4 spelling :(

The associated family of a minimal surface gives a tangible counterexample. The Weierstrass representation lets you cook up a conformally parameterized minimal surface from a meromorphic pair $f \sqrt{dz}$, $g \sqrt{dz}$.

The parameterization is then given by $$F(x,y) = Re\int_0^{x+iy} (f^2 - g^2, i(f^2 + g^2), 2 fg) ~dz$$

The normal map of $F$ can be obtained by thinking of $g/f$ as a map to the Riemann sphere, and the metric induced by $F$ is just $4(|f|^2 + |g|^2)^2 |dz|^2$. From this data, you can cook up the Gauss and mean curvatures, and it happens to be true that if $f \sqrt{dz}$ and $g \sqrt{dz}$ are meromorphic, you get a minimal surface.

But then consider what happens if you multiply both $f$ and $g$ by $e^{i \theta}$ --- the normal map and the metric are both unchanged, and $e^{i\theta} f \sqrt{dz}, e^{i\theta} g \sqrt{dz}$ are still quite meromorphic, so you get a new minimal surface which is isometric to your old one. This means you have made a new surface whose principle principal curvatures agree with your old one!

I think the moral here is that even knowing the metric and the complete set of principle principal curvatures isn't enough to reconstruct a surface --- the curvature directions are also vital data.

To see all this in action, here is a video with strange music showing the helicoid transforming into the catenoid, which starts with the Weierstrass data for the catenoid and then multiplies by $e^{i \theta}$, with $\theta$ increasing as the movie progresses. Every one of the surfaces is isometric to the catenoid! But they do have different second fundamental forms.

3 added 167 characters in body

(on second thought, this might not quite answer your question; but it does show that extrinsic curvatures aren't sufficient to reconstruct a surface)

The associated family of a minimal surface gives a tangible counterexample. The Weierstrass representation lets you cook up a conformally parameterized minimal surface from a meromorphic pair $f \sqrt{dz}$, $g \sqrt{dz}$.

The parameterization is then given by $$F(x,y) = Re\int_0^{x+iy} (f^2 - g^2, i(f^2 + g^2), 2 fg) ~dz$$

The normal map of $F$ can be obtained by thinking of $g/f$ as a map to the Riemann sphere, and the metric induced by $F$ is just $4(|f|^2 + |g|^2)^2 |dz|^2$. From this data, you can cook up the Gauss and mean curvatures, and it happens to be true that if $f \sqrt{dz}$ and $g \sqrt{dz}$ are meromorphic, you get a minimal surface.

But then consider what happens if you multiply both $f$ and $g$ by $e^{i \theta}$ --- the normal map and the metric are both unchanged, and $e^{i\theta} f \sqrt{dz}, e^{i\theta} g \sqrt{dz}$ are still quite meromorphic, so you get a new minimal surface which is isometric to your old one. This means you have made a new surface whose principle curvatures agree with your old one!

Here

I think the moral here is that even knowing the metric and the complete set of principle curvatures isn't enough to reconstruct a surface --- the curvature directions are also vital data.

To see all this in action, here is a video with strange music showing the helicoid transforming into the catenoid, which starts with the Weierstrass data for the catenoid and then multiplies by $e^{i \theta}$, with $\theta$ increasing as the movie progresses. Every one of the surfaces is isometric to the catenoid! But they do have different second fundamental forms.

2 deleted 170 characters in body

(on second thought, this might not quite answer your question; but it does show that extrinsic curvatures aren't sufficient to reconstruct a surface)

The associated family of a minimal surface gives a tangible counterexample. The Weierstrass representation lets you cook up a conformally parameterized minimal surface from a meromorphic pair $f \sqrt{dz}$, $g \sqrt{dz}$.

The parameterization is then given by $$F(x,y) = Re\int_0^{x+iy} (f^2 - g^2, i(f^2 + g^2), 2 fg) ~dz$$

The normal map of $F$ can be obtained by thinking of $g/f$ as a map to the Riemann sphere, and the metric induced by $F$ is just $4(|f|^2 + |g|^2)^2 |dz|^2$. From this data, you can cook up the Gauss and mean curvatures, and it happens to be true that if $f \sqrt{dz}$ and $g \sqrt{dz}$ are meromorphic, you get a minimal surface.

But then consider what happens if you multiply both $f$ and $g$ by $e^{i \theta}$ --- the normal map and the metric are both unchanged, and $e^{i\theta} f \sqrt{dz}, e^{i\theta} g \sqrt{dz}$ are still quite meromorphic, so you get a new surface which is isometric to your old one, and with the same principle curvatures too!

Here is a video with strange music showing the helicoid transforming into the catenoid, which starts with the Weierstrass data for the catenoid and then multiplies by $e^{i \theta}$, with $\theta$ increasing as the movie progresses. Every one of the surfaces in the video has is isometric to the same Gaussian curvaturecatenoid! But they do have different second fundamental forms. The principle curvatures stay the same, but the curvature directions are doing all sorts of crazy twisting!

I think one moral of all of this is that not just the principle curvatures are important; we also have to track the principle curvature directions.

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