# Is the principal curvature of a cylinder positive or negative according to the second fundamental form?

First off, what is the name of the tensor associated with the second fundamental form? For the first fundamental form, I believe we call the associated tensor, "the metric tensor."

Principal curvatures are defined as the eigenvalues of the aforementioned tensor, call it $\bf b$. Well, if we diagonalize $\bf b$ then we obtain ${\bf b} = -\frac{1}{R} {\bf e}_{\theta} \otimes {\bf e}_{\theta}$, where $R$ is the radius of the cylinder, and $\theta$ is defined as usual. This implies that $\kappa_1 = 0$ and $\kappa_2 = -\frac{1}{R}$. But I am pretty sure the nonzero principal curvature of a cylinder is supposed to be positive.

This raises the question: How do we define positive and negative principal curvatures? I have seen definitions about the sum of triangles on the surface having a sum of angles greater or less than $180^o$, but don't find this rigorous. I think it has something to do with how the unit normal changes along the eigenvectors of $\bf b$, no?

We also know that $\frac{d{\bf n}}{ds} = - {\bf b} {\bf t}$, where $\bf t$ is the unit tangent along some curve in the surface and $s$ is the arc length parameter.

Principal curvatures depend on a continuous choice of unit normal field; changing the sign of the unit normal field changes the signs of the principal curvatures. That is, principal curvatures have no intrinsic sign. If you have only a cylinder in Euclidean $3$-space (with no choice of normal field), it makes no sense to say the principal curvatures are non-negative (or are non-positive).
Still assuming you're talking about surfaces in $3$-space, the product of the principal curvatures—a.k.a., the Gaussian curvature—does have an intrinsic sign, and indeed, an intrinsic definition by the Theorema Egregium. One version of the definition is related to angular defects of triangles, but this is clumsy to make precise.