My confusion comes from the answer of this question: Is there a way to relate convexity to Gaussian curvature?

In the first paragraph of Willie's answer, he wrote

The answer is yes. Convexity implies that for every point $p$ on the boundary $\partial \Omega$, we can find a plane $\Pi$ through $p$ such that $\partial \Omega$ lies "on the one side" of $\Pi$. This implies that the second fundamental form is signed (either positive-semi-definite or negative-semi-definite). For surfaces in $\mathbb{R}^3$, the Gaussian curvature is the determinant of the second fundamental form, and hence is the product of the eigenvalues (the principal curvatures), and hence is always positive definite.

However, if the second fundamental form is negative-semidefinite, then because the matrix representation of the second fundamental form for a surface in $\mathbb{R}^3$ is $2 \times 2$, won't the Gaussian curvature also be non-negative?

Thanks for all the help; I am new to differential geometry and my apologies if this question is too elementary.


1 Answer 1


That’s precisely the point: For a surface, Gaussian curvature does not depend on the choice of the unit normal. However, changing that choice changes the sign of the second fundamental form — hence changing a positive-definite symmetric matrix to negative-definite and vice-versa.

  • $\begingroup$ By changing the “choice”, are you referring to choosing the additive inverse of the originally chosen normal vector? $\endgroup$
    – Robert
    Feb 13, 2023 at 9:01
  • $\begingroup$ Precisely, yes. $\endgroup$ Feb 13, 2023 at 14:23

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