Positively curved Einstein four-manifolds

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

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

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

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,

Here trace-freeness of both matrices gives the middle identity and removes the scalar term. Polarization proves the operator equality.

By Lemma 2.2, 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,

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 2.1 puts Z in the first summand.

Let

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

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

To differentiate this equation, note first that \nabla N=2B_F(T)Z. For each derivative index j, the covariant commutation formula reads

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 2.3 gives P_GB_F(T)P_G=Q_{5/2}(T). On G, the total plus generators are I_s+\Sigma_s, so

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

The scalar contributions sum to 13, while the two curvature contributions sum to -20Q_{5/2}(T), proving (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

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

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

This is globally an identity of smooth functions. Put v_\varepsilon=\sqrt{v^2+\varepsilon^2}. Direct differentiation gives

The refined derivative estimate in (2) yields

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 (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

Define operators on covariant two-tensors by

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

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

In this contraction the minus curvature block is orthogonal to the plus block. Substituting their expansions gives

The final equality uses \sum t_s=0 and \sum t_s^2=6v^2. Consequently

The pointwise upper operator bound, inserted into the positive |S|^2 term of the Bochner identity, now yields

For smooth a, the closed-manifold Poisson statement in Appendix A.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

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 A.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 (2) give

Indeed, \sigma\le1, v^2/A_\varepsilon\le1, and e\ge5|p|^2/3. Multiply by f_\varepsilon and integrate:

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 3.2 to f_\varepsilon and use the preceding energy bound:

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

When T\not\equiv0, the common factor is positive, proving (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 (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. The indicated probability measure is invariant under geodesic flow. Indeed, in cotangent coordinates the Hamiltonian flow of kinetic energy preserves canonical volume: its divergence vanishes by equality of the mixed second derivatives of the Hamiltonian. Applying this observation to energy shells gives invariance of the measure on the unit-energy level. In orthonormal fiber coordinates this is precisely Riemannian base volume times spherical measure.

The plus and minus contributions to A are orthogonal conjugates of T/2 and U/2, using the normal bases I_s\dot\gamma and J_s\dot\gamma. The same description applies to their derivatives in parallel frames. The mixed contractions have zero spherical average: they are invariant bilinear pairings between the inequivalent types (2,0) and (0,2), or between (5/2,1/2) and (1/2,5/2). Consequently,

Here the second factor 1/4 in the derivative identity is the spherical average of a squared directional derivative in four dimensions. Integrating the equation for \Delta T yields

since \sigma\le1 and v<2. The exchanged identity gives the same bound for \bar e. The stated formulas follow.□

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 4.3. 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

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,

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

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

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

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

Lemma 4.5 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

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

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

Expanding the four powers gives the displayed rational coefficients.□

Proof of Proposition 4.1. For the sine-squared probability weight defining B, its time variance is t^2(1/12-1/(2\pi^2)). The variance identity and (8) give

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,

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 4.4 and 4.6 therefore show that the coefficient of s_0 saved from the round volume is at least

The first three rational bounds above give \eta_+>2/25>3/40, proving (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,

and

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

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

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

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

Direct differentiation gives H'''(z)=L(z)z^{-5/2} and \Phi'''(z)=V(z). For x=\sqrt{z/2}, the polynomial inequalities

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

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:

To verify these inequalities exactly, put

For each of

the required inequality, after multiplication by its positive B_\pm^2, is P_\pm(x)>0, where

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

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

The degrees and strict lower bounds for the resulting coefficients are

For direct reproduction, the first two polynomial inputs are

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

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 (5) and v+b<2, 2\le s_0<4. Equations (6) and (7) give

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 (6) then gives

Combining this with v^2<(2-b)^2 yields

Integrating (3) for b, using \bar\sigma\le1 and \bar e\ge5|\nabla b|^2/3, gives

The integrability of \bar e/b proved with (3) implies c=\sqrt b\in H^1 by regularization. Put

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,

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 (4), extended by Sobolev approximation, therefore gives

Equivalently,

The distribution of b is a probability measure on [0,2] satisfying the two moment constraints. Lemma 4.8 contradicts (9). Thus \tau=0, and (6) gives equality of the two second moments.□

Proof. The product rule gives four inequivalent, multiplicity-one summands

Skewing the derivative slots takes values in

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

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

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

Hence

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),

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 5.2,

The last term in its expansion is the last term of (10). Since \mathcal P preserves the type splitting and X=Y+D_0, the other terms are

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.

The identity term. Integration by parts and (1) give

This will be multiplied by the coefficient 4 in \mathcal P.

The derivative-slot curvature terms. Write A_{ikjl}=(T_D)_{ikjl}. The two terms to integrate are

Integrate the derivative k on the second paired tensor. The derivative of m cancels the second term exactly. Moreover \nabla_kA_{ikjl}=0, by divergence freedom and skewness in i,k. The result is

This calculation uses neither symmetry of X nor parallelness of T_D, and the factor 1/2 comes only from skewing the two covariant derivatives.

The curvature in the derivative slot has the contraction

To see its normalization, substitute the quaternion expansions: the contraction over i,k removes all mixed plus/minus terms and gives \frac12\sum_s t_s(1+t_s)\delta_{lh}=3v^2\delta_{lh}. The matrix-slot curvature gives the quadratic form of

Since t_s^2=N_{ss}+2v^2, its compression is

After division by 6, the full T_D contribution is therefore

For U_D, the same integration gives 3\mathbf E[mb^2e]. Its matrix-slot curvature contraction is zero, because the minus forms are orthogonal to the plus forms defining the curvature action on F.

The nonparallel matrix-slot coefficient. Put A=B_F(T) temporarily. One integration by parts gives

The operator \nabla_iA is symmetric, so the last term is

Here \nabla_iA=B_F(Z_i) and \Delta A=6B_F(T)-6B_F(N), since B_F is a linear parallel construction. Also, pairing B_F(T)Q_{5/2}(T)Z with Z gives |Q_{5/2}(T)Z|^2, because the latter vector belongs to G. After division by 6, the contribution is

Since \langle B_F(Z_i)Z,Z\rangle =\operatorname{tr}(Z_i\sum_jZ_j^2), multiplication by -5/3 produces exactly the middle term in (10). In particular, the variation of B_F(T) has been retained explicitly; \mathcal P is not assumed parallel.

The prescribed trace component. By Lemmas 5.1 and 5.2,

The characteristic polynomial of a trace-free 3\times3 matrix gives B_F(T)T=N and B_F(T)N=v^2T. Thus

Using \Delta T=6(T-N), the two trace subtractions, before division by 6, are

Since \nabla N=2B_F(T)Z and \langle T,Z_i\rangle=6vp_i, their normalized contribution is

Multiplying the preceding contributions by their coefficients in \mathcal P and collecting gives G_v as stated.

Finally, a global Lipschitz weight m\ge m_0>0 can be approximated uniformly and in H^1 by smooth positive weights with a common positive lower bound. The first derivatives converge in L^2, so their quadratic products in the square cost converge in L^1; the uniformly bounded reciprocals of the weights pass to the same limit. All other terms use only the weight and its first derivative. Hence the identity and inequality extend to every weight in the stated class.□

Proof. Expanding the diagonal and off-diagonal matrix entries in (2) gives

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

Substitution into the definition of G_v gives (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 (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

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

For 0\le r\le1, the function \sqrt r(1-r/5) is increasing and has maximum 4/5. This proves (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},

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

This operator acts only on the plus factors and commutes with P_-. On the entire high-plus subspace,

the last inequality follows from x+y\le2. It follows in operator order that P_-AP_-\le A there. Thus, for every real a,

where the last equality is Lemma 5.6. 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 (14). Finally, for any unit real a, the same equality and the lower bound for A give

In particular, subsequent upper estimates may replace |p-q|^2 by (|p|+|q|)^2 in (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

This gives the displayed factors after dividing each coefficient by the corresponding multinomial coefficient. Equivalently, substitute

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 7.1 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

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 7.1. Thus each bound is an inequality between rational coefficients. The lemma proves the required signs.□

Proof. Use the vertices

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,

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 (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

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:

Use

For \mathcal A, apply s_{ij} from Lemma 7.1 to the three entries of

where

These are the original entries of (21) at \sigma=1, expressed only in the affine polynomials V,W.

For \mathcal B, put

Apply s_{ij} to H_*, to its exchange in V,W, and to

For example, on CBE the triples at (i,j)=(8,0),(8,1),(8,2) are

For the two scalar rows, apply s_{ij} to 10h d_0V-\kappa_* and, respectively, to

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 7.1 and the stated domain coverage prove the assertions.□

Proof. Use v=1-n+t, b=1-n-t. The full domain becomes

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

For n<1, put z=t^2/(1-n)^2\in[0,1]. The degree-four Bernstein coefficients in z are

Define

Each quotient q_i=h_i/d_i is a polynomial. Applying the interval conversion formula to it on each of the four intervals

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 7.1 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

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. Suppose otherwise. Proposition 4.9 gives \mathbf E v^2=\mathbf E b^2, so Proposition 6.1 applies. Its integrable total expression \mathcal I has nonnegative integral and satisfies

where

The zero-set conventions established with (3) apply to the quotients. For v>0,

so R_v\le1/\sqrt v, with equality only at v=1. Cauchy–Schwarz therefore gives r^2\le j.

This inequality is strict almost everywhere on the set where p or q is nonzero. Indeed a Lipschitz function has zero gradient almost everywhere on each of its level sets: at a differentiability and density point of a level set, a nonzero differential would contradict density. Thus, outside a null set, an active gradient p has v>0 and v\ne1, and similarly for q. At such a point

Terms with vanishing gradients are read as zero. It follows that \mathcal I\le0 almost everywhere, and that \mathcal I<0 almost everywhere where either gradient is nonzero. Since \mathbf E\mathcal I\ge0, both gradients vanish almost everywhere. Connectedness and continuity make v,b constant. Both constants are positive, so the separate gaps (5) imply v+b\ge2, contradicting v+b\le x+y<2.□

Proof. Proposition 3.4 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 4.3. 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,

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,

In particular,

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 (8), \mathcal E|\mathcal R|^2=3+3s_0/2=15/2. For any ordered product of k\ge2 curvature matrices,

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

The term k=1 is exactly 3\operatorname{tr}G. Therefore

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

Since a=t/N, Lemma 4.5 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

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 8.3. Write the first bound in Lemma 8.5 relative to t^3\sin\theta/\theta. Its exponential factor is

Integrate Lemma 8.4, 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

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 8.2 gives \mathbf E v^2=1. The equality case of Proposition 3.3 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

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

This follows by differentiating \langle Y,iz\rangle=0. Substitution in the curvature formula yields

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 4.3 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 8.6 and Lemmas 8.7–8.8 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 1.1. 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 8.9.

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, Proposition 3.25]. 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 8.9 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, Theorem 3.19]. 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, Section 2.2, property (f)]. 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 3.2. Fix 2<s<4 and minimize

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

After setting u=m_s^{1/(s-2)}w, it becomes

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

The strong Dirichlet existence theorem [16, Theorem 9.15] 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, Theorem 9.11 and Lemma 9.16].

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

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

Since \operatorname{div}(\operatorname{Hess}_0F) =\frac34\,d(\Delta F+4F), differentiation gives

Pairing with dF and integrating yields

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

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

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

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

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

gives U_0=h\operatorname{Id} and the transport equations

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

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

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,

In coordinate form,

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

In particular, derivatives of the varying potential have already disappeared before the radial integral is taken. The transport formula now gives

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

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

Since the transport equation at the center gives U_2(0)=-\frac12(LU_1)(0), we obtain

The lower coefficients have zero signed diagonal trace. Multiplication by the heat prefactor (4\pi t)^{-2} now proves the assertion.□

Proof of Proposition 3.4. By Hodge theory, harmonic forms represent de Rham cohomology. The scalar kernel is one-dimensional, and the one-form kernel vanishes by Lemma A.1. 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 3.4 with the topological signature and Euler characteristic without using simple connectivity.

The potential-square supertrace is

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

On plus two-forms only the plus generators act, and each speed-two spin-one generator has square trace -8. Therefore

The scalar connection is flat. Hence \operatorname{str}\sum\Omega_{ki}^2=24(b^2-v^2), and Lemma A.2 gives the signed constant density

Integrate this against d\operatorname{vol}_g and compare with the constant alternating heat trace from Lemma A.1:

Reverse orientation to get the analogous formula with n_-,b,v. Adding and subtracting the two formulas gives exactly (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.□

Proof of Lemma 4.3. For a unit-speed geodesic of length L, the index form on normal fields with zero endpoint values is

It is nonnegative on a minimizing segment, by the second variation of energy. Summing it on three parallel normal unit fields multiplied by \sin(\pi t/L) gives

so a minimizing segment has length at most \pi. It cannot contain an interior conjugate point: a nonzero Jacobi field vanishing at the initial point and that interior point, extended by zero, has index zero but is not in the nullspace of the index form. The latter assertion follows by integration by parts against a field whose value pairs nontrivially with the jump in its derivative. A nonnegative quadratic form, however, pairs a zero-energy vector to zero against every vector, a contradiction.

The ordinary Dirichlet Poincaré inequality and \mathcal R<3\mathrm{Id} show that a nonzero field with zero endpoints has positive index when L\le\pi/\sqrt3. Thus no conjugate point occurs at such a distance. We show that no cut point occurs at a smaller distance either. Let c be the infimum of all cut radii; compactness and normal balls give c>0. Suppose c<\pi/\sqrt3 and choose cut segments with lengths tending to c, all below the conjugate-radius bound. Each endpoint has two distinct minimizing segments from its starting point. To see this, extend a cut segment a little and take limits of minimizers to the extended endpoints. If the limiting minimizer were the original segment alone, local invertibility of the exponential map would force the extended segment to remain minimizing, contrary to its cut time.

After passing to a subsequence, compactness gives two length-c minimizing segments with common endpoints p,q. Their initial vectors cannot coalesce: the endpoint map

is a local diffeomorphism at every vector in question, so two converging branches in the same inverse-function neighborhood would coincide. This map also supplies two distinct smooth local geodesic branches for nearby pairs of endpoints.

Their terminal unit velocities must be opposite. If these velocities are u_1,u_2 with u_2\ne-u_1, move q initially in direction -(u_1+u_2). The first length variation of both branches is -(1+\langle u_1,u_2\rangle)<0. Both branches would then have length less than c. Every geodesic of length less than the global cut infimum is minimizing and lies in the injectivity domain, so two distinct such branches are impossible. Moving p instead proves that their initial velocities are opposite as well. The first segment followed by the reverse of the second is therefore a smooth closed geodesic.

Parallel transport around this closed geodesic fixes the tangent and preserves orientation. Its action on the three-dimensional normal space belongs to \mathrm{SO}(3), and hence fixes a nonzero vector. Let V be the resulting periodic parallel normal field. Move both endpoints along geodesics with initial velocities V(p),V(q), so their accelerations at the variation origin vanish. The local geodesic branches vary smoothly. Their variation fields J are normal: their tangential components are affine and have zero boundary values. For either branch, W=V-J has zero boundary values and I(J,W)=0, while I(W,W)\ge0 because the unvaried segment is minimizing. Thus

The first length variations vanish, and these negative index values are the second length variations, since there are no endpoint acceleration or tangential terms. Both branch lengths therefore become less than c. They remain distinct in their disjoint inverse-function neighborhoods, and lengths below c force them to be minimizing. This is the same contradiction as above, proving the injectivity bound.

Finally, a nontrivial free homotopy class would have a shortest representative. Indeed, constant-speed minimizing sequences have a uniform Lipschitz bound; compactness gives a uniformly convergent subsequence, and sufficiently close loops are freely homotopic by the short normal geodesics between corresponding points. A uniform normal ball radius also gives a positive lower bound on lengths of nontrivial loops. Local geodesic replacement shows that the minimizing loop is a nonconstant smooth closed geodesic. Its normal holonomy again fixes a nonzero parallel field, whose periodic variation has strictly negative second length variation. The variation preserves its free homotopy class, a contradiction. Hence the fundamental group is trivial.□