Surface integral of the Gaussian curvature covering the vertex of a cone

Question

The Gaussian curvature of a cone is undefinable at the vertex, and vanishes elsewhere on the cone. However, the cone is an ideal surface which must not be excluded from quantum mechanics where the quantum state must be defined at any point on the surface including the vortex. Thus we must at least have a well defined surface integral of the Gaussian curvature covering the vertex. Does it have a definite definition from mathematics?

Answer 1

The Gaussian curvature can be defined not just as an ordinary function, but also as a kind of "generalized function". In your example in your question, you can define the Gaussian curvature as a Dirac delta function which is zero away from the cone point and has value $2 \pi - \alpha$ at the cone point, where $\alpha$ is the total angle of the cone.

More formally, what one is doing is to define an integrand or measure, for purposes of integrating along the surface. In the example given in your question, this measure evaluates to zero on any measurable set disjoint from the cone point, and it evaluates to $2 \pi - \alpha$ on any measurable set containing the cone point. You can then plug that measure into any appropriate integration.

This works as well for more general surfaces embedded in $\mathbb{R}^3$ that are smoothly embedded except at some finite set of "cone points". Away from the cone points the measure is $K \, dA$ where $K$ is the Gaussian curvature, and $dA$ is the ordinary area form coming from the coordinate system restricted to the surface. To this measure one then adds a Dirac measure for each cone point, with value $2 \pi - \alpha$ where $\alpha$ is the total angle of the cone point. Interestingly, this works whether $\alpha < 2 \pi$ (which is how we ordinarily visualize a cone point) or $\alpha > 2 \pi$.