Facets of the r-stable n,k-hypersimplex

Let $k, n$ and $r$ be positive integers with $k<n$ and $r\leq\lfloor\frac{n}{k}\rfloor$. We determine the facets of the $r$-stable $n,k$-hypersimplex. As a result, it turns out that the $r$-stable $n,k$-hypersimplex has exactly $2n$ facets for every $r<\lfloor\frac{n}{k}\rfloor$. We then utilize the equations of the facets to study when the $r$-stable hypersimplex is Gorenstein. For every $k>0$ we identify an infinite collection of Gorenstein $r$-stable hypersimplices, consequently expanding the collection of $r$-stable hypersimplices known to have unimodal Ehrhart $\delta$-vectors.


Introduction
The r-stable n, k-hypersimplex was introduced in [2]. First we recall the definition of the rstable n, k-hypersimplex. Let [n] = {1, 2, . . . , n}. The characteristic vector of a subset I of [n] is the (0, 1)-vector ǫ I = (ǫ 1 , . . . , ǫ n ) for which ǫ i = 1 for i ∈ I and ǫ i = 0 for i / ∈ I. Fix integers k with 0 < k < n and let r be an integer with 1 ≤ r ≤ n k . We label the vertices of a regular n-gon embedded in R 2 in a clockwise fashion from 1 to n. A subset S ⊂ [n] is called r-stable if, for each pair i, j ∈ S, the path of shortest length from i to j about the n-gon uses at least r edges. The r-stable n, k-hypersimplex, denoted by ∆ stab(r) n,k , is the convex polytope in R n which is the convex hull of the characteristic vectors of all r-stable k-subsets of [n]. In particular the 1-stable n, k-hypersimplex ∆ stab(1) n,k is just the n, k-hypersimplex ∆ n,k ([9, p.75]). These convex polytopes form the nested chain ∆ n,k ⊃ ∆ stab(2) n,k ⊃ ∆ stab(3) n,k ⊃ · · · ⊃ ∆ stab(⌊ n k ⌋) n,k .
A regular unimodular triangulation of ∆ n,k which can be restricted to a regular unimodular triangulation of each polytope ∆ stab(r) n,k in the above chain is studied in [2]. In the present paper, we utilize the regular unimodular triangulation to compute the facets of ∆ stab(r) n,k and then study when ∆ stab(r) n,k is Gorenstein. In section 2, we recall the details of the regular unimodular triangulation of ∆ stab(r) n,k . In section 3, we compute the facets of ∆ stab(r) n,k for r < n k (Theorem 3.1). As a result, it turns out that ∆ stab(r) n,k has exactly 2n facets for every r < n k (Corollary 3.25). Finally in section 4, we classify 1 ≤ r < n k for which ∆ stab(r) n,k is Gorenstein (Theorem 4.8). We conclude that the Ehrhart δ-vector of ∆ stab(r) n,k is unimodal if it is Gorenstein (Corollary 4.10).
2. The Regular Unimodular Triangulation of ∆ stab(r) n,k In [7], Lam and Postnikov compare four different triangulations of the hypersimplex and show that they are identical. One construction of this triangulation, known as the circuit triangulation, is introduced in [7]. In [2], it is shown that the circuit triangulation restricts to a triangulation of each r-stable n, k-hypersimplex. For the purposes of this paper it will be helpful to recall the details of this construction.
Fix 0 < k < n, and let G n,k be the labeled, directed graph with the following vertices and edges. The vertices of G n,k are all the vectors ǫ I ∈ R n where I is a k-subset of [n]. We think of the indices of a vertex of G n,k modulo n. Now suppose that ǫ and ǫ ′ are two vertices of G n,k such that for some i ∈ [n] (ǫ i , ǫ i+1 ) = (1, 0) and ǫ ′ is obtained from ǫ by switching the order of ǫ i and ǫ i+1 . Then the directed and labeled edge ǫ i → ǫ ′ is an edge of G n,k . Hence, an edge of G n,k corresponds to a move of a single 1 in a vertex ǫ one spot to the right, and such a move can be done if and only if the next spot is occupied by a 0.
We are interested in the circuits of minimal length in the graph G n,k . Such a circuit is called a minimal circuit. Suppose that ǫ is a vertex in a minimal circuit of G n,k . Then the minimal circuit can be thought of as a sequence of edges in G n,k that moves each 1 in ǫ into the position of the 1 directly to its right (modulo n). It follows that a minimal circuit in G n,k has length n. An example of a minimal circuit in G 6,3 is provided in Figure 1. Notice that for a fixed initial vertex of the minimal circuit the labels of the edges form a permutation ω = ω 1 ω 2 · · · ω n ∈ S n , the symmetric group on n elements. Moreover, the permutations corresponding to two different choices of initial vertex will always be equivalent modulo cyclic shifts ω 1 · · · ω n ∼ ω n ω 1 · · · ω n−1 . By convention, we associate a minimal circuit in G n,k with the permutation in S n corresponding to the lexicographically maximal choice of starting vertex. This corresponds to picking the permutation consisting of the labels of the edges of the circuit for which ω n = n.
Theorem 2.1. [7,Lam and Postnikov] A minimal circuit in the graph G n,k is uniquely determined by the permutation ω modulo cyclic shifts. A permutation ω ∈ S n such that ω n = n corresponds to a minimal circuit in the graph G n,k if and only if the inverse permutation ω −1 has exactly k − 1 descents.
We let (ω) denote the minimal circuit in G n,k corresponding to the permutation ω ∈ S n with ω n = n. Let v (ω) denote the set of vertices of (ω), and let σ (ω) denote the convex hull of v (ω) . Notice that σ (ω) will always be an (n − 1)-simplex. Lam and Postnikov] The collection of simplices σ (ω) given by the minimal circuits in G n,k are the maximal simplices of a triangulation of the hypersimplex ∆ n,k . We call this triangulation the circuit triangulation.
Denote the circuit triangulation of ∆ n,k by ∇ n,k , and let max ∇ n,k denote the set of maximal simplices of ∇ n,k . To simplify notation we will write ω to denote the simplex σ (ω) ∈ max ∇ n,k . induced by ∇ n,k , and let max ∇ r n,k denote the set of maximal simplices of ∇ r n,k . Notice that we have the nesting of triangulations ∇ n,k ⊃ ∇ 2 n,k ⊃ ∇ 3 n,k ⊃ · · · ⊃ ∇ ⌊ n k ⌋ n,k .
In the coming section we will utilize this nesting of triangulations to compute the facets of ∆ stab(r) n,k .
3. The Facets of ∆ stab(r) n,k It is a well-known fact that the facets of the n, k-hypersimplex ∆ n,k ⊂ R n−1 are given by the supporting hyperplanes Let H denote the hyperplane in R n and consider the affine isomorphism Notice that ϕ(Z n−1 ) = H ∩ Z n . Under the affine isomorphism ϕ we see that the supporting hyperplanes of ∆ n,k ⊂ R n are given by intersecting the hyperplane H with the hyperplanes , and In the following we compute the facets of the r-stable n, k-hypersimplex ∆ stab(r) n,k . To do so, we compute the (n − 1)-dimensional hyperplanes that, when intersected with H, give the supporting hyperplanes of ∆ n,k ⊂ R n . Our main result is the following theorem.
Computing Facets with a Nesting of Triangulations. As we noted in section 2, we will use the nested triangulations ∇ r n,k to prove Theorem 3.1. To do so, we will use the following theorem of Lam and Postnikov [7].
Theorem 3.2. [7, Lam and Postnikov] Two simplices u and ω are adjacent simplices (i.e. they share a facet) of max ∇ n,k if and only if there exists some i ∈ [n] such that u i − u i+1 = ±1 mod n and ω is obtained from u by switching u i with u i+1 .
We should note that since our goal is to study (n − 1)-dimensional polytopes embedded in the hyperplane H ⊂ R n , we will refer to the (n − 2)-dimensional flats corresponding to supporting hyperplanes of these polytopes under the embedding ϕ simply as the supporting hyperplanes of the polytope. We do this because the simplices of the circuit triangulation are best described in R n , and we will use these simplices to compute the facets of ∆ stab(r) n,k . Remark 3.3. If u and ω are adjacent simplices in max ∇ n,k then the supporting hyperplane, say H u,ω , of the shared facet is spanned by the common vertices of u and ω. Moreover, since ∇ n,k is a triangulation of ∆ n,k then u and ω must lie in opposite closed halfspaces defined by H u,ω .
Example 3.4. Here is an example of two adjacent maximal simplices in the triangulation max ∇ 10,3 of ∆ 10,3 . Consider the simplices ω and u in max ∇ 10,3 given by the circuits in Figures 2 and 3, respectively.
With Theorem 3.2 in hand, we are now ready to compute the facets of ∆ stab(r) n,k . For now, we assume ∆ stab(r) n,k is (n − 1)-dimensional. We will soon see that this is almost always true. In the following, for a convex polytope P ⊂ R N of dimension d, we will let ∂P denote the boundary of P , and relint(P ) denote the relative interior of P .
Once more, we remark that we only wish to study (n − 1)-dimensional polytopes embedded in the hyperplane H ⊂ R n . For this reason, we refer to the (n − 2)-dimensional flats in R n that correspond to the supporting hyperplanes of these polytopes under the affine isomorphism ϕ simply as the supporting hyperplanes of the polytope. This will be useful because of the nice description of the simplices in the circuit triangulation in R n .
, and is therefore contained in a facet of ∆ stab(r−1) n,k . Thus, H F is a supporting hyperplane of ∆ stab(r−1) n,k . We can think of the facets of ∆ stab(r) n,k that share supporting hyperplanes with facets of ∆ stab(r−1) n,k as having preserved (supporting) hyperplanes, and those facets that intersect the relative interior of ∆ stab(r−1) n,k as having new (supporting) hyperplanes. Hence, to determine the facets of ∆ stab(r) n,k it suffices to determine the set of preserved hyperplanes and the set of new hyperplanes. In the following, we let for ℓ ∈ [n] and 1 ≤ r ≤ n k . The preserved supporting hyperplanes. We first determine the set of preserved hyperplanes of ∆ stab(r) n,k . Proposition 3.6. Let 1 < r < n k . The supporting hyperplanes H ℓ for ℓ ∈ [n] of ∆ n,k are also supporting hyperplanes of ∆ stab(r) n,k . That is, the hyperplane H ℓ is preserved for every ℓ ∈ [n] and 1 < r < n k .
Proof. First notice that ∆ stab(r) n,k clearly lies in the closed halfspace x i ≥ 0. This is because ∆ n,k lies in this closed halfspace. Hence, to prove the claim it suffices to identify a simplex ω ∈ max ∇ r n,k such that H ℓ is a supporting hyperplane of ω.
Fix i ∈ [n]. When 1 < r < n k we may construct a minimal circuit in the graph G n,k that corresponds to a simplex in max ∇ r n,k with supporting hyperplane H ℓ . To do this, we first construct an initial vertex, say ǫ ℓ , for this circuit in the following way: (5) Make all other entries 0. For ℓ = 2, this produces the following vertex: Notice that for the vertex ǫ ℓ to be r-stable we need at least (r − 1) 0's between each pair of consecutive 1's. This is guaranteed by our construction for all pairs of 1's except for 1 k and 1 1 . However, the total number of entries accounted for by the k 1's and the 0's between the pairs Here, we count r entries for each 1 t with t = 2, k, since this counts each such 1 and the r − 1 zeros following it. Then we count 2r entries for 1 2 and the 2r − 1 zeros that follow it. Finally we add 1 for the entry filled by 1 k . Since we need at least r − 1 zeros between 1 k and 1 1 then we want at least r − 1 entries between these two 1's. In other words, we want Now notice that since r < n k then r+1 ≤ n k . Moreover, n k = n−α k for some α ∈ {0, 1, . . . , k−1}. Hence, and so k(r + 1) + α ≤ n.
Hence, k(r + 1) ≤ n. Finally, notice that Thus, there are at least (r − 1) zeros between 1 1 and 1 k , and we conclude that ǫ ℓ is r-stable. Now we construct a minimal circuit in G n,k , which we will denote (ω ℓ ), starting with initial vertex ǫ ℓ . To do so, make the following sequence of moves.
(3) Repeat step (2) r − 1 more times. (4) Move 1 k until it rests in entry ℓ − 1. It is clear that this produces a minimal circuit in G n,k since each 1 t has moved precisely enough times to replace 1 t+1 . Moreover, every vertex in the circuit is r-stable. To see this, notice that 1 2 first makes r moves of its necessary 2r moves. Since 1 2 must make a total of 2r moves to replace 1 3 then the pair {1 2 , 1 3 } is still r-stable. Moreover, there are now 2r − 1 0's between 1 1 and 1 2 . So 1 1 may move once and not violate r-stability. Now there are r 0's between 1 k and 1 1 . So 1 k may move once without violating r-stability. Now there are r 0's between 1 k−1 and 1 k . So 1 k−1 may move once without violating r-stability, and so on. Hence, steps (2) and (3) preserve r-stability in the circuit (ω ℓ ). Finally, r-stability is also preserved in step (4) since all 1 t 's for t = k have already assumed the position of 1 t+1 . In particular, 1 1 is in the entry that was originally occupied by 1 2 , and this entry is (r − 1) entries away from entry ℓ − 1. Hence, the minimal circuit (ω ℓ ) uses only r-stable vertices.
It follows that ω ℓ ∈ max ∇ r n,k . Moreover, since r > 1, the simplex ω ℓ has only one vertex satisfying x ℓ = 1, and this is the vertex following ǫ ℓ in the circuit (ω ℓ ). Hence, all other vertices of ω ℓ satisfy x ℓ = 0. So H ℓ is a supporting hyperplane of ω ℓ . Since ∆ stab(r) n,k lies in the closed halfspace Here is an example of the circuit constructed in the proof of Proposition 3.6. Let n = 10, k = 3, r = 2, and ℓ = 4. Then the vertex ǫ ℓ is The minimal circuit (ω ℓ ) produced by the algorithm in the proof of Proposition 3.6 is depicted in Figure 4. Notice ω ℓ = 341752869(10) ∈ max ∇ 2 10,3 , and the only vertex in (ω ℓ ) that satisfies x ℓ = 1 is the vertex immediately following ǫ ℓ .
The following theorem on the dimension of ∆ stab(r) n,k follows from the proof of Proposition 3.6.
In [2], it is shown that ∆ stab(r) n,k is a unimodular (n − 1)-simplex for n ≡ 1 mod k and r = n k . This result, together with Theorem 3.8, tells us that every polytope in the chain is (n − 1)-dimensional except for (possibly) the smallest polytope in the chain for n ≡ 1 mod k. Thus, we see that most of the r-stable hypersimplices are (n − 1)-dimensional. The following Proposition will be useful in an inductive argument we use to prove Theorem 3.1. is not a supporting hyperplane of ∆ stab(r) n,k . That is, H ℓ,r−1 is not a preserved hyperplane for any (n − 1)-dimensional r-stable hypersimplex.
Proof. Suppose for the sake of contradiction that H ℓ,r−1 is a supporting hyperplane of ∆ stab(r) n,k . Since ∆ stab(r) n,k is (n − 1)-dimensional then there exists an (n − 1)-simplex ω ∈ max ∇ r n,k such that H ℓ,r−1 is a supporting hyperplane of ω. In other words, every vertex in (ω) satisfies ℓ+r−2 i=ℓ x i = 1 except for exactly one vertex, say ǫ ⋆ . Since all vertices in (ω) are (0, 1)-vectors, this means all vertices other than ǫ ⋆ have exactly one entry in the set of entries {ℓ, ℓ + 1, ℓ + 2, . . . , ℓ + r − 2} being 1 and all other entries in this set are 0. Similarly, the vertex ǫ ⋆ has a 0 in all entries in the set {ℓ, ℓ + 1, ℓ + 2, . . . , ℓ + r − 2}. Since (ω) is a minimal circuit this means that the move preceding the vertex ǫ ⋆ in (ω) results in the only 1 in the set of entries {ℓ, ℓ + 1, ℓ + 2, . . . , ℓ + r − 2} leaving this set of entries. Similarly, the move following the vertex ǫ ⋆ in (ω) results in a single 1 moving into the set of entries {ℓ, ℓ + 1, ℓ + 2, . . . , ℓ + r − 2}. Suppose that Hence, neither the vertex preceding or following the vertex ǫ ⋆ is r-stable. For example, in the vertex following ǫ ⋆ there is a 1 in entries ℓ and ℓ+r−1. This contradicts the fact that ω ∈ max ∇ r n,k . Hence, H ℓ,r−1 is not a supporting hyperplane of ∆ stab(r) n,k . To see why Proposition 3.9 will be useful suppose that Theorem 3.1 holds for ∆ stab(r−1) n,k for some 1 < r < n k . Then Propositions 3.6 and 3.9 tell us that the collection of preserved hyperplanes for ∆ In this way, we will be able to prove Theorem 3.1 by induction on r.
The new supporting hyperplanes. We begin with a helpful definition. • u ∈ max ∇ r n,k , • ω ∈ max ∇ r−1 n,k \ max ∇ r n,k , and • ω uses exactly one vertex that is not r-stable, and this is the only vertex by which u and ω differ. We then say that the ordered pair of simplices (u, ω) is an r-supporting pair of the hyperplane H u,ω , where H u,ω is the hyperplane spanned by the common vertices of u and ω.
Then H F is the supporting hyperplane of a pair of adjacent simplices u and ω that form an r-supporting pair for the hyperplane H u,ω , and That is, there exists some α ∈ F such that α ∈ relint ∆ stab(r−1) n,k . Now recall that ∇ r−1 n,k is a triangulation of ∆ stab(r−1) n,k that restricts to a triangulation ∇ r n,k of ∆ is (n − 1)dimensional we may assume, without loss of generality, that α lies in the relative interior of an (n − 2)-dimensional simplex in the triangulation of ∂∆ stab(r) n,k ∩ relint ∆ stab(r−1) n,k induced by ∇ r n,k and ∇ r−1 n,k \∇ r n,k . Therefore, there exists some u ∈ max ∇ r n,k such that H F is a supporting hyperplane of u and α ∈ u ∩ H F , and there exists some ω ∈ max ∇ r−1 n,k \ max ∇ r n,k such that α ∈ ω ∩ H F . Since ∇ r−1 n,k is a triangulation of ∆ stab(r−1) n,k it follows that u ∩ H F = ω ∩ H F . Hence, u and ω are adjacent simplices that share the supporting hyperplane H F , and they form an r-supporting pair (u, ω) with H u,ω = H F . Suppose (u, ω) is an r-supporting pair. It will be helpful to understand the vertex of ω that is not r-stable. To do so, we will use the following definition.
Definition 3.13. Let ǫ ∈ R n be a vertex of ∆ n,k . A pair of 1's in ǫ is an ordered pair of two coordinates of ǫ, (i, j), such that ǫ i = ǫ j = 1, and ǫ t = 0 for all i < t < j (modulo n). A pair of 1's is called an r-stable pair if there are at least r − 1 0's separating the two 1's.
Suppose a hyperplane K has the r-supporting pair (u, ω). We would now like to understand the possible (r − 1)-stable but not r-stable vertices that can be used by the simplex ω. We call this vertex the key vertex of the r-supporting pair (u, ω).
Proposition 3.15. Suppose (u, ω) is an r-supporting pair, and let ǫ be the key vertex of this pair. Then ǫ has precisely one (r − 1)-stable but not r-stable pair, say (i, j), this pair is followed by an (r + 1)-stable pair (j, t), and all other pairs of 1's in ǫ are r-stable.
Proof. Consider the minimal circuit (ω) in the graph G n,k associated with the simplex ω. Think of the key vertex ǫ as the initial vertex of this circuit, and recall that each edge of the circuit corresponds to a move of exactly one 1 to the right by exactly one entry. Hence, in the circuit (ω) the vertex following ǫ differs from ǫ by a single right move of a single 1. Since ǫ is the only vertex in (ω) that is (r − 1)-stable but not r-stable then the move of this single 1 to the right by one entry must eliminate all pairs that are (r − 1)-stable but not r-stable. Moreover, this move cannot introduce any new (r − 1)-stable but not r-stable pairs. Since a single 1 can be in at most two pairs, and this 1 must move exactly one entry to the right, then this 1 must be in entry j in the pairs (i, j) and (j, t) where (i, j) is (r − 1)-stable but not r-stable, and (j, t) is (r + 1)-stable. Moreover, since the move of the 1 in entry j can only change the stability of the pairs (i, j) and (j, t) then it must be that all other pairs are r-stable.
Example 3.16. Consider the key vertex of the r-supporting pair (u, ω) from Example 3.11. This is the vertex (1, 0, 1, 0, 0, 0, 1, 0, 0, 0). We saw in Example 3.14 that this vertex has one 2-stable but not 3-stable pair of 1's, (1,3), and the other two pairs of 1's are 3-stable. Notice now that the pair of 1's (3, 7) is in fact 4-stable. Hence, the key vertex of the r-supporting pair (u, ω) indeed satisfies Proposition 3.15. Moreover, it is easy to see from the circuit (ω) depicted in Figure 2 that we must move the 1 in entry 3 of the key vertex to ensure that no other vertices in the circuit contain an (r − 1)-stable but not r-stable pair. Proposition 3.15 says that the key vertex of an r-supporting pair (u, ω) for H u,ω contains an (r − 1)-stable but not r-stable pair followed by an (r + 1)-stable pair, and all other pairs in the vertex are r-stable. We can construct all such possible key vertices by picking the (r − 1)-stable but not r-stable pair, (ℓ, ℓ + r − 1), then picking the next pair (ℓ + r − 1, j) to be (r + 1)-stable, and then distributing the remaining 1's across the remaining entries of the vertex in an r-stable fashion. With this in mind, consider the following proposition.
The r-supporting pairs (u, ω) and (u ′ , ω ′ ) have respective key vertices (1, 1, 0, 0, 1, 0, 0, 0, 0) and (1, 1, 0, 0, 0, 1, 0, 0, 0), and these vertices have the same (r − 1)-stable but not r-stable pair, namely (1,2). Hence, ω and ω ′ are simplices in max ∇ 9,3 \ max ∇ 2 9,3 that are in r-supporting pairs for which the key vertices share the same (r − 1)-stable but not r-stable pair. It is then easy to see that the vertices of the circuits (ω) and (ω ′ ) that are not the key vertices satisfy x 1 + x 2 = 1. Hence, Remark 3.19. Notice that Proposition 3.15 is not an "if and only if" statement. It is not necessary that each vertex that has an (r − 1)-stable but not r-stable pair followed by an (r + 1)-stable pair, and has all other pairs being r-stable needs to be the key vertex of an r-supporting pair of a supporting hyperplane of ∆ stab(r) n,k . However, Proposition 3.17 says that if two such vertices happen to agree on their (r − 1)-stable but not r-stable pair, and both happen to be vertices of simplices in r-supporting pairs then they are in r-supporting pairs of the same hyperplane and that hyperplane is H ℓ,r for some ℓ ∈ [n].
Example 3.20. Let n = 10, k = 3, and r = 3. In this case, the possible key vertices for rsupporting pairs come in two types. (1, 0, 0, 0, 0, 1, 0, 0, 1, 0) (0, 1, 0, 0, 0, 1, 0, 0, 0, 1) (0, 1, 0, 0, 0, 0, 1, 0, 0, 1) The Type 1 vertices are all equivalent modulo cyclic shifts, as are the Type 2 vertices. Notice all of these vertices consist of an (r − 1)-stable but not r-stable pair, followed by an (r + 1)-stable pair, and have all other pairs being r-stable. The (r − 1)-stable but not r-stable pair has its first 1 highlighted in red in the above table. Notice now that only the Type 1 vertices are used as key vertices in r-supporting pairs. To see this, recall that the first vertex in the Type 1 column is the key vertex in the r-supporting pair (u, ω) described in Example 3.11. Similar to this example, every other Type 1 vertex is the key vertex in an r-supporting pair that uses the simplex u. This is easily verified by switching the order of the moves around each vertex in the minimal circuit (u), one at a time. On the other hand, notice that 10 3 = 3, and 10 ≡ 1 mod 3. Hence, by is a unimodular 9-simplex. Therefore, {u} = max ∇ 3 10,3 , and each Type 1 vertex corresponds to a simplex in max ∇ 2 10,3 \ max ∇ 3 10,3 that shares exactly one face with u. Hence, no Type 2 vertex can be in a simplex adjacent to u, and therefore these vertices cannot be in r-supporting pairs. Proof. By Proposition 3.12 the hyperplane H F is given by some r-supporting pair (u, ω). By Proposition 3.15 ω has a unique vertex that is (r − 1)-stable but not r-stable with a unique (r − 1)stable but not r-stable pair, say (ℓ, ℓ + r − 1), for some ℓ ∈ [n]. By Proposition 3.17 it follows that H F = H ℓ,r . So far, we have shown that the facets of ∆ stab(r) n,k that intersect the relative interior of ∆ stab(r−1) n,k are supported by hyperplanes H ℓ,r . We now show that H ℓ,r is a supporting hyperplane of ∆ stab(r) n,k for all ℓ ∈ [n]. Proof. First we note that the result is clearly true for r = 1. So in the following we assume r > 1. Now notice that ∆ stab(r) n,k is in the closed halfspace given by ℓ+r−1 i=ℓ x i ≤ 1. This is because all vertices of ∆ stab(r) n,k are r-stable. Hence, it suffices to show that H ℓ,r supports an (n − 1)-simplex ω ∈ max ∇ r n,k . Since 1 < r < n k or n = kr + 1 then by Theorem 3.8 and [2, Lemma 2.7] we know that ∆ stab(r) n,k is (n − 1)-dimensional. So there is at least one (n − 1)-dimensional simplex in ∇ r n,k . We also know that n ≥ kr + 1. With this in mind, we would like to construct a minimal circuit in G n,k that uses only r-stable vertices and has H ℓ,r as a supporting hyperplane. To do so, we first construct an initial vertex, say ǫ ℓ , for this minimal circuit. Construct the vertex ǫ ℓ as follows: (3) Make all other entries 0.
It follows that ǫ ℓ ∈ R n is r-stable. To see this, first notice that we have constructed ǫ ℓ such that 1 t and 1 t+1 are separated by r − 1 0's for all t ≤ k − 1. Then notice that the k 1's and the 0's between 1 t and 1 t+1 for all t ≤ k − 1 account for k + (k − 1)(r − 1) = kr − (r − 1) = kr − r + 1 entries of the vertex. Since n ≥ kr + 1 then there are at least r entries (all filled with 0's) between 1 k and 1 1 . Hence, ǫ ℓ is r-stable.
From this, it is easy to see that ω ℓ ∈ max ∇ 2 9,3 . Moreover, thinking of ǫ ℓ as the initial vertex of the circuit, only the third vertex in the circuit does not satisfy x 2 + x 3 = 1, and this is precisely the vertex preceding the first move of 1 1 . are supported by the hyperplanes H ℓ,r for ℓ ∈ [n]. We are now ready to prove Theorem 3.1.
Proof of Theorem 3.1. First recall that Theorem 3.1 is known to be true for r = 1. Now let 1 < r < n k . By Theorem 3.8 we know that ∆ stab(r) n,k is (n − 1)-dimensional. We proceed by induction on r. First let r = 2. By Proposition 3.6 we know that H ℓ is a supporting hyperplane of ∆ stab(2) n,k for all ℓ ∈ [n]. By Proposition 3.9 we know that for every ℓ ∈ is a unimodular (n − 1)-simplex whenever n ≡ 1 mod k. Notice also that Proposition 3.21 only requires ∆ stab(r) n,k to be (n − 1)-dimensional and r > 1. (It does not require that r < n k .) Also, Proposition 3.22 holds for n = kr + 1. Moreover, since ∆ stab(⌊ n k ⌋) n,k is an (n − 1)-simplex then it has precisely n facets. Therefore, by n,k must be H ℓ,r for ℓ ∈ [n] whenever n k > 1. In the case that r = n k = 1, then ∆ stab(⌊ n k ⌋) n,k = ∆ k+1,k which clearly has supporting hyperplanes H ℓ,1 .
The following is an immediate consequence of Theorem 3.1.
Corollary 3.25. Fix 0 < k < n. All but (possibly) the smallest polytope in the nested chain n,k has 2n facets. This is an interesting geometric property since the number of vertices of these polytopes form a strictly decreasing sequence. In the special case when n ≡ 1 mod k, by Scholium 3.24 we completely understand the facets of all polytopes in this chain, and the "possibly" is not necessary. We remark that this technique may be interesting to apply to any pair of nested d-dimensional polytopes Q ⊂ P for which there is triangulation of P that restricts to a triangulation of Q.

The Gorenstein r-stable Hypersimplices
One application for the equations of the facets of a rational convex polytope is to determine whether or not the polytope is Gorenstein [5]. First we recall the definition of a Gorenstein polytope. Let P ⊂ R N be a rational convex polytope of dimension d, and for an integer q ≥ 1 let qP := {qα : α ∈ P }. Let x 1 , x 2 , . . . , x N , and z be indeterminates over some field K. Given an integer q ≥ 1, let A(P ) q denote the vector space over K spanned by the monomials x α 1 1 x α 2 2 · · · x α N N z q for (α 1 , α 2 , . . . , α N ) ∈ qP ∩ Z N . Since P is convex we have that A(P ) p A(P ) q ⊂ A(P ) p+q for all p and q. It then follows that the graded algebra is finitely generated over K = A(P ) 0 . We call A(P ) the Ehrhart Ring of P , and we say that P is Gorenstein if A(P ) is Gorenstein.
We now recall the combinatorial criterion given in [4] for a integral convex polytope P to be Gorenstein. Let ∂P denote the boundary of P and let relint(P ) = P − ∂P . We say that P is of standard type if d = N and the origin in R d is contained in relint(P ). When P ⊂ R d is of standard type we define its polar set The polar set P ⋆ is again a convex polytope of standard type, and (P ⋆ ) ⋆ = P . We call P ⋆ the dual polytope of P .
Remark 4.1. An elementary fact in the theory of convex polytopes is the existence of an inclusionreversing bijection between the faces of P and those of P ⋆ . This bijection is given as follows. Suppose (α 1 , α 2 , . . . , α d ) ∈ R d , and K is the hyperplane in R d defined by the equation d i=1 α i x i = 1. Then (α 1 , α 2 , . . . , α d ) is a vertex of P ⋆ if and only if K ∩ P is a facet of P .
Notice that Remark 4.1 implies that the dual polytope of a rational polytope is always rational. However, it need not be that the dual of an integral polytope is always integral. If P is an integral polytope with integral dual we say that P is reflexive. This idea plays a key role in the following combinatorial characterization of Gorenstein polytopes. Fix an integer point α ∈ q(relint(P )) ∩ Z d , and let Q denote the integral polytope qP − α ⊂ R d . Then the polytope P is Gorenstein if and only if the polytope Q is reflexive.
Since the facets of ∆ stab(r) n,k for r < n k are given in Theorem 3.1, we now wish to apply Remark 4.1 and Theorem 4.2 to determine exactly which of these r-stable hypersimplices are Gorenstein. We remark that we need not consider the case of ∆ stab(⌊ n k ⌋) n,k when n ≡ 1 mod k since this polytope is a unimodular (n − 1)-simplex and is therefore trivially Gorenstein. Hence, we only consider the case when r < n k . Recall that we computed the facets ∆ stab(r) n,k ⊂ R n as an (n−1)-dimensional polytope embedded in the hyperplane H under the affine isomorphism ϕ defined in section 3. Therefore, we must first apply the inverse isomorphism ϕ −1 to ensure that ∆ stab(r) n,k is full-dimensional. Also recall that ϕ Z n−1 = H ∩ Z n . Hence, we have the isomorphism of Ehrhart Rings as graded algebras We now give a description of the facets, and their associated closed halfspaces, for P stab(r) n,k in terms of those defining ∆ stab(r) n,k . In the following, it will be convenient to let T = {ℓ, ℓ+1, ℓ+2, . . . , ℓ+r−1} for ℓ ∈ [n]. We also let T c denote the complement of T in [n]. Notice that for a fixed r < n k and ℓ ∈ [n], the set T is precisely the set of summands in the defining equation of the hyperplane H ℓ,r . We first consider the facets of ∆ stab(r) n,k with supporting hyperplanes H ℓ,r . For these supporting hyperplanes we have two cases.
(1) First suppose that ℓ = n. Then (2) Next suppose that ℓ = n. Then since Hence, Therefore, the facets of P The codegree of P stab(r) n,k . Given the above description of P stab(r) n,k , we would now like to determine the smallest positive integer q for which qP stab(r) n,k contains a lattice point in its relative interior. To do so, recall that for a lattice polytope P of dimension d we can define the (Ehrhart) δ-polynomial of P . If we write this polynomial as δ P (z) = δ 0 + δ 1 z + δ 2 z 2 + · · · + δ d z d then we call the coefficient vector δ(P ) = (δ 0 , δ 1 , δ 2 , . . . , δ d ) the δ-vector of P . We let s denote the degree of δ P (z), and we call q = (d + 1) − s the codegree of P . It is a consequence of Ehrhart Reciprocity that q is the smallest positive integer such that qP contains a lattice point in its relative interior [1]. Hence, we would like to compute the codegree of P stab(r) n,k . To do so requires that we first prove two Lemmas. In the following let q = n k . Our first goal is to show that there is at least one integer point in relint qP stab(r) n,k for r < n k . We then show that q is the smallest positive integer for which this is true. Recall that q = n+α k for some α ∈ {0, 1, . . . , k − 1}. Also recall that for a fixed 0 < k < n we have the nesting of polytopes  Proof. It suffices to show that (x 1 , x 2 , . . . , x n−1 ) = (1, 1, . . . , 1) satisfies the set of inequalities i∈T c x i > (k − 1)q, for n ∈ T . We do this in two cases. First suppose that α = 0. Then k divides n and q = n k . Clearly, (i) is satisfied. To see that (ii) is also satisfied simply notice that n − 1 k < n k , To see that (iii) is satisfied recall that #T = r and r < n k = q. Finally, to see that (iv) is satisfied notice that #T c = n − r. So we must show that n − r > (k − 1)q. To see this, consider the following with 0's. Now add 1 to each entry of this lattice point. If the resulting point is (x 1 , x 2 , . . . , x n ) then replace the entry x n = 1 with the value Call the resulting vertex ǫ α . Next consider the hyperplane x i = kq , and the affine isomorphism Notice that by our construction of ǫ α , the point ϕ −1 (ǫ α ) is simply ǫ α with the last coordinate deleted.
Lemma 4.4. Fix 0 < k < n such that q = n k = n+α k for 2 ≤ α ≤ k − 1. Then for every r < n k the lattice point Proof. It suffices to show that (x 1 , x 2 , . . . , x n−1 ) = ϕ −1 (ǫ α ) satisfies the set of inequalities i∈T x i < q, for n / ∈ T , and (iv) i∈T c x i > (k − 1)q, for n ∈ T , for r = n k − 1. It is clear that (i) is satisfied. To see that (ii) is satisfied notice that Here, we count n − 1, one for each coordinate of the point, and then we add α since exactly α coordinates are occupied by 2's and all other coordinates are occupied by 1's. It then follows that Hence, (ii) is satisfied. To see that (iii) is satisfied first notice that for T with n / ∈ T i∈T if T contains no entry with value 2, r + 1 otherwise.
This is because we have chosen the 2's to be separated by at least r − 1 0's. Thus, since k does not divide n we have that i∈T x i ≤ r + 1 = n k < q.
In particular, ∆ stab(r) n,k is not Gorenstein.
Proof. First recall the method by which we constructed the lattice point ǫ α . Notice that, since the 1 st entry of ǫ α is 0, this method works just as well if we first construct an r-stable vertex with k − 1 1's in entries (n + 1) − r, (n + 1) − 2r, (n + 1) − 3r, . . . , (n + 1) − (k − 1)r, as opposed to the entries n − r, n − 2r, n − 3r, . . . , n − (k − 1)r. Hence, if we produce a second vertex using this placement of the k − 1 1's, say ζ α , then ϕ (ζ α ) −1 also lies in the relative interior of qP stab(r) n,k . The details of the proof are analogous to the case of ǫ α . i∈T c x i = (k − 1)q − (n − r), for n ∈ T . Recall that by Theorem 4.2 the polytope ∆ stab(r) n,k will be Gorenstein if and only if the the polytope Q is reflexive. In other words, we must apply Remark 4.1 to the facets of Q to determine when all vertices of Q ⋆ will be integral. This is the content of the following theorem. is Gorenstein if and only if n = kr + k.
Proof. First recall that by Corollary 4.7 we need only consider those polytopes described in Lemma 4.3. For these polytopes we must determine when the all vertices of Q ⋆ are integral. We do so by means of the inclusion-reversing bijection described in Remark 4.1. First consider the hyperplanes of type (a) in the above list. These are x i = −1 for i ∈ [n − 1]. Equivalently, we may write such a hyperplane as Recall that q = n k = n+α k for some α ∈ {0, 1}. Hence, this hyperplane is equivalently expressed as Therefore, for the corresponding vertex in Q ⋆ to be integral it must be that α = 0. It remains to consider the hyperplanes corresponding to the sets of indices T . First consider those of type (c) in the above list. We may equivalently write these hyperplanes as i∈T 1 q − r x i = 1.
Hence, the corresponding vertex of Q ⋆ is integral, and we conclude that, for r < n k , the polytope ∆ stab(r) n,k is Gorenstein if and only if n = kr + k. Remark 4.9. Recall that for n ≡ 1 mod k the polytope ∆ stab(⌊ n k ⌋) n,k is a unimodular (n−1)-simplex [2]. Hence, it is trivially Gorenstein. Therefore, for every r ≥ 1 we have two Gorenstein r-stable n, k-hypersimplices, one of which is a (n − 1)-simplex, and they are given by n = kr + k when 0 < k < n, and r = n k when n ≡ 1 mod k. Thinking of this result in regards to the nested chain of polytopes ∆ n,k ⊃ ∆ stab(2) n,k ⊃ ∆ stab(3) n,k ⊃ · · · ⊃ ∆ stab(⌊ n k ⌋) n,k it follows from Theorem 3.8 that the smallest polytope in this chain is Gorenstein for n ≡ 1 mod k, and the second smallest polytope is Gorenstein when n divides k. In both cases, no larger polytope in the chain is Gorenstein.
We also have the following corollary to Theorem 4.8.
Corollary 4.10. Let r ≥ 1. The r-stable n, k-hypersimplices ∆ stab(r) n,k for n = kr + k when 0 < k < n, and r = n k when n ≡ 1 mod k have unimodal δ-vectors. is Gorenstein for n = kr + k when 0 < k < n, and r = n k when n ≡ 1 mod k. By [3, Theorem 1] we conclude that the δ-vector of ∆ stab(r) n,k is unimodal for n = kr + k when 0 < k < n, and r = n k when n ≡ 1 mod k. Theorem 4.8 is interesting since it demonstrates that the Gorenstein property is quite rare amongst the class of r-stable hyperpsimplices, and also because it extends the collection of r-stable hypersimplices known to have unimodal δ-vectors given [2].