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Rolling a sheet into a circular cylinder bends it without changing its internal lengths and angles. Fitting that same sheet smoothly onto a sphere requires distortion of its first fundamental form. Gaussian curvature distinguishes these situations: it measures the intrinsic curvature of a smooth surface, even though its simplest definition uses how the surface bends in Euclidean space.

The Gaussian curvature at a point is the determinant of the shape operator, or the product of the two principal curvatures:Here $I=Ed{u}^2+2Fdudv+Gd{v}^2$ is the first fundamental form and $II=ed{u}^2+2fdudv+gd{v}^2$ is the second fundamental form, with $S=-dN$. Regularity makes $EG-F^2>0$. Reversing the unit normal negates $e,f,g$ and both principal curvatures, leaving Gaussian curvature unchanged. ^[ref]

The sign describes bending at the point: positive Gaussian curvature means the principal curvatures have the same nonzero sign, as at an elliptic point; negative Gaussian curvature means opposite signs, as on a saddle; zero Gaussian curvature means at least one principal curvature vanishes. Zero Gaussian curvature at a single point does not imply that a neighborhood is a plane. ^[ref]

Four worked models make the signs concrete:

The patches use a common length unit, with $R=1$ for the sphere and circular cylinder. The marked red and blue curves have principal directions at the black point; the arrow is the chosen unit normal. For physical lengths, principal curvature and mean curvature have units of inverse length, while Gaussian curvature has units of inverse squared length. Rescaling a surface by $\lambda >0$ divides its principal curvatures by $\lambda$ and its Gaussian curvature by ${\lambda }^2$. Writing the saddle as $h=(x^2-y^2)/(2L)$ restores an explicit length scale: $K(0)=-1/L^2$.

The Theorema Egregium says that Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. A local isometry therefore preserves Gaussian curvature, even if the shape operator and the principal curvatures change. ^[ref]

Here is the outline, rather than a full proof. The Gauss formula splits second derivatives of a parametric surface into tangential Christoffel symbols and the second fundamental form. The Christoffel symbols are determined by the first fundamental form and its first derivatives. Commuting the next partial derivatives gives the Gauss equation, which identifies $\det S$ with the intrinsic sectional curvature computed from the Riemann curvature tensor. Thus the apparently extrinsic product $k_1{k}_2$ is determined by the Riemannian metric. ^[ref]

One can see the distinction directly by rolling a plane strip with coordinates $(u,v)$ into the circular cylinder $X(u,v)=(R\cos (u/R),R\sin (u/R),v)$. Both have $I=d{u}^2+d{v}^2$, so the map is a local isometry, and is one-to-one on an open strip of width less than $2\pi R$. Extending $u$ over all real numbers wraps around periodically; it does not give a global isometry between the entire plane and the entire circular cylinder. A nonempty open patch of a sphere cannot be flattened by a local isometry into a plane, because its Gaussian curvature $1/R^2$ cannot become zero. The obstruction concerns preservation of the first fundamental form, rather than a particular drawing or choice of seam.

Parametrize the graph by $X(u,v)=(u,v,f(u,v))$. The first fundamental form has coefficients $E=1+f_u^2$, $F=f_u{f}_v$, $G=1+f_v^2$. With unit normal $(-f_u,-f_v,1)/\sqrt{1+|\nabla f{|}^2}$, the second fundamental form has coefficients $e=f_{uu}/\sqrt{1+|\nabla f{|}^2}$, $f_{\mathrm{II}}=f_{uv}/\sqrt{1+|\nabla f{|}^2}$, and $g=f_{vv}/\sqrt{1+|\nabla f{|}^2}$. Consequently the Gaussian curvature of a graph surface is

A compact embedded surface has a highest point in the vertical direction. At that point its tangent plane is horizontal, so it is locally a graph with a local maximum of $f$. Its Hessian matrix is negative semidefinite, giving $\det D^{2}f\ge 0$ and hence $K\ge 0$. Thus the Gaussian curvature cannot be everywhere negative.

An embedded ring torus, with major radius $R$ larger than minor radius $r>0$, is a compact surface without boundary. Its standard parametrization is $((R+r\cos \theta )\cos \varphi ,(R+r\cos \theta )\sin \varphi ,r\sin \theta )$, and the Gaussian curvature of a ring torus isSince the denominator is positive, curvature is positive on the outer side and negative on the inner side, so this example changes sign.

The tangent vectors ofareHence the first fundamental form has coefficientsPutThe upward unit normal is $N=(-f_u,-f_v,1)/W$, so the second fundamental form has coefficientsThus the two forms are

The Gaussian curvature isSince $EG-F^2=W^2$, the graph formula isas in Gaussian curvature of a graph surface.

For the final claim, fix a point of $\gamma$ and make a rigid motion taking $P$ to the plane $z=0$. The common tangent plane is horizontal, so locally $\Sigma$ is the graph $z=f(u,v)$. Along the projected curve $c(s)$ of tangency,Differentiating the second identity givesBecause $c$ is a smooth curve, $c^{\prime }(s)\ne 0$, so the Hessian is singular. Its determinant is zero, and the graph formula gives $K=0$ at every point of $\gamma$. This is tangency to a plane along a curve forces zero Gaussian curvature.

At $p$, the surface $\tilde{S}$ cannot cross the boundary $S=\partial \mathcal{R}$, because it lies inside $\mathcal{R}$. Consequently the two surfaces are tangent at $p$. Apply a proper Euclidean motion of Euclidean three-space so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward unit normal is $N=e_3$. Locally writewith the region $\mathcal{R}$ on the side $z\ge f(x,y)$. ThenThus $g-f$ has a local minimum at zero and its Hessian matrix is positive semidefinite. The fundamental forms of a graph surface at a horizontal tangent plane giveTaking the trace of the positive-semidefinite difference proves the mean-curvature comparison at tangential contact:

There is no analogous conclusion for $K$. In the same local coordinates takeThen $g\ge f$, so the second graph lies on the inward side of the first, while the Gaussian curvature of a graph surface givesHence the mean-curvature inequality holds butUsing smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the counterexample under the global hypotheses.