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!
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.