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A NONSPECTRAHEDRAL HYPERBOLICITY CONE
a plain-text rewrite
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This document rewrites the paper "A nonspectrahedral hyperbolicity cone"
into plain text. The goal is that a motivated high-school student who
knows polynomials, matrices, and a little calculus can follow every step.
Every theorem from the original is here. The big proofs are spelled out.
Some LaTeX dressing is trimmed. A few paragraphs are added to explain
the ideas behind each step.
The main result in one sentence: there is a polynomial p in 23 real
variables, homogeneous of degree 20, whose hyperbolicity cone cannot be
written as {x : L(x) >= 0} for any finite real symmetric linear matrix
pencil L.
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CONTENTS
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0. The one-paragraph summary
1. Background
1.1 Cones
1.2 Symmetric matrices and positive semidefiniteness
1.3 Hyperbolic polynomials
1.4 Hyperbolicity cones and examples
1.5 Spectrahedra
1.6 The Generalized Lax conjecture and its history
2. The counterexample
2.1 The biquadratic form b
2.2 The positive linear map Phi_y
2.3 The polynomial p and the main theorem
2.4 The reduced polynomial q
3. The form b has no nonzero dominated square (Lemma 1)
4. p is hyperbolic; two slices of its cone
5. What a hypothetical pencil would force
6. Rank-one limits and a norm identity
7. The tangent obstruction: finishing the proof
8. Removing a determinant factor
9. Appendix: a path-containment example
10. What this does and does not prove
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0. THE ONE-PARAGRAPH SUMMARY
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A hyperbolicity cone is a certain kind of convex cone: the set of x such
that a fixed homogeneous polynomial, evaluated along the line through x
in direction e, has only nonnegative roots. Examples include the cone of
positive semidefinite matrices and the Lorentz cone.
The Generalized Lax conjecture says: every hyperbolicity cone is a
spectrahedron, i.e. can be cut out by finitely many linear matrix
inequalities L(x) >= 0 using the same variables, no extra ones.
This paper constructs one hyperbolicity cone that is not spectrahedral.
The mechanism:
(a) There is a 3-variable biquadratic form b(z,y) with a rigid
property: b >= 0, b is nonzero, but the only bilinear form l with
l(z,y)^2 <= b(z,y) for all z,y is l = 0. (Lemma 1.)
(b) A positive linear map Phi_y (built from b) turns this scalar
rigidity into a matrix inequality in 23 variables.
(c) A degree-20 polynomial p(X,Z,y) = det((det X) Z - Phi_y(adj X)) has
a hyperbolicity cone K.
(d) If K were spectrahedral, one manipulates that pencil into a
nonzero bilinear form l with l^2 <= b, contradicting (a).
Nothing from analysis on infinite-dimensional spaces, nothing from
randomness, no numerical approximation. Every step is exact.
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1. BACKGROUND
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1.1 CONES
A subset C of a real vector space is a cone if
- x in C and t >= 0 imply tx in C, and
- x, y in C imply x + y in C.
Together these imply convexity: for t in [0,1], (1-t)x + ty = x + t(y-x)
is a nonnegative combination of x and y.
Examples: the positive orthant {x in R^n : x_i >= 0}; the PSD cone; the
Lorentz cone.
1.2 SYMMETRIC MATRICES AND POSITIVE SEMIDEFINITENESS
Write Sym^N for the real vector space of N x N symmetric matrices; its
dimension is N(N+1)/2.
A in Sym^N is positive semidefinite, written A >= 0, if x^T A x >= 0
for every x in R^N. It is positive definite, A > 0, if the inequality
is strict for every nonzero x.
Every A in Sym^N has real eigenvalues. Moreover:
A >= 0 <=> every eigenvalue of A is >= 0.
A > 0 <=> every eigenvalue of A is > 0.
A >= 0 <=> A = B^T B for some B (possibly rectangular).
In a basis where A is diagonal, all of this is obvious.
1.3 HYPERBOLIC POLYNOMIALS
Fix a real homogeneous polynomial p of degree d >= 1 in m variables, and
a vector e in R^m with p(e) != 0.
p is hyperbolic with respect to e if, for every x in R^m, the
one-variable polynomial
t |--> p(t e - x)
has all real roots.
Write the roots, with multiplicity, as lambda_1(x), ..., lambda_d(x).
Homogeneity gives the identity
p(x + t e) = (t + lambda_1(x)) (t + lambda_2(x)) ... (t + lambda_d(x)).
So "hyperbolic with respect to e" means: along every real line in
direction e, the polynomial has real roots.
Example: p(X) = det X on Sym^N, e = I. Then p(tI - X) = det(tI - X), the
characteristic polynomial of X, whose roots are the eigenvalues of X.
Since eigenvalues of symmetric matrices are real, det is hyperbolic with
respect to I.
Example: p(x0, x1, x2) = x0^2 - x1^2 - x2^2, e = (1, 0, 0). Then
p(t e - x) = (t - x0)^2 - x1^2 - x2^2,
whose roots are t = x0 +/- sqrt(x1^2 + x2^2). These are both real.
1.4 HYPERBOLICITY CONES AND EXAMPLES
The closed hyperbolicity cone of (p, e) is
K(p, e) = { x in R^m : lambda_j(x) >= 0 for j = 1, ..., d }.
Equivalently, x belongs to K iff the polynomial t |--> p(t e - x) has no
strictly positive real root (roots are allowed to be zero, so the cone
is closed).
A fundamental fact, due to Garding (1959), is that K(p, e) is always a
convex cone.
Examples:
- p(X) = det X, e = I: K is the cone of positive semidefinite
matrices (including singular ones).
- p(x0, x1, x2) = x0^2 - x1^2 - x2^2, e = (1, 0, 0): K is the
Lorentz cone {x0 >= sqrt(x1^2 + x2^2)}.
1.5 SPECTRAHEDRA
Let L : R^m -> Sym^N be a linear map. A spectrahedron is a set
K = { x in R^m : L(x) >= 0 }.
If we fix a basis of Sym^N, we can write L(x) = sum_i x_i A_i for fixed
symmetric matrices A_i, one for each coordinate x_i.
So a spectrahedron is exactly a set you get by asking finitely many
quadratic forms (the principal minors of L(x), or equivalently the
eigenvalues of L(x)) to be nonnegative, in a way that depends linearly
on x.
Examples:
- The PSD cone itself: take L(x) = x as a matrix (m = N(N+1)/2).
- The Lorentz cone in R^3:
L(x0, x1, x2) = [[x0 - x1, -x2], [-x2, x0 + x1]].
This L >= 0 iff trace = 2 x0 >= 0 and det = x0^2 - x1^2 - x2^2 >= 0.
- The cube [-1,1]^3:
L(x) = diag(1 - x1, 1 + x1, 1 - x2, 1 + x2, 1 - x3, 1 + x3).
1.6 THE GENERALIZED LAX CONJECTURE AND ITS HISTORY
A hyperbolicity cone is defined by a single polynomial. A spectrahedron
is defined by a matrix pencil. They look very different, but they often
turn out to be the same kind of object. The question is whether they
always are.
The Generalized Lax conjecture (GLC):
Every hyperbolicity cone is a spectrahedron, using exactly the
same variables, with no extra (auxiliary) variables.
This is a conjecture about the original coordinates. A weaker statement
("every hyperbolicity cone is a spectrahedral shadow") would allow extra
variables. The paper's theorem concerns the strictly stronger form.
History, in brief:
1958 Lax asks his original conjecture for polynomials in 3
variables.
1959 Garding proves that hyperbolicity cones are convex.
2005 Lewis, Parrilo, and Ramana show that the 3-variable Lax
conjecture is equivalent to a determinantal representation.
2007 Helton and Vinnikov prove the 3-variable case.
2011 Branden produces hyperbolic polynomials none of whose positive
powers has a definite determinantal representation.
This does not refute GLC, because a different pencil with
extra factors might still define the same cone.
2016 Kummer, for the specialized Vamos polynomial, exhibits a
7x7 representation of the cone even though no positive power
of the polynomial itself works.
~2018 Raghavendra, Ryder, Srivastava, Weitz prove exponential lower
bounds on the size of spectrahedral representations for some
cones.
2026 Netzer proves the case for hyperbolic cubics in 5 variables.
2026 Kummer and Netzer prove the case for strictly hyperbolic
polynomials (those for which t |--> p(t e - x) has distinct
roots for every x not parallel to e).
2026 Scheiderer, Netzer, Sanyal, and others study spectrahedral
shadows and local sums-of-squares obstructions. These are
related but different tools.
2026 Gonzalez Nevado claims a proof of the full GLC. The appendix
of this paper refutes one of the intermediate assertions in
that proof.
The counterexample in this paper is not covered by the Kummer-Netzer
strict-hyperbolicity theorem, because the polynomial fails the strict
hypothesis at a specific point w = (I, 0, 0). This is discussed in
section 4.
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2. THE COUNTEREXAMPLE
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Throughout this section, indices i, j, k range over {1, 2, 3}, and are
read cyclically mod 3: so e_4 = e_1, e_0 = e_3, etc.
2.1 THE BIQUADRATIC FORM b
For z, y in R^3, define
b(z, y) = sum_i z_i^2 y_i^2
- 2 sum_{i<j} z_i y_i z_j y_j
+ sum_i z_i^2 y_{i+1}^2. (*)
This is bilinear in (z, y): for fixed y it is a quadratic form in z, and
vice versa. It is the "coefficient-one Choi-Lam biquadratic form".
Two immediate facts about b, proved in section 3 (Lemma 1):
- b(z, y) >= 0 for every z, y.
- b(e_1, e_1) = 1, where e_i is the coordinate basis vector of R^3.
There is also an important hidden fact:
For every bilinear form l with l(z,y)^2 <= b(z,y) for all z, y,
we must have l = 0.
That surprising rigidity is the scalar seed of the whole argument.
Define a 3x3 symmetric matrix Q(y), quadratic in y, by
Q(y)_{ii} = y_i^2 + y_{i+1}^2
Q(y)_{ij} = -y_i y_j (for i != j).
Then b(z, y) = z^T Q(y) z for all z, y. Expanding confirms this
matches (*).
2.2 THE POSITIVE LINEAR MAP Phi_y
Fix y in R^3. Consider a 4x4 symmetric matrix written in the block form
M = [ a r^T ]
[ r T ]
with a in R, r in R^3, T in Sym^3. Define
Phi_y(M) = [ tr(Q(y) T) -r^T Q(y) ]
[ -Q(y) r a Q(y) ].
This is another 4x4 symmetric matrix. It depends on y through Q(y),
so it is quadratic in y. Because Q(y) is symmetric, the entries of
Phi_y(M) are correctly arranged for Phi_y(M) to be symmetric.
Some facts about Phi_y:
- Phi_y is linear in its matrix argument M.
- Phi_y(s y) = s^2 Phi_y(y) for real s. In particular Phi_{-y} = Phi_y
and Phi_0 = 0.
- Phi_y is order-preserving: M >= 0 implies Phi_y(M) >= 0.
Equivalently Phi_y is a "positive map" (see section 4).
The order-preservation is the content of the key identity
u^T Phi_y(v v^T) u = b(v_0 u' - u_0 v', y). (**)
for unit "double vectors" u = (u_0, u') and v = (v_0, v') in R^4. The
right side is nonnegative by Lemma 1. Every PSD matrix is a nonnegative
combination of rank-one matrices v v^T, so this identity upgrades to
"Phi_y >= 0" (see section 4 for details).
2.3 THE POLYNOMIAL p AND THE MAIN THEOREM
For any 4x4 real matrix X, let adj X denote the adjugate matrix, defined
by adj X = (det X) X^{-1} when X is invertible, and extended by
polynomiality when X is singular. The entries of adj X are (up to sign)
3x3 minors of X; adj X is always a real symmetric matrix when X is.
Define the polynomial p on the 23-dimensional space
Sym^4 x Sym^4 x R^3 (dimensions 10 + 10 + 3 = 23)
by
p(X, Z, y) = det( (det X) Z - Phi_y(adj X) ). (***)
Here the 4x4 matrix (det X) Z has entries of polynomial degree 5: 4 in
X (from det X) plus 1 in Z. The entries of Phi_y(adj X) also have degree
5: 3 in X (from adj X) plus 2 in y. So the entries of the matrix in
(***) all have degree 5. The determinant of a 4x4 matrix of degree-5
entries has degree 20.
The main theorem:
THEOREM 1.
The polynomial p in (***) is homogeneous of degree 20 and hyperbolic
with respect to e = (I_4, I_4, 0), with p(e) = 1. Its closed
hyperbolicity cone K = K(p, e) is not spectrahedral. In other words,
for every finite N >= 1 and every real linear map
L : Sym^4 x Sym^4 x R^3 -> Sym^N,
we have
K != { (X, Z, y) : L(X, Z, y) >= 0 }.
Two quick checks (both proved later):
- p(e) = 1. At (X, Z, y) = (I, I, 0): det X = 1, adj X = I,
Phi_0(anything) = 0, so the matrix in (***) is I. Determinant 1.
- At w = (I, 0, 0) and e = (I, I, 0), the line t e - w gives
X = (t - 1) I, Z = t I, y = 0. Then det X = (t - 1)^4, adj X =
(t - 1)^3 I, Phi_0 = 0, so
p(t e - w) = (t - 1)^16 t^4.
This has a multiple root at t = 1 (multiplicity 16), so p is NOT
strictly hyperbolic. See section 4.
2.4 THE REDUCED POLYNOMIAL q
The polynomial p has a removable factor det X. Section 8 shows that one
can write
p = (det X) q, q homogeneous of degree 16,
where q is a genuine polynomial on the same space, even at singular X.
COROLLARY 2.
The quotient q has degree 16, is hyperbolic with respect to the same
e = (I_4, I_4, 0), has q(e) = 1, and its closed hyperbolicity cone
is exactly K. So K is not spectrahedral either.
So the abstract may quote either "degree 20" (the convenient form p)
or "degree 16" (the reduced form q). Both are non-spectrahedral.
We do not claim that 23 variables or degree 16 is minimal. The
construction is not optimized for size.
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3. THE FORM b HAS NO NONZERO DOMINATED SQUARE (LEMMA 1)
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LEMMA 1.
The form b in (*) is nonnegative and nonzero. If a real bilinear
form l : R^3 x R^3 -> R satisfies l(z,y)^2 <= b(z,y) for every z, y,
then l = 0 identically.
Idea of the proof. The statement has two halves. The first ("b >= 0") is
a warm-up: write b as z^T M(z) z and check the principal minors of M.
The second ("only l = 0 survives") is the crucial rigidity. The trick is
to look at the zeros of b: at any (z, y) with b(z, y) = 0, the bound
l(z,y)^2 <= 0 forces l(z, y) = 0. There are enough zeros of b to first
kill all the off-diagonal coefficients of l, then a one-parameter curve
of zeros kills the diagonal too.
PROOF.
Step 1: b is PSD in y for every fixed z. Write a_i = z_i^2. Then b is
the quadratic form in y with 3x3 matrix
M(z) = [ a1 + a3 -z1 z2 -z1 z3 ]
[ -z1 z2 a2 + a1 -z2 z3 ]
[ -z1 z3 -z2 z3 a3 + a2 ]
Its diagonal entries are nonnegative. Its 2x2 principal minors, with
indices {i, i+1}, are:
det M(z)[{i, i+1}] = a_i^2 + a_{i-1} (a_i + a_{i+1}) >= 0.
Its full determinant:
det M(z) = a1^2 a2 + a2^2 a3 + a3^2 a1 - 3 a1 a2 a3 >= 0
by the AM-GM inequality (the three terms on the left have product
(a1 a2 a3)^3 and there are exactly three of them). Since all
principal minors are nonnegative, M(z) >= 0, so b(z, y) >= 0.
Nonvanishing: b(e1, e1) = 1.
Step 2: Zeros of b force l = 0. Write l(z, y) = sum_{i,j} c_{ij} z_i y_j
with real coefficients c_{ij}. We have l(z,y)^2 <= b(z,y); so any
(z, y) with b(z, y) = 0 satisfies l(z, y) = 0.
(a) Off-diagonal coefficients with "cyclic" indices vanish. Take
z = e_i, y = e_{i-1}. Then b(e_i, e_{i-1}) = 0 as a direct check
from (*) shows. So l(e_i, e_{i-1}) = c_{i, i-1} = 0.
(b) Off-diagonal coefficients of the "opposite" cyclic order vanish.
For each sign vector sigma in {-1, +1}^3, we have
b(sigma, sigma) = 3 - 6 + 3 = 0. So
0 = sum_i c_{ii} + sum_{i<j} (c_{ij} + c_{ji}) sigma_i sigma_j.
Averaging over the eight sigma's with weights sigma_p sigma_q
gives c_{pq} + c_{qp} = 0 for each p < q. Combining with (a),
c_{i,i-1} = 0, forces c_{i,i+1} = 0 too (since c_{i,i+1} =
-c_{i+1,i} = 0 by (a) at i+1 -> i). So the only possible
surviving coefficients are the diagonal ones c_{ii}.
So l(z, y) = c_1 z_1 y_1 + c_2 z_2 y_2 + c_3 z_3 y_3.
(c) Diagonal coefficients vanish. Fix a cyclic index i and let
z = e_i + s e_{i-1},
y = e_{i-1} + s e_i (s in R).
A direct substitution into (*) gives b(z, y) = s^4, and
l(z, y) = s (c_i + c_{i-1}).
For s != 0, l^2 <= b gives (c_i + c_{i-1})^2 s^2 <= s^4, i.e.
(c_i + c_{i-1})^2 <= s^2. Letting s -> 0 gives c_i + c_{i-1} = 0
for every cyclic i. These linear relations force
c_1 = c_2 = c_3 = 0.
Therefore l = 0 identically. This proves Lemma 1.
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4. p IS HYPERBOLIC; TWO SLICES OF ITS CONE
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First, a warm-up: the rank-one identity (**) from section 2.2.
For u = (u_0, u') and v = (v_0, v') in R^4, block multiplication in the
definition of Phi_y gives
u^T Phi_y(v v^T) u = (v_0 u' - u_0 v')^T Q(y) (v_0 u' - u_0 v')
= b(v_0 u' - u_0 v', y).
By Lemma 1 the right side is >= 0, so Phi_y(v v^T) >= 0 for every rank-
one matrix. Since every PSD matrix is a nonnegative combination of rank-
one matrices, Phi_y is order-preserving: M >= 0 implies Phi_y(M) >= 0.
Phi_y is what is called a "positive map" (from Sym^4 to itself).
PROPOSITION 3 (hyperbolicity).
The polynomial p in (***) is homogeneous of degree 20 on the 23-
dimensional space Sym^4 x Sym^4 x R^3, satisfies p(e) = 1, and is
hyperbolic with respect to e = (I_4, I_4, 0).
PROOF.
Homogeneity and p(e) = 1 have already been checked in section 2.3.
Fix real symmetric X, Z and real y, and consider the one-variable
polynomial t |--> p(t e - (X, Z, y)). We must show its roots are all
real.
Work first with complex t of the form t = a + i eta, with eta > 0. Set
H(t) = t I_4 - Z - Phi_{-y}( (t I_4 - X)^{-1} ).
(We use Phi_{-y} = Phi_y, but writing -y will be more convenient.)
Claim: Im H(t) >= eta I_4 > 0.
Indeed: diagonalize X by an orthogonal matrix. For any real a and
any eta > 0,
Im ( (a + i eta) I - X )^{-1}
= -eta ( (a I - X)^2 + eta^2 I )^{-1} < 0.
For a complex symmetric matrix A, the entrywise imaginary part
Im A equals the Hermitian imaginary part (A - A^*)/(2i), which is
Hermitian. Complex linearity of Phi_{-y} and positivity of Phi_{-y}
(section 2.2) then give
Im H(t) = eta I_4 - Im Z - Im(Phi_{-y}((tI - X)^{-1}))
= eta I_4 + Phi_{-y}( -Im((tI - X)^{-1}) )
= eta I_4 + Phi_{-y}( eta ((aI - X)^2 + eta^2 I)^{-1} )
>= eta I_4 > 0.
Here Im Z = 0 because Z is real. And -Im((tI - X)^{-1}) is
positive definite, so Phi_{-y} of it is positive semidefinite.
Any complex matrix H with Im H > 0 is invertible. Suppose H v = 0
for nonzero v. Then v^* (Im H) v = Im(v^* H v) = 0, contradicting
Im H > 0. So H(t) is invertible.
Since t I_4 - X is also invertible for eta > 0, the algebraic
identity adj(tI - X) = det(tI - X) * (tI - X)^{-1} holds, and the
definition of p gives
p( t e - (X, Z, y) ) = det(tI - X)^4 * det(H(t)).
Because H(t) is invertible, p(t e - (X, Z, y)) != 0.
Thus the polynomial has no roots with positive imaginary part, and
(since its coefficients are real) no roots with negative imaginary
part. All roots are real. Its leading coefficient is p(e) = 1 by
homogeneity, and its degree in t is 20. So it has exactly 20 real
roots. Therefore p is hyperbolic with respect to e. QED.
PROPOSITION 4 (two cone slices).
Let K = K(p, e). For real X, Z and y:
(i) If X > 0, then (X, Z, y) in K <=> Z >= Phi_y(X^{-1}).
(ii) (X, Z, 0) in K <=> X >= 0 and Z >= 0.
Moreover e is an interior point of K.
PROOF.
First, an identity for the roots. Let lambda_1(x), ..., lambda_20(x)
be the roots of t |--> p(t e - x). Homogeneity gives
p(x + t e) = prod_{j=1}^{20} (t + lambda_j(x)).
So x in K iff p(x + t e) != 0 for every strictly positive real t.
(The strict inequality on t allows zero roots, hence includes the
boundary of K.)
Step 1: proof of (i). Assume X > 0. For t >= 0, set
F(t) = Z + t I_4 - Phi_y( (X + t I_4)^{-1} ).
Notice that F(t) is what appears in (***) when we substitute
(X, Z, y) -> (X + t I, Z + t I, y) = (X, Z, y) + t e:
the determinant of F(t) times det(X + t I)^4 is p((X, Z, y) + t e).
If F(0) = Z - Phi_y(X^{-1}) >= 0, then for t > 0 we have
X + tI > X > 0, hence (X + tI)^{-1} < X^{-1}, hence
X^{-1} - (X + tI)^{-1} > 0. Positivity of Phi_y then gives
F(t) = F(0) + t I_4 + Phi_y( X^{-1} - (X + t I)^{-1} )
>= t I_4 > 0.
Hence det F(t) != 0 for every t > 0. Multiplying by the nonzero
factor det(X + tI)^4, we get p((X,Z,y) + t e) != 0 for t > 0.
So (X, Z, y) in K.
Conversely, suppose F(0) is not PSD. Its least eigenvalue is < 0.
For t > 0, the same positivity of Phi_y on PSD inputs and the bound
Phi_y((X + tI)^{-1}) <= t^{-1} Phi_y(I_4) give F(t) > 0 for all
sufficiently large t. Continuity of the least eigenvalue yields
some t > 0 at which F(t) is singular. Since X + tI > 0 and det(X +
tI) != 0, the identity p((X,Z,y) + t e) = det(X + tI)^4 det F(t)
gives p((X,Z,y) + t e) = 0, a positive real root. So (X, Z, y) is
not in K. This proves (i) in both directions.
Step 2: proof of (ii). When y = 0, Phi_0 = 0. The polynomial (***)
reduces to
p(t e - (X, Z, 0)) = det(tI - X)^4 det(tI - Z).
Its roots are the eigenvalues of X (each of multiplicity four) and
the eigenvalues of Z. All roots are >= 0 iff X >= 0 and Z >= 0.
Step 3: e is interior. The set X > 0, Z - Phi_y(X^{-1}) > 0 is open
in the full 23-dimensional space, contains e = (I, I, 0), and is
contained in K by (i). So e is interior. QED.
Remark on strict hyperbolicity. A hyperbolic polynomial is "strictly
hyperbolic" if the roots of t |--> p(t e - x) are distinct when x is
not a scalar multiple of e. The Kummer-Netzer theorem (2026) says
cones of strictly hyperbolic polynomials are spectrahedral. This
paper's polynomial is not strictly hyperbolic: at w = (I, 0, 0), we
computed p(t e - w) = (t - 1)^16 t^4, which has a repeated root at
t = 1. So the Kummer-Netzer theorem does not apply. This is not a
loophole in their theorem; it is the precise reason our polynomial
escapes it.
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5. WHAT A HYPOTHETICAL PENCIL WOULD FORCE
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The two slices in Proposition 4 are the constraints any pencil must
satisfy. This section extracts them into a strong structural statement
about any hypothetical pencil L.
Recall:
- A linear map between symmetric matrix spaces is "positive" if it
sends PSD matrices to PSD matrices.
- A positive map is "unital" if it sends I to I.
PROPOSITION 5 (pencil reduction).
Let K = K(p, e) for the polynomial (***). Suppose K has a
representation by a finite homogeneous real symmetric linear pencil
L(X, Z, y) : Sym^4 x Sym^4 x R^3 -> Sym^N. Then there exist integers
a, c >= 1, positive unital linear maps
D : Sym^4 -> Sym^a, D(I_4) = I_a,
E : Sym^4 -> Sym^c, E(I_4) = I_c,
and a linear map B : R^3 -> R^{a x c}, such that for every X > 0,
Z in Sym^4, and y in R^3, the equivalences
Z >= Phi_y(X^{-1}) <=> E(Z) >= B(y)^T D(X)^{-1} B(y) (5.1)
hold. Here B(y)^T is the transpose, and D(X)^{-1} exists because D
is positive unital and X > 0 force D(X) > 0.
PROOF.
Step 1: remove the pencil's common kernel. Split
L(X, Z, y) = L_1(X) + L_2(Z) + L_3(y).
Idea. If A, H are symmetric and A + H >= 0 and A - H >= 0
simultaneously, then ker A is contained in ker H. (Check: for w in
ker A, both w^T (A + H) w and w^T (A - H) w are >= 0, and both equal
+/- w^T H w plus zero. So w^T H w = 0. A PSD matrix annihilates
every vector on which its quadratic form vanishes.)
Since e is interior to K, for every ambient vector x there is a
small epsilon > 0 with L(e) +/- epsilon L(x) >= 0. The sublemma
above then gives ker L(e) inside ker L(x) for every x. So ker L(e)
is a "common kernel" of the whole pencil. Compress L to the
orthogonal complement of ker L(e). The represented set is unchanged,
and after compression L(e) > 0.
Why nonzero complement? Otherwise the pencil is identically zero,
and then -I, I, 0 (i.e. X = -I, Z = I, y = 0) is accepted, which is
false by (4.ii).
Step 2: decompose L_1 and L_2 into positive maps. By (4.ii), (X, 0, 0)
is in K for every X >= 0. A pencil representing K must have
L(X, 0, 0) >= 0 there. Since L_2(0) = 0 and L_3(0) = 0 (linearity),
L(X, 0, 0) = L_1(X). So L_1(X) >= 0 for every X >= 0; L_1 is a
positive map. The same argument with X = 0 gives L_2(Z) >= 0 for
every Z >= 0.
Set U = ker L_1(I_4). Positivity at I_4 +/- epsilon X (for X PSD
and small epsilon) and the common-kernel sublemma show every L_1(X)
annihilates U. In a block decomposition of the ambient space as
U^perp (+) U:
L_1(X) = [ D(X) 0 ]
[ 0 0 ]
with D(X) acting on U^perp and vanishing on U. Both summands of the
ambient space are nonzero:
- If U^perp = 0, then L_1 = 0 and the pencil accepts (-I, I, 0),
contradicting (4.ii).
- If U = 0, then L_1(I) > 0, so L_1(I) - epsilon L_2(I) > 0 for
small epsilon > 0, and the pencil accepts (I, -epsilon I, 0),
also contradicting (4.ii).
Set a = dim U^perp, c = dim U, and let E(Z) be the lower diagonal
block of L_2(Z). Both D and E are positive maps.
Step 3: normalize D and E to be unital. We have D(I) > 0 (no
kernel) and E(I) > 0 (compression of L(e) > 0 to U). Apply the
fixed block-diagonal congruence diag(D(I)^{-1/2}, E(I)^{-1/2}) to
the whole pencil. This preserves PSD (a congruence) and makes
D(I) = I_a and E(I) = I_c.
Step 4: the off-block G vanishes identically. Write
L_2(Z) = [ F(Z) C(Z) ]
[ C(Z)^T E(Z) ]
L_3(y) = [ A(y) B(y) ]
[ B(y)^T G(y) ]
in the same block decomposition. Fix y and set Z_0 = Phi_y(I_4).
By (4.i), (I, s^2 Z_0, s y) lies in K for every real s. So the
bottom block of L(I, s^2 Z_0, s y) is
s^2 E(Z_0) + s G(y) >= 0 for all s in R.
Divide by |s| and let s -> 0 from both signs: G(y) >= 0 and
G(y) <= 0. So G(y) = 0 for every y.
Step 5: rescale the pencil along the quadratic curve. Fix X > 0,
Z, y, and consider L(X, s^2 Z, s y) for s != 0. Apply the
congruence diag(I_a, s^{-1} I_c). The result is
M_s(X, Z, y) = [ D(X) + s A(y) + s^2 F(Z) B(y) + s C(Z) ]
[ B(y)^T + s C(Z)^T E(Z) ]
Congruence preserves PSD. By (4.i) the equivalence
M_s(X, Z, y) >= 0 <=> Z >= Phi_y(X^{-1}) (5.2)
holds for every real s != 0.
Step 6: take the limit. As s -> 0, the matrices M_s converge entrywise
to
M(X, Z, y) = [ D(X) B(y) ]
[ B(y)^T E(Z) ].
Two consequences of (5.2):
(a) If Z >= Phi_y(X^{-1}), then M_s >= 0 for all s != 0 (by
(5.2)), and the limit M is PSD.
(b) Conversely suppose M >= 0. Positivity and unitality give
D(X) >= lambda_min(X) I_a > 0. For any delta > 0, the
matrix M(X, Z + delta I, y) = M + diag(0, delta I_c) is
strictly positive definite (a vector with nonzero lower
component gets positive contribution from the delta term;
a vector with zero lower component gets positive contribution
from D(X)). As s -> 0 the entries of M_s converge to those
of M, so the smallest eigenvalue of M_s converges to the
positive smallest eigenvalue of M. Hence for each fixed
delta > 0 and small enough nonzero s, M_s(X, Z + delta I, y)
> 0. By (5.2),
Z + delta I >= Phi_y(X^{-1}). Letting delta -> 0 and using
closedness of the PSD cone gives Z >= Phi_y(X^{-1}).
Combining (a) and (b),
Z >= Phi_y(X^{-1}) <=> M(X, Z, y) >= 0. (5.3)
Step 7: Schur complement. Because D(X) > 0, we may use the standard
Schur complement identity: for block matrices
[ P Q ] >= 0 (with P > 0) <=> R - Q^T P^{-1} Q >= 0.
[ Q^T R ]
With P = D(X), Q = B(y), R = E(Z), this converts (5.3) into
Z >= Phi_y(X^{-1}) <=> E(Z) >= B(y)^T D(X)^{-1} B(y). (5.4)
That is (5.1). QED.
================================================================================
6. RANK-ONE LIMITS AND A NORM IDENTITY
================================================================================
The maps D, E, B from Proposition 5 are now fixed: D and E are positive
unital maps between symmetric matrix spaces of sizes a and c, and B is
linear in y.
For a unit vector v in R^4, define the orthogonal projection
P(v) = orthogonal projection of R^a onto ker D(I - v v^T).
For a unit vector u in R^4, define
R(u) = orthogonal projection of R^c onto ker E(I - u u^T).
Both are PSD matrices whose squares equal themselves. If D(I - v v^T)
has full rank, P(v) = 0. If v has a "generic" direction, the kernel can
be small; if v has a "special" direction, the kernel can be large.
PROPOSITION 6 (norm identity).
Let a, c >= 1, let D, E be positive unital linear maps, and let B be
linear in y, satisfying (5.1). Define P(v) and R(u) as above. Then
for every pair of unit vectors u, v in R^4 and every y in R^3,
b( v_0 u' - u_0 v', y ) = || P(v) B(y) R(u) ||_{op}^2. (6.1)
Here the operator norm || . ||_{op} is the Euclidean operator norm
for an a x c matrix, i.e. the largest singular value.
PROOF.
Fix u, v, y. For t >= 1, set
X_t = v v^T + t (I - v v^T),
Z_t = u u^T + t (I - u u^T).
These are positive definite (eigenvalues t and 1). By positivity and
unitality, D(X_t) >= I_a and E(Z_t) >= I_c. So all inverse square
roots below exist. Define
A_t = Z_t^{-1/2} Phi_y(X_t^{-1}) Z_t^{-1/2}
C_t = D(X_t)^{-1/2} B(y) E(Z_t)^{-1/2}.
The matrix A_t is PSD (Phi_y is positive, so a congruence by
Z_t^{-1/2} preserves PSD). For real r, congruence by Z_t^{-1/2}:
r Z_t >= Phi_y(X_t^{-1}) <=> r I >= A_t
<=> r >= lambda_max(A_t).
Likewise, linearity of E and congruence by E(Z_t)^{-1/2} give:
E(r Z_t) >= B(y)^T D(X_t)^{-1} B(y)
<=> r I >= C_t^T C_t
<=> r >= || C_t ||_{op}^2.
By (5.1) the two thresholds r agree. Therefore
lambda_max(A_t) = || C_t ||_{op}^2 for all t >= 1. (6.2)
Now take t -> infinity.
Left side. X_t^{-1} = v v^T + t^{-1}(I - v v^T), so X_t^{-1} -> v v^T,
and Phi_y(X_t^{-1}) -> Phi_y(v v^T). Also Z_t^{-1/2} = u u^T + t^{-1/2}
(I - u u^T), so Z_t^{-1/2} -> u u^T. Consequently
A_t -> u u^T Phi_y(v v^T) u u^T = alpha u u^T,
where alpha = u^T Phi_y(v v^T) u = b(v_0 u' - u_0 v', y) >= 0.
The largest eigenvalue of alpha u u^T is alpha. Continuity of
lambda_max gives lambda_max(A_t) -> alpha.
Right side. Unitality gives D(X_t) = I + (t - 1) D(I - v v^T). Since
D(I - v v^T) is PSD, diagonalize it. For an eigenvector with
eigenvalue mu >= 0, the factor (1 + (t-1) mu)^{-1/2} tends to 1 if
mu = 0, and to 0 if mu > 0. So
(I + (t - 1) H)^{-1/2} -> projection onto ker H,
in operator norm, for any fixed PSD H. Applying this with
H = D(I - v v^T) and H = E(I - u u^T):
D(X_t)^{-1/2} -> P(v),
E(Z_t)^{-1/2} -> R(u).
Therefore C_t -> P(v) B(y) R(u) in operator norm. Taking limits in
(6.2) gives (6.1). QED.
Immediate consequence: at u = v, the left side of (6.1) is
b(0, y) = 0. So
P(v) B(y) R(v) = 0 for every v in S^3 and every y in R^3. (6.3)
The next section differentiates (6.1) near u = v.
================================================================================
7. THE TANGENT OBSTRUCTION: FINISHING THE PROOF
================================================================================
PROOF OF THEOREM 1.
Let p, e, K be as in section 2.3. Proposition 3 gives degree 20, p(e)
= 1, hyperbolic. Suppose K has a real symmetric linear pencil
representation of some finite size. Proposition 5 gives D, E, B
with (5.1). Proposition 6 gives the norm identity (6.1) for the
projections P(v) and R(u).
We show (6.1) is impossible.
Setup. Let S^3 be the unit sphere in R^4. Let
Omega = { u in S^3 : u_0 != 0 }.
This is an open subset of S^3 (dense). For u in Omega, define
A(u) = E(I - u u^T) in Sym^c.
The rank of A(u) takes only finitely many values in {0, 1, ..., c}, as
u varies over Omega. Let rho be the maximum, and choose v in Omega
where rank A(v) = rho. The set { u : rank A(u) = rho } is open and
nonempty, and contains v.
Smoothness of R near v. If rho > 0, select rho independent columns
of A(v) and let W(u) consist of the same columns of A(u). Those
columns remain linearly independent on a neighborhood of v in Omega,
and by maximality of rho they span the range of A(u) throughout that
neighborhood (any larger span would mean higher rank). Symmetry of
A(u) then lets us write
R(u) = I - W(u) ( W(u)^T W(u) )^{-1} W(u)^T.
This is a smooth matrix function of u on the neighborhood. If
rho = 0, then R(u) = I on Omega trivially, also smooth.
So R is smooth near v, and (6.1) becomes an identity of smooth
functions of u near v.
Evaluating (6.1) at u = v. As noted after (6.1),
P(v) B(y) R(v) = 0 for every y in R^3. (7.1)
Differentiating at u = v. Let h = (h_0, h') be any vector in v^perp
(the tangent space to S^3 at v). Define the path
u_s = (v + s h) / sqrt(1 + s^2 ||h||^2), s in R.
For small s, u_s is on S^3 and stays in the neighborhood where R
is smooth. By direct computation,
v_0 u_s' - (u_s)_0 v' = s * (v_0 h' - h_0 v') / sqrt(1 + s^2 ||h||^2).
Define J h = v_0 h' - h_0 v' (linear in h). Then v_0 u_s' - (u_s)_0
v' = s J h / sqrt(1 + s^2 ||h||^2).
Since b is quadratic in its first argument, plugging u = u_s into
(6.1):
b( s J h / sqrt(...), y ) = s^2 b(J h, y) / (1 + s^2 ||h||^2).
Right side of (6.1): R(u_s) = R(v) + s dR_v(h) + O(s^2), where dR_v
is the differential of R at v. Multiply on the left by P(v) and on
the right by B(y):
P(v) B(y) R(u_s)
= P(v) B(y) R(v) + s P(v) B(y) dR_v(h) + O(s^2)
= s P(v) B(y) dR_v(h) + O(s^2) by (7.1).
Taking squared norms and dividing by s^2, and letting s -> 0:
b(J h, y) = || P(v) B(y) dR_v(h) ||_{op}^2. (7.2)
This holds for every h in v^perp and every y in R^3.
J is an isomorphism v^perp -> R^3. Indeed, J h = 0 means v_0 h' -
h_0 v' = 0. Since v_0 != 0 (v is in Omega), we get h = (h_0 / v_0) v.
Combined with h in v^perp (i.e. h . v = 0), this forces h = 0.
So J is injective; dimensions match (3 = 3); so J is bijective.
Define F : R^3 x R^3 -> R^{a x c} by
F(z, y) = P(v) B(y) dR_v(J^{-1} z).
This is bilinear in (z, y), because B and dR_v are linear.
Entrywise domination. Fix bases of R^a and R^c. Write F(z, y) =
(f_{kl}(z, y))_{k=1..a, l=1..c}. Each f_{kl} is a bilinear form
in (z, y) for the same reason F is. For any matrix M and any entry,
| M_{kl} | <= ||M||_{op}
(the operator norm is at least the largest singular value, which
dominates every entry). Therefore, for each (k, l),
f_{kl}(z, y)^2 <= || F(z, y) ||_{op}^2
= || P(v) B(y) dR_v(J^{-1} z) ||_{op}^2.
By the definition of F and (7.2) with h = J^{-1} z:
f_{kl}(z, y)^2 <= b(z, y).
Lemma 1 then forces f_{kl} = 0 for every k, l. So F = 0 identically.
Consequently b(z, y) = || F(z, y) ||_{op}^2 = 0 for every z, y in
R^3. But b(e_1, e_1) = 1 (section 3). Contradiction.
Therefore no finite spectrahedral representation of K exists. QED.
================================================================================
8. REMOVING A DETERMINANT FACTOR
================================================================================
PROOF OF COROLLARY 2.
For i = 1, 2, 3, define the 4x4 matrix
K_i = [ 0 e_i^T ]
[ - e_i 0 ],
with e_i the i-th coordinate basis vector of R^3. Let
C = [ K_1 K_2 K_3 ] in R^{4 x 12}.
Here "U tensor V" denotes the block matrix with blocks U_{ij} V; for
instance, the 3x3 identity tensor X (where X is 4x4) is the 12 x 12
block diagonal matrix with three blocks equal to X.
Direct block multiplication in the definition of Phi_y (section 2.2)
shows that, for T in Sym^4,
Phi_y(T) = sum_{i,j=1}^3 Q(y)_{ij} * K_i T K_j^T
= C * ( Q(y) tensor T ) * C^T.
Now define
q(X, Z, y) = det of the 16x16 block matrix
[ I_3 tensor X C^T ]
[ C ( Q(y) tensor I_4 ) Z ]
The block sizes: I_3 tensor X is 12 x 12; C is 4 x 12, so C^T is
12 x 4; Q(y) tensor I_4 is 12 x 12; Z is 4 x 4. So the whole matrix
is 16 x 16, polynomial in (X, Z, y).
Compute. For X invertible, use the Schur complement formula on the
(1,1) block I_3 tensor X:
q(X, Z, y) = det(I_3 tensor X) *
det( Z - C (Q tensor I_4) (I_3 tensor X)^{-1} C^T ).
Now det(I_3 tensor X) = (det X)^3, and (I_3 tensor X)^{-1} =
I_3 tensor X^{-1}. Also (A tensor B)(F tensor G) = AF tensor BG for
block matrices of the appropriate sizes. So
C (Q tensor I_4) (I_3 tensor X^{-1}) C^T
= C (Q tensor X^{-1}) C^T
= Phi_y(X^{-1}).
Thus
q(X, Z, y) = (det X)^3 det( Z - Phi_y(X^{-1}) ). (8.1)
On the other hand, from (***) and adj X = (det X) X^{-1}:
p(X, Z, y) = (det X)^4 det( Z - Phi_y(X^{-1}) ). (8.2)
Dividing (8.2) by (8.1):
p(X, Z, y) = (det X) * q(X, Z, y) (8.3)
for every invertible X. Both sides are polynomials (in X, Z, y),
and they agree on a nonempty open set (X invertible is dense).
Therefore (8.3) is a polynomial identity everywhere.
Consequences:
- Degrees. p has degree 20; det X has degree 4; so q has degree 16.
- q(e) = 1. Since det(e_X) = det(I_4) = 1 and p(e) = 1, (8.3) gives
q(e) = 1.
- q is hyperbolic. The polynomial q(t e - (X, Z, y)) is (by (8.3))
a factor of p(t e - (X, Z, y)): specifically,
p = det(tI_4 - X) * q. Since p(t e - x) has all real roots,
q(t e - x) does too (a factor of a real-rooted polynomial is
real-rooted). So q is hyperbolic with respect to e.
Cone identity. Let K_q = K(q, e). From (8.3):
K = { X >= 0 } intersect K_q. (8.4)
We claim K_q already forces X >= 0, hence K_q = K.
Idea. q is even in y: since Q(y) is quadratic in y, Q(-y) = Q(y),
so q(X, Z, -y) = q(X, Z, y). Hyperbolicity cones are convex (Garding,
1959). So if (X, Z, y) is in K_q, then (X, Z, -y) is too, and their
midpoint (X, Z, 0) is also in K_q.
Slice at y = 0. From (8.1), q(X, Z, 0) = (det X)^3 det Z, whose
closed hyperbolicity cone in the slice {y = 0} is
{ (X, Z, 0) : X >= 0, Z >= 0 }.
Therefore: for any (X, Z, y) in K_q, the midpoint (X, Z, 0) is in
K_q, hence X >= 0 and Z >= 0. In particular every element of K_q has
X >= 0.
Combining with (8.4): K_q = K.
Since K is not spectrahedral (Theorem 1), K_q is not spectrahedral
either. QED.
================================================================================
9. APPENDIX: A PATH-CONTAINMENT EXAMPLE
================================================================================
The paper "The generalized Lax conjecture is true ..." by Gonzalez
Nevado (version 1, 2026) claims a positive solution of the Generalized
Lax conjecture. The paper's proof invokes an intermediate assertion:
that a fixed rigidly convex set is contained throughout a smooth
real-zero deformation, from a product of normalized tangent linear
factors to a multiple of the defining polynomial.
The following one-variable example violates that intermediate
assertion under its stated hypotheses.
Set a > 0, b > 0, and let eta > 0 be arbitrary. In one variable x,
define
g(x) = ( 1 - x/a ) ( 1 + x/b )
c(t) = 1 + eta sin^2( pi t ) ( t in [0, 1] )
H_t(x) = g( c(t) x ).
Basic facts:
- The zero set of g is the two-point compact smooth set {-b, a}.
- The normalized tangent linear factors of g at these points are
(1 + x/b) and (1 - x/a), whose product is g.
- H_t(0) = 1 for every t.
- H_t has two distinct real roots, -b/c(t) and a/c(t). So H_t is
real-zero for every t, with compact smooth zero set.
- H_t depends smoothly on t.
- H_t even admits a smooth monic real symmetric determinantal
representation:
H_t(x) = det [ [ 1 - c(t) x/a, 0 ],
[ 0, 1 + c(t) x/b ] ].
- At t = 0 and t = 1, c(t) = 1, so H_0 = H_1 = g. Both endpoints
equal g, and the final polynomial is g times the constant cofactor
"1" (which is the allowed form).
The difficulty: the closed rigidly convex set of H_t is
[ -b/c(t), a/c(t) ],
which is strictly smaller than the closed rigidly convex set
[ -b, a ] of g, for every 0 < t < 1 (since c(t) > 1 there). So during
the deformation, the rigidly convex set of H_t does not contain the
rigidly convex set of the endpoint.
This violates the all-intermediate-times conclusion of the cited
Theorem 52.
The example does not refute an "endpoint-only" version of the
assertion, because at t = 0 and t = 1 the polynomials coincide with g.
The parameters a, b > 0 give an open family of normalized real
quadratics with negative leading coefficient, so the failure is not
confined to a symmetric or repeated-root case. The parameter eta may
be arbitrarily small. The cited Theorem 52 states no restriction to
two or more variables, no lower bound on matrix size, and no
requirement that the endpoint cofactor be nonconstant. Under its
stated hypotheses, the example violates the relevant assertion.
The main theorem (nonspectrahedrality of K) is independent of this
example: it follows directly from the construction and obstruction
in sections 2-7.
================================================================================
10. WHAT THIS DOES AND DOES NOT PROVE
================================================================================
What this proves:
- There is a real polynomial p, homogeneous of degree 20 in 23 real
variables, hyperbolic with respect to e = (I_4, I_4, 0), whose
closed hyperbolicity cone K is not equal to {x : L(x) >= 0} for
any finite real symmetric linear pencil L.
- The polynomial p divides by det X to a degree-16 polynomial q with
the same cone. Either p or q disproves the Generalized Lax
conjecture in its original variables (no auxiliary variables).
- There is a one-variable counterexample to a specific path-
containment assertion used in one published proof of the
Generalized Lax conjecture.
What this does NOT prove:
- It does not say anything about spectrahedral shadows, i.e.
representations of K as a linear image of {y : L(y) >= 0} with
extra auxiliary variables. Shadows are strictly more permissive,
and there are known positive results (Netzer-Sanyal 2015 and
others) that the theorem here does not contradict.
- It does not claim that 23 variables or degree 16 is minimal. A
smaller or lower-degree counterexample may exist.
- It does not contradict the Kummer-Netzer theorem on strictly
hyperbolic polynomials, because p is not strictly hyperbolic (it
has a repeated root at t = 1 for w = (I, 0, 0)).
- It does not contradict the ordinary Lax conjecture for polynomials
in three variables (Helton-Vinnikov 2007, Lewis-Parrilo-Ramana
2005); the counterexample lives in 23 variables.
- It does not contradict Branden's earlier obstruction. Branden
showed some polynomials have no positive power that is a definite
determinant. That is a different statement. Kummer's specialized
Vamos polynomial is the explicit example: no positive power of
the polynomial is a definite determinant, yet the cone itself has
a 7x7 spectrahedral representation. The distinction between "the
polynomial is a determinant" and "the cone is a spectrahedron"
is the gap that this paper closes.
What the arguments rely on:
- Garding's 1959 theorem that hyperbolicity cones are convex, used
in section 8 to take midpoints in K_q.
- The Choi-Lam theorem (1977) that the coefficient-one biquadratic
form is extremal, used in the historical sense; here proved
directly as Lemma 1.
- Standard facts about positive semidefinite matrices: spectral
theorem, Schur complements, invertibility of matrices with
positive-definite imaginary part.
================================================================================
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================================================================================
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END
================================================================================