3. From the geometric inequality to a numerical criterion

We selected orthonormal Euclidean directions \(e_1,\ldots,e_\ell\), set

\[ x=(\nu_1,\ldots,\nu_\ell)^{\mathsf T}, \qquad \nu_i=\ip{\nu}{e_i}, \]

and began with an arbitrary positive function

\[ \tag{3.1}\label{eq:direct-G-model} g^2=G(\nu_1,\ldots,\nu_\ell)=G(x). \]

The aim was to check directly whether

\[ \abs{A}^2g^2+g\Delta_M g\geq c\abs{A}^2 \]

could hold with \(c\gt{}0\).

Since \(M\) is minimal, its Gauss-map components satisfy

\[ \Delta_M\nu_i=-\abs{A}^2\nu_i, \qquad 1\leq i\leq\ell. \]

Write

\[ G_i=\frac{\partial G}{\partial\nu_i}, \qquad G_{ij}=\frac{\partial^2G}{\partial\nu_i\partial\nu_j}. \]

Applying the chain rule to (3.1) gives

\[ \tag{3.2}\label{eq:direct-coordinate-identity} \class{key-formula-math}{\abs{A}^2g^2+g\Delta_M g =W\abs{A}^2- \sum_{i,j=1}^\ell\alpha_{ij}\ip{\nabla\nu_i}{\nabla\nu_j},} \]

where

\[ \tag{3.3}\label{eq:direct-W} W=G-\frac{1}{2}\sum_{i=1}^\ell\nu_iG_i, \qquad \alpha_{ij}=-\frac{1}{2}G_{ij}+\frac{G_iG_j}{4G}. \]

The remaining task was to find a constant \(s\) such that

\[ \class{key-formula-math}{\sum_{i,j=1}^\ell\alpha_{ij}\ip{\nabla\nu_i}{\nabla\nu_j} \leq s\abs{A}^2.} \]

The remaining term in (3.2) was organized in a frame adapted to the selected directions. Put

\[ e_i^\top=e_i-\nu_i\nu, \qquad 1\leq i\leq\ell, \]

and let

\[ \pi=\operatorname{span}\{e_1^\top,\ldots,e_\ell^\top\} \subset T_pM. \]

At an interior point \(\abs{x}\lt{}1\), the vectors \(e_1^\top,\ldots,e_\ell^\top\) are linearly independent. To keep this exploratory reduction as simple as possible, we imposed the additional ansatz that \(\pi\) is spanned by principal-curvature directions. We may then choose an orthonormal principal frame \(\tau_1,\ldots,\tau_n\) such that \(\pi=\operatorname{span}\{\tau_1,\ldots,\tau_\ell\}\), and write

\[ \tag{3.4}\label{eq:active-principal-frame} A(\tau_a,\tau_b)=\lambda_a\delta_{ab}, \qquad 1\leq a,b\leq n. \]

Equation (3.4) gives

\[ \tag{3.5}\label{eq:gradient-nu-principal-frame} \nabla\nu_i =-\sum_{a=1}^\ell \lambda_a\ip{e_i}{\tau_a}\tau_a, \qquad \sum_{a=1}^\ell \ip{e_i}{\tau_a}\ip{e_j}{\tau_a} =\delta_{ij}-\nu_i\nu_j. \]

Consequently,

\[ \sum_{i,j=1}^\ell\alpha_{ij}\ip{\nabla\nu_i}{\nabla\nu_j} =\sum_{a=1}^\ell B_{aa}\lambda_a^2, \]

where the coefficients required by the numerical calculation are given directly by

\[ \tag{3.6}\label{eq:B-entrywise} B_{aa} =-\frac{1}{2}\sum_{i,j=1}^\ell G_{ij}\ip{e_i}{\tau_a}\ip{e_j}{\tau_a} +\frac{1}{4G} \left(\sum_{i=1}^\ell G_i\ip{e_i}{\tau_a}\right)^2. \]

For simplicity, assume \(n\gt{}\ell\) and \(\lambda_{\ell+1}=\cdots=\lambda_n\). By minimality, the sharp choice of \(s\) is the smallest number such that

\[ \sum_{a=1}^\ell B_{aa}\lambda_a^2 \leq s\left( \sum_{a=1}^\ell\lambda_a^2 +\frac{(\lambda_1+\cdots+\lambda_\ell)^2}{n-\ell} \right). \]

Define

\[ \tag{3.7}\label{eq:threshold-polynomials} p(t)=\prod_{a=1}^\ell(t-B_{aa}), \qquad P(t)=(n-\ell)p(t)+tp'(t). \]

The sharp value of \(s\) is the largest real root of \(P\); thus we take

\[ \tag{3.8}\label{eq:s-largest-root} s=\max\{t\in\R\colon P(t)=0\}. \]

Hence

\[ \abs{A}^2g^2+g\Delta_M g\geq (W-s)\abs{A}^2. \]

Numerical verification routine

A numerical verification routine takes

\[ x\in\left\{\abs{x}\leq 1\colon G(x)\gt{}0\right\} \]

and samples the coefficients \(\ip{e_i}{\tau_a}\) subject to the second identity in (3.5). It evaluates \(G,G_i,G_{ij}\), constructs \(W,\alpha,B_{aa},p,P,s\) by (3.3)–(3.8), and computes

\[ \mathfrak m_{n,\ell}=W-s. \]

The candidate passes the sampled test precisely when the sampled minimum of \(\mathfrak m_{n,\ell}\) is positive.

Warning 3.1

The numerical experiments were used only to identify which test functions were promising. Even when a candidate passed every sampled test, the required uniform inequality still had to be verified by rigorous mathematical calculation.

3. From the geometric inequality to a numerical criterion

Gaoming Wang

Assistant Professor