Regulating the space of all surfaces by a cut on curvature

Question

Consider compact connected surfaces embedded in $\mathbb{R}^3$, with boundary given by a fixed unit circle. It seems the collection of all such surfaces is some infinite-dimensional space, comprising mostly of very 'jagged' surfaces. If I try to regulate this space by only considering surfaces whose gaussian curvature is bounded in norm by some $\kappa_0 > 0$, does the space become finite-dimensional in some topology? It's not quite clear to me what the natural topology to use here is, a candidate is the Hausdorff distance (though this seems a bit too degenerate), or an idea along the Earth-movers distance?

Answer 1

tl;dr: Even among surfaces of rotation with fixed area, the space of surfaces of bounded curvature is infinite-dimensional.

---

Theorem 2.5 of A Symplectic Look at Surfaces of Revolution asserts (in particular) a bijective correspondence between

• Smooth, functions $\varphi$ defined on $[0, 1]$, positive on $(0, 1)$ and satisfying $$ \varphi(1) = 0,\qquad \varphi'(1) = -2,\qquad \varphi(0) = 1. $$
• Isometry classes of smooth, rotation-invariant Riemannian metrics of area $2\pi$ on a disk, bounded by a unit circle. In the construction of the paper, if $\tau$ denotes a real parameter for $[0, 1]$, the Gaussian curvature of the metric induced by $\varphi$ is $K = -\frac{1}{2}\varphi''(\tau)$.

For example, the quadratic polynomial $\varphi_{0}(\tau) = 1 - \tau^{2}$ correspods to a round unit hemisphere. The boundary circle is at $\tau = 0$. (If it's of interest, the metric embeds in Euclidean three-space as a surface of rotation if and only if $|\varphi'| \leq 2$.)

Now, if $\psi$ denotes an arbitrary smooth function in $[0, 1]$ that vanishes to order $2$ at both endpoints and satisfies $\sup |\psi''| < 2$, then the function $\varphi = \varphi_{0} + \psi$ induces a metric of metric of bounded, positive curvature. The space of $\psi$ is infinite-dimensional (e.g., take any smooth function absolutely bounded by $1$, integrate twice, and add a suitable cubic polynomial to ensure second-order vanishing at $\tau = 1$); distinct $\psi$ induce non-isometric metrics.