The Lumenary · Research pathways

Geometric analysis

Name: ______________________

Date: ______________________

You will learn to

Derive the minimal surface equation as the Euler–Lagrange equation of area, verify that the catenoid and Scherk's surface are minimal, and prove by calibration that minimal graphs minimise area
Explain with the catenoid why a minimal surface need not minimise area, and state the solution of the Plateau problem by Douglas and Radó
Define harmonic maps and state the Eells–Sampson theorem, and describe the Yamabe problem and its solution
Explain the Ricci flow equation, solve it on round spheres, prove a maximum principle for a discrete heat flow, and explain why singularities and surgery arise
Survey mean curvature and curve-shortening flow, and separate established results from open problems

1.
Under Ricci flow $\partial_t g = -2\,\mathrm{Ric}(g)$ , a round $n$ -sphere of radius $r_0$ has $g(t) = (r_0^2 - 2(n-1)t)\,g_{S^n}$ . At what time does a round 3-sphere of radius $r_0 = 4$ shrink to a point?
2.
Which of these surfaces in $\mathbb{R}^3$ are minimal (mean curvature identically zero)? (Choose all that apply.)
- A.A plane
- B.A catenoid
- C.A helicoid
- D.A sphere of radius $r$
- E.A circular cylinder of radius $r$
3.
Two coaxial rings of radius $1$ are placed at $x = \pm 0.5$ . Two catenoids span them, with $a \approx 0.848$ (fat) and $a \approx 0.235$ (thin). Which statement is correct?
- A.Both are minimal surfaces, but the thin one has larger area and is unstable.
- B.Only the fat one is a minimal surface, since it has less area.
- C.Both minimise area among all surfaces spanning the rings.
- D.By the calibration theorem, the fat catenoid minimises area among all surfaces spanning the rings.
4.
In the calibration proof that minimal graphs minimise area, $\omega(X, Y) = \det(\nu, X, Y)$ for the unit vector field $\nu$ . Which property gives the inequality $\int_S \omega \le \mathrm{Area}(S)$ for every surface $S$ ?
- A. $|\nu| = 1$ , so $\omega(X, Y) = \nu \cdot (X \times Y) \le 1$ for orthonormal $X, Y$ .
- B. $d\omega = 0$ .
- C.The minimal surface equation, directly.
- D.Convexity of the domain $\Omega$ .
5.
A smooth embedded closed curve in the plane encloses area $4\pi$ and evolves by curve-shortening flow. At what time does it shrink to a point?
6.
The Lean theorem round_sphere_ricci_flow proves, for $\rho(t) = r_0^2 - 2(n-1)t$ and $T = r_0^2/(2(n-1))$ : $\rho' = -2(n-1)$ , $\rho(0) = r_0^2$ , $\rho(T) = 0$ and $\rho(t) > 0$ for $t < T$ . What does it not establish?
- A.That the round metric has $\mathrm{Ric} = (n-1)g_{S^n}$ , so that Ricci flow reduces to this ODE.
- B.That $\rho$ solves $\rho' = -2(n-1)$ .
- C.That the extinction time is $r_0^2 / (2(n-1))$ .
- D.That the sphere has positive size before the extinction time.
7.
For which compact targets $N$ does the Eells–Sampson theorem guarantee that every continuous map from a compact Riemannian manifold into $N$ is homotopic to a harmonic map? (Choose all that apply.)
- A.A flat torus $\mathbb{R}^2/\mathbb{Z}^2$
- B.A closed hyperbolic surface (curvature $-1$ )
- C.The round 2-sphere
- D.The round projective plane
8.
Bernstein's theorem says an entire minimal graph over $\mathbb{R}^2$ is a plane. For which $n$ is every entire minimal graph over $\mathbb{R}^n$ a hyperplane?
- A.Exactly for $n \le 7$ .
- B.For all $n$ .
- C.Only for $n = 2$ .
- D.Exactly for $n \le 3$ .
9.
Write heat_steps(u, lam, k): apply $k$ steps of the scheme $u_i \mapsto \lambda u_{i-1} + (1 - 2\lambda) u_i + \lambda u_{i+1}$ to the list u, with periodic indices (so u[-1] is the neighbour of u[0]). With $\lambda \le \tfrac12$ the maximum never increases; with $\lambda = 1$ it can overshoot.
Computational — Practice

Answer key — Geometric analysis

1.
With $n = 3$ the rate is $2(n - 1) = 4$ , so $\rho(t) = 16 - 4t$ vanishes at $T = 4$ . The reason the sphere shrinks: its Ricci curvature $(n-1)g_{S^n}$ is positive and does not depend on the radius, so the metric loses a fixed amount per unit time.
2.
The plane, catenoid and helicoid are minimal. The sphere ( $H = 2/r$ ) and the cylinder ( $H = 1/r$ ) are not: at each point they bend to the same side in all directions. A minimal surface is saddle-shaped at every point where it is not flat.
3.
Both catenoids solve $yy'' = 1 + y'^2$ , so both are minimal. The experiment gives areas $\approx 5.99$ (fat) and $\approx 6.85$ (thin), against $2\pi \approx 6.28$ for two discs; here the fat catenoid has least area. The thin catenoid is an unstable critical point: deform it slightly and the area can decrease.
4.
For an orthonormal pair $X, Y$ tangent to $S$ , $\omega(X, Y) = \nu \cdot (X \times Y) \le |\nu|\,|X \times Y| = 1$ , so $\omega \le dA$ on $S$ . On the graph, $X \times Y = \nu$ and equality holds. Closedness then gives $\int_S \omega = \int_G \omega$ , and together $\mathrm{Area}(S) \ge \mathrm{Area}(G)$ .
5.
$A(t) = A(0) - 2\pi t = 4\pi - 2\pi t$ , which vanishes at $t = 2$ . By Grayson's theorem the curve does reach zero area as a round point at exactly this time, whatever its initial shape. For a circle of radius 2 this is $r(t)^2 = 4 - 2t$ .
6.
The theorem is about the real function $\rho$ : its derivative, initial value, zero and sign. The geometric step — that a round metric $\rho\, g_{S^n}$ has Ricci tensor $(n-1)g_{S^n}$ independent of $\rho$ , so Ricci flow becomes $\rho' = -2(n-1)$ — is not formalised. Saying exactly which part is machine-checked is part of the honesty of formal mathematics.
7.
Eells–Sampson requires the target to have non-positive sectional curvature: the flat torus and hyperbolic surfaces qualify, the round sphere and projective plane do not. Then the harmonic map heat flow converges; with positive curvature it may bubble, and harmonic representatives can fail to exist.
8.
Bernstein's theorem holds for $n \le 7$ and fails for $n \ge 8$ . The dividing line is the same as for regularity of area-minimising hypersurfaces, which are smooth in ambient dimension at most 7 and can have singularities like the Simons cone from dimension 8.
9.
```
def heat_steps(u, lam, k):
    u = [float(x) for x in u]
    n = len(u)
    for _ in range(k):
        u = [lam * u[i - 1] + (1 - 2 * lam) * u[i] + lam * u[(i + 1) % n]
             for i in range(n)]
    return u
```
For $0 \le \lambda \le \tfrac12$ each new value is a convex combination of old values, so it lies between the old minimum and maximum — the discrete maximum principle proved in Lean (discrete_max_principle). The total is conserved because the weights of each old value also sum to 1. With $\lambda = 1$ the middle weight is $-1$ , and the alternating pattern $0, 1, 0, 1, \dots$ jumps to values $2$ and $-1$ : the scheme is unstable.