ = B ∪ V (ℓ 0 ) ∪ {y} and observe by the same reasoning as above that B ′ has more than ⌊ |B ′ |−1 2 ⌋ internal points, which provokes a contradiction.Suppose instead that ℓ ̸ = ℓ 1 .Then there is a skewering sequence and observe that the internal points of B ′ include all the internal points of B, as well as one endpoint of each α i , one endpoint of ℓ 0 , and y.Therefore, the number of internal points in B ′ is more than which violates Property ∆ 2 .Thus, V (C ℓ ) must have Property ∆ 1 relative to ℓ. □ Lemma 6.3.Suppose that P satisfies Property ∆ 1 relative to the boundary line ℓ and that for each proper subset Q ⊂ P , the poset of noncrossing partitions NC(Q) is rank-symmetric.Then NC ℓ (P ) is rank-symmetric.
Proof.We proceed by induction on the number of internal points of P .First, if P has no internal points, then since ℓ was assumed to be a boundary line, we can see that NC ℓ (P ) is isomorphic to NC n−1 , which is rank-symmetric.Now, suppose that the claim is true for any configuration with up to k − 1 internal points which satisfies the lemma's hypotheses, and let P be a configuration with k internal points such that for all proper subsets Q ⊂ P , we know that NC(Q) is rank-symmetric.Let u and v be the endpoints of ℓ, and let P v denote the complement P − {v}.
We will compare the interval [π ℓ , 1] ⊂ NC(P ) to the noncrossing partition lattice NC(P v ), which we know is rank-symmetric by assumption.Define the map ϕ v : NC ℓ (P ) → NC(P v ) by removing v from each partition in the domain and observe that ϕ v is always injective, but typically not surjective.Our goal is to show that NC ℓ (P ) is rank-symmetric; since NC(P v ) is assumed to be rank-symmetric and ϕ v is injective, it suffices to show that the complement NC(P v ) − ϕ v (NC ℓ (P )) is a union of centered, disjoint, rank-symmetric intervals.
Let W ⊂ P −{u, v} be the set of all points w with the property that int({u, v, w}) is nonempty and note that since P satisfies Property ∆ 1 relative to ℓ, we must have |int({u, v, w})| = 1.Then NC(P v ) − ϕ v (NC ℓ (P )) is the collection of all partitions σ of P v with a block which contains both u and a point w ∈ W , but not the unique point in int({u, v, w}).Rephrasing this characterization in the language of skewering trees, let T be the collection of skewering trees for P v such that the initial line ℓ 0 has endpoints u and w for some w ∈ W and the positive side of ℓ 0 is chosen to be the one which does not include v; then a partition σ ∈ NC(P v ) lies in NC(P v ) − ϕ v (NC ℓ (P )) if and only if it belongs to the skewering interval for some skewering tree in T .Together with Lemmas 4.13 and 4.14, this tells us that the skewering intervals for trees in T form a collection of centered and disjoint intervals whose union is NC(P v ) − ϕ v (NC ℓ (P )).All that remains is to show that each such skewering interval is rank-symmetric.
Fix a skewering tree T ∈ T and consider the skewering interval NC(P v , T ).By Lemma 4.11, we have a poset isomorphism By our assumption that every proper subset of P has a rank-symmetric lattice of noncrossing partitions, we know that NC(P v ∩ C − ℓ0 ) is rank-symmetric.For the middle term in the product, we claim that the set P v ∩ C + ℓ0 satisfies Property ∆ 1 relative to the boundary line ℓ 0 .To see this, recall that the endpoints of ℓ 0 are the extremal point u and some w ∈ W , and define w ′ be the unique point of P which lies in the interior of Conv({u, v, w}.If there were points x, x ′ , x ′′ ∈ P v ∩ C + ℓ0 such that both x ′ and x ′′ lie in the convex hull Conv({u, w, x}), then {u, v, w, w ′ , x, x ′ , x ′′ } would be a set of seven points, at least three of which are internal (w ′ , x ′ , and x ′′ ).This would violate our assumption that P has Property ∆ 1 relative to ℓ, so it must be the case that P v − C + ℓ0 satisfies Property ∆ 1 relative to ℓ 0 , and by our inductive hypothesis, we know that NC ℓ0 (P v ∩ C + ℓ0 ) is rank-symmetric.Generalizing this argument, we now show that P v ∩ C α satisfies Property ∆ 1 relative to α for each non-initial line α in T .Let α 1 ⊣ • • • ⊣ α m be a skewering sequence in T such that α 1 = ℓ 0 and α m = α.If there are points x, x ′ , x ′′ ∈ P v ∩C α such that x ′ and x ′′ lie in the triangle formed by x and the endpoints of α, and if w and w ′ are defined as above, then {x, x ′ , x ′′ , w ′ , v} ∪ V (α 1 ) ∪ • • • ∪ V (α m ) is a set of 2m + 5 points, of which at least m + 2 points must be internal -see Figure 19 for an illustration.Since this set contains both u and v, this violates our assumption that P satisfies Property ∆ 1 relative to ℓ, so we may conclude that P v ∩C α satisfies Property ∆ 1 relative to α, as desired.By the inductive hypothesis, NC α (P v ∩ C α ) is rank-symmetric.
Finally, we have that NC(P v , T ) is a product of rank-symmetric posets and is therefore rank-symmetric itself, which completes the proof.□ Theorem 6.4 (Theorem B).Let P ⊂ C be a set of n distinct points which satisfies Property ∆ 2 .Then NC(P ) is a rank-symmetric graded lattice.
Proof.We proceed by induction on the number of internal points for P .When P has no internal points, NC(P ) is isomorphic to the classical noncrossing partition lattice NC n , which is rank-symmetric.Now, suppose that every configuration with fewer than |int(P )| internal points has a rank-symmetric lattice of noncrossing partitions; we will show that NC(P ) is rank-symmetric as well.By Theorem 3.10, there is a sequence of ∆ 2 -moves which transforms P into a convex configuration, for which the lattice of noncrossing partitions is isomorphic to NC n and thus ranksymmetric.The only remaining step is to prove that if m : P → m(P ) is a ∆ 2 -move, then NC(P ) is rank-symmetric if and only if NC(m(P )) is rank-symmetric.
In the proof of Theorem A, we showed that when P satisfies Property ∆ 1 , the block-switching map BS m is a rank-preserving bijection between pre-m-noncrossing partitions of P and post-m-noncrossing partitions of m(P ).If P is only assumed to satisfy Property ∆ 2 , the block switching map might not be a bijection; there may be partitions π ∈ NC(P ) such that BS m (π) is not in NC(m(P )), or elements σ ∈ NC(m(P )) such that BS −1 m (σ) does not lie in NC(P ).To complete the proof, we must show that these two collections are rank-symmetric subposets of NC(P ) and NC(m(P )) respectively.By symmetry, it suffices to examine the first collection.
Suppose that the move m takes a point z ∈ P across a line ℓ ∈ A with endpoints e(ℓ) = {w 1 , w 2 } and let F m (P ) denote the noncrossing partitions of P which fail to be accounted for by the block-switching map BS m ; that is, F m (P ) = {π ∈ NC(P ) | BS m (π) ̸ ∈ NC(m(P ))}.
If F m (P ) is empty, then there is nothing to prove.Otherwise, for each π ∈ F m (P ), there are points x, y ∈ P such that y, z ∈ int({w 1 , w 2 , x}), and these five points are divided into three distinct blocks of π (each of which might contain other points) as follows: x and z belong to one block, w 1 and w 2 belong to another, and y lies in a third -see Figure 20.Using this characterization, we will prove that (1) F m (P ) is a union of skewering intervals, (2) each of these skewering intervals is centered and rank-symmetric, and (3) intersections of the skewering intervals are centered and rank-symmetric.On the left and right, the minimum and maximum elements for a skewering interval are given.The intersection of these intervals is another interval, albeit one which does not arise from a skewering tree; its minimum and maximum elements are displayed in the center.

Figure 1 .
Figure 1.Some examples of configurations which satisfy Property ∆ 1

Figure 2 .
Figure 2. Some examples of configurations which satisfy Property ∆ 2

Example 2 . 4 .Figure 3 .
Figure 3.The lattice of noncrossing partitions for a particular arrangement of four points in C

Figure 5 .
Figure 5.The Boolean lattice Bool n−1 arises as the set of noncrossing partitions for a configuration of n collinear points.

2 Figure 7 .Definition 3 . 6 .
Figure 7.If m : P → m(P ) is a move which takes z across a non-separating edge between extremal points w 1 and w 2 , and if the triple {a 1 , a 2 , m(z)} (indicated with dashed lines) has more than one point in its interior, then the quadrilateral {a 1 , a 2 , w 1 , w 2 } (indicated with solid lines) has at least three points in its interior.

Figure 8 .Lemma 3 . 8 .
Figure8.If w 2 is an internal point and m is a move which takes z across the line containing w 1 and w 2 , then there are extremal points a 1 and a 2 such that the convex hull of a 1 , a 2 , and w 1 (depicted with dashed lines) has w 2 , z, and m(z) in its interior.

Figure 9 .
Figure 9.This configuration of 15 points has five internal points for which the corresponding regions have a convex hull (outlined with dashed blue lines in the upper left image) where no side of the convex hull contains a side of a region.By extending each side of the convex hull into a line, each of the five points determines a cone (shaded red in the upper right image) which contains at least one extremal point.Selecting one extremal point from each cone (highlighted red in the bottom image) yields a set of extremal points which contains the starting internal points and is at most as numerous, thus violating Property ∆ 2 .

Figure 10 .
Figure 10.The block-switching map BS m , compared to the induced map m * for a fixed move m.In this example, BS m takes an element of NC(P ) to an element of NC(m(P )).
Figure11.A skewering tree for a configuration of 14 points which satisfies Property ∆ 2 , but not Property ∆ 1 .The tree consists of two skewering sequences: ℓ 0 ⊣ α 1 ⊣ α 2 and ℓ 0 ⊣ β 1 .For the sake of visual clarity, only the cores of the four lines have been drawn.

ℓ 1 ℓ 2 ℓFigure 12 .
Figure12.If ℓ intersects the convex hull of ℓ 1 and ℓ 2 (denoted in blue), then the set of all endpoints has half of its elements in the interior (shown here as unfilled red dots).

Lemma 4 .
13.Each skewering interval NC(P, T ) is a centered subposet of NC(P ).Proof.Let T be a skewering tree for P .Then the partition 0T has rank |T | and 1T has rank |P − (|T | + 1), so the interval [ 0T , 1T ] is centered.□ Finally, we examine the ways in which skewering intervals can intersect.Lemma 4.14.If P has Property ∆ 2 , then distinct maximal skewering trees with the same initial line and choice of positive side yield disjoint skewering intervals.

Lemma 5 . 1 . 2 Figure 17 .
Figure 17.If the region R z has a side which belongs to a line through an interior point v, then z and v must both belong to a common triangle.

Theorem 5 . 4 (
Figure 18.An atom in NC(P ) with the corresponding coatom

Figure 19 .
Figure19.If α 1 ⊣ α 2 ⊣ α 3 is a skewering sequence for P and the points in the cell for α 3 do not satisfy Property ∆ 1 relative to α 3 (illustrated on the left in blue), then there is a set of points containing u and v (illustrated on right in red) which demonstrate that P does not satisfy Property ∆ 1 relative to the line with endpoints u and v.

2 Figure 20 .
Figure20.For the depicted partition π of P , if m moves z across the line containing w 1 and w 2 , then BS m (π) is not an element of NC(m(P )) since it has w 1 , w 2 , and x together in a single block without the interior point y.
Figure21.On the left and right, the minimum and maximum elements for a skewering interval are given.The intersection of these intervals is another interval, albeit one which does not arise from a skewering tree; its minimum and maximum elements are displayed in the center.