Combinatorics of stratified hyperbolic slices

1 Stratification

We start off by defining hyperbolic slices and showing how we stratify these sets. Then we look closer at the example from the introduction. We finish this section by showing that the collection of strata, partially ordered by inclusion, is a lattice.

Example 1.4

Let d = 5 and s = 2 and let

f = ( t + π ) ( t + 2 ) t ( t − 1 , 23456789123456789 ) ( t − e ) .

If we map the last three coefficients of the polynomials in H₂(f), we get the three-dimensional Figure 1 in the introduction. The polynomials with no repeated roots make up the interior, and the strata H2u(f) with u ≠ (1^d) lie on the pieces of the boundary as indicated by Figure 2. For instance, if u = (2, 1, 2), then the compositions (2, 3) and (3, 2) are the compositions smaller than u; thus H2u(f) corresponds to a closed one-dimensional piece of the boundary (including its endpoints).

Figure 2

Strata H2u(f) in H₂(f)

Next we show that the poset of strata is a lattice. If a and b are elements of a lattice, we denote their join by a ∨ b and their meet by a ∧ b.

Proof. We prove this by explicitly constructing the join of two compositions u and v of d. Having done so, the existence and uniqueness of the meet is also established since the meet of u and v is just the join of all compositions smaller than both u and v.

If a and b are two compositions of d, note that a ≤ b if and only if

So let M = {u₁, u₁ + u₂, . . . , d} ∪ {v₁, v₁ + v₂, . . . , d} and construct the tuple m = (m₁, m₂, . . . , m_l) containing all the distinct elements of M where m_i < m_i+1 for any i ∈ [l − 1]; note that m_l = d.

Consider the composition w = (m₁, m₂−m₁, m₃−m₂, . . . , m_l−m_l−1) of m_l = d. Since {u₁, u₁+u₂, . . . , d} ⊆ M and {v₁, v₁+v₂, . . . , d} ⊆ M, we have u ≤ w and v ≤ w. By construction, M is the unique minimal set that contains both {u₁, u₁ + u₂, . . . , d} and {v₁, v₁ + v₂, . . . , d}, hence w is the join of u and v. □

From Lemma 1.5 we immediately get that the strata of H_s(f), partially ordered by inclusion, form a lattice. To see this we determine the meet of two strata Hsu(f) and Hsv(f) of H_s(f). The meet of two strata is contained in their intersection since the partial order is given by inclusion. Also, by definition of the strata we have

so the intersection is a stratum of H_s(f). Thus we have that

and we have shown the following:

2 Vandermonde varieties

To describe the combinatorial structure of the lattice of strata we need to establish some geometric properties of our strata. In particular, we will show that Arnold’s, Givental’s and Kostov’s work on so-called "Vandermonde varieties" (see [1], [9] and [10]) implies that Hsu(f) is either empty, or a point or of dimension ℓ(u) − s, and that in the latter case the polynomials with composition u make up the relative interior.

To see the connection to their work, let the symmetric group Sym([d]) act on ℝ[x₁, . . . , x_d] by permuting the variables. It is well known that the elementary symmetric polynomials generate the ring of invariants; see Chapter 7.1 in [6]. The induced action on ℝ^d permutes the coordinates of the points in ℝ^d, and the orbit space ℝ^d/Sym([d]) can be identified with the image of ℝ^d under the mapping Π : ℝ^d → H, where

and e_i(x) is the i^th elementary symmetric polynomial.

Thus the orbit space can be identified with the monic hyperbolic polynomials of degree d, and by restricting Π to the set K_d := {x ∈ ℝ^d | x₁ ≤⋅ ⋅⋅ ≤ x_d} we obtain a bijection between K_d and Π(ℝ^d). So we see that hyperbolic slices can be thought of as sections of the orbit space obtained by intersecting it with certain affine hyperplanes.

Similarly, if u = (u₁, . . . , u_l) and h∈Hsu(f) has the roots a₁ ≤⋅ ⋅⋅ ≤ a_l , then

where a_u := (a₁, . . . , a₁, . . . , a_l , . . . , a_l) and a_i is repeated u_i times. So if we define the real algebraic set

then Hsu(f) is the image of Vsu(f)∩Kl under the mapping

Another set of generators for the invariant ring ℝ[x₁, . . . , x_d]^Sym([d]) is the set of all power sums p_i(x) := ∑j∈[d]xji with i ∈ [d]. Newton’s identities express the first s power sums as polynomials in the first s elementary symmetric polynomials, hence we can equivalently define Vsu(f) as {x ∈ ℝ^l | p_i(x_u) = c_i for every i ∈ [s]}, where c₁, . . . , c_s ∈ ℝ are obtained from −f₁, . . . , (−1)^s f_s using Newton’s identities. Such sets are referred to as Vandermonde varieties, since the Jacobian of the first s power sums is a constant multiple of a Vandermonde determinant.

To make use of the previous work on Vandermonde varieties, we first need to establish that Π_u is a homeomorphism. Note that, as with H_s(f), we equip the sets Vsu(f),Kl and Hsu(f) with the subspace topology of the Euclidean topology.

Let B_ε(a) denote the real open ball about a ∈ ℝ^k with radius ε > 0, and let Bε(a)¯ denote its closure. Similarly, let D_ε(z) denote the complex open ball about z ∈ ℂ^k with radius ε, and let Dε(Z)¯ denote its closure.

Proof. Note that Π_u is a bijection and a polynomial mapping, thus it is a continuous bijection. To see that the inverse map is continuous and that Hsu(f) is closed in ℝ^d−s, we show that the images of closed sets in Vsu(f)∩Kl are closed in ℝ^d−s.

Let S be a closed subset of Vsu(f)∩Kl; then since Vsu(f) and K_l are closed in ℝ^l , so is S. Let

be the inclusion x1,…,xl↦xu=x1,…,x1,…,xl,…,xl, where x_i is repeated u_i times. Then Π_u(x) = (Π ∘ ι_u)(x) and clearly ι_u(S) is a closed subset of ℂ^d.

Let h = t^d +h₁t^d−1 +⋅ ⋅ ⋅+h_d ∉ Π_u(S) have the roots a = (a₁, . . . , a_d) and let h_i = f_i for all i ∈ [s]. Let ε > 0 be such that D_ε(σ(a))∩ ι_u(S) is empty for every σ ∈ Sym([d]). If b₁, . . . , b_k are the distinct roots of h with respective multiplicities v₁, . . . , v_k, then there is a δ > 0 such that any polynomial g of degree d with |h_i − g_i | ≤ δ for all i ∈ [d] has exactly v_i zeroes in D_ε(b_i); this statement is proved in [18]. Thus, since D_ε(σ(a)) ∩ ι_u(S) is empty for any σ ∈ Sym([d]), then so is B_δ(h) ∩ Π_u(S), and therefore Π_u(S) is closed in ℝ^d−s. □

Proof. Firstly, we can use Newton’s identities to define Vsu(f) in terms of the first s power sums in d variables. Then the proof that Vsu(f)∩Kl is contractible or empty can be found in [10, Theorem 1.1]. By Lemma 2.1 the map Πu:Vsu(f)∩Kl→Hsu(f) is a homeomorphism, thus Hsu(f) is contractible if it is nonempty. □

To see how this proposition can be used to further describe our strata we need some more definitions. First note that as Hsu(f) is the image of a semialgebraic set under a polynomial mapping, it is semialgebraic. Thus the dimension of the stratum is the maximum integer n such that Hsu(f) contains an open set which is homeomorphic to an open set of ℝⁿ; see [4, Chapter 2.8].

Proof. Suppose that h∈Hsu(f) has the distinct roots a = (a₁, . . . , a_k) and composition v = (v₁, . . . , v_k); then k ≤ ℓ(u) ≤ s. If k = 1, then v = (d) and there is only one solution to the equation

and so we have Hsv(f)=Hkv(f)={h}. If k > 1, then as previously mentioned, Vsv(f) can be defined as

where p_i is the i^th power sum in d variables and c₁, . . . , c_k ∈ ℝ are obtained from the numbers −f₁, . . . , (−1)^k f_k using Newton’s identities.

The map F : ℝ^k → ℝ^k where F(x) = (p₁(x_v), . . . , p_k(x_v)) is a continuously differentiable function whose Jacobian matrix is ivjxji−1i,j≤k and so its determinant is

Since all the a_i’s are distinct, the determinant is nonzero at a. Thus the Jacobian matrix is invertible and by the inverse function theorem, F is invertible on some neighbourhood U of F(a) = (c₁, . . . , c_k). By Proposition 2.2, Vkv(f)∩Kk is contractible and since a is isolated in this set, it must be the only point there. Therefore we have Hsv(f)=Hkv(f)={h}.

So for any composition w ≤ u that occurs in Hsu(f) we have that Hsu(f) is a point. Since there are only finitely many compositions smaller than or equal to u,Hsu(f) contains only finitely many points. But since Hsu(f) is contractible it can contain at most one point. □

Let

denote the projection that forgets the last r coordinates. This map will help us describe the relative interior.

Proof. Firstly we consider the case when l = 1; then u = (d) and s = 0. So for any a ∈ ℝ we have that (t−a)d=td−datd−1+⋯+(−a)d∈H0u(f). Thus P^d−1((t − a)^d) = t^d − dat^d−1 and so the map

is essentially just mapping a to −da. This is naturally a homeomorphism and since, by Lemma 2.1, Π_u is a homeomorphism, then so is P^d−l. Lastly, since the image of P^d−l is all of ℝ, the image is closed in ℝ.

Next suppose that l ≥ 2. By Lemma 2.4, the polynomials of Hsu(h) are uniquely determined by their first l coefficients, thus P^d−l is a bijection between Hsu(h) and Pd−lHsu(f). Also, the topology on Hsu(f) is the subspace topology of the product topology on ℝ^d−s with respect to the projections on each coordinate. Thus, by definition, the map P^d−l is continuous.

To see that the inverse is continuous and that Pd−lHsu(f) is closed, we will show that the images of closed subsets of Hsu(f) are closed in ℝ^l−s. So let S be a closed subset of Hsu(f); then S is also a closed subset of ℝ^d−s since, by Lemma 2.1, Hsu(f) is closed in ℝ^d−s.

Let g be a point in the closure of P^d−l(S). Then for any ε > 0, the closed ball B¯ ε(g) meets P^d−l(S) and so the inverse image

is nonempty. It is also closed since B¯ ε(g)×ℝd−l and S are closed in ℝ^d−s.

Since Π_u is a homeomorphism, the set

is a closed subset of Vsu(f)∩Kl, and since Vsu(f)∩Kl is closed in ℝ^l, so is M. We also have that

So if a ∈ M, then e₁(a_u)≤ g₁ − ε and e₂(a_u)≥ g₂ − ε since l ≥ 2. Thus, by Newton’s identities, we have

and so M is bounded. Since M is closed and bounded, it is compact.

Since Π_u and P^d−l are continuous, Pd−l°  Πu is continuous. And since the continuous image of a compact set is compact, we have that Pd−l°  Πu(M) is compact. Thus

and so P^d−l(S) is closed in ℝ^l−s. Therefore P^d−l is a closed map and thus P^d−l is a homeomorphism. Lastly, by setting S=Hsu(f), we see that Pd−lHsu(f) is closed in ℝ^l−s. □

We see from Lemma 2.5 that when ℓ(u) ≥ s, the largest dimension that Hsu(f) can have is ℓ(u) − s. Thus we say that the maximal dimension of Hsu(f) is max{ℓ(u) − s, 0}.

Proof. If s = 0 and l = ℓ(u), the map Pd−l∘Πu:Kl→Pd−lH∩u(f) is a homeomorphism by Lemma 2.1 and Lemma 2.5. So since the dimension of K _l is l and its interior points are the points with no repeated coordinates, the dimension of Pd−lH0u(f)⊆ℝl is l and its interior points are the polynomials with composition u.

Next suppose that s > 0 and let Afir denote the affine hyperplane of ℝ^r defined by fixing the i^th coordinate to be equal to f_i. Then

Thus if there is a polynomial h∈Hsu(f) with composition u, then P^d−l(h) lies in the interior of Pd−lH0u(f) and therefore also in the interior of Pd−lH0u(f) and Pd−lH0u(f) must be of dimension max{l − s, 0}. Since P^l−s is a homeomorphism, Hsu(f) is maximal dimensional and h is in its relative interior.

For the reverse inclusion, suppose that l > s so that Hsu(f) is at least one-dimensional. If its relative interior contains a polynomial g with v(g) < u, then P^d−l(g) lies in the interior of the (l−s)-dimensional set Pd−lHsu(f)⊆ ℝ^l−s. Thus P^d−l(g) lies in the interior of the one-dimensional set Pd−lHl−1u(g).

By Lemma 2.4 there are only finitely many polynomials in Hl−1u(g) with a smaller composition than u. Thus there are two polynomials p₋ and p₊ in Hl−1u(g) with composition u, and a δ > 0 such that

Since P^d−l(p₋) and P^d−l(p₊) are interior points of Pd−lH0u(f), there is an ε > 0 such that BεPd−lp− and B_ε(P^d−l(p₊)) are contained in the interior of Pd−lH0u(f). And since P^d−l(g) is in the boundary of Pd−lH0u(f), the ball B_ε(P^d−l(g)) must contain a polynomial q = t^d + q₁t^d−1 + ⋅⋅ ⋅ + q_l t^d−l that is not in Pd−lH0u(f).

Thus Aq1l∩⋯∩Aql−1l is a line that passes through q and the two balls B_ε(P^d−l(p₋)) and B_ε(P^d−l(p₊)). But if q separates the nonempty sets

and

then

is nonempty but not contractible. This contradicts Proposition 2.2; therefore g cannot be in the relative interior of Hsu(f). □

Thus if ℓ(u) > s, then the stratum Hsu(f) is of maximal dimension if and only if it contains a polynomial with composition u. We can use this observation to determine all the possibilities for the dimension of Hsu(f). It is worth mentioning that a similar observation can be found in [3], Proposition 5.

Proof. If s = 0, then any composition occurs and thus by Theorem 2.3, any stratum is maximal dimensional. Similarly, if s = 1, then for any composition u, the polynomial −e₁(x_u)− f₁ has a real zero with l = ℓ(u) distinct coordinates ordered increasingly. To see this pick l real numbers a₁, . . . , a_l such that a₁/u₁ < ⋅ ⋅⋅ < a_l/u_l and

let

then

Thus the composition u occurs and Hsu(f) is of maximal dimension.

Next we suppose that s ≥ 2. If ℓ(u) ≤ s or Hsu(f) does not contain a polynomial with at least s distinct roots, then Hsu(f)=∪ w≤u and ℓ(w)=s−1Hsw(f). By Lemma 2.4, Hsw(f) contains at most one point when ℓ(w) = s − 1. Since there are only finitely many compositions of length s − 1, the stratum Hsu(f) contains only finitely many polynomials and since Hsu(f) is contractible it contains at most one polynomial.

So suppose that h∈Hsu(f) has k ≥ s distinct roots and v(h) < u. Let v ≤ u be a composition that covers v(h). Then ℓ(v) = k + 1 and so by Proposition 2.5, Hkv(h) is at most one-dimensional. Since v(h) < v we can write h as ∏i=1k+1t−aivi and without loss of generality we may assume that a₁ < ⋅ ⋅⋅ < a_k = a_k+1.

As in the proof of Lemma 2.4, if we define Vkv(f) as x∈ℝk+1∣p1xu=c1,…,pkxu=ck, then the Jacobian matrix of the defining polynomials is ivjxji−1i≤k,j≤k+1 and the determinant of the leftmost k × k submatrix is

Since the first k coordinates of a = (a₁, . . . , a_k+1) are distinct, the determinant does not vanish at a∈Vkv(h). So by Proposition 3.3.10 in [4], a is a nonsingular point of a one-dimensional irreducible component, say V, of Vkv(h). Thus a lies in an open neighbourhood U of V where U is a one-dimensional manifold.

By Lemma 2.4, the one-dimensional manifold U intersects the hyperplane H = {x ∈ ℝ^k+1 | x_k = x_k+1} only once. So U must meet the open halfspace H⁺ := {x ∈ ℝ^k+1 | x_k < x_k+1} and thus there is a point in Vkv(h) with no repeated coordinates. So Hkv(h)⊆Hsu(f) contains a polynomial g with composition v.

We can apply the same argument to the polynomial g if v < u, and keep doing this inductively until we find a polynomial with composition u. Then by Theorem 2.6, Hsu(f) is maximal dimensional. □

Proof. By Theorem 2.7 we may suppose that Hsu(f) is maximal dimensional and at least one-dimensional. We will prove the statement by induction on the dimension of the strata. If Hsu(f) is one-dimensional, then it is connected by Proposition 2.2. Thus it contains at most two relative boundary points, and any open ball about a point of Hsu(f) must contain infinitely many points from the relative interior. Thus Hsu(f) is the closure of its relative interior.

Now suppose that the statement is true for all (n − 1)-dimensional strata, where n − 1 ≥ 1. Suppose that Hsu(f) is n-dimensional and h∈Hsu(f). By Proposition 2.2, Hsu(f) is connected, so for any ε > 0 the ball B_ε(h) contains infinitely many polynomials from Hsu(f). Since there are only finitely many polynomials in Hsu(f) with at most s distinct roots, there is a g∈Bε(h)∩Hsu(f) with at least s + 1 distinct roots.

Then by Theorem 2.7 the stratum Hs+1u(g) is (n − 1)-dimensional, and by the inductive hypothesis g is in the closure of its relative interior. So by Theorem 2.6, any open ball about g contains a polynomial with composition u. Thus B_ε(h) also contains a polynomial with composition u, and so h is in the closure of the relative interior of Hsu(f). □

3 Combinatorial structure

In this section we explore the combinatorial structure of the lattice of strata and show that this lattice is graded, atomic and coatomic. Then we use this to determine which compositions actually occur in a given hyperbolic slice. Due to Theorem 2.7 we may assume that H _s(f) is of dimension d − s > 0 for this section since the one-point lattice trivially satisfies the main combinatorial properties that follow. We begin with the notion of a graded poset.

Proof. As observed in [14, Proposition 4.1], a stratum of H _s(f) is compact when s ≥ 2 since we can rewrite the s first elementary symmetric polynomials in terms of the s first power sums, and the second power sum defines a sphere.

For the second statement, note that when s = 2, the equations

define a nonempty intersection of a hyperplane, whose normal vector is (1, 1, . . . , 1), and a (d − 1)-sphere. This intersection can either be a (d−2)-sphere in the hyperplane, or it can be one of the two points ±c2d(1,1,…,1).

That is, if v = (d) and the stratum Hsv(f) is nonempty, then there can be no other point in H_s(f) since there is no other point in H₂(f). So since H _s(f) is at least one-dimensional we have H S v ( f ) = ∅ , and Hsu(f) contains the empty stratum since v is smaller than any composition u. □

For s = 0 and s = 1, the hyperbolic slices will have all the main combinatorial properties we are establishing. But the argument is different from the other cases, so we will restrict s to be at least 2 for now. This allows us to use Lemma 3.2, which will be very helpful. Before we proceed note that we use the convention that the empty set has dimension −1.

Proof. If a stratum Hsu(f) is strictly contained in Hsv(f), then dimHsu(f)<dimHsv(f) by Lemma 3.2 the lattice of strata contains the empty set as its minimal element. Thus any maximal chain in the lattice of strata has length at most dim(H_s(f)) + 1 = d − s + 1.

Conversely, suppose that Hsu(f) is strictly contained in Hsv(f), where dimHsu(f)<dimHsv(f)−1; we show that there is a stratum Hsw(f) with Hsu(f)⊊Hsw(f)⊊Hsv(f).

If Hsu(f) is empty, Hsv(f) is at least one-dimensional and by Theorem 2.6 its relative interior consists of the polynomials with composition v. But by Lemma 3.2 it is compact and so it must contain a nonempty stratum Hsw(f) in its relative boundary. Then by Lemma 3.2, both Hsv(f) and Hsw(f) contain the empty stratum Hsu(f).

If h∈Hsu(f) and there are no polynomials in Hsu(f) with at least s distinct roots, then Hsu(f)={h} by Theorem 2.7. Since Hsu(f) is zero-dimensional, Hsv(f) is at least two-dimensional, and by Lemma 3.2 and Proposition 2.2 it is compact and contractible. Thus its relative boundary is at least one-dimensional and connected.

If Hsu(f) is not contained in a larger stratum of Hsv(f), it must be an isolated part of the relative boundary of Hsv(f). Since the relative boundary is one-dimensional and connected, Hsu(f) must be the whole boundary. But this is impossible since Hsu(f) is zero-dimensional, thus there is a stratum Hsw(f) which is strictly contained in Hsv(f) and that strictly contains Hsu(f).

Lastly, if there is an h∈Hsu(f) with at least s distinct roots, then we may assume that u = v(h) by Theorem 2.7. Also, we then have that all compositions greater than u occur in H _s(f). Since

we have ℓ(u) < ℓ(v) − 1 and so there is a composition w with u < w < v. By Theorem 2.7, Hsw(f) is of dimension ℓ(w) − s and therefore Hsu(f)⊊Hsw(f)⊊Hsv(f).

Thus any maximal chain will be at least of length d − s + 1, and so any maximal chain has length d − s + 1. Also, by the above argument any stratum of dimension n ≥ 0 covers a stratum of dimension n − 1, thus its rank must be n + 1. □

Next up is the notion of atomic lattices.

Proof. By Proposition 3.3, an n-dimensional stratum Hsu(f) contains an (n − 1)-dimensional stratum Hsv(f). Also, by Lemma 3.2 a nonempty stratum Hsu(f) is compact and so its relative boundary is nonempty but not contractible. But by Proposition 2.2, Hsv(f) is contractible and so it cannot be the whole relative boundary of Hsu(f).

Since Hsv(f) is closed, RelBdHsu(f)\Hsv(f) is relatively open in RelBdHsu(f) with the subspace topology. Thus RelBdHsu(f)\Hsv(f) is (n−1)-dimensional and so there must be another (n−1)-dimensional stratum in Hsu(f). □

Proof. By convention the empty set is the join of an empty set of atoms, and an atom is naturally the join of itself. Also, by Proposition 3.3, the lattice is graded and a stratum’s rank is its dimension plus one, so the atoms are the zero-dimensional strata.

If Hsu(f) is an n-dimensional stratum and n > 0, then by Lemma 3.5 there are two distinct (n−1)-dimensional strata Hsv(f) and Hsw(f), contained in Hsu(f). Since Hsv(f) and Hsw(f) are distinct, by Proposition 3.3 any stratum that contains both must be at least n-dimensional. Since Hsu(f) is n-dimensional and contains both Hsv(f) and Hsw(f), it must be the join of Hsv(f) and Hsw(f). By induction, both Hsv(f) and Hsw(f) are joins of atoms, and since Hsu(f) is the join of Hsv(f) and Hsw(f) it must also be a join of atoms. □

Lastly, we look at a sort of converse of atomic lattices called coatomic lattices.

Proof. Let Hsu(f) be an n-dimensional stratum. If n > 0 then by Theorem 2.7 it is maximal dimensional. Thus there is a polynomial with composition u, and since ℓ(u) − s = n < d − s − 1 we have ℓ(u) < d − 1, so u is covered by at least two distinct compositions, say v and w, of length ℓ(u) + 1. So again by Theorem 2.7, Hsv(f) and Hsw(f) must be (n + 1)-dimensional.

If n = 0, then Hsu(f)={h} and since d−s > 1, the set H_s(f) is at least two-dimensional. By Proposition 3.3, h is contained in a two-dimensional stratum Hsw(f) and in a one-dimensional stratum Hsv(f)⊂Hsw(f). By Theorem 2.6, Hsv(f) is in the one-dimensional relative boundary of Hsw(f) and h is in the relative boundary of Hsv(f).

According to Corollary 2.8, Hsw(f) is the closure of its relative interior, so there is a connected component C of RelIntHsw(f) such that h is in its closure C. Thus, starting from h, we can traverse its boundary clockwise or counter-clockwise. But since h is one of the relative boundary points of Hsv(f), at most one of the directions consists immediately of polynomials whose composition is v. Thus there must be some other one-dimensional stratum for which h is a boundary point.

Lastly, if n = −1 then Hsu(f) is empty. Since H_s(f) is at least one-dimensional and, by Proposition 3.6, the lattice of strata is atomic, it must contain at least two atoms. Thus the empty stratum is contained in at least two zero-dimensional strata. □

Proof. The argument is analogous to the proof of Proposition 3.6, just start the induction from the (d − s − 1)-dimensional strata and use Lemma 3.8 instead of Lemma 3.5 for the induction step. □

Proof. The only case that remains to check is when s ≤ 1. As we saw in the proof of Theorem 2.7, all compositions occur when s ≤ 1, so the lattice is isomorphic to the lattice of compositions. The lattice of compositions is isomorphic to the face lattice of a (d − 2)-dimensional simplex S. To see this, note that any k + 1 vertices in S determine a k-dimensional face of S.

Similarly there are d − 1 compositions of d of length 2. So if vi1,…,vik+1 are k + 1 distinct compositions of length 2, then their first parts are distinct. We may assume that they are ordered such that v1i1<⋯<v1ik+1 and so by the argument in Lemma 1.5 their join is

which is a composition of length k + 2. Thus any bijection from the set of length 2 compositions to the set of vertices of S induces an isomorphism of lattices. Lastly, by items (i) and (v) of Theorem 2.7 in [19], the face lattice of a simplex is graded, atomic and coatomic. □

Similar to the case when s ≤ 1, when s = 2 it can be shown that the lattice of strata is isomorphic to the face lattice of a simplex. Although this does not need to be true in general, as can be seen by intersecting H₂(f) from Example 1.4 with the affine hyperplane defined by fixing the third coefficient to be 0. This gives us a lattice which is isomorphic to the face lattice of a quadrilateral.

Thus there are examples for which the lattice of strata is not isomorphic to the face lattice of a simplex; however, it would be interesting to find an answer to the following question for general d and s.

Proof. Let w ∈ W; then w ≥ v ∨ u for some v, u ∈ U. Thus both v and u occur in H_s(f) and so Hsw(f) contains at least two polynomials. By Theorem 2.7, Hsw(f) is maximal dimensional and therefore there is a polynomial with composition w. Thus all compositions computed in the algorithm occur in H_s(f).

Suppose that a composition u with ℓ(u) > s occurs in H_s(f). Then by Theorem 2.7, Hsu(f) is at least one-dimensional and by Theorem 3.10, Hsu(f) is the join of at least two distinct atoms Hsv(f)={h} and Hsw(f)={g}. We may assume that v and w are the compositions of h and g respectively. Then v ∨ w ∈ V and v ∨ w ≤ u, thus u ∈ W and it was not left out by the algorithm. □

4 A short note on concavity

We will discuss some previously established results on the boundary of hyperbolic slices which, if Question 3.11 is answered in the affirmative, would imply that hyperbolic slices always look like partially deflated polytopes. To do this we start by quickly introducing discriminants and subdiscriminants; more information on these objects can be found in [8] and in Chapter 4 of [2].

Let Δ_d denote the discriminant of a real, monic, univariate polynomial of degree d. If g=td+g1td−1+⋯+gd∈ ℝ[t] has the roots a₁, . . . , a_d, the discriminant is defined as

This is a real polynomial, symmetric in the roots of g, so it may be written in the elementary symmetric polynomials evaluated at (a₁, . . . , a_d). That is, the discriminant may be written as a polynomial in the coefficients of the univariate polynomial g, and it vanishes whenever the corresponding univariate polynomial has a repeated root.

Let Z(Δ_d) ⊆ ℝ^d denote the real algebraic set given by the discriminant. Then the points of Z(Δ_d) that correspond to polynomials with a repeated real root split the space of real coefficients into ⌊d/2⌋ regions, each of which is characterised by the number of real roots that the polynomials have.

Similarly, we denote by Δ_d,k the k^th subdiscriminant of a real, monic, univariate polynomial of degree d. The subdiscriminant is defined as

and can also be written as a real polynomial in the coefficients of the univariate polynomial g. We can see that the r first subdiscriminants (including the discriminant) vanish whenever the corresponding univariate polynomial has at most d − r distinct roots.

So if h=td+h1td−1+⋯+hd lies in the real algebraic set Z(Δ_d,0, . . . , Δ_d,k) ⊆ ℝ^d given by the first k subdiscriminants, then h has at most d − k distinct roots. If h has exactly d − k distinct roots, then it was shown in Proposition 1.3.4 of [12] that a neighbourhood of h in the tangent space of Z(Δ_d,0, . . . , Δ_d,k) at h lies in the region which locally about h has the maximal number of real roots. That is, let T_h be the tangent space of Z(Δ_d,0, . . . , Δ_d,k) at h, where we consider the tangent space as having been translated to h and not centred at the origin. Then there is a neighbourhood N ⊂ ℝ^d of h such that T_h ∩ N consists of polynomials that have exactly n real roots and no other polynomial in N has more than n real roots.

Naturally the strata of H_s(f) labelled by compositions of length d − k are subsets of Z(Δ_d,0, . . . , Δ_d,k). Now suppose that h is hyperbolic and has composition u. Then if the tangent space of Hsu(h) at h is well defined, it must be included in Th∩Ah1∩⋯∩Ahs where Ahi⊂ℝd is the affine hyperplane defined by fixing the i^th coordinate to be equal to h_i.

The maximal number of real roots of a polynomial in N is d since h is hyperbolic, thus Th∩N∩Ah1∩⋯∩Ahs⊂ H_s(h) and so the tangent space of Hsu(h) at h, intersected with N, must also lie in H _s(h). Thus the strata of H _s(h) are, in a sense, concave towards H_s(h).

Index of notation

• [s] = {1, 2, . . . , s}

• f = t d + f 1 t d − 1 + ⋯ + f d

• H_s(f) = {h = t^d + h₁t^d−1 + ⋅⋅ ⋅ + h_d | h is hyperbolic and h_i = f_i for all i ∈ [s]}

• v(h) denotes the composition of h

• H s u ( f ) = h ∈ H s ( f ) ∣ v ( h ) ≤ u

• ℓ(u) denotes the length of the composition u

• For x ∈ ℝ^ℓ(u) let xu=x1,…,x1,…,xℓ(u),…,xℓ(u), where x_i is repeated u_i times.

• e_i(x) is the i^th elementary symmetric polynomial in d variables.

• p_i(x) is the i^th power sum in d variables.

• V s u ( f ) = x ∈ ℝ l ∣ − e 1 x u − f 1 = 0 , … , ( − 1 ) s e s x u − f s = 0

• K_l = {x ∈ ℝ^l | x₁ ≤⋅ ⋅⋅ ≤ x_l}

• For x ∈ ℝ^l let Πu(x)=td−e1xutd−1+⋯+(−1)dedxu

• B_ε(a) is the open ball around a ∈ ℝⁿ of radius ε.

• u ∨ v denotes the join and u ∧ v denotes the meet of two elements u and v of a lattice.