==== TABLE OF CONTENTS ====
A-Nonspectrahedral-Hyperbolicity-Cone//01-history.tex
A-Nonspectrahedral-Hyperbolicity-Cone//01-introduction.tex
A-Nonspectrahedral-Hyperbolicity-Cone//02-form.tex
A-Nonspectrahedral-Hyperbolicity-Cone//03-hyperbolic.tex
A-Nonspectrahedral-Hyperbolicity-Cone//04-pencil.tex
A-Nonspectrahedral-Hyperbolicity-Cone//05-norm.tex
A-Nonspectrahedral-Hyperbolicity-Cone//06-tangent.tex
A-Nonspectrahedral-Hyperbolicity-Cone//07-comparison.tex
A-Nonspectrahedral-Hyperbolicity-Cone//main.tex
A-Nonspectrahedral-Hyperbolicity-Cone//reduced-polynomial.tex
A-Nonspectrahedral-Hyperbolicity-Cone//tangent-variation.tex
A-Nonspectrahedral-Hyperbolicity-Cone//references.bib
===========================
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//01-history.tex ====
G\aa rding's theory establishes the convexity of hyperbolicity
cones~\cite{garding-1959}. The original Lax conjecture asks for a
definite determinantal representation of homogeneous hyperbolic polynomials in three
variables~\cite{lax-1958}. Helton and Vinnikov proved the corresponding
definite determinantal representation theorem~\cite{helton-vinnikov-2007};
Lewis, Parrilo, and Ramana established its equivalence with Lax's
formulation~\cite{lewis-parrilo-ramana-2005}. In higher dimension, Br\"and\'en
constructed a real-zero polynomial none of whose positive powers admits
a definite determinantal representation~\cite[Theorem~3.3]{branden-2011}.
That obstruction does not exclude a different determinant, with additional
factors, from defining the same cone. The geometric conjecture concerns this remaining
possibility. Kummer exhibited the distinction concretely: the cone of
the specialized V\'amos polynomial has a $7\times7$ representation,
although no positive power of that polynomial has a definite
determinantal representation~\cite[Section~3]{kummer-2016-vamos}.
Raghavendra, Ryder, Srivastava, and Weitz proved exponential
lower bounds on the matrix size of spectrahedral representations for
suitable hyperbolicity cones~\cite{raghavendra-ryder-srivastava-2018}.
Recent positive results include Netzer's theorem for
hyperbolic cubics in five variables~\cite[Theorem~3.9]{netzer-2026}.
Kummer and Netzer prove spectrahedrality for cones of strictly
hyperbolic polynomials~\cite[Theorem~1]{kummer-netzer-2026-strict}.
Here \emph{strictly hyperbolic} means that $p(te-x)$ has distinct roots
whenever $x\notin\R e$. The polynomial~\eqref{eq:polynomial} fails this
hypothesis: at $w=(I_4,0,0)\notin\R e$ it gives
\[
p(te-w)=(t-1)^{16}t^4.
\]
For the reduced polynomial of Corollary~\ref{cor:reduced}, the same line
gives $q(te-w)=(t-1)^{12}t^4$. Thus neither defining polynomial satisfies
the strict-case hypothesis.
A \emph{spectrahedral shadow} is a linear image of a set defined by an
affine real symmetric linear matrix inequality.
Netzer and Sanyal proved that a hyperbolicity cone is a spectrahedral
shadow when every nonzero boundary point is a smooth point of the
defining polynomial~\cite[Theorem~1.1]{netzer-sanyal-2015}.
Our result concerns a direct representation in the original variables;
it makes no assertion about representations using auxiliary variables.
The scalar ingredient is the coefficient-one Choi--Lam biquadratic
form~\cite[Section~4]{choi-lam-1977}, which reduces the coefficient of
the extra cyclic squared monomials from two in Choi's earlier
nonnegative form~\cite{choi-1975} to one. Its inability to dominate a nonzero
bilinear square follows from the extremality proved in
\cite[Theorem~4.4]{choi-lam-1977}; we give a short proof of precisely this
property. It is called \emph{weak extremality} by Blekherman, Rai\c{t}\u{a},
Shankar, and Sinn~\cite[Definition~2.7]{blekherman-raita-shankar-sinn-2022},
whose characterization uses Taylor expansions of dominated squares at
zeros~\cite[Theorem~3.2]{blekherman-raita-shankar-sinn-2022}.
The correspondence between nonnegative biquadratic forms and
positive maps on real symmetric matrices is classical; see
\cite[Section~3]{ha-2013}. Positive maps have also been used to obtain
certificates of conic polynomial stability~\cite{dey-gardoll-theobald-2021}.
Here the construction turns the scalar obstruction into an obstruction
to every finite pencil describing one hyperbolicity cone. A related
general strategy appears in Scheiderer's local sums-of-squares
obstructions to spectrahedral shadows~\cite[Section~4]{scheiderer-2018}.
Our proof extracts bilinear square minorants from a differentiated
operator-norm identity; it does not invoke a criterion for shadows.
Gonz\'alez Nevado states a positive solution of the geometric conjecture
in~\cite[Theorem~53, pp.~16--17]{gonzalez-nevado-2026}.
Appendix~\ref{app:opposition} records a counterexample to the
path-containment assertion used in that argument. The construction
and proof in the present paper are independent of that comparison.
==== END A-Nonspectrahedral-Hyperbolicity-Cone//01-history.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//01-introduction.tex ====
\section{Introduction}
\label{sec:introduction}
A homogeneous polynomial $p\in\R[x_1,\ldots,x_m]$ of degree $d\geq1$ is
\emph{hyperbolic with respect to} $e\in\R^m$ if $p(e)\ne0$ and every root of
$t\mapsto p(te-x)$ is real for every $x\in\R^m$.
Write these roots, with multiplicity, as $\lambda_1(x),\ldots,\lambda_d(x)$.
The closed hyperbolicity cone is
\[
\Lambda_+(p,e)=\{x\in\R^m:\lambda_j(x)\geq0\text{ for }1\leq j\leq d\}.
\]
We write $\Sym^N$ for the real symmetric $N\times N$ matrices and
$A\succeq0$ (respectively, $A\succ0$) for positive semidefiniteness
(respectively, positive definiteness).
A cone is \emph{spectrahedral} if it has the form
\begin{equation}
K=\{x\in\R^m:L(x)\succeq0\},\qquad
L(x)=\sum_{i=1}^m x_iA_i,\quad A_i\in\Sym^N,
\label{eq:spectrahedral}
\end{equation}
for some finite $N\geq1$.
The geometric Generalized Lax conjecture asserts that every hyperbolicity
cone is spectrahedral. We give a counterexample.
\subsection{The explicit polynomial}
All indices in $\{1,2,3\}$ in the following construction are cyclic.
For $z,y\in\R^3$, let
\begin{equation}
b(z,y)=\sum_{i=1}^3 z_i^2y_i^2
-2\sum_{1\leq i0$. Orthogonal diagonalization of $X$ gives
\[
\operatorname{Im}(tI_4-X)^{-1}
=-\eta\bigl((aI_4-X)^2+\eta^2I_4\bigr)^{-1}\prec0.
\]
For a complex symmetric matrix, the entrywise imaginary part equals
the Hermitian imaginary part $(A-A^*)/(2i)$ and is real symmetric.
Complex linearity and positivity of $\Phi_{-y}$ consequently show
that
\[
H=tI_4-Z-\Phi_{-y}\bigl((tI_4-X)^{-1}\bigr)
\quad\text{satisfies}\quad
\operatorname{Im}H\succeq\eta I_4\succ0.
\]
Such an $H$ is invertible: if $Hw=0$ for a nonzero complex vector
$w$, then $0=\operatorname{Im}(w^*Hw)=w^*(\operatorname{Im}H)w>0$.
Also $tI_4-X$ is invertible. Equation~\eqref{eq:inverse-formula}
therefore implies $p(te-(X,Z,y))\ne0$.
The coefficients of this polynomial in $t$ are real, so conjugation
also excludes roots in the lower half-plane. Finally, its leading
coefficient is $p(e)=1$ by homogeneity; it has degree $20$ for every
real input. All its roots are therefore real, as required.
\end{proof}
\begin{proposition}\label{prop:slices}
Let $K=\Lambda_+(p,e)$. For real symmetric $X,Z$ and $y\in\R^3$,
\begin{align}
X\succ0:\qquad
(X,Z,y)\in K
&\ \Longleftrightarrow\
Z\succeq\Phi_y(X^{-1}),
\label{eq:slice-positive}\\
(X,Z,0)\in K
&\ \Longleftrightarrow\
X\succeq0\ \text{and}\ Z\succeq0.
\label{eq:slice-zero}
\end{align}
Moreover, $e$ lies in the interior of $K$ in the full ambient space.
\end{proposition}
\begin{proof}
If $\lambda_1(x),\ldots,\lambda_{20}(x)$ are the roots of
$t\mapsto p(te-x)$, homogeneity gives
\[
p(x+te)=\prod_{j=1}^{20}(t+\lambda_j(x)).
\]
Thus $x\in K$ if and only if $p(x+te)\ne0$ for every real $t>0$.
The strict inequality on $t$ allows zero roots and hence includes
the boundary of $K$.
Suppose $X\succ0$ and set, for $t\ge0$,
\[
F(t)=Z+tI_4-\Phi_y\bigl((X+tI_4)^{-1}\bigr).
\]
If $F(0)\succeq0$, order preservation yields
\[
F(t)=F(0)+tI_4+
\Phi_y\bigl(X^{-1}-(X+tI_4)^{-1}\bigr)
\succeq tI_4\succ0\qquad(t>0).
\]
Equation~\eqref{eq:inverse-formula} then shows that there is no
positive root, so $(X,Z,y)\in K$.
Conversely, if $F(0)\not\succeq0$, its least eigenvalue is negative.
For $t>0$, positivity gives
\[
0\preceq\Phi_y\bigl((X+tI_4)^{-1}\bigr)
\preceq t^{-1}\Phi_y(I_4).
\]
Consequently $F(t)\succ0$ for all sufficiently large $t$.
Continuity of its least eigenvalue gives a $t>0$ at which $F(t)$
is singular. Since $X+tI_4\succ0$,
Equation~\eqref{eq:inverse-formula} now gives a positive root of
$p((X,Z,y)+te)$. This proves \eqref{eq:slice-positive} in both
directions, including equality in the matrix inequality.
When $y=0$, the polynomial definition directly gives
\[
p(te-(X,Z,0))=\det(tI_4-X)^4\det(tI_4-Z),
\]
also for singular $X$ and $Z$. Its roots are the eigenvalues of $X$,
each repeated four times, and the eigenvalues of $Z$.
They are all nonnegative exactly when both matrices are positive
semidefinite. This proves \eqref{eq:slice-zero}.
Finally, $X\succ0$ and $Z-\Phi_y(X^{-1})\succ0$ define an open
subset of the full ambient space. It contains $e$ and, by
\eqref{eq:slice-positive}, is contained in $K$. Thus $e$ is an
interior point.
\end{proof}
==== END A-Nonspectrahedral-Hyperbolicity-Cone//03-hyperbolic.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//04-pencil.tex ====
\section{What a semidefinite representation would imply}
\label{sec:pencil}
The two slices in Proposition~\ref{prop:slices} constrain every pencil
representing $K$. We now turn those constraints into an exact block
matrix inequality whose off-diagonal block is linear in $y$.
A positive linear map preserves positive semidefiniteness; it is
called \emph{unital} if it sends the identity to the identity.
\begin{proposition}\label{prop:pencil}
Let $K=\Lambda_+(p,(I_4,I_4,0))$ for the polynomial
in~\eqref{eq:polynomial}. Suppose that $K$ admits a representation by a finite homogeneous real
symmetric linear pencil. Then there are integers $a,c\geq1$, positive
linear maps
\[
D:\Sym^4\longrightarrow\Sym^a,
\qquad E:\Sym^4\longrightarrow\Sym^c,
\qquad D(I_4)=I_a,\quad E(I_4)=I_c,
\]
and a linear map $B:\R^3\longrightarrow\R^{a\times c}$ such that, for
every $X\succ0$, $Z\in\Sym^4$ and $y\in\R^3$,
\begin{equation}\label{eq:schur-equivalence}
Z\succeq\Phi_y(X^{-1})
\quad\Longleftrightarrow\quad
E(Z)\succeq B(y)^\top D(X)^{-1}B(y).
\end{equation}
\end{proposition}
\begin{proof}
Write the hypothetical pencil as
\[
L(X,Z,y)=L_1(X)+L_2(Z)+L_3(y).
\]
We first remove the pencil's common kernel and normalize
two positive blocks.
We then rescale the pencil and prove that its limit represents the
same inequality.
\emph{Compression and normalization.}
We use the elementary fact that
\[
A+H\succeq0,\quad A-H\succeq0
\quad\Longrightarrow\quad \ker A\subseteq\ker H.
\]
Indeed, if $w\in\ker A$, the two nonnegative quadratic forms on $w$
sum to zero. A positive semidefinite matrix annihilates every vector
on which its quadratic form vanishes, so both $(A+H)w$ and $(A-H)w$
are zero.
Since $e$ is interior to $K$, for every ambient vector $x$ there is
an $\varepsilon>0$ with $L(e)\pm\varepsilon L(x)\succeq0$.
The preceding fact shows that $\ker L(e)$ is a common kernel of the
whole pencil. Compressing to its orthogonal complement preserves the
represented cone and makes $L(e)\succ0$. This complement is nonzero:
otherwise the pencil would vanish identically, whereas
$(-I_4,I_4,0)\notin K$ by~\eqref{eq:slice-zero}.
The same slice shows that $L_1$ and $L_2$ are positive maps. Set
$U=\ker L_1(I_4)$. Positivity at $I_4\pm\varepsilon X$, for small
$\varepsilon>0$, and the same kernel argument show that every
$L_1(X)$ annihilates $U$. Thus, in $U^\perp\oplus U$,
\[
L_1(X)=\begin{pmatrix}D(X)&0\\0&0\end{pmatrix}.
\]
Both summands are nonzero. If $U^\perp=0$, then $L_1=0$ and the
pencil accepts $(-I_4,I_4,0)$, contradicting~\eqref{eq:slice-zero}.
If $U=0$, then $L_1(I_4)\succ0$, so
$L_1(I_4)-\varepsilon L_2(I_4)\succ0$ for sufficiently small
$\varepsilon>0$. This would accept $(I_4,-\varepsilon I_4,0)$,
again contradicting that slice.
Let $a=\dim U^\perp$, $c=\dim U$, and let $E(Z)$ be the lower
diagonal block of $L_2(Z)$. Both $D$ and $E$ are positive maps.
Moreover, $D(I_4)\succ0$ by the definition of $U$, and
$E(I_4)\succ0$ because it is the compression of $L(e)\succ0$ to
$U$. Apply the fixed block diagonal congruence
\[
\operatorname{diag}\bigl(D(I_4)^{-1/2},E(I_4)^{-1/2}\bigr)
\]
and retain the notation for the resulting blocks. We now have
$D(I_4)=I_a$ and $E(I_4)=I_c$.
\emph{Rescaling the pencil.}
Write the other blocks as
\[
L_2(Z)=\begin{pmatrix}F(Z)&C(Z)\\C(Z)^\top&E(Z)\end{pmatrix},
\qquad
L_3(y)=\begin{pmatrix}A(y)&B(y)\\B(y)^\top&G(y)\end{pmatrix}.
\]
Fix $y$ and put $Z_0=\Phi_y(I_4)$. By
\eqref{eq:slice-positive} and $\Phi_{sy}=s^2\Phi_y$, the point
$(I_4,s^2Z_0,sy)$ lies in $K$ for every real $s$. Its lower pencil
block therefore satisfies
\[
s^2E(Z_0)+sG(y)\succeq0 \qquad(s\in\R).
\]
Dividing by $|s|$ and taking limits through positive and negative
$s$ gives $G(y)\succeq0$ and $-G(y)\succeq0$. Hence $G(y)=0$.
For fixed $X\succ0$, $Z$ and $y$, and any real $s\ne0$, congruence
of $L(X,s^2Z,sy)$ by $\operatorname{diag}(I_a,s^{-1}I_c)$ gives
\begin{equation}\label{eq:scaled-pencil}
M_s(X,Z,y)=
\begin{pmatrix}
D(X)+sA(y)+s^2F(Z)&B(y)+sC(Z)\\
B(y)^\top+sC(Z)^\top&E(Z)
\end{pmatrix}.
\end{equation}
Congruence preserves positive semidefiniteness for either sign of
$s$. The exact slice~\eqref{eq:slice-positive} gives
\begin{equation}\label{eq:scaled-test}
M_s(X,Z,y)\succeq0
\quad\Longleftrightarrow\quad Z\succeq\Phi_y(X^{-1}),
\qquad s\ne0.
\end{equation}
As $s\to0$, these matrices converge to
\[
M(X,Z,y)=
\begin{pmatrix}D(X)&B(y)\\B(y)^\top&E(Z)\end{pmatrix}.
\]
It follows immediately that $Z\succeq\Phi_y(X^{-1})$ implies
$M(X,Z,y)\succeq0$.
\emph{Recovering the inequality from the limit.}
For the converse, suppose that $M(X,Z,y)\succeq0$. Positivity and
unitality give
\[
D(X)\succeq\lambda_{\min}(X)I_a\succ0.
\]
For every $\delta>0$,
\[
M(X,Z+\delta I_4,y)
=M(X,Z,y)+\operatorname{diag}(0,\delta I_c)
\succ0.
\]
To see strict positivity, a vector with nonzero lower component has
strictly positive contribution from the added term. A nonzero
vector with lower component zero has strictly positive quadratic
form under $D(X)$.
Consequently, for each fixed $\delta>0$, the matrices
$M_s(X,Z+\delta I_4,y)$ are positive definite for sufficiently small
nonzero $s$. Equation~\eqref{eq:scaled-test} yields
$Z+\delta I_4\succeq\Phi_y(X^{-1})$. Letting $\delta\downarrow0$
proves the converse. We have therefore shown
\[
Z\succeq\Phi_y(X^{-1})
\quad\Longleftrightarrow\quad M(X,Z,y)\succeq0.
\]
Taking the Schur complement of the positive definite block $D(X)$
gives~\eqref{eq:schur-equivalence}.
\end{proof}
The maps and their finite dimensions are now fixed by the
hypothetical pencil. We next recover the scalar form $b$ from
\eqref{eq:schur-equivalence}; this will force a nonzero bilinear square
dominated by $b$ and contradict its defining obstruction.
==== END A-Nonspectrahedral-Hyperbolicity-Cone//04-pencil.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//05-norm.tex ====
\section{Rank-one limits and a norm identity}
\label{sec:norm}
Let $D,E,B$ be the maps supplied by Proposition~\ref{prop:pencil}.
Thus $D$ and $E$ are positive maps with output sizes $a,c\geq1$ and
are \emph{unital}, meaning $D(I_4)=I_a$ and $E(I_4)=I_c$,
and $B(y)$ is an $a\times c$ matrix.
For unit vectors $u,v\in\R^4$, define the orthogonal projections
\begin{equation}
P(v)=\operatorname{proj}_{\ker D(I-vv^\top)},
\qquad
R(u)=\operatorname{proj}_{\ker E(I-uu^\top)}.
\label{eq:projections}
\end{equation}
These act on $\R^a$ and $\R^c$, respectively.
The next proposition recovers the scalar form~\eqref{eq:form} from the
Schur threshold~\eqref{eq:schur-equivalence}.
\begin{proposition}
\label{prop:norm}
Let $a,c\geq1$, let $D:\Sym^4\to\Sym^a$ and
$E:\Sym^4\to\Sym^c$ be positive unital linear maps, and let
$B:\R^3\to\R^{a\times c}$ be linear. Assume that
\eqref{eq:schur-equivalence} holds for every $X\succ0$,
$Z\in\Sym^4$ and $y\in\R^3$, with $\Phi_y$ defined
by~\eqref{eq:phi}. Define $P(v)$ and $R(u)$ by~\eqref{eq:projections}.
For every pair of unit vectors $u=(u_0,u'),v=(v_0,v')\in\R^4$
and every $y\in\R^3$,
\begin{equation}
b(v_0u'-u_0v',y)
=\norm{P(v)B(y)R(u)}_{\op}^{2}.
\label{eq:norm-identity}
\end{equation}
Here the operator norm uses the Euclidean structures on the two block spaces.
\end{proposition}
\begin{proof}
Fix $u,v,y$. For $t\geq1$, set
\[
X_t=vv^\top+t(I-vv^\top),
\qquad
Z_t=uu^\top+t(I-uu^\top).
\]
Both matrices are at least $I$. Positivity and unitality therefore give
$D(X_t)\succeq I$ and $E(Z_t)\succeq I$, so all inverse square roots below
exist. Put
\[
A_t=Z_t^{-1/2}\Phi_y(X_t^{-1})Z_t^{-1/2},
\qquad
C_t=D(X_t)^{-1/2}B(y)E(Z_t)^{-1/2}.
\]
The matrix $A_t$ is positive semidefinite. For every real $r$, congruence
by $Z_t^{-1/2}$ gives
\[
rZ_t\succeq\Phi_y(X_t^{-1})
\quad\Longleftrightarrow\quad
r\geq\lambda_{\max}(A_t).
\]
Likewise, linearity of $E$ and congruence by $E(Z_t)^{-1/2}$ give
\[
\begin{split}
E(rZ_t)\succeq B(y)^\top D(X_t)^{-1}B(y)
&\quad\Longleftrightarrow\quad rI\succeq C_t^\top C_t\\
&\quad\Longleftrightarrow\quad r\geq\norm{C_t}_{\op}^{2}.
\end{split}
\]
Equation~\eqref{eq:schur-equivalence} identifies these two closed rays in
$r\in\R$, including their endpoints. Consequently,
\begin{equation}
\lambda_{\max}(A_t)=\norm{C_t}_{\op}^{2}.
\label{eq:threshold-norm}
\end{equation}
This applies to rectangular $C_t$ and includes a zero threshold.
The explicit formulas
\[
X_t^{-1}=vv^\top+t^{-1}(I-vv^\top),
\qquad
Z_t^{-1/2}=uu^\top+t^{-1/2}(I-uu^\top)
\]
show that
\[
A_t\longrightarrow
uu^\top\Phi_y(vv^\top)uu^\top
=\alpha uu^\top,
\qquad
\alpha=u^\top\Phi_y(vv^\top)u\geq0.
\]
The sign follows from positivity of $\Phi_y$. Thus the largest eigenvalue
of this limit is $\alpha$, also when $\alpha=0$. Continuity of the largest
eigenvalue gives $\lambda_{\max}(A_t)\to\alpha$.
For the other side, unitality gives
\[
D(X_t)=I+(t-1)D(I-vv^\top).
\]
If $H\succeq0$ is a fixed finite-dimensional matrix, diagonalizing $H$
shows that
\begin{equation}
[I+(t-1)H]^{-1/2}
\longrightarrow\operatorname{proj}_{\ker H}
\quad\text{in operator norm}.
\label{eq:kernel-limit}
\end{equation}
Indeed, the factor on an eigenvector of eigenvalue $\mu\geq0$ is
$(1+(t-1)\mu)^{-1/2}$, which is one when $\mu=0$ and tends to zero
otherwise. Applying~\eqref{eq:kernel-limit} to $D(I-vv^\top)$ and
$E(I-uu^\top)$ yields
\[
D(X_t)^{-1/2}\longrightarrow P(v),
\qquad
E(Z_t)^{-1/2}\longrightarrow R(u).
\]
Hence $C_t\to P(v)B(y)R(u)$ in operator norm. The dimensions $a,c$ are
fixed throughout; the limits include zero and identity kernel projections.
Taking limits in~\eqref{eq:threshold-norm} and using the rank-one
identity~\eqref{eq:rank-one} proves~\eqref{eq:norm-identity}.
\end{proof}
At $u=v$, the left side of~\eqref{eq:norm-identity} vanishes, so
$P(v)B(y)R(v)=0$ for every $y$.
We next study the identity as $u$ varies near $v$.
==== END A-Nonspectrahedral-Hyperbolicity-Cone//05-norm.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//06-tangent.tex ====
\section{The tangent obstruction}\label{sec:tangent}
We finish the proof by taking a first-order limit in the norm identity
at a point where the kernel projection varies smoothly. This produces
a bilinear matrix whose entries must vanish by Lemma~\ref{lem:form}.
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Proposition~\ref{prop:hyperbolic} establishes the polynomial's degree,
normalization, and hyperbolicity. Suppose its closed hyperbolicity cone
has a real symmetric pencil representation of some finite size.
Fix the maps \(D,E,B\) supplied by Proposition~\ref{prop:pencil}, with
their fixed finite dimensions \(a,c\). Proposition~\ref{prop:norm} gives
\eqref{eq:norm-identity} for the projections in \eqref{eq:projections}.
We shall show that this identity is impossible.
Write \(S^3=\{u\in\R^4:\norm{u}=1\}\) and consider the symmetric matrix
\[
A(u)=E(I-uu^\top),\qquad
u\in\Omega:=\{u\in S^3:u_0\ne0\}.
\]
Its ranks take only finitely many values, so their maximum \(\rho\) is
attained at some \(v\in\Omega\). If \(\rho>0\), select \(\rho\) independent
columns of \(A(v)\), and let \(W(u)\) contain the same columns of \(A(u)\).
These columns remain independent on a neighborhood of \(v\) in
\(\Omega\). By maximality of \(\rho\), they span the range of \(A(u)\)
throughout that neighborhood. Symmetry of \(A(u)\) then gives
\[
R(u)=I-W(u)\bigl(W(u)^\top W(u)\bigr)^{-1}W(u)^\top .
\]
This is a smooth matrix function. If \(\rho=0\), then \(R(u)=I\) on
\(\Omega\), so it is smooth in this case as well.
Keep \(v\) fixed. At \(u=v\), Equation~\eqref{eq:norm-identity} gives
\[
P(v)B(y)R(v)=0\qquad(y\in\R^3),
\]
because \(b(0,y)=0\). For \(h=(h_0,h')\in v^\perp\), define
\[
Jh=v_0h'-h_0v',
\qquad
u_s=\frac{v+sh}{\sqrt{1+s^2\norm{h}^2}}.
\]
For small \(s\), this path stays in the neighborhood where \(R\) is
smooth, and its velocity at zero is \(h\). Moreover,
\[
v_0u_s'-(u_s)_0v'
=\frac{s\,Jh}{\sqrt{1+s^2\norm{h}^2}}.
\]
Thus, for \(s\ne0\), homogeneity of \(b\) and
\eqref{eq:norm-identity} give
\[
\frac{b(Jh,y)}{1+s^2\norm{h}^2}
=
\norm{P(v)B(y)\frac{R(u_s)-R(v)}{s}}_{\op}^{2}.
\]
The matrix difference quotient converges to the differential
\(dR_v(h)\). Passing to the limit by continuity of the operator norm
yields
\begin{equation}\label{eq:tangent-identity}
b(Jh,y)=\norm{P(v)B(y)dR_v(h)}_{\op}^{2}
\qquad(h\in v^\perp,\ y\in\R^3).
\end{equation}
Only \(R\) has been differentiated; the projection \(P(v)\) remains
fixed. Figure~\ref{fig:local-tangent} illustrates this local variation.
\input{figures/tangent-variation}
The linear map \(J:v^\perp\to\R^3\) is an isomorphism. Indeed, if
\(Jh=0\), then \(v_0\ne0\) gives
\(h=(h_0/v_0)v\); since \(h\perp v\), this forces \(h=0\).
Both spaces have dimension three.
Consequently the matrix
\[
F(z,y)=P(v)B(y)dR_v(J^{-1}z),\qquad z,y\in\R^3,
\]
has entries that are bilinear in \(z,y\), since \(B\) and the differential
\(dR_v\) are linear. In fixed orthonormal bases,
each entry \(\ell_{ij}(z,y)\) satisfies
\[
\ell_{ij}(z,y)^2
\le \norm{F(z,y)}_{\op}^{2}
=b(z,y)
\qquad(z,y\in\R^3).
\]
Lemma~\ref{lem:form} forces every entry to vanish identically.
Equation~\eqref{eq:tangent-identity} would therefore make \(b\)
identically zero, contradicting \(b(e_1,e_1)=1\).
This excludes the hypothetical finite pencil and completes the proof.
\end{proof}
==== END A-Nonspectrahedral-Hyperbolicity-Cone//06-tangent.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//07-comparison.tex ====
\section{A path-containment assertion}
\label{app:opposition}
Theorem~53 (pp.~16--17) of version~1 of Gonz\'alez Nevado's
\cite{gonzalez-nevado-2026} states a positive solution of the geometric
Generalized Lax conjecture. Its proof invokes Theorem~52 (p.~16), which asserts
containment of a fixed rigidly convex set throughout a smooth real-zero
deformation from a product of normalized tangent linear factors to a
multiple of the defining polynomial. The following example violates
that intermediate assertion under its stated hypotheses.
A real polynomial $g$ with $g(0)=1$ is \emph{real-zero} if the roots of
$s\mapsto g(sx)$ are real for every real vector $x$. Its closed
rigidly convex set is the closure of the component of $\{g\ne0\}$
containing the origin. Fix $a,b,\eta>0$ and, in one variable, put
\[
g(x)=(1-x/a)(1+x/b),\qquad
c(t)=1+\eta\sin^2(\pi t),\qquad
H_t(x)=g(c(t)x),\quad 0\le t\le1.
\]
The zero set of $g$ is the compact smooth set $\{-b,a\}$.
The normalized tangent linear factors at these two points are
$1+x/b$ and $1-x/a$, whose product is $g$. Every $H_t$ has value $1$
at the origin and two distinct real roots,
\[
-\frac{b}{c(t)}\quad\hbox{and}\quad\frac{a}{c(t)}.
\]
Thus the path depends smoothly on $t$ and consists of real-zero
polynomials with compact smooth zero sets. It even admits the smooth
monic real symmetric determinantal representation
\[
H_t(x)=\det\begin{pmatrix}
1-c(t)x/a&0\\
0&1+c(t)x/b
\end{pmatrix}.
\]
Both endpoints equal $g$, so the final polynomial is $g$ times the
constant cofactor $1$, as allowed by the stated hypotheses.
Nevertheless the rigidly convex set of $H_t$ is
\[
\left[-\frac{b}{c(t)},\frac{a}{c(t)}\right],
\]
which is strictly smaller than $[-b,a]$ whenever $00$ describe the open family of normalized real
quadratics with negative leading coefficient, so the failure is not
confined to a symmetric or multiple-root example. The parameter $\eta$
may be arbitrarily small. The cited Theorem~52 states no restriction
to two or more variables, no lower bound on pencil size, and no
requirement that the endpoint cofactor be nonconstant.
This example refutes the all-intermediate-times conclusion of that
theorem. Its endpoints coincide, so it does not refute an endpoint-only
statement. The nonspectrahedrality result of this paper follows from
the construction and obstruction proved above.
==== END A-Nonspectrahedral-Hyperbolicity-Cone//07-comparison.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//main.tex ====
\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[margin=1.05in]{geometry}
\usepackage{amsmath,amssymb,amsthm,mathtools}
\usepackage{microtype}
\usepackage{xcolor}
\usepackage{tikz}
\usepackage[colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black]{hyperref}
\usepackage{bookmark}
\usepackage[numbers,sort&compress]{natbib}
\pdftrailerid{}
\hypersetup{pdftitle={A nonspectrahedral hyperbolicity cone},pdfauthor={OpenAI}}
\newcommand{\R}{\mathbb R}
\newcommand{\Sym}{\mathbb S}
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\newcommand{\op}{\mathrm{op}}
\DeclareMathOperator{\adj}{adj}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\proj}{proj}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\numberwithin{equation}{section}
\setlength{\emergencystretch}{1.5em}
\title{A nonspectrahedral hyperbolicity cone}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
We construct a homogeneous polynomial of degree $16$ in $23$ real variables
whose hyperbolicity cone has no representation by a finite homogeneous
real symmetric linear matrix inequality. This disproves the geometric Generalized Lax
conjecture.
\end{abstract}
\tableofcontents
\input{sections/01-introduction}
\input{sections/02-form}
\input{sections/03-hyperbolic}
\input{sections/04-pencil}
\input{sections/05-norm}
\input{sections/06-tangent}
\input{sections/reduced-polynomial}
\appendix
\input{sections/07-comparison}
\bibliographystyle{plainnat}
\bibliography{references}
\end{document}
==== END A-Nonspectrahedral-Hyperbolicity-Cone//main.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//reduced-polynomial.tex ====
\section{Removing a determinant factor}\label{sec:reduced}
The polynomial $p$ makes the positive-map construction explicit through
a $4\times4$ determinant. We now remove a factor that is unnecessary for
its hyperbolicity cone. A larger block determinant shows directly that
the quotient remains polynomial at singular $X$.
\begin{proof}[Proof of Corollary~\ref{cor:reduced}]
For the standard coordinate vectors $e_i$ of $\R^3$, put
\[
K_i=\begin{pmatrix}0&e_i^\top\\-e_i&0\end{pmatrix}\in\R^{4\times4},
\qquad C=\begin{pmatrix}K_1&K_2&K_3\end{pmatrix}\in\R^{4\times12}.
\]
Here $U\otimes V$ denotes the block matrix with blocks $U_{ij}V$.
Direct block multiplication in~\eqref{eq:phi} gives, for $T\in\Sym^4$,
\[
\Phi_y(T)=\sum_{i,j=1}^3 Q(y)_{ij}K_iTK_j^\top
=C\bigl(Q(y)\otimes T\bigr)C^\top.
\]
Define the polynomial
\begin{equation}\label{eq:reduced-polynomial}
q(X,Z,y)=\det\begin{pmatrix}
I_3\otimes X&C^\top\\
C\bigl(Q(y)\otimes I_4\bigr)&Z
\end{pmatrix}.
\end{equation}
For invertible $X$, taking a Schur complement yields
\[
q(X,Z,y)=(\det X)^3\det\bigl(Z-\Phi_y(X^{-1})\bigr).
\]
Comparison with~\eqref{eq:inverse-formula} proves the polynomial identity
\begin{equation}\label{eq:reduced-factor}
p(X,Z,y)=(\det X)q(X,Z,y)
\end{equation}
first for invertible $X$ and then everywhere by continuity.
Since $p$ and $\det X$ are homogeneous of degrees $20$ and $4$, this
identity makes $q$ homogeneous of degree $16$; also $q(e)=1$.
For each real $(X,Z,y)$, the polynomial $q(te-(X,Z,y))$ is a factor of
the real-rooted polynomial $p(te-(X,Z,y))$. Hence $q$ is hyperbolic
with respect to $e$.
Write $K_q=\Lambda_+(q,e)$. The factorization~\eqref{eq:reduced-factor}
gives
\begin{equation}\label{eq:cone-intersection}
K=\{(X,Z,y):X\succeq0\}\cap K_q,
\end{equation}
because the roots contributed by $\det(tI_4-X)$ are the eigenvalues of
$X$. It remains to show that $K_q$ already forces $X\succeq0$.
The polynomial $q$ is even in $y$, so $(X,Z,y)\in K_q$ implies
$(X,Z,-y)\in K_q$. By convexity of hyperbolicity
cones~\cite{garding-1959}, their midpoint $(X,Z,0)$ also belongs to $K_q$.
But~\eqref{eq:reduced-polynomial} gives
\[
q(X,Z,0)=(\det X)^3\det Z,
\]
whose closed hyperbolicity cone in this slice is exactly
$\{(X,Z,0):X\succeq0,\ Z\succeq0\}$.
Thus $X\succeq0$ throughout $K_q$, and~\eqref{eq:cone-intersection}
implies $K_q=K$. Theorem~\ref{thm:main} now supplies nonspectrahedrality.
\end{proof}
==== END A-Nonspectrahedral-Hyperbolicity-Cone//reduced-polynomial.tex ====
==== BEGIN A-Nonspectrahedral-Hyperbolicity-Cone//tangent-variation.tex ====
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[x=1cm,y=1cm,font=\small]
\definecolor{tangentblue}{RGB}{35,92,135}
\node[anchor=west,font=\footnotesize] at (0,3.02)
{Local sphere section};
\coordinate (O) at (1.65,1.45);
\coordinate (V) at (2.75,1.45);
\coordinate (Us) at (2.6469,1.9149); %
\draw[black!28,line width=0.6pt] (O) circle[radius=1.1];
\draw[tangentblue,line width=1.6pt]
(2.6962,1.1101) arc[start angle=-18,end angle=48,radius=1.1];
\draw[black!40,line width=0.5pt] (O) -- (V);
\draw[black!40,line width=0.5pt] (O) -- (Us);
\fill[black!55] (O) circle[radius=0.025];
\node[anchor=north,font=\scriptsize] at (1.65,1.36) {$0$};
\draw[->,line width=0.8pt] (V) -- (2.75,2.65);
\node[anchor=west,font=\footnotesize] at (2.86,2.49) {$h\perp v$};
\fill (V) circle[radius=0.035];
\node[anchor=north west] at (2.83,1.42) {$v$};
\fill[tangentblue] (Us) circle[radius=0.037];
\node[anchor=south east,text=tangentblue] at (2.54,1.96) {$u_s$};
\node[anchor=north,font=\scriptsize] at (1.65,0.22)
{$S^3\cap\operatorname{span}\{v,h\}$};
\node[anchor=west,font=\footnotesize] at (4.65,3.02)
{Only $R$ is differentiated};
\node[anchor=west] at (4.65,2.49)
{$\mathcal U\subset\Omega,\qquad \operatorname{rank}A(u)=\rho\quad(u\in\mathcal U).$};
\node[anchor=west] at (4.65,1.79)
{$\displaystyle u_s=\frac{v+sh}{\sqrt{1+s^2\|h\|^2}},
\qquad u_{s=0}=v,\quad \left.\frac{du_s}{ds}\right|_{s=0}=h.$};
\node[anchor=west] at (4.65,1.02)
{$P(v)\text{ stays fixed.}$};
\node[anchor=west,text=tangentblue] at (4.65,0.43)
{$R(u_s)=R(v)+s\,dR_v(h)+o(s)\qquad(s\to0).$};
\end{tikzpicture}
\caption{A schematic local tangent step, for a fixed nonzero $h\in v^\perp$.
The highlighted arc lies in a chosen constant-rank neighborhood $\mathcal U$;
the expansion of $R$ is used only for small $s$ with $u_s\in\mathcal U$.
The pictured section belongs to the parameter sphere, whereas $P(v)$
and $R(u_s)$ act on the separate block spaces $\mathbb R^a$ and
$\mathbb R^c$.}
\label{fig:local-tangent}
\end{figure}
==== END A-Nonspectrahedral-Hyperbolicity-Cone//tangent-variation.tex ====
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