==== TABLE OF CONTENTS ====
Hyperbolicity-Cones-Without-Semidefinite-Lifts//01-introduction.tex
Hyperbolicity-Cones-Without-Semidefinite-Lifts//02-cone.tex
Hyperbolicity-Cones-Without-Semidefinite-Lifts//03-hankel.tex
Hyperbolicity-Cones-Without-Semidefinite-Lifts//04-separation.tex
Hyperbolicity-Cones-Without-Semidefinite-Lifts//05-obstruction.tex
Hyperbolicity-Cones-Without-Semidefinite-Lifts//appendix-certificate.tex
Hyperbolicity-Cones-Without-Semidefinite-Lifts//main.tex
Hyperbolicity-Cones-Without-Semidefinite-Lifts//references.bib
===========================

==== BEGIN Hyperbolicity-Cones-Without-Semidefinite-Lifts//01-introduction.tex ====
\section{Introduction}

A real homogeneous polynomial $p$ on a finite-dimensional real vector
space is \emph{hyperbolic with respect to} $e$ if $p(e)>0$ and, for every
$x$, all roots of $t\mapsto p(te-x)$ are real. Its closed hyperbolicity
cone is
\[
 K(p,e)=\{x:\text{every root of }t\mapsto p(te-x)
                         \text{ is nonnegative}\}.
\]
G\aa rding proved that this is a convex cone \cite{Garding1959}.
The determinant on real symmetric matrices, with direction the identity,
gives the positive semidefinite cone as a basic example. Hyperbolicity
cones also support logarithmic self-concordant barriers, as shown by
G\"uler \cite{Guler1997}, and hence form a natural setting for convex
optimization beyond semidefinite programming.

A \emph{spectrahedron} is a set defined by a finite affine linear matrix
inequality. A \emph{spectrahedral shadow}, or a set with a finite
semidefinite lift, is a set $S\subset\R^d$ of the form
\begin{equation}\label{eq:lift-definition}
 S=\left\{x\in\R^d:\exists u\in\R^a,
 A_0+\sum_{i=1}^d x_iA_i+\sum_{j=1}^a u_jB_j\succeq0\right\},
\end{equation}
where $a$ and the common size of the real symmetric matrices are finite.
Equality in this definition is exact; taking the closure of the displayed
projection is not part of the definition. The \emph{generalized Lax
conjecture} asserts that every hyperbolicity cone is a spectrahedron.
The weaker \emph{Projected Lax Conjecture}, formulated by Netzer and
Sanyal \cite{NetzerSanyal2015}, asks whether every hyperbolicity cone is
a spectrahedral shadow.

\begin{theorem}\label{thm:main}
There exists a real homogeneous hyperbolic polynomial whose closed
hyperbolicity cone has no finite semidefinite lift.
\end{theorem}

Thus both conjectures have a negative resolution. The assertion excludes
every finite choice in \eqref{eq:lift-definition}, with arbitrary real
coefficients and any number of auxiliary variables. The construction is
an existence argument in high dimension, not a lower bound on the size
of a particular representation.

\subsection{Context and earlier obstructions}

The original Lax question concerned homogeneous polynomials in three
variables \cite{Lax1958}. The determinantal theorem of Helton and
Vinnikov \cite{HeltonVinnikov2007}, together with the equivalence
explained by Lewis, Parrilo, and Ramana \cite{LewisParriloRamana2005},
gave an affirmative answer: a ternary hyperbolic polynomial has a
definite symmetric determinantal representation. The distinction between
a representation of a polynomial and a representation of its cone
becomes essential in higher dimension. Br\"and\'en
\cite{Branden2011} constructed a real-zero polynomial for which no
positive integral power has a definite determinantal representation.
This does not exclude another polynomial with the same positivity
region. Indeed, Kummer \cite{Kummer2016} proved that the related
specialized V\'amos polynomial has a spectrahedral hyperbolicity cone.
Nor does failure of a specified sums-of-squares relaxation, such as
those studied by Kummer, Plaumann, and Vinzant
\cite{KummerPlaumannVinzant2015}, exclude all finite lifts.

Positive representation results cover substantial classes.
Br\"and\'en \cite{Branden2014} proved spectrahedrality for elementary
symmetric hyperbolicity cones. Building on the representability
criteria of Helton and Nie \cite{HeltonNie2010}, Netzer and Sanyal
\cite[Theorem~1.1]{NetzerSanyal2015} proved that a hyperbolicity cone
is a spectrahedral shadow when each nonzero boundary point is a smooth
point of its defining hyperbolic polynomial. Scheiderer
\cite{Scheiderer2025} subsequently obtained second-order-cone lifts
for hyperbolicity cones with Nash-smooth boundary. Recent preprints
prove spectrahedrality for
five-variable hyperbolic cubics \cite{Netzer2026} and for strictly
hyperbolic polynomials \cite{KummerNetzer2026}. These results concern
special families or impose regularity hypotheses; they do not yield
an exact finite lift for every hyperbolicity cone.

For general convex semialgebraic sets, Scheiderer
\cite{Scheiderer2018} established nonexistence of semidefinite lifts
by local positivity obstructions. His finite-dimensional
sum-of-squares pullback criterion and local arguments explain how a
putative lift constrains functions near a smooth point. He explicitly
asked whether these methods could settle the hyperbolicity-cone
question \cite[Section~5.4]{Scheiderer2018}. The argument here is
closely related to that approach, but tests analytic Gram identities
directly by a finite positive moment functional. Moment and
sum-of-squares semidefinite methods are classical
\cite{Lasserre2001}; our functional need not be integration against a
measure. The general relationship between cone lifts and positive
factorizations is developed by Gouveia, Parrilo, and Thomas
\cite{GouveiaParriloThomas2013}. We do not need to invoke their
factorization theorem.

\subsection{The construction and proof strategy}

We begin with a matrix-valued homogeneous quadratic map
$Q:\R^m\to\Sym_r$ that is positive semidefinite at every real input.
Here $\Sym_r$ denotes the real symmetric $r$-by-$r$ matrices.
Section~\ref{sec:cone} associates to $Q$ a homogeneous polynomial
$p_Q$ and proves its hyperbolicity by a symmetric determinant identity
on each line in the input space. It also identifies an exact affine
section and projection of its closed cone:
\[
 E_Q=\{(A,t,y): A\succeq0,\ \ran Q(y)\subseteq\ran A,
                    \ t\ge\tr(A^\dagger Q(y))\}.
\]
The symbol $A^\dagger$ denotes the Moore--Penrose inverse. It is enough
to choose $Q$ so that $E_Q$ has no finite lift.

The difficulty is to force such an obstruction while keeping $Q(y)$
positive semidefinite for \emph{every} $y$. Sections~\ref{sec:hankel}
and~\ref{sec:separation} address this constraint. Let $H$ be the
36-dimensional space of quadratic forms in eight variables. Each
linear functional $y$ on quartic forms determines a Hankel matrix
$Y(y)$ on $H$ by $Y(y)(f,g)=y(fg)$. We construct a rank-twenty
positive semidefinite Hankel matrix whose kernel has strong
multiplication properties. After normalizing translations and dilations,
its geometry yields a twenty-dimensional subspace $W\subset H$ and a
finite set of projective classes of nonzero symmetric forms on $W$.
The class of every nonzero limit of inverses of compressions to $W$
arising in the separation argument belongs to this set. The seed matrix
is established by the finite-field certificate in
Appendix~\ref{app:certificate}; the passage from its rank computations
to a real positive semidefinite matrix is part of the proof.

For a sufficiently large integer $N$, finiteness lets us choose a tuple
$w=(w_1,\ldots,w_N)\in W^N$ with $\sum_i w_i^TCw_i\ne0$ for every
representative $C$ of one of these classes. We then choose
$U\subset w^\perp\subset H^N$ so that convergence to $w$ of vectors
$(I_N\otimes Y(y))u$, $u\in U$, would force one of these pairings
to vanish. Thus $w$ lies outside the closure of those Hankel images.
A symmetric form $g$ can be chosen nonnegative on every such image
and negative at $w$. Using a basis matrix also denoted by $U$, we set
\[
 G(y)=I_N\otimes Y(y),\qquad Q(y)=U^\top G(y)gG(y)U.
\]
Nonnegativity on the Hankel images proves that $Q$ is admissible.
The construction uses $m=330$ and $r=20N-1$, where one may take
$N>10\cdot36^2$.

Section~\ref{sec:obstruction} turns the negative direction of $g$
into a local obstruction. Evaluation functionals at points $(1,s)$,
$s\in\R^7$, give a degree-four polynomial patch $x(s)$ in $E_Q$.
A positive definite moment functional $\Lambda$, applied to
$x(c+\eps s)$, produces points violating the trace inequality of
$E_Q$ by a negative term of order $\eps^4$. At the same time their
$A$ coordinates are bounded below by a positive multiple of
$\eps^4 I$.

If a finite lift existed, semialgebraic selection would provide analytic
lift coordinates and analytic Gram factors on an open part of this
patch. Applying $\Lambda$ to their Taylor polynomials almost preserves
the lift inequality. A mixing argument on the common support of the
feasible pencil values repairs the small error without a strict
feasibility assumption. It yields feasible points within
$O(\eps^{16})$ of the tested patch, too close to remove its order-four
violation. The finite positive extension and finite-jet approximation
lemmas isolate reusable parts of this obstruction. Together with the
finite inverse-limit construction, they supply the proof of
Theorem~\ref{thm:main}.
==== END Hyperbolicity-Cones-Without-Semidefinite-Lifts//01-introduction.tex ====

==== BEGIN Hyperbolicity-Cones-Without-Semidefinite-Lifts//02-cone.tex ====
\section{A determinant construction and its affine section}\label{sec:cone}
The first step is to place the obstruction in a tractable affine
section. We prove hyperbolicity for every admissible quadratic matrix
and identify the section exactly, including its singular boundary.

Let $Q:\R^m\to\Sym_r$ have homogeneous quadratic entries and satisfy
$Q(y)\succeq0$ for every $y$. Put $n=r+1$. For
$W=\left(\begin{smallmatrix}a&v^\top\\v&T\end{smallmatrix}\right)
\in\Sym_n$, define
\begin{equation}\label{eq:Phi}
 \Phi_y(W)=
 \begin{pmatrix}
  \tr(Q(y)T)&-v^\top Q(y)\\
  -Q(y)v&aQ(y)
 \end{pmatrix}.
\end{equation}
The polynomial used throughout the paper is
\begin{equation}\label{eq:pQ}
 p_Q(X,Z,y)=
 \det\bigl((\det X)Z-\Phi_y(\operatorname{adj}X)\bigr),
 \qquad X,Z\in\Sym_n,\quad y\in\R^m.
\end{equation}
The adjugate in this formula is the classical adjugate, so the formula
is polynomial even when $X$ is singular.

\begin{proposition}\label{prop:hyperbolic}
The polynomial $p_Q$ is homogeneous of degree $n(n+1)$ and hyperbolic
with respect to $e=(I_n,I_n,0)$, with $p_Q(e)=1$.
Fix $y$ and choose vectors $b_1,\ldots,b_n\in\R^r$ with
$Q(y)=\sum_{j=1}^n b_jb_j^\top$. Define
\[
 B_j=\begin{pmatrix}0&b_j^\top\\-b_j&0\end{pmatrix},
 \qquad
 \mathcal B=\begin{pmatrix}B_1\\\vdots\\B_n\end{pmatrix}.
\]
Then the closed hyperbolicity cone $K_Q$ of $p_Q$ is characterized,
at this value of $y$, by
\begin{equation}\label{eq:block-cone}
 (X,Z,y)\in K_Q
 \quad\Longleftrightarrow\quad
 \begin{pmatrix}I_n\otimes X&\mathcal B\\
                 \mathcal B^\top&Z\end{pmatrix}\succeq0.
\end{equation}
\end{proposition}

\begin{proof}
The entries of the matrix in \eqref{eq:pQ} have degree $n+1$, and
$p_Q(e)=1$, which proves the degree assertion. A Gram factorization
as in the statement exists since $n>r$; zero vectors may be added.
Direct multiplication gives
\[
 \Phi_y(W)=\sum_{j=1}^n B_j^\top W B_j.
\]
For every real $\eta$ we have the polynomial identity
\begin{equation}\label{eq:line-determinant}
 p_Q(X,Z,\eta y)=
 \det\begin{pmatrix}I_n\otimes X&\eta\mathcal B\\
                     \eta\mathcal B^\top&Z\end{pmatrix}.
\end{equation}
Indeed, when $X$ is invertible the right side is
\[
 (\det X)^n
 \det\bigl(Z-\eta^2\Phi_y(X^{-1})\bigr),
\]
which equals the left side by homogeneity of $Q$ and the adjugate
identity. Equality everywhere follows because both sides are
polynomials in $X,Z,\eta$.

Write $D_y(X,Z)$ for the symmetric matrix on the right of
\eqref{eq:block-cone}. Substituting $tI_n-X,tI_n-Z,-y$ in
\eqref{eq:line-determinant} gives
\[
 p_Q(te-(X,Z,y))=\det\bigl(tI_{n(n+1)}-D_y(X,Z)\bigr).
\]
Its roots are real, being the eigenvalues of a real symmetric matrix.
They are all nonnegative exactly when $D_y(X,Z)\succeq0$, proving both
hyperbolicity and \eqref{eq:block-cone}.
\end{proof}

The Gram factorization in Proposition~\ref{prop:hyperbolic} is chosen
separately for each $y$. In particular, \eqref{eq:block-cone} is not
asserted to be a linear matrix inequality jointly in $X,Z,y$.
The closed cone $K_Q$ is convex by G\aa rding's theorem
\cite{Garding1959}.

For a positive semidefinite matrix $A$, let $A^\dagger$ denote its
Moore--Penrose inverse. Define
\begin{equation}\label{eq:EQ}
 E_Q=\left\{(A,t,y):
 \begin{array}{l}
 A\in\Sym_r,\ A\succeq0,\quad \ran Q(y)\subseteq\ran A,\\[2pt]
 t\geq\tr\bigl(A^\dagger Q(y)\bigr)
 \end{array}\right\}.
\end{equation}

\begin{proposition}\label{prop:epigraph}
The set $E_Q$ is the image of the affine section
$K_Q\cap\{X=\diag(1,A):A\in\Sym_r\}$ under the projection
$(X,Z,y)\mapsto(A,Z_{00},y)$. Consequently, if $K_Q$ is a
spectrahedral shadow, then so is $E_Q$.
\end{proposition}

\begin{proof}
For a symmetric block matrix
$\left(\begin{smallmatrix}H&C\\C^\top&Z\end{smallmatrix}\right)$,
positive semidefiniteness is equivalent to
\[
 H\succeq0,\qquad \ran C\subseteq\ran H,\qquad
 Z-C^\top H^\dagger C\succeq0.
\]
To see the boundary assertion directly, positivity forces every
vector in $\ker H$ to annihilate $C$, by testing the quadratic form
on $(u,\lambda v)$. The asserted range condition then follows, and
a triangular congruence reduces the matrix to the direct sum of
$H$ and $Z-C^\top H^\dagger C$.

Apply this criterion to \eqref{eq:block-cone} with $X=\diag(1,A)$.
Its range condition says $b_j\in\ran A$ for every $j$, equivalently
$\ran Q(y)\subseteq\ran A$. Moreover,
\[
 \mathcal B^\top(I_n\otimes X)^\dagger\mathcal B
 =\Phi_y(\diag(1,A^\dagger))
 =\diag\bigl(\tr(A^\dagger Q(y)),Q(y)\bigr).
\]
It follows that every point of the affine section projects into
$E_Q$. Conversely, given $(A,t,y)\in E_Q$, take
$Z=\diag(t,Q(y))$ to obtain a point of that section. This proves
the asserted equality with no closure operation. Affine sections
and coordinate projections preserve the existence of an affine
semidefinite lift, proving the final assertion.
\end{proof}

We will construct $Q$ for which $E_Q$ has no such lift. The rest of
the argument therefore concerns \eqref{eq:EQ}, not the much larger
determinant in \eqref{eq:pQ}.
==== END Hyperbolicity-Cones-Without-Semidefinite-Lifts//02-cone.tex ====

==== BEGIN Hyperbolicity-Cones-Without-Semidefinite-Lifts//03-hankel.tex ====
\section{Hankel matrices and finitely many inverse limits}
\label{sec:hankel}

We construct a positive semidefinite Hankel matrix of rank $20$ and a
$20$-dimensional subspace $W$ with a finiteness property: certain nonzero
limits of inverses of compressions to $W$ have only finitely many
projective directions. The construction has two parts. A rank calculation,
proved in Appendix~\ref{app:certificate}, supplies a point at which a
kernel-projection map has sufficiently large derivative. We then divide
out translations and dilations, choose a finite fiber of the resulting
map, and use that fiber to classify the inverse limits.

\subsection{The rank stratum and the seed matrix}

Let $x=(x_0,x_1,\ldots,x_7)$ and put
\[
 H=\R[x]_2=L\oplus P,
 \qquad
 L=\langle x_0^2,x_0x_1,\ldots,x_0x_7\rangle,
 \qquad
 P=\R[x_1,\ldots,x_7]_2.
\]
Subscripts denote homogeneous degree. Monomials are our coordinate bases,
and these bases are orthonormal whenever we take an orthogonal complement
or compression. Setting $x_0=1$ identifies $H$ with polynomials of degree
at most two in $s=(s_1,\ldots,s_7)$. In this identification, $L$ consists
of the constant and linear terms and $P$ consists of the quadratic terms.
Write
\begin{equation}
\label{eq:hankel-dimensions}
 h=\dim H=36,\qquad d_*=20,\qquad k=h-d_*=16,
 \qquad q=d_*-\dim L=12.
\end{equation}

Let $\mathcal H\subset\Sym(H)$ be the space of Hankel matrices
\begin{equation}
\label{eq:hankel-functional}
 Y(p,p')=y(pp'),\qquad p,p'\in H,
 \qquad y\in\R[x]_4^*.
\end{equation}
Multiplication $\Sym^2H\to\R[x]_4$ is onto, so $y\mapsto Y$ is
injective and $\dim\mathcal H=\binom{11}{4}=330$.
We also call $Y$ a moment matrix. This terminology refers only to
\eqref{eq:hankel-functional}; no representing measure is assumed.
In the chart $x_0=1$, the functional $y$ acts on polynomials in $s$ of
degree at most four.

For subspaces of polynomial spaces, a product such as $JH_0$ denotes
the linear span of all products of one element of each space. We use
the following algebraic input.

\begin{lemma}[Seed matrix]
\label{lem:seed}
There is a positive semidefinite matrix $T\in\mathcal H$ of rank $d_*=20$
with the following properties. Its restriction to $L$ is positive
definite. If
\[
 J=\ker T,\qquad K=\operatorname{proj}_P J,
 \qquad H_0=\{p\in H:T(p,L)=0\},
\]
then $\dim K=k=16$, multiplication
\[
 \Sym^2K\longrightarrow\R[x_1,\ldots,x_7]_4
\]
is injective, and
\begin{equation}
\label{eq:pure-cubic-spanning}
 K\langle x_1,\ldots,x_7\rangle
   =\R[x_1,\ldots,x_7]_3.
\end{equation}
Finally, $JH_0$ has codimension $9$ in $\R[x]_4$.
\end{lemma}

The proof in Appendix~\ref{app:certificate} begins with two cyclic
orbits of real points and deforms their moment matrix inside the
rank-$20$ stratum. All rank conditions needed for that deformation are
verified there by explicit minors.

Let $\mathcal X$ be the set of projective classes $[Y]\in\mathbf P(\mathcal H)$
such that $\rank Y=d_*$, $\ker Y\cap L=0$, and the projected kernel
$K_Y=\operatorname{proj}_P(\ker Y)$ satisfies the two multiplication
conditions of Lemma~\ref{lem:seed}. Positivity is not part of the
definition of $\mathcal X$. Define
\[
 \pi:\mathcal X\longrightarrow\Gr(k,P),
 \qquad \pi([Y])=K_Y.
\]

\begin{lemma}
\label{lem:hankel-stratum}
The set $\mathcal X$ is a smooth semialgebraic manifold of dimension
\[
 D=330-1-\frac{k(k+1)}2=193.
\]
Before projectivization, the tangent space at a representative $Y$ is
the space of $Y'\in\mathcal H$ whose restriction to
$\ker Y\times\ker Y$ is zero.
\end{lemma}

\begin{proof}
Put $J_Y=\ker Y$. Taking highest homogeneous parts in the affine chart
identifies $J_Y$ with $K_Y$. Injectivity of multiplication on
$\Sym^2K_Y$ therefore implies injectivity on $\Sym^2J_Y$: a relation
among products in $J_Y$ would have a relation among their highest parts.
By duality, restriction of Hankel forms gives a surjection
\[
 \mathcal H\longrightarrow\Sym(J_Y).
\]
Choose a complement of $J_Y$ in $H$. The compression of $Y$ to this
complement is invertible: a vector in the complement orthogonal for
$Y$ to that complement is orthogonal for $Y$ to all of $H$, and hence
lies in $J_Y$. In a neighborhood of $Y$, rank $d_*$ is thus equivalent
to the vanishing of a $k$-by-$k$ symmetric Schur complement. At $Y$,
the derivative of this Schur complement is restriction to $J_Y$.
The implicit function theorem gives a smooth manifold of codimension
$k(k+1)/2=136$ in $\mathcal H$, with the asserted tangent space.
The condition $\ker Y\cap L=0$ and the multiplication conditions on
$K_Y$ are open rank conditions. Finally, projectivization removes the
nonzero scalar parameter, giving dimension $193$.
\end{proof}

\subsection{Removing translations and dilations}

The fibers of $\pi$ contain the orbits of affine translations and
positive dilations in the $s$ variables. We give explicit coordinates
for removing these eight parameters. Because $J_Y\cap L=0$, its
elements, in the affine chart, have a unique description
\begin{equation}
\label{eq:kernel-graph}
 J_Y=\{p(s)+b(p)(s)+a(p):p\in K_Y\},
\end{equation}
where $b:K_Y\to\R[s]_1$ takes values in homogeneous linear forms and
$a:K_Y\to\R$ is linear. These coefficient maps depend continuously
and semialgebraically on $[Y]$.

Pushforward of moments by $s\mapsto s+t$ transforms a kernel
polynomial by $f(s)\mapsto f(s-t)$. Thus translation changes the
coefficient maps by
\begin{equation}
\label{eq:translation-kernel-coefficients}
 b_t(p)=b(p)-\partial_t p,
 \qquad
 a_t(p)=a(p)-b(p)(t)+p(t),
\end{equation}
where $\partial_t p$ is the directional derivative of $p$ in direction
$t$. Pushforward by $s\mapsto\lambda s$, with $\lambda>0$, gives
\begin{equation}
\label{eq:dilation-kernel-coefficients}
 b\longmapsto\lambda b,
 \qquad a\longmapsto\lambda^2 a.
\end{equation}
Both actions are congruences on moment matrices. They preserve Hankel
structure, rank, and $\pi$, and preserve positive semidefiniteness when
it is present.

\begin{lemma}
\label{lem:kernel-affine-action}
For every $[Y]\in\mathcal X$, the map
\[
 \Delta_{K_Y}:\R^7\longrightarrow
   \operatorname{Hom}(K_Y,\R[s]_1),
 \qquad t\longmapsto(p\mapsto\partial_t p)
\]
is injective, and the coefficient maps $a,b$ in
\eqref{eq:kernel-graph} are not both zero. The seven infinitesimal
translations and the infinitesimal dilation give eight independent
variations of the kernel graph. These assertions, including the
independence of the variations, also hold over $\mathbf C$ under the
same rank and multiplication hypotheses.
\end{lemma}

\begin{proof}
If $t\ne0$ belongs to the kernel of $\Delta_{K_Y}$, a linear change of
the pure variables makes $t$ the first coordinate direction. All
quadratics in $K_Y$ are then independent of that coordinate, so their
products with linear forms cannot span its cube. This contradicts
\eqref{eq:pure-cubic-spanning}.

Suppose next that $a=b=0$, so $J_Y=K_Y$ in the chart. The kernel
condition and \eqref{eq:pure-cubic-spanning} force all moments of
degree three to vanish. Multiplying that spanning identity by linear
forms shows that $K_YP=\R[s]_4$, so all moments of degree four vanish
as well. With coordinates ordered by degrees $0,1,2$, the moment
matrix consequently has the form
\[
 \begin{pmatrix}
  *&*&*\\ *&*&0\\ *&0&0
 \end{pmatrix},
\]
where the degree-zero block has size one and the degree-one block
has size seven. The last block row has rank at most one, and the
first eight rows have rank at most eight. Thus $\rank Y\le9$, a
contradiction.

For independence of the infinitesimal variations, suppose a translation
in direction $t$ and a dilation with infinitesimal coefficient $\alpha$
have zero combined variation of the graph. Equations
\eqref{eq:translation-kernel-coefficients} and
\eqref{eq:dilation-kernel-coefficients} give
\[
 \alpha b(p)=\partial_t p,
 \qquad 2\alpha a(p)=b(p)(t)
 \quad(p\in K_Y).
\]
If $\alpha=0$, injectivity of $\Delta_{K_Y}$ gives $t=0$.
If $\alpha\ne0$, put $c=t/\alpha$. Then
$b(p)=\partial_c p$ and $a(p)=p(c)$, so every polynomial in the graph
is $p(s+c)$. A translation would therefore produce the already
excluded case $a=b=0$. This proves independence. Each argument is
algebraic and applies equally over $\mathbf C$.
\end{proof}

Equip the spaces of coefficient maps with the Euclidean inner
products induced by the monomial bases. We normalize $[Y]$ in two
steps. First choose the unique translation for which
$b\perp\ran\Delta_{K_Y}$. Explicitly, its parameter is
\[
 t=(\Delta_{K_Y}^*\Delta_{K_Y})^{-1}\Delta_{K_Y}^*b.
\]
Next choose the unique positive dilation for which
\begin{equation}
\label{eq:kernel-normalization}
 \|b\|^2+\|a\|^2=1.
\end{equation}
For the translated coefficients, this amounts to solving
$\lambda^2\|b\|^2+\lambda^4\|a\|^2=1$. The left side is strictly
increasing from zero to infinity, by Lemma~\ref{lem:kernel-affine-action}.

\begin{lemma}
\label{lem:hankel-normalization}
The normalized matrices, together with their data $(K_Y,b,a)$, form
a semialgebraic set $\mathcal M$ of dimension $185$. Normalization is
continuous, and each $[Y]\in\mathcal X$ is uniquely recovered from
its normalized data and its seven translation parameters and positive
dilation parameter. Moreover, $\mathcal M$ has a compact ambient
semialgebraic space in which the projection to $\Gr(k,P)$ is
continuous.
\end{lemma}

\begin{proof}
The translation formula is continuous because $\Delta_{K_Y}$ has
constant full column rank. The unique positive solution for $\lambda$
is continuous and semialgebraic as well. Dilation preserves the
orthogonality condition on $b$, so the two normalizations are compatible.
Conversely, start with normalized data, undo its positive dilation,
and then undo its translation. Applying the two normalization steps
recovers the prescribed parameters. This gives a semialgebraic
homeomorphism between $\mathcal X$ and
$\mathcal M\times\R^7\times\R_{>0}$, and hence
$\dim\mathcal M=193-8=185$.

To specify a compact ambient space, retain the projective matrix
$[Y]$ and the subspace $K_Y$, and extend $a,b$ by zero on $K_Y^\perp$
in $P$. The extended maps have the same norms as the original maps.
Condition \eqref{eq:kernel-normalization} places their pair on a
fixed unit sphere. The product of this sphere with
$\mathbf P(\mathcal H)$ and $\Gr(k,P)$ is compact, and projection
to the last factor is continuous. We take the closure of
$\mathcal M$ in this product.
\end{proof}

\subsection{Selecting a finite fiber}

We now show that the positive semidefinite part of $\mathcal X$ has
enough distinct projected kernels to avoid both infinite normalized
fibers and limits at the boundary of $\mathcal M$. We use the standard
dimension and fiber-dimension theorems for semialgebraic sets, as well
as the fact that the frontier $\overline S\setminus S$ of a
semialgebraic set has smaller dimension than $S$; see
\cite{BCR1998}. A zero-dimensional semialgebraic set is finite.

At any $T$ representing a point of $\mathcal X$, put $J=\ker T$ and
$H_0=\{p:T(p,L)=0\}$. The kernel of the derivative of the
kernel-projection map, before projectivizing, is
\begin{equation}
\label{eq:projection-tangent}
 \{Y'\in\mathcal H:Y'(J)\subset T(L)\}
       =(JH_0)^\perp.
\end{equation}
Here the right side is an annihilator in $\R[x]_4^*$, identified
with $\mathcal H$. To verify the identity, continue a vector
$j\in J$ along a differentiable curve of kernels. Its first variation
$j'$ must satisfy
\[
 Y'j+Tj'=0.
\]
The projected kernel has zero first variation precisely when the
pure part of each $j'$ lies in $K$. Subtracting an element of $J$
then makes $j'$ belong to $L$, without changing this equation.
Thus the condition is $Y'J\subset T(L)$. Conversely, that condition
allows all the corrections to be chosen in $L$; it also implies
$Y'|_{J\times J}=0$, so Lemma~\ref{lem:hankel-stratum} gives an
actual rank-stratum tangent. Finally,
$H_0=T(L)^\perp$, which proves the annihilator description.

At the seed matrix of Lemma~\ref{lem:seed}, the space in
\eqref{eq:projection-tangent} has dimension $9$. The rank stratum
before projectivizing has dimension $194$, so the derivative of
$\pi$ has rank $194-9=185$. Projectivization removes one kernel
direction, namely scaling, and leaves this derivative rank unchanged.
Near the seed matrix, rank remains $20$ and its nonzero eigenvalues
remain positive. Its $L$ block remains positive definite as well.
Consequently the $\pi$-image of matrices with these positivity
properties has dimension at least $185$. It has dimension at most
$185$ because $\pi$ factors through $\mathcal M$.

The seed's final condition can also be retained in this neighborhood.
Indeed, an invertible $321$-row product minor for $JH_0$ remains
invertible nearby. On the other hand, scaling, translations, and
dilation always give nine independent tangent directions in
\eqref{eq:projection-tangent}: the eight affine directions have
independent kernel variations by Lemma~\ref{lem:kernel-affine-action},
whereas scaling has zero kernel variation. Thus $\dim(JH_0)\le321$
throughout this neighborhood.

\begin{proposition}
\label{prop:finite-normalized-fiber}
The matrix $T$ in Lemma~\ref{lem:seed} may be chosen so that, for
$K=\operatorname{proj}_P\ker T$,
\begin{enumerate}
 \item the fiber $\mathcal M_K$ of $\mathcal M\to\Gr(k,P)$ is finite;
 \item no point of $\overline{\mathcal M}\setminus\mathcal M$ projects
 to $K$.
\end{enumerate}
\end{proposition}

\begin{proof}
The set of subspaces over which the fiber of $\mathcal M$ has positive
dimension has dimension at most $184$, by the fiber-dimension theorem
and $\dim\mathcal M=185$. The frontier of $\mathcal M$ has dimension
at most $184$, so its projection to $\Gr(k,P)$ has dimension at most
$184$ as well. The positive semidefinite image just constructed has
dimension $185$. We can therefore choose $K$ in that image outside
both exceptional sets, and choose a corresponding positive
semidefinite $T$ in the neighborhood under consideration. Its fiber
is nonempty and zero-dimensional, hence finite.
\end{proof}

Fix this $T$ and $K$ for the rest of the paper, and set
\begin{equation}
\label{eq:fixed-complements}
 R=K^\perp\subset P,\qquad W=L\oplus R.
\end{equation}
Thus $H=W\oplus K$ is an orthogonal decomposition, $\dim R=q=12$,
and $\dim W=d_*=20$.

Each normalized class $[\bar Y]\in\mathcal M_K$ determines a
nonzero symmetric matrix on $W$, up to scale. Let $\bar A$ be the
compression of a representative $\bar Y$ to $W$. The kernel of $\bar Y$
is a graph over $K$ with lower terms in $L$, so $W$ is a complement of
that kernel. The same nondegeneracy argument used in
Lemma~\ref{lem:hankel-stratum} shows that $\bar A$ is invertible,
even if $\bar Y$ is indefinite. Define $C_{\bar Y}\in\Sym(W)$ to
have block form
\begin{equation}
\label{eq:finite-inverse-directions}
 C_{\bar Y}=
 \begin{pmatrix}0&0\\0&(\bar A^{-1})_{R,R}\end{pmatrix}
 \quad\hbox{on }W=L\oplus R,
 \qquad
 \mathcal C=\{[C_{\bar Y}]:[\bar Y]\in\mathcal M_K\}.
\end{equation}
If its $R,R$ block were zero, the invertible operator $\bar A^{-1}$
would map $R$ into $L$, which is impossible because $12>8$.
Thus $\mathcal C$ is a finite nonempty set of projective classes of
nonzero symmetric matrices on $W$.

\subsection{The inverse-limit property}

To connect the normalized fiber to arbitrary limiting sequences, we
first need a rank correction that remains small even when some
matrix entries diverge. All norms below are fixed Euclidean operator
norms; convergence is equivalent in any choice of such norms.

\begin{lemma}[Uniform rank correction]
\label{lem:hankel-rank-correction}
Use the fixed decomposition $H=W\oplus K$ from
\eqref{eq:fixed-complements}. Suppose $Y_i\in\mathcal H$ have blocks
\[
 Y_i=\begin{pmatrix}A_i&B_i\\B_i^T&D_i\end{pmatrix},
\]
where $A_i$ is invertible, $\sup_i\|A_i^{-1}\|<\infty$, and
\[
 A_i^{-1}B_i\longrightarrow0,
 \qquad D_i-B_i^TA_i^{-1}B_i\longrightarrow0.
\]
There are Hankel perturbations $\Delta_i\to0$ such that
$Y_i^\sharp=Y_i+\Delta_i$ has rank $d_*$. Its $W$ compression
$A_i^\sharp$ satisfies
\[
 (A_i^\sharp)^{-1}-A_i^{-1}\longrightarrow0,
\]
and $\ker Y_i^\sharp\to K$, or equivalently
$\ran Y_i^\sharp\to W$.
\end{lemma}

\begin{proof}
On matrices whose $W$ block is invertible, write
$\mathcal S(Y)=D-B^TA^{-1}B$ for the Schur complement on $K$.
For a variation $\Delta$ with blocks $(a,b,d)$ and $F=A^{-1}B$,
\begin{equation}
\label{eq:schur-differential}
 D\mathcal S_Y(\Delta)
   =d-b^TF-F^Tb+F^TaF.
\end{equation}
At $Y_i$, these derivatives tend to the restriction map
$\Delta\mapsto\Delta_{K,K}$ on Hankel space. That restriction map
is onto $\Sym(K)$ because multiplication on $\Sym^2K$ is injective.
Choose a fixed linear right inverse
$\mathcal E:\Sym(K)\to\mathcal H$.

We record why an implicit-function argument is uniform in $i$.
Put $M=\sup_i\|A_i^{-1}\|$. On a fixed sufficiently small ball of
perturbations, the inverse of the perturbed $W$ block obeys
\[
 (A_i+a)^{-1}=(I+A_i^{-1}a)^{-1}A_i^{-1},
 \qquad \|(A_i+a)^{-1}\|\le2M.
\]
Moreover,
\[
 (A_i+a)^{-1}(B_i+b)
   =(I+A_i^{-1}a)^{-1}(A_i^{-1}B_i+A_i^{-1}b)
\]
is uniformly bounded there. Differentiating the inverse and this
last expression gives
\[
 \delta(A^{-1})=-A^{-1}(\delta A)A^{-1},
 \qquad
 \delta F=A^{-1}(\delta B-(\delta A)F).
\]
Together with \eqref{eq:schur-differential}, these identities give a
uniform bound for the second derivatives of $\mathcal S$ on that
ball. This bound uses neither boundedness of $A_i$ nor boundedness
of $B_i$.

Define $f_i(z)=\mathcal S(Y_i+\mathcal E z)$ for
$z\in\Sym(K)$ in a fixed small ball. We have $f_i(0)\to0$ and
$Df_i(0)\to I$. Shrinking the fixed ball if needed, the uniform
second-derivative bound gives
$\|Df_i(z)-I\|\le1/2$ there for all sufficiently large $i$.
The map $z\mapsto z-f_i(z)$ is therefore a contraction on the
closed ball of radius $2\|f_i(0)\|$: its value at zero has norm
$\|f_i(0)\|$, and its Lipschitz constant is at most $1/2$.
Its fixed point $z_i$ satisfies $f_i(z_i)=0$ and $z_i\to0$.
Set $\Delta_i=\mathcal E z_i$.

The $W$ block remains invertible, and its Schur complement is zero,
so $\rank Y_i^\sharp=d_*$. The inverse identity gives
$(A_i^\sharp)^{-1}-A_i^{-1}\to0$. The displayed formula for the
perturbed cross block also gives
$(A_i^\sharp)^{-1}B_i^\sharp\to0$. The kernel of $Y_i^\sharp$ is
the graph with columns
$\binom{-(A_i^\sharp)^{-1}B_i^\sharp}{I_K}$, proving the final
assertion. Discarding finitely many indices has no effect on any
of the conclusions.
\end{proof}

\begin{lemma}[Finite inverse limits]
\label{lem:inverse-limits}
Let $T,K,R,W$, and $\mathcal C$ be chosen as above. Suppose a
sequence $Y_i\in\mathcal H$ has a group of $d_*$ eigenvalues bounded
away from zero in absolute value, and the span of the corresponding
eigenvectors tends to $W$. Suppose all remaining eigenvalues tend
to zero. If the inverses of the compressions $A_i=(Y_i)_{W,W}$
converge to a nonzero matrix $C$, then
\[
 [C]\in\mathcal C.
\]
\end{lemma}

\begin{proof}
We first put the sequence in the situation of
Lemma~\ref{lem:hankel-rank-correction}. Split off the distinguished
eigenvalues to write $Y_i=Y_i^{\mathrm{large}}+E_i$, where
$\rank Y_i^{\mathrm{large}}=d_*$ and $E_i\to0$. After passage to a
subsequence, orthonormal frames for the large eigenspaces have
$W$ components $S_i$ converging to an orthogonal matrix and $K$
components tending to zero. In particular $S_i$ is invertible, and
\begin{equation}
\label{eq:spectral-graph}
 Y_i^{\mathrm{large}}=
 \begin{pmatrix}I\\G_i\end{pmatrix}
 A_i^0
 \begin{pmatrix}I&G_i^T\end{pmatrix},
 \qquad G_i\longrightarrow0,
 \qquad \sup_i\|(A_i^0)^{-1}\|<\infty.
\end{equation}
Indeed $A_i^0=S_i\operatorname{diag}(\lambda_{i,1},\ldots,
\lambda_{i,d_*})S_i^T$, and the inverse eigenvalues are bounded.

Write the blocks of $E_i$ as $(a_i,b_i,d_i)$ in the same order as
in the rank-correction lemma. Then
\[
 A_i=A_i^0+a_i,
 \qquad B_i=A_i^0G_i^T+b_i,
 \qquad D_i=G_iA_i^0G_i^T+d_i.
\]
The inverses $A_i^{-1}$ are bounded, and
\[
 A_i^{-1}B_i=G_i^T+A_i^{-1}(b_i-a_iG_i^T)\longrightarrow0.
\]
With $e_i=b_i-a_iG_i^T$, direct block elimination gives
\[
 D_i-B_i^TA_i^{-1}B_i
 =d_i-G_ib_i-b_i^TG_i^T+G_ia_iG_i^T-e_i^TA_i^{-1}e_i
 \longrightarrow0.
\]
These formulas explain explicitly why divergence of the large
eigenvalues causes no difficulty. Apply
Lemma~\ref{lem:hankel-rank-correction}. The corrected Hankel matrices
have exact rank $d_*$, their kernels tend to $K$, and their
compression inverses have the same nonzero limit $C$. We replace
$Y_i$ by these corrected matrices for the rest of the proof.

For large $i$, the class $[Y_i]$ belongs to $\mathcal X$: its kernel
is close to $K\subset H$, and the required product conditions are
open at $K$. Put $K_i=\pi([Y_i])$. Then $K_i\to K$, and the lower
coefficient maps $a_i^{\mathrm{ker}},b_i^{\mathrm{ker}}$ of the kernel
graph tend to zero.
The translations used in normalization consequently tend to zero,
by the explicit left-inverse formula for $\Delta_{K_i}$. The
translated coefficient maps still tend to zero, so the normalizing
dilation factors tend to infinity. Let their reciprocals be
$\eps_i\to0$.

The normalized data lie in the compact ambient space of
Lemma~\ref{lem:hankel-normalization}. Pass to a convergent
subsequence. Its limit projects to $K$, and
Proposition~\ref{prop:finite-normalized-fiber} excludes a limit in
the frontier of $\mathcal M$. It is therefore a member
$[\bar Y]\in\mathcal M_K$. Choose representatives $Z_i$ of the
normalized matrices converging to a representative $\bar Y$.
Up to nonzero scalar multiples, the original matrices are obtained
from $Z_i$ by dilation by $\eps_i$, followed by translations whose
parameters tend to zero.

We compute the inverse limit first before those translations.
Set $R_i=K_i^\perp\subset P$ and $W_i=L\oplus R_i$. Use orthonormal
coordinates on $W_i$ converging to the fixed coordinates on $W$,
and keep the coordinates on $L$ unchanged. The compressions
$\bar A_i=(Z_i)_{W_i,W_i}$ then converge to the invertible
compression $\bar A$ of $\bar Y$ to $W$. Dilation acts on monomial
columns by
\[
 D_{\eps}=\diag(1,\eps I_7,\eps^2 I_{28})
\]
on $H$, according to degrees $0,1,2$. Since it preserves $W_i$,
the compression of the dilated matrix is
\[
 A_i^{\mathrm{dil}}
   =D_{\eps_i,W_i}\bar A_iD_{\eps_i,W_i}.
\]
Consequently
\begin{equation}
\label{eq:dilated-inverse-limit}
 \eps_i^4(A_i^{\mathrm{dil}})^{-1}
 =\diag(\eps_i^2,\eps_i I_7,I_q)\,
   \bar A_i^{-1}\,
   \diag(\eps_i^2,\eps_i I_7,I_q)
 \longrightarrow C_{\bar Y}.
\end{equation}

It remains to justify changing $W_i$ to $W$ and restoring the
translations. After dilation, the kernel coefficients of $Z_i$
are $\eps_i b_i^{\mathrm{norm}}$ and
$\eps_i^2 a_i^{\mathrm{norm}}$. The normalized coefficients are
bounded. Write the dilated kernel in $W_i\oplus K_i$ as
$\{(F_i k,k):k\in K_i\}$; the preceding coefficient bounds give
$F_i\to0$. Its orthogonal complement, which is the range of the
dilated matrix, has graph embedding
\[
 \iota_i:W_i\longrightarrow H,\qquad
 \iota_i w=(w,-F_i^*w).
\]
Thus $\iota_i$ tends to the inclusion of $W$ after the chosen
coordinate identifications, and the full dilated matrix is
$\iota_iA_i^{\mathrm{dil}}\iota_i^T$.

Let $\mathsf B_i$ denote the monomial matrix of the remaining translation.
Its parameter tends to zero, so $\mathsf B_i\to I_H$. Compression to $W$
after this translation therefore gives
\[
 M_iA_i^{\mathrm{dil}}M_i^T,
 \qquad M_i=\operatorname{proj}_W\mathsf B_i\iota_i\longrightarrow I_W.
\]
The actual compression $A_i$ differs from this matrix by a nonzero
scalar $\alpha_i$, coming from the choice of projective
representative. Equation~\eqref{eq:dilated-inverse-limit} yields
\[
 \alpha_i\eps_i^4 A_i^{-1}
   =M_i^{-T}\bigl(\eps_i^4(A_i^{\mathrm{dil}})^{-1}\bigr)M_i^{-1}
   \longrightarrow C_{\bar Y}\ne0.
\]
Hence $[A_i^{-1}]\to[C_{\bar Y}]$. Since $A_i^{-1}\to C\ne0$
also gives $[A_i^{-1}]\to[C]$, we obtain
$[C]=[C_{\bar Y}]\in\mathcal C$.
\end{proof}

We retain the positive semidefinite matrix $T$, the subspaces
$K$, $R=K^\perp\subset P$, and $W=L\oplus R$, and the finite set
$\mathcal C$ from \eqref{eq:finite-inverse-directions}. These are the
geometric data used in the next construction.
==== END Hyperbolicity-Cones-Without-Semidefinite-Lifts//03-hankel.tex ====

==== BEGIN Hyperbolicity-Cones-Without-Semidefinite-Lifts//04-separation.tex ====
\section{Separation from Hankel images}
\label{sec:separation}

Retain the Hankel matrix $T$, the spaces $K$, $R=K^\perp\subset P$,
and $W=L\oplus R$ from the preceding section.  In particular,
$\dim W=d_*=20$ and $H=W\oplus K$ orthogonally.  Let $\mathcal C$
be the finite set of projective classes of nonzero symmetric operators
on $W$ supplied by Lemma~\ref{lem:inverse-limits}.
We will construct a vector $w$ and a subspace $U$ such that $w$ is
outside the closure of all images of $U$ obtained by applying one
Hankel matrix componentwise.  A quadratic form separating $w$ from
those images will then give the required positive semidefinite
quadratic matrix $Q$.

Fix an integer $N>10h^2$, where $h=\dim H=36$, and put
\begin{equation}
\label{eq:separation-dimensions}
 r=d_*N-1.
\end{equation}
We equip $H^N$ with the sum of the Euclidean inner products on its
factors.  For a subspace $F\subset H$, write $P_F$ for orthogonal
projection onto $F$; the notation $P_F^N$ denotes its componentwise
action on $H^N$.
The two subspaces in the third condition below will represent
directions in which eigenvalues diverge or have finite nonzero
limits.  Their different projection constraints arise by taking
limits of bounded Hankel images of unit tuples.

\begin{lemma}[Choice of the vector and subspace]
\label{lem:generic-wu}
There exist a unit vector $w=(w_1,\ldots,w_N)\in W^N$ and an
$r$-dimensional subspace $U\subset w^\perp\subset H^N$ with the
following properties.
\begin{enumerate}
\item The vectors $w_1,\ldots,w_N$ span $W$, and
\begin{equation}
\label{eq:inverse-pairings}
 \sum_{i=1}^N w_i^T Cw_i\ne0
 \qquad\text{whenever }[C]\in\mathcal C.
\end{equation}
\item The map
\begin{equation}
\label{eq:projection-U}
 P_W^N\big|_U:U\longrightarrow w^\perp\cap W^N
\end{equation}
is an isomorphism.
\item Suppose that $F_\infty,F_0\subset H$ are orthogonal subspaces,
$C_0$ is a symmetric operator on $F_0$, and
$\dim(F_\infty\oplus F_0)>d_*$.  No nonzero tuple
$u=(u_1,\ldots,u_N)\in U$ satisfies
\begin{equation}
\label{eq:forbidden-projections}
 P_{F_\infty}u_i=0,\qquad
 P_{F_0}u_i=C_0P_{F_0}w_i
 \qquad(1\le i\le N).
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
For a nonzero symmetric operator $C$ on $W$, the polynomial
$\sum_i w_i^T Cw_i$ on $W^N$ is nonzero: a vector with just one
nonzero component already detects a nonzero value of the quadratic
form of $C$.  Its nonvanishing therefore defines a dense open subset
of $W^N$.  The spanning condition is also dense and open, since
$N\ge\dim W$.  There are only finitely many classes in
$\mathcal C$, and changing a representative does not affect
nonvanishing.  We may consequently choose $w$ satisfying all these
conditions and then normalize it to unit length.

Fix this $w$ and consider the Grassmannian
\[
 \mathcal G=\Gr(r,w^\perp),\qquad
 \dim w^\perp=hN-1.
\]
Inside $w^\perp$ we have the orthogonal decomposition
\[
 w^\perp=(w^\perp\cap W^N)\oplus K^N.
\]
The $r$-planes complementary to $K^N$ form a nonempty open subset
of $\mathcal G$.  They are exactly the subspaces satisfying
\eqref{eq:projection-U}.

We show that the subspaces violating the third property have smaller
dimension than $\mathcal G$.  Fix dimensions for $F_\infty$ and
$F_0$, and set $j=\dim F_\infty+\dim F_0>d_*$.  The triples
$(F_\infty,F_0,C_0)$ form a semialgebraic parameter space of
dimension at most $4h^2$.  For example, orthogonal projections onto
the two spaces and the extension of $C_0$ by zero give matrix
coordinates with this bound.  For each fixed triple,
\eqref{eq:forbidden-projections} prescribes the component of every
$u_i$ in $F_\infty\oplus F_0$.  Its solution set in $H^N$ is
therefore an affine space of dimension $(h-j)N$.

For a fixed nonzero $u\in w^\perp$, the subspaces in
$\mathcal G$ containing $u$ form a Grassmannian of codimension
\[
 (hN-1)-r=(h-d_*)N
\]
in $\mathcal G$.  Apply the semialgebraic dimension bounds to the
incidence relation consisting of a parameter triple, a nonzero
solution $u\in w^\perp$ of
\eqref{eq:forbidden-projections}, and an $r$-plane containing $u$.
Its image in $\mathcal G$ has dimension at most
\begin{align*}
 \dim\mathcal G+4h^2+(h-j)N-(h-d_*)N
 &=\dim\mathcal G+4h^2-(j-d_*)N\\
 &<\dim\mathcal G.
\end{align*}
There are finitely many possible pairs of dimensions for
$F_\infty,F_0$.  The union of their exceptional images still has
smaller dimension and cannot contain the nonempty open set of
complements to $K^N$.  Choosing $U$ outside that union proves both
remaining properties.
\end{proof}

Fix $w$ and $U$ as in Lemma~\ref{lem:generic-wu}.  Notice that the
$P^N$ component of $w$ is nonzero: otherwise all its components would
lie in $L$ and could not span $W$.  We now use the third property of
the lemma to control every possible limit of Hankel images, including
limits produced by unbounded matrices.

\begin{lemma}[Separation from the closure]
\label{lem:separation}
For $w$ and $U$ as in Lemma~\ref{lem:generic-wu}, let
\[
 \mathcal S_U=
 \{(I_N\otimes Y)u: u\in U,\ Y\text{ a Hankel matrix on }H\}.
\]
Then $w\notin\overline{\mathcal S_U}$.
\end{lemma}

\begin{proof}
Suppose, to the contrary, that there are tuples $u^{(n)}\in U$ and
Hankel matrices $Y_n$ such that
\[
 (I_N\otimes Y_n)u^{(n)}\longrightarrow w.
\]
After omitting finitely many terms the tuples are nonzero.  Replacing
$u^{(n)}$ by $u^{(n)}/\|u^{(n)}\|$ and $Y_n$ by
$\|u^{(n)}\|Y_n$ leaves the products unchanged.  We may thus assume
that $\|u^{(n)}\|=1$.  Pass to a subsequence for which
$u^{(n)}\to u\in U$, where $\|u\|=1$.

Choose an orthonormal eigenbasis $e_{1,n},\ldots,e_{h,n}$ for each
$Y_n$, with eigenvalues $\lambda_{1,n},\ldots,\lambda_{h,n}$.
After a further subsequence, each eigenvector converges and each
eigenvalue converges in the extended real line.  The limiting
eigenvectors form an orthonormal basis of $H$.  Let $F_\infty$ be
the span of those corresponding to eigenvalues whose absolute
values tend to infinity, and let $F_0$ be the span of those
corresponding to finite nonzero limits.  All remaining eigenvalues
tend to zero.  Define $C_0$ on $F_0$ by taking the reciprocal of
each corresponding eigenvalue limit in this limiting basis.

For each component of the tuples, the spectral identity is
\begin{equation}
\label{eq:spectral-components}
 \langle e_{\ell,n},Y_nu_i^{(n)}\rangle
 =\lambda_{\ell,n}\langle e_{\ell,n},u_i^{(n)}\rangle.
\end{equation}
When $\lambda_{\ell,n}\to0$, the right side tends to zero because
$u^{(n)}$ is bounded.  It follows that each $w_i$ belongs to
$F:=F_\infty\oplus F_0$, and the spanning property of $w$ gives
$W\subset F$.  When $|\lambda_{\ell,n}|\to\infty$, divide
\eqref{eq:spectral-components} by $\lambda_{\ell,n}$ to see that
the corresponding component of $u_i$ is zero.  For finite nonzero
limits, the same identity gives
\[
 P_{F_\infty}u_i=0,\qquad
 P_{F_0}u_i=C_0P_{F_0}w_i
 \qquad(1\le i\le N).
\]
Since $u\ne0$, Lemma~\ref{lem:generic-wu} excludes $\dim F>d_*$.
Together with $W\subset F$ this yields $F=W$.  Moreover $F_0$ is
nonzero.  Otherwise $F_\infty=W$ and $P_W^Nu=0$, contradicting
the injectivity in \eqref{eq:projection-U}.

Extend $C_0$ by zero on $F_\infty$ to obtain a nonzero symmetric
operator $C$ on $W$.  We next verify that $C$ is an inverse limit
covered by Lemma~\ref{lem:inverse-limits}.  The two groups defining
$F$ consist of exactly $d_*$ eigenvalues, bounded away from zero
in absolute value, and the span of their eigenvectors converges to $W$.
All remaining eigenvalues tend to zero.  Let
\[
 A_n=P_WY_n\big|_W
\]
be the compression to $W$.  To identify its inverse, first keep only
the $d_*$ eigenvalues in the two groups.  In a fixed orthonormal
basis of $W$, their contribution to $A_n$ is
\[
 A_n^{\mathrm{large}}=S_n\Lambda_nS_n^T,
\]
where the columns of $S_n$ are the projected eigenvectors and
$\Lambda_n$ is the diagonal matrix of these eigenvalues.
The matrices $S_n$ converge to an orthogonal matrix, while
$\Lambda_n^{-1}$ converges to the diagonal matrix of the
reciprocal finite limits and zeros for the divergent eigenvalues.
Consequently
\[
 (A_n^{\mathrm{large}})^{-1}\longrightarrow C.
\]
The contribution of the remaining eigenvalues tends to zero in
norm.  Since the displayed inverses are bounded, the identity
\[
 A_n^{-1}
 =\bigl(I+(A_n^{\mathrm{large}})^{-1}
                  (A_n-A_n^{\mathrm{large}})\bigr)^{-1}
    (A_n^{\mathrm{large}})^{-1}
\]
shows that $A_n$ is invertible for all sufficiently large $n$ and
$A_n^{-1}\to C$.  Lemma~\ref{lem:inverse-limits} now gives
$[C]\in\mathcal C$.

The projection identities above say $P_Wu_i=Cw_i$.  Since $u\in U$
and $U\subset w^\perp$, we obtain
\[
 0=\langle w,u\rangle
   =\sum_{i=1}^N\langle w_i,P_Wu_i\rangle
   =\sum_{i=1}^N w_i^TCw_i.
\]
This contradicts \eqref{eq:inverse-pairings} and proves the lemma.
\end{proof}

The preceding separation has a useful quadratic consequence.  The
set $\mathcal S_U$ is closed under multiplication by any real
scalar.  Hence its closed set of directions on the unit sphere
contains neither $w$ nor $-w$.  Compactness gives a number
$\delta>0$ such that
\[
 \|P_{w^\perp}z\|\ge\delta\|z\|
 \qquad(z\in\mathcal S_U).
\]
Here the projections act on $H^N$, and $w$ has unit length.  Define
the symmetric operator
\[
 g=\delta^{-2}P_{w^\perp}-P_{\R w}.
\]
For $z\in\mathcal S_U$, its quadratic form satisfies
$z^Tgz\ge\|z\|^2-\|P_{\R w}z\|^2\ge0$, whereas $w^Tgw=-1$.
We have therefore obtained
\begin{equation}
\label{eq:g-properties}
 w^Tgw<0,\qquad
 \bigl((I_N\otimes Y)u\bigr)^T
 g\bigl((I_N\otimes Y)u\bigr)\ge0
 \quad(u\in U,\ Y\text{ Hankel}).
\end{equation}

We can now define the quadratic matrix used in the rest of the
proof.  Use the monomial coordinates to identify
$\R[x_0,\ldots,x_7]_4^*$ with $\R^m$, where $m=330$.
For such a functional $y$, let $Y(y)$ be its Hankel matrix on $H$,
so $Y(y)(p,p')=y(pp')$, and set
\[
 G(y)=I_N\otimes Y(y).
\]
Choose a basis of $U$ and also write $U$ for the $hN$-by-$r$ matrix
whose columns are these basis vectors.  Define
\begin{equation}
\label{eq:constructed-Q}
 Q(y)=U^TG(y)gG(y)U.
\end{equation}
The entries of $G(y)$ are linear in $y$, so $Q$ is a real symmetric
$r$-by-$r$ matrix with homogeneous quadratic entries.  For every
$a\in\R^r$, the vector $Ua$ belongs to the subspace $U$, and
\eqref{eq:g-properties} gives
\[
 a^TQ(y)a=(G(y)Ua)^Tg(G(y)Ua)\ge0.
\]
Thus $Q(y)\succeq0$ for every real $y$.  The data retained for the
next section are $T$, $w$, $U$, and $g$, together with this
admissible map $Q$; the strict inequality at $w$ in
\eqref{eq:g-properties} will produce the obstruction to a finite
semidefinite lift.
==== END Hyperbolicity-Cones-Without-Semidefinite-Lifts//04-separation.tex ====

==== BEGIN Hyperbolicity-Cones-Without-Semidefinite-Lifts//05-obstruction.tex ====
\section{A local obstruction to a semidefinite lift}
\label{sec:obstruction}

We prove that the quadratic matrix map constructed in
Section~\ref{sec:separation} has an epigraph set $E_Q$ with no finite
semidefinite lift.  The proof has two parts.  First, a polynomial patch in
$E_Q$ admits a positive moment functional whose application at small scales
violates the defining inequality of $E_Q$ to order four.  Second, any finite
lift would force the resulting points to lie much closer to $E_Q$ than this
violation allows.  The second part is a local sum-of-squares obstruction,
in the sense of the methods developed by Scheiderer
\cite[Section~4]{Scheiderer2018}; here we give a direct proof using finite
Taylor polynomials of Gram factors.

We retain the spaces $H=L\oplus P$, $R=K^\perp\subset P$, and
$W=L\oplus R$, the positive semidefinite Hankel matrix $T$, and the data
$N,w,U,g$ of the preceding sections.  In particular,
\[
 \dim H=36,\qquad \dim L=8,\qquad \dim R=12,\qquad
 \rank T=20,\qquad r=20N-1.
\]
Write $J=\ker T$.  We use that $J\cap L=0$ and that the projection of $J$
to $P$ is $K$.  As in Section~\ref{sec:separation}, the symbol $U$ also
denotes a $36N$-by-$r$ matrix whose columns form a basis of the subspace
$U\subset H^N$.  By Lemma~\ref{lem:generic-wu}, orthogonal projection
induces an isomorphism
\begin{equation}
 \label{eq:obstruction-U-projection}
 U\longrightarrow w^\perp\cap W^N.
\end{equation}
Moreover, $w\in W^N$ has components spanning $W$, and
$w^Tgw<0$.  The admissible quadratic matrix map is
$Q(y)=U^TG(y)gG(y)U$, as defined in~\eqref{eq:constructed-Q}.

To keep track of repeated blocks, for an operator $S$ on $H$ we write
$S_N=I_N\otimes S$, an operator on $H^N$.  The same convention applies to
maps between subspaces of $H$.  We order the coordinates of $H^N$ as
$L^N\oplus P^N$ when using two-by-two block matrices.  Subscripts $LL$,
$LP$, and $PP$ on an operator on $H^N$ refer to these blocks.

\subsection{A polynomial patch and its moment deformation}

Let $v(s)\in H$, for $s\in\R^7$, be the column of homogeneous quadratic
monomials evaluated at $(1,s)$.  Thus its entries have affine degrees zero,
one, and two.  Let $y_s$ be evaluation at $(1,s)$ on homogeneous quartics,
and define
\begin{equation}
 \label{eq:patch-matrices}
 V(s)=I_N\otimes v(s)\in\R^{36N\times N},\qquad
 M(s)=U^TV(s)\in\R^{r\times N}.
\end{equation}
The moment identity gives $G(y_s)=V(s)V(s)^T$.  In the following lemma,
$E_Q$ is the set defined in Proposition~\ref{prop:epigraph}.

\begin{lemma}[Polynomial patch]
 \label{lem:polynomial-patch}
There is an open ball $\Omega_0\subset\R^7$ about zero such that $M(s)$
has rank $N$ for every $s\in\Omega_0$.  The polynomial map
\begin{equation}
 \label{eq:polynomial-patch}
 x(s)=\bigl(M(s)M(s)^T,\ \tr(gG(y_s)),\ y_s\bigr)
\end{equation}
has degree at most four and takes $\Omega_0$ into $E_Q$.  Its values satisfy
the trace inequality defining $E_Q$ with equality.
\end{lemma}

\begin{proof}
Decompose $w=w_L+w_P$ in $L^N\oplus P^N$.  Since the components of $w$
span $W$ and $R\ne0$, we have $w_P\ne0$; also $w_P\in R^N$.  Given
$\ell\in L^N$, the vector
\[
 \ell-\frac{\langle\ell,w_L\rangle}{\|w_P\|^2}w_P
\]
belongs to $w^\perp\cap W^N$.  By
\eqref{eq:obstruction-U-projection}, it is the projection to $W^N$ of
some vector of $U$.  Thus projection from $U$ to $L^N$ is surjective.
The columns of $V(0)$ form a basis for the $N$ constant-coordinate
directions in $L^N$.  If $M(0)a=0$, then $V(0)a$ is orthogonal to $U$;
surjectivity onto $L^N$ forces $V(0)a=0$, and hence $a=0$.  Therefore
$\rank M(0)=N$, and the same holds on a sufficiently small open ball.

Put $\Theta(s)=V(s)^TgV(s)$.  The definition of $Q$ gives
\[
 Q(y_s)=M(s)\Theta(s)M(s)^T.
\]
Consequently
\[
 \ran Q(y_s)\subset\ran M(s)=\ran(M(s)M(s)^T).
\]
Full column rank of $M(s)$ gives
\[
 M(s)^T(M(s)M(s)^T)^\dagger M(s)=I_N,
\]
and hence
\[
 \tr\bigl((M(s)M(s)^T)^\dagger Q(y_s)\bigr)
 =\tr\Theta(s)=\tr(gG(y_s)).
\]
This proves membership in $E_Q$.  Finally, $M(s)$ has degree at most two,
while $G(y_s)$ and $y_s$ have degree at most four, proving the degree
bound.
\end{proof}

For a linear functional $\Lambda$ on polynomials in seven variables through
degree four, its moment matrix is the symmetric matrix indexed by monomials
of degree at most two, with $(\alpha,\beta)$ entry
$\Lambda(s^{\alpha+\beta})$.  Under the affine chart $x_0=1$, this is
precisely a Hankel matrix on $H$.  Applying $\Lambda$ to a vector or matrix
of polynomials always means applying it entrywise.

\begin{lemma}[A deformation outside $E_Q$]
 \label{lem:moment-violation}
There exist a linear functional
$\Lambda:\R[s_1,\ldots,s_7]_{\le4}\to\R$ and an open ball
$\Omega\subset\Omega_0$ about zero with the following properties:
\begin{enumerate}
 \item $\Lambda(1)=1$, and its moment matrix on polynomials of degree at
 most two is positive definite;
 \item for every $c\in\Omega$, writing
 \begin{equation}
  \label{eq:tested-patch}
  (A_{c,\eps},t_{c,\eps},y_{c,\eps})
  =\Lambda_s\bigl(x(c+\eps s)\bigr),
 \end{equation}
 there are constants $a_c>0$ and $\kappa_c>0$ such that, as
 $\eps\downarrow0$,
 \begin{gather}
  \label{eq:deformation-lower-bound}
  A_{c,\eps}\succeq a_c\eps^4 I_r,\\
  \label{eq:deformation-violation}
  t_{c,\eps}-\tr\bigl(A_{c,\eps}^{-1}Q(y_{c,\eps})\bigr)
  =-\kappa_c\eps^4+o(\eps^4).
 \end{gather}
 All three entries of~\eqref{eq:tested-patch} remain bounded as
 $\eps\downarrow0$.
\end{enumerate}
\end{lemma}

\begin{proof}
We first allow the moment matrix to be an arbitrary Hankel matrix $S$
near $T$, and normalize its constant moment at the end.  Let $B_c$ be the
matrix on $H$ determined by
$B_cv(s)=v(s+c)$.  This is an invertible matrix, depends polynomially on
$c$, and satisfies $B_0=I_H$.  In the affine chart, $L$ consists of the
constant and linear monomials, and $P$ consists of the quadratic monomials.
On this degree decomposition of $H$,
put
\[
 D_\eps=\diag(1,\eps I_7,\eps^2I_{28}).
\]
The functional with moment matrix $S$, applied at $c+\eps s$, has moment
matrix $B_cD_\eps S D_\eps B_c^T$.  Its repeated matrix is therefore
\begin{equation}
 \label{eq:translated-scaled-moment}
 G=B_{c,N}D_{\eps,N}S_ND_{\eps,N}B_{c,N}^T.
\end{equation}
For the corresponding tested data, $A=U^TGU$ and $t=\tr(gG)$.
Whenever $A$ is invertible, the quantity to be made negative is
\begin{equation}
 \label{eq:trace-residual}
 t-\tr(A^{-1}Q(y))
 =\tr\bigl(g\,[G-GU(U^TGU)^{-1}U^TG]\bigr).
\end{equation}

We now analyze the matrix inside this trace at small $\eps$.  Projection
of the column space of $B_{c,N}^TU$ to $L^N$ remains surjective for $c$
near zero.  We can therefore choose a continuously varying basis matrix
$U_c$ for this column space in the form
\begin{equation}
 \label{eq:Uc-blocks}
 U_c=\begin{pmatrix}I_{8N}&0\\ F_c&E_c\end{pmatrix},
 \qquad
 F_c\in\R^{28N\times8N},\quad
 E_c\in\R^{28N\times(12N-1)}.
\end{equation}
For completeness, keep one invertible $8N$-by-$8N$ minor of the projection
matrix and normalize its columns to have $L^N$ components $I_{8N}$.
From each remaining column subtract its $L^N$ component expressed in
these normalized columns.  The resulting columns have zero $L^N$
component and give the block $E_c$.  All these operations depend
continuously on $c$.  At $c=0$, projection of the
columns of $E_0$ to $R^N$ is injective: a vector $(0,E_0z)$ with zero
$R^N$ projection would belong to $U\cap K^N$, which is zero by
\eqref{eq:obstruction-U-projection}.

Let $D_{\eps,L,N}$ denote the restriction of $D_{\eps,N}$ to $L^N$.
For $\eps>0$, multiplication of $D_{\eps,N}U_c$ on the right by the
invertible block diagonal matrix with blocks $D_{\eps,L,N}^{-1}$ and
$\eps^{-2}I_{12N-1}$ gives
\begin{equation}
 \label{eq:rescaled-U}
 \widehat U_{c,\eps}
 =\begin{pmatrix}
 I_{8N}&0\\
 \eps^2F_cD_{\eps,L,N}^{-1}&E_c
 \end{pmatrix}.
\end{equation}
This matrix extends continuously to $\eps=0$, where the lower-left block
is zero.  The Gram matrix
$\widehat U_{0,0}^TT_N\widehat U_{0,0}$ is positive definite.  Indeed,
if a vector $(\ell,E_0z)$ belongs to $J^N$, then $E_0z\in K^N$.
Injectivity of its $R^N$ projection gives $z=0$, and $J\cap L=0$ then
gives $\ell=0$.  Thus the column space of $\widehat U_{0,0}$ meets
$\ker T_N$ trivially.

It follows that, for $(c,\eps,S)$ near $(0,0,T)$, the Gram matrix in
\begin{equation}
 \label{eq:rescaled-residual}
 \widehat C_{c,\eps}(S)
 =S_N-S_N\widehat U_{c,\eps}
  \bigl(\widehat U_{c,\eps}^TS_N\widehat U_{c,\eps}\bigr)^{-1}
  \widehat U_{c,\eps}^TS_N
\end{equation}
is invertible.  In particular, the displayed residual is continuous in all
three arguments, including at $\eps=0$.  Changing a basis of the columns
does not change the residual in~\eqref{eq:trace-residual}.  After removing
the outer translation factors, define
\[
 C_{c,\eps}(S)=D_{\eps,N}\widehat C_{c,\eps}(S)D_{\eps,N}.
\]
The exact relation to the matrix in~\eqref{eq:trace-residual} is
\begin{equation}
 \label{eq:restored-residual}
 G-GU(U^TGU)^{-1}U^TG
 =B_{c,N}C_{c,\eps}(S)B_{c,N}^T.
\end{equation}
The matrix $C_{c,\eps}(S)$ annihilates $U_c$.  If a symmetric block matrix
$C$ annihilates the
first $8N$ columns $\binom{I_{8N}}{F_c}$ of $U_c$, its blocks satisfy
$C_{PL}=-C_{PP}F_c$ and $C_{LL}=F_c^TC_{PP}F_c$.  Hence, with
\[
 \iota_c=\begin{pmatrix}-F_c^T\\I_{28N}\end{pmatrix}
 :P^N\longrightarrow H^N,
\]
we obtain the exact identity
\begin{equation}
 \label{eq:order-four-residual}
 C_{c,\eps}(S)
 =\eps^4\iota_c\bigl(\widehat C_{c,\eps}(S)\bigr)_{PP}\iota_c^T
 \qquad(\eps>0).
\end{equation}
Combining~\eqref{eq:restored-residual} and
\eqref{eq:order-four-residual}, the limit of
\eqref{eq:trace-residual} divided by $\eps^4$ is
\begin{equation}
 \label{eq:continuous-leading-coefficient}
 \tr\bigl(gB_{c,N}\iota_c
       (\widehat C_{c,0}(S))_{PP}\iota_c^TB_{c,N}^T\bigr).
\end{equation}
This expression is continuous in $c$ and $S$.  Its sign can therefore be
computed first at the singular moment matrix $T$.

At $(c,\eps,S)=(0,0,T)$, the matrix in
\eqref{eq:rescaled-residual} is positive semidefinite of rank
$20N-r=1$.  To see this rank directly, set
$Z=T_N^{1/2}\widehat U_{0,0}$.  The residual equals
$T_N^{1/2}(I-P_Z)T_N^{1/2}$, where $P_Z$ is the orthogonal projection
onto the $r$-dimensional column space of $Z$, a subspace of
$\ran T_N^{1/2}$.  It annihilates $L^N$, the embedded columns
$\binom{0}{E_0}$, and
$J^N$.  Its range is consequently contained in
\[
 P^N\cap(J^N)^\perp\cap(\ran E_0)^\perp
 =R^N\cap(\ran E_0)^\perp.
\]
Because projection of $E_0$ to $R^N$ has rank $12N-1$, this last space is
one-dimensional.  It contains $w_P$, since $w$ is orthogonal to every
column of $U_0$.  Therefore
\[
 \bigl(\widehat C_{0,0}(T)\bigr)_{PP}
 =\alpha w_Pw_P^T\qquad\text{for some }\alpha>0.
\]
The other columns of $U_0$ give $w_L=-F_0^Tw_P$, so
$\iota_0w_P=w$.  Restoring the translation in
\eqref{eq:order-four-residual}, whose matrix is the identity at $c=0$,
shows that the limiting coefficient in~\eqref{eq:trace-residual} is
\[
 \alpha\tr(gww^T)=\alpha w^Tgw<0.
\]

Choose a positive definite Hankel matrix $S$ sufficiently close to $T$.
Such a choice is possible by adding a small positive multiple of the
degree-two moment matrix of a nondegenerate Gaussian measure on $\R^7$.
By the continuity just proved, the limiting coefficient remains negative
for every $c$ in an open ball about zero.  Its constant moment is positive.
Normalize the corresponding functional by its value at $1$ and call the
result $\Lambda$.  Scaling the functional by a positive number $\rho$
scales $A,t,y$ by $\rho$, and scales $Q(y)$ by $\rho^2$.  It therefore
scales~\eqref{eq:trace-residual} by $\rho$, preserving the sign.  Shrinking
the ball to lie in $\Omega_0$ proves~\eqref{eq:deformation-violation}.

Let $S_\Lambda\succ0$ be the normalized moment matrix.  For fixed $c$,
the matrix $B_{c,N}^TU$ has full column rank.  Since the smallest diagonal
entry of $D_{\eps,N}$ is $\eps^2$ for $0<\eps\le1$, formula
\eqref{eq:translated-scaled-moment} gives, for $z\in\R^r$,
\[
 z^TA_{c,\eps}z
 \ge \lambda_{\min}(S_\Lambda)\eps^4
       \|B_{c,N}^TUz\|^2
 \ge a_c\eps^4\|z\|^2
\]
with $a_c>0$.  Finally, the tested coordinates are polynomials in $\eps$,
so they remain bounded.  This completes the proof.
\end{proof}

The preceding lemma supplies points outside $E_Q$ whose violation has a
controlled leading term.  We next show that a semidefinite lift would
force any such moment deformation of a polynomial patch to be arbitrarily
well approximated, to a prescribed finite order, by feasible points.

\subsection{Moment tests of analytic lift data}

We first record why moments given only through degree four can be used to
test Taylor polynomials of higher degree.  The assertion concerns finite
positive moment matrices and does not require a representing measure.

\begin{lemma}[Finite positive extension]
 \label{lem:positive-moment-extension}
Let $n\ge1$ and $j\ge0$ be integers, let $z=(z_1,\ldots,z_n)$, and let the
linear functional
$\lambda:\R[z]_{\le2j}\to\R$ satisfy $\lambda(p^2)>0$ for every
nonzero $p\in\R[z]_{\le j}$.  For every integer $b\ge j$, there is an
extension $\widetilde\lambda:\R[z]_{\le2b}\to\R$ such that
$\widetilde\lambda(p^2)>0$ for every nonzero
$p\in\R[z]_{\le b}$.
\end{lemma}

\begin{proof}
It suffices to extend from $j$ to $j+1$.  Index the enlarged moment matrix
first by monomials of degree at most $j$, and then by monomials homogeneous
of degree $j+1$.  It has the form
\[
 \begin{pmatrix}M_j&C\\C^T&D\end{pmatrix},\qquad M_j\succ0.
\]
The mixed block $C$ uses moments of degree at most $2j+1$.  Its entries
of degree at most $2j$ are already fixed, and we may assign arbitrary
values to all moments of degree $2j+1$.  Every entry of $D$ has degree
exactly $2j+2$, so none of these moments has yet been assigned.  Give them
the values $a\int z^\gamma\,d\gamma_n(z)$, where $\gamma_n$ is a
nondegenerate Gaussian probability measure and $a>0$.  Then $D=aD_0$,
where $D_0\succ0$: the integral of the square of any nonzero homogeneous
polynomial is strictly positive.  Choosing $a$ sufficiently large makes
$aD_0-C^TM_j^{-1}C$ positive definite.  The Schur complement proves
positivity of the enlarged moment matrix.  Each degree layer is assigned
as a functional on monomials, so all repeated Hankel entries agree.
Iteration proves the assertion.
\end{proof}

The next lemma provides simultaneous analytic lift variables and Gram
factors.  Only a nonempty open part of the polynomial patch is needed.

\begin{lemma}[Local analytic Gram factors]
 \label{lem:analytic-lift}
Let $C\subset\R^a$ have an exact representation
\[
 C=\{x\in\R^a:\text{there exists }u\in\R^\nu
                         \text{ with }\mathcal L(x,u)\succeq0\},
\]
where $\nu\ge0$ and $\mathcal L$ is an affine symmetric pencil of size
$\ell\ge1$.
Let $\Omega\subset\R^n$ be nonempty, open, and semialgebraic, and let
$p:\R^n\to\R^a$ be polynomial with $p(\Omega)\subset C$.
There is a nonempty open subset $\Omega'\subset\Omega$ and real analytic
maps $u:\Omega'\to\R^\nu$ and
$F:\Omega'\to\R^{\ell\times\ell}$ such that
\begin{equation}
 \label{eq:analytic-Gram-identity}
 \mathcal L(p(s),u(s))=F(s)F(s)^T\qquad(s\in\Omega').
\end{equation}
\end{lemma}

\begin{proof}
Consider the semialgebraic set
\[
 \mathcal A=\{(s,u,F):s\in\Omega,\quad
                  \mathcal L(p(s),u)=FF^T\}.
\]
Its projection to $\Omega$ is surjective, since every positive
semidefinite matrix has a real square Gram factor.  A semialgebraic set
admits a finite stratification by real analytic semialgebraic manifolds;
see \cite[Proposition~9.1.8]{BCR1998}, or
\cite[Section~2.2.2]{LRT2025}.  On at least one stratum, the differential
of the projection has rank $n$ at some point.  Otherwise Sard's theorem
on each of the finitely many strata would imply that the whole image has
Lebesgue measure zero, contradicting that it contains $\Omega$.
The analytic submersion theorem at such a point gives a local analytic
section of the projection.  Its $u$ and $F$ coordinates give
\eqref{eq:analytic-Gram-identity}.
\end{proof}

\begin{lemma}[Feasible approximation after a moment test]
 \label{lem:moment-near-feasibility}
Let $C$, $p$, and $\Omega$ be as in Lemma~\ref{lem:analytic-lift}.
Fix an integer $b\ge\max\{1,\deg p\}$ and a linear functional
$\lambda:\R[z_1,\ldots,z_n]_{\le2b}\to\R$ satisfying
$\lambda(1)=1$ and $\lambda(q^2)\ge0$ for all
$q\in\R[z]_{\le b}$.  There is a nonempty open subset
$\Omega'\subset\Omega$ such that, for each $c\in\Omega'$, the points
\[
 x_\eps=\lambda_z\bigl(p(c+\eps z)\bigr)
\]
admit points $\widetilde x_\eps\in C$, for all sufficiently small
$\eps>0$, with
\begin{equation}
 \label{eq:near-feasibility}
 \|\widetilde x_\eps-x_\eps\|=O(\eps^b).
\end{equation}
The implied constant may depend on $c$ and on the displayed data.
\end{lemma}

\begin{proof}
Choose analytic maps $u,F$ on $\Omega'$ as in
Lemma~\ref{lem:analytic-lift}, and fix $c\in\Omega'$.  For a real analytic
map $f$, write its degree-$b$ Taylor polynomial at $c$, evaluated at
$c+\eps z$, as
\[
 f_{c,b}(\eps,z)=\sum_{|\alpha|\le b}
  \frac{\partial^\alpha f(c)}{\alpha!}\eps^{|\alpha|}z^\alpha.
\]
Set $u_\eps=\lambda_z(u_{c,b}(\eps,z))$ and
\[
 P_\eps=\mathcal L(x_\eps,u_\eps),\qquad
 H_\eps=\lambda_z\bigl(F_{c,b}(\eps,z)F_{c,b}(\eps,z)^T\bigr).
\]
For every $v\in\R^\ell$, the scalar $v^TH_\eps v$ is the value of
$\lambda$ on a sum of squares of polynomials of degree at most $b$.
Thus $H_\eps\succeq0$.  The Taylor identity
\eqref{eq:analytic-Gram-identity} implies that $P_\eps$ and $H_\eps$
have the same coefficients through degree $b$ in $\eps$.  Here
$\deg p\le b$ ensures that the polynomial patch is retained exactly,
and $\lambda(1)=1$ retains the constant term of the affine pencil.
Both tested expressions are polynomials in $\eps$, so
\begin{equation}
 \label{eq:approximate-pencil}
 \|P_\eps-H_\eps\|=O(\eps^{b+1}),\qquad
 P_\eps\succeq-O(\eps^{b+1})I_\ell.
\end{equation}

To turn this approximate positivity into exact feasibility, let
\[
 K_0=\bigcap_{\mathcal L(x,u)\succeq0}\ker\mathcal L(x,u),
 \qquad S_0=K_0^\perp.
\]
Every selected matrix $\mathcal L(p(s),u(s))$ annihilates $K_0$.
Its Taylor coefficients at $c$ therefore do so as well.  Since
$P_\eps$ is obtained by applying $\lambda$ to its Taylor polynomial
through degree $b$, we have
\begin{equation}
 \label{eq:exact-common-kernel}
 P_\eps K_0=0.
\end{equation}
In particular, any negative eigenvalues of $P_\eps$ occur on $S_0$.

There is a feasible pencil matrix $P_*=\mathcal L(x_*,u_*)$ whose
restriction to $S_0$ is positive definite.  Indeed, choose a feasible
matrix of maximum rank.  For any other feasible matrix $P$, the average
$(P_*+P)/2$ is feasible, and
$\ker(P_*+P)=\ker P_*\cap\ker P$.  Maximality of the rank therefore
forces $\ker P_*\subset\ker P$, proving $\ker P_*=K_0$.
If $S_0=0$, equation~\eqref{eq:exact-common-kernel} already gives
$P_\eps=0$, so $x_\eps\in C$ and there is nothing to prove.

Otherwise choose $\mu>0$ with $P_*|_{S_0}\succeq\mu I_{S_0}$ and put
$\delta_\eps=\eps^b$.  For sufficiently small positive $\eps$,
\[
 \bigl((1-\delta_\eps)P_\eps+\delta_\eps P_*\bigr)|_{S_0}
 \succeq\bigl(\mu\eps^b-C_1\eps^{b+1}\bigr)I_{S_0}
 \succeq0.
\]
Both summands annihilate $K_0$, so the mixed matrix is positive
semidefinite on all of $\R^\ell$.  By affineness of the pencil it is the
pencil value at
\[
 \widetilde x_\eps=(1-\delta_\eps)x_\eps+\delta_\eps x_*,\qquad
 \widetilde u_\eps=(1-\delta_\eps)u_\eps+\delta_\eps u_*.
\]
Thus $\widetilde x_\eps\in C$.  The vector $x_\eps$ is polynomial in
$\eps$ and hence bounded near zero, giving~\eqref{eq:near-feasibility}.
\end{proof}

\subsection{The contradiction}

\begin{theorem}
 \label{prop:no-lift}
For the admissible quadratic matrix map $Q$ of
\eqref{eq:constructed-Q}, the set $E_Q$ is not a spectrahedral shadow.
Consequently its associated entire closed hyperbolicity cone $K_Q$ is not
a spectrahedral shadow.
\end{theorem}

\begin{proof}
Suppose that $E_Q$ has a finite affine semidefinite lift.  Take the
polynomial patch $x$, the functional $\Lambda$, and the open ball
$\Omega$ from Lemmas~\ref{lem:polynomial-patch}
and~\ref{lem:moment-violation}.  By
Lemma~\ref{lem:positive-moment-extension}, extend $\Lambda$ through
degree $32$ so that it is positive definite on squares of polynomials of
degree at most $b=16$.  This extension leaves all tested patch coordinates
unchanged, since their degrees are at most four, and preserves
$\Lambda(1)=1$.

Apply Lemma~\ref{lem:moment-near-feasibility} with $C=E_Q$, $p=x$,
and $b=16$.  Fix a center $c$ in the resulting nonempty open subset of
$\Omega$, and abbreviate the tested data by
$x_\eps=(A_\eps,t_\eps,y_\eps)$.  There are feasible points
\[
 \widetilde x_\eps
 =(A_\eps+\Delta A_\eps,\ t_\eps+\Delta t_\eps,
                         y_\eps+\Delta y_\eps)\in E_Q
\]
with all three perturbations of norm $O(\eps^{16})$.
By~\eqref{eq:deformation-lower-bound},
$A_\eps\succeq a_c\eps^4 I_r$.  Therefore, for sufficiently small
$\eps>0$,
\[
 A_\eps+\Delta A_\eps\succeq\tfrac12a_c\eps^4 I_r,
 \qquad
 \|A_\eps^{-1}\|+
 \|(A_\eps+\Delta A_\eps)^{-1}\|=O(\eps^{-4}).
\]
The inverse identity gives
\begin{align*}
 &(A_\eps+\Delta A_\eps)^{-1}-A_\eps^{-1}\\
 &\hspace{1cm}=-(A_\eps+\Delta A_\eps)^{-1}
          \Delta A_\eps A_\eps^{-1}=O(\eps^8)
\end{align*}
in operator norm.  The vectors $y_\eps$ are bounded, and $Q$ is
quadratic, so
\[
 Q(y_\eps)=O(1),\qquad
 Q(y_\eps+\Delta y_\eps)-Q(y_\eps)=O(\eps^{16}).
\]
Combining these estimates and using that $r$ is fixed yields
\begin{align*}
 &t_\eps+\Delta t_\eps
  -\tr\bigl((A_\eps+\Delta A_\eps)^{-1}
                         Q(y_\eps+\Delta y_\eps)\bigr)\\
 &\qquad=t_\eps-\tr\bigl(A_\eps^{-1}Q(y_\eps)\bigr)
                +O(\eps^8)\\
 &\qquad=-\kappa_c\eps^4+o(\eps^4)<0
\end{align*}
for all sufficiently small positive $\eps$.  Since
$A_\eps+\Delta A_\eps$ is positive definite, this contradicts the
trace inequality defining the feasible point $\widetilde x_\eps\in E_Q$.
Thus $E_Q$ has no finite semidefinite lift.  Proposition~\ref{prop:epigraph}
identifies $E_Q$ as a projection of an affine section of $K_Q$, so the
same conclusion holds for $K_Q$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
The construction in Section~\ref{sec:separation} gives finite dimensions
$m=330$ and $r=20N-1$ and a quadratic matrix map $Q$ satisfying
$Q(y)\succeq0$ for every $y\in\R^m$.
Proposition~\ref{prop:hyperbolic} shows that its polynomial $p_Q$ is
hyperbolic with respect to the specified direction.
Theorem~\ref{prop:no-lift} shows that the entire closed hyperbolicity cone
$K_Q$ has no finite semidefinite lift.  These are all the assertions of
Theorem~\ref{thm:main}.
\end{proof}
==== END Hyperbolicity-Cones-Without-Semidefinite-Lifts//05-obstruction.tex ====

==== BEGIN Hyperbolicity-Cones-Without-Semidefinite-Lifts//appendix-certificate.tex ====
\section{A rank certificate for the Hankel seed}
\label{app:certificate}

We prove Lemma~\ref{lem:seed}.  Recall that
\[
 H=\R[x_0,\ldots,x_7]_2=L\oplus P,
 \qquad \dim H=36,\quad \dim L=8,\quad \dim P=28,
\]
where $L$ consists of the quadratic monomials divisible by $x_0$.
We seek a positive semidefinite Hankel matrix $T$ of rank $20$ whose
restriction to $L$ is positive definite.  Its projected kernel
$K=\operatorname{proj}_P(\ker T)$ must have independent pairwise
products and span all pure cubics after multiplication by pure linear
forms.  The remaining requirement is
\[
 \dim\bigl((\ker T)\cdot H_0\bigr)=321,
 \qquad H_0=\{p\in H:T(p,L)=0\},
\]
inside the $330$-dimensional quartic space.

Our starting matrix is the moment matrix of twenty points arranged in
two cyclic orbits.  At that matrix the last product space has dimension
$302$.  We exhibit a tangent direction for which nineteen independent
product directions appear to first order.  All nonvanishing assertions
are certified below by explicit minors over $\mathbb F_7$; we then
explain why these certificates produce a real positive semidefinite
matrix with the required properties.

\subsection{Cyclic orbit matrices and the required rank increase}
\label{subsec:orbit-seed}

For the algebraic construction work first over $\mathbb C$, keeping
$x_0$ as the homogenizing coordinate.  Give the eight variables the
respective weights
\begin{equation}
 \omega=(0,5,1,9,2,8,4,6)\pmod {10}.
 \label{eq:certificate-weights}
\end{equation}
Let $\zeta$ be a primitive tenth root of unity.  For
$a=(1,a_1,\ldots,a_7)$, take the two orbits of $(1,\ldots,1)$ and $a$
under the diagonal action
$x_j\mapsto\zeta^{\omega_j}x_j$.
Give every orbit point weight $1/10$, and let $T(a)$ be the resulting
quadratic moment matrix.  If $m$ is a quartic monomial, its moment is
\begin{equation}
 t_a(m)=
 \begin{cases}
  1+m(a),&\operatorname{wt}(m)=0,\\
  0,&\operatorname{wt}(m)\ne0.
 \end{cases}
 \label{eq:orbit-moments}
\end{equation}
In particular $T(a)_{u,v}=t_a(uv)$ for quadratic monomials $u,v$.
Formula~\eqref{eq:orbit-moments} has integer coefficients in the
parameters and also defines the matrices used in the finite-field
calculation; no tenth roots of unity in $\mathbb F_7$ are needed.

Write $I_i$ for the ordered quadratic monomials of weight $i$ and set
\[
 C_i(a)=
 \begin{pmatrix}
  1&\cdots&1\\
  u_1(a)&\cdots&u_{|I_i|}(a)
 \end{pmatrix},
 \qquad (u_1,\ldots,u_{|I_i|})=I_i.
\]
The only nonzero block in column weight $i$ of $T(a)$ is
$C_{-i}(a)^TC_i(a)$.  If every $C_i(a)$ has rank two, then
$\rank T(a)=20$ and its kernel is the direct sum of the nullspaces
$J_i=\ker C_i(a)$.  Thus $J=\ker T(a)$ has dimension $16$.

Let $D$ denote a matrix whose columns are the quartic products of a
basis of $J$ with a basis of $H_0$.  For a Hankel variation $T_1$
satisfying
\begin{equation}
 T_1|_{J\times J}=0,
 \label{eq:certificate-tangent}
\end{equation}
write $D_1$ for the first-order product variation obtained by
differentiating the equations for bases of $J$ and $H_0$.  We give
explicit formulas below and later realize $T_1$ as the velocity of
a smooth arc.  We shall establish, simultaneously with the two
product conditions on $K$,
\begin{equation}
 \rank D=302,
 \qquad
 \rank\begin{pmatrix}D&0\\D_1&D\end{pmatrix}\ge623
 =2\cdot302+19.
 \label{eq:certificate-target-ranks}
\end{equation}
These ranks do not depend on the choices of differentiable bases.
Indeed, a change of the derivatives of those bases changes $D_1$ by
a matrix of the form $AD+DB$, which is removed by block row and column
operations in the second matrix of
\eqref{eq:certificate-target-ranks}.

We first record the upper bounds that give the target ranks their
meaning.  They apply in characteristic zero wherever $T$ has rank
$20$, $J\cap L=0$, and $K$ spans all pure cubics after multiplication
by pure linear forms.  The seven translations and the dilation in
the chart $x_0=1$ have independent derivatives on the kernel graph,
by the proof of Lemma~\ref{lem:kernel-affine-action}, which uses
only these hypotheses and works over $\mathbb C$ as well as over
$\R$.  Recall the obstruction to a dependence involving dilation:
it would allow a translation making the kernel equal to the homogeneous space $K$.
Cubic spanning would then annihilate all chart moments of degrees
three and four, forcing rank at most $8+1=9$ instead of $20$.

A translation or dilation preserves the projected kernel $K$.
Continue each kernel vector with its highest homogeneous part
fixed, so that its first derivative lies in $L$.  Differentiating
the kernel equation therefore gives
$T'(0)J\subset T(L)$, so its quartic functional annihilates $JH_0$.
Scaling $T$ gives one further independent annihilator, since scaling
has zero kernel derivative whereas the eight affine motions have
independent kernel derivatives.  Hence, throughout the rank-$20$
locus under consideration,
\begin{equation}
 \rank D\le330-9=321.
 \label{eq:certificate-upper-321}
\end{equation}

At an orbit matrix, scaling belongs to the twenty-dimensional family
obtained by varying the individual weights of its twenty points.
Rank $20$ means
that their evaluation functionals are independent already on
quadratics, and hence on quartics by multiplication by $x_0^2$.
Each such quartic evaluation annihilates $JH_0$, and all weight
variations have zero kernel derivative.  The eight affine motions
remain independent modulo these twenty directions.  Thus at the
orbit matrix
\begin{equation}
 \rank D\le330-20-8=302.
 \label{eq:certificate-upper-302}
\end{equation}
In particular, attaining rank at least $302$ there is an open condition
equivalent to equality.

\subsection{Bases and first derivatives modulo seven}
\label{subsec:certificate-bases}

All entries in the remainder of this subsection and the next are in
$\mathbb F_7$.  Use the specialization
\begin{equation}
 a=(1,6,6,0,3,0,2,5).
 \label{eq:certificate-point}
\end{equation}
Index monomials by nondecreasing strings of variable indices, ordered
lexicographically, both globally and within each weight.  For example,
$23$ means $x_2x_3$ and $0011$ means $x_0^2x_1^2$.
Every row and column number below starts at one within its indicated
block.

For $C_i$, take pivot columns
\begin{equation}
 E_i=(1,2)\quad\text{except that}\quad E_0=E_8=(1,3).
 \label{eq:certificate-pivots}
\end{equation}
Use the nullspace basis having an identity matrix on the remaining
coordinates, in their order.  The quadratic blocks and resulting columns
of $J_i$ are as follows.  A digit string in the third column records
one entire column vector in the order of $I_i$.
\begin{center}
\begin{tabular}{c l l}
\toprule
$i$&$I_i$&columns of $J_i$\\
\midrule
0&$00,11,23,45,67$&$61000,\ 00610,\ 40201$\\
1&$02,17,34$&$421$\\
2&$04,22,56,77$&$4210,\ 2401$\\
3&$15,24,36$&$601$\\
4&$06,13,44,57$&$6010,\ 0601$\\
5&$01,26,37$&$511$\\
6&$07,12,46,55$&$4210,\ 2401$\\
7&$14,27,35$&$151$\\
8&$05,33,47,66$&$6100,\ 3031$\\
9&$03,16,25$&$601$\\
\bottomrule
\end{tabular}
\end{center}
Each specified two-column pivot is invertible.  Thus every $C_i$ has
rank two and $T$ has rank $20$.  The block on $L$ is also invertible:
in the order $00,01,\ldots,07$ its determinant is $6$.

For later calculations, the ordered list of weight-zero quartics is
\begin{equation}
\begin{gathered}
0000,0011,0023,0045,0067,0126,0137,0225,0334,0447,\\
0466,0556,0577,1111,1123,1145,1167,1244,1257,1346,\\
1355,2233,2247,2266,2345,2367,3356,3377,4446,4455,\\
4567,4777,5557,5666,6677.
\end{gathered}
\label{eq:certificate-zero-quartics}
\end{equation}
Define $T_1$ by setting its quartic moments outside weight zero equal
to zero and its moments in the order
\eqref{eq:certificate-zero-quartics} equal to
\begin{equation}
 (2,0,0,0,3,6,1,4,1,6,0,3,0,5,2,2,0,0,4,0,6,
 \underbrace{0,\ldots,0}_{14}).
 \label{eq:certificate-tangent-moments}
\end{equation}
The weight-zero block of products in $\Sym^2J$ has seventeen columns.
Its transpose times the vector
\eqref{eq:certificate-tangent-moments} is the zero vector.  Products
of every other weight pair trivially with $T_1$, so
\eqref{eq:certificate-tangent} holds.

We specify a basis $N_i$ for the weight-$i$ part of $H_0$.  There is
one coordinate of $L$ in each weight except $3$ and $7$.  Denote its
within-block index by $L_i$; thus $L_0=1$ corresponds to $00$.
If weight $-i$
has no $L$ coordinate, let $N_i$ be the identity matrix.  Otherwise
let $N_i$ be the nullspace basis of the single row $T_{L_{-i},I_i}$,
using pivot $L_i$ and identity on the free indices.  Its pivot is
nonzero because $T|_L$ is invertible.

The following formulas specify derivatives of both bases.  Put
$J_i'$ equal to zero off the pivot rows $E_i$, and set
\begin{equation}
 (J_i')_{E_i}=
 -\bigl(T_{E_{-i},E_i}\bigr)^{-1}
       (T_1)_{E_{-i},I_i}J_i.
 \label{eq:certificate-J-derivative}
\end{equation}
If $N_i$ is an identity matrix, put $N_i'=0$.  In the other cases,
put $N_i'$ equal to zero off row $L_i$, and set
\begin{equation}
 (N_i')_{L_i}=
 -\bigl(T_{L_{-i},L_i}\bigr)^{-1}
       (T_1)_{L_{-i},I_i}N_i.
 \label{eq:certificate-N-derivative}
\end{equation}
These are the differentiated kernel equations with the free
coordinates held fixed.  For $J_i'$, tangency ensures that solving
the two selected equations also solves all the equations
$TJ_i'+T_1J_i=0$: the image of $T_1J_i$ lies in the image of $T$,
and the two selected rows give coordinates on that image in the
relevant weight block.  The formulas for $N_i'$ follow by
differentiating $T(L,N_i)=0$.

\subsection{The explicit minors}
\label{subsec:certificate-minors}

We now give the minors that establish the two product conditions on
$K$ and the rank increase in
\eqref{eq:certificate-target-ranks}.  All product matrices use the
following convention.  In weight $w$, multiply a column $u$ from
block $i$ by a column $v$ from block $w-i$, adding
\[
 u_{jj'}v_{kk'}
 \quad\text{to row}\quad \operatorname{sort}(jj'kk').
\]
Order columns lexicographically by the weight $i$, the column number
of $u$, and the column number of $v$.  For products from
$\Sym^2J$, retain only pairs for which the first index
$(i,\text{column number})$ is at most the second.  Thus each
unordered product occurs once, with its ordinary polynomial
coefficients.

Let $G_1$ be this matrix for $J\cdot J$, restricted to quartic rows
with no zero index.  It is the product matrix for $\Sym^2K$.
Let $G_2$ be the product matrix of $J$ with the seven coordinate
vectors $x_0x_j$, $1\le j\le7$, restricted to rows with exactly one
zero index.  Removing that zero identifies it with the product map
from $K$ times pure linear forms to pure cubics.  Rows and columns
are numbered anew after the restriction to a weight and to these
row sets.  In all the tables, a colon denotes the inclusive range.

The following square minors use all columns of $G_1$ and all rows of
$G_2$, respectively.  The specified omissions determine the other
index set completely.
\begin{center}
\small
\begin{tabular}{c l r r l r r}
\toprule
&\multicolumn{3}{c}{$G_1$}&\multicolumn{3}{c}{$G_2$}\\
$w$&omit rows&size&det.&omit columns&size&det.\\
\midrule
0&$15,19{:}22$&17&3&$8,10,11$&8&6\\
1&$9,11,12,15{:}20$&11&4&$9{:}11$&8&6\\
2&$13,15,19{:}22$&16&4&$8,10,11$&9&2\\
3&$3,10,14{:}20$&11&5&$8,10$&8&6\\
4&$13,15,18,19,21,22$&16&6&$10{:}12$&9&6\\
5&$9,11,13,15{:}20$&11&5&$8,10,11$&8&4\\
\bottomrule
\end{tabular}
\end{center}
The number of columns of $G_1$ in weights $0,\ldots,5$ is
$17,11,16,11,16,11$; the number of rows of $G_2$ is
$8,8,9,8,9,8$.  Thus these determinants give precisely the full
column and full row ranks needed in these weights.

For $D$, use first factors from $J$ and second factors from $H_0$,
with the preceding product order.  Form $D_1$ by replacing the first
factor by its derivative in
\eqref{eq:certificate-J-derivative}, replacing the second by its
derivative in \eqref{eq:certificate-N-derivative}, and adding the
two resulting product columns.  In weights $0,\ldots,5$, the
dimensions of $D$ are respectively
\[
 35\times48,\quad31\times42,\quad35\times47,\quad
 31\times43,\quad35\times47,\quad31\times42.
\]
The next table specifies a square submatrix $E=D[R,S]$ in each
weight by listing the omitted rows and columns; $R,S$ always
denote the retained indices.
\begin{center}
\small
\setlength{\tabcolsep}{4pt}
\begin{tabular}{c l l r r}
\toprule
$w$&omit rows&omit columns&size&$\det E$\\
\midrule
0&$13,34,35$&$5,9,11,33,34,36{:}38,40{:}43,45{:}48$&32&6\\
1&$24,26,31$&$7,9,10,28,30,32,33,36{:}42$&28&1\\
2&$32,34,35$&$12,14{:}16,18,19,35,38,40,41,43{:}47$&32&6\\
3&$26,31$&$14,16,17,19,20,35{:}43$&29&3\\
4&$33,34,35$&$17,20,21,23{:}25,27,28,41{:}47$&32&5\\
5&$19,26,31$&$17,18,20,22,23,25,26,36{:}42$&28&3\\
\bottomrule
\end{tabular}
\end{center}

To obtain the bordered minors, use the indices $R,S$ in both
diagonal positions of $\left(\begin{smallmatrix}D&0\\D_1&D
\end{smallmatrix}\right)$.  Add rows $R'$ in its second block row
and columns $S'$ in its first block column.  Set
$A=D[R,S']$ and $B=D[R',S]$.  Eliminating the two copies of $E$
leaves the corner
\begin{equation}
\begin{split}
 Z={}&D_1[R',S']-D_1[R',S]E^{-1}A
       -BE^{-1}D_1[R,S']\\
    &\hspace{27mm}+BE^{-1}D_1[R,S]E^{-1}A.
\end{split}
\label{eq:certificate-corner}
\end{equation}
The corner matrices are the following, with rows separated by
semicolons when their entries are listed.
\begin{center}
\begin{tabular}{c l l l r}
\toprule
$w$&$R'$&$S'$&$Z$&$\det Z$\\
\midrule
0&$34$&$38$&$(5)$&5\\
1&$24,31$&$38,40$&$(4,2;\ 2,4)$&5\\
2&$34,35$&$45,46$&$(6,4;\ 2,0)$&6\\
3&$26,31$&$37,39$&$(3,6;\ 0,5)$&1\\
4&$33,35$&$43,46$&$(6,4;\ 1,2)$&1\\
5&$26,31$&$37,38$&$(2,6;\ 2,0)$&2\\
\bottomrule
\end{tabular}
\end{center}
In particular the resulting bordered minor has size $2|R|+1$ in
weight zero and $2|R|+2$ in each of weights one through five.
The bordered determinant is, up to its row and column ordering
sign, $(\det E)^2\det Z$, as follows directly by eliminating the
two diagonal copies of $E$.  Thus the table proves the asserted
bordered lower bounds without requiring an upper bound on the
rank of $D$ in the finite field.

For completeness, every entry in these tables can be checked with
the following fixed elimination rule.  At column $n$, interchange
row $n$ with the first row at or below it whose entry $t$ in that
column is nonzero.  Divide the new row by $t$ and subtract the
appropriate multiples from every other row, always reducing modulo
seven.  The determinant is the product of the pivots, with one
minus sign for each interchange.  Appended columns give the
inverse actions in \eqref{eq:certificate-corner} by the same rule.
For the $G_2$ minor in weight zero, the signed pivots are
\[
 (-2,-2,-5,-1,4,-1,1,5).
\]
For $E$ in weight zero they are
\begin{equation}
\begin{gathered}
 1,4,1,4,-4,-1,-4,-5,-5,-4,-4,6,-4,4,6,-3,\\
 -3,-1,6,6,4,5,-5,2,1,5,2,5,6,4,4,4.
\end{gathered}
\label{eq:certificate-pivot-example}
\end{equation}
Together with the monomial order, the basis rules, and the moment
vectors above, this specifies the finite calculation using matrices
with at most thirty-five quartic rows in each weight.

\subsection{Characteristic zero and simultaneous real data}
\label{subsec:certificate-lifting}

We now pass from the explicit modular minors to actual orbit
matrices and tangent directions in characteristic zero.  First lift
\eqref{eq:certificate-point} to the same integer vector.  The chosen
pivots of the $C_i$, the entries used as pivots for $N_i$, and the
$2\times2$ blocks in
\eqref{eq:certificate-J-derivative} are all invertible modulo seven.
Hence the corresponding bases and inverse formulas are defined over
$\mathbb Z_{(7)}$, the rational numbers whose denominators are not
divisible by seven.

There is one point to address before lifting the bordered minors:
the first variation must remain tangent exactly, not just modulo
seven.  The weight-zero part of $\Sym^2J$ has dimension seventeen.
Its seventeen product columns are independent, since their
restriction to the pure rows contains the nonzero $17\times17$
minor of $G_1$ above.  Consequently tangency of a weight-zero
functional gives seventeen independent equations on its thirty-five
moment coordinates.  Choose as pivot coordinates the seventeen
quartic rows of that minor.  Lift the remaining eighteen coordinates
of \eqref{eq:certificate-tangent-moments} to integers and solve
these seventeen equations over $\mathbb Z_{(7)}$.  The solution
reduces to precisely the displayed moment vector.  It therefore
defines an exact characteristic-zero tangent whose basis
derivatives reduce to
\eqref{eq:certificate-J-derivative} and
\eqref{eq:certificate-N-derivative}.  Every nonzero determinant in
the preceding tables thus proves that its defining rational
function is nonzero in characteristic zero.

It remains to obtain all weights and a real positive semidefinite
matrix simultaneously.  We spell out the genericity argument to
make clear that the tangent need not be chosen separately for
different weights.  The permutation
\[
 (x_2,x_3),\quad(x_4,x_5),\quad(x_6,x_7)
 \quad\text{interchanged, with }x_0,x_1\text{ fixed},
\]
negates every weight and preserves the family of orbit matrices.
The nonzero minors for weights $1,2,3,4$ therefore also give nonempty
rank conditions for weights $9,8,7,6$, respectively.

More formally, take the Zariski open subset of the parameter space
$\mathbb C^7$ on which all $C_i$ have rank two, $T|_L$ is
invertible, and the seventeen weight-zero products are independent.
This is a nonempty irreducible parameter space.  The solutions of
the seventeen tangency equations form a rank-eighteen algebraic
vector bundle over it: locally an invertible seventeen-column
minor solves for seventeen moment coordinates in terms of the
other eighteen.  Its total space is irreducible.  Each condition
certified above is a nonempty Zariski open subset of this space,
as is its image under the weight-negating permutation, after
restricting to the common nonempty parameter domain.  Their finite
intersection is nonempty.  The same reasoning for the minors of
$G_1$ and $G_2$, which depend only on $a$, gives both product
conditions on $K$.

The resulting lower bounds in weights $0,\ldots,9$ are
\begin{equation}
\begin{array}{c|rrrrrrrrrr|r}
w&0&1&2&3&4&5&6&7&8&9&\text{sum}\\\hline
\rank G_1&17&11&16&11&16&11&16&11&16&11&136\\
\rank G_2&8&8&9&8&9&8&9&8&9&8&84\\
\rank D&32&28&32&29&32&28&32&29&32&28&302\\
\rank\left(\begin{smallmatrix}D&0\\D_1&D\end{smallmatrix}\right)
 &65&58&66&60&66&58&66&60&66&58&623
\end{array}
\label{eq:certificate-rank-sums}
\end{equation}
Here $136=\dim\Sym^2K$ and $84$ is the dimension of the pure
cubic space.  The first two rows prove the required injectivity
and surjectivity.  Equation~\eqref{eq:certificate-upper-302}
then turns the third row into equality, and the fourth row gives
\eqref{eq:certificate-target-ranks}.

Impose now the real structure in which $x_0,x_1$ are real and the
pairs $(x_2,x_3)$, $(x_4,x_5)$, $(x_6,x_7)$ are conjugate.  Its
fixed parameter set
\[
 a_1\in\R,\qquad a_3=\overline{a_2},\quad
 a_5=\overline{a_4},\quad a_7=\overline{a_6}
\]
is a real form of $\mathbb C^7$ and is Zariski dense.  One way to
see this is to write each conjugate pair as $(u+iv,u-iv)$: this is
an invertible complex linear change of coordinates from a real
coordinate space, and a polynomial vanishing on that real space
is zero.  The weight-zero tangent equations at a fixed real
parameter are preserved by the real structure.  Their solution
space is therefore the complexification of its real fixed
subspace, which is likewise Zariski dense.  A nonempty open
condition in the tangent bundle is attained over a nonempty open
set of parameters, and within each such fiber it is an open
condition on a linear space.  We can thus choose simultaneously
a parameter in the real form and a real tangent satisfying every
rank condition in \eqref{eq:certificate-rank-sums}.

For these parameters all twenty orbit points are real in the
corresponding real coordinates.  Indeed multiplication by
$\zeta^{\omega_j}$ respects the conjugate pairs and multiplies
the weight-five coordinate by $\pm1$.  Since every point has
positive weight $1/10$, the moment matrix is real positive
semidefinite.  Its rank is $20$, and its nonsingular restriction
to $L$ is positive definite.
Choose ordinary real coordinates on this fixed real form, keeping
$x_0$ and the pure-variable span, and relabel them $x_0,\ldots,x_7$.
The induced polynomial change of coordinates preserves $L\oplus P$
and multiplication, and hence the total product and first-variation
ranks above. Thus these are real Hankel data in the conventions of
Section~\ref{sec:hankel}.

\subsection{Deformation to the seed matrix}

We have obtained a real positive semidefinite orbit matrix $T$ and
a real Hankel tangent $T_1$ satisfying
\eqref{eq:certificate-target-ranks}.  Independence of the products
in $\Sym^2K$ implies independence of those in $\Sym^2J$ by taking
highest chart degree.  Lemma~\ref{lem:hankel-stratum}, or its
Schur-complement proof, therefore shows that the rank-$20$ Hankel
locus is smooth near $T$, with tangent condition
\eqref{eq:certificate-tangent}.  There is a real smooth arc
$T(t)$ on this locus with $T(0)=T$ and $T'(0)=T_1$.
For sufficiently small $t$, its twenty nonzero eigenvalues remain
positive, its restriction to $L$ remains positive definite, and
the two product conditions on $K(t)$ remain valid.

Choose smooth local kernel and $H_0$ bases along the arc and let
$D(t)$ be their product matrix.  Apply fixed invertible row and
column operations that put $D(0)$ in rank normal form
\[
 D(0)=\begin{pmatrix}I_{302}&0\\0&0\end{pmatrix}.
\]
Writing the transformed derivative in the same blocks as
$\left(\begin{smallmatrix}A&B\\C&F\end{smallmatrix}\right)$,
elementary block elimination gives
\[
 \rank\begin{pmatrix}D(0)&0\\D'(0)&D(0)\end{pmatrix}
 =604+\rank F.
\]
Thus $\rank F\ge19$.  Eliminate the invertible leading
$302\times302$ block of $D(t)$.  Its remaining Schur complement
is $tF+o(t)$, since both off-diagonal blocks vanish at $t=0$.
A nonzero $19\times19$ minor of $F$ remains nonzero after division
by $t^{19}$ for every sufficiently small nonzero $t$.  Hence
$\rank D(t)\ge321$ there.  The upper bound
\eqref{eq:certificate-upper-321} applies along the arc and gives
equality.  Taking one such $T(t)$ proves Lemma~\ref{lem:seed}.

\paragraph{Reproducible calculation.}
The accompanying source file \texttt{verify\_certificate.py}
reconstructs the monomial blocks, bases, differentiated products,
and every displayed minor using exact integer arithmetic modulo
seven.  It also verifies the ranks in weights six through nine
at the very same specialization
\eqref{eq:certificate-point} and tangent
\eqref{eq:certificate-tangent-moments}.  This supplies an
independent reproduction of the finite calculation; the explicit
minors and the characteristic-zero argument above establish the
lemma without relying on the program.
==== END Hyperbolicity-Cones-Without-Semidefinite-Lifts//appendix-certificate.tex ====

==== BEGIN Hyperbolicity-Cones-Without-Semidefinite-Lifts//main.tex ====
\newcommand{\DoNotLoadEpstopdf}{}
\documentclass[11pt]{article}
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\usepackage{amsmath,amssymb,amsthm,mathtools}
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\hypersetup{pdftitle={Hyperbolicity Cones Without Semidefinite Lifts},pdfauthor={OpenAI}}
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\numberwithin{equation}{section}
\title{Hyperbolicity Cones Without Semidefinite Lifts}
\author{OpenAI}
\date{October 5, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove that not every hyperbolicity cone is a spectrahedral shadow:
some closed hyperbolicity cones admit no finite affine semidefinite lift,
even with arbitrary real coefficients and any finite number of auxiliary
variables. This disproves the Projected Lax Conjecture and hence the
generalized Lax conjecture.
\end{abstract}
\input{01-introduction}
\input{02-cone}
\input{03-hankel}
\input{04-separation}
\input{05-obstruction}
\appendix
\input{appendix-certificate}
\bibliographystyle{abbrv}
\bibliography{references}
\end{document}
==== END Hyperbolicity-Cones-Without-Semidefinite-Lifts//main.tex ====

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