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 3, 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 a space 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 2 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 4 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 \[T_D+U_D\le 2(x+y)\operatorname{Id}<4\operatorname{Id}\] 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 closed-manifold Poisson statement in Appendix 9.1 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 functions are dense in \(H^1\) by the local mollification argument recalled in Appendix 9.1. Since \(K\) is bounded above and below by positive constants, this approximation also passes the weighted energy to the limit and proves the assertion on \(H^1\). ◻

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 9 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. 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 Lemma 14. 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. The area formula therefore bounds \(V_M\) by the integral of \(D\) over this set, 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 12. 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 \[\eta_+=\frac{9\pi}{48}\int_0^1 f(u)(1-u^2/2)\sin^3(\pi u)\,du.\] The first three rational bounds above give \(\eta_+>2/25>3/40\), proving [eq:7]. ◻

Proof. 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 \(2\). It follows that \(\Phi(z)\ge16(11/20)^2z^2\). The moment constraint and \(J(\mu)\ge0\) now give the claim about \(d\).

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 moment condition 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. By [eq:5] and \(v+b<2\), \(2\le s_0<4\). 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 \(v^2<(2-b)^2\) 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 \[P-\mathbf E b\ge\frac49\mathbf E|\nabla\sqrt b|^2.\] The integrability of \(\bar e/b\) proved with [eq:3] implies \(c=\sqrt b\in H^1\) by regularization. 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. 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 \(-4\). Symmetry in the derivative slots gives the same value for \(\sum I_s^{(i)}\Sigma_s\). 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 \((3,1)\), \[\mathcal P\ge4-2x-2y>0.\] 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 \(a\), its reduced density on \(A\) is \(\frac12|a|^2\mathrm{Id}_A\). Indeed, the three traceless Hermitian operators on \(A\) are scalar multiples of \(-iI_s\), and each has zero expectation because \(a^{\mathsf T}I_sa=0\). Its trace 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 29 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 29. 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 below.

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 \[\kappa_* =\kappa(V,W),\qquad \bar\kappa_* =\kappa(W,V).\] For clarity, the derivative of the exchanged polynomial with respect to its first variable is \((\partial_2\kappa)(W,V)\), and its derivative with respect to its second variable is \((\partial_1\kappa)(W,V)\).

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 29 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 29 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 29 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. ◻

Proof. Proposition 11 gives \(n_-=0\) and \(\chi=2+n_+\), \(\tau=n_+\). Solving its two identities yields the displayed volume and moment formulas. Positivity of volume gives \(n_+<4\); the nonzero-block gap gives \(s_0\ge1\), while \(v<2\) gives \(s_0<4\). The finite nonnegative integer \(n_+\) is therefore \(1\) or \(2\). Substitution gives the stated values in the second case. ◻

Proof. The open ball of radius \(\pi/\sqrt3\) lies in the injectivity domain at every center by Lemma 14. Its volume is the corresponding integral of \(D\) and is at most \(V_M\). Averaging over centers gives the bound. Increasing open radii justify the endpoint; it does not change the integral. ◻

Proof. For each integer \(N\ge2\), put \(a=t/N\). Let \(L_0\) be the scalar Dirichlet tridiagonal matrix on the interior nodes \(1,\ldots,N-1\), with diagonal entries \(2\) and adjacent off-diagonal entries \(-1\). Put \(G=a^2L_0^{-1}\). Its entries are nonnegative; explicitly, \[G_{ij}=a^2\frac{\min(i,j)(N-\max(i,j))}{N},\qquad \operatorname{tr}G=\frac{a^2(N^2-1)}6\longrightarrow\frac{t^2}{6}.\] Write \(\mathcal R_N\) for the block diagonal matrix with blocks \(\mathcal R(ia)\) and define \(H_N=(G^{1/2}\otimes\mathrm{Id})\mathcal R_N (G^{1/2}\otimes\mathrm{Id})\). For fixed \(t<\pi/\sqrt3\) and large \(N\), \[0\prec H_N\preceq q_t\mathrm{Id},\qquad q_t<1, \qquad \|H_N\|\le\frac{3a^2}{2-2\cos(\pi/N)},\qquad \operatorname{tr}H_N=3\operatorname{tr}G.\] In particular, \[-\log\det(\mathrm{Id}-H_N) =\sum_{k\ge1}\frac{\operatorname{tr}H_N^k}{k} \le\frac{3\operatorname{tr}G}{1-q_t}.\] This bound is uniform over all geodesics and large \(N\) at the fixed time. It justifies taking expectations in the series and gives a uniform positive lower bound for the determinant ratios, independent of their increasing matrix dimensions.

By [eq:8], \(\mathcal E|\mathcal R|^2=3+3s_0/2=15/2\). For any ordered product of \(k\ge2\) curvature matrices, \[|\operatorname{tr}(\mathcal R_1\cdots\mathcal R_k)| \le3^{k-2}|\mathcal R_1|\,|\mathcal R_k|.\] Cauchy–Schwarz and invariance of geodesic-flow measure bound its expectation by \((5/6)3^k\). Neither commutation nor independence of the matrices is needed. Expanding the block trace of \(((G\otimes\mathrm{Id})\mathcal R_N)^k\), its scalar coefficients are products of nonnegative entries of \(G\). Their sum is \(\operatorname{tr}(G^k)\), so \[\mathcal E\operatorname{tr}H_N^k \le\frac56\,3^k\operatorname{tr}(G^k)\quad(k\ge2).\] The term \(k=1\) is exactly \(3\operatorname{tr}G\). Therefore \[\mathcal E\log\det(\mathrm{Id}-H_N) \ge\frac56\log\det(\mathrm{Id}-3G)-\frac12\operatorname{tr}G.\] Jensen’s inequality supplies a lower bound for the expected determinant by the exponential of this expression.

The flat block determinant is \(N^3\), and the normalization is exactly \[\det(\mathrm{Id}-H_N)=\frac{\det\mathcal L_N}{N^3}.\] Since \(a=t/N\), Lemma 16 gives \(\det(\mathrm{Id}-H_N)\to D(t)/t^3\), uniformly over geodesics at this fixed \(t\): on the compact unit tangent bundle the curvature coefficients and their first derivatives are uniformly bounded. The scalar version gives \[\det(\mathrm{Id}-3G)\longrightarrow\frac{\sin\theta}{\theta}.\] Passing to these limits proves the first assertion. Equivalently, the uniform logarithmic bound permits passing averaged logarithms directly; applying Jensen before the limit already suffices.

Finally, in the scalar positive logarithmic series keep only the \(k=2,3\) terms and, in each, the largest eigenvalue of \(G\), which tends to \(t^2/\pi^2\). Subtracting \(3\operatorname{tr}G\) and passing to the limit yields the second assertion. ◻

Proof of Proposition 35. Write the first bound in Lemma 37 relative to \(t^3\sin\theta/\theta\). Its exponential factor is \[\exp\left[\frac16\left( -\log\frac{\sin\theta}{\theta}-\frac{\theta^2}{6}\right)\right] \ge1+\frac16\left[\frac12\left(\frac\theta\pi\right)^4 +\frac13\left(\frac\theta\pi\right)^6\right].\] Integrate Lemma 36, first up to any \(T<\pi/\sqrt3\) and then let \(T\) increase to the endpoint. All integrands are nonnegative, so monotone convergence suffices; no uniform endpoint bound for a logarithm is required. The changes of variable \(\theta=\sqrt3t\), \(u=\theta/\pi\) give \[\frac{V_M}{\pi^2} \ge\frac29(\pi^2-4) +\frac{\pi^3}{27}\int_0^1 u^2(u^4/2+u^6/3)\sin(\pi u)\,du.\] The last two rational integral bounds and their constant term make this strictly greater than \(1343/1000>4/3\), proving the assertion. ◻

Proof. In the nonzero one-block case, the preceding exclusion gives \(n_+=1\), so Lemma 34 gives \(\mathbf E v^2=1\). The equality case of Proposition 10 yields \(\nabla T=0\) and \(v=\sigma=1\), hence eigenvalues \(2,-1,-1\). The exchanged orientation is identical. If both blocks vanish, the curvature-operator decomposition gives constant sectional curvature one. ◻

Proof. The first assertion follows immediately from the curvature-operator decomposition. In the second case choose at one point the orthogonal complex structure \(I\) corresponding to the eigenline of \(T\) with eigenvalue \(2\). On an orthonormal pair \(X,Y\), the sectional curvature is \[K(X,Y)=\tfrac12\bigl(1+3\langle IX,Y\rangle^2\bigr).\] Sectional curvatures determine an algebraic curvature tensor by polarization. Only this choice at a single tangent space is needed; no global complex structure is assumed.

For comparison, let \(g_H\) be the quotient metric on \(S^5/S^1=\mathbb{CP}^2\) obtained by isometric horizontal lift. At \(z\in S^5\), the unit vertical field is \(iz\). For horizontal lifts of base vector fields, the horizontal part of the sphere connection is the lifted base connection by the metric formula, while \[(\nabla_XY)^{\mathrm{vert}}=-\langle iX,Y\rangle iz.\] This follows by differentiating \(\langle Y,iz\rangle=0\). Substitution in the curvature formula yields \[K_{g_H}(X,Y)=1+3\langle iX,Y\rangle^2.\] To track the factor three, the vertical bracket is \([X,Y]^{\mathrm{vert}}=-2\langle iX,Y\rangle iz\) and invariant horizontal lifts satisfy \(\nabla_{iz}Y=iY\). The differentiated vertical term and the vertical bracket therefore contribute, respectively, one and two copies of the same squared inner product. Equivalently, the extra horizontal term in the sphere curvature expression is \(3\langle iX,Y\rangle iY\), whose pairing with \(X\) is \(-3\langle iX,Y\rangle^2\) relative to the base expression. Scaling \(g_H\) by two gives the required sectional curvatures.

The quotient metric has parallel curvature: a unitary transformation that fixes a complex line and is minus the identity on its orthogonal complement induces an isometry fixing that point with tangent differential \(-\operatorname{Id}\). It acts by \(-1\) on the covariant five-tensor \(\nabla R\), so \(\nabla R=0\). Transitivity gives this conclusion at every point. The round sphere likewise has parallel curvature. ◻

Proof. Lemma 14 shows that \(M\) is simply connected. Both model spaces are simply connected as well. For the Hopf quotient this can be seen directly: local sections lift any loop to a path in \(S^5\). Join its endpoints in the circle fibre to close the path, contract the resulting loop in \(S^5\), and project the contraction. The added fibre path projects to a constant, so the original loop contracts.

Choose a tangent-space isometry matching the full curvature tensor at one point. On sufficiently small normal balls the exponential maps give an isometry: in parallel frames the Jacobi equations have identical coefficients because both curvature tensors are parallel. At a new point, the differential again matches curvature and defines the same construction on a new normal ball. Compactness supplies a common positive radius for these local constructions on the two manifolds.

An isometry germ is determined by its value and differential through exponential maps. Hence it continues uniquely along each path. Subdivide a homotopy square into small pieces lying in normal balls; uniqueness on overlaps shows that continuation is invariant under homotopy. Simple connectivity therefore gives a global local isometry. Continue the inverse germ in the same way on the model. The two compositions agree with the identity near the original points and then everywhere by continuation. They are inverse global isometries. ◻

Proof. Strict sectional positivity gives a positive Einstein constant, so the metric can be normalized to \(\operatorname{Ric}=3g\). Proposition 38 and Lemmas 39–40 identify that normalized metric with the unit round sphere or twice the Hopf quotient metric. Undoing the positive normalization proves the stated oriented conclusion. ◻

Completion of Theorem 1. Strict sectional positivity makes the Ricci tensor positive definite, so the Einstein constant is positive. If \(M\) is orientable, choose an orientation and apply Corollary 41.

Otherwise let \(p:\widetilde M\to M\) be the orientation double cover. It is connected, closed and oriented; for the standard construction and connectedness see [20]. The pulled-back metric \(\widetilde g=p^*g\) has the same Einstein and strict sectional-positivity properties. The nontrivial deck map \(\tau\) switches the two local orientations over each point, so it is a free orientation-reversing involution, and \(p\circ\tau=p\) makes it an isometry of \(\widetilde g\). After positive scaling, Corollary 41 identifies this cover with one of its two standard models. Conjugating \(\tau\) by the model isometry preserves its freeness and its orientation sign.

The \(\mathbb{CP}^2\) case is impossible. Its integral cohomology ring is \(\mathbb Z[h]/(h^3)\) with \(|h|=2\), so \(h^2\) generates \(H^4(\mathbb{CP}^2;\mathbb Z)\) [20]. Any self-diffeomorphism sends \(h\) to \(\pm h\) and therefore fixes \(h^2\). It has degree \(+1\), contradicting the required orientation reversal.

On the unit round \(S^4\subset\mathbb R^5\), distance determines ambient inner products by \(\langle x,y\rangle=\cos d(x,y)\). An isometry therefore preserves these inner products: its values on the five coordinate unit vectors determine an \(A\in O(5)\), and comparison with their images gives \(\tau(x)=Ax\) for every \(x\in S^4\). Since \(\tau^2=\operatorname{id}\), we have \(A^2=I\). An orthogonal involution has an orthogonal decomposition into its \(+1\) and \(-1\) eigenspaces. A nonzero \(+1\) eigenvector would give a fixed point on the sphere, so freeness forces \(A=-I\). The scaled base is consequently the round quotient \(S^4/\{x\sim -x\}=\mathbb{RP}^4\).

The antipodal map has degree \((-1)^5=-1\) on \(S^4\), so it reverses orientation [20]. Its quotient is a closed connected smooth manifold and inherits the Einstein and positive-sectional-curvature properties from the round sphere through the local isometry. Thus this third model does occur, and it is the only nonorientable case. ◻

Proof of Proposition 9. Fix \(2<s<4\) and minimize \[Q_s(w)= \frac{\mathbf E(w^2+\tfrac12|\nabla w|^2)} {(\mathbf E|w|^s)^{2/s}}\] under \(\mathbf E|w|^s=1\). A minimizing sequence is bounded in \(H^1\); weak compactness, compact inclusion into \(L^s\), and lower semicontinuity give a minimizer. Taking its absolute value does not increase the energy, so choose \(w\ge0\). The ordinary Sobolev embedding makes the minimum \(m_s\) positive, and the constant test function gives \(m_s\le1\). The Euler equation is \[-\Delta w+2w=2m_s w^{s-1}.\] After setting \(u=m_s^{1/(s-2)}w\), it becomes \[-\Delta u+2u=2u^{s-1}.\]

We record the regularity needed to use a negative power of \(u\). We first pass from the weak minimizer to a strong solution. Choose a smooth coordinate ball \(B\) and a cutoff \(\eta\in C_c^\infty(B)\). For \(z=\eta u\) the weak equation gives \[(\Delta_g-2)z =-2\eta u^{s-1}+2\langle d\eta,du\rangle+(\Delta_g\eta)u =:F\in L^{r_0}(B), \qquad r_0=\min\{2,4/(s-1)\}>4/3.\] The strong Dirichlet existence theorem [16] gives a solution \(w_D\in W^{2,r_0}(B)\cap W^{1,r_0}_0(B)\) with this right side. The domain and coefficients are smooth, the operator is uniformly elliptic, and its zeroth-order coefficient is \(-2\). In dimension four, \(r_0>4/3\) implies \(W^{2,r_0}\subset W^{1,q}\) for some \(q>2\); compatibility of Sobolev traces therefore gives \(w_D\in H^1_0(B)\). The difference \(z-w_D\in H^1_0(B)\) solves \((\Delta_g-2)(z-w_D)=0\) weakly. Testing by \(z-w_D\) shows that it vanishes. Thus \(u\) is locally \(W^{2,r_0}\), and the strong interior estimates and integrability improvement now apply [16].

Initially \(u\in L^4\), while \(u^{s-2}\in L^r\) with \(r=4/(s-2)>2\). If \(u\in L^q\), the equation has right side in \(L^p\) with \(1/p=1/q+1/r\). The \(W^{2,p}\) estimate and Sobolev embedding improve \(1/q\) by a fixed positive amount smaller than \(1/2-1/r\). Initially \(p=4/(s-1)>1\). Iteration reaches \(p>2\), giving boundedness; at a borderline exponent \(p=2\), use any finite larger \(q\) before the next iteration.

After boundedness, the equation gives \(u\in W^{2,p}\) for every finite \(p\). Since \(s-1>1\), the derivative of \(t^{s-1}\) is bounded on the bounded nonnegative range of \(u\). Differentiating the equation once, and applying the same local regularity argument to each first coordinate derivative, gives \(u\in W^{3,p}\) for every finite \(p\). In particular, \(u\in C^{2,\alpha}\) for some \(\alpha>0\). For \(C\ge2\) we have \[(-\Delta_g+C)u=(C-2)u+2u^{s-1}\ge0.\] The classical strong maximum principle gives \(u>0\), since \(u\ge0\) and \(u\not\equiv0\). A direct barrier proof uses a positive exponential radial function on an annulus tangent to the zero set: the weak maximum principle forces a nonzero inward derivative at the tangency point, contradicting nonnegativity and differentiability. Compactness now gives a positive minimum. The nonlinearity is smooth on this positive range, so further differentiation and elliptic regularity give smoothness of \(u\).

Put \(F=u^{-1}\). Its equation is \[F\Delta F-2|\nabla F|^2+2F^2=2F^{4-s}.\] Since \(\operatorname{div}(\operatorname{Hess}_0F) =\frac34\,d(\Delta F+4F)\), differentiation gives \[\begin{aligned} \operatorname{div}(F^{-3}\operatorname{Hess}_0F) &=\frac34F^{-4} \bigl[F\,d\Delta F+(\Delta F)dF -4\operatorname{Hess}F(\nabla F,\,\cdot\,)+4F\,dF\bigr]\\ &=\frac32(4-s)F^{-1-s}\,dF . \end{aligned}\] Pairing with \(dF\) and integrating yields \[-\mathbf E F^{-3}|\operatorname{Hess}_0F|^2 =\frac32(4-s)\mathbf E F^{-1-s}|dF|^2.\] Both sides have opposite signs, so \(dF=0\). The positive constant solution of the equation for \(u\) is \(u=1\). The normalization of \(w\) then gives \(w=1\) and \(m_s=1\).

Thus the claimed inequality holds with \(4\) replaced by every \(s\in(2,4)\) and outer exponent \(2/s\). For fixed \(f\in H^1\), the ordinary inclusion \(H^1\subset L^4\) and dominated convergence give \(\|f\|_{L^s(d\nu)}\to\|f\|_{L^4(d\nu)}\) as \(s\uparrow4\). This proves the stated critical inequality. ◻

Proof. The middle complex Laplacian is \[dd^*+2(d^+)^*d^+ =dd^*+d^*d+d^*\!\star d.\] The last term is zero by \(d^2=0\) and the formula for the codifferential. On self-dual two-forms, Hodge star exchanges \(dd^*\) and \(d^*d\), so \(2P_+dd^*\) is their sum. This identifies the last Laplacian; the scalar one is immediate. Their elliptic principal symbols are consequently those of the metric Laplacian.

Commuting covariant derivatives in the one-form Hodge Laplacian gives the Ricci endomorphism, equal to \(3\operatorname{Id}\). For the plus potential, write a self-dual form as \(\eta=\sum_s\eta_s I_s/\sqrt2\). Its Hodge energy is \(2\mathbf E|d^*\eta|^2\), which equals \[\mathbf E\left|\sum_s I_s\nabla\eta_s\right|^2 =\mathbf E\left( |\nabla\eta|^2+\frac12\langle C\nabla\eta,\nabla\eta\rangle \right),\] where \(C=\sum I_s\otimes\Sigma_s\) now uses the spin-one generators on the plus coefficients. The equality follows by expanding the quaternion products. Integrating the \(C\) term and taking its skew derivative commutator gives the potential \[-\frac12\sum_s(1+t_s)\Sigma_s^2.\] In the spin-one representation, \(\Sigma_s^2=4(e_se_s^{\mathsf T}-\operatorname{Id})\). The potential is therefore \(4\operatorname{Id}-2T\), as asserted.

The scalar kernel consists of the constants. A harmonic one-form satisfies \[0=\mathbf E\bigl(|\nabla\eta|^2+3|\eta|^2\bigr),\] so the middle kernel vanishes. The last kernel has dimension \(n_+\). On each positive eigenspace of the complex Laplacian, the complex differential commutes with the Laplacian, and its adjoint divided by the eigenvalue is a contracting homotopy. That finite-dimensional complex has zero alternating dimension. Summing the heat factors over the positive spectrum leaves only the kernel contribution \(1+n_+\). ◻

Proof. Fix the source point of the kernel. In normal coordinates \(z\) around it, use radially parallel orthonormal frames on each of the three bundles and set \(\mu=\sqrt{\det g}\), \(h=\mu^{-1/2}\). The ansatz \[(4\pi t)^{-2}e^{-|z|^2/(4t)} \sum_{j\ge0}t^jU_j(z)\] gives \(U_0=h\operatorname{Id}\) and the transport equations \[\left(z\cdot\partial+j+\frac12z\cdot\partial\log\mu\right)U_j =-LU_{j-1}.\] Here the radial connection coefficient is zero. The scalar part follows also from \(\Delta|z|^2=8+2z\cdot\partial\log\mu\). Solving the radial equation gives \[U_j(z)=-h(z)\int_0^1 s^{j-1} \bigl(h^{-1}LU_{j-1}\bigr)(sz)\,ds.\] These transport solutions compute the actual local asymptotics. Indeed, truncate at any large \(N\) and apply a cutoff inside a normal ball. The resulting kernel has initial delta mass, and the transport equations leave a residual uniformly \(O(t^{N-2})\); derivatives of the cutoff give exponentially small terms. The heat evolution of a metric connection with bounded potential has a bounded sup norm on bounded time intervals, by the maximum principle for the squared norm. Duhamel integration therefore makes the difference from the true kernel \(O(t^{N-1})\). Compactness permits a uniform normal-ball radius and uniform constants in the source point. Taking \(N\) large justifies the diagonal coefficient calculation below.

In these frames write the connection as \(\partial_i+A_i\). Every \(A_i\) is skew-symmetric and radial gauge gives \[A_i(z)=\frac12z^k\Omega_{ki}(0)+O(|z|^2).\] For example, the exact radial-gauge identity \(A_i(z)=\int_0^1s z^k\Omega_{ki}(sz)\,ds\) follows by contracting the curvature equation with \(z^k\) and integrating along the radius. The signed ranks and potentials satisfy, at every point, \[\operatorname{str}\operatorname{Id}=1-4+3=0, \qquad \operatorname{str}E=-12+\operatorname{tr}(4\operatorname{Id}-2T)=0.\] In coordinate form, \[L=-g^{ij}\bigl[ (\partial_i+A_i)(\partial_j+A_j) -\Gamma^k_{ij}(\partial_k+A_k)\bigr]+E.\] Apply this to \(U_0=h\operatorname{Id}\). All purely scalar terms have zero supertrace. Every term linear in \(A_i\) or a coordinate derivative of \(A_i\) has zero ordinary trace in each bundle, by skewness. The potential has zero supertrace pointwise. Thus the exact surviving expression is \[\operatorname{str}(LU_0)(z) =-h(z)g^{ij}(z)\operatorname{str}(A_i(z)A_j(z)).\] In particular, derivatives of the varying potential have already disappeared before the radial integral is taken. The transport formula now gives \[\begin{aligned} \operatorname{str}U_1(z) &=h(z)\int_0^1 g^{ij}(sz) \operatorname{str}(A_i(sz)A_j(sz))\,ds\\ &=\frac1{12}z^kz^\ell \operatorname{str}\sum_i\Omega_{ki}(0)\Omega_{\ell i}(0) +O(|z|^3). \end{aligned}\] The coefficient is \(\frac14\int_0^1s^2\,ds=\frac1{12}\). Scalar and metric corrections do not change this quadratic term.

At the center, \(A_i(0)=0\), \(\Gamma(0)=0\), and \(\sum_i\partial_iA_i(0)=\frac12\sum_i\Omega_{ii}(0)=0\) as an endomorphism. Consequently \[U_1(0)=c\operatorname{Id}-E(0),\qquad (LU_1)(0)=-\sum_i\partial_i^2U_1(0)+E(0)U_1(0),\] where \(c\) is the same scalar for all three bundles. Thus there are no additional connection terms at the center. Taking the Laplacian of the signed quadratic expression gives \[\operatorname{str}\sum_i\partial_i^2U_1(0) =\frac16\operatorname{str}\sum_{ki}\Omega_{ki}^2, \qquad \operatorname{str}(EU_1(0))=-\operatorname{str}E^2.\] Since the transport equation at the center gives \(U_2(0)=-\frac12(LU_1)(0)\), we obtain \[\operatorname{str}U_2(0) =\frac12\operatorname{str}E^2 +\frac1{12}\operatorname{str}\sum_{ki}\Omega_{ki}^2.\] The lower coefficients have zero signed diagonal trace. Multiplication by the heat prefactor \((4\pi t)^{-2}\) now proves the assertion. ◻

Proof of Proposition 11. By Hodge theory, harmonic forms represent de Rham cohomology. The scalar kernel is one-dimensional, and the one-form kernel vanishes by Lemma 42. Hodge star gives the corresponding dimensions in degrees four and three. On harmonic two-forms, the intersection form is positive definite on the self-dual part and negative definite on the anti-self-dual part. Thus its signature is \(n_+-n_-\), and the alternating sum of the Betti numbers is \(2+n_++n_-\). This identifies the integers in Proposition 11 with the topological signature and Euler characteristic without using simple connectivity.

The potential-square supertrace is \[\operatorname{str}E^2 =-\operatorname{tr}(9\operatorname{Id}_{\Lambda^1}) +\operatorname{tr}\bigl((4\operatorname{Id}_{\Lambda^+}-2T)^2\bigr) =12+24v^2.\] On one-forms, the connection curvature has plus and minus generators \(I_s,J_s\). Orthogonality of the two-form bases, together with \(I_s^2=J_s^2=-\operatorname{Id}\), gives \[\operatorname{tr}_{\Lambda^1}\sum_{ki}\Omega_{ki}^2 =-4\sum_s\bigl((1+t_s)^2+(1+u_s)^2\bigr) =-24(1+v^2+b^2).\] On plus two-forms only the plus generators act, and each speed-two spin-one generator has square trace \(-8\). Therefore \[\operatorname{tr}_{\Lambda^+}\sum_{ki}\Omega_{ki}^2 =-8\sum_s(1+t_s)^2=-24(1+2v^2).\] The scalar connection is flat. Hence \(\operatorname{str}\sum\Omega_{ki}^2=24(b^2-v^2)\), and Lemma 43 gives the signed constant density \[\frac1{16\pi^2}(6+10v^2+2b^2).\] Integrate this against \(d\operatorname{vol}_g\) and compare with the constant alternating heat trace from Lemma 42: \[16\pi^2(1+n_+) =\operatorname{Vol}(M)\, \mathbf E(6+10v^2+2b^2).\] Reverse orientation to get the analogous formula with \(n_-,b,v\). Adding and subtracting the two formulas gives exactly [eq:6].

On anti-self-dual forms the potential is \(4\operatorname{Id}-2U\). If \(U=0\), the energy of a harmonic anti-self-dual form is the integral of \(|\nabla\eta|^2+4|\eta|^2\), so its kernel is zero. The assertion for \(T=0\) follows in the same way. ◻