1. Divisors on Riemann surfaces All the Riemann surfaces in this note are assumed to be connected and compact.

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1 1. Divisors on Riemann surfaces All the Riemann surfaces in this note are assumed to be connected and compact. Let X be a Riemann surface of genus g 0 and K(X) be the field of meromorphic functions on X. The free abelian group generated by points of X is denoted by Div(X). Elements of Div(X) are called divisors on X. They are of the form (1.1) D n 1 P n k P k for some n 1,, n k Z and for some points P 1,, P k X and for some k 1. Or equivalently, D can be written as D P X n P P, where n : X Z is a function so that n P 0 except a finite number of P X. Given a divisor D of the form (1.1), its degree denoted by deg D is defined to be (1.2) deg D P X n P. The map deg : Div(X) Z sending a divisor D to its degree deg D is a group homomorphism. Let f be a nonconstant meromorphic function on X and P be a point of X. In a neighborhood of P, we write f g/h with f, g holomorphic. We set ord P f ord P g ord P h. We say that f has a zero of order ord P f if ord P f > 0 and that f has a pole of order ord P f if ord P f < 0. If P is neither a zero nor a pole, we set ord P f 0. We define a divisor (f) P X(ord P f)p. If f g/h with g, h holomorphic and relatively prime, the divisors of zero (f) 0 is defined to be (f) 0 (ord P g)p P and the divisor of poles is defined to be (f) P h)p. P X(ord Clearly, these divisors are well-defined as long as we require that g, h are relatively prime, and (f) (f) 0 (f). It is clear that the map ( ) : K(X) Div(X) sending f to its associated divisor (f) is a group homomorphism. A divisor D is called principal if there exists f K(X) such that D (f). Lemma 1.1. Let f K(X). Then deg(f) 0. Proof. If f is a constant function, (f) 0 (f) 0. Hence deg(f) 0 is obvious. Assume that f is not a constant function. Since X is compact, there are only finitely many zeros and poles of f on X. Write (f) 0 n 1 P n k P k and (f) m 1 Q m l Q l where 1

2 2 P 1,, P k, Q 1,, Q l X and n i, m j 1. Since a nonconstant meromorphic function f on X is a holomorphic mapping f : X P 1, the degree of f is defined to be deg f e P. P f 1 (Q) Here e P is the ramification index of f at P (or the multiplicity). When we choose Q 0, we get deg f n n k. When we choose Q, we obtain deg f m m l. Then deg(f) (n n k ) (m m l ) deg f deg f 0. A meromorphic one-form ω on X is also called an abelian differential. Given an abelian differential ω, we can also define its associated divisor (ω) P X(ord P ω)p. A divisor of this form is called a canonical divisor. In the next lemma, we are going to see that any two canonical divisors are linearly equivalent. Lemma 1.2. Let ω 1 and ω 2 be two abelian differentials on X with ω 1 0. Then there is a unique geomorphic function f on X such that ω 2 fω 1. Hence (ω 1 ) is linearly equivalent to (ω 2 ). Proof. Let φ : U X V C be a local chart on X. Write ω i φ(z) g i (z)dz for some meromorphic function g i on V. Set h g 2 /g 1. Then f h φ is a meomorphic function on U. Then f can be glued to a global meromorphic function on X such that ω 2 fω 1. This implies that (ω 2 ) (f) + (ω 1 ). In other words, (ω 2 ) (ω 1 ) (f) is a principal divisor. Hence (ω 1 ) and (ω 2 ) are linearly equivalent. Notice that by deg(f) 0, we have deg(ω 2 ) deg(f) + deg(ω 1 ) deg(ω 1 ). This formula tells us that every canonical divisor has the same degree. Now, let us compute the degree of a canonical divisor on a compact Riemann surface. In order to do this, we need the following lemma. Lemma 1.3. Let f : X Y be holomorphic mapping between Riemann surfaces and ω be a meromorphic one-form on Y. For a point p X, we have ord p (f ω) (1 + ord f(p) ω)e p (f) 1. Proof. Let (ψ, U) and (ϕ, V ) be local charts around p and f(p) with f(u) V, and ψ(p) 0, and ϕ(f(p)) 0 such that the local representation F ϕ f ψ 1 of f is given by the local normal form z z n where m e p (f). Assume that ω is represented by ω U (w) (c k w k + c k+1 w k+1 + )dw in the local chart (ϕ, V ) with c k 0 and k ord f(p) ω. Hence the local representation of f ω in local chart (ψ, U) is represented by F ω V F (c k w k + c k+1 w k+1 + )df w (c k z nk + c k+1 z n(k+1) + )dz n (c k z nk + c k+1 z n(k+1) + )nz n 1 dz (nc k z nk+n 1 + nc k+1 c n(k+1)+n 1 + )dz.

3 This implies that ord p (f ω) nk + n 1 (ord f(p) ω + 1)e p (f) 1 which proves our assertion. Proposition 1.1. Let X be a compact Riemann surface of genus g and K X be a canonical divisor. Then deg K X 2g 2. Proof. Let f be a nonconstant meromorphic function on X. Then f : X P 1 is a nonconstant holomorphic mapping and thus a ramified covering of P 1. Denote by d the degree of f. The Riemann-Hurwitz formula implies b p (f) 2g 2 + 2d. Let us define a one-form ω on P 1 as follows. Write P 1 C { }. Let z be the coordinate on C and w 1/z be a coordinate around. Define ω dz. Then ω dw/w 2 as a pole of order two around and no other zeros or poles. Hence (ω) 2. Denote η f ω. Then η is a meromorphic one form on X. Now let us compute the degree of (η). deg(η) (ord p η) 3 (ord p f ω) ((1 + ord f(p) ω)e p (f) 1) ((1 + ord f(p) ω)e p (f) 1) + ((1 + ord f(p) ω)e p (f) 1), f(p), f(p) We know ord q ω 0 if q and ord q ω 2 if q. Hence we have deg(η) (e p (f) 1) + ( e p (f) 1), f(p), f(p), f(p) (e p (f) 1) + 2 p f 1 ( ), f(p), f(p),f(p) e p (f) (e p (f) + 1) (e p (f) + 1),f(p), f(p) b p (f) 2d 2g 2 + 2d 2d 2g 2. Since any two canonical divisors have the same degree, we obtain that deg K X 2g 2. Let f : X Y be a nonconstant holomorphic mapping between Riemann surfaces. Given a point q Y, we define the pull back divisor f (q) by f (q) e p (f)p.

4 4 In general, given a divisor D n qq, the pull back divisor is defined to be f (D) n q f (q). Lemma 1.4. Let f : X Y be a nonconstant holomorphic map between Riemann surfaces. (1) f : Div(Y ) Div(X) is a group homomorphism. (2) The pull back of principal divisor is principal. In fact, if h is a meromorphic function on Y, then f (h) (h f). (3) If X and Y are compact, so that divisors have degrees, we have deg f D deg f deg D. Proof. It follows from the definition that f is a group homomorphism. Let h be a meromorphic function on Y. Then h f is a meromorphic function on X. Write (h) (ord q h)q. Then f (h) (ord q h)f q (ord q h)e p (f)p ord p (h f)p (h f). Here we use the fact that e p (h f) e p (f) ord f(p) h. Thus (b) is proved. Since f is linear, if D n qq, then f D n q f q n q e p (f)p. Hence deg f D n q e p (f) n q Since e p(f) deg f, we find deg f D deg f n q deg f deg D. e p (f). If f : X Y is a nonconstant holomorphic mapping, the ramification divisor of f is the divisor on X defined by R f b p (f)p. The branched divisor of f is the divisor on Y defined by B f b p (f) q. Proposition 1.2. Let f : X Y be nonconstant holomorphic map between Riemann surfaces. Let ω be a nonzero merormorphic one forms on Y. Then (f ω) f (ω) + R f.

5 5 Proof. The proof of this equation is based on the fact that ord p f ω (1 + ord f(p) ω)e p (f) 1 (ord f(p) ω)e p (f) + (e p (f) 1). We obtain (f ω) (ord p f ω)p (ord f(p) ω)e p (f)p + p (f) 1)p (e (ord q ω)e p (f)p + R f (ord q ω)f q + R f f (ω) + R f. Corollary 1.1. As above, when X and Y are compact, then 2g(X) 2 (deg f)(2g(y ) 2) + deg(r f ), where g(x) and g(y ) are genus of X and Y respectively. Proof. Computing the degree of (f ω) and using deg(ω) 2g(Y ) 2, we find deg(f ω) deg f (ω) + deg R f deg f deg ω + deg R f (deg f)(2g(y ) 2) + deg R f. Since ω is a meromorphic one form on Y, f ω is a meromorphic one form on X. Since the degree of any canonical divisor on X has degree 2g 2, deg(f ω) 2g 2. We prove our assertion. Usually we denote deg R f by B. Corollary 1.1 gives us the second proof of the Riemann- Hurwitz formula: g(x) (deg f)(g(y ) 1) B 2.