We want to link our definition to the classical definition of Gaussian curvature. Suppose we have a surface in given by
We give the metric induced from the Euclidean metric of . Then the metric in terms of the coordinates and is given by
and the subscripts and mean partial differentiation with respect to and .
Let be the unit normal to the surface . Choose a point in the surface , and let be a curve cut out from by a plane through containing . Let
be the equations defining with the parameter equal to the arc length of . We now compute the curvature of the curve . Let and denote respectively the first and second order derivatives of with respect to . Since is a unit vector, the curvature of is given by the length of . Since is a plane curve on the same plane as , the vector is parallel to . Hence the curvature of at is given by , which by the chain rule is equal to
where the primes for and mean differentiation with respect to . Let
Since is the arc length of , we can write
The expression makes the quotient of two quadratic forms whose variables are the components of the tangent vector at which belongs to . As we change this tangent vector, or equivalent, as we change , the value of changes, and we have two extremal values. The product of these two extremal values is known as the Gaussian curvature. We let
By differentiating with respect to and , we conclude that the extremal values of satisfy the two equations
The vanishing of the determinant of the coefficients of and yields a quadratic equation in . The product of the two roots of in this equation can be read off from the coefficients of the equation. So we conclude the Gaussian curvature is given by
Another way to see it is that the Gaussian curvature is the product of the eigenvalues of the matrix
with respect to the matrix
and so is equal to the quotient of the determiannts of the two matrices.
Now we verify that the Gaussian curvature agrees with the notion of curvature defined as a measure of the failure of the commutativity of partial covariant differentiation. We define covariant differentiation of a vector field on as the orthogonal projection of the usual differentiation onto the tangent planes of . First, let us check that this is a torsion-free connection compatible with the metric. Suppose we have a vector field along a curve of parametrized by and assume that the covariant derivative of along the curve vanishes. Then is proportional to and
and is constant. Next, we have to verify that the Christoffel symbols with respect to a local coordinate system are symmetric in the two covariant indices. The vectors and are respectively the vectors and . The total covariant derivative of is the projection of onto the tangent plane of , i.e. . Relabel the two variables respectively as . Then
Clearly the Christoffel symbols are symmetric in and .
Let us now compute the curvature from the connection. We have seen that corresponds to . Then corresponds to the orthogonal projection onto the tangent plane of . Now
So corresponds to the orthogonal projection of onto the tangent plane of . Since is a unit vector, it is always perpendicular to its derivatives. So equals its own orthogonal projection onto the tangent plane of . We assume without loss of generality that at , the two vectors and are orthonormal. Then is the inner product of with . So
which is equal to
with respect to . In particular, with respect to an orthonormal frame is
which is the Gaussian curvature. Incidentally, this verification proves the theorem of Gauss that the Gaussian curvature of a surface in depends only on the first fundamental form.