Proof. The first Bianchi identity says that the traces of the two diagonal blocks of the curvature operator agree. Their sum is half the scalar curvature, hence is \(6\). Contraction identifies the off-diagonal block with the trace-free Ricci tensor: its nine entries give the coefficients in the independent symmetric trace-free matrices \(I_sJ_t\). Thus the Einstein condition makes this block zero, and each diagonal block has trace \(3\). Expanding in the orthonormal two-form bases gives the displayed curvature tensor.

A unit simple two-form has self-dual and anti-self-dual parts of norm \(1/\sqrt2\). Conversely, a unit two-form whose two parts have these norms has exterior square zero, and is simple: in its orthogonal normal form \(\alpha=a\,e^1\wedge e^2+b\,e^3\wedge e^4\), the condition \(\alpha\wedge\alpha=0\) is \(ab=0\). Therefore all pairs of plus and minus parts of norm \(1/\sqrt2\) occur. The smallest sectional curvature is \(1-(x+y)/2\), proving the equivalence.

For the norm comparison, write the eigenvalues of \(T\) as \(-x,a,x-a\). Minimality of \(-x\) gives \(-x\le a\le2x\), and hence \[\frac32x^2 \le x^2+a^2+(x-a)^2 \le6x^2.\] Division by \(6\) gives \(v\le x\le2v\). The proof for \(U\) is identical.

Contracting the differential Bianchi identity shows that the curvature tensor is divergence-free when Ricci is parallel. Project its last pair of indices onto \(\Lambda^+\) or \(\Lambda^-\); these projections are parallel. Subtracting the parallel scalar-curvature contribution proves divergence freedom of each Weyl block. In a quaternionic basis, the resulting constraint on \(\nabla T\) is exactly the stated identity. ◻

Proof. The generators are skew-Hermitian and satisfy \([\Sigma_1,\Sigma_2]=2\Sigma_3\), cyclically. Diagonalize \(\Sigma_3/(2i)\) and take the usual conjugate raising and lowering operators, which shift its weight by one. If a unit highest vector has weight \(j\), commuting the raising and lowering operators gives the squared lowering coefficient at weight \(j-r\) as \[(r+1)(2j-r).\] Nonnegativity and termination of the chain imply that \(2j\) is a nonnegative integer and give the listed weights and Casimir. Taking invariant orthogonal complements decomposes an arbitrary representation. Multiplying the weight lists, and successively removing the chain of highest remaining weight, proves the tensor product rule with multiplicity one. Commuting actions operate on the corresponding multiplicity spaces. The same chain description shows that an intertwiner between single irreducible copies is a scalar.

The actions on the tangent space and on \(\Lambda^+\) give the first two types. The symmetric square of the spin-one representation is spin two plus the scalar trace, giving the third. Finally, these splittings are defined by invariant algebraic operators. The connection induced by the orthonormal frame bundle preserves every such operator. Thus the projections are parallel even though a chosen local frame, or a curvature eigenbasis, need not be parallel. ◻

Proof. In a diagonal basis, \(a_s+x\ge0\) and every \(\Sigma_s^2\) is nonpositive. Thus \[\sum_s a_s\Sigma_s^2 \le -x\sum_s\Sigma_s^2 =4xj(j+1)\operatorname{Id},\] which proves the bound.

To identify \(Q_2\), diagonalize \(S=\operatorname{diag}(d_1,d_2,d_3)\). The speed-two matrix action gives \(|\Sigma_sS|^2=8(d_j-d_k)^2\), with \(s,j,k\) cyclic; these three derivatives are mutually orthogonal. Consequently, allowing \(A\) to be nondiagonal in this basis, \[\begin{aligned} \langle Q_2(A)S,S\rangle &=-\frac13\sum_s A_{ss}(d_j-d_k)^2\\ &=-\frac13\sum_s A_{ss} \bigl(2\operatorname{tr}(S^2)-3d_s^2\bigr) =\operatorname{tr}(AS^2). \end{aligned}\] Here trace-freeness of both matrices gives the middle identity and removes the scalar term. Polarization proves the operator equality.

By Lemma 4, spin two occurs once in \(\operatorname{End}(H_j)\cong H_j\otimes H_j\). Hence an equivariant operator family linear in \(A\) is unique up to scalar. The same is true with an irreducible minus factor when the operators commute with its action. Test the scalar at \(A=\operatorname{diag}(-1,-1,2)\) on a highest vector. At weight \(m\), \[Q_j(A)=\frac{j(j+1)-3m^2}{j(2j-1)},\] so its highest-weight eigenvalue is \(-1\). The \(F\) highest vector also has \(B_F(A)\) eigenvalue \(-1\). A product of highest vectors is the highest vector of the stretched coupling, proving the claimed compression. For the same reason, compression of a single generator from a spin-\(\ell\) factor to a stretched spin-\(j\) product is \(\ell/j\) times the total generator. ◻

Proof. The derivative space \(T^*M\otimes F\) has types \((5/2,1/2)\) and \((3/2,1/2)\). The divergence takes values in \(T^*M\otimes\Lambda^+\), whose plus spins are at most \(3/2\). It is nonzero: a nonzero covector tensored with \(\operatorname{diag}(1,-1,0)\) has nonzero divergence. It therefore vanishes on the first summand and is injective on the second, by irreducibility. Lemma 3 puts \(Z\) in the first summand.

Let \[C=\sum_s I_s\otimes\Sigma_s\] on \(T^*M\otimes F\). Expansion of the combined Casimir gives \(C=-4\) on \(G\) and \(C=6\) on its complement. In particular \(P_G=(6-C)/10\) is a parallel orthogonal projector. The bundle curvature on \(F\) is \[F_{ik}=\frac12\sum_s(1+t_s)I_{s,ik}\Sigma_s\] at a point of diagonalization. Taking the covariant divergence of \(CZ=-4Z\), using skewness of \(I_s\) to take a commutator and \(\sum_{ik}I_{s,ik}I_{r,ik}=4\delta_{sr}\), gives \[-4\Delta T=\sum_s(1+t_s)\Sigma_s^2T =-24T+24B_F(T)T=-24(T-N).\]

To differentiate this equation, note first that \(\nabla N=2B_F(T)Z\). For each derivative index \(j\), the covariant commutation formula reads \[\Delta Z_j =\nabla_j\Delta T+\nabla^i(F_{ij}T) +F_{ij}\nabla^iT +\operatorname{Ric}_{ja}Z_a.\] The divergence of \(F\) vanishes because it is the induced action of the divergence-free Riemann curvature. Thus the differentiated equation contributes \(6Z-12B_F(T)Z\), the Ricci term contributes \(3Z\), and the remaining bundle term is \(-\sum_s(1+t_s)I_s\Sigma_sZ\). Its constant part is \(-CZ=4Z\).

The left side already lies in \(G\), since \(P_G\) is parallel and \(P_GZ=Z\). We may therefore compress the right side to \(G\), although its individual summands need not preserve \(G\). Lemma 5 gives \(P_GB_F(T)P_G=Q_{5/2}(T)\). On \(G\), the total plus generators are \(I_s+\Sigma_s\), so \[40Q_{5/2}(T) =P_G\sum_s t_s(I_s+\Sigma_s)^2P_G.\] The \(I_s^2\) term vanishes by \(\sum_s t_s=0\), and the \(\Sigma_s^2\) term compresses to \(24Q_{5/2}(T)\). Hence \[P_G\sum_s t_s I_s\Sigma_sP_G=8Q_{5/2}(T).\] The scalar contributions sum to \(13\), while the two curvature contributions sum to \(-20Q_{5/2}(T)\), proving [eq:1]. For a nondiagonal \(T\) the same calculation uses the symmetric contraction \(\sum_{r,s}T_{rs}(I_r+\Sigma_r)(I_s+\Sigma_s)\): its \(I_rI_s\) part is zero by symmetry and trace-freeness. Thus the computation is tensorial and entails no differentiation of a moving eigenframe. ◻

Proof. Write \(D_i=Z_{ii}\) and \(a_i=I_iZ_{jk}\). Quaternion multiplication turns the three divergence equations into \(D_i+a_j-a_k=0\). Trace-freeness and the orthogonal basis \(s,\widetilde s\) give uniquely \[p=\frac16\sum_i s_iD_i,\qquad k=-\frac1{6\sqrt3}\sum_i\widetilde s_iD_i,\qquad D_i=s_ip-\sqrt3\,\widetilde s_i k.\] Since \(s_j-s_k=\sqrt3\,\widetilde s_i\) and \(\widetilde s_j-\widetilde s_k=-\sqrt3\,s_i\), the vectors \(s_i k+\widetilde s_i p/\sqrt3\) have exactly the differences required of the \(a_i\). Their difference from \(a_i\) is therefore a common vector \(l\), proving the formulas and their converse. This also establishes their full algebraic scope.

Orthogonality of \(s,\widetilde s\) gives \[\sum_i|Z_{ii}|^2=6|p|^2+18|k|^2,\qquad 2\sum_{j<k}|Z_{jk}|^2=4|p|^2+12|k|^2+6|l|^2.\] The factor two in the second sum is the Hilbert–Schmidt multiplicity of off-diagonal entries. Division by six proves the norm identity. For \(Z=\nabla T\), \(\nabla|T|^2=2v\sum_i s_iZ_{ii}=12vp\); comparison with \(|T|^2=6v^2\) identifies the parameter \(p\) with \(\nabla v\). ◻

Proof. On \(\{v>0\}\), pair the equation for \(\Delta T\) with \(T\) to obtain \[\Delta(v^2)=12(v^2-\sigma v^3)+2e.\] This is globally an identity of smooth functions. Put \(v_\varepsilon=\sqrt{v^2+\varepsilon^2}\). Direct differentiation gives \[\Delta v_\varepsilon =\frac{6(v^2-\sigma v^3)}{v_\varepsilon} +q_\varepsilon,\qquad q_\varepsilon=\frac e{v_\varepsilon} -\frac{v^2|p|^2}{v_\varepsilon^3}.\] The refined derivative estimate in [eq:2] yields \[\frac25\frac e{v_\varepsilon} \le q_\varepsilon\le\frac e{v_\varepsilon}.\] These assertions hold almost everywhere, including the zero set: every smooth component of \(T\) has derivative zero almost everywhere on its zero set, hence \(Z=0\) almost everywhere on \(\{T=0\}\). One can see this elementary fact at density points of a level set; a nonzero derivative would instead make that scalar level set a local hypersurface.

The first term in \(\Delta v_\varepsilon\) is uniformly bounded in absolute value by \(6(v+v^2)\). Its integral is the negative of the integral of \(q_\varepsilon\). Consequently \(\mathbf E(e/v_\varepsilon)\) is uniformly bounded. Monotone convergence proves the asserted \(L^1\) bound for \(e/v\). Off the zero set, \(q_\varepsilon\) is dominated by \(e/v\) and tends to \((e-|p|^2)/v\); on the zero set both are zero almost everywhere. Dominated convergence therefore gives an \(L^1\) limit, rather than only a weak limit of measures. The zeroth-order term converges in \(L^1\) as well. Since \(v_\varepsilon\to v\) uniformly, their Laplacians converge distributionally, proving [eq:3] without a residual singular measure.

Finally, approximate a bounded Lipschitz test function uniformly and in \(H^1\) by smooth functions, using local mollification and a partition of unity. Uniform convergence controls its pairing with the \(L^1\) right side, and \(H^1\) convergence controls its gradient pairing with \(\nabla v\). This proves the asserted testing rule. ◻

Proof. First let \(\phi\) be smooth and write \(S=\operatorname{Hess}_0\phi\). The integrated Bochner identity and \(\operatorname{Ric}=3g\) give \[\mathbf E\bigl[(\Delta\phi)^2\bigr] =\mathbf E\left(4|\nabla\phi|^2+\frac43|S|^2\right).\] Define operators on covariant two-tensors by \[(T_D)_{ikjl}=\frac12\sum_s t_s I_{s,ik}I_{s,jl}, \qquad (U_D)_{ikjl}=\frac12\sum_s u_s J_{s,ik}J_{s,jl},\] where \(ij\) is the row index and \(kl\) is the column index. These tensorial definitions may be evaluated after pointwise diagonalization. They annihilate metric trace. On the symmetric trace-free tensors, the nine matrices \(I_iJ_j\) form an orthogonal basis, with \(T_D\) and \(U_D\) eigenvalues \(t_i\) and \(u_j\). For example, \(T_D\) acts by \(-\frac12\sum_s t_s I_s(\,\cdot\,)I_s\); the quaternion relations give the eigenvalue \(t_i\) on \(I_iJ_j\). Thus \[T_D+U_D\le 2(x+y)\operatorname{Id}\le4\operatorname{Id}\] on this space.

The integrated curvature contraction has a useful sign and a precise coefficient. Since \(T_D\) annihilates trace, replace \(S\) by \(\operatorname{Hess}\phi\) in its quadratic form. Integrate the derivative \(k\), use \(\nabla_k(T_D)_{ikjl}=0\), and use skewness in \(i,k\) to obtain \[\begin{aligned} \mathbf E\langle T_DS,S\rangle &=-\mathbf E (T_D)_{ikjl}\phi_l\nabla_k\nabla_i\phi_j\\ &=\frac12\mathbf E (T_D)_{ikjl}R_{ikja}\phi_a\phi_l. \end{aligned}\] In this contraction the minus curvature block is orthogonal to the plus block. Substituting their expansions gives \[\frac12\sum_{ikj}(T_D)_{ikjl}R_{ikja} =\frac12\sum_s t_s(1+t_s) \sum_j I_{s,jl}I_{s,ja} =3v^2\delta_{la}.\] The final equality uses \(\sum t_s=0\) and \(\sum t_s^2=6v^2\). Consequently \[\mathbf E\langle(T_D+U_D)S,S\rangle =3\mathbf E\bigl[(v^2+b^2)|\nabla\phi|^2\bigr].\] The pointwise upper operator bound, inserted into the positive \(|S|^2\) term of the Bochner identity, now yields \[\mathbf E\bigl[(\Delta\phi)^2\bigr]\ge\mathbf E K|\nabla\phi|^2.\]

For smooth \(a\), the Poisson equation on a closed connected manifold gives a unique smooth zero-mean solution of \(\Delta\phi=a-\mathbf E a\). Integration by parts and Cauchy–Schwarz with the pointwise weight \(K\) give \[\begin{aligned} \operatorname{Var}(a) &=-\mathbf E\langle\nabla a,\nabla\phi\rangle\\ &\le \left(\mathbf E\frac{|\nabla a|^2}{K}\right)^{1/2} \left(\mathbf E K|\nabla\phi|^2\right)^{1/2}\\ &\le \left(\mathbf E\frac{|\nabla a|^2}{K}\right)^{1/2} \operatorname{Var}(a)^{1/2}. \end{aligned}\] Division proves the assertion when the variance is nonzero; the zero case is immediate. Smooth approximation in \(H^1\) completes the proof: \(K\) is bounded above and below by positive constants, so the weighted energy also converges. ◻

Proof. Put \(A_\varepsilon=v^2+\varepsilon^2\) and \(f_\varepsilon=A_\varepsilon^{1/6}\). These are smooth functions. The equation for \(\Delta(v^2)\) and [eq:2] give \[\begin{aligned} \Delta f_\varepsilon =A_\varepsilon^{-5/6} \left[ 2(v^2-\sigma v^3)+\frac13e -\frac59\frac{v^2}{A_\varepsilon}|p|^2 \right] \ge 2A_\varepsilon^{-5/6}(v^2-v^3). \end{aligned}\] Indeed, \(\sigma\le1\), \(v^2/A_\varepsilon\le1\), and \(e\ge5|p|^2/3\). Multiply by \(f_\varepsilon\) and integrate: \[\mathbf E|\nabla f_\varepsilon|^2 \le 2\mathbf E\frac{v^3-v^2}{A_\varepsilon^{2/3}} \le 2\mathbf E v^{5/3}.\] Their \(L^2\) norms are uniformly bounded as well. Thus \(f_\varepsilon\) is bounded in \(H^1\), and its uniform limit \(f=v^{1/3}\) belongs to \(H^1\) by weak compactness. This bound is obtained directly from the regularized equation; it requires no positive lower bound on \(v\).

Apply Proposition 10 to \(f_\varepsilon\) and use the preceding energy bound: \[\begin{aligned} \bigl(\mathbf E f_\varepsilon^4\bigr)^{1/2} &\le \mathbf E\left[ A_\varepsilon^{1/3} +\frac{v^3-v^2}{A_\varepsilon^{2/3}}\right]\\ &=\mathbf E\frac{\varepsilon^2+v^3}{A_\varepsilon^{2/3}}. \end{aligned}\] The term \(\varepsilon^2/A_\varepsilon^{2/3}\) is at most \(\varepsilon^{2/3}\), while \(v^3/A_\varepsilon^{2/3}\le v^{5/3}\). Dominated convergence therefore gives \[\bigl(\mathbf E f^4\bigr)^{1/2} \le\mathbf E(f^2v) \le\bigl(\mathbf E f^4\bigr)^{1/2} \bigl(\mathbf E v^2\bigr)^{1/2}.\] When \(T\not\equiv0\), the common factor is positive, proving [eq:5].

If \(\mathbf E v^2=1\), equality holds in Cauchy–Schwarz. Thus \(v=\alpha f^2\) almost everywhere for a constant \(\alpha>0\). It follows that \(v\) takes only the values \(0\) and \(\alpha^3\) almost everywhere, and hence everywhere by continuity. Connectedness and \(T\not\equiv0\) force the positive constant value on all of \(M\). The moment condition gives \(v=1\). Equation [eq:3] then becomes \(0=6(1-\sigma)+e\). Both terms are nonnegative, so \(\sigma=1\) and \(e=0\). Smoothness gives \(\nabla T=0\) everywhere. ◻

Proof. Simple connectivity gives \(b_1(M)=0\), and Poincaré duality gives \(b_3(M)=0\). Hodge theory identifies \(b_2=n_++n_-\) and the signature with \(n_+-n_-\), proving the first two identities. The Chern–Gauss–Bonnet and signature formulas for an Einstein four-manifold are \[8\pi^2\chi=\int_M\left(|T|^2+|U|^2+\frac{R^2}{24}\right) \,d\operatorname{vol},\qquad 12\pi^2\tau=\int_M(|T|^2-|U|^2)\,d\operatorname{vol},\] where \(R=12\) is the scalar curvature and the Weyl norms are the Hilbert–Schmidt operator norms used here; compare [20] and [11]. Substitution of \(|T|^2=6v^2\) and \(|U|^2=6b^2\) proves [eq:6]. ◻

Proof. Fix a center \(p\) and take all vectors of length less than \(\pi\) whose radial geodesics have no conjugate point at or before their endpoints. This is an open set: the Jacobi determinant divided by \(t^3\), continued as \(1\) at zero, has a positive minimum on each such radial interval. Every point is the exponential image of a minimizing vector of length at most \(\pi\) by the diameter bound. A minimizing vector is in this open set unless it is critical for \(\exp_p\), has length \(\pi\), or is zero.

The exponential map is smooth with bounded derivative on the closed radius-\(\pi\) ball. The area formula with multiplicity states that the integral of its absolute Jacobian over a measurable set equals the integral of the number of preimages over its image. In particular, critical vectors and the boundary sphere have images of zero four-dimensional volume: their respective Jacobian integral and domain volume are zero. On the open preconjugate set the determinant is positive. For a vector \(t\xi\), the Cartesian Jacobian of \(\exp_p\) is \(D(t,\xi)/t^3\) on this set. The area formula therefore bounds \(V_M\) by the integral of this Jacobian, with multiplicities only increasing that integral. Nonconjugate cut vectors are included in the set; no identification of cut points with conjugate points is used. Polar coordinates give the claimed bound at each center, since the area of the unit three-sphere is \(2\pi^2\). Average over centers and use Tonelli’s theorem. ◻

Proof. Define matrices without using inverses by \[Y_0=0,\quad Y_1=\mathrm{Id},\qquad Y_{i+1}=(2\mathrm{Id}-a^2\mathcal R(ia))Y_i-Y_{i-1}.\] The position \(aY_i\) and backward velocity \(Y_i-Y_{i-1}\) evolve by first changing velocity by \(-a\mathcal R(ia)(aY_i)\) and then changing position by \(a\) times the new velocity. On the exact Jacobi solution one step has error \(O(a^2)\), while the update amplifies errors by at most \(1+Ca\). Iteration proves convergence of position and velocity, uniformly on the fixed interval. The constants are uniform under the stated coefficient bounds.

Suppose first that \(t\) is preconjugate. Away from zero the exact Jacobi matrix is invertible with uniformly bounded inverse on compact time intervals, so the converging \(aY_i\) are invertible there. Near zero use the identity, still without inverses, \[Y_i=i\mathrm{Id}-a^2\sum_{r<i}(i-r)\mathcal R(ra)Y_r.\] For sufficiently small fixed \(ia\) it gives \(\max_{r\le i}|Y_r|\le Ci\) by absorbing the error, and then \(Y_i/i=\mathrm{Id}+O((ia)^2)\). Thus all the early matrices are invertible as well. All inverses needed for elimination have now been obtained independently of that elimination.

The Schur blocks are \[S_i=Y_{i+1}Y_i^{-1} =2\mathrm{Id}-a^2\mathcal R(ia)-S_{i-1}^{-1}.\] They are symmetric, starting with \(S_1=2\mathrm{Id}-a^2\mathcal R(a)\). Near zero they differ from \(((i+1)/i)\mathrm{Id}\) by \(O((ia)^2+a^2)\); away from zero, convergence of position and velocity gives \(S_i=\mathrm{Id}+O(a)\). Choose the small initial interval first, and then \(N\) large. All Schur blocks are positive definite. Block elimination proves positivity of \(\mathcal L_N\) and gives \[\det\mathcal L_N=\prod_{i=1}^{N-1}\det S_i=\det Y_N.\] The desired limit follows. The determinant identity holds for arbitrary coefficients by continuity, since invertible Schur blocks form a dense set; convergence therefore gives the limit without the preconjugacy assumption as well. ◻

Proof. Compare \(\mathcal L_N\) with the matrix for \(\mathcal R=\mathrm{Id}\), whose scalar sine-mode eigenvalues are \(2-2\cos(k\pi/N)-a^2\). They are positive for large \(N\) because \(t<\pi\). Normalize \(\mathcal L_N\) by the inverse square root of this comparison matrix. The resulting positive matrix has trace \(3\) on every diagonal sine-mode block, since \(\operatorname{tr}A(ia)=0\).

Split off its first three-dimensional mode and write the normalized matrix as \(\left(\begin{smallmatrix}P&C\\C^*&Q\end{smallmatrix}\right)\). Its Schur complement is positive and bounded above by \(Q\). Thus \[\det\begin{pmatrix}P&C\\C^*&Q\end{pmatrix} =\det P\,\det(Q-C^*P^{-1}C) \le\det P\,\det Q\le\det P.\] The last inequality is arithmetic–geometric mean, since \(\operatorname{tr}Q=\dim Q\). The first-mode compression tends to \(\mathrm{Id}-\beta B\), because its normalized sine vector has entries \(\sqrt{2/N}\sin(\pi i/N)\) and \[\frac{a^2}{2-2\cos(\pi/N)-a^2}\longrightarrow\beta.\] Lemma 16 gives the determinant ratio limit \(D(t)/\sin^3t\). Positive compressions have a positive semidefinite limit, and no inverse of that limiting compression is used. This proves the first comparison.

The eigenvalues of \(B\) lie in \([-1,2]\) and sum to zero, so \(|B|^2\le6\). Before conjugacy the determinant loss is \[1-\det(\mathrm{Id}-\beta B) =\frac{\beta^2}{2}|B|^2+\beta^3\det B.\] Only \(\det B<0\) is unfavorable. Write the eigenvalues then as \(-r,a_1,a_2\), where \(a_1,a_2\ge0\) and \(a_1+a_2=r\). One has \[|\det B|\le\frac r6|B|^2,\qquad r\le1,\qquad\beta r\le2.\] The first inequality follows from \((a_1-a_2)^2\ge0\); the last follows from \(\beta a_i\le1\). These inequalities give the displayed loss bound. If \(\det B\ge0\) the same bound is immediate. After conjugacy its left side is \(1\), so \(|B|^2\le6\) gives it directly. ◻

Proof. The elementary bounds \(a_0<\pi<b_0\) and the alternating sine expansion through the negative seventh-degree term give \(S(z)\le\sin(\pi z)\) on this interval. Also \[S(z)\ge z\left(a_0-\frac{b_0^3}{24} -\frac{b_0^7}{322560}\right)\ge\frac95z\ge0.\] Expanding the four powers gives the displayed rational coefficients. ◻

Proof of Proposition 13. For the sine-squared probability weight defining \(B\), its time variance is \(t^2(1/12-1/(2\pi^2))\). The variance identity and [eq:8] give \[\mathcal E|B|^2\ge\mathcal E|A|^2 \left(1-\frac32t^2\left(\frac1{12}-\frac1{2\pi^2}\right)\right).\] Indeed, parallel identification along a geodesic and Cauchy–Schwarz show \(\mathcal E|A(s)-A(r)|^2\le(s-r)^2\mathcal E|A'|^2\). With \(u=t/\pi\) the factor in parentheses is at least \(1-u^2/2\). For \(0\le u<1\), furthermore, \[f(u)\le\min\{1,\beta^2\max(1,3-\beta)\},\qquad \beta=\frac{u^2}{1-u^2}.\] At \(u=1\) we give the capped comparison function its continuous value \(1\). For \(u\le5/8\), comparison with \(\beta^2(3-\beta)\) reduces to \(6-13u^2+6u^4\ge0\). On this interval \(f\) is increasing and \(f(5/8)=7511875/8388608<1\), which also verifies the cap by \(1\). The comparison function \(\min\{1,\beta^2\max(1,3-\beta)\}\), as a function of \(u\), is nondecreasing and remains \(1\) after it first reaches \(1\); at \(u=5/8\) its value is \(57500/59319>15/16\), which handles the rest of the interval. Lemmas 15 and 17 therefore show that the coefficient of \(s_0\) saved from the round volume is at least \(\eta_+\). The three rational bounds above give \(\eta_+>2/25>3/40\), proving [eq:7]. ◻

Proof. The two Weyl gaps give \(s_0\ge2\), while \(v+b\le2\) gives \(v^2+b^2\le4\). If \(s_0=4\), continuity forces equality at every point. The only possible norm pairs would then be \((v,b)=(2,0)\) and \((0,2)\). Connectedness makes the pair constant, contradicting that neither block vanishes identically. Hence \(2\le s_0<4\). Equations [eq:6] and [eq:7] give \[\chi\le2(1-3s_0/40)(1+s_0)<7, \qquad \frac{|\tau|}{\chi} =\frac23\frac{|\mathbf E v^2-\mathbf E b^2|}{1+s_0} \le\frac23\frac{s_0-2}{1+s_0}<\frac4{15}.\] Here the first polynomial is increasing on \([2,4]\) and equals \(7\) at \(4\). The integers \(\chi=2+n_++n_-\) and \(\tau=n_+-n_-\) have the same parity. If \(\tau\ne0\), the only possibility is \(\chi=5\), \(|\tau|=1\). Exchange the blocks so that \(P=\mathbf E b^2\) is the smaller moment. The ratio in [eq:6] then gives \[\mathbf E v^2=\frac{13P+3}{7}.\] Combining this with \(v^2\le(2-b)^2\) yields \[P\ge1,\qquad28\mathbf E b+6P\le25.\]

Integrating [eq:3] for \(b\), using \(\bar\sigma\le1\) and \(\bar e\ge5|\nabla b|^2/3\), gives \[P-\mathbf E b\ge\frac49\mathbf E|\nabla\sqrt b|^2.\] To justify the square root at zeros, put \(c_\varepsilon=\sqrt{b+\varepsilon}\). Then \(|\nabla c_\varepsilon|^2=|q|^2/(4(b+\varepsilon))\) is dominated off \(\{b=0\}\) by the integrable function \(|q|^2/(4b)\le3\bar e/(20b)\), and is zero almost everywhere on that set. The gradient vectors converge in \(L^2\) by dominated convergence to \(q/(2\sqrt b)\) off the zero set and to zero on it. Together with uniform convergence of \(c_\varepsilon\), this identifies the weak derivative of \(c=\sqrt b\in H^1\) and gives \(|\nabla c|^2=|q|^2/(4b)\) under our zero-set convention. Put \[G(c)=c+\frac{c^5}{40}-\frac{c^9}{1152},\qquad H(b)=G(c).\] The range of \(c\) is bounded, and \(G\) has bounded derivative there, so the Sobolev chain rule applies, including at \(c=0\). On \(0\le b\le2\), \[G'(c)=1+\frac{b^2}{8}-\frac{b^4}{128} \in\left[0,\frac{\sqrt{4+b^2}}2\right] \subseteq\left[0,\frac{\sqrt K}2\right].\] For the first upper bound, put \(z=b^2/4\in[0,1]\) and square the positive quantity \(1+z/2-z^2/8\); its square is \(1+z-z^3/8+z^4/64\le1+z\). The variance inequality [eq:4], extended by Sobolev approximation, therefore gives \[\operatorname{Var}(H(b)) \le\frac14\mathbf E|\nabla\sqrt b|^2 \le\frac9{16}(P-\mathbf E b).\] Equivalently, \[J:=16(\mathbf E H(b))^2-\mathbf E\Phi(b)\ge0, \qquad \Phi(b)=16H(b)^2+9b(1-b). \tag{9}\] The distribution of \(b\) is a probability measure on \([0,2]\) satisfying the two moment constraints. Lemma 19 contradicts [eq:9]. Thus \(\tau=0\), and [eq:6] gives equality of the two second moments. ◻

Proof of Lemma 19. The feasible measures form a nonempty weakly compact convex set. Suppose the continuous functional \(J(\mu)=16(\int H\,d\mu)^2-\int\Phi\,d\mu\) has a nonnegative maximum, attained at \(\mu\), and set \(d=\int H\,d\mu\). First \(d\ge11/20\). Indeed, \[1+z^2/40-z^4/1152\ge1+(31/1440)z^2,\] and \[25+\frac{31}{45}z^2-\frac{346}{25}z \ge\frac{17}{225}>0\qquad(0\le z\le2).\] The last quadratic is decreasing on this interval and has the stated value at \(2\). Squaring the first bound in the definition of \(H\) gives \[\Phi(z)-\frac{121}{25}z^2 \ge z\left(25+\frac{31}{45}z^2-\frac{346}{25}z\right) \ge\frac{17}{225}z\ge0.\] Thus \(\Phi(z)\ge16(11/20)^2z^2\). The moment constraint and \(J(\mu)\ge0\) give \(16d^2\ge16(11/20)^2\), and \(H\ge0\) gives \(d\ge11/20\).

For another feasible measure \(\nu\), direct expansion with \(\ell=32dH-\Phi\) gives \[J((1-t)\mu+t\nu)-J(\mu) =t\left(\int\ell\,d\nu-\int\ell\,d\mu\right) +16t^2\left(\int H\,d\nu-d\right)^2.\] For \(t\downarrow0\), maximality shows that \(\mu\) maximizes the linear functional \(\int\ell\). This conclusion does not require concavity of \(J\).

Put \(g_1(z)=28z+6z^2-25\) and \(g_2(z)=1-z^2\). The measure with mass \(9/32\) at \(2\) and the rest at \(0\) has respective constraint expectations \(-5/2\) and \(-1/8\), so both constraints are strictly feasible. The compact convex set of triples \[\left\{\left(\int g_1\,d\nu,\int g_2\,d\nu,\int\ell\,d\nu\right): \nu\text{ a probability measure on }[0,2]\right\}\] is disjoint from the open set of triples with first two coordinates negative and third coordinate greater than \(\int\ell\,d\mu\). Separation gives coefficients \(-\alpha,-\gamma,\beta\), with \(\alpha,\gamma,\beta\ge0\). The last coefficient cannot be zero: otherwise the strictly feasible triple would violate separation. Normalize \(\beta=1\). Evaluating on \(\mu\) and on point masses shows that \(\mu\) is supported on the global maxima of \[F_d(z)=32dH(z)-\Phi(z) -\alpha(28z+6z^2)-\gamma(1-z^2), \qquad \alpha,\gamma\ge0.\] For completeness, separation bounds \(\int(\ell-\alpha g_1-\gamma g_2)\,d\nu\) by \(\int\ell\,d\mu\) for every \(\nu\); for \(\mu\) the reverse inequality follows from its nonpositive constraints. Equality and then point masses give precisely the asserted support condition.

We next verify that \(F_d'''(z)>0\) on \((0,2]\). Set \[L(z)=\frac38+\frac{3z^2}{64}-\frac{35z^4}{1024},\qquad V(z)=\frac{24}{5}-\frac{16z^2}{15}-\frac{7z^4}{48} +\frac{7z^6}{1152}.\] Direct differentiation gives \(H'''(z)=L(z)z^{-5/2}\) and \(\Phi'''(z)=V(z)\). For \(x=\sqrt{z/2}\), the polynomial inequalities \[L(2x^2)>0,\qquad D_*(x):=\frac{88}{5}L(2x^2)-\frac{17}{3}x^5V(2x^2)>0 \qquad(0\le x\le1)\] are certified below. Since \(32d\ge88/5\) and \(4\sqrt2<17/3\), they imply \(32dL(z)-z^{5/2}V(z)>0\) when \(V(z)\ge0\); when \(V(z)<0\) it is immediate. The constraint polynomials have zero third derivative, proving the claim.

Also \(F_d'(0+)=+\infty\). Since \(F_d''\) is strictly increasing, \(F_d'\) can cross from positive to negative at most once. There is at most one interior local maximum; an interval of maxima is impossible. The endpoint \(0\) is not a maximum, so the support contains at most one interior point and the endpoint \(2\). A point mass is infeasible: its first constraint would give \(z\ge1\), and then \(28z+6z^2\ge34\). Thus the support is \(\{w,2\}\), with weight \(t\) at \(2\). Necessarily \(0<w<1\), and the moment constraints become \[t_-(w):=\frac{1-w^2}{4-w^2}\le t\le t_+(w):=\frac{25-28w-6w^2}{80-28w-6w^2}.\] The denominators are positive. Nonemptiness reduces to \((2-w)(10-37w)\ge0\), hence \(0<w\le10/37\).

For fixed \(w\), \(J\) is a convex quadratic in \(t\), so it is bounded above by the larger of its two endpoint values. Those values are negative: \[16t(1-t)\bigl(H(2)-H(w)\bigr)^2 >9\bigl((1-t)(w^2-w)+2t\bigr) \qquad(t=t_-(w),\ t_+(w)).\] To verify these inequalities exactly, put \[x=\sqrt{w/2}\le\frac{37}{100},\qquad R_*(x)=1+\frac{x^4}{10}-\frac{x^8}{72}.\] For each of \[(A_-,B_-)=(1-w^2,4-w^2),\qquad (A_+,B_+)=(25-28w-6w^2,80-28w-6w^2),\quad w=2x^2,\] the required inequality, after multiplication by its positive \(B_\pm^2\), is \(P_\pm(x)>0\), where \[P_\pm(x)=32A_\pm(B_\pm-A_\pm) \bigl(R_*(1)-xR_*(x)\bigr)^2 -9\bigl((B_\pm-A_\pm)(w^2-w)+2A_\pm\bigr)B_\pm.\] This uses \(H(2)-H(w)=\sqrt2(R_*(1)-xR_*(x))\). The endpoint \(w=10/37\), at which the weight interval can collapse, is included in the following strict certificates.

Here are all the coefficient bounds used in this proof. If \(p(r+(s-r)y)=\sum_{j=0}^n a_jy^j\), define \[h_i=\sum_{j=0}^i a_j\frac{\binom{i}{j}}{\binom nj} \quad(0\le i\le n).\] Binomial expansion gives \(p(r+(s-r)y)=\sum_i h_i\binom ni y^i(1-y)^{n-i}\); thus positive \(h_i\) give positivity on \([r,s]\). If the original coefficients are \(p(x)=\sum p_kx^k\), they are obtained explicitly from \[a_j=(s-r)^j\sum_{k=j}^n p_k\binom{k}{j}r^{k-j}.\] The degrees and strict lower bounds for the resulting coefficients are

For direct reproduction, the first two polynomial inputs are \[L(2x^2)=\frac38+\frac3{16}x^4-\frac{35}{64}x^8, \qquad V(2x^2)=\frac{24}{5}-\frac{64}{15}x^4-\frac73x^8+\frac7{18}x^{12}.\] For example the least coefficient of \(L(2x^2)\) is \(1/64\); the least coefficient of \(D_*\) over its four intervals is \(2459668689/5793382400>2/5\). For \(P_-\) and \(P_+\) the respective least coefficients are \[\begin{aligned} \frac{58454914732902747377800095783827783501884053} {50000000000000000000000000000000000000000000}&>1,\\ \frac{132428032291164570287168821949356607977047} {770000000000000000000000000000000000000}&>1. \end{aligned}\] All inputs and endpoints in these coefficient calculations are rational. This proves the derivative and endpoint signs. The strict endpoint inequalities say exactly that \(J(t_-(w)),J(t_+(w))<0\), contradicting its assumed nonnegative maximum. ◻

Proof. The product rule gives four inequivalent, multiplicity-one summands \[\mathcal H=(3,1)\oplus(3,0)\oplus(2,1)\oplus(2,0).\] Skewing the derivative slots takes values in \[\Lambda^2T^*M\otimes F=(3,0)\oplus(2,0)\oplus(1,0)\oplus(2,1).\] For the actual Hessian \(X=\nabla\nabla T\), its skew difference is \([\nabla_i,\nabla_j]T=F_{ij}T\), where \(F_{ij}\) is the curvature of the bundle \(F\). It is essential that the tensor on which this curvature acts is the same Weyl block \(T\) that occurs in \(F_{ij}\). The minus rotation algebra acts trivially on \(F\), so there is no minus component. The dependence on \(T\) is an equivariant linear term and an equivariant term in its symmetric square. The latter has plus spins \(4,2,0\); its only common type with \(\Lambda^+\otimes F=(3,0)\oplus(2,0)\oplus(1,0)\) is \((2,0)\). The linear term has that type as well. Thus the skew commutator has neither a \((3,0)\) nor a \((2,1)\) component. This conclusion is specific to the Hessian of \(T\), not to the derivative of an arbitrary section of \(G\).

Skewing is injective on each of those two domain summands. The \((3,0)\) summand is entirely skew: its stretched plus coupling makes the two derivative plus spin halves symmetric, while their minus coupling to spin zero is alternating. To treat \((2,1)\), suppose that a tensor of this type is symmetric in its full derivative slots. Its derivative minus pair has spin \(1\) and is symmetric, so its derivative plus pair must also be symmetric and have spin \(1\). The inner Bianchi constraint makes the coupling operator \(\sum I_s^{(j)}\Sigma_s\) equal to \(-4\). To see that symmetry gives the same value for the outer coupling, let \(S\) interchange the derivative slots. Then \(C^{(i)}=SC^{(j)}S\); hence \(SX=X\) and \(C^{(j)}X=-4X\) imply \(C^{(i)}X=-4X\). The total plus Casimir would therefore be \[-8-24+2(-4-4)=-48,\] whereas total spin \(2\) requires \(-24\). This contradiction proves injectivity. Moreover, the two images lie in distinct target isotypic components, so they cannot cancel one another. The corresponding components of \(X\) must vanish separately.

For the trace calculation, let \(K:F\to T^*M\otimes T^*M\otimes F\) be \(K(S)_{ij}=\delta_{ij}S\). The adjoint of the restricted trace is \(\operatorname{Tr}^*=P_{G}^{(j)}K\), where the projector acts on the inner derivative and matrix slots. Now \[\operatorname{Tr}K=4\mathrm{Id}, \qquad \operatorname{Tr}\bigl(C^{(j)}K(S)\bigr) =\sum_{i,s}(I_s)_{ii}\Sigma_sS=0.\] Thus \(\operatorname{Tr}\operatorname{Tr}^*=12\mathrm{Id}/5\). By equivariance, trace vanishes on \((3,1)\) and is supported on the unique \((2,0)\) summand. The displayed nonzero composition makes its restriction to that five-dimensional summand an isomorphism onto \(F\). It therefore determines the whole of \(D_0\), including its skew part, and gives both the formula and the norm identity. One can also see explicitly that the trace component retains the curvature commutator. At a point, its adjoint is \[(\operatorname{Tr}^*S)_{ij} =\frac35\delta_{ij}S+\frac1{10}\sum_s I_{s,ij}\Sigma_sS.\] Hence \[(D_0)_{ij}-(D_0)_{ji} =\frac1{12}\sum_s I_{s,ij}\Sigma_s\Delta T =\frac12\sum_s(1+t_s)I_{s,ij}\Sigma_sT.\] For the last equality use \(\Delta T=6(T-N)\) and, in a pointwise diagonal frame, \(\Sigma_sN=-t_s\Sigma_sT\). The invariant version is \(\Sigma_rN=-\sum_sT_{rs}\Sigma_sT\); no eigenframe has been differentiated. Thus the displayed skew difference is exactly the bundle-curvature action.

Finally, the type projections are invariant expressions in the combined Casimirs, and trace is a metric contraction. They are parallel and do not depend on any curvature eigenframe. The spin labels describe tensor representations of the oriented frame group; no spin structure is used. ◻

Proof. The total plus generators preserve each joint type. Similarly, \(\sum_su_s(J_s^{(i)}+J_s^{(j)})^2=4U_D\), so \(U_D\) vanishes on minus spin zero and is \(Q_1(U)\) on minus spin one. Its latter spectrum is the spectrum of \(U\). The displayed combined formula for \(\mathcal P\) therefore preserves the joint types. On \((3,1)\), \[\mathcal P\ge4-2x-2y\ge0.\] On \((2,0)\) the denominator in \(Q_2\) is \(24\), so under the trace identification \(\widetilde Q=(2/5)Q_2(T)=(2/5)B_F(T)\) and \(U_D=0\). This proves the stated trace action. ◻

Proof. By Lemma 22, \[\mathbf E\left[ \frac{m}{6}\left\langle \mathcal P\left(Y+\frac{(L_m)_Y}{2m}\right), Y+\frac{(L_m)_Y}{2m}\right\rangle\right]\ge0.\] The last term in its expansion is the last term of [eq:10]. Since \(\mathcal P\) preserves the type splitting and \(X=Y+D_0\), the other terms are \[\frac16\mathbf E\bigl[ m\langle\mathcal PX,X\rangle+\langle\mathcal PX,L_m\rangle -m\langle\mathcal PD_0,D_0\rangle-\langle\mathcal PD_0,L_m\rangle \bigr].\] We compute these terms for smooth \(m\) first. Compression may be omitted inside any of the displayed inner products, because both arguments belong to \(\mathcal H\).

Proof. Expanding the diagonal and off-diagonal matrix entries in [eq:2] gives \[\frac{B}{v}\ \longleftrightarrow\ \begin{pmatrix} 4\sigma/3&-4\rho/\sqrt3&0\\ -4\rho/\sqrt3&-4\sigma&-1\\ 0&-1&0 \end{pmatrix}, \qquad \frac{C_N}{v^2}\ \longleftrightarrow\ \begin{pmatrix} 4/3&0&-\rho/\sqrt3\\ 0&-4&-\sigma\\ -\rho/\sqrt3&-\sigma&0 \end{pmatrix}.\] Here one uses \(\sum s_i\widetilde s_i=0\), \(\sum\widetilde s_i^3=-6\rho\), and \(N/v^2=\operatorname{diag}(\sigma s_i+\rho\widetilde s_i)\). The energy metric in these coordinates is \(H=\operatorname{diag}(5/3,5,1)\). If \(\mathsf B\) denotes the matrix of \(B\), the matrix of \(E_B\) is \(\mathsf B H^{-1}\mathsf B\): \(B_F(T)\) compressed to \(G\) is the self-adjoint operator whose quadratic form is \(B\). Using \(\sigma^2+\rho^2=1\), this gives \[\frac{E_B}{v^2}\ \longleftrightarrow\ \begin{pmatrix} 16/15&0&4\rho/(5\sqrt3)\\ 0&21/5&4\sigma/5\\ 4\rho/(5\sqrt3)&4\sigma/5&1/5 \end{pmatrix}.\] Substitution into the definition of \(G_v\) gives [eq:11] by matrix addition. The invariant expression for \(G_v\) is defined everywhere; on the zero set of \(T\), all its derivative parameters vanish almost everywhere, as established for the norm identity. ◻

Proof. By homogeneity, assume \(|a|=1\); the case \(Z_a=0\) is immediate. Put \(\nu=|Z_a|/\sqrt6\), and diagonalize the normalized matrix \(s=Z_a/\nu\). Apply the algebraic parametrization [eq:2] with this \(s\), rather than with \(T/v\), and denote its vector parameters by \(\widehat p,\widehat k,\widehat l\). The diagonal entries of \(Z_a=\nu s\) imply \(\widehat p_a=\nu\) and \(\widehat k_a=0\). The part corresponding to the single component \(\widehat p_a a\) has energy \(e_1=5\nu^2/3\). It is orthogonal in the energy metric to the remaining parameters. In the displayed matrix for \(B/v\), it has no cross term with those parameters, because \(\widehat k_a=0\) and the \(\widehat p,\widehat l\) entry is zero. Its contribution to \(\frac16\sum_j\operatorname{tr}(sZ_j^2)\) has absolute value at most \(4e_1/5\).

For any symmetric trace-free \(3\times3\) matrix \(S\), its characteristic polynomial gives \[|(S^2)_0|^2=\frac{|S|^4}{6}.\] Since \(|s|=\sqrt6\), Cauchy–Schwarz yields \(|\operatorname{tr}(sS^2)|\le |S|^2\). Applied to the remaining tensor, this bounds its contribution by \(e-e_1\). Therefore \[\left|\frac{\operatorname{tr}(Z_a\sum_jZ_j^2)}6\right| \le\nu\left(e-\frac{e_1}{5}\right) =\sqrt{\frac35}\,e^{3/2} \sqrt{\frac{e_1}{e}}\left(1-\frac{e_1}{5e}\right).\] For \(0\le r\le1\), the function \(\sqrt r(1-r/5)\) is increasing and has maximum \(4/5\). This proves [eq:13]. ◻

Proof. Complexify at the point, writing the outer tangent representation as \(A\otimes B\), with \(A,B\) the plus and minus spin halves, and \(G_{\mathbb C}=H_{5/2}\otimes B'\). For a real vector \(a\), define the operator \(R_a\) on \(A\) by \(\operatorname{tr}(R_aD)=\langle a,(D\otimes\operatorname{Id}_B)a\rangle\) for every operator \(D\) on \(A\). This is the partial trace over \(B\) of the rank-one operator \(a a^*\). We claim that \(R_a=\frac12|a|^2\operatorname{Id}_A\). Indeed, the three operators \(-iI_s\) form a basis of the traceless Hermitian operators on \(A\), and each has zero expectation because \(a^{\mathsf T}I_sa=0\). The trace of \(R_a\) is \(|a|^2\), which fixes the scalar part.

The complete vector is \(a\otimes Z\), a product between the outer factor and the entire inner factor. Consequently, for every operator \(O\) acting only on \(A\otimes H_{5/2}\), \[\langle a\otimes Z,O(a\otimes Z)\rangle =\frac{|a|^2}{2} \langle Z,((\operatorname{Tr}_A O)\otimes\mathrm{Id}_{B'})Z\rangle.\] This partial trace retains every correlation of the inner plus factor with \(B'\); neither \(a\) nor \(Z\) is assumed to be a product within its own plus and minus factors.

The operator \(\operatorname{Tr}_A P_+\) is scalar on the irreducible space \(H_{5/2}\). Its trace is the dimension \(7\) of spin \(3\), whereas \(H_{5/2}\) has dimension \(6\). Thus it equals \(7\mathrm{Id}/6\), proving the first identity.

The family \(\operatorname{Tr}_A(P_+\widetilde Q(T)P_+)\) is equivariant and linear in the spin-\(2\) variable \(T\). Since spin \(2\) occurs once in \(\operatorname{End}(H_{5/2})\), it is a scalar multiple of \(Q_{5/2}(T)\). To determine the scalar, take \(T=\operatorname{diag}(-1,-1,2)\) and the inner highest weight \(5/2\). The outer weight \(1/2\) gives total weight \(3\) and \(Q_3(T)\)-eigenvalue \(-1\). The outer weight \(-1/2\) projects to total weight \(2\), where that eigenvalue is \((12-3\cdot2^2)/15=0\). The normalized outer trace is therefore \(-1/2\). Since \(Q_{5/2}(T)\) has eigenvalue \(-1\) on its highest weight, the second identity has coefficient \(1/2\), as asserted. ◻

Proof. Let \(P_-\) be the projection onto total minus spin \(1\), so the highest Hessian projection is \(P_+P_-\). On that type, \(-U_D\le y\), and hence \(\mathcal P\le A\), where \[A=4+y-\frac52\widetilde Q.\] This operator acts only on the plus factors and commutes with \(P_-\). On the entire high-plus subspace, \[A\ge4+y-2x\ge0;\] the last inequality follows from \(x+y\le2\). It follows in operator order that \(P_-AP_-\le A\) there. Thus, for every real \(a\), \[\langle\mathcal P(a\otimes Z)_Y,(a\otimes Z)_Y\rangle \le \langle AP_+(a\otimes Z),P_+(a\otimes Z)\rangle =6|a|^2\left[\frac7{12}(4+y)e-\frac54B\right],\] where the last equality is Lemma 26. This operator argument is valid for arbitrary correlations after the plus projection and does not discard a mixed minus-factor term.

Almost everywhere, \(dm=-\frac34M(p-q)\). Substituting \(a=dm\), dividing by \(24m\), and using \(m=M^2\) gives [eq:14]. Finally, for any unit real \(a\), the same equality and the lower bound for \(A\) give \[\frac7{12}(4+y)e-\frac54B \ge \frac7{12}(4+y-2x)e\ge0.\] In particular, subsequent upper estimates may replace \(|p-q|^2\) by \((|p|+|q|)^2\) in [eq:14] without reversing the inequality. ◻

Proof. For an interval, expand \(z^j=z^j(z+(1-z))^{N-j}\) and compare coefficients in the Bernstein basis. On a triangle, expand \[z^dw^e=z^dw^e(z+w+(1-z-w))^{N-d-e}.\] This gives the displayed factors after dividing each coefficient by the corresponding multinomial coefficient. Equivalently, substitute \[(v,b)=\frac{u_1p_1+u_2p_2+u_3p_3}{u_1+u_2+u_3}\] and multiply by \((u_1+u_2+u_3)^N\). The normalized coefficient of \(u_1^iu_2^ju_3^{N-i-j}\) is \([p]_{ij}\). The positivity statements follow from the nonnegative partition of unity, also for matrix quadratic forms. ◻

Proof. Apply the interval formula of Lemma 31 in the degrees and on the intervals indicated below. Every coefficient has at least the displayed positive lower bound. In the first two rows the bound applies on each of the two intervals separately.

For the first two rows the coefficients are in fact strictly greater than \(1/2\). For example, their respective minima on the two intervals are \[\left(\frac{3971}{3200},\frac{1063}{1600}\right), \qquad \left(\frac{101027}{20000},\frac{283729}{400000}\right).\] All entries in this table are obtained by substituting the displayed formulas for \(A_0,L_0,C_0\) and for \(A,G,H,L,Z_*\) into the coefficient sum in Lemma 31. Thus each bound is an inequality between rational coefficients. The lemma proves the required signs. ◻

Proof. Use the vertices \[\begin{gathered} O=(0,0),\quad S=(1,0),\quad C=(1,1),\quad U=(2,0),\quad W_0=(0,2),\\ A=(1/2,1/2),\quad B=(3/2,1/2),\quad E=(1,1/2). \end{gathered}\] A three-letter word specifies the vertices of an ordered triangle. The triangles \(OSC\), \(OCW_0\), and \(SCU\) cover \(\mathcal D\). The triangles \(OAS\), \(ASE\), \(ACE\), \(SBE\), \(CBE\), and \(SUB\) cover \(\mathcal D_+\); the latter is also the single triangle \(OCU\), or the union \(OSC\cup SCU\). All edges are included. The two subdivisions are shown in Figure 1.

For a matrix row, the last two columns below bound, respectively, \[\min([r]_{ij},[h']_{ij}),\qquad [r]_{ij}[h']_{ij}-[s']_{ij}^{2}.\] Thus the last column is the determinant of each coefficient matrix. For a scalar row, the first bound applies directly to its coefficient. The bounds apply to every nonzero coefficient; a zero matrix exception means that all three of its entries vanish.

Here are explicit expansion instructions for all entries of the table. They also fix the coefficient ordering. Write \(a=V-1\), \(s=W-1\) in [eq:15] and form \(\kappa(V,W)\) and its two partial derivatives. A partial derivative is obtained by reducing the appropriate exponent of each monomial and multiplying by that exponent. Differentiate before substituting the affine expressions \(V,W\). Put \[\kappa_* =\kappa(V,W),\qquad \bar\kappa_* =\kappa(W,V).\] Thus the partial derivatives of the exchanged function, before affine substitution, are \[\partial_V\bar\kappa_*(V,W)=(\partial_2\kappa)(W,V),\qquad \partial_W\bar\kappa_*(V,W)=(\partial_1\kappa)(W,V).\]

Ordinary coefficient arrays are multiplied by convolution: \[(PQ)_{de}=\sum_{f=0}^d\sum_{g=0}^e P_{fg}Q_{d-f,e-g}.\] Use \[\begin{aligned} m_*&=\left(1-\frac38(V-W)\right)^2, &d_0&=2-V-W,\\ P_{k,*}&=300m_*+160h d_0, &P_{l,*}&=35m_*V,\\ Y_*&=m_*(42+3W^2+10V^2) -h\bigl[6(V^2+W^2-V-W)-26d_0\bigr]. \end{aligned}\] For \(\mathcal A\), apply \(s_{ij}\) from Lemma 31 to the three entries of \[\begin{aligned} &(P_{k,*}+30hV) \bigl\{V(A_1,Y_*,A_{12})-\kappa_*(5,1,0)\bigr\}\\ &\hspace{22mm}-2V^2(P_{k,*}^2,P_{l,*}^2,P_{k,*}P_{l,*}), \end{aligned}\] where \[A_1=5Y_*+m_*(300V+140V^2)+160h d_0V, \qquad A_{12}=m_*(75V+35V^2)+40h d_0V.\] These are the original entries of [eq:21] at \(\sigma=1\), expressed only in the affine polynomials \(V,W\).

For \(\mathcal B\), put \[\begin{aligned} K_*&=4+V^2+W^2,\qquad \Theta_* =\Theta(V,W),\\ X_*&=m_*(-70+100V-25V^2-5W^2) +\frac53h\bigl(6(a^2+s^2)+32a d_0\bigr),\\ H_*&=8VK_*\Theta_*(1+V)\\ &\quad-(1+6V+V^2) \left\{VD+K_*\left[V\bigl(X_*+(\partial_1\kappa)(V,W)\bigr) +\frac{2\kappa_*}{3}\right]\right\}. \end{aligned}\] Apply \(s_{ij}\) to \(H_*\), to its exchange in \(V,W\), and to \[S_*(V,W)\left\{ \frac{K_*}{2}\bigl((\partial_2\kappa)(V,W) +(\partial_2\kappa)(W,V)\bigr)-D\right\}.\] For example, on \(CBE\) the triples at \((i,j)=(8,0),(8,1),(8,2)\) are \[\left(\frac{11041}{810},\frac{4063}{810},\frac{60557}{16200}\right), \quad \left(\frac{4328}{405},\frac{839}{405},\frac{3977}{2700}\right), \quad \left(\frac{191}{81},\frac{191}{81},-\frac{1189}{1350}\right).\]

For the two scalar rows, apply \(s_{ij}\) to \(10h d_0V-\kappa_*\) and, respectively, to \[\begin{aligned} c(V,W)\bigg\{&D\left(1-\frac{d_0^2}{4}\right)(V-W)^2 +30h d_0(V^2a^2+W^2s^2)\\ &+6(Va\kappa_*+Ws\bar\kappa_*) +12d_0^2(10h d_0W-\bar\kappa_*)\bigg\}-\Theta_*^2. \end{aligned}\] Together with the coefficient-sum formula, these prescriptions produce the displayed rational bounds by addition and multiplication of finite arrays.

The zero exceptions are as follows, with the third homogeneous index always \(N-i-j\). For \(\mathcal B\), they are the vertex \(C\) in \(ACE\) and \(CBE\), namely \((0,10,0)\) and \((10,0,0)\), and also \((9,1,0)\) in \(CBE\). For \(L_v\) the only zero is \((0,6,0)\), the vertex \(C\) in \(OCU\). For \(F+12c\delta^2L_b\) on \(OSC\), the zeros are exactly \(i+j\le2\). On \(SCU\) they are exactly \(9-j\le2\), together with \((i,j)=(0,6)\). There are no other exceptions.

Every nonzero matrix coefficient has positive diagonal entries and positive determinant, and each exceptional matrix is zero. Thus all coefficient matrices are positive semidefinite. The scalar coefficients are all nonnegative. Lemma 31 and the stated domain coverage prove the assertions. ◻

Proof. Use \(v=1-n+t\), \(b=1-n-t\). The full domain becomes \[0\le n\le1,\qquad |t|\le1-n.\] Write \(F_*(n,t)=F(1-n+t,1-n-t)\) for the pullback to these coordinates. It is even in \(t\), and direct expansion gives \(F_*=\sum_{j=0}^4 f_j(n)t^{2j}\), where \[\begin{aligned} 240f_0={}&n^3(1-n)\bigl(53511n^4-151593n^3 +222537n^2-83607n+10256\bigr),\\ 60f_1={}&n\bigl(19389n^5-418926n^4+523719n^3 -289776n^2+67782n+10564\bigr),\\ 60f_2={}&-1260129n^4+1370598n^3-451164n^2+23442n+74,\\ 10f_3={}&-12200n^2+2601n+75,\qquad f_4=-\frac{23}{5}. \end{aligned}\] For \(n<1\), put \(z=t^2/(1-n)^2\in[0,1]\). The degree-four Bernstein coefficients in \(z\) are \[h_i(n)=\sum_{j=0}^i \frac{\binom ij}{\binom4j}f_j(n)(1-n)^{2j}, \qquad 0\le i\le4.\] Define \[d_0(n)=n^3(1-n),\qquad d_1(n)=n(1-n),\qquad d_2(n)=d_3(n)=d_4(n)=1-n.\] Each quotient \(q_i=h_i/d_i\) is a polynomial. Applying the interval conversion formula to it on each of the four intervals \[[0,1/4],\quad[1/4,1/2],\quad[1/2,3/4],\quad[3/4,1]\] gives the following lower bounds for every coefficient:

The complete prescription for each entry is to insert the displayed \(f_j\) into \(h_i\), cancel the indicated polynomial factor, and apply the coefficient sum in Lemma 31 in the listed degree. In particular, the identities \(h_i=d_iq_i\) hold as polynomial identities, including at the endpoints.

Since every \(d_i\) is nonnegative on \([0,1]\), every \(h_i\) is nonnegative. Therefore \[F_*=\sum_{i=0}^4h_i(n)\binom4i z^i(1-z)^{4-i}\ge0 \qquad(n<1).\] At \(n=0\), the coordinate \(z=t^2\) still covers the entire permitted interval \(|t|\le1\); the vanishing of \(h_0,h_1\) causes no difficulty. At \(n=1\), the domain contains only \(t=0\), and directly \(F_*(1,0)=F(0,0)=f_0(1)=0\). Thus no numerical division by a vanishing factor is required, and all of \(\mathcal D\) is covered. ◻