# Are principal curvatures intrinsic when both are nonzero?

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?

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

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