Contradiction in definitions of Gaussian curvature and principal curvatures?

Question

Let us call the first fundamental form $I$, the second fundamental form $II$, the principal curvatures $k_1$ and $k_2$, and the Gaussian curvature $G$.

Wikipedia states that $G = k_1k_2 = \tfrac{det(II)}{det(I)}$. However, it also states that $k_1$ and $k_2$ are the eigenvalues of $II$. This is a contradiction, because that means that $k_1k_2 = det(II)$ according to basic linear algebra, and hence these two definitions imply that $det(II) = \tfrac{det(II)}{det(I)}$ by the transitive property. How do we resolve this discrepancy?

Answer 1

The first line of the Principal curvature article says

the two principal curvatures at a given point of a surface are the eigenvalues of the shape operator at the point.

The shape operator has matrix $I^{-1}I\!I$, hence $\det{(I\!I)}/\det{(I)} = k_1k_2$, as you say.

Now, the article does not say that they are the eigenvalues of $I\!I$ in general: it says

Let $M$ be a surface in Euclidean space with second fundamental form $I\!I(X,Y)$. Fix a point $p \in M$, and an orthonormal basis $X_1$, $X_2$ of tangent vectors at $p$. Then the principal curvatures are the eigenvalues of the symmetric matrix $$ \left[I\!I_{ij}\right] = \begin{bmatrix} I\!I(X_1,X_1)&I\!I(X_1,X_2)\\ I\!I(X_2,X_1)&I\!I(X_2,X_2) \end{bmatrix}.$$

In an orthonormal basis, $I(X_i,X_j)=\delta_{ij}$ is the identity matrix, so if you work in an orthonormal basis, the principal curvatures are eigenvalues of the matrix of $I\!I$ in that basis.