If we start with a surface in $\mathbb{R}^3$ with nonzero principal curvatures $\kappa_1,\kappa_2\neq 0$ is it ever possible to choose an isometric embedding of that surface in $\mathbb{R}^3$ with different principal curvatures? We might choose a different normal vector, "inverting" the surface and giving principal curvatures $-\kappa_1,-\kappa_2$ but could we for example double one principal curvature while halving the other?

Gauss's Theorema Egregium only implies that $\kappa_1 \kappa_2$ is an intrinsic invariant. If the surface is flat and $\kappa_1 \kappa_2=0$ then we can achieve any pair of principal curvatures $\kappa_1,\kappa_2$ such that $\kappa_1 \kappa_2=0$ with some isometric local embedding. Is it possible to change the principal curvatures by choosing a new isometric embedding when the surface isn't flat?

  • $\begingroup$ @TedShifrin Thanks, I think this title is better. $\endgroup$
    – subrosar
    Commented Oct 4, 2021 at 21:25
  • $\begingroup$ @TedShifrin Clarified that I do want them to be isometric. $\endgroup$
    – subrosar
    Commented Oct 4, 2021 at 21:33

1 Answer 1


The obvious example one thinks of is the $1$-parameter family of minimal surfaces interpolating between the catenoid and the helicoid. These are locally — but not globally — isometric. However, because they're minimal, having equal curvatures (and mean curvature $0$) in fact forces the principal curvatures to be equal.

Here's the next "obvious" example. If we puncture a unit sphere at the north and south poles, we can take the standard embedding or we can take various embeddings as different surfaces of revolution (still symmetric about the $z$-axis and still having $K=1$). By Minding's Theorem, these surfaces are (locally) isometric. (You can also check directly from the formulas below.) However, the principal curvatures are quite different, as you can easily check.

(Take the profile curve $(f(u),g(u))$ with $f(u)=a\cos u$, $|a|\ne 1$, $g(u) = \int\sqrt{1-a^2\sin^2u}\,du$.)

Constant Positive Curvature


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