Relation between (the $2 \pi$ in) Gauss-Bonnet and in Cauchy's differentation formula

Question

When I 1st saw Gauss-Bonnet, I was wondering if this $2 \pi$ had any relation with the $2 \pi$ in Cauchy's differentiation formula. Maybe a better question is to ask about the relation between Gauss-Bonnet and Cauchy's differentiation formula. I recall the $2 \pi$ (or $\tau$) in Cauchy's differentiation formula is to do with homotopy with a circle in the original Cauchy's integral formula. I guess Gauss-Bonnet will have something to do with circles or spheres. So far I know only some basic version of Gauss-Bonnet and not yet its generalisations.

Okay so not exactly any question yet ummm...

1. Where does the $2 \pi$ come from in Gauss-Bonnet? Is it something to do with circles or spheres (Or $S^n$)?

2. Is the $2 \pi$ in Gauss-Bonnet related to the $2 \pi$ in Cauchy's differentiation formula?

3. What's the relation of Gauss-Bonnet with Cauchy's differentiation formula?

Answer 1

The Gauss-Bonnet theorem can be seen as a limiting continuous case of a simpler theorem, to do with surfaces made out glued-together polygons. To each point $p$ on such a surface, we can measure the angle $\theta(p)$ "visible" around $p$. If $p$ lies on a face or an edge we have $\theta(p) = 2\pi$, but on the corners things are different: if $p$ lies on the corner of a cube then $\theta(p) = 3 \pi / 2$ for example, while a point on the corner of a tetrahedron would have the visible angles adding to $\theta(p) = \pi$. We could also imagine gluing together "too many" polygons along their edges, to get a point with $\theta(p) > 2 \pi$.

Define the angle defecit at a point to be $\delta(p) = 2\pi - \theta(p)$, so $\delta(p) = 0$ for everything except corner points, where it measures "deviation" from being flat. Then we have a discrete analogue of the Gauss-Bonnet theorem: for any polygonal surface $S$, $$ \sum_{p \in S} \delta(p) = 2 \pi \chi(S),$$ where $\chi(S)$ is the Euler characteristic of $S$. We can quickly check this for the tetrahedron for example: $\delta(p) = \pi$ at the four corner points, so we have $\sum_{p \in S} \delta(p) = 4 \pi$. On the other hand, the Euler characteristic of a tetrahedron is $2$, since it is homeomorphic to a sphere. Of course the wonderful thing is that no matter how you assemble polygons to make a (orientable) surface, the sum of the angle defecits only depends on the underlying topology. You can read more about this in the wonderful book Mostly Surfaces by Richard Evan Schwartz.

The differential geometry version of Gauss-Bonnet is some kind of limiting version of the above: the Gaussian curvature $K$ is the infinitesimal version of the angle defecit, and we have $$ \int_{S} K \, dA = 2 \pi \chi(S).$$ The interpretation of the $2 \pi$ here being an angle also makes sense from dimensional analysis: area has units of $[\mathrm{length}]^2$ while Gaussian curvature has units of $[\mathrm{length}]^{-2}$, so the integral of one over the other should be dimensionless. (A good rule of thumb is to always assume that a $2 \pi$ appearing in a dimensionless quantity is an angle).

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As for Cauchy's integral formula, I feel that the spirit of the question is the following: why is it true that $\oint_C \frac{dz}{z} = 2 \pi i$ for any closed curve $C$ encircling the origin counter-clockwise, and where does the $2 \pi$ come from? Again the $2 \pi$ appearing is an angle, and the quantity being counted is called the winding number: if $C$ were to wrap around the origin twice, we would get $4 \pi i$. Imagine standing at the origin and watching the curve $C$ being traced out: when it closes in on itself, you may have turned around a net $n$ times, and the result of the integral will be $2 \pi n i$.

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I cannot myself draw a direct connection between them further than this: The $2 \pi$ appearing in each is an angle, and they are both theorems relating complicated things (differential geometry, or contour integration) to simpler topological things (Euler characteristic, or winding number).

Answer 2

In some sense, the answer is yes, there is a relation, passing through Riemann-Roch theorem (two corollaries of it, actually). Indeed, one can prove Gauss-Bonnet (at least for Riemann surfaces$^1$) with Cauchy and RR.

Proof: On a Riemannian surface we can always find local coordinates, in which the metric tensor is conformal and the Gaussian curvature take the particularly easy form

$$\mathbf{g}=\frac12\lambda^2(z,\overline{z}) (dz\otimes d\overline{z}+d\overline{z}\otimes dz)=\lambda^2(dx\otimes dx+dy\otimes dy)\\ K=-\frac1{\lambda^2}\Delta\log(\lambda)$$

By partition of unity, we can glue them to a global construction. The volume form $dS$ takes the form $dS=\lambda^2 dx\wedge dy$, and thus we obtain

$$KdS=-\Delta \log(\lambda)dx\wedge dy$$

Using Dolbeaut's operators we can write it as $$KdS=2id(\partial \log(\lambda))$$ Now, given a differential meromorphic $1-$form $\omega$ (which exists thanks to RR), one can prove that, writing locally $\omega=fdz$, the expression $$\varphi=\frac{\lambda}{|f|}$$ defines a meromorphic function. Since in particular $\ln(|f|)$ is harmonic, we can write $$KdS=2id(\partial\log(\varphi))$$

Now, to the Gauss-Bonnet theorem: writing $X_\varepsilon$ as the riemann surface minus small "discs" $D_{k,\varepsilon}$ around the singularities $z_k$ of $\varphi$, we have

$$\int_X KdS=\lim_{\varepsilon\to 0}\int_{X_\varepsilon}2id(\partial \log(\varphi))=\lim_{\varepsilon\to 0} 2i\int_{\partial X_\varepsilon}\partial \log(\varphi)=\lim_{\varepsilon\to 0}\sum _k\int_{\partial D_{k,\varepsilon}}\partial \log(\varphi) $$

Now, near the singularities of $\varphi$, i.e. the poles or zeros of $f$, we can write $\varphi$ as $\frac{\psi}{|z|^m}$. Thus, by Cauchy's integral formula $$\lim_{\varepsilon\to 0}\int_{\partial D_{k,\varepsilon}}\partial \log(\varphi)=\lim_{\varepsilon\to 0}-m_k\int_{\partial D_{k,\varepsilon}}\partial \log(|z|)=\lim_{\varepsilon\to 0}-\frac {m_k}{2}\int_{\partial D_{k,\varepsilon}}\frac{dz}{z}=-\pi im_k$$

Since, by RR, the degree of the canonical divisor is $2g-2$ and $\sum m_k=-\text{deg}(\omega)$ Thus we get

$$\int_X KdS=-2\pi \text{deg}(\omega)=2\pi (2-2g)=2\pi \chi(X)$$

$^1:$ One can actually prove, though it is much harder, that every orientable surface admits such a local coordinates system, called isothermal coordinates. The usual proofs relies on the solution of the Beltrami equation