6. The local repair and the final profile

6.1 The linear model near a zero

Once the balanced \(2/m\)-power had been shown to work away from its zeros, the remaining question was local. Near the prescribed direction \(p_1\), put

\[ t=1-\ip{x}{p_1}, \qquad P(x)=t^\alpha P_{\mathrm{reg}}(x), \qquad P_{\mathrm{reg}}(p_1)\gt{}0. \]

The other pole factors are smooth and positive there, so the leading behavior is determined by the single exponent \(\alpha\). This reduced the first question to a direct one-pole calculation: which powers retain a positive first-order gap as \(t\downarrow0\)?

The fixed-\(k\) one-pole calculation ruled out exponents that were too small. If \(C=P_{\mathrm{reg}}(p_1)\gt{}0\), its leading term is

\[ \tag{6.1}\label{eq:local-alpha-expansion} W-s =Ct^{\alpha-1} \left( \frac{\alpha\bigl((n-1)\alpha-(n-2)\bigr)}{2n} +O(t) \right). \]

Thus a strictly positive leading margin requires

\[ \alpha\gt{}\frac{n-2}{n-1}. \]

The natural dimension-independent choice is therefore \(\alpha=1\), corresponding to the linear model

\[ \phi(t)\sim ct \qquad(t\downarrow0). \]

A direct substitution of this model restores the positive near-zero margin that was lost by the pure \(t^{2/m}\) power.

6.2 The logarithmic transition and the rank threshold

The next question was whether the local exponent \(1\) could be connected to the successful outer exponent \(2/m\). This is a different \(k\downarrow0\) calculation from the fixed-\(k\) zero model above. An arbitrary smooth cutoff need not work, because its second derivative appears in the first-order threshold. For a positive profile \(\phi\), we therefore introduced its logarithmic slope and curvature

\[ \xi=\frac{\dd\log\phi}{\dd\log t}, \qquad \eta=\frac{\dd\xi}{\dd\log t}. \]

Thus \(\xi=1\) is the linear model, \(\xi=a\) is a power \(t^a\), and \(\xi=0\) is a plateau. Applying the small-parameter calculation to the one-pole transition gives a second-order differential inequality for \((\xi,\eta)\). Here the ambient dimension \(n\) disappears; only the active rank \(\ell\) remains. The two endpoint regimes give the necessary conditions

\[ \eta\gt{}-\xi^2+\frac{3\ell-4}{2(\ell-1)}\xi-\frac{1}{2} \quad(\ell\geq2), \qquad \eta\gt{}\frac{\ell-3}{2(\ell-2)}\xi-\frac{1}{2}\xi^2 \qquad(\ell\geq3). \]

Setting \(\eta=0\) shows that a terminal power \(t^a\) must satisfy

\[ \tag{6.2}\label{eq:rank-threshold} a\gt{}\frac{\ell-3}{\ell-2} \qquad(\ell\geq3). \]

The rank dependence became transparent in this ODE.

  1. Rank two. The radial condition, which is the full logarithmic condition in rank two, becomes
\[ \eta+\xi^2-\xi+\frac{1}{2}\gt{}0. \]

It allows \(\xi\) to decrease from \(1\) all the way to \(0\) in a finite logarithmic interval. After rescaling, this transition can be placed in an arbitrarily small pole cap. Outside the caps every factor is constant, so \(P\) itself is constant and the rank-two construction becomes especially simple.

  1. Rank three. A convenient sufficient condition for the full one-pole ODE is
\[ 0\lt{}\xi\leq1, \qquad \eta\geq-c\xi^2, \qquad 0\lt{}c\lt{}\frac{7}{32}. \]

This condition prevents \(\xi\) from reaching \(0\) in finite logarithmic time, explaining the failure of the earlier linear-to-plateau attempts. It does, however, allow \(\xi\) to decrease from \(1\) to any fixed positive exponent, in particular to \(2/m\). The logarithmic transition has a fixed finite length once \(m\) is fixed; shrinking its scale places the entire transition in an arbitrarily small physical neighborhood of the prescribed direction.

  1. Rank four. Condition (6.2) requires a terminal exponent strictly greater than \(1/2\). Thus the balanced exponent \(2/m\) has zero limiting margin when \(m=4\), and it lies below the admissible range when \(m\gt{}4\). This local obstruction already shows why the same test-function design encounters a genuine difficulty in a rank-four sheeting problem.

6.3 The final profile and its verification

For the rank-three construction, choose a small scale \(T\gt{}0\) and a smooth nondecreasing profile such that

\[ \phi_T(t)= \begin{cases} T^{2/m-1}t, & 0\leq t\leq T,\\ C_m t^{2/m}, & t\geq e^L T, \end{cases}, \]

where \(L\lt{}\infty\) is the logarithmic length of the chosen transition. It may be arranged that throughout the transition

\[ \frac{2}{m}\leq\xi\leq 1, \qquad \eta\geq-\frac{1}{8}\xi^2. \]

The resulting multipole tilt function is

\[ \class{key-formula-math}{G_{\mathbf p}(\nu) =\left( 1-\nu_1^2-\nu_2^2-\nu_3^2 +k\prod_{j=1}^m\phi_T\bigl(1-\ip{\nu}{p_j}\bigr) \right)^{1/2}.} \]

To guard against floating-point error in the very small sampled minima, all entries in the last column were computed using 90-digit arithmetic.

Pole configuration \(m\) \(T\) \(k\) Sampled \(\min(W-s)\)
Coordinate triple 3 \(1.35\times10^{-4}\) \(5.53\times10^{-5}\) \(4.52\times10^{-5}\)
Tetrahedral four 4 \(2.47\times10^{-6}\) \(1.20\times10^{-8}\) \(5.24\times10^{-6}\)
Five-point cluster 5 \(3.90\times10^{-11}\) \(4.10\times10^{-15}\) \(1.17\times10^{-16}\)
Five points with a close pair 5 \(1.77\times10^{-12}\) \(6.40\times10^{-16}\) \(6.39\times10^{-11}\)
Octahedral six 6 \(8.28\times10^{-10}\) \(1.20\times10^{-18}\) \(2.78\times10^{-9}\)
Table 6.1. Finite-\(k\) numerical verification of the final power-tail profile.

All sampled margins were positive. This supported the candidate without proving a uniform lower bound, so the rigorous proof used three regions.

  • Near a zero, freeze the nonvanishing factors and use the direct linear one-pole calculation.

  • Away from the zeros, use the strict \(k\downarrow0\) gap from the previous section.

  • In a transition annulus, rescale and use the logarithmic ODE; this is laborious but no longer exploratory.

Fix \(T\) first to control the pole caps and transition errors, then choose \(kT^{2/m-1}\ll1\).

Gaoming Wang
Gaoming Wang
Assistant Professor

My research interests include Geometric Analysis and Partial Differential Equations.