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Let $S$ be a disc endowed with a spherical metric (constant positive Gaussian curvature). What is known about its smooth isometric immersions in $\mathbb R^3$? By [GS20], an immersion is uniquely determined by its value along any curve, no matter how short, but clearly, not any assignment along a curve would be consistent with a smooth isometric immersion. The question, therefore, is how much freedom remains?

References:

[GS20] Rigidity of Nonnegatively Curved Surfaces Relative to a Curve M. Ghomi and J. Spruck Int. Math. Res. Not.  2020  5387-5400  (2020)

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  • $\begingroup$ I have not worked out the details, nor have I studied [GS]. But the uniqueness theorem and the fact that the isometric embedding equation is strictly elliptic for the standard embedding indicates to me that there might be a straightforward proof using the inverse function theorem that any small smooth deformation of the boundary can be extended to an isometric embedding of the surface. Doing this for an arbitrary isometric embedding of the boundary seems much more difficult to me, since the PDE can become a degenerate elliptic PDE. It's hard for me to come up with a plausible theorem. $\endgroup$
    – Deane Yang
    Commented Aug 15 at 15:41
  • $\begingroup$ I suggest consulting Ghomi and/or Spruck about this. $\endgroup$
    – Deane Yang
    Commented Aug 15 at 15:42
  • $\begingroup$ Their argument is quite straightforward and applies to general positively curved surfaces; indeed, there is a local argument which is then more or less supplements by a clopen argument to show that two immersions that agree on a segment agree everywhere. Again, this is a uniqueness clause and not an existence clause. $\endgroup$
    – Raz Kupferman
    Commented Aug 16 at 19:10

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Actually, it's not enough to specify the isometric immersion along a curve. You have to specify more data than that. For example, you can reflect the unit sphere across its equator, and that will give you a second isometric immersion that agrees with the original along the equator.

A correct local statement is this: Let $C\subset S^2\subset\mathbb{R}^3$ be a connected real analytic curve in the unit $2$-sphere, let $\iota:C\to\mathbb{R}^3$ be any real-analytic isometric immersion, and let $\nu:C\to S^2$ be a real-analytic mapping such that $\nu\cdot \mathrm{d}\iota \equiv 0$. Then there is a connected open neighborhood $U$ of $C$ in $S^2$ and an isometric immersion $f:U\to\mathbb{R}^3$ such that $f$ restricted to $C$ is $\iota$ and the oriented normal to $f(U)$ at $x\in C$ is equal to $\nu(x)$.

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  • $\begingroup$ Thanks Robert Bryant. . The above uniqueness clause is modulo a rigid transformation. I am actually looking for a non-local answer. For example, how much flexibility is there is deforming isometrically a hemisphere? Or for which submanifolds of the sphere a smooth isometric immersion is necessarily also an embedding? $\endgroup$
    – Raz Kupferman
    Commented Aug 15 at 13:13
  • $\begingroup$ Then you should have a look at the discussion in Hilbert and Cohn-Vossen Geometry and the Imagination ($\S32$, Property 10), "...the spherical surface can be bent [i.e., isometrically deformed] as soon as any arbitrarily small portion is removed; it is in fact, sufficient even to slit the sphere open along an arbitrarily small segment of a great circle." The two paragraphs following the cited one give an explicit 'physical' construction of such deformations using the connection with surfaces of constant mean curvature. In particular, a hemisphere is easily deformable. $\endgroup$
    – Robert Bryant
    Commented Aug 15 at 13:28
  • $\begingroup$ @RobertBryant, Is this the statement: There exist deformations of the boundary that can be extended to isometric deformations of the surface that have constant mean curvature? $\endgroup$
    – Deane Yang
    Commented Aug 15 at 15:30
  • $\begingroup$ @DeaneYang: No. What H&C-V do is use the fact that a surface of non-zero constant mean curvature has a parallel surface of constant Gauss curvature: If a surface $X:M^2\to \mathbb{R}^3$ has constant mean curvature $H\not=0$ with oriented normal $\nu$, then $X+1/(2H) \nu:M^2\to \mathbb{R}^3$ has constant Gauss curvature $K=4H^2$. You can make surfaces of nonzero constant mean curvature by dipping a wire loop in a soap solution and blowing on the minimal surface to 'inflate' it. For most wire loops, the parallel surfaces of constant Gauss curvature made this way will not be spherical. $\endgroup$
    – Robert Bryant
    Commented Aug 15 at 18:52
  • $\begingroup$ @RobertBryant, wow. That’s really cool. Thanks. $\endgroup$
    – Deane Yang
    Commented Aug 15 at 19:52

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