Theorema Egregium

Remarks

The key step is that $K$ can be written, in any local parametrisation, through $E, F, G$ of the first fundamental form and their partial derivatives of order at most $2$. This follows from the Gauss equation among the Gauss–Codazzi equations, which expresses $eg - f^2$ through the Christoffel symbols, and these depend only on $E, F, G$. A local isometry has, in suitable parametrisations, the same functions $E, F, G$, hence the same $K$. Example: the map $(u, v) \mapsto (\cos u, \sin u, v)$ wraps the plane onto the unit cylinder as a local isometry, and indeed both have $K = 0$. Consequence: no open piece of a sphere, where $K > 0$, is isometric to a piece of the plane, so every flat map of a region of the Earth distorts some distances. The converse fails: a map that preserves $K$ need not be a local isometry. For example, the map $(x, y, 0) \mapsto (2x, y, 0)$ of the plane keeps $K = 0$ but doubles some lengths.

Going deeper

For a student

Gauss called this result remarkable because $K$ was defined through the normal and the second fundamental form, which depend on how the surface sits in space. The theorem shows that $K$ is intrinsic, and it is the starting point of Riemannian geometry, where curvature is defined from the metric alone. In orthogonal coordinates, where $F = 0$, one has $$K = -\frac{1}{2\sqrt{EG}} \left( \frac{\partial}{\partial u} \frac{G_u}{\sqrt{EG}} + \frac{\partial}{\partial v} \frac{E_v}{\sqrt{EG}} \right).$$ The mean curvature is not intrinsic: the plane and the cylinder are locally isometric but have $H = 0$ and $H \ne 0$.