# Is the geodesic curvature of a 1D curve in a 2D space equal to the trace of the curve's extrinsic curvature tensor?

I believe my question to be most likely trivial for a differential geometer, however, I have been having problems with this --- partly, I suspect, due to the different nomenclature used by theoretical physicists (like me) and mathematicians.

Let there be a one-dimensional curve in a two-dimensional space. (The 2D space is not embedded in any higher dimensions.) Now, I know a curve has no intrinsic curvature, however, it does have an extrinsic notion of curvature in the 2D space: for example, the geodesic curvature $$k_g$$. I also know that, in general, given an ($$n$$-1)-dimensional hypersurface in an $$n$$-dimensional manifold, one can define the extrinsic curvature tensor $$K_{\mu \nu}$$ (second fundamental form) of the hypersurface. My question is, for a 1D hypersurface (i.e., a curve) in a 2D manifold, is the trace of its extrinsic curvature tensor equal to its geodesic curvature? I.e., is $$K=h^{\mu \nu}K_{\mu \nu}$$ equal to $$k_g$$ for the curve, with $$h^{\mu \nu}$$ the raised-indices metric tensor of the curve?

The motivation for this question is the following: I am using the Gauss-Bonnet theorem in my work and I know of two different but seemingly equivalent formulations for a two-dimensional manifold $$M$$ with one-dimensional boundary $$\partial M$$:

$$\hspace{0.7cm} (1) \hspace{0.7cm} \chi (M) = \frac{1}{4\pi} \int_M \sqrt{g} R d^{2} x + \frac{1}{2\pi} \int_{\partial M} \sqrt{h} K dx$$ ; $$\hspace{0.7cm} (2) \hspace{0.7cm} \chi (M) = \frac{1}{4\pi} \int_M \sqrt{g} R d^{2} x + \frac{1}{2\pi} \int_{\partial M} \sqrt{h} k_g dx$$

where $$\chi (M)$$ is the Euler characteristic of the manifold, $$R$$ is the Ricci scalar (scalar curvature) of the manifold, $$g$$ is the determinant of the metric tensor on the manifold, and $$h$$ is the determinant of the metric tensor on the boundary. Clearly, by comparing the boundary terms of formulae (1) and (2), it is implied that $$K=k_g$$ for a (closed) curve in a 2D space, but I have been unable to prove this so far.

Any help would be much appreciated! :)

## 1 Answer

I think this is a notational confusion. There is really no trace to take when you have a $$2$$-tensor on a $$1$$-dimensional manifold (i.e., in coordinates, you have a $$1\times 1$$ matrix).

But let's look at the definitions. If $$e_1,e_2$$ is an adapted orthonormal frame, so $$e_1$$ is tangent to the curve and $$e_2$$ normal, then the geodesic curvature is $$k_g = e_1'\cdot e_2$$ (where $$'$$ denotes differentiation with respect to arclength of the curve). The second fundamental form is $$f ds\otimes ds$$, with $$f=-e_2'\cdot e_1=e_1'\cdot e_2$$. So, yes, $$f=k_g$$, assuming the correct choice of normal vector.

• Thank you very much for your answer @TedShifrin
– cwr1
Commented Jun 5, 2023 at 14:25
• ... I just have a comment and a query. Comment: for a 1d hypersurface in a 2d manifold, I know of two different formulations for the extrinsic curvature tensor, namely a 1x1 matrix, as you mention, and alternatively a 2x2 matrix, where the indices "run over" the full manifold's coordinates (in which case a trace makes more sense, perhaps). See, for example, Eq. (2.192) of link. Query: could you please elaborate on your caveat, "assuming the correct choice of normal vector"?
– cwr1
Commented Jun 5, 2023 at 15:07
• I've never seen any mathematician treat a covariant tensor on a submanifold as a covariant tensor on the ambient manifold. Calculate this for me for an example — say, a circle of radius $a$ in the plane. There are always two unit normal vectors, and the second fundamental form depends on the choice; similarly, geodesic curvature will depend on the orientation of the surface. Commented Jun 5, 2023 at 19:23