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LEVEL 3 OF 3 · Hyperbolicity cones without semidefinite lifts
An Exact Semidefinite Lift of a Nonspectrahedral Hyperbolicity Cone
expertly designed by an internal OpenAI model · released 2026-10-05
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IntroductionLet \(\mathbb S^n\) denote the real symmetric \(n\times n\) matrices. A spectrahedron is a set defined by a finite affine linear matrix inequality; a spectrahedral shadow is a linear projection of such a set. Thus auxiliary variables may make a semidefinite representation possible even when no representation exists in the original variables. We study this distinction for one explicit hyperbolicity cone. For a homogeneous real polynomial \(p\) with \(p(e)>0\), hyperbolicity with respect to \(e\) means that \(t\mapsto p(te-x)\) has only real roots for every \(x\). We use the closed-cone convention \[K(p,e)=\{x:\text{every root of }t\mapsto p(te-x) \text{ is nonnegative}\}.\] In particular, zero roots are allowed. The convexity of hyperbolicity cones goes back to Gårding (Gårding 1959). In three variables, the Helton–Vinnikov theorem yields definite symmetric determinantal representations, and hence spectrahedral cones (Helton and Vinnikov 2007; Lewis et al. 2005). Allowing auxiliary variables leads to a broader representation question. Netzer and Sanyal proved semidefinite liftability when every nonzero boundary point is a smooth point of the defining hyperbolic polynomial (Netzer and Sanyal 2015). More recently, Scheiderer obtained second-order cone lifts under a Nash-smooth boundary hypothesis (Scheiderer 2025). Our result instead concerns an explicit cone and gives its lift by a bounded-degree algebraic construction. The cone and the main resultFor \(y=(y_1,y_2,y_3)\in\mathbb R^3\), define \[ Q(y)=\begin{pmatrix} y_1^2+y_2^2&-y_1y_2&-y_1y_3\\ -y_1y_2&y_2^2+y_3^2&-y_2y_3\\ -y_1y_3&-y_2y_3&y_3^2+y_1^2 \end{pmatrix}. \tag{1}\] The associated biquadratic form \(z^TQ(y)z\) is the coefficient-one Choi–Lam form (Choi and Lam 1977, sec. 4). For \(a\in\mathbb R\), \(r\in\mathbb R^3\), and \(T\in\mathbb S^3\), set \[ \Phi_y\!\begin{pmatrix}a&r^T\\r&T\end{pmatrix} =\begin{pmatrix} \mathop{\mathrm{tr}}(Q(y)T)&-r^TQ(y)\\ -Q(y)r&aQ(y) \end{pmatrix}. \tag{2}\] This is a linear map from \(\mathbb S^4\) to itself for each fixed \(y\). On the space \(\mathcal V=\mathbb S^4\times\mathbb S^4\times\mathbb R^3\), define \[ p(X,Z,y)=\det\!\left((\det X)Z-\Phi_y(\mathop{\mathrm{adj}}X)\right), \qquad e=(I_4,I_4,0), \tag{3}\] where \(\mathop{\mathrm{adj}}X\) denotes the classical adjugate. Each matrix entry inside this determinant is homogeneous of degree five, and \(p(e)=1\); hence \(p\) is homogeneous of degree twenty. We write \(K=K(p,e)\). The slice representation in Section 2 will in particular verify that \(p\) is hyperbolic with respect to \(e\). Identify \(\mathcal V\) with \(\mathbb R^{23}\) by listing the upper triangular entries of \(X\), then those of \(Z\), each in lexicographic order of \((i,j)\), followed by \(y_1,y_2,y_3\). Theorem 1. There exist fixed matrices \(A_1,\ldots,A_{23},B_1,\ldots,B_{307} \in\mathbb S^{100}\) such that, for every \(x=(X,Z,y)\in\mathcal V\), \[x\in K\quad\Longleftrightarrow\quad \text{there exists }v\in\mathbb R^{307}\text{ with } \sum_{\nu=1}^{23}x_\nu A_\nu+ \sum_{j=1}^{307}v_jB_j\succeq0.\] The equivalence holds on the entire closed cone, including points with singular \(X\). The sizes in Theorem 1 are explicit bounds, with no minimality assertion. The matrices are specified by a finite coefficient rule in Section 4. The cone in (3) was constructed in the companion manuscript (OpenAI 2026), whose Theorem 1.1 proves that it has no homogeneous semidefinite representation in its original twenty-three coordinates. This also excludes affine representations. Indeed, if a cone containing \(0\) equals \(\{x:A_0+L(x)\succeq0\}\), where \(L\) is linear, then \(A_0\succeq0\). For each point \(x\) of the cone and \(t>0\), we have \(A_0/t+L(x)\succeq0\); letting \(t\to\infty\) gives \(L(x)\succeq0\). Conversely, \(L(x)\succeq0\) implies \(A_0+L(x)\succeq0\); thus the same cone has the homogeneous representation \(L(x)\succeq0\). We use the companion result only for this comparison: the proof of Theorem 1 below is self-contained. The companion also establishes hyperbolicity and, in Proposition 3.2, describes the part of \(K\) with \(X\succ0\). Our slice calculation recovers those two facts as part of the lift construction. Proof strategyThe starting point is a family of planes in \(\mathbb R^3\): \[y=Y_\theta(\eta,\xi) =\eta(\cos\theta,\sin\theta,0)+\xi(0,0,1).\] They cover all values of \(y\). On each plane, the matrix \(Q(y)\) has a Gram factorization linear in \((\eta,\xi)\) and trigonometric of degree three in \(\theta\). This gives a \(20\times20\) linear pencil \(L_\theta\) whose positive semidefinite locus is exactly the corresponding slice of \(K\), including its boundary. The angular parameter does not itself furnish a finite linear matrix inequality. We instead introduce a real linear functional on a finite-dimensional space of trigonometric polynomials with coefficients linear in the slice variables. We require it to be nonnegative on \(q^TL_\theta q\) for every vector \(q\) of trigonometric polynomials of degree at most two. These requirements form one finite matrix inequality. Evaluation at a single angle provides a feasible functional for every point of \(K\). This use of a finite matrix to encode positivity against a space of test polynomials belongs to the moment and sum-of-squares approach to semidefinite representations (Lasserre 2001, 2009). Here exactness must hold at one fixed degree. The main issue is the converse: an arbitrary feasible functional need not be evaluation at an angle, or integration against a measure. The degree-two matching identity in Section 3 supplies enough test vectors to recover the matrix inequalities defining each slice. Its construction rotates a fixed complex Gram factor by a \(2\times2\) unitary matrix. The rotation preserves the Gram matrix and reduces a cubic trigonometric pairing to a first harmonic. Matching its value and first derivative then identifies the pairing exactly. This bounded-degree construction is the principal additional algebraic ingredient of the lift. Section 4 applies it to prove exactness for every feasible functional and completes the proof of Theorem 1. Exact semidefinite representations on angular slicesWe first construct a matrix pencil on each plane \(y=Y_\theta(\eta,\xi)\). Its determinant will agree with \(p\) on that plane. This will identify cone membership with positive semidefiniteness even when \(X\) is singular. For \(\theta\in\mathbb R\), set \[c=\cos\theta,\qquad s=\sin\theta,\qquad w=e^{i\theta}, \qquad d(\theta)=(c,s,0),\qquad k=(0,0,1),\] and write \[Y_\theta(\eta,\xi)=\eta d(\theta)+\xi k \qquad (\eta,\xi\in\mathbb R).\] These planes cover \(\mathbb R^3\). We first factor \(Q\) on each plane, using complex notation to keep the formulas short. For a complex matrix \(F=(f_1,f_2)\in\mathbb C^{3\times2}\), define its realification by \[\widetilde F=(\operatorname{Re}f_1,\operatorname{Im}f_1, \operatorname{Re}f_2,\operatorname{Im}f_2).\] Thus \(\widetilde F\widetilde F^T=\operatorname{Re}(FF^*)\), where \(*\) denotes conjugate transpose. Define \[ U(\theta)=\frac1{\sqrt2} \begin{pmatrix} -w^{-1}&-w\\ sw^{-2}&sw^2\\ cw^{-2}&-cw^2 \end{pmatrix},\qquad V_0=\frac1{\sqrt2} \begin{pmatrix}0&0\\ i&i\\1&1\end{pmatrix}, \qquad F_\theta(\eta,\xi)=\eta U(\theta)+\xi V_0. \tag{4}\] Lemma 2 (Angular Gram factorization). For every \(\theta,\eta,\xi\in\mathbb R\), \[ \widetilde F_\theta(\eta,\xi) \widetilde F_\theta(\eta,\xi)^T =Q\bigl(Y_\theta(\eta,\xi)\bigr). \tag{5}\] In particular, \(Q(y)\succeq0\) for every \(y\in\mathbb R^3\). Proof. The three coefficients in the quadratic expansion of \(\operatorname{Re}(F_\theta F_\theta^*)\) are \[\begin{align*} \operatorname{Re}(UU^*)&= \begin{pmatrix}1&-sc&0\\-sc&s^2&0\\0&0&c^2\end{pmatrix}, &\operatorname{Re}(V_0V_0^*)&= \begin{pmatrix}0&0&0\\0&1&0\\0&0&1\end{pmatrix},\\ \operatorname{Re}(UV_0^*+V_0U^*)&= \begin{pmatrix}0&0&-c\\0&0&-s\\-c&-s&0\end{pmatrix}. \end{align*}\] For the \((2,3)\) entry of the last matrix, the multiplication gives \(s\cos(2\theta)-c\sin(2\theta)=-s\). The sum of these matrices with coefficients \(\eta^2,\xi^2,\eta\xi\) is exactly \(Q(\eta c,\eta s,\xi)\). Every real \(y\) has this form, so the factorization also proves positive semidefiniteness. ◻ The slice pencilFor a real column \(r\in\mathbb R^3\) and a real vector \(u=(u_0,u')\in\mathbb R\times\mathbb R^3\), define \[ B(r)=\begin{pmatrix}0&r^T\\-r&0_{3\times3}\end{pmatrix}, \qquad J_u=\begin{pmatrix}u'&-u_0I_3\end{pmatrix}. \tag{6}\] Then \(B(r)u=J_u^Tr\). If \(r_1,\ldots,r_4\) are the columns of \(\widetilde F_\theta(\eta,\xi)\), direct multiplication and Lemma 2 give \[ \sum_{j=1}^4 B(r_j)^TWB(r_j) =\Phi_{Y_\theta(\eta,\xi)}(W) \qquad(W\in\mathbb S^4). \tag{7}\] Indeed, for \(W=\left(\begin{smallmatrix}a&v^T\\v&T\end{smallmatrix}\right)\), the individual summands have blocks \(r_j^TTr_j\), \(-r_j^Tv\,r_j^T\), \(-r_jr_j^Tv\), and \(a r_jr_j^T\). Summing uses \(\sum_jr_jr_j^T=Q(Y_\theta(\eta,\xi))\). Consequently, for every \(y\in\mathbb R^3\), \[ u^T\Phi_y(W)u=\mathop{\mathrm{tr}}(J_uWJ_u^TQ(y)). \tag{8}\] Here we used that the angular planes cover \(\mathbb R^3\). Write \(\mathcal X,\mathcal Z\in\mathbb S^4\) for the matrix variables on a slice, and define \[ L_\theta(\mathcal X,\mathcal Z,\eta,\xi) =\begin{pmatrix} I_4\otimes\mathcal X&\mathcal B_\theta(\eta,\xi)\\ \mathcal B_\theta(\eta,\xi)^T&\mathcal Z \end{pmatrix}, \qquad \mathcal B_\theta(\eta,\xi)= \begin{pmatrix}B(r_1)\\B(r_2)\\B(r_3)\\B(r_4)\end{pmatrix}. \tag{9}\] For fixed \(\theta\), this is a homogeneous real symmetric linear pencil of size twenty. Its coefficients are trigonometric polynomials in \(\theta\) of degree at most three. Lemma 3. For every \(\theta,\eta,\xi\in\mathbb R\) and \(\mathcal X,\mathcal Z\in\mathbb S^4\), \[ (\mathcal X,\mathcal Z,Y_\theta(\eta,\xi))\in K \quad\Longleftrightarrow\quad L_\theta(\mathcal X,\mathcal Z,\eta,\xi)\succeq0. \tag{10}\] In particular, \(p\) is hyperbolic with respect to \(e\). If \(\mathcal X\succ0\), the conditions in (10) are also equivalent to \(\mathcal Z\succeq\Phi_{Y_\theta(\eta,\xi)}(\mathcal X^{-1})\). Proof. For invertible \(\mathcal X\), the block determinant formula and (7) show that \[\begin{align*} \det L_\theta &= (\det\mathcal X)^4 \det\!\left(\mathcal Z- \Phi_{Y_\theta(\eta,\xi)}(\mathcal X^{-1})\right)\\ &=p(\mathcal X,\mathcal Z,Y_\theta(\eta,\xi)). \end{align*}\] The second equality uses \(\mathop{\mathrm{adj}}\mathcal X=(\det\mathcal X)\mathcal X^{-1}\) and linearity of \(\Phi_y\) in its matrix argument. Both sides are polynomials in the slice variables, so the identity also holds for singular \(\mathcal X\). Moreover, \(L_\theta(I_4,I_4,0,0)=I_{20}\). Thus \[ p\!\left(te-(\mathcal X,\mathcal Z,Y_\theta(\eta,\xi))\right) =\det\!\left(tI_{20}-L_\theta(\mathcal X,\mathcal Z,\eta,\xi)\right). \tag{11}\] The roots are the real eigenvalues of this symmetric pencil evaluated at the specified point. Since every \(y\) lies on some angular plane, this proves hyperbolicity. Nonnegativity of all these eigenvalues, including zero eigenvalues, proves (10). For \(\mathcal X\succ0\), the final assertion is the Schur complement criterion applied to (9). ◻ A degree-two matching identityWrite \[Q(y,z)=\tfrac12\bigl(Q(y+z)-Q(y)-Q(z)\bigr)\] for the symmetric bilinear polarization of \(Q\), so that \(Q(y,y)=Q(y)\). For \(S\in\mathbb S^3\) and \(F,H\in\mathbb C^{3\times2}\), put \[\langle F,H\rangle_S =\operatorname{Re}\operatorname{tr}(F^*SH) =\operatorname{tr}(\widetilde F^T S\widetilde H).\] This is a symmetric real bilinear form. A complex trigonometric polynomial of degree at most \(j\) is a linear combination of \(e^{i\ell\theta}\) for \(-j\leq\ell\leq j\). The finite lift will test \(L_\theta\) against trigonometric polynomial vectors built from the columns of a matrix \(\widetilde C(\theta)\). A constant Gram matrix for \(\widetilde C\) makes the coefficient of \(\mathcal X\) in the quadratic contribution of the top blocks independent of \(\theta\). Requiring \(\langle F_\theta,C\rangle_S=\operatorname{tr}(SQ(Y_\theta,z))\) makes the mixed contribution linear in \(Y_\theta\). The next lemma achieves both properties with entries of degree at most two. The matrix \(S\) in the pairing may be indefinite. Lemma 4 (Degree-two matching). Let \(S\in\mathbb S^3\) and let \(z\in\mathbb R^3\) satisfy \((z_1,z_2)\ne(0,0)\). There is a matrix \(C(\theta)\in\mathbb C^{3\times2}\) whose entries are trigonometric polynomials of degree at most two such that, for all \(\theta,\eta,\xi\in\mathbb R\), \[ \begin{aligned} \widetilde C(\theta)\widetilde C(\theta)^T&=Q(z),\\ \langle F_\theta(\eta,\xi),C(\theta)\rangle_S &=\operatorname{tr}\bigl(SQ(Y_\theta(\eta,\xi),z)\bigr). \end{aligned} \tag{12}\] Proof. Write \(z=\tau d(\alpha)+\rho k\), where \(\tau>0\), and set \[w_0=e^{i\alpha},\qquad R_0=\tau U(\alpha)+\rho V_0, \qquad N=SR_0.\] By Lemma 2, \(\widetilde R_0\widetilde R_0^T=Q(z)\). We will set \(C(\theta)=R_0M(\theta)\) with \(M(\theta)\) unitary, so that this Gram matrix remains fixed. We choose the rotation to make \(\langle V_0,C\rangle_S\) constant and \(\langle U(\theta),C(\theta)\rangle_S\) a linear combination of \(\cos\theta\) and \(\sin\theta\). After those frequency bounds are established, values and a derivative at \(\alpha\) will identify the pairings. Choosing the unitary rotation. Write \(N_j\) for the \(j\)th row of \(N\) and introduce the two-component row \[v=N_3-iN_2.\] Subscripts \(+\) and \(-\) denote the first and second components of a row. Set \[g=v_++\overline{v_-},\qquad h=v_+-\overline{v_-},\qquad \kappa(\theta)=(w_0/w)^2,\qquad \delta=|g|^2+|h|^2.\] We seek a matrix of the form \[M(\theta)=\begin{pmatrix}a&b\\-\overline b&\overline a\end{pmatrix}.\] Writing \(v^M=vM\), we impose the two conditions \[ v^M_++\overline{v^M_-}=g, \qquad v^M_+-\overline{v^M_-}=h\kappa. \tag{13}\] The sum on the left controls the pairing with \(V_0\), while the difference multiplies \(w^3\) in the pairing with \(U\). These conditions therefore keep the first pairing constant and lower the cubic term of the second to frequency one, since \(w^3\kappa=w_0^2w\). The expansion below will also bound its remaining frequencies. The two left sides of (13) are \(ga+\overline h\,\overline b\) and \(ha-\overline g\,\overline b\), respectively. Thus, for \(\delta>0\), we choose \(a(\theta)\) and \(b(\theta)\) by solving \[ \begin{pmatrix}g&\overline h\\h&-\overline g\end{pmatrix} \begin{pmatrix}a\\\overline b\end{pmatrix} =\begin{pmatrix}g\\h\kappa\end{pmatrix}. \tag{14}\] The coefficient matrix \(A\) satisfies \(A^*A=\delta I_2\), so the system has a unique solution. Explicitly, \[ a=\frac{|g|^2+|h|^2\kappa}{\delta},\qquad \overline b=\frac{gh(1-\kappa)}{\delta}. \tag{15}\] Since the right side of (14) has squared norm \(\delta\), we have \(|a|^2+|b|^2=1\). For \(\delta=0\), take \(a=1\) and \(b=0\); then \(g=h=0\) and the same system is satisfied. In both cases \(M\) is unitary, \(M(\alpha)=I_2\), and \(a,\overline b\) involve only the frequencies \(0\) and \(-2\). Thus \(C=R_0M\) has degree at most two and \(C(\alpha)=R_0\). Moreover, \[\widetilde C\widetilde C^T =\operatorname{Re}(R_0MM^*R_0^*) =\operatorname{Re}(R_0R_0^*)=Q(z),\] which proves the first identity in (12). Reducing the frequencies of the pairings. The formula for \(V_0\) and (13) yield \[\sqrt2\,\langle V_0,C\rangle_S =\operatorname{Re}(v^M_++v^M_-)=\operatorname{Re}g,\] so this pairing is constant. Set \(m=N_3+iN_2\), \(n=N_1\), and write \(m^M=mM\), \(n^M=nM\). Expanding \(c\) and \(s\) in powers of \(w\), the formula for \(U\) gives \[\begin{align*} \sqrt2\,\langle U(\theta),C(\theta)\rangle_S =\operatorname{Re}\bigl[{} &\tfrac12w^3v^M_+ +w(\tfrac12m^M_+-n^M_+)\\ &-\tfrac12w^{-3}v^M_- +w^{-1}(-\tfrac12m^M_--n^M_-) \bigr]. \end{align*}\] Inside the real part, we may replace each negative-power term by its complex conjugate. The resulting expression is \[ \operatorname{Re}\left[ \frac{w^3}{2}(v^M_+-\overline{v^M_-}) +w\left(\frac{m^M_+}{2}-n^M_+ -\frac{\overline{m^M_-}}{2}-\overline{n^M_-}\right) \right]. \tag{16}\] The first term has frequency one because \(w^3\kappa=w_0^2w\). In the second term, the expression in parentheses has only frequencies \(0\) and \(-2\): for any fixed row \(r\), \[(rM)_+=r_+a-r_-\overline b, \qquad \overline{(rM)_-}=\overline{r_+}\,\overline b+\overline{r_-}a.\] Consequently (16) is a real linear combination of \(\cos\theta\) and \(\sin\theta\), as required. Identifying the pairings. At \(\theta=\alpha\), polarization of Lemma 2 within the plane spanned by \(d(\alpha)\) and \(k\) gives \[\begin{align*} \langle V_0,R_0\rangle_S &=\operatorname{tr}\bigl(SQ(k,z)\bigr),\tag{17}\\ \langle U(\alpha),R_0\rangle_S &=\operatorname{tr}\bigl(SQ(d(\alpha),z)\bigr). \tag{18}\end{align*}\] Both sides of the desired identity \(\langle V_0,C(\theta)\rangle_S=\operatorname{tr}(SQ(k,z))\) are constant, so (17) proves it for every \(\theta\). For the pairing with \(U\), both \(\langle U(\theta),C(\theta)\rangle_S\) and \(\operatorname{tr}(SQ(d(\theta),z))\) are linear combinations of \(\cos\theta\) and \(\sin\theta\). They agree at \(\alpha\) by (18); it remains to compare their derivatives there. The fixed Gram matrix of \(C\) gives \[\langle C(\theta),C(\theta)\rangle_S =\operatorname{tr}(SQ(z)), \qquad \langle R_0,C'(\alpha)\rangle_S=0.\] Constancy of the pairing with \(V_0\) also gives \(\langle V_0,C'(\alpha)\rangle_S=0\). Since \(R_0=\tau U(\alpha)+\rho V_0\) and \(\tau>0\), it follows that \[ \langle U(\alpha),C'(\alpha)\rangle_S=0. \tag{19}\] Next differentiate the identity \[\langle\tau U(\theta)+\rho V_0, \tau U(\theta)+\rho V_0\rangle_S =\operatorname{tr}\bigl(SQ(\tau d(\theta)+\rho k)\bigr),\] which follows from Lemma 2. Evaluating at \(\alpha\) and dividing by \(2\tau\) gives \[\langle U'(\alpha),R_0\rangle_S =\operatorname{tr}\bigl(SQ(d'(\alpha),z)\bigr).\] Together with (19), this is the required derivative identity. A real linear combination of \(\cos\theta\) and \(\sin\theta\) is determined by its value and derivative at a single angle. Hence the pairing with \(U\) has the desired value for all \(\theta\). Linearity in \(\eta,\xi\) completes the proof. ◻ A finite semidefinite liftWe now replace the angular family of pencils by one finite matrix condition. Evaluation on an angular slice will provide a feasible lift for each point of \(K\). Conversely, the functions supplied by Lemma 4 will turn positivity of the finite matrix into the inequalities that characterize the slice, including its singular boundary. The matrix conditionLet \(\mathcal T_j\) be the real vector space of trigonometric polynomials of degree at most \(j\), with basis \[1,\cos\theta,\sin\theta,\ldots,\cos(j\theta),\sin(j\theta).\] Let \(\mathcal E\) be the space of real functions of \((\theta,\mathcal X,\mathcal Z,\eta,\xi)\) that are linear homogeneous in the \(22\) coordinates of \((\mathcal X,\mathcal Z,\eta,\xi)\in \mathbb S^4\times\mathbb S^4\times\mathbb R^2\), with coefficients in \(\mathcal T_7\). Thus \(\dim\mathcal E=22\cdot15=330\). Our lifting variable is a real linear functional \(\Lambda:\mathcal E\to\mathbb R\); on vectors and matrices we apply it entrywise. For an output point \((X,Z,y)\), impose \[ \Lambda(\mathcal X)=X,\qquad \Lambda(\mathcal Z)=Z,\qquad \Lambda\bigl(Y_\theta(\eta,\xi)\bigr)=y. \tag{20}\] We also require \[ \Lambda\bigl(q(\theta)^T L_\theta(\mathcal X,\mathcal Z,\eta,\xi)q(\theta)\bigr)\geq0 \qquad\text{for every }q\in(\mathcal T_2)^{20}. \tag{21}\] The expression to which \(\Lambda\) is applied lies in \(\mathcal E\): the entries of \(L_\theta\) have trigonometric degree at most three, so its product with two entries of \(q\) has degree at most seven. Condition (21) is a single finite semidefinite condition, using the finite test-space construction familiar from moment relaxations (Lasserre 2001, 2009). Indeed, set \((f_1,\ldots,f_5)= (1,\cos\theta,\sin\theta,\cos2\theta,\sin2\theta)\) and define \(H(\Lambda)\in\mathbb S^{100}\) by \[ H(\Lambda)_{(i,a),(j,b)} =\Lambda\bigl(f_a f_b (L_\theta)_{ij}\bigr), \qquad 1\leq i,j\leq20,\quad 1\leq a,b\leq5. \tag{22}\] If \(q_i=\sum_a t_{i,a}f_a\), the left side of (21) is \(t^TH(\Lambda)t\). Hence (21) is equivalent to \(H(\Lambda)\succeq0\). Every entry of this real symmetric matrix is linear homogeneous in the \(330\) coordinates of \(\Lambda\). Proposition 5. For \((X,Z,y)\in\mathbb S^4\times\mathbb S^4\times\mathbb R^3\), the following are equivalent:
Proof. Suppose first that \((X,Z,y)\in K\). Choose \(\theta_0,\eta_0,\xi_0\) with \(y=Y_{\theta_0}(\eta_0,\xi_0)\), and let \(\Lambda\) be evaluation at \((\theta_0,X,Z,\eta_0,\xi_0)\). The output equations hold, and Lemma 3 gives \(L_{\theta_0}(X,Z,\eta_0,\xi_0)\succeq0\), which implies (21). This evaluation functional is a finite feasible witness even when \(X\) is singular. For the converse, suppose that \(\Lambda\) satisfies the stated conditions. Taking constant \(q\) supported on one of the four top blocks of \(L_\theta\) in (21) gives \(X\succeq0\). We next obtain a family of scalar inequalities from which the remaining slice conditions will follow. Fix \(u\in\mathbb R^4\), \(D\in\mathbb S^4\), and \(z\in\mathbb R^3\) with \((z_1,z_2)\neq(0,0)\), and apply Lemma 4 with \(S=J_uDJ_u^T\in\mathbb S^3\). Neither \(D\) nor \(S\) is required to be positive semidefinite. Let \(b_1(\theta),\ldots,b_4(\theta)\) be the columns of the resulting real matrix \(\widetilde C(\theta)\). Choose the five four-dimensional blocks of \(q\) to be \[q(\theta)= \begin{pmatrix} -DJ_u^Tb_1(\theta)\\ \vdots\\ -DJ_u^Tb_4(\theta)\\ u \end{pmatrix}.\] The degree bound in Lemma 4 puts this vector in \((\mathcal T_2)^{20}\). Write \(r_j\) for the columns of \(\widetilde F_\theta(\eta,\xi)\), as in the definition of \(L_\theta\). Using \(B(r_j)u=J_u^Tr_j\), the top blocks and the cross terms of \(q^TL_\theta q\) are respectively \[\begin{align*} \sum_{j=1}^4 b_j^T J_uD\mathcal X DJ_u^Tb_j &=\operatorname{tr}\bigl(J_uD\mathcal X DJ_u^TQ(z)\bigr),\\ -2\sum_{j=1}^4 b_j^TJ_uDJ_u^Tr_j &=-2\langle F_\theta(\eta,\xi),C(\theta)\rangle_S\\ &=-2\operatorname{tr}\bigl(J_uDJ_u^T Q(Y_\theta(\eta,\xi),z)\bigr). \end{align*}\] These identities hold before applying \(\Lambda\). The first identity removes the angular dependence from the coefficient of \(\mathcal X\); the last expression is linear in \(Y_\theta(\eta,\xi)\). Consequently, applying (20) and (21) gives \[ u^TZu+ \operatorname{tr}\bigl(J_uDXDJ_u^TQ(z)\bigr) -2\operatorname{tr}\bigl(J_uDJ_u^TQ(y,z)\bigr)\geq0. \tag{23}\] For fixed \(u,D\), the left side is a polynomial in \(z\), so the inequality extends by continuity to every \(z\in\mathbb R^3\). Only this scalar inequality is extended; no continuity of the functions \(C\) is needed. Thus (23) holds for all \(u\in\mathbb R^4\), \(D\in\mathbb S^4\), and \(z\in\mathbb R^3\). If \(X\) were positive definite, setting \(z=y\) and \(D=X^{-1}\) in (23) would give \(Z\succeq\Phi_y(X^{-1})\). For singular \(X\), the same family of tests also supplies the necessary range condition. Fix slice parameters \(y=Y_{\theta_0}(\eta_0,\xi_0)\) and, for the rest of the proof, let \(r_j\) denote the columns of \(\widetilde F_{\theta_0}(\eta_0,\xi_0)\) and set \(B_j=B(r_j)\). For \(v\in\ker X\), take \(D=t vv^T\) and \(z=y\) in (23), where \(t\in\mathbb R\) is arbitrary. Since \(DXD=0\), this gives \[u^TZu-2t\,(J_uv)^TQ(y)(J_uv)\geq0 \qquad(t\in\mathbb R).\] Its coefficient of \(t\) must vanish. The Gram factorization of \(Q(y)\) therefore yields \[0=(J_uv)^TQ(y)(J_uv) =\sum_{j=1}^4(v^TB_ju)^2.\] As \(u\) and \(v\in\ker X\) are arbitrary, it follows that \[ \operatorname{range}B_j\subseteq(\ker X)^\perp =\operatorname{range}X\qquad(1\leq j\leq4). \tag{24}\] Let \(X^\dagger\) be the symmetric matrix that inverts the positive eigenvalues of \(X\) and is zero on \(\ker X\). In particular, \(X^\dagger XX^\dagger=X^\dagger\). Taking \(D=X^\dagger\) and \(z=y\) in (23), and using (8) and (7), gives \[ Z\succeq\Phi_y(X^\dagger) =\sum_{j=1}^4B_j^TX^\dagger B_j. \tag{25}\] We now verify positivity of the slice pencil directly by completing a square using (24) and (25). For \(a_1,\ldots,a_4,u\in\mathbb R^4\), its quadratic form equals \[\begin{align*} &\sum_{j=1}^4a_j^TXa_j +2\sum_{j=1}^4a_j^TB_ju+u^TZu\\ &\quad=\sum_{j=1}^4 (a_j+X^\dagger B_ju)^TX(a_j+X^\dagger B_ju) +u^T\left(Z-\sum_{j=1}^4B_j^TX^\dagger B_j\right)u \geq0. \end{align*}\] Here the cross terms agree because \(XX^\dagger B_j=B_j\) by (24). Thus \(L_{\theta_0}(X,Z,\eta_0,\xi_0)\succeq0\), and Lemma 3 gives \((X,Z,y)\in K\). ◻ The converse has used an arbitrary feasible functional \(\Lambda\). In particular, the proof requires no representation of \(\Lambda\) by a measure and no closure operation on the projected feasible set. The size of the liftProof of Theorem 1. Choose the displayed trigonometric basis of \(\mathcal T_7\) and the upper-triangular coordinate functions of \(\mathcal X\) and \(\mathcal Z\), together with \(\eta,\xi\), to specify the \(330\) coordinates of \(\Lambda\). The output equations (20) prescribe precisely the following \(23\) distinct coordinates: \[\begin{gather*} \Lambda(\mathcal X_{ij}),\quad \Lambda(\mathcal Z_{ij})\qquad(1\leq i\leq j\leq4),\\ \Lambda(\eta\cos\theta),\quad \Lambda(\eta\sin\theta),\quad \Lambda(\xi). \end{gather*}\] They are independent members of the chosen coordinate system on \(\mathcal E^*\). Substitute the corresponding entries of \(X,Z,y\) for these coordinates in (22), and use the remaining \(330-23=307\) coordinates as auxiliary variables. The result is a fixed real symmetric \(100\times100\) matrix pencil, homogeneous in the output and auxiliary variables. Proposition 5 proves that its projection is exactly \(K\). Hence the requested representation has \(N=100\), \(m=307\), and zero constant term. ◻
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