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We have the usual formula for the euler characteristic in differential geometry $$\chi = \frac{1}{4\pi}\int_{M}d^{2}\sigma g^{1/2}R + \frac{1}{2\pi}\int_{\partial M}ds k$$ where we define the geodesic curvature $k$ by $$k = \pm t^{a}n_{b}\nabla_{a}t^{b}$$

I want to calculate the geodesic curvature for the flat disk, and for a disk with the metric of a hemisphere. I know how to do this easily using cellular homology, where we may conclude that $$\chi(D^{2}) = 1$$ since the higher homology groups vanish due to the disk being contractible. However, how would I calculate this using differential geometry. I don't have much experience at all with doing calculations in differential geometry, so it would be preferable if one could give the details to the calculation.

I understand that the Ricci scalar is trivial for the case of the flat disk, but I read that we may conclude $t^{a}\nabla_{a}t^{b} = -n^{b}$ for this case which I am unsure of how to conclude.

In addition, how would one compute the geodesic curvature for the case of the disk with the hemisphere metric?

It would be preferable if someone could tell me how to compute the geodesic curvature in general as well. The Ricci scalar is rather straightforward to compute given the metric, but I'm unsure of how to handle the tangent and normal vectors $t^{a},n_{b}$.

Thanks!

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Yes, you mean the (integral of the) geodesic curvature of the boundary circle, either in the flat disk or in the upper hemisphere. In the case of the flat disk, all the curvature of the circle is geodesic curvature, and the integral is $2\pi$. In the case of the upper hemisphere, the boundary circle is a geodesic, and so its geodesic curvature is $0$.

I would recommend you study some undergraduate materials on curves and surfaces before going to all this tensor/GR notation in higher dimensions. For obvious reasons, I recommend my notes, available in .pdf form.

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  • $\begingroup$ Yes I did mean the boundary circle. Sorry if that isn't clear. I took a differential geometry course, but haven't done this stuff in a while. Could you please explain how you would calculate the curvature given a curve in general? Did you arrive at your conclusion for the flat disk, because the curvature was constant? And the conclusion was arrived for the case of the hemisphere metric, since the geodesics for $S^{2}$ in general are the great circles? $\endgroup$
    – user238194
    Jul 2, 2015 at 21:01
  • $\begingroup$ How would we conclude that $t^{a}\nabla_{a}t^{b} = -n^{b}$? $\endgroup$
    – user238194
    Jul 2, 2015 at 21:09
  • $\begingroup$ First of all, note that there's index lowering going on, so $t^a$ may be components of the tangent, but $n_b$ are not the components of the normal unless you've got a coordinate system that gives rise to orthonormal basis vectors (which we can do in a flat space). Again, I suggest you go back to the elementary viewpoint and understand it, and then interpret the fancy notation. $\endgroup$
    – Ted Shifrin
    Jul 2, 2015 at 21:42
  • $\begingroup$ I do understand the basic differential geometry, although I have more experience with differential topology, and yes I do understand the abstract index notation as well. In this case, the $t^{a}$ is the tangent vector to the boundary and $n^{a}$ is the outward unit pointing normal that is orthogonal to $t^{a}$ by hypothesis. But yes, you are correct since $n_{b} = g_{ab}n^{a}$, so $n_{b}$ are clearly not the components of the normal $\endgroup$
    – user238194
    Jul 2, 2015 at 21:48
  • $\begingroup$ So, in the case of the disk, compute with a coordinate system (locally defined around the boundary circle) with $\partial/\partial x^1 = t$ and $\partial/\partial x^2 = n$. $\endgroup$
    – Ted Shifrin
    Jul 2, 2015 at 22:05

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