# degree of line bundle as an integral

We consider the compact Riemann surface $\mathbb{P}^1$ (my notation for the Riemann sphere). I make an open cover $\mathcal{U} = \{U_1, U_2\}$ where $U_1 = \{z \in \mathbb{C} | |z| < 1+ \epsilon\}$ and $U_2 = \{z \in \mathbb{P}^1 | |z| > 1- \epsilon\}$. Let $L \to X$ be the line bundle on $\mathbb{P}^1$ that is determined by some transition function $g_{12} : U_1 \cap U_2 \to \mathbb{C}^*$. In other words, if $h_1: L_{U_1} \to U_1 \times \mathbb{C}$ and $h_2: L_{U_2} \to U_2 \times \mathbb{C}$ are the local trivialisations of this line bundle (and the line bundle is fully determined by the transtion function), then $h_1 \circ h_2^{-1} (x, t) = (x, g_{12}(x) t)$.

Now I want to show that

$\deg L = \frac{1}{2\pi i} \int_{|z| = 1} \frac{g_{12}'(z)}{g_{12}(z)}dz$.

Of course, $\deg L = deg(D)$ where $D$ is the divisor of a meromorphic section $s$ of $L$ over $X$.

I'm not sure where to begin...the $2\pi i$ reminds me of the residue theorem, and degree feels like Riemann-Roch but I don't know where to go....may I please have some help?

This is question 29.2 of Forster's Lectures on Riemann Surfaces.

## 1 Answer

(Edited to make this work in the complex-analytic setting - see comment below.)

Let $s$ be a meromorphic section of $L$, described by the expressions $$U_1 \to \mathbb C, \ \ \ \ z \mapsto a(z)$$ $$U_2 \to \mathbb C, \ \ \ \ z \mapsto b(z)$$ where $a(z) = g_{12}(z) b(z)$ on the overlap $U_1 \cap U_2$.

For convenience, we'll assume that $a(z)$ and $b(z)$ have no zeroes or poles on the contour $\{ |z| = 1 \}$. (If this isn't true, then we can just redefine our choice of $s$ by multiplying or dividing by linear functions, so this isn't a problem.) So the zeroes and poles of $s$ can be partitioned into two sets: those that lie in the region $\{ |z| < 1 \}$, and those that lie in the region $\{ |z| > 1 \}$.

In the region $\{ |z| < 1 \} \subset U_1$, the zeroes and poles of the section $s$ coincide with the zeroes and poles of $a(z)$. By the argument principle, the number of zeros minus the number of poles of $a(z)$ within this region is given by the contour integral $$\frac{1}{2\pi i} \oint_{\{ |z| = 1\}}\frac{a'(z)}{a(z)} dz,$$ where the circular contour $\{ |z| = 1 \}$ is traversed counter-clockwise.

Similarly, in the region $\{ |z| > 1 \} \subset U_2$, $s$ is trivialised by $b(z)$, and the number of zeros minus the number of poles of $b(z)$ within this region is given by $$- \frac{1}{2\pi i} \oint_{\{ |z| = 1\}}\frac{b'(z)}{b(z)} dz.$$ [To derive this carefully, we would need to switch to the coordinate $w := 1 / z$, so that the point at infinity is covered by our coordinate choice, and apply the argument principle using $w$ as the coordinate. The minus sign comes from the fact that an anticlockwise traversal of the circle $\{ |w| = 1 \}$in $w$-space corresponds to a clockwise traversal of the circle $\{ |z| = 1 \}$ in $z$-space.]

The upshot is that the degree of $s$ is given by $$\frac{1}{2\pi i} \oint_{\{ |z| = 1\}}\frac{a'(z)}{a(z)} dz - \frac{1}{2\pi i} \oint_{\{ |z| = 1\}}\frac{b'(z)}{b(z)} dz.$$

Finally, since $a(z) = g_{12}(z)b(z)$ on the contour $\{ |z| = 1\}$, it's easy to show that $$\frac{a'(z)}{a(z)} - \frac{b'(z)}{b(z)} = \frac{g'_{12}(z)}{g_{12}(z)},$$ which completes the proof.

• Thank you for your excellent explanation! But could you please tell me how exactly your map from $U_1 \to \mathbb{C}$ is defined, considering $g_{12}$ is only defined on $U_1 \cap U_2$ but not on all of $U_1$? – Acton Jul 1 '18 at 12:11
• @Acton Ah, I was thinking in "algebraic geometry mode", where all holomorphic functions on open sets can be considered as meromorphic functions on the whole space. Let me edit this so that it works in the analytic case... – Kenny Wong Jul 1 '18 at 12:14
• @Acton Done - sorry about that! – Kenny Wong Jul 1 '18 at 12:34