On Intersections of the Hyperbolicity Domain With Straight Lines

We consider the family of polynomials xn + a1xn−1 + a2xn−2 + · · · + an, ai ∈ R, and its hyperbolicity domain Πn, i.e. the set of values of the coefficients ai for which the polynomial is with real roots only. We prove that for 0 ≤ k ≤ n − 2 there exist generic straight lines in Rn Oa1 . . . an intersecting Πn along k segments and two half-lines.


The Hyperbolicity Domain
A real non-constant polynomial in one variable is (strictly) hyperbolic if it has only real (and distinct) roots.Consider the general family of monic real degree n polynomials P(x, a) := x n + a 1 x n−1 + a 2 x n−2 + • • • + a n .
Its hyperbolicity domain Π n ⊂ R n Oa 1 . . .a n is the set of values of the coefficients a i for which the polynomial is hyperbolic.In what follows we allow a slight abuse of terminology by identifying the points of the domain Π n with the corresponding hyperbolic polynomials.We set Π n,0 := Π n ∩ {a 1 = 0} ⊂ R n−1 Oa 2 . . .a n and Π * n := Π n,0 ∩ {a 2 = −1}.The set Π n,0 is of interest to us because a polynomial P can be transformed by the shift x → x − a 1 /n into one with a 1 = 0.The sets Π n and Π n,0 are invariant under the one-parameter groups of quasi-homogeneous dilatations x → e τ x, a j → e jτ a j , τ ∈ R. (1) Therefore a hyperbolic polynomial P with a 1 = 0, a 2 0 can be transformed into one with a 1 = 0, a 2 = −1 (If a 1 = 0 and a 2 0, then a 2 < 0 because , where x i are the roots of P. The only hyperbolic polynomial with a 1 = a 2 = 0 is x n ).
Definition 1 We use only the letters x and y to denote independent variables.We denote by Δ[P] the discriminant of the family of polynomials P, i.e. the set of values of the coefficients a i for which P has a multiple root in x.A similar notation is used for the discriminants of all other polynomial families used in the paper.The discriminant Δ[P] is defined as the zero set of a polynomial in the coefficients a i .It can be considered in the context of real or of complex coefficients.
In the present paper we are interested in pencils of real monic polynomials of the form tQ + (1 − t)R, t ∈ R, degQ =degR = n.We consider the question how many times can such a pencil intersect the hyperbolicity domain.A classical result (see Obrechkoff, 1963Obrechkoff, /2003) ) says that all polynomials of the pencil are strictly hyperbolic if and only if both Q and R are strictly hyperbolic and their roots interlace.
Remark 2 (1) The discriminant Δ[P] is defined as the set of zeros of the determinant D of the Sylvester matrix S (P, P ) of the polynomials P and P .The matrix is of size (2n − 1) × (2n − 1).Suppose that all coefficients a i are affine functions μ i s + ν i of a (real or complex) parameter s.The first column of S (P, P ) does not depend on s while in any of its other columns there are at least two entries which are affine functions of s.Therefore it is to be expected that typically degD = 2n − 2. One can show that the coefficient of s 2n−2 in D is a non-trivial homogeneous degree 2n − 2 polynomial in the coefficients μ i .Therefore a generic straight line intersects Δ[P] at 2n − 2 points.In the case when all coefficients are real part (or none or all) of the points are real, the remaining ones form complex pairs.In the context of Π n,0 (i.e. if one sets a 1 = 0) one has degD = 2n − 3 (because two columns of S (P, P ) are independent of s).In the same way for a 1 = 0, a 2 = −1 (i.e. in the context of Π * n ) one has degD = 2n − 4. Hence in these cases a generic line intersects Δ[P] at 2n − 3 and 2n − 4 points respectively.

Examples
In the present subsection we show how the hyperbolicity domain looks like for small values of n.We discuss what can be the intersection of a straight line with the hyperbolicity domain.
(3) The domain Π 3,0 is the part of R 2 delimited by the semi-cubic parabola 4a 3 2 + 27a 2 3 = 0 and containing the half-line a 2 ≤ 0, see Figure 2. Any generic line is neither vertical nor horizontal hence intersects it along a halfline or a half-line and a segment, see lines 1 and 3 on Figure 2. Within the class of non-generic lines there is the subclass of "least non-generic" ones.It consists of the vertical lines intersecting the domain along segments (line 5), the vertical lines not intersecting it at all (line 6), the horizontal lines not passing through the origin (line 4, they intersect the domain along half-lines), the non-horizontal lines passing through the origin (line 7, they intersect the domain along a half-line and at the origin) and the lines tangent to the semi-cubic parabola, but not at the origin (line 2).Finally there are the "most non-generic" lines-the two coordinate axes.The right part of Figure 3 shows the Oa 2 a 3 -plane and the semi-cubic parabola MON.A straight line S T belonging to the Oa 2 a 3 -plane and as shown on Figure 3 intersects Π 4,0 along a segment and a half-line.The segment contains inside itself a point at which the line S T is tangent to the border of Π 4,0 ; this point is in fact S T ∩ OP.Hence when the line S T is shifted slightly while remaining parallel to itself so that the a 2 -and a 3 -coordinates of its points remain the same while the a 4 -coordinate decreases, the intersection S T ∩ Π 4,0 will consist of two segments and a half-line.
We are not giving the complete description of the possible cases of intersection of a line with Π 4,0 .However we prove the following proposition: Proposition 3 Any straight line not intersecting Π 4,0 either belongs to a plane {a 2 = const} or its intersection with the half-space {a 2 ≤ 0} belongs to the set {a 2 ≤ 0, a 3 = 0, a 4 < 0}.
Remark 4 Suppose that a straight line belongs to a plane {a 2 = r}.If r > 0, then any such straight line does not intersect Π 4,0 because Π 4,0 ⊂ {a 2 ≤ 0}.If r ≤ 0, then the line does not intersect Π 4,0 exactly when it avoids the compact set Π 4,0 ∩ {a 2 = r}.This set is obtained from Π * 4 by a linear transformation, see formula (1).Hence it is necessary and sufficient for the line to avoid the (rectilinear) triangle ABC, see Figure 3.
Proof of Proposition 3. If a line does not belong to a plane {a 2 = const}, then it can be parametrized as follows: (2) If γ > 0 or γ = 0 < δ, then the absolute values of the a 3 -and a 4 -coordinates of a point of the line increase at most linearly in s.Consider a half-disk {a 2 = −1, a 2 3 + a 2 4 ≤ ε, a 4 ≥ 0 }, where ε > 0 is so small that the half-disk belongs to Π * 4 .Consider the image of the half-disk under the one-parameter group of diffeomorphisms defined by formula (1).This image belongs to Π 4,0 .Set s = e t .Then the a 3 -and a 4 -coordinates of the image of each point of the interior of the half-disk increase respectively as s 3/2 and s 2 .Hence for s sufficiently large the image of the half-disk contains the point of the straight line.
For γ = δ = 0 and α 0 or α = 0 β the reasoning is much the same.For s large enough the point of the line belongs to the border of the image of the half-disk, but is not at its centre (i.e.not on the a 2 -axis), hence the point belongs to the interior of Π 4,0 .
For γ < 0, α = 0 β or γ = 0 > δ, α = 0 β the point of the line belongs to the border of the image of the domain, but is not on the a 2 -axis, hence the point belongs to the interior of Π 4,0 .
Hence the line might not intersect Π 4,0 only for α = β = 0, i.e. when the straight line belongs to the Oa 2 a 4 -plane.The line must not intersect the set {a 2 ≤ 0, a 3 = 0, a 4 ≥ 0}.Indeed, in this case it will intersect Π 4,0 because it will either intersect OP (the negative a 2 -half-axis) or the arc OC (the a 4 -coordinate of a point of the arc OC increases as s 2 while the a 4 -coordinate of the point of the line increases at most linearly in s).Hence the intersection of the line with the half-space {a 2 ≤ 0} belongs to the set {a 2 ≤ 0, a 3 = 0, a 4 < 0}.

The Basic Result
The main result of the paper is the following theorem: Theorem 5 For n ≥ 3 the following statements hold true: (1) For a 1 = 0 a generic straight line in R n−1 Oa 2 . . .a n intersects Π n,0 along k segments and a half-line, where 0 ≤ k ≤ n − 2, and for any such k there exists such a generic straight line.
(2) For 0 ≤ k ≤ n − 2 there exist generic straight lines in R n Oa 1 . . .a n intersecting Π n along k segments and two half-lines.
To prove or disprove the following conjecture would also be of interest: Conjecture 6 Any straight line in R n which is not parallel to any of the coordinate hyperplanes intersects the interior of Π n .

Proof of Theorem 5
For n = 3 part (1) of the theorem can be checked directly, see part (3) of Subsection 1.2.
We prove part (1) by induction on n.Suppose that it is proved for n − 1 and that n ≥ 4. The domain Π * n is compact (as x k and all coefficients a i are bounded).This domain contains the polynomial U In a neighbourhood of the point U ∈ R n (respectively U ∈ R n−1 ) one can present any polynomial in the form Lemma 7 In a neighbourhood of the point U ∈ R n (respectively U ∈ R n−1 ) one can choose as local coordinates the quantities (λ 1 , λ 2 , . . ., λ n−1 , u) (respectively (λ 1 , λ 2 , . . ., λ n−1 )).
The lemma is proved in the next section.
Set y := (x − b)/v, where v > 0 is a parameter.Hence v n−1 U = y n−1 (nb + vy).To choose v > 0 small means to magnify the local picture around the point U.
Consider the family of polynomials in y ] can be regarded as a family of hypersurfaces in the space Oc 2 . . .c n−1 analytically depending on v.
By inductive assumption for any 0 ≤ k ≤ n − 3 there exist generic straight lines in R n−2 Oc 2 . . .c n−1 intersecting the hyperbolicity domain of the family of polynomials V along k segments and a half-line.Suppose that for any such k such a generic line L k is fixed.Genericity of the lines implies that the intersections of these lines with the discriminant Δ[V] are transversal.Hence the intersections of these lines (when considered as relative to the family v n−1 Ũ, not V) will be transversal for v ≥ 0 small enough as well.
For any v > 0 small enough the straight line L k defines a straight line M k in Oa 2 . . .a n .Indeed, the coefficients a i are affine functions (depending on v as on a parameter) of the coefficients c i .One can introduce a parameter w ∈ R and express the coordinates of a point of the line L k in the form Thus the coordinates of a point of the line M k can also be expressed by affine functions (depending on v) of w.
The lines L k and the coordinates c i can be considered also as complex ones.Hence each line L k intersects transversally the complex discriminant Δ[V] at 2n − 5 complex points out of which 2k + 1 are real.Genericity implies that all these points are finite.
Fix v small enough.The line M k intersects the domain Π * n along k segments whose endpoints are close to the point U and along another segment only one of whose endpoints is close to U. It is true that the line L k intersects the hyperbolicity domain of the family of polynomials V along a half-line, but the presentation of a polynomial in the form (x − u)G(x, λ) (respectively (x − λ 1 )G) is defined only locally, in a neighbourhood of the point U.It implies the existence of the endpoints of the k segments and of the half-line as intersection points of the line L k with the discriminant Δ[V].However the line M k a priori intersects the compact domain Π * n only along segments and/or points, not along a half-line.Hence its intersection with Π * n consists of k + 1 segments.The line M k belongs to a hyperplane {a 2 = const}.It has 2n − 4 intersection points with Δ[V] out of which 2n − 5 are close to the point U and 2k + 2 are real.Perturb the position of M k so that it no longer belongs to a hyperplane of the form {a 2 = const}.Such a generic perturbation introduces a single "distant" point which is real (otherwise it would be a complex conjugate pair).This point is the endpoint of a half-line of the intersection of M k with Π n,0 .
To finish the proof of the induction step for part (1) one should prove that there exist generic lines intersecting Π n,0 along half-lines and no segments.Consider a monic strictly hyperbolic polynomial P with roots , where x i < z i < x i+1 , i = 1, . .., n − 2. The interlacing of the roots implies that for any t ∈ R the polynomial S t := P + tQ has at least n − 2 real roots ξ i such that any interval (x i , x i+1 ) and any interval (z i , z i+1 ), i = 1, . .., n − 3 contains exactly one of the roots ξ i .
There exists t 0 such that the rightmost root of the polynomial S t 0 is a double one.For t > t 0 , S t has a complex conjugate pair of roots.Indeed, as S t = S t 0 + (t − t 0 )Q and as Q > 0 for x > z n−2 , the polynomial S t is positive for x > z n−2 .On the other hand it has at most n − 1 and at least n − 2 roots that are distinct and ≤ z n−2 .For any t ∈ R the polynomial S t is of degree n.Hence for t > t 0 the polynomial S t has exactly n − 2 real distinct roots.
For t < t 0 a similar reasoning shows that S t has two distinct real roots that are > z n−2 .As S t has at least n−3 distinct real roots that are < z n−2 , it has at least n − 1 distinct real roots.Hence it has n real roots (because degS t = n).Its roots that are not greater than z n−2 are distinct hence all its roots are distinct.This means that the straight line defined by the family of polynomials S t has the Property (A): it has a double root only for t = t 0 , it has n distinct real roots for t < t 0 and exactly n − 2 distinct real roots for t > t 0 .
The roots z i are required only to interlace with the roots x 1 , . .., x n−2 .Hence any nearby line defines a family S t which satisfies the Property (A) for some t * close to t 0 .This means that the straight line defined by the family S t is generic.This line belongs to Oa 2 . . .a n because the coefficients of x n−1 are 0 in P and in Q.
Part (2) of the theorem follows from part (1).Indeed, suppose that a generic line M k ⊂ Π n,0 as above is constructed (−1 ≤ k ≤ n − 3).Perturb this line so that it no longer belongs to a hyperplane of the form {a 1 = const}.Such a generic perturbation introduces a single "distant" point which (not being a complex conjugate pair) is real.This point is the endpoint of a half-line of the intersection of M k with Π n .

Proof of Lemma 7
The following formulas connect the quantities a i with the quantities u and λ i : Hence the Jacobian matrix of this change of variables equals To find its determinant add the first row multiplied by u to the second one, then the second row multiplied by u to the third one, . . .then the (n − 1)-st row multiplied by u to the n-th one.The matrix becomes Its entry in position (n, 1) equals −G(u, λ), so its determinant is (−1) n+1 G(u, λ).The root u is not close to b, therefore G(u, λ) 0.

Figure 1 .
Figure 1.The hyperbolicity domain for n = 2 and the lines intersecting it

Figure 2 .
Figure 2. The hyperbolicity domain for n = 3, a 1 = 0, and the lines intersecting it

Figure 3 .
Figure 3.The hyperbolicity domain for n = 4, a 1 = 0, and the lines intersecting it 1 and u and λ 1 are close to −(n − 1)b.