Calculating euler number of disk

Question

I'm trying to do exercise 3.1 from Polchinski, which should be a rather easy differential geometry problem. I have to calculate the euler number defined by $$\chi = \frac{1}{4\pi}\int_{M}d^{2}\sigma g^{1/2}R + \frac{1}{2\pi}\int_{\partial M}ds k$$ where the geodesic curvature $k$ is defined by $$k = \pm t^{a}n_{b}\nabla_{a}t^{b}$$ I have to calculate $\chi$ for the flat disk and the disk with the metric of a hemisphere. It is easy for me to do this using cellular homology, which easily gives the euler characteristic of $\chi = 1$, (since the closed disk is contractible, and the higher homology groups vanish).

However, I do not have that much experience doing calculations in differential geometry. I understand that the Ricci scalar vanishes for the case of the flat disk, but how would one calculate the scalar curvature in this case? In particular, how should the calculation be carried out for the disk with the metric of the hemisphere? If it's possible, please include the details in the calculation.

Thanks

Answer 1

Let's do this explicitly for both cases. For these examples, the classical formula for the geodesic curvature $k_g$ suffices. Let $\gamma(t)$ be a curve in a surface $S \subset \mathbb{R}^3$, and let $n(t)$ be the unit normal to $S$ at the point $\gamma(t)$. Then $$ k_g = \frac{\ddot{\gamma}(t) .(n(t) \times \dot{\gamma}(t))}{|\dot{\gamma}(t)|^3} $$

First the case where we take the disk with the flat metric and unit radius. Then $R = 0$ and we only have to worry about the geodesic curvature $k_g$. Here our surface is $\mathbb{R}^2$ and we parametrize the boundary of our disk as $\gamma(t) = (\text{cos}(t), \text{sin}(t),0)$, and $n = (0,0,1)$ throughout. Then a simple calculation shows that $k_g = 1$. Therefore $$\chi(\text{flat disk}) = \frac{1}{2\pi}\int_{0}^{2\pi} dt \,\,1 = 1$$

Now lets take the representation of the disk as a hemisphere with the round metric. In fact let's be more general: Let $S^2$ be the sphere of radius 1. The $z=a$ plane where $a \in (-1,1)$ splits $S^2$ into two parts. We can take our disk to be the upper part and call it $D_a$ (note that $D_0$ is the hemisphere). To compute $\chi(D_a)$ we note that in general both terms will be present, since $\partial D_a$ is only a geodesic (and thus $k_g = 0$) for $a=0$. Therefore both terms play a role and we compute them separately. The first term is easy: $R = 2$ still, (since as was mentioned by Jerry, the pullback operator to $D_a$ is trivial) so the first term reads $$\frac{1}{4\pi} \int_{D_a} d\theta \,d\phi\, \text{sin}(\theta) \,\, 2 = 1-a .$$

For $k_g$, note that since $\partial D_a$ is a circle of radius $(1-a^2)^{1/2}$ centered at $(0,0,a)$ (drawing a picture is helpful) it can be parametrized by $$\gamma(t) = ((1-a^2)^{1/2} \,\text{cos}(t), (1-a^2)^{1/2} \,\text{sin}(t), \,a)$$ and $$n(t) = ((1-a^2)^{1/2} \text{cos}(t), (1-a^2)^{1/2} \text{sin}(t), a).$$ You can then compute $$k_g = \frac{a}{(1-a^2)^{1/2}}.$$ The second term is then $$\frac{1}{2\pi} (\int_{\partial D_a})\frac{a}{(1-a^2)^{1/2}} = a .$$

Therefore $$\chi(D_a) = (1-a) + a = 1 = \chi(\text{flat disk})$$ as we expected.

Answer 2

You don't need to use the metric of the hemisphere. This is because the pullback of arbitrary forms onto the submanifold is the trivial pullback operator. All you need to do is apply the projection operator. Therefore, the extrinsic curvature tensor is just $K_{ab} = - \gamma_{a}{}^{c}\gamma_{b}{}^{d}\nabla_{c}n_{d}$, where $\gamma_{ab}$ is the metric of the submanifold (or if you prefer, $\gamma_{a}{}^{b}$ is the projection operator onto the submanifold from the enveloping space), and $\nabla$ is the connection of the enveloping space.

It depends on what you mean by $t^{a}$ and $n^{a}$, but it should be easy enough to prove that this experession is equivalent to yours, if you take the correct contraction and leverage a few identities.

Answer 3

Disk: The Ricci scalar is trivially zero because the disk flat. $t^b\nabla_b t^a$ measures the "acceleration" of the boundary's tangent vector. You can convince yourself this is proportional to the normal vector, $t^b\nabla_b t^a=- n^a$. No differential geometry required here, just calculate the acceleration vector of a particle traveling along a unit circle at unit speed. Contracting both sides with $n$ gives $k=1$, i.e. $$\chi=\underbrace{\frac{1}{4\pi}\int_D\sqrt{g}R\,\mathrm{d}^2x}_{0}+\underbrace{\frac{1}{2\pi}\int_0^{2\pi}\mathrm{d}\theta}_1=1$$

Hemisphere: The Ricci curvature of a general 2-sphere is $2/r^2$. For the unit 2-sphere this is $R=2$. The measure is $\sqrt{g}\mathrm{d}^2x=\sin\theta\,\mathrm{d}\theta\mathrm{d}\phi$. We inegrate this over $0\le\theta<\pi/2$ and $0\le\phi<2\pi$. That integral gives $4\pi$. The boundary of a hemisphere is a great circle, a geodesic, so $t^b\nabla_b t^a=0$ by the geodesic equation, and we have $k=0$. Putting it all together, we obtain $$\chi=\frac{1}{4\pi}\underbrace{\int_0^{2\pi}\int_0^{\pi/2}2\sin\theta\,\mathrm{d}\theta\mathrm{d}\phi}_{4\pi}+\underbrace{\frac{1}{2\pi}\int k\,\mathrm{d}s}_0=1$$