Is the knowledge of Gaussian and mean curvature (and thus of the principal curvatures) sufficient to characterize a surface uniquely?

If not, is there another geometric quantity one can add to obtain a unique characterization?

  • $\begingroup$ It is interesting to compare your question with this one and the answers to it. $\endgroup$ Apr 17, 2014 at 4:36

1 Answer 1


The Bonnet theorem (see also here) states, roughly speaking, that if we are given the data $\tilde{I}$ and $\tilde{II}$ that are proposed to be the first and the second fundamental forms of the surface, then provided they satisfy the equations of Gauss and Codazzi (and Ricci in a more general setting) there exist a surface in the Euclidean space such that its first and second fundamental forms are $I=\tilde{I}$ and $II=\tilde{II}$ respectively. Such a surface is unique up to a rigid motion of the embracing space.

The principal curvatures $\kappa_1,\,\kappa_2$ are the eigenvalues of the shape operator $I^{-1}II$. Knowing them is equivalent to knowing the Gauss $K=\kappa_1 \cdot \kappa_2$ and the mean $H=\tfrac{1}{2}\left( \kappa_1+\kappa_2 \right)$ curvatures. This knowledge is insufficient because the problem would be underdetermined: given the eigenvalues of the matrix $I^{-1}II$, there is a lot of freedom to choose the matrices $I$ and $II$.

Proofs of the Bonnet theorem can be found in Spivak's "A Comprehensive Introduction to Differential Geometry", and, of course, in many other textbooks on Differential Geometry. An advanced reader may enjoy Werner Greub's paper "Gauss-Codazzi tensor fields and the Bonnet immersion theorem" (1977).

  • $\begingroup$ Notice that the Wikipedia article gives currently an imprecise statement of the Bonnet theorem. $\endgroup$ Apr 16, 2014 at 10:22
  • $\begingroup$ Thanks! If I understand correctly, the following is true: given the first fundamental form and Gaussian curvature, for any fixed H, I can find a unique surface (in the sense of the Bonnet theorem) and all of those surfaces have the same intrinsic geometry, as they have the same first fundamental form. Correct? (btw: are there global restrictions on mean curvature like there is the Gauss-Bonnet theorem for Gaussian curvature?) $\endgroup$
    – madison54
    Apr 16, 2014 at 11:25
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    $\begingroup$ @madison54 not quite: the given data must satisfy the integrability conditions aka the Gauss-Codazzi equations. If you succeed then the intrinsic geometries will be identical. (The total square of the mean curvature leads to the Willmore functional). $\endgroup$ Apr 16, 2014 at 12:39
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    $\begingroup$ Do you have a pair of examples that shows that the problem is undetermined if you only know the principle curvatures? $\endgroup$
    – Elle Najt
    Feb 27, 2019 at 20:40
  • $\begingroup$ @Lorenzo I am not too sure about having a couple of examples ready in my mind, but please look at this question to see if you can convince yourself that such examples exist. $\endgroup$ Feb 28, 2019 at 7:20

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