The Canonical Construction of $\mathcal I$ Using Differential Geometry

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We will construct a highly symmetric leaf (an interface leaf) as a special subset in some $\mathcal F$-completion, this time furnishing the leaf with additional geometric data including a metric and curvature.

We start by asking the following optimization question:

Fix $n=3$ and consider a surface of revolution $S$ and an embedding $e:S \hookrightarrow X^3$ for $X^3=[0,1]^3$ with points $p,q$ elements of $\partial X^3$ where $\partial X^3=X^3-(0,1)^3$ for $\mathrm {sup}~ \mathrm{dist}(p,q)=\sqrt{3}$.   

What is $\rho_{\mathrm{max}}=\mathrm{max} \lbrace \mathrm{vol}(S) \rbrace_{p,q}$ assuming $S$ must remain a surface of revolution and have constant positive Gaussian curvature?


An abstract surface of revolution with constant positive Gaussian curvature (to be embedded/optimized) within $X^3.$


In other words, what is the volume of the largest surface of revolution with constant Gaussian curvature that can be embedded in $X^3$ with a pair of antipodal corners as cone points?

Let $a = \frac{2}{\sqrt{3}}$. For a positive $b$ and $0 < u < a/b$, define the function 

$$\phi(u) = au - bu^2 = u(a - bu),$$

which gives a derivative 

$$\phi'(u) = a - 2bu.$$

Due to geometric considerations, there exists a unique positive $b$ such that

$$\int_0^{a/b} \sqrt{\frac{1 - \frac{1}{4} \phi'(u)^2}{\phi(u)}} \, du = \sqrt{3}.$$

The maximum volume of a surface with constant Gaussian curvature, enclosed within a unit cube and featuring cone points at a pair of opposite corners, is given by

$$\pi \int_0^{a/b} \sqrt{\bigl(1 - \tfrac{1}{4} \phi'(u)^2\bigr) \phi(u)} \, du.$$

Numerical approximation methods are necessary for further analysis.

Metric and Gaussian Curvature:

Given a smooth function $\phi(u)$ of class $C^2$ on a real interval, consider the metric

$$g_\phi = \frac{1}{\phi(u)} \, du^2 + \phi(u) \, dv^2.$$

For this metric, the Gaussian curvature is 

$$K_\phi = -\frac{1}{2} \phi''(u).$$

Thus, surfaces of constant Gaussian curvature correspond to cases where $\phi(u)$ is a quadratic polynomial.

A surface can embed as a surface of revolution in three-dimensional Euclidean space only if $|\phi'(u)| \leq 2$. If this surface arises from rotating the curve $y = \xi(x)$ about the $x$-axis, then the relationships are:

  • $\phi(u) = \xi(x)^2$,
  • $du = \xi(x) \sqrt{1 + \xi'(x)^2} \, dx$,
  • $\xi'(x) = \frac{\phi'(u)}{\sqrt{4 - \phi'(u)^2}}$.

Here, $|\phi'(u)| = 2$ corresponds to a vertical tangent on the graph of $y = \phi(x)$.

Geometric Problem:

Consider the unit cube $[0, 1]^3$ with a diagonal $\ell$ connecting the corners $(0, 0, 0)$ and $(1, 1, 1)$. The goal is to find the "momentum profile"

$$\phi(u) = au - bu^2$$

where $b$ determines the Gaussian curvature, such that:

  • The rotated graph $y = \xi(x)$ fits entirely within a cube corner.
  • The surface closes at the opposite corner, satisfying $\xi(\sqrt{3}) = 0$.

The condition for fitting within the cube ensures the slope of the generator (viewed along $\ell$) does not exceed $1/\sqrt{2}$. At $u = 0$ (where $x = 0$), this slope restriction becomes:

$$\frac{1}{\sqrt{2}} = \xi'(0) = \frac{\phi'(0)}{\sqrt{4 - \phi'(0)^2}} = \frac{a}{\sqrt{4 - a^2}}.$$

Solving this gives $a = \frac{2}{\sqrt{3}}$. Thus, the momentum profile is $\phi(u) = u(a - bu)$, with $b$ determined by additional conditions.

Surface Properties:

The relationship between $x$ and $u$ is governed by the differential equation:

$$\frac{dx}{du} = \frac{1}{\xi(x) \sqrt{1 + \xi'(x)^2}} = \frac{1}{\sqrt{\phi(u)}} \sqrt{1 - \tfrac{1}{4} \phi'(u)^2}.$$

The surface length is equal to the diagonal of the cube, $\sqrt{3}$, if and only if

$$x(a/b) = \int_0^{a/b} \sqrt{\frac{1 - \frac{1}{4} \phi'(u)^2}{\phi(u)}} \, du = \sqrt{3}.$$

This implies a unique positive $b$ exists, though finding it explicitly is challenging.

Volume Calculation:

The volume enclosed by the surface of revolution is:

$$\pi \int_0^{\sqrt{3}} \xi(x)^2 \, dx = \pi \int_0^{a/b} \phi(u) \frac{dx}{du} \, du = \pi \int_0^{a/b} \sqrt{\bigl(1 - \tfrac{1}{4} \phi'(u)^2\bigr) \phi(u)} \, du.$$

Obtaining the Interface Construction, $\mathcal I$:

To obtain the $\mathcal I$ from the above arguments, we simply use our construction to place $4$ total copies of the interface leaf in $X^3$. Here is what that looks like:



Here is another way to view $\mathcal I$:


Thus, if we index the leaves in our $\mathcal F$-completion by enclosed volume, then we can view $\mathcal I$ as a constant time slice in our $\mathcal F$-completion - perhaps the above depiction corresponds to what we calculated already, so the volume is around $1/2$ for each of the four surfaces. 



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