A negatively pinched Kähler threefold without bounded holomorphic coordinates

Proof. Suppose that F=(F_1,F_2,F_3) is such a map, and choose B\ge1 so that |F_\ell|\le B on M_\phi for every component. Since M_\phi is open in \C^3, its coordinates (z_1,z_2,w) are global. The function

is holomorphic and nowhere zero on \B.

Fix z\in\B and put \delta=1-|z|. The restriction F_\ell(\,\cdot\,,0) is bounded and holomorphic on the entire base ball. For j=1,2, the disk \{z+\zeta e_j:|\zeta|<\delta\} is contained in that ball. Cauchy’s estimate, applied on disks with radii tending to \delta, gives

For this fixed z, the fiber in M_\phi is the disk of radius e^{-\phi(z)/2}. Cauchy’s estimate on this disk gives

In particular, the base estimate requires no uniform fiber radius near z; it is taken entirely on the zero section.

The six terms in the determinant expansion, together with (2.7) and (2.8), yield

and hence

Because \B is simply connected and d has no zeros, it has a global holomorphic logarithm \ell. For example, a primitive of the closed holomorphic one-form d^{-1}\partial d, with a suitable constant, satisfies e^\ell=d. The holomorphic function H=2\ell has \operatorname{Re}H=\log|d|^2, so (2.9) contradicts (2.4).□

Proof of Theorem 1.1. Choose \phi from Proposition 2.1, and then choose \lambda as in Proposition 2.2. The domain M=M_\phi\subset\C^3 is contractible and carries the asserted smooth complete Kähler metric with two uniform negative real-sectional bounds. Lemma 2.3 excludes a bounded holomorphic map to \C^3 with nowhere-vanishing Jacobian. A biholomorphism to a bounded domain would be such a map: the chain rule applied to its holomorphic inverse makes its Jacobian invertible everywhere.□

Proof. For each fixed r_N<1, the Taylor polynomials of H converge uniformly on r_NS. The representing identities therefore give

Integrating the proposed inequality contradicts the assumed limit.□

Proof. Put \rho=D_0/\sqrt{k}\le1. Maximality gives a covering by caps of radius \rho, and separation gives disjoint interiors for caps of radius \rho/2. Their measures yield 1\le N_k\rho^2 and N_k\rho^2/4\le1, proving (3.3).

We also need a local count whose constant does not depend on D_0. About the centers within distance u/\sqrt{k} of a fixed line, place caps of radius 1/(3\sqrt{k}). Their interiors are disjoint, and they lie inside the cap of radius (u+1/3)/\sqrt{k} about that line. Comparing measures bounds their number by 9(u+1/3)^2 when the latter radius is at most one. Otherwise their number is at most 9k\le9(u+1/3)^2, by comparison with the whole space. Thus the local count is at most C(1+u)^2 in all cases.

Since

the contribution from the annulus n\le\sqrt{k}d(\xi,p)<n+1 is at most C(n+2)^2e^{-n^2/2}. Summing these bounds from n=\lfloor u\rfloor onward proves (3.4): the polynomial factor is absorbed by a weaker Gaussian exponent, and the range 0\le u\le2 is absorbed by the constant. Setting u=0 and using homogeneity gives the first assertion in (3.5), including points outside the unit ball. Finally, (3.1) gives

The triangle inequality proves the stated integral bound.□

Proof. The projective radius of a patch is \sqrt{1-e^{-2R_0/k}}\le\sqrt{2R_0/k}, so the separation of the centers makes the patches disjoint. Formula (3.7) follows from |\xi_1|^2=e^{-2y/k} in (3.1).

If \xi\in\mathcal U_{k,p} and p'\ne p is another center, the metric triangle inequality gives

The peak from p is exactly e^{-y}e^{ik\theta}. Summing all other peaks by (3.4) proves (3.8). Outside every patch, each center satisfies

The final inequality uses k\ge2R_0 and 1-e^{-v}\ge v/2 for 0\le v\le1. Another application of (3.4) proves (3.9).□

Proof. The strict inequality 2(\cosh\alpha-1)>\alpha^2 allows a choice

Since a<2, this choice also satisfies s_1(1-e^{-\alpha})<e^{-\alpha}. Choose a small \delta>0 and a smooth cutoff \chi, equal to one on [0,\alpha], between zero and one, and zero for y\ge\alpha+\delta. For sufficiently small \delta,

Choose a smooth compactly supported function \nu, supported strictly after \alpha+2\delta, with 0\le\nu\le1, and with

Such a function is obtained by approximating the indicator of a sufficiently long interval beginning just after \alpha+2\delta. Then f=-s_1\chi+(N/P)\nu has the required plateau and weighted integral. The two supports are disjoint, so \|f\|_\infty\le\max\{s_1,N/P\}<1.

Define

It satisfies b'-b=f and b(0)=0. Beyond the support of f, the integral is exactly zero by the weighted cancellation, so b also has compact support. Near zero, b(y)=-s_1(e^y-1), which is smooth up to the boundary of the half-line. Smoothness here does not require extending f by zero to negative arguments. Finally choose R_0 larger than both supports, and enlarge it further until aCe^{-cR_0}\le1/2 in (3.9).□

Proof. Proceed by induction, beginning with W_0=1. The restriction S_p is a smooth periodic function. To check the central circle of a patch, use the smooth coordinates

All terms of (4.2) are smooth in these coordinates, including at \zeta=0. They do not depend on the transverse polar angle \gamma, which would be singular there. Their dependence on \theta is periodic, so it defines a smooth function along the full central circle. At the outer boundary, f and b vanish in a collar of R_0; hence the correction is identically zero in a collar of the patch boundary. Its extension by zero is therefore smooth. Summing the finitely many corrections proves the induction.□

Proof. At stage zero the phase assertions are immediate. The polynomial identity for \sigma follows by global phase averaging: every nonconstant homogeneous holomorphic polynomial has nonzero positive phase frequency.

Assume the assertions through stage j-1, and write k=k_j. The central restriction has a finite expansion

Since b'-b=f, the bracket in (4.2) is

All integers m=k+s are positive because \ell_{j-1}<k. The patch and y are invariant under global phase rotation, whereas \theta increases by the rotation angle. Hence the correction has frequencies in

They omit zero. Multiplying by a function of y and extending by zero across the invariant patch boundary do not alter these frequencies. The bandwidth of W_j is thus at most k+\ell_{j-1}=\ell_j, and its orbitwise mean remains one.

For the full moment calculation, use an adapted monomial \xi_1^n\xi_2^l, with nonnegative integers n,l. If l>0, averaging in the transverse phase \gamma gives zero. If l=n=0, averaging in \theta gives zero because all m=k+s are positive. Suppose then that l=0 and n\ge1. As b is real, the correction is

Since \xi_1^n=e^{-ny/k}e^{in\theta}, only the conjugate frequency with k+s=n can survive the \theta integration. Formula (3.7) gives exactly

Here a coefficient outside the previous bandwidth is zero. The factor 2/k in surface measure combines with the 1/2 from the real part to leave 1/k. Both endpoint terms vanish because b(0)=b(R_0)=0. The argument applies to every degree n, without a degree cutoff. Unitary changes of coordinates preserve holomorphic polynomials, so each correction annihilates every polynomial in the original coordinates. Adding the corrections proves (4.4) and completes the induction.□

Proof. Choose the largest 1\le l\le n such that k_l|u|\le1, or put l=0 if no such index exists. The comparisons between successive densities give

When l\ge1, logarithmic differentiation and the phase derivative bound show that the last ratio is at most \exp(K_*k_l|u|)\le e^{K_*}. For l=0 it equals one. If l=n, this already proves the assertion. Otherwise the maximality of l, including the case l=0, gives

It follows that R^{n-l}\le R(1+k_n|u|)^\beta, proving (4.6).□

Proof. Choose a fixed smooth compactly supported function \chi:\mathbb R\to\mathbb R which equals one on [-2,2], and put

For normalized convolution on the circle, the frequency assumption implies s''=s*K_q''. We claim that, for every positive integer N,

where C_N depends only on N,\chi.

Indeed the coefficients of K_q'' are

There are O(q) nonzero coefficients, each of size O(q^2), so direct summation gives |K_q''|\le Cq^3. An N-th finite difference of a_h is an iterated integral of the N-th derivative of q^2g(v/q), and hence has size at most C_Nq^{2-N}. Only O_N(q) such differences can be nonzero, since g is compactly supported and q\ge1. Their absolute sum is therefore at most C_Nq^{3-N}. Multiplication of the Fourier series by (1-e^{iu})^N produces these differences, up to an index shift. Using |1-e^{iu}|\ge2|u|/\pi on [-\pi,\pi] gives the second estimate

Combining it with the direct estimate proves (4.7).

Choose an integer N>\beta+1. Positivity, the assumed growth bound, and (4.7) now give

□

Proof. Besides (4.9), use

Differentiating (4.2) and taking absolute values gives

Since k\ge1, the assumed normalized bounds imply the stated estimate.□

Completion of the proof of Theorem 4.2. Fix K_*>2C_2/c_0. The constant C_1 in (4.8) is now fixed as well. We give a finite list of requirements on Q, all independent of the induction stage.

The horizontal derivative estimate at stage j-1, integrated along the radial arc just described, will imply

Indeed the logarithmic derivative along a unit horizontal arc is at most K_*\sqrt{k_{j-1}}, and the arc length is at most D_R/\sqrt{k_j}. At the initial stage the ratio is exactly one. Since k_j\ge Q, put

Choose a threshold Q_0 such that every integer Q\ge Q_0 satisfies all earlier degree and patch requirements, and also

These conditions are compatible: H(Q)\to1 and F=1-4c_0<1-2c_0.

We now fix any integer Q\ge Q_0 and induct on j. Assume the claimed estimates through stage j-1; at j=1 the preceding density is the constant one. Equation (4.8), obtained from earlier stages only, and the horizontal comparison (4.10) establish the normalized hypotheses of Lemma 4.6. Moreover,

At any point at most one patch correction is nonzero. Hence

which proves positivity and the required multiplicative comparison.

For Y=Z/\sqrt{k_j}, the previous derivative bound gives |YW_{j-1}|\le(K_*/\sqrt Q)W_{j-1}. For Y=T/k_j, it gives the smaller coefficient K_*/Q. Together with Lemma 4.6, (4.11), and the choice of K_*, these yield

Outside the patches the correction vanishes. Its smooth zero extension at their boundaries gives the same estimate there. Multiplying by \sqrt{k_j} or k_j proves the two derivative bounds and completes the induction.

Lemma 4.3 supplies the exact moments, rotation mean and bandwidth independently of these inequalities. In particular \int_SW_j\,d\sigma=1; positivity therefore makes W_j\,d\sigma a probability measure.□

Proof. Under the global phase rotation \xi\mapsto e^{i\omega}\xi, homogeneity gives P_k(e^{i\omega}\xi)=e^{ik\omega}P_k(\xi). Thus U_k(r\xi) has only phase frequencies in k\mathbb Z. By Lemma 4.3, the phase bandwidth of W_{j-1} is \ell_{j-1}<k and its constant coefficient on each orbit is one. Averaging the product over the phase therefore gives

Each patch is phase-invariant. Its radial variable y is fixed under rotation, while \theta and \gamma both increase by \omega. Write the central restriction as S_p(\theta)=W_{j-1}(e^{i\theta}p) =\sum_s\widehat S_p(s)e^{is\theta}. In the patch increment of (4.2), the frequencies are \pm(k+s) with |s|\le\ell_{j-1}<k. Pairing with a frequency in k\mathbb Z is possible only for s=0. Since \widehat S_p(0)=1 and S_p' has zero constant coefficient, the surviving part of the increment is exactly e^{2y/k}f(y)\cos(k\theta). Integrating this phase average proves (5.4).□

Proof. Put

We first compute the integral for a single peak:

At \rho=1, the unregularized logarithm has only one singularity, at (y,t')=(\alpha,0) modulo 2\pi. Near that point,

Its logarithm is locally integrable. For 0<\epsilon\le1, the regularized logarithm is bounded above uniformly, and its negative part is bounded by the negative part of the unregularized logarithm plus \log2. Dominated convergence therefore applies in (5.6).

For q<1, the mean and cosine moment of \log|1-qe^{it'}|^2 are 0 and -q, respectively, by the logarithm series. For q>1, factoring out q^2 gives mean 2\log q and cosine moment -q^{-1}. Using the cancellation \int_0^{R_0}e^{-y}f(y)\,dy=0, we obtain

The last equality uses f=-s_1 on [0,\alpha].

We now compare this model with (5.4). For k=k_j put \rho=r^k, so homogeneity gives U_k(r\xi)=F_\epsilon(\rho P_k(\xi)). Off the patches, |aP_k|\le1/2 by Lemma 3.4. If v=\rho P_k(\xi), 0\le\rho\le1, and s=|1-av|^2, then 1/4\le s\le9/4 and

The derivative in s of the first term has magnitude at most 16\epsilon^2, and |s-1|\le C|v|. Hence

The off-patch set is phase-invariant, and the phase mean of the unregularized logarithm is zero. By the L^1 estimate of Lemma 3.2, its total contribution is thus at most

Here C_{\rm off} is independent of D_0,k,\rho and 0<\epsilon\le1.

On a patch, the measure formula of Lemma 3.3 writes the combined contribution as

For fixed \epsilon,R_0, replace P_k(\xi) by e^{-y}e^{ik\theta} using the single-peak estimate, and replace e^{-2y/k} by one. The bounded derivative of F_\epsilon and the bound |e^{-2y/k}-1|\le2R_0/k show that the result differs from (2/k)I_\epsilon(\rho) by at most

All constants are uniform over patches and 0\le\rho\le1. The change from k\theta to t' in the angular average is exact, as k is an integer.

Choose \epsilon\in(0,1] so small that

By continuity of (5.6) in \rho, choose \eta\in(0,1/2) such that I_\epsilon(\rho)\le-3\Delta/4 for 1-\eta\le\rho\le1. Next choose D_* large enough for patch disjointness and for the first error in (5.9) to be at most \Delta/16 inside its factor 2/k. For each D_0\ge D_*, take k_* large enough for all the geometric estimates and for the remaining error inside this factor to be at most \Delta/16. Summing the patches and adding (5.8) gives, in particular,

Thus d_2=\Delta/(2D_0^2) is a valid choice.□

Proof. Include the density and degree thresholds in Q_*, and require Q\ge4. By Lemma 4.3, the increment W_\nu-W_{\nu-1} has frequencies within \ell_{\nu-1} of \pm k_\nu. Since \ell_{\nu-1}\le2k_{\nu-1}, each such frequency has absolute value at least (Q-2)k_{\nu-1}. Summing the increments with \nu>j shows that every frequency in W_N-W_j has absolute value at least (Q-2)k_j.

For fixed \xi,r, write the logarithmic profile as

The uniform polynomial bound gives |A|\le aC_P. Differentiating the logarithm shows that

Two integrations by parts bound its mth Fourier coefficient by B_\epsilon/m^2 for m\ne0. To track these coefficients as functions of \xi, define

Homogeneity and a change of variable give c_m(e^{i\omega}\xi)=e^{imk_j\omega}c_m(\xi). Thus c_m has global phase frequency mk_j, and U_{k_j}(r\xi)=\sum_{m\in\mathbb Z}c_m(\xi). The coefficient bound makes this series uniformly convergent. Its truncation to |m|\le Q-3 has uniform error at most 2B_\epsilon/(Q-3) and only frequencies of absolute value less than (Q-2)k_j. Phase averaging therefore makes its integral against W_N-W_j vanish. Both densities are positive probabilities by Theorem 4.2; hence \|W_N-W_j\|_{L^1(\sigma)}\le2. It follows that

Choose Q_* so that this is at most d_2/2, and apply (5.5).□

Proof. Fix 0<r_0<r_1<1. Around each point of |z|\le r_0, the complex polydisk of radius (r_1-r_0)/(2\sqrt2) lies in |z|<r_1. Cauchy estimates there, together with Lemma 3.2, give

for every fixed derivative order m. Here and below D^m denotes any real partial derivative of total order m. The values of P_k remain in a fixed bounded disk. Every positive-order derivative of F_\epsilon(P_k) is a sum of products containing at least one positive-order derivative of P_k or its conjugate. The bounded derivatives of F_\epsilon therefore give the same exponential estimate for D^mU_k. For m=0, use (5.3). Since \sum_j r_1^{Q^j}<\infty, the series converges in every local C^m norm.

Each summand vanishes at zero and is plurisubharmonic by (5.2). Termwise differentiation is justified by the local C^2 convergence, and hence

□

Proof. Put

For every j\le N, Bernoulli’s inequality gives

Thus all of the first N modes satisfy the hypothesis of (5.10) at this same radius. For j=N+m>N, on the other hand,

By the uniform polynomial bound and (5.3),

independently of N. Because W_N is a positive probability, the same bound controls the weighted tail, without requiring a uniform pointwise bound on W_N.

The base potential contributes only one logarithm:

Combining this with (5.10) and (6.4) yields

By Theorem 4.2, the measures \mu_N=W_N\,d\sigma are positive representing probabilities. Apply Lemma 3.1 to these measures, the radii r_N, and the continuous function u(z)=\phi(z)+4\log(1/(1-|z|)). The limit (6.5) is exactly its hypothesis, so the required minorant cannot exist.□

Proof. After a unitary change of coordinates, put the center at (r,0), where 0\le r<1, and let \delta=1-r. Rotating P_k preserves its degree and its uniform homogeneous bound. Use the chart T_r of (2.1), on the fixed complex polydisk

The denominators in T_r have modulus at least 1-\rho. We estimate the polynomial in the two ranges of k\delta.

High degrees: k\delta>1. The ball automorphism identity gives, on D_\rho,

It follows that 1-|T_r(\zeta)|\ge c\delta, with a fixed c>0. Thus

Cauchy estimates in this fixed polydisk bound every fixed-order holomorphic derivative at zero, including order zero, by C_m e^{-ck\delta}.

Low degrees: k\delta\le1. Consider the polydisk in the original coordinates

It may extend outside the ball, but P_k is an entire homogeneous polynomial. The bound of Lemma 3.2 therefore holds there as well. In fact

Cauchy estimates on the concentric half-polydisk give

Meanwhile the chart displacements satisfy

Both numerical coefficients are less than 1/2. As \delta\le1/k, the chart image lies in the half-polydisk just used. Integrating the gradient along the straight segment from (r,0) to T_r(\zeta) gives

This segment stays in the half-polydisk by convexity. Apply Cauchy estimates on D_\rho to the holomorphic difference P_k\circ T_r-P_k(r,0). Every positive-order derivative at zero is bounded by C_m\sqrt{k\delta}. We make no such assertion for the zeroth-order value in this range.

Composition and summation. In both ranges the values of P_k\circ T_r stay in a fixed bounded disk. Each positive-order real derivative of U_k\circ T_r=F_\epsilon(P_k\circ T_r) is a sum of products containing differentiated factors of P_k\circ T_r or its conjugate. For 1\le m\le4, the preceding bounds and the smooth composition estimate give

Indeed every term contains at least one differentiated factor, and a product of several factors obeys the same bound because the displayed scale factor is at most one. This proves the estimate for all real derivatives, hence for all pure and mixed complex derivatives of these orders.

Let J be the largest index with Q^J\delta\le1, or put J=0 if there is no such index. The low-degree contribution is bounded by the geometric sum

For the remaining degrees, the first value of Q^j\delta exceeds one, so

Both bounds are uniform in the chart center. Termwise differentiation is legitimate by Lemma 6.1. Multiplication by the fixed amplitude L preserves these bounds.

Finally, the base potential transforms as

Its positive-order derivatives through order four at zero are uniformly bounded. The last term is pluriharmonic, so the complex Hessian of \psi\circ T_r is exactly the standard ball Hessian in these coordinates. The derivatives of \phi\circ T_r needed for the pulled-back Hessian components and their first and second derivatives have orders two, three and four. The bounds just proved, together with h\ge I from Lemma 6.1, establish the assertion.□

Proof of Proposition 2.1. Choose a,\alpha,s_1,f,b,R_0 as in Lemma 3.4. Proposition 5.2 first chooses \epsilon,\eta and then permits any sufficiently large spacing D_0. Fix such a D_0 and its degree threshold k_*, and fix d_3=d_2/2>0. The density theorem holds for every Q above a threshold depending only on the fixed patch data and spacing. Its derivative constant K_* is fixed before that threshold is chosen. We may therefore choose one integer Q satisfying the density threshold, Q\ge k_*, and the Fourier-error threshold of Lemma 5.3.

Now fix a finite amplitude L satisfying (6.2). Lemma 6.1 constructs the smooth potential with h\ge I; Lemma 6.2 proves the required obstruction; and Lemma 6.3 supplies its finite uniform chart bounds. The chart constant is allowed to depend on these fixed choices, including L. The metric parameter \lambda in Proposition 2.2 is chosen only afterward from these bounds, so none of the choices above depends on \lambda.□

Proof. The domain and a smooth positive metric. The function t=e^\phi|w|^2 is smooth on \B\times\C, so M_\phi=\{t<1\} is open in \C^3. The map

is a smooth real diffeomorphism from M_\phi to \B\times\mathbb D, where \mathbb D=\{u\in\C:|u|<1\}. Its inverse is (z,u)\mapsto(z,e^{-\phi(z)/2}u). Thus M_\phi is contractible; no holomorphic trivialization is needed for this conclusion.

The potential \mathcal K_\lambda is smooth at every point of M_\phi, including its zero section. Write

The base covectors together with \vartheta form a global smooth complex coframe. Differentiating the potential gives

For a complex tangent vector u=(u_z,u_w) this means

Since h\ge0, the first term controls every nonzero horizontal projection through G\ge\lambda I; when u_z=0, the second term is positive unless u_w=0. Thus \widehat g is positive definite. At the zero section its matrix is particularly simple:

The mixed blocks vanish. Being a positive complex Hessian of a smooth real potential, \widehat g defines the stated Kähler metric.

Away from the zero section one can write s=\log t and q=x(1+x)=t/(1-t)^2. Then \partial s=\vartheta/w, so (7.1) also reads

We will use this expression for the curvature computation. Formula (7.1), rather than the coordinate s, defines the metric across w=0.

Two functions with bounded gradient. For a real smooth function f, the convention g=2\operatorname{Re}\widehat g gives

Indeed, for a real vector X with (1,0) part u, g(X,X)=2\widehat g(u,u) and df(X)=2\operatorname{Re}\partial f(u), and the identity follows by taking dual norms.

Globally on M_\phi,

In the coframe of (7.1), the Hermitian metric is block diagonal. Therefore

This calculation includes w=0, where both sides are zero. For the base potential, direct inversion of its complex Hessian gives

The covector \partial\psi has only base components in the same coframe, and G\ge\lambda I. Combining these facts with (7.4) yields

Compact containment of metric balls. Fix p\in M_\phi and a finite radius R\ge0. Integrating (7.5) along arbitrary piecewise smooth paths and taking the infimum of their lengths shows that d_g(p,q)\le R implies

This argument does not require a minimizing geodesic. Set

Every such q=(z,w) satisfies t(q)\le\tau_R and |z|\le\rho_R. Let m_R be the minimum of \phi on the compact base ball \{|z|\le\rho_R\}. Then

Consequently the metric ball is contained in

This set is closed and bounded in \C^3, hence compact, and the strict bounds \rho_R<1 and \tau_R<1 place it entirely inside M_\phi. In particular, a bounded-length curve cannot escape through the base boundary, through a fiber boundary, or through unbounded fiber coordinates. The zero section is an interior locus with the smooth positive metric (7.2).

The same compact-containment argument applies to sublevel sets of \mathcal K_\lambda. Both \psi and -\log(1-t) are nonnegative, so \mathcal K_\lambda\le R implies \psi\le R/\lambda and -\log(1-t)\le R. These bounds place the sublevel set in a compact set of the form (7.6); the sublevel set is closed there, hence compact. Positivity of its complex Hessian was proved above, so \mathcal K_\lambda is a strictly plurisubharmonic exhaustion.

Geodesic completeness. Suppose a unit-speed geodesic has a finite maximal forward time T. Its image lies in the compact set K_T. On this set the smooth positive metric has a uniform positive Euclidean lower bound, so the geodesic’s Euclidean velocity is bounded. The Christoffel symbols are also bounded on K_T, and the geodesic equation then bounds its Euclidean acceleration. Both position and velocity have limits as time tends to T, with limiting position in K_T\subset M_\phi. Local existence for the smooth geodesic equation extends the geodesic beyond T, a contradiction. The same argument in reverse time proves geodesic completeness.□

Proof. Write F(s)=-\log(1-e^s). Direct differentiation gives

The centered-chart expression for \psi differs from -\log(1-|z|^2) by a pluriharmonic function and a constant. Its derivatives of type (2,1) therefore vanish at the center, and its derivatives of type (2,2) are S_I.

For brevity in the derivative tables, put

Before normalizing the vertical vector, the only possibly nonzero third derivatives of type (2,1), up to symmetry, are

The fourth derivatives of type (2,2), up to symmetry and conjugation, are

To check that these lists are complete, apply the chain rule to F(s) and use (8.1). In the four-horizontal derivative, the two pairings of mixed second derivatives of s give qS_h; the third pairing contains s_{ik}s_{\bar j\bar l}=0. All other terms contain a vanished first or pure second horizontal derivative. In particular, \mathcal K_{ik\bar v}=\mathcal K_{vv\bar j}=0, \mathcal K_{ik\bar v\bar v}=0, and the fourth derivatives with one horizontal and three vertical indices vanish. No vanishing of pure third horizontal derivatives is needed.

Substitution into (8.3) gives (8.5) and (8.6). For clarity, all remaining curvature types in the normalized frame are

The last identity follows because its unnormalized value is

and the four vertical normalization factors multiply to q^{-2}. The tensors S_V and S_{D+V}-S_D-S_V give precisely the vertical and balanced two-horizontal blocks in (8.11). The Kähler symmetries and conjugation now account for every component, proving (8.4).

The lower bound h\ge cI implies

The assumed component-derivative bounds for h, applied to (8.5) and (8.6), prove (8.7) with constants independent of \lambda\ge1 and x>0.

Finally, G=\lambda I+xh commutes with h at the center. On an eigendirection of h with eigenvalue d, the ratio of D to h is

The subtracted fraction is nonnegative, and it is at most x when \lambda\ge d. This proves (8.8).□

Proof. The real squared area of the pair is 4a_{\widehat g}, because g(U_i,U_j)=2\operatorname{Re}\widehat g(u_i,u_j). In particular a_{\widehat g}>0 for a real linearly independent pair. Write

The Kähler symmetries in the expansion on U_j=u_j+\bar u_j give the real numerator

Division by the real squared area proves (8.14). Expanding the right-hand side of (8.16) gives the same expression A-\operatorname{Re}B for T. Finally, if c_P=P(u_1,u_2), direct contraction of (8.2) gives

□

Proof. Complex Cauchy–Schwarz gives P(u_1,u_1)P(u_2,u_2)\ge|P(u_1,u_2)|^2, proving a_P\ge b_P^2. Equality holds for V because its complex rank is one. Let A and B be the two-by-two real Gram matrices with entries \operatorname{Re}P(u_i,u_j) and \operatorname{Re}V(u_i,u_j), respectively. Then

For positive definite A, put C=A^{-1/2}BA^{-1/2}. The last expression equals (\det A)\operatorname{tr}C, whereas \sqrt{a_Pa_V}=(\det A)\sqrt{\det C}. The inequality \operatorname{tr}C\ge2\sqrt{\det C} proves (8.18), and continuity gives the semidefinite case. If both determinants are positive, equality means that C is a scalar matrix, which is exactly the claimed proportionality. The displayed formula for m_P is linear in the entries of P. Its nonnegativity for positive semidefinite P then also proves monotonicity.□

Proof. In (8.16), the tensor T_4 pairs two horizontal blocks of \Xi. The tensor T_3 pairs one horizontal block and one mixed block, since it has exactly three horizontal indices. Cauchy–Schwarz and (8.7) give (8.21) and |\mathcal H_{T_4}|\le Cx\|\Xi_{HH}\|^2. Now use (8.19). These estimates remain valid when one of the projected real areas vanishes.□

Proof. If a_V=0, then b_V=0, because the vertical factor has complex dimension one. If a_h=0, the two horizontal projections are linearly dependent over \mathbb R, since h is positive definite. Their pairing under every Hermitian form is therefore real, so b_E=0. Thus a_ha_V=0 implies b_Eb_V=0, and

Here the last inequality uses the positive definiteness of h+V and the real linear independence of the pair.

Suppose now that A=a_h>0 and U=a_V>0. The Hermitian Gram determinant gives 0\le\tau\le1. Make a real change of basis in the plane so that its horizontal projections, denoted u_H,v_H, are real-orthonormal for h. Both ratios |b_h|/\sqrt{a_h} and |b_E|/\sqrt{a_h} are unchanged by this operation. The squared h-norms of u_H+i v_H and u_H-i v_H are 2(1+\tau) and 2(1-\tau), in some order. The absolute difference of their squared E-norms is 4|b_E| in the new basis. Comparing one norm from above and the other from below gives

For either order of the two norms, the displayed right side is an upper bound, since 2-\beta\ge1. The last inequality follows from

it is strict unless \tau=1. This argument does not require \beta\ge0.

The mixed Gram inequality (8.18) and the Hermitian Gram inequality yield

Since U=b_V^2, it follows that

which proves (9.1).

It remains to examine equality when h\le E\le2h. The zero-area case already treated shows that A,U>0. Set s=\sqrt{A+3b_h^2} and B=|b_E|. The inequalities above, with \beta=0, give

Equality forces s=2\sqrt U, B=s, and equality in the bounds used to reach this expression. The strictness in (9.2) gives \tau=1. Consequently

Since a_E\ge B^2 and m_E\ge2\sqrt{a_EU}, these identities force a_E=B^2 and equality in the mixed Gram inequality. The horizontal real E-Gram matrix and the vertical real Gram matrix are therefore proportional.

In the real h-orthonormalized basis used above, \tau=1 says that the horizontal projections have the form

Their real E-Gram matrix is a positive scalar multiple of the identity. Proportionality implies that the vertical projections have equal positive norms and real inner product zero. The sign b_Eb_V<0 selects the opposite complex orientation, so these projections have the form

Direct substitution now gives

The other mixed block vanishes as well. Under any real change of basis in the plane, \Xi is multiplied by the determinant of that change. The vanishing therefore holds in the original basis.□

Proof. Increase the prospective threshold so that \lambda\ge\max(1,C). Then (8.8) gives

If the lemma were false, there would be a sequence \lambda\to\infty of permitted data and real two-planes for which F=-\mathcal H_R\le0. We suppress the sequence index. By (8.22),

The contracted estimates (8.20) apply to the allowed error tensors by their component types and symmetries.

The regime q/\lambda\to0. Suppose first that a subsequence has this property. Monotonicity and linearity of the mixed Gram term give m_D\ge c(1+x)m_I, while

Discarding nonnegative terms in (9.3), we get

Absorb the first product into 2a_V, and the last into \frac c2(1+x)m_I. Since q/(1+x)=x, this gives

Here (1+x)^2+x=O(1+q)=o(\lambda), because \lambda\to\infty. Every coefficient is eventually positive, whereas a_I+a_V+m_I=a_{I+V}>0. This contradicts F\le0.

The regime with \lambda/q bounded. If the preceding regime has no subsequence, we may pass to one on which q/\lambda is bounded below by a positive constant. Then \lambda/q is bounded and x\to\infty. The term qS_h has four horizontal slots, whereas the vertical term S_V already has coefficient one. To balance these two blocks, multiply horizontal vector components by q^{1/4}. In the resulting plane choose a real-orthonormal pair for \operatorname{Re}(I+V). The inverse transformation and a real basis change give a basis of the original plane; the latter change multiplies F by a positive factor. Thus its sign is unchanged. The rescaled pairs belong to a fixed compact Stiefel manifold, and their quantities satisfy a_{I+V}=1.

In the rescaled variables, the leading part of (9.3) is

The transformed tensors, denoted \widetilde T_4,\widetilde T_3, satisfy

Indeed, each horizontal slot supplies a factor q^{-1/4}.

The lower bound on D gives E\ge h. For an eigenvalue d\in[c,C] of h, the corresponding ratio is

Since

we have the uniform bound

The constants c_1>0,C_1<\infty depend only on c,C.

Pass to a subsequence on which h,E,\lambda/q, and the normalized pair converge. The limiting forms satisfy h\le E\le2h. The error contributions tend to zero by (9.5). Lemma 9.1 shows that the two limiting summands in (9.4) are nonnegative. Since F\le0, both must vanish. In particular P_h(E)=0 at the limit, so a_h,a_V>0 there and the mixed blocks of the limiting \Xi vanish. The comparison

follows from the corresponding real Gram-matrix bounds. Hence a_I,a_h,a_V are bounded below by positive constants along this subsequence.

We now improve the error estimate on these particular planes. By the contraction formula (8.16), a tensor with exactly three horizontal indices can pair only one horizontal-horizontal block with one mixed block. Thus

Here \Xi_{HH} denotes the block (\Xi^{i\bar j}), and \Xi_{\mathrm{mix}} comprises the blocks (\Xi^{i\bar0}) and (\Xi^{0\bar i}). The first block is bounded because the normalized pairs form a compact set, and the mixed blocks tend to zero by the equality case just proved. The explicit tensor factor q^{-1/4} is what makes this an o(x^{-1/2}) estimate for the evaluated contribution. No rate of convergence of the planes is being assumed.

Apply (9.1) before taking the limit, with

This choice is permitted by (9.6), including when \beta<0. The positive lower bounds for a_I,a_h,a_V, and boundedness of the normalized pairs, imply

along the subsequence, for some c_2>0,C_2<\infty. For x\ge1, if \lambda\ge x^{3/2}, then

If instead \lambda\le x^{3/2}, then

Thus (9.8) is bounded below by a positive constant times x^{-1/2} for large x. The evaluated \widetilde T_4 contribution is O(x^{-1})=o(x^{-1/2}), and (9.7) gives the same little-o comparison for \widetilde T_3. It follows that F>0 eventually, a contradiction.

The two regimes exhaust the possible failure sequences. This proves the asserted finite threshold \lambda_0, uniformly over all permitted matrices, error tensors, and real planes.□

Proof of Proposition 2.2. The geometric conclusions concerning the domain and completeness were proved in Lemma 7.1. It remains to obtain both uniform real sectional-curvature bounds. By Proposition 8.1, the errors of the actual metric have the types and bounds in Lemma 9.2, with C_{\mathrm e} depending only on the chart bounds of Proposition 2.2. Choose any finite \lambda\ge\lambda_0. Then -\mathcal H_R>0 for every point and every real two-plane. We show that the sectional-curvature quotient is bounded away from zero and from negative infinity.

First restrict x to a bounded interval [0,X]. Normalize the pairs for \operatorname{Re}(I+V). The data consisting of x, the matrix h, the component arrays of T_4,T_3, and the normalized pair range over a closed bounded subset of a finite-dimensional space: impose cI\le h\le CI, the stated component types and symmetries, and the closed norm bounds C_{\mathrm e}x,C_{\mathrm e}\sqrt q. This set is compact. The formulas for G,D,R are continuous on it. Since Lemma 9.2 applies to this entire set, -\mathcal H_R has a strictly positive minimum and a finite maximum. The metric G+V is uniformly positive definite and bounded there, so a_{G+V} is bounded above and away from zero. Formula (8.14) therefore gives both a strictly negative upper curvature bound and a finite lower curvature bound on 0\le x\le X.

To cover the end x\to\infty, use the horizontal rescaling from the preceding proof. Here \lambda remains fixed: the mixed form D/\sqrt q will converge to h, rather than to an arbitrary form between h and 2h as allowed in the preceding contradiction argument. The rescaled Hermitian metric is

For the fixed finite \lambda,

uniformly in the permitted data. Because G commutes with h, the formula for D also gives

Consequently

The rescaled tensor identity now implies

in component norm, uniformly over all allowed data. Indeed, the term \lambda S_I is divided by q, the term qS_h becomes exactly S_h, the mixed term converges by (9.10), and the error tensors satisfy (9.5).

For the limiting metric h+V and tensor -S_{h+V},

For an (I+V)-real-orthonormal pair, a_{h+V}\ge\min(c,1)^2, so these area denominators are uniformly bounded away from zero. The forms h+V also range over a compact bounded family. The metric and tensor estimates (9.9)–(9.11) therefore imply uniform convergence of the corresponding sectional-curvature quotients over all real planes. For sufficiently large X_\lambda<\infty, they give, for example,

Combining this with the compact-range bounds at X=X_\lambda produces finite constants 0<A\le B for which

for every point and every real two-plane. The tensor identity at x=0, given in (8.12), is computed in a smooth normalized fiber frame; equivalently the same bounds extend to the zero section by smoothness of the metric. This completes Proposition 2.2.□