Gaussian curvature invariant under an isometry

Question

This question concerns the transformation of one surface into another. I note the relevant postings but lack the knowledge/expertise to use them. What prior concepts do I need to grasp to understand on an intuitive level why Gaussian curvature is invariant under an isometry?

Answer 1

Well, in many ways, it depends on what definition of isometry you're using. As I understand it, an isometry usually means a mapping that preserves the metric, i.e., some means of understanding distance between points and lengths of vectors on a surface - so, a place to start is to understand exactly what the metric is. In an elementary differential geometry course (that is, one where things are mostly done with $\mathbb{R}^{3}$ in mind), this is usually seen in the form of the quantities $E, F,$ and $G$, where:

$$E = x_{u} \cdot x_{u}$$ $$F = x_{u} \cdot x_{v}$$ $$G = x_{v} \cdot x_{v}$$

where $x(u, v)$ is a coordinate patch and $\cdot$ denotes the usual Euclidean dot product. A standard theorem one usually covers in a differential geometry course is that the Gaussian curvature $K$ can be expressed solely in terms of $E, F, G$ and partial derivatives of $E, F,$ and $G$ with respect to $u$ and $v$. Since isometries preserve the metric, i.e. $E, F, G$, it can therefore be seen that isometries preserve the Gauss curvature. This can be understood in more broad, conceptual terms, of course: if a mapping preserves the notion of distance between points and lengths of vectors between surfaces, it would seem natural that the Gaussian curvature, a quantity intimately tied to the notion of length, would also be preserved. This intuition, of course, is very much predicated on your definition of Gauss curvature, as there are many definitions and though all are equivalent, many are very different in presentation and give different insights into the meaning of Gauss curvature. However, that the Gaussian curvature is dependent only on $E, F, G$ and appropriate derivatives is not a trivial result. Any standard intro differential geometry text should cover this result and these ideas; I suggest you look for information in a good intro text accordingly.

Answer 2

To grasp the meaning of Gaussian curvature intuitively, it helps to keep in mind the following fact. At a critical point of a function $f(x,y)$, the Gaussian curvature of the graph of $f$ in $\mathbb{R}^3$ is the determinant of the Hessian matrix of $f$ (i.e., the matrix of second derivatives of $f$).

The fact that this quantity is invariant under isometries is so surprising that when Gauss discovered it, he called it the theorema egregium ("total, or amazing, theorem").

All of the proofs in the literature are somewhat technical. For a fairly compact proof, see lecture notes at http://u.math.biu.ac.il/~katzmik/egreglong.pdf (the material on theorema egregium is in chapter 11).