1 16.01. Lecture 1. Complex Algebraic Curves in C2 and Real Algebraic Curves

Home , Algebraic curve, Complex torus, Curve, Projective line

1 16.01. Lecture 1. Complex algebraic curves in C2 and real algebraic curves

Let P (x, y) ∈ C[x, y] be a polynomial in 2 variables with complex coeﬃcients. P has no repeated factors if it cannot be written as

P (x, y) = Q(x, y)2R(x, y) for non-constant polynomial Q.

Deﬁnition 1.1. Let P ∈ C[x, y] be a non-constant polynomial with no repeated factors. The complex algebraic curve deﬁned by P is

2 C = {(x, y) ∈ C | P (x, y) = 0}. We want to assume that curves are given by polynomials with no repeated factors because of the following

Theorem 1.2 (Hilbert’s Nullstellensatz). If P (x, y) and Q(x, y) are polynomials in C[x, y] then 2 2 {(x, y) ∈ C | P (x, y) = 0} = {(x, y) ∈ C | Q(x, y) = 0} if and only if there exist m, n ∈ N such that P divides Qm and Q divides P n. In other words, if and only if P and Q have the same irreducible factors possibly occurring with different multiplicities. It follows from Theorem 1.2 that two polynomials P , Q with no repeated factors define the same complex algebraic curve C if and only if there exists λ ∈ C∗ such that P = λQ. In general, one can define complex algebraic curves as equivalence classes of non- constant polynomials P ∈ C[x, y] where P ∼ Q if and only if P and Q have the same irreducible factors. A polynomial with repeated factors defines a curve with multiplicities attached, for P = x2y the line {y = 0} has multiplicity one while {x = 0} comes into C = {P (x, y) = 0} with multiplicity two. Definition 1.3. The degree of a curve C = {P (x, y) = 0} is the degree of the polynomial P r s P (x, y) = cr,sx y , i.e. d = max{r + s | cr,s 6= 0}. Definition 1.4. A point (a, b) ∈ C is a singular point (or singularity) of C, if ∂P ∂P (a, b) = 0, (a, b) = 0. ∂x ∂y The set of singular points is denoted by Sing(C). C is called non-singular if Sing(C) = ∅. Definition 1.5. A curve defined by

αx + βy + γ = 0, where α, β, γ ∈ C and (α, β) 6= (0, 0), is a line.

1 We will be mostly interested in curves in projective plane. They are given by homogeneous polynomials

Deﬁnition 1.6. A non-zero polynomial P ∈ C[x1, . . . , xn] is homogeneous of degree d if d P (λx1, . . . , λxn) = λ P (x1, . . . , xn), for all λ ∈ C. Equivalently P has the form

X r1 rn P (x1, . . . , xn) = ar1,...,rn x1 . . . xn . r1+...+rn=d

Lemma 1.7. If P ∈ C[x, y] is homogeneous of degree d then it factors as a product of linear polynomials d Y P (x, y) = (αix + βiy), i=1 for some αi, βi ∈ C. Proof. We have

d d X X X x P (x, y) = a xrys = a xryd−r = yd a ( )r. r,s r,d−r r,d−r y r+s=d r=0 r=0

Pd r Polynomial Pe(t) := r=0 ar,d−rt ∈ C[t] is a complex polynomial in one variable, hence it can be factorised as

d e X r Y Pe(t) = ar,d−rt = ae (t − γi), r=0 i=1 where e = max{r | ar,d−r 6= 0} is the degree of Pe. Then

e e d x d Y x d−e Y P (x, y) = y Pe( ) = y ae (( ) − γi) = aey (x − γiy). y y i=1 i=1

Since P (x, y) is a polynomial it has a ﬁnite Taylor expansion

X ∂i+jP (x − a)i(y − b)j P (x, y) = (a, b) ∂xi∂yj i!j! i,j≥0 at any point (a, b) ∈ C2. Deﬁnition 1.8. The multiplicity of curve C = {P (x, y) = 0} at point (a, b) ∈ C is the ∂mP smallest m ∈ N such that ∂xi∂yj (a, b) 6= 0. The polynomial

X ∂mP (x − a)i(y − b)j (a, b) (1) ∂xi∂yj i!j! i+j=m

2 is then homogeneous of degree m and so by Lemma 1.7 can be written as a product of polynomials of the form

α(x − a) + β(y − b) where (α, β) ∈ C2 \{0}. The lines deﬁned by these linear polynomials are tangent lines to C at (a, b). The point (a, b) is non-singular if and only if its multiplicity is 1; in this case C has just one tangent line ∂P ∂P (a, b)(x − a) + (a, b)(y − b) = 0. ∂x ∂y

A point (a, b) ∈ C is a double point (respectively triple point, etc.) if the multiplicity of C at (a, b) is two (respectively three, etc.). A singular point is ordinary if polynomial (1) has no repeated factors, i.e. if C has m distinct tangent lines at (a, b).

Cubic curves

2 3 2 2 3 C1 = {y − x − x = 0},C2 = {y − x = 0} have double points at the origin. Indeed, if Ci = {Pi(x, y) = 0} then ∂P ∂P 1 = −3x2 − 2x, 2 = −3x2, ∂x ∂x ∂P ∂P 1 = 2y, 2 = 2y, ∂y ∂y ∂2P ∂2P 1 = −6x − 2, 2 = −6x, ∂x2 ∂x2 ∂2P ∂2P 1 = 0, 2 = 0, ∂x∂y ∂x∂y ∂2P ∂2P 1 = 2, 2 = 2. ∂y2 ∂y2

Polynomials (1) are respectively

x2 y2 x2 y2 − 2 + 0xy + 2 = (y − x)(y + x), 0 + 0xy + 2 = y2. 2! 2! 2! 2!

It follows that only for C1 the double point at the origin is ordinary. 4 4 2 2 2 The curve C3 = {(x + y ) − x y = 0} has a singular point of multiplicity four at 4 4 2 2 2 2 2 the origin which is not ordinary. The curve C4 = {(x + y − x − y ) − 9x y } has an ordinary singular point of multiplicity 4 at the origin. Historically, real algebraic curves were studied first. Let P (x, y) ∈ R[x, y] be a non-constant polynomial in 2 variables with real coefficients. A real algebraic curve defined by P is

2 C = {(x, y) ∈ R | P (x, y) = 0}.

3 The study of real algebraic curves originates in ancient Greece. Ancient Greeks had sophisticated geometrical methods but a relatively primitive understanding of algebra. To them a circle was not deﬁned by an equation

(x − a)2 + (y − b)2 = r2 but it was the locus of all points having equal distance from a fixed point (a, b). Generalising the above point of view we arrive at the definition of conchoid. 2 More precisely, consider a fixed point q ∈ R and a fixed constant a ∈ R>0. The conchoid of C with respect to q and with parameter a is the locus of all points p ∈ R2 such that the line through p and q meets C at a distance a from p. The classical example is the conchoid of Nicomedes (225 B.C.) which is the conchoid of a line with respect to a point not on a line. If the line is

x = b and q = (0, 0) is the origin then the conchoid is

(x2 + y2)(x − b)2 = a2x2.

It can be used to trisect an angle α. Let α be the angle AOB and let D be any point

E C

D L

on |AO|. Line perpendicular to |AD| in point D intersects line |OB| in point L. We consider the conchoid of line DL with respect to point O with parameter 2|OL| (i.e. |DE| = 2|OL|). Line parallel to |OA| and passing through L intersect the conchoid in point C. Then the angle AOC is one third of the angle AOB.

4 One can also consider a path of a ﬁxed point on a circle rolling along a line. The obtained curve, cycloid, was discovered by Bernoulli around 1700. It is a path that a particle takes when sliding from one point to another on a vertical plane. This discovery led to the development of the theory of variations.

5 2 19.01. Lecture 2. Complex projective spaces and projective transformations

Complex algebraic curves, as defined in the previous lecture, are never compact. For many purposes it is useful to compactify complex algebraic curves by adding points ”infinity”. For example, curves 2 2 C1 = {y = x − 1},C2 = {y = cx} intersect when c 6= ±1. If c = ±1, C1 and C2 do not intersect but they are asymptotic as x and y tend to infinity. We would like to “add points at infinity” to say that C1 and C2 intersect, possibly at infinity, for any value of c ∈ C. To do so, we need the notion of a complex projective space P2. Recall that a topological space X is compact if any open cover of X has a finite subcover. A metric space (X, d) is compact if (X, d) is complete (any Cauchy sequence has a limit) and universally bounded (for any ε > 0 there exists a finite open cover of X with balls of radius ε). X is a Hausdorff space if it satisfies axiom T2, i.e. if for any pair of points x, y of X there exist open subsets U, V ⊂ X such that x ∈ U, y ∈ V and U ∩ V = ∅. Remark 2.1. Recall important properties and topological condition for compactness (i) A subset of Rn or Cn is compact if and only if it is closed and bounded (Heine-Borel theorem). (ii) If f : X → Y is a continuous map between topological spaces and X is compact, then f(X) is compact.

(iii) If X is compact and f : X → R is a continuous map, then f is bounded and attains its bounds. (iv) A closed subset of a compact space is compact. (v) A compact subset of a Hausdorff space is closed. (vi) A finite union of compact spaces is compact. The idea behind constructing P2 is to view (x, y) ∈ C2 as a one-dimensional complex subspace of C3 spanned by (x, y, 1). Every one-dimensional complex subspace of C3 not contained in the plane S = {(x, y, z) ∈ C3 | z = 0} is uniquely defined by a point (x0, y0, 1). One-dimensional subspaces contained in S provide“points at infinity”. Definition 2.2. Complex projective space Pn of dimension n is the set of complex one- dimensional subspaces of Cn+1. When n = 1 we have the projective line P1, for n = 2 we get the projective plane P2. To define topology on Pn, we define a quotient map Cn+1 \{0} → Pn. First, note that any one-dimensional complex subspace U of Cn+1 is spanned by u ∈ Cn+1 \{0} and that u, v ∈ Cn+1 \{0} span the same U if and only if u = λv, for some λ ∈ C∗. Therefore, Pn is the set of equivalence classes for the equivalence relation ∼ on Cn+1 \{0} such that u ∼ v if and only if there exists λ ∈ C∗ such that u = λv.

6 n+1 n Deﬁnition 2.3. Any non-zero (x0, . . . , xn) ∈ C represents an element x of P . We call (x0, . . . , xn) homogeneous coordinates and write

x = [x0 : ... : xn].

Then n n+1 P = {[x0 : ... : xn] | (x0, . . . , xn) ∈ C \{0}} and [x0 : ... : xn] = [y0 : ... : yn] ∗ if and only if there exists λ ∈ C such that xj = λyj, for j = 0, . . . , n.

We make Pn into topological space. We deﬁne Π: Cn+1 \{0} → Pn as

Π(x0, . . . , xn) = [x0 : ... : xn] and endow Pn with the quotient topology, i.e. U ⊂ Pn is open if and only if Π−1(U) ⊂ Cn+1 \{0} is. Remark 2.4. It follows that

(i) B ⊂ Pn is closed if and only if Π−1(B) ⊂ Cn+1 \{0} is.

(ii)Π: Cn+1 \{0} → Pn is continuous.

(iii) Map f : Pn → X is continuous if and only if f ◦ Π: Cn+1 \{0} → X is. More generally, if A ⊂ Pn is any subset then f : A → X is continuous if and only if f ◦ Π:Π−1(A) → X is.

n We deﬁne subsets Ui ⊂ P as

Ui = {[x0 : ... : xn] | xi 6= 0}.

∗ Note that it is well-deﬁned: if xi 6= 0 then λxi 6= 0, for λ ∈ C . Moreover,

−1 n+1 Π (Ui) = {(x0, . . . , xn) ∈ C \{0} | xi 6= 0}

n+1 n is an open subset of C \{0}, hence Ui ⊂ P is open. n Deﬁne ϕ0 : U0 → C by

x1 xn ϕ0[x0 : ... : xn] = ( ,... ). x0 x0

It is well-deﬁned with inverse (y1, . . . , yn) 7→ [1 : y1 : ... : yn].

x1 xn Deﬁnition 2.5. Coordinates ( ,..., ) of [x0 : ... : xn] ∈ U0 are called inhomogeneous x0 x0 coordinates.

7 −1 n Since the map ϕ0 ◦ Π:Π (U0) → C is continuous, so is ϕ0. The inverse of ϕ0 is the n n+1 composition of Π with the continuous map C → C \{0},(y1, . . . , yn) 7→ (1, y1, . . . , yn). It follows that ϕ0 is a homeomorphism. ' n Similarly, ϕj : Uj −→ C are homeomorphisms. n The complement P \ Un is the hyperplane

n−1 {[x0 : ... : xn] | xn = 0}' P .

n n−1 n Then P = Un ∪ P and Un ' C . n n Any point of P lies in some Ui, hence {Ui | i = 0, . . . , n} is an open cover of P . It n −1 n follows that f : P → X is continuous if and only if f ◦ ϕi : C → X is continuous, for n −1 i = 0, . . . , n. Furthermore, f : X → P is continuous if and only if f (Ui) ⊂ X is open, −1 n for any i = 0, . . . , n, and ϕi ◦ f : f (Ui) → C is continuous, for any i.

Proposition 2.6. The projective space Pn is compact.

2n+1 n+1 2 2 Proof. Let S = {(x0, . . . , xn) ∈ C | |x0| +...+|xn| = 1} be an 2n+1-dimensional sphere. It is a closed bounded subset of Cn+1, hence it is compact by Heine-Borel theorem. 2n+1 n The restriction Π|S2n+1 : S → P is continuous, hence its image is compact, see 2.1(iii). We shall show that Π(S2n+1) = Pn. n 2 2 Let now [x0 :√... : xn] ∈ P√be an arbitrary√ point and√ let λ = |x0| +√... + |xn| √. Then [x0 : ... : xn] = [ λx0 : ... : λxn] and [ λx0 : ... : λxn] = Π|S2n+1 ( λx0,..., λxn). It follows that Π|S2n+1 is surjective.

Deﬁnition 2.7. A projective transformation of Pn is a bijection f : Pn → Pn such that n+1 n+1 there exists a linear isomorphism ϕ: C → C such that f ◦ Π = Π ◦ ϕ|Cn+1\{0}:

f n / n PO PO Π Π ϕ Cn+1 \{0} / Cn+1 \{0}

Lemma 2.8. A projective transformation f : Pn → Pn is continuous.

n+1 n+1 Proof. By deﬁnition f ◦ Π = Π ◦ ϕ|Cn+1\{0}, for some ϕ: C → C . As both Π and ϕ are continuous, so is f ◦ Π. It follows from 2.4(iii) that f is continuous.

Deﬁnition 2.9. A hyperplane in Pn is the image under Π of V \{0}, for a linear subspace V ⊂ Cn+1 of dimension n.

n Lemma 2.10. Given n + 2 distinct points p0, . . . , pn and q of P , no n + 1 of which lie on a hyperplane, there exists a unique projective transformation f mapping pi to [0 : ... : 0 : 1 : 0 ..., 0], where 1 is in the i’th place, and mapping q to [1 : ... : 1].

n+1 Proof. Let u0, . . . , un and v be some elements of C \{0} such that Π(ui) = pi and Π(v) = q. By assumption, u0, . . . , un are linearly independent, hence they form a basis n+1 of C and there exists a unique linear transformation ψ mapping hu0, . . . , uni to the n+1 standard basis he0, . . . , eni of C . Vector v does not lie in a linear subspace generated

8 by any n of u0, . . . , un, hence ψ(v) does not lie in a linear subspace generated by any n of e0, . . . , en. In other words, ψ(v) = (λ0, . . . , λn) and λi 6= 0, for all i. Let ϕ be the composition of ψ with the linear transformation with matrix Diag(1/λ0,..., 1/λn). Then ϕ deﬁnes the required projective transformation f.

Proposition 2.11. The projective space Pn is Hausdorﬀ.

n Proof. Let p and q be two distinct points of P . If p and q both lie in U0 then ϕ0(p) and n ϕ0(q) have disjoint open neighbourhoods V and U in C . Then p and q have disjoint −1 −1 neighbourhoods ϕ0 (V ), ϕ0 (U). If p and q do not lie in U0 then we can ﬁnd p1, . . . , pn such that p, p1, . . . , pn, q satisfy assumption of Lemma 2.10. Thus, there exists a projective transformation f such that f(p) = [1 : 0 : ... : 0] and f(q) = [1 : ... : 1], i.e. f(p) and f(q) lie in U0. By the above argument, there exist disjoint open neighbourhoods U 0 and V 0 of f(p) and f(q). Then f −1(U 0) and f −1(V 0) are disjoint open neighbourhoods of p and q.

9 19.01. Homework I

The homework in due on Thursday the 2nd of February. Please hand in your solutions before the lecture.

1. Show that the subset of C2 consisting of points of the form

2 3 (t , t + 1), t ∈ C is a complex algebraic curve.

2. The Hesse pencil of cubic curves consist of curves

3 3 3 Cλ = {[x : y : z] | x + y + z + 3λxyz = 0}.

2 For which values λ ∈ C is the curve Cλ in P non-singular? List singular points and their multiplicities when they exist.

3. Show that the complex line in P2 through the points [0 : 1 : 1] and [t : 0 : 1] meets the projective curve C = {x2 + y2 = z2} in two points [0 : 1 : 1] and [2t : t2 − 1 : t2 + 1]. Show that there is a bijection from the complex line deﬁned by {y = 0} to C given by

[t : 0 : 1] 7→ [2t, t2 − 1, t2 + 1], [1 : 0 : 0] 7→ [0 : 1 : 1].

Deduce that the complex solutions to Pythagoras’ equation

x2 + y2 = z2

are x = 2λµ, y = λ2 − µ2, z = λ2 + µ2, for λ, µ ∈ C.

10 Solutions to Homework I

1. Any point (t2, t3 + 1) in C2 satisﬁes the equation (t2)3 = (t3 + 1 − 1)2 = (t3 + 1)2 − 2(t3 + 1) + 1. In other words, any point (t2, t3 + 1) lies on a curve

2 3 2 3 2 C = {(x, y) ∈ C | x − y + 2y − 1 = x − (y − 1) = 0}. To show that any point on C is of the form (t2, t3 + 1) for some t ∈ C we consider ψ : C → C ψ(t) = (t2, t3 + 1) and check that ϕ: C → C defined as y−1 if (x, y) 6= (0, 1), ϕ(x, y) = x 0 if (x, y) = (0, 1) is inverse to ψ. First, we check that ϕ is well-defined. If (x, y) ∈ C and x = 0 then (y − 1)2 = 0 hence y = 1. Thus, for points (x, y) 6= (0, 1) in C the quotient (y − 1)/x is defined. For t ∈ C \{0}, we have t3 ϕψ(t) = ϕ(t2, t3 + 1) = = t t2 while for t = 0: ϕψ(0) = ϕ(0, 1) = 0.

Finally, for (x, y) ∈ C \{(0, 1)} we have y − 1 (y − 1)2 (y − 1)3 x3 (y − 1)3 ψϕ(x, y) = ψ( ) = ( , + 1) = ( , + 1) = (x, y). x x2 x3 x2 (y − 1)2 The last equality follows from the fact that if (x, y) ∈ C and y = 1 then x3 = 0, i.e. (y−1)3 (x, y) = (0, 1). Hence, for (x, y) ∈ C \{(0, 1)}, y − 1 6= 0 and (y−1)2 = y − 1. For (0, 1) ∈ C we clearly have ψϕ(0, 1) = ψ(0) = (0, 1).

2. Assume that [a : b : c] is a singular point of C = {[x : y : z] | x3 + y3 + z3 + 3λxyz = 0}. Then we have ∂P P (a, b, c) = a3 + b3 + c3 + 3λabc = 0, (a, b, c) = 3a2 + 3λbc = 0, ∂x ∂P ∂P (a, b, c) = 3b2 + 3λac = 0, (a, b, c) = 3c2 + 3λab = 0. ∂y ∂z

11 Since the above equations remain unchanged under any permutation of the set {a, b, c}, points [a : c : b], [b : a : c], [c : b : a], [b : c : a] and [c : a : b] are also singular points of C. a b Since either a or b or c in non-zero, we can assume that c 6= 0, i.e. that [ c , c : 1] is a singular point of C. Putting a = a , b = b , we get e c e c

3 3 2 ea + eb + 1 + 3λeaeb = 0, ea + λeb = 0, 2 eb + λea = 0, 1 + λeaeb = 0. From the last equality it follows that

λ 6= 0 and 1 eb = − . λea Putting it into the second and the third equation we get 1 1 1 a2 − λ = a2 − = 0, + λa = 0, e e 2 2 e λea ea λ ea i.e.

3 3 3 ea − 1 = 0, 1 + λ ea = 0.

3 Since ea = 1, we get that 1 + λ3 = 0.

3 2 2 πi Let now λ = −1, i.e. λ ∈ {−1, −ε3, −ε3} for ε3 = e 3 . Then points

2 2 [λ : −1 : λ], [λε3 : −1 : λε3], [λε3, −1 : λε3] and points obtained by any permutation of coordinates are the singular points of C. If λ = −1, curve C has 3 singular points:

[−1 : −1 : −1] = [1 : 1 : 1], 2 2 2 [ε3 : 1 : ε3] = [1 : ε3 : ε3] = [ε3 : ε3 : 1] 2 2 2 [ε3 : 1 : ε3] = [ε3 : ε3 : 1] = [1 : ε3 : ε3].

If λ = −ε3, curve C has 3 singular points:

2 [ε3 : 1 : ε3] = [1 : ε3 : 1], 2 [1 : 1 : ε3] = [ε3 : ε3 : 1], 2 [ε3 : 1 : 1] = [1 : ε3 : ε3].

12 2 If λ = −ε3 curve C has 3 singular points:

2 2 [ε3 : 1 : ε3] = [1 : ε : 1], 2 2 [ε3 : 1 : 1] = [1 : ε3 : ε3], 2 2 [1 : 1 : ε3] = [ε3 : ε3 : 1].

∂2P Each of the above singular point is of multiplicity two since ∂x2 = 6x doesn’t vanish on any of them.

3. A projective line is given by an equation

Ax + By + Cz = 0.

It passes through points [0 : 1 : 1] and [t : 0 : 1] if and only of

B + C = 0, At + C = 0.

It follows that line L through [0 : 1 : 1] and [t : 0 : 1] is

Lt = {Ax + Aty − Atz = 0} = {x + ty − tz = 0}.

If t = 0, L0 meets C = {[x : y : z] | x2 + y2 = z2} at points [x : y : z] such that x = 0 and y2 = z2, i.e. we have

L0 ∩ C = {[0 : 1 : 1], [0 : −1 : 1]}.

For t 6= 0 point [a : b : c] lies in Lt ∩ C if and only if x + ty = tz, x2 + y2 = z2

Comparing the square of the ﬁrst equation with the second equation multiplied by t gives

x2 + 2txy + t2y2 = t2x2 + t2y2,

i.e.

x(x(1 − t2) + 2ty) = 0.

It follows that we have two solutions, x = 0, ty = tz, i.e. the point [0 : 1 : 1] and x = 2t, y = (t2 −1), tz = x+ty = 2t+t3 −t = t(t2 +1), i.e the point [2t : t2 −1, t2 +1]. Thus

2 2 Lt ∩ C = {[0 : 1 : 1], [2t : t − 1 : t + 1]}.

Let A = {y = 0}.

13 For a ∈ A \{[1 : 0 : 0]} let La be the line via [0 : 1 : 1] and a. Then

La ∩ C = {[0 : 1 : 1], q} for q 6= [0 : 1 : 1]. Hence, the map ϕ0 : A \{[1 : 0 : 0]} → C, ϕ0(a) = q, can be extended to a map ϕ: A → C which sends [1 : 0 : 0] to [0 : 1 : 1].

On the other hand, let c ∈ C \{[0 : 1 : 1]}. Then line Lc via [0 : 1 : 1] and c intersects A in some point, hence we get a map ψ0 : C \{[0 : 1 : 1]} → A. The line tangent to C at [0 : 1 : 1] is given by y + z = 0 and it intersects A at the point [1 : 0 : 0]. Thus, ψ0 can be extended to a map ψ : C → A.

For a ∈ A \{[1 : 0 : 0]}, a is a point on Lϕ(a). Since the intersection points of two lines in unique, we have ψϕ(a) = 0. Similarly, c ∈ C \{[0 : 1 : 1]} lies on the line Lψ(c), hence ϕψ(c) = c. As ϕ([1 : 0 : 0]) = [0 : 1 : 1] and ψ([0 : 1 : 1]) = [1 : 0 : 0], it follows that ϕ and ψ are mutually inverse bijections. Let now [λ : 0 : µ] be an arbitrary point of A. Then

λ λ2 λ2 ϕ([λ : 0 : µ]) = [2 : − 1 : + 1] = [2λµ : λ2 − µ2 : λ2 + µ2] µ µ2 µ2 is an arbitrary point on C. It follows that any complex solution to

x2 + y2 = z2 is given by x = 2λµ, y = λ2 − µ2, z = λ2 + µ2 for some λ, µ ∈ C.

14 3 23.01. Lecture 3. Complex projective curves in 2 CP

Recall from Deﬁnition 1.6 that P ∈ C[x, y, z] is homogeneous if P (λx, λy, λz) = λdP (x, y, z), for any λ ∈ C∗. On the last lecture we have deﬁned a projective plane P2 as

2 3 P = {[x : y : z] | (x, y, z) ∈ C \{0}}, with [x : y : z] = [u : v : w] if and only if there exists λ ∈ C∗ such that x = λu, y = λv and z = λz. It follows that if P is a homogeneous polynomial then P (x, y, z) = 0 if and only if P (u, v, w) = 0. In other words, the set of zeroes of a homogeneous polynomial P is a well-deﬁned subset of P2.

Deﬁnition 3.1. Let P ∈ C[x, y, z] be a non-constant homogeneous polynomial. The projective curve deﬁned by P is

2 C = {[x : y : z] ∈ P | P (x, y, z) = 0}. (2)

Deﬁnition 3.2. The degree of a projective curve C ⊂ P2 is the degree of P . Curve C is irreducible if polynomial P is. An irreducible projective curve D = {Q(x, y, z) = 0} is a component of C if Q divides P .

Deﬁnition 3.3. A point [a : b : c] ∈ C as in (2) is singular if ∂P ∂P ∂P (a, b, c) = 0, (a, b, c) = 0, (a, b, c) = 0. ∂x ∂y ∂z The set of singular points of C is denoted by Sing(C). The curve C is non-singular if Sing(C) = ∅. ∂mP The multiplicity of a singular point [a : b : c] is min{m | ∂xi∂yj ∂zm−i−j (a, b, c) 6= 0}.

2 2 2 2 3 Consider P1 = x + y − z and P2 = y z − x . Then ∂P ∂P 1 = 2x, 2 = −3x2, ∂x ∂x ∂P ∂P 2 = 2y, 2 = 2yz, ∂y ∂y ∂P ∂P 1 = −2z, 2 = y2. ∂z ∂z

2 Since [0 : 0 : 0] is not a point of P , curve C1 = {P1(x, y, z) = 0} is non-singular. Curve C2 = {P2(x, y, z) = 0} has singular point [0 : 0 : 1].

Lemma 3.4. Let C = {P = 0} ⊂ P2 be an algebraic projective curve and let f : P2 → P2 be a projective transformation. Then f(C) is an algebraic curve and p ∈ C is a singular point if and only if f(p) ∈ f(C) is.

15 Proof. By definition there exists a linear isomorphism ϕ: C3 → C3 defining f. Let ψ = ϕ−1. It defines g : P2 → P2 such that fg = Id = gf. Then

2 2 2 f(C) = {f(p) ∈ P | p ∈ C} = {q ∈ P | g(q) ∈ C} = {q ∈ P | P ◦ g(q) = 0}. As g is linear in the homogeneous coordinates of q, the composition P g is a homogeneous polynomial, hence f(C) is an algebraic curve. ∂P A point p ∈ C is singular if and only if any ﬁrst order derivative of P , ∂v , for v = αx + βy + γz vanishes. It happens if and only if any ﬁrst order derivative of P ◦ g vanishes, i.e. if and only if f(p) is a singular point of f(C).

Deﬁnition 3.5. A projective curve deﬁned by a linear equation

αx + βy + γz = 0 where (α, β, γ) ∈ C3 \{0} is a projective line. The tangent line to a projective curve C as in (2) at a non-singular point [a : b : c] is the line ∂P ∂P ∂P (a, b, c)x + (a, b, c)y + (a, b, c)z = 0. ∂x ∂y ∂z

We endow C ⊂ P2 with the induced topology. Then

Lemma 3.6. A projective curve C ⊂ P2 as in (2) is compact and Hausdorﬀ.

Proof. Since P2 is compact, see Proposition 2.6, it suffices to check that C ⊂ P2 is closed (use Remark 2.1(iv)). By Remark 2.4(i), C is closed if and only if Π−1(C) ⊂ C3 \{0} is closed. The set Π−1(C) = {(x, y, z) ∈ C3 \{0} | P (x, y, z) = 0} is the preimage under P of 0 ∈ C. Hence, Π−1(C) is closed, as P is continuous. Any closed subset of a Hausdorff space is Hausdorff, hence Proposition 2.11 and the above imply that C is Hausdorff. Complex algebraic curves as in Definition 1.1 are sometimes called affine to distinguish them from projective curves defined in 3.1. Any affine curve C yields a projective curve Ce obtained by “adding points at infinity”. More precisely, we can identify C2 with the open set 2 U = {[x : y : z] | z 6= 0} ⊂ P 2 x y via ϕ: U → C , ϕ[x : y : z] = ( z , z ) with inverse (x, y) 7→ [x : y : 1]. Let Pe(x, y, z) be a non-constant homogeneous polynomial of degree d. Under the identification of U with C2 the restriction of Ce = {Pe = 0} to U is given by polynomial P ∈ C[x, y] defined as P (x, y) = Pe(x, y, 1). Conversely, given a polynomial Q ∈ C[x, y] of degree d one can define a homogeneous d x y 2 polynomial Qe ∈ C[x, y, z] via Qe(x, y, z) = z Q( z , z ). If CeQ = {Qe = 0} ⊂ P is the projective curve defined by Qe, then the intersection CeQ ∩ U ⊂ U, under identification 2 U ' C , is equal to CQ = {Q = 0}.

16 P r s P r s d−r−s If Q = r+s≤d ar,sx y , then Qe = r+s≤d ar,sx y z . Since d is the degree of Q, there exists ar,s 6= 0 such that r + s = d. It follows that Qe is not divisible by z, i.e. the “line at infinity” z = 0 is not contained in CeQ. In this way we get a bijective correspondence between affine curves in C2 and projective curves in P2 not containing the line at infinity. If Ce is non-singular, then so is its restriction C = Ce ∩ U. However, Ce might have singular points even if C does not. More precisely, we have

Lemma 3.7. Let [a : b : c] be a point of the projective curve

2 Ce = {[x : y : z] ∈ P | Pe(x, y, z) = 0}.

a b If c 6= 0 then [a : b : c] is a non-singular point of Ce if and only if ( c , c ) is a non-singular point of the aﬃne curve

2 C = {(x, y) ∈ C | Pe(x, y, 1) = 0}.

Moreover, the intersection of C2 ' U = {[x : y : z] | z 6= 0} and the projective tangent line a b 2 at [a : b : c] to Ce is the tangent line at ( c , c ) to C in C . In the proof and many times afterwards we shall need Euler’s relation: ∂R ∂R ∂R x + y + z = (deg R)R. (3) ∂x ∂y ∂z

To prove it we note that for a homogeneous polynomial R ∈ C[x, y, z] of degree m, we have R(λx, λy, λz) = λmR(x, y, z). Diﬀerentiating the identity with respect to λ, we get

∂R ∂R ∂R x + y + z = mλm−1R. ∂x ∂y ∂z

Putting λ = 1, we get (3).

a b Proof of Lemma 3.7. Note that ( c , c ) is a singular point of C if and only if a b ∂P a b ∂P a b P ( , , 1) = ( , , 1) = ( , , 1) = 0. c c ∂x c c ∂y c c

In view of (3), the above condition is equivalent to

a b ∂P a b ∂P a b ∂P a b P ( , , 1) = ( , , 1) = ( , , 1) = ( , , 1) = 0, c c ∂x c c ∂y c c ∂z c c

a b i.e. to the condition that [ c : c : 1] = [a : b : c] is a singular point of Ce. The projective tangent line, deﬁned as ∂P ∂P ∂P x (a, b, c) + y (a, b, c) + z (a, b, c) = 0, ∂x ∂y ∂z

17 intersects C2 ' U in ∂P ∂P ∂P x (a, b, c) + y (a, b, c) + (a, b, c) = 0. (4) ∂x ∂y ∂z

For simplicity, let us assume that c = 1. Euler’s relation (3) implies that

∂P ∂P ∂P (a, b, c) = −a (a, b, c) − b (a, b, c). ∂z ∂x ∂y

The “missing” term deg P ·P (a, b, c) vanishes as [a : b : c] is a point of the curve. The above equality together with equation (4) and the assumption c = 1 imply that the intersection of the projective tangent line with U is ∂P ∂P (x − a) (a, b, 1) + (y − b) (a, b, 1) = 0, ∂x ∂y i.e. it is the aﬃne tangent line.

18 4 26.01. Lecture 4. Bezout theorem and basic properties of resultants

We shall study some algebraic properties of complex algebraic curves. We shall first describe how two projective curves C and D in P2 can intersect. If C is of degree n and D is of degree m and C and D have no common components, then C ∩ D has at most mn points. It has exactly nm points if every point p of C ∩ D is a non-singular point both of C and D and if tangent lines to C and D at p are distinct. We will in fact show a more general result about the number of points in C ∩ D once we have defined the intersection multiplicity Ip(C,D). It if defined to be infinity if p lies on a common component of C and D. Otherwise Ip(C,D) ∈ N and Ip(C,D) = 0 if and only if p∈ / C ∩ D. Moreover, we shall see that Ip(C,D) = 1 if and only if p is a non-singular point of both C and D and the tangent lines to C at p and to D at p are distinct. We aim to prove Theorem 4.1 (Bezout’s theorem). Let C and D be two projective curves of degree n and m in P2 which have no common components. Then X X Ip(C,D) = Ip(C,D) = nm. p∈C∩D p∈P2

We need some preparations to define Ip(C,D) and to prove Theorem 4.1. On this lecture we shall define a resultant of a polynomial and prove some of its properties. Definition 4.2. Let

n m P (x) = a0 + a1x + ... + anx ,Q(x) = b0 + b1x + ... + bmx ∈ C[x] be polynomials with an 6= 0 and bm 6= 0. The resultant RP,Q of P and Q is the determinant of the m + n by m + n matrix   a0 a1 . . . an 0 0 ...... 0  0 a0 a1 . . . an 0 ...... 0            MP,Q =  0 0 ... 0 a0 a1 ...... an     b0 b1 ...... bm 0 ...... 0     0 b0 b1 ...... bm 0 ... 0      0 ... 0 b0 b1 ...... bm If

n P (x, y, z) = a0(y, z) + a1(y, z)x + ... + an(y, z)x , m Q(x, y, z) = b0(y, z) + b1(y, z)x + ... + bm(y, z)x , then the resultant RP,Q(y, z) is deﬁned as the determinant of an analogous matrix of polynomials. Note that RP,Q(y, z) is a polynomial in y, z such that RP,Q(b, c) is the resultant of P (x, b, c) and Q(x, b, c) provided an(b, c) 6= 0 and bm(b, c) 6= 0.

19 Lemma 4.3. Let P (x), Q(x) be polynomials in x. Then P (x) and Q(x) have a non- constant common factor if and only if RP,Q = 0. Proof. Let

n m P (x) = a0 + a1x + ... + anx ,Q(x) = b0 + b1x + ... + bmx . P and Q have non-constant common factor R(x) if and only if P (x) = R(x)ϕ(x),Q(x) = R(x)ψ(x). It happens if and only if there exist non-zero polynomials

n−1 m−1 ϕ(x) = α0 + α1x + ... + αn−1x , ψ(x) = β0 + β1x + ... + βm−1x . of degrees at most n − 1 and m − 1 such that P (x)ψ(x) = Q(x)ϕ(x). Comparing the coeﬃcients in front of xj in the above equality, we get

a0β0 = b0α0,

a0β1 + a1β0 = b0α1 + b1α0,

a0β2 + a1β1 + a2β0 = b0α2 + b1α1 + b2α0, ...

anβm−1 = bmαn−1.

In other words, such (α0, . . . , αn−1, β0, . . . , βm−1) exist if and only if

[β0, . . . , βm−1, −α0,..., −αn−1]MP,Q = 0, i.e. if and only if det MP,Q = 0. By definition of RP,Q the existence of (α0, . . . , αn−1, β0, . . . , βm−1) is equivalent to the vanishing of RP,Q. To prove the next statement we need to recall some facts from algebra. Let R be a commutative ring. Then R[x] denotes the ring of polynomials in x with coefficients in R. A polynomial f(x) ∈ R[x] is primitive if the only common factors in R of all coefficients of f are units. It follows that every monic polynomial is primitive. As before, f(x) is irreducible if f(x) = g(x)h(x) implies that either g(x) or h(x) is constant. Recall that R is a unique factorisation domain if it is an integral domain (product of non-zero elements is non-zero) and every non-zero non-unit element can be written as a product of prime elements, uniquely up to order and units. If R is an UFD then so is R[x]. a The field of fractions K of R is a field with elements b , for a, b ∈ R, b 6= 0. Lemma 4.4. Let R be a unique factorisation domain and K its field of fractions. Any f(x) ∈ R[x] can be written as

f(x) = λf1(x) . . . fk(x), where λ ∈ R and f1(x), . . . , fk[x] are primitive in R[x] and irreducible in K[x].

20 Corollary 4.5. Let R and K be as above. Then f(x), g(x) ∈ R[x] have a non-constant common factor as elements in R[x] if and only if they have a non-constant common factor as elements of K[x]. Proof. It follows from the fact that the factorisation in R[x] can be viewed as factorisation in K[x]. Lemma 4.6. Let P (x, y, z) and Q(x, y, z) be non-constant homogeneous polynomials such that P (1, 0, 0) 6= 0 6= Q(1, 0, 0). Then P (x, y, z) and Q(x, y, z) have a non-constant common factor if and only if RP,Q(y, z) is identically zero. Let P (x, y, z) be a homogeneous polynomial of degree d. Then

d X i P (x, y, z) = ai(y, z)x , i=0 and deg(ai) = d − i. In particular, deg(ad) = 0. Since a non-constant homogeneous polynomial a satisfies a(0, 0) = 0, condition P (1, 0, 0) 6= 0 is equivalent to ad 6= 0. In other words, the degree of P as a polynomial in C[y, z][x] is equal to the degree of P as a polynomial in C[x, y, z]. Proof of Lemma 4.6. Without loss of generality we can assume that P (1, 0, 0) = 1 = Q(1, 0, 0). Then we can regard P and Q as monic (i.e. coefficient in front of xn in P is one, similarly the coefficient in front of xm in Q) polynomials in x with coefficients in the field C(y, z) of rational functions, i.e. functions f(y, z) , g(y, z) where f, g ∈ C[y, z] and g(y, z) is not identically zero. It follows from Lemma 4.3 that RP,Q vanishes if and only if P and Q have non-constant common factor when considered as elements of C(y, z)[x]. By Corollary 4.5, it happens if and only if P and Q have non-constant common factor when considered as elements of C[y, z][x] ' C[x, y, z]. Lemma 4.7. Let P (x, y, z) and Q(x, y, z) be homogeneous polynomials of degrees n, respectively m. Then RP,Q(y, z) is a homogeneous polynomial of degree nm.

Proof. By deﬁnition, RP,Q(y, z) is the determinant of n + m by n + m matrix whose ijth entry rij(y, z) is a homogeneous polynomial of degree

n + i − j if 1 ≤ i ≤ m, d = ij i − j if m + 1 ≤ i ≤ n + m,

(Note that 0 is a homogeneous polynomial of an arbitrary degree). Then RP,Q(y, z) is a sum of terms of the form

n+m Y ± riσ(i)(y, z), i=1

21 where σ is a permutation of {1, . . . , n + m}. Each non-zero term is a homogeneous polynomial of degree

n+m m m+n X X X diσ(i) = (n + i − σ(i)) + (i − σ(i)) i=1 i=1 i=m+1 m+n X = nm + (i − σ(i)) = nm. i=1

Lemma 4.8. If P (x) = (x − λ1) ... (x − λn), Q(x) = (x − µ1) ... (x − µm), where λ1, . . . , λn, µ1, . . . , µm ∈ C, then Y RP,Q = (µj − λi). 1≤i≤n,1≤j≤m In particular, RP,QT = RP,QRP,T . More generally, if P, Q, R ∈ C[x, y, z], then

RP,QT (y, z) = RP,Q(y, z)RP,T (y, z).

Proof. If we regard P and Q as homogeneous polynomials in x, λ1, . . . , λn and x, µ1, . . . , µm, then the same argument as in the proof of Lemma 4.8 shows that RP,Q is a homogeneous polynomial of degree nm in variables λ1 . . . , λn, µ1 . . . µm. By Lemma Q 4.6, RP,Q vanishes if λi = µj. It follows that RP,Q is divisible by (µi − λj), i.e. Q RP,Q = λ (µi − λj). To calculate λ we evaluate the polynomial RP,Q(λ1, . . . , λn, µ1, . . . , µm) at a point λi 6= 0 and µ1 = ... = µm = 0. Then RP,Q is the determinant of a matrix such that the last n rows have only one non-zero element, 1. In the row m + 1’st the element 1 is in the m + 1’st column, in the row m + 2’nd it is in the m + 2’nd column and so on. In Q m other words, the matrix is upper diagonal and its determinant is the product (−λ1) of elements on the diagonal. Hence, λ = 1 and Y RP,Q = (µi − λj).

It follows that RP,QT = RP,QRP,T . If P , Q, T are polynomials in x, y, z then

RP,QT (b, c) = RP,Q(b, c)RP,T (b, c) for inﬁnitely many b, c ∈ C. It follows that

RP,QT (y, z) = RP,Q(y, z)RP,T (y, z).

22 27.01. Workshop I

1. Find the singular points and the tangent lines at the singular points of each of the following curves in C2: • y3 − y2 + x3 − x2 + 3y2x + 3x2y + 2xy = 0, • x4 + y4 − x2y2 = 0.

2. Find the singular points and the multiplicities of the singular points of the following projective curves:

• xy4 + yz4 + xz4 = 0, • y2z = x(x − z)(x − λz), for λ ∈ C.

3. Let C be a non-singular projective curve of degree two in P2 deﬁned by a polynomial with rational coeﬃcients. Use the following steps to obtain an algorithm which decides whether or not C has any rational points, i.e. whether there is a point of C which can be represented by rational homogeneous coordinates

(a) Use the theory of the diagonalisation of quadratic forms to show that there is a projective transformation defined by a matrix with rational coefficients taking C to the curve defined by ax2 + by2 = z2, for some a, b ∈ Q \{0}. (b) Show that by making additional diagonal transformation we can assume that a and b are integers with no square factors, i.e. each is product of distinct primes. Show that we may also assume that |a| ≥ |b|. (c) Show that if C has a rational point then b is a square modulo p for every prime p dividing a. Deduce from the Chinese reminder theorem that b is a square modulo a, so there are integers m, a1, such that |m| ≤ |a|/2 and

2 m = b + aa1.

2 2 2 2 (d) Show that if m = b + aa1 and ax + by = z then

2 2 2 2 2 2 2 a1(z − by ) + b(my − z) x = (mz − by) x . Deduce that C has a rational point if and only if the same is true about the curve deﬁned by 2 2 2 a1x + by = z .

(e) Show that if |a| > 1 then |a1| < |a|, and thus the problem is reduced to one of the same form in which |a| + |b| is smaller. Deduce that the argument can be repeated until either b fails to be a square modulo a or we reach the situation |a| = |b| = 1, in which case the curve has a rational point if and only if at least one of a and b is positive.

23 Solutions to workshop I

1. • P (x, y) = y3 − y2 + x3 − x2 + 3y2x + 3x2y + 2xy Point (a, b) is a singular point if P and its derivatives ∂P ∂P = 3x2 − 2x + 3y2 + 6xy + 2y = 3y2 − 2y + 6xy + 3x2 + 2x ∂x ∂y vanish at (a, b) i.e. if

 3 2 3 2 2 2  b − b + a − a + 3ab + 3a b + 2ab = 0, 3a2 − 2a + 3b2 + 6ab + 2b = 0,  3b2 − 2b + 6ab + 3a2 + 2a = 0.

Multiplying the ﬁrst equation by 3 and subtracting a times the second equation and b times the third equation we get:

3b3 − 3b2 + 3a3 − 3a2 + 9ab2 + 9a2b + 6ab − 3a3 + 2a2 − 3ab2 − 6a2b − 2ab − 3b3 + 2b2 − 6ab2 − 3a2b − 2ab = − b2 − a2 − 2ab = −(a − b)2 = 0.

Thus, we have a = b. Then the ﬁrst equation gives

a3 − a2 + a3 − a2 + 3a3 + 3a3 + 2a2 = 8a3 = 0

Thus, the only singular point of C = {P (x, y) = 0} is (0, 0). We have

∂2P ∂2P ∂2P = 6x − 2 + 6y, = 6y + 6x + 2, = 6y − 2 + 6x. ∂x2 ∂x∂y ∂y2

∂2P ∂2P ∂2P It follows that ∂x2 (0, 0) = −2, ∂x∂y (0, 0) = 2 and ∂y2 (0, 0) = −2. Thus tangent lines are linear factors of the polynomial −2 2 −2 x2 + xy − y2 = −(x − y)2. 2! 1!1! 2! Thus C has one tangent line x − y = 0 at the point (0, 0). • P (x, y) = x4 + y4 − x2y2. Point (a, b) is a singular point if P and its derivatives ∂P ∂P = 4x3 − 2xy2, = 4y3 − 2x2y, ∂x ∂y vanish at (a, b) i.e. if

 4 4 2 2  a + b − a b = 0, 4a3 − 2ab2 = 0,  4b3 − 2a2b = 0.

24 Clearly (0, 0) is a singular point. To find other singular points we multiply the second equation by a/2 to get 2a4 = a2b2. Then the first equation gives 2a4 + 2b4 − 2a2b2 = a2b2 + 2b4 − 2a2b2 = b2(2b2 − a2) = 0, √ √ i.e. b = 0 or ( 2b − a) = 0 or ( 2b + a) = 0. 4 The first possibility gives√ point (0, 0) as the first equation yields a = 0. In the remaining cases, a = ± 2b, the first equation gives 4b4 + b4 − 2b4 = 3b4 = 0, i.e. again b = 0 and a = 0. It follows that (0, 0) is the only singular point of C = {P (x, y) = 0}. The second derivatives ∂2P ∂2P ∂2P = 12x2 − 2y2, = −4xy, = 12y2 − 2x2, ∂x2 ∂x∂y ∂y2 vanish at (0, 0). The third derivatives ∂3P ∂3P ∂3P ∂3P = 24x, = −4y = −4x, = 24y ∂x3 ∂x2∂y ∂x∂y2 ∂y3 vanish at (0, 0). The fourth derivatives ∂4P ∂4P ∂4P ∂4P ∂4P = 24, = 0, = −4, = 0, = 24 ∂x4 ∂x3∂y ∂x2∂y2 ∂x∂y3 ∂y4 are not all zero at (0, 0). It follows that tangent lines at (0, 0) are linear factors of the polynomial 24 4 24 x4 − x2y2 + y4 = x4 − x2y2 + y4. 4! 2!2! 4! First, we note that √ w w2 3w2 w i 3w t2 − tw + w2 = t2 − 2t + + = (t − )2 − ( )2 = √ 2 √4 4 2 √ 2 √ w i 3w w i 3w 1 + i 3 1 − i 3 (t − − )(t − + ) = (t − w( ))(t − w( )). 2 2 2 2 2 2 √ √ 1+i 3 1−i 3 Putting λ1 = 2 and λ2 = 2 we get 4 2 2 4 2 2 2 2 x − x y + y = (x − λ1y )(x − λ2y ) = p p p p (x − λ1y)(x + λ2y)(x − λ2y)(x + λ2y). Thus the tangent lines at (0, 0) are p p p p x − λ1y = 0, x + λ2y = 0, x − λ2y = 0, x + λ2y = 0.

25 2. • Consider the curve C = {P (x, y, z) = 0} for

P (x, y, z) = xy4 + yz4 + xz4.

A point [a : b : c] is a singular point of C if partial derivatives

∂P ∂P ∂P = y4 + z4, = 4xy3 + z4, = 4yz3 + 4xz3 ∂x ∂y ∂z

vanish at (a, b, c). It follows from Euler’s relation that then also P (a, b, c) = 0. We thus want to ﬁnd solutions of

 4 4  b + c = 0, 4ab3 + c4 = 0,  4bc3 + 4ac3 = 0.

From the last equation it follows that either a + b = 0 or c = 0. We consider the case a + b = 0 first. Then the first and the second equations imply that −4b4 − b4 = 0, i.e. b = 0 hence also a = 0. Then c4 = 0 thus we get a = b = c = 0 which is not a point in P2. Let us now look at the case c = 0. The first equation implies that b = 0. Then a is arbitrary, thus [1 : 0 : 0] is a singular point of C. To find its multiplicity we calculate

∂2P ∂2P ∂2P = 0, = 4y3, = 4z3, ∂x2 ∂x∂y ∂x∂z ∂2P ∂2P ∂2P = 12xy2, = 4z3, = 12yz2 + 12xz2 ∂y2 ∂y∂z ∂z2

and note that all of them vanish at [1 : 0 : 0]. The same is true about third derivatives ∂3P ∂3P ∂3P ∂3P = 0, = 0, = 0 = 24xy, ∂x3 ∂x2∂y ∂x2∂z ∂y3 ∂3P ∂3P ∂3P ∂3P = 12y2, = 0, = 24yz + 24xz, = 12z2, ∂y2∂x ∂y2∂z ∂z3 ∂z2∂x ∂3P ∂3P = 12z2, = 0 ∂z2∂y ∂x∂y∂z

The multiplicity of [1 : 0 : 0] is four as

∂4P = 24x 6= 0. ∂y4

26 • Let us now consider the curve C = {P (x, y, z) = 0} with

P (x, y, z) = y2z − x(x − z)(x − λz).

First, we look for singular points of C such that z = 1. As we know these are the singular points of the aﬃne curve given by

Pe(x, y) = y2 − x(x − 1)(x − λ).

∂Pe If (x0, y0) is a singular point then ∂y = 2y vanishes at y0, thus y0 = 0. It follows that x0 is the common zero of Q(x) = x(x−1)(x−λ) and its derivative 0 Q (x). Such a x0 exists only if λ = 0 or λ = 1. Thus, – If λ = 0 then [0 : 0 : 1] is a singular point of C. – If λ = 1 then [1 : 0 : 1] is a singular point of C. – If λ 6= 0, 1 then C ∩ {z 6= 0} is smooth. It remains to consider the intersection C ∩ {z = 0}. From z = 0 it follows that x = 0, hence C ∩ {z = 0} = [0 : 1 : 0]. As ∂P = (λ + 1)x2 − 2λxz − y2 ∂z does not vanish at [0 : 1 : 0], it is never a singular point of C. We have

P (x, y, z) = y2z − (x2 − xz)(x − λz) = y2z − x3 + (λ + 1)x2z − λxz2

hence ∂P ∂P ∂P = −3x2 + 2(λ + 1)zx − λz2, = 2yz, = (λ + 1)x2 − 2λxz ∂x ∂y ∂z and ∂2P ∂2P ∂2P = −6x + 2(λ + 1)z, = 0, = 2(λ + 1)x − 2λz, ∂x2 ∂x∂y ∂x∂z ∂2P ∂2P ∂2P = 2z, = 2y, = −2λx. ∂y2 ∂y∂z ∂z2

∂2P ∂2P Thus ∂y2 (0, 0, 1) = 2 = ∂y2 (1, 0, 1), i.e. singular points [0 : 0 : 1] and [1 : 0 : 1] are of multiplicity two.  a b c  3. The curve C is given by a symmetric matrix A =  b d e  ∈ M3×3(Q). A c e f point [u : v : w] ∈ P2 lies on C if and only if (u, v, w)A(u, v, w)T = 0. There exists a change of basis (u, v, w) → (x, y, z) of Q3 such that (u, v, w)A(u, v, w)T = 0 ⇔ (x, y, z)B(x, y, z)T = 0

27  n 0 0  for B =  0 m 0 . If one of the m, n, k is zero, say k then [0 : 0 : 1] is a 0 0 k singular point of C. It follows that n, m, k ∈ Q \ 0. Dividing by −k we can thus write C = {n0x2 + m0y2 = z2} for n0, m0 ∈ Q \{0}. If n0 = α/β and m0 = γ/δ then projective transformation (x, y, z) 7→ (x/β, y/δ, z) takes C to a curve {αβx2 + γδy2 = z2}. Finally, if αβ = p2ε and γδ = q2η then the projective transformation (x, y, z) 7→ (x/p, y/q, z) takes C to the curve {εx2 + ηy2 = z2}. Exchanging x and y if necessary we can thus assume that C = {ax2 + by2 = z2}, for square-free integers a, b such that |b| ≤ |a|. If x, y and z are integers and Qr 2 a = i=1 pi then b is congruent to (z/y) modulo any pi.

Let now m ∈ [0, a − 1] be such that m is congruent to z/y modulo any pi. Such an m exists and is unique by the Chinese remainder theorem. It follows that b is congruent to m2 modulo a hence we have

2 m = b + aa1. Moreover, as m2 is congruent to (a−m)2 modulo m we can assume that |m| ≤ |a|/2. 2 2 2 2 We thus have m = b + aa1 and ax + by = z . It follows that

2 2 2 2 2 2 2 a1(z − by ) + b(my − z) x − (mz − by) x = 2 2 2 2 2 2 2 2 2 2 2 2 a1(z − by ) + b(m y − 2myz + z )x − (m z − 2mbyz + b y )x = 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a1(z − by ) + bm x y − 2bmx yz + bx z − m x z + 2mbx yz − b x y = 2 2 2 2 2 2 2 2 2 2 2 2 a1(z − by ) + b(b + aa1)x y + bx z − (b + aa1)x z − b x y = 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a1(z − by ) + b x y + aa1bx y + bx z − bx z − aa1x z − bx y = 2 2 2 2 2 2 2 2 2 2 a1(z − by ) + aa1x (by − z ) = a1(ax ) − aa1x (ax ) = 0. In other words [(z2 − by2):(my − z)x :(mz − by)x] is a rational point of the curve

0 2 2 2 C = {a1x + by = z }.

Since the role of a and a1 can be exchanged it follows that C has a rational point if and only if C0 does. If |a| > 1 then |a| ≤ |a|2/2, hence 3|a|2 |aa | = |m2 − b| ≤ |a|2/4 + |a| ≤ < |a|2, 1 4

28 2 2 2 i.e. |a1| < |a|. We have thus reduced the question whether C = {ax + by = z } has a rational point to the question whether C0 = {a0x2 + b0y2 = z2} has a rational point with |a0| + |b0| < |a| + |b|. Continuing this way we either get to the case when b is not a square modulo a or |a| = |b| = 1. If |a| = |b| = 1 the curve C has a rational point if and only if at least one of a, b is positive.

29 5 30.01. Lecture 5. Weak Bezout theorem and its applications

Using Lemmas from the last lecture we show Theorem 5.1. Any two projective curves in C and D in P2 intersect in at least one point. Proof. Let C and D be defined by homogeneous polynomials P and Q. By Lemma 4.7 the resultant RP,Q(y, z) is a homogeneous polynomial of degree nm in y and z. Therefore, by Lemma 1.7 RPQ is either identically zero or it is a product of nm linear polynomials. 2 In either case there exists (b, c) ∈ C \ 0 such that RP,Q(b, c) = 0. This means that the resultant of polynomials P (x, b, c), Q(x, b, c) in x is zero, i.e. they have a common root a. Then [a : b : c] ∈ C ∩ D. Theorem 5.2 (Weak version of Bezout’s theorem). If two projective curves C and D in P2 of degrees n and m have no common components then they intersect in at most nm points. Proof. Suppose that C and D have at least nm + 1 points of intersection. We shall show that then they have a common component. Choose any set S of nm + 1 distinct points in C ∩ D. Then we can choose a point in P2 which does not lie on C nor on D nor on any line which passes through any two points of S. By applying projective transformation we can assume that this point is [1 : 0 : 0]. Then C and D are given by P (x, y, z), Q(x, y, z) respectively, and P (1, 0, 0) 6= 0 6= Q(1, 0, 0). By Lemma 4.7 the resultant RP,Q(y, z) is a homogeneous polynomial of degree nm. If RP,Q(b, c) = 0 then RP (x,b,c),Q(x,b,c) = 0, i.e. polynomials P (x, b, c), Q(x, b, c) have a common root, see Lemma 4.3. Thus, there exists a such that P (a, b, c) = 0 = Q(a, b, c), i.e. [a : b : c] ∈ C ∩ D. If [a : b : c] ∈ S then P (a, b, c) = 0 = Q(a, b, c) and (b, c) 6= (0, 0), because [1 : 0 : 0] ∈/ S. It follows that bz − cy is a factor of RP,Q(y, z). Moreover, if [α, β, γ] is a different point in S then βz − γy is not a scalar multiple of bz − cy (if it was then [a : b : c], [α : β : γ] and [1 : 0 : 0] would lie on bz = cy). It shows that RP,Q has at least nm + 1 distinct linear factors, hence it is identically zero. By Lemma 4.6, C and D have a common component. We shall now use the above theorems to prove some properties of algebraic curves. Corollary 5.3. (i) A non-singular projective curve C in P2 is irreducible. (ii) An irreducible projective curve C in P2 has at most finitely many singular points. Proof. (i) Let 2 C = {[x : y : z] ∈ P | P (x, y, z)Q(x, y, z) = 0} be a reducible curve in P2. By Theorem 5.1 there is at least one [a : b : c] ∈ P2 such that P (a, b, c) = 0 = Q(a, b, c). As ∂P Q ∂P ∂Q ∂P Q ∂P ∂Q ∂P Q ∂P ∂Q = Q + P , = Q + P , = Q + P ∂x ∂x ∂x ∂y ∂y ∂ ∂z ∂z ∂z [a : b : c] is a singular point of C.

30 (ii) Let C be defined by a homogeneous polynomial P (x, y, z) of degree n. Without loss of generality we may assume that [1 : 0 : 0] ∈/ C, i.e. that P (1, 0, 0) 6= 0 so the coefficient in front of xn in P is not equal to zero. This ensures that Q(x, y, z) = ∂P ∂x (x, y, z) is a homogeneous polynomial of degree n − 1 which is not identically zero, hence defines a curve in D in P2. Any singular point of C belongs to the intersection C∩D. By Theorem 5.2 |C∩D| ≤ n(n − 1) is finite. The statement follows.

Deﬁnition 5.4. A conic is a curve of degree two in C2 or P2.

Corollary 5.5. Any irreducible projective conic C in P2 is equivalent under projective transformation to the conic x2 = yz and in particular is non-singular.

Proof. By Corollary 5.3, C has finitely many singular points. Hence, by applying a suitable projective transformation, we may assume that [0 : 1 : 0] is a non-singular point of C and that the tangent line to C at [0 : 1 : 0] is z = 0. Then C must be defined by a polynomial of the form ayz + bx2 + cxz + dz2. 2 ∂ Indeed, the coefficient in front of y must vanish and ∂x (0, 1, 0) = 0, i.e. the coefficient in front of xy must be zero. Since C is irreducible, both a and b are non-zero. Then √ [x, y, z] 7→ [ bx, ay + cx + dz, −z] takes C to the conic x2 = yz. Since this conic is non-singular, so is C. Let C be a non-singular conic defined by x2 = yz. There is a homeomorphism

1 2 2 f : P → C, f([x : y]) = [xy : y : x ], with inverse [x : y] if y 6= 0, g : C → 1, g([x : y : z]) = P [z : x] if z 6= 0.

Note that if [x : y : z] ∈ C then x2 = yz and if y 6= 0 and z 6= 0 then x 6= 0 and

[x : y] = [x2, xy] = [yz : xy] = [z : x].

Thus, by Corollary 5.5 any irreducible conic is homeomorphic to P1.

Proposition 5.6. It two projective curves C and D of degrees n in P2 intersect in exactly n2 points and if exactly nm of these points lie on an irreducible curve E of degree m < n then the remaining n(n − m) points lie on a curve of degree at most n − m.

31 Proof. Let C, D and E be deﬁned by homogeneous polynomials

P (x, y, z),Q(x, y, z),R(x, y, z).

Choose a point [a : b : c] on E which does not lie on C ∩ D. Then the curve of degree n deﬁned by C0 = λP (x, y, z) + µQ(x, y, z) = 0, with λ = Q(a, b, c), µ = −P (a, b, c) meets E in at least nm + 1 points, namely in [a : b : c] and nm of points of C ∩ D which lie on E. Then by Theorem 5.2 C0 and E must have a common component which must be E, as E is irreducible. Thus

λP (x, y, z) + µQ(x, y, z) = R(x, y, z)S(x, y, z), for some homogeneous polynomial S of degree n − m. Hence, if [u : v : w] ∈ C ∩ D then either R(u, v, w) of S(u, v, w) vanishes. Therefore, the n(n − m) points of C ∩ D which do not lie on E must lie on {S(x, y, z) = 0}.

2 A hexagon in P is determined by six distinct points p1, . . . , p6 – vertices and lines joining p1 to p2, p2 to p3 and so on till p6 to p1. The side opposite the line joining p1 to p2 is the line joining p4 to p5 and so on. A hexagon is inscribed in a conic if p1, . . . , p6 lie on a conic.

Corollary 5.7 (Pascal’s mystic hexagon). The pairs of opposite sides of a hexagon inscribed in an irreducible conic in P2 meet in 3 collinear points.

Proof. Let the successive sides of the hexagon be lines given by polynomials L1,...,L6. Two projective curves deﬁned by L1L2L5 and L2L4L6 intersect in the six vertices of the hexagon and three points of intersection of the opposite sides. By assumption, the ﬁrst 6 points lie on a curve of degree 2 ≤ 3. Thus, by Proposition 5.6, the remaining 3 points lie of a curve of degree at most 3 − 2 = 1.

32 6 02.02. Lecture 6. Intersection multiplicity and the proof of Bezout Theorem

To prove the strong form of Bezout’s theorem we must first define the intersection 2 multiplicity Ip(C,D) at a point p = [a : b : c] of two curves C and D in P . We shall define the intersection multiplicity by using the resultant of the polynomials defining C and D in a suitable coordinate system. In order to show that the definition is independent of the choice we show that Ip(C,D) is uniquely determined by the properties listed below

Theorem 6.1. There exists a unique intersection multiplicity Ip(C,D) deﬁned for all projective curves C, D in P2 such that

(i) Ip(C,D) = Ip(D,C).

(ii) Ip(C,D) = ∞ if p lies on a common component of C and D and otherwise Ip(C,D) ∈ Z≥0.

(iii) Ip(C,D) = 0 if and only if p∈ / C ∩ D. (iv) Two distinct lines meet with intersection multiplicity one at their unique point of intersection

(v) If C1 = {P1(x, y, x) = 0}, C2 = {P2(x, y, z) = 0} and C = {P1(x, y, z)P2(x, y, z) = 0} then Ip(C,D) = Ip(C1,D) + Ip(C2,D), for any projective curve D and any point p.

(vi) If C = {P (x, y, z) = 0} and D = {Q(x, y, z) = 0} with deg P = n and deg Q = m and E = {PR + Q = 0} with deg R = m − n then

Ip(C,D) = Ip(C,E).

Moreover if C and D have no common component and we choose projective coordinates so that the conditions

(a) [1 : 0 : 0] does not belong to C ∪ D,

(b) [1 : 0 : 0] does not lie on any line containing two distinct points of C ∩ D,

(c) [1 : 0 : 0] does not lie on the tangent line to C or D and any point of C ∩ D, are satisﬁed then the intersection multiplicity Ip(C,D) of C and D at any

p = [a : b : c] ∈ C ∩ D

k is the largest integer k such that (bz − cy) divides the resultant RP,Q(y, z).

33 Proof. To simplify the notation we shall write Ip(P,Q) instead of Ip(C,D) for curves C = {P = 0}, D = {Q = 0}. First we show that conditions (i)-(vi) determine Ip(P,Q) completely. Since the conditions are independent of the choice of coordinates we might assume that p = [0 : 0 : 1]. We may assume that Ip(P,Q) is ﬁnite, greater than zero and that P and Q are irreducible. Thus, Ip(P,Q) = k, for some k > 0. By induction on k we may assume that any intersection multiplicity strictly less than k can be calculated using only conditions (i)-(vi). Consider polynomials P (x, 0, 1) and Q(x, 0, 1). Assume that they are of degree r and s respectively. By (i) we can assume that r ≤ s. There are 2 cases to consider r = 0 First, we consider an example. Assume P (x, y, z) has degree 3. Then it is a linear 3 3 3 2 2 2 2 2 2 3 combination of x , y , z , x y, x z, y z, xy , xz , yz and xyz. P (x, 0, 1) = ax3 x + 2 ax2zx + axz2 x + az3 . Since P (0, 0, 1) = 0, we know that az3 = 0. Because P (x, 0, 1) is constant (since its degree is zero!), we know that it is identically zero, hence ax3 = 3 2 2 2 2 ax2z = axz2 = az3 = 0. It follows that P (x, y, z) is spanned by y , x y, y z, xy , yz and xyz, i.e. P (x, y, z) = yR(x, y, z). The above works for P of any degree, hence P (x, y, z) = yR(x, y, z). Moreover, by dividing the monomials into those which do or do not contain powers of y, we can present Q as Q(x, y, z) = Q(x, 0, z) + yS(x, y, z). Because Q(0, 0, 1) = 0, we have Q(x, 0, z) = xqT (x, z), for some q > 0, where T (x, z) is a polynomial such that T (0, 1) 6= 0. It implies that [0 : 0 : 1] does not lie on the curve {T (x, z) = 0}, hence Ip(y, T (x, z)) = 0, by (iii). Condition (v) gives Ip(P,Q) = Ip(y, Q) + Ip(R,Q). q q q By (vi), Ip(y, Q) = Ip(y, x T ) = Ip(y, x ) + Ip(y, T ) = Ip(y, x ). Condition (v) q implies that Ip(y, x ) = qIp(y, x) = q, by (iv). It follows that

Ip(P,Q) = Ip(R,Q) + q,

for some q > 0. By induction, Ip(R,Q) < Ip(P,Q) is uniquely determined by conditions (i)-(vi). r > 0 We want to construct a polynomial S such that S(x, 0, 1) = Q(x, 0, 1)−xs−rP (x, 0, 1) has degree t < s. First, we note that multiplying P and Q by scalars, we can assume that Q(x, 0, 1) and P (x, 0, 1) are monic. We deﬁne S(x, y, z) = zn+s−rQ(x, y, z) − xs−rzmP (x, y, z). Since P and Q are irreducible and distinct, S is not identically zero. By (vi) we have n+s−r n+s−r Ip(P,S) = Ip(P, z Q) = Ip(P, z ) + Ip(P,Q) = Ip(P,Q), n+s−r where Ip(P, z ) = 0 because p does not lie on {z = 0}. Now, we can replace P and Q with P and S and continue till we reach the situation r = 0.

34 This completes the uniqueness part of the proof. To prove existence, we deﬁne Ip(C,D) as follows:

• If p lies on a common component of C and D then Ip(C,D) = ∞,

• If p does not belong to C ∩ D then Ip(C,D) = 0, • If p ∈ C ∩ D but does not lie on a common component, first remove all common components of C and D and then choose coordinates such that (a)-(c) are satisfied. k If p = [a : b : c] in these coordinates then Ip(C,D) is the largest k such that (bz−cy) divides the resultant RP,Q(y, z). It remains to show that (i)-(vi) are satisfied:

(i) is a direct consequence that interchanging two rows changes the sign of the determinant, RP,Q = ±RQ,P . (ii) follows from the definition and Lemma 4.6. (iii) If p = [a : b : c] ∈ C ∩ D then P (x, b, c) and Q(x, b, c) have a common root, hence their resultant vanishes by Lemma 4.3. By Lemma 4.6, RP,Q(y, z) vanishes when y = b and z = c, i.e. it is divisible by bz − cy, which shows that Ip(C,D) > 0 (because by (a), (b, c) 6= (0, 0)). (iv) is a straightforward computation of 2 by 2 determinant (v) Follows from Lemma 4.8 (vi) holds because determinant is unchanged by the addition of a scalar multiple of one row to another. The resultant of P and PR + Q is the determinant of a matrix (sij) obtained from the matrix (rij) defining the resultant of P and Q by addition of scalar multiples of the first n rows to the last m rows. Let us consider an example P = x + y, Q = x2 + xy + yz, R = x + z. Then

 y 1 0  RP,Q = det  0 y 1  yz y 1

while RP,Q+PR is a determinant of a matrix with last row z[y, 1, 0] + 1[0, y, 1] + [yz, y, 1]. Indeed RP + Q = x(x + y) + z(x + y) + x2 + xy + yz = 2x2 + x(2y + z) + 2yz. n−m In general, if R(x, y, z) = ρ0(y, z) + xρ1(y, z) + ... + x ρn−m(y, z), then rij if i ≤ m, sij = Pi−n rij + k=i−m ρi−n−krkj if i > m.

Now, we are ready to prove

35 Theorem (4.1 Bezout’s theorem). Let C and D be two projective curves of degree n and m in P2 which have no common components. Then X X Ip(C,D) = Ip(C,D) = nm. p∈C∩D p∈P2 Proof. Let C and D be curves with no common component. We may choose coordinates so that conditions (a)-(c) are satisﬁed. Let C = {P (x, y, z) = 0}, D = {Q(x, y, z) = 0} be polynomials in this coordinate system. Then by Lemmas 4.6 and 4.7, RP,Q(y, z) is a non- zero homogeneous polynomial of degree nm. Hence, by Lemma 1.7 it can be expressed as a product of linear factors

k Y e RP,Q(y, z) = (ciz − biy)i i=1 where e1 + ... + ek = nm. It means that the resultant RP,Q(bi, ci) = 0, i.e. polynomials P (x, bi, ci), Q(x, bi, ci) have a common root ai. Then pi = [ai : bi : ci] belongs to C ∩ D. On the other hand, if p = [a : b : c] ∈ C ∩ D then (b, c) 6= (0, 0) because [1 : 0 : 0] ∈/ C ∩ D. It follows that bz −cy is a factor of RP,Q(y, z), i.e. p = pi, for some i. Any two points [a : b : c], [α : β : γ] of C ∩D give diﬀerent factors of RP,Q(y, z). Indeed, if bz −cy was proportional to βz −γy then points [a : b : c], [α : β : γ] and [1 : 0 : 0] would be collinear which contradicts (c). To sum up, C ∩ D = {[ai : bi : ci] | 1 ≤ i ≤ k}.

By the deﬁnition of Ip(C,D), we get

k X X Ip(C,D) = ei = nm. p∈C∩D i=1

Lemma 6.2. If p ∈ C ∩ D is a singular point of C then Ip(C,D) > 1. Proof. We may assume that C and D have no common component, and hence we may choose coordinates such that p = [0 : 0 : 1] and the conditions (a)-(c) of Theorem 6.1 are satisﬁed. 2 We wish to show that y divides the resultant RP,Q(y, z), where C = {P = 0} and D = {Q = 0}. We have

∂P ∂P ∂P (0, 0, 1) = (0, 0, 1) = (0, 0, 1) = 0, ∂x ∂y ∂z hence P (x, y, z) cannot have terms xzn−1, yzn−1 and zn. It follows that P (x, y, z) is a sum of monomials of degree at least two in x and y:

n P (x, y, z) = a0(y, z) + xa1(y, z) + ... + x an(y, z)

36 2 where y divides a0 and y divides a1. Also Q(0, 0, 1) = 0, so

m Q(x, y, z) = b0(y, z) + xb1(y, z) + ... + x bm(y, z) and b0 is divisible by y. It follows that the first column of the matrix defining RP,Q(y, z) is divisible by y. We want to check that either the first column is divisible by y2 or that both the first and the second column are divisible by y. For this, we write

m−1 2 m−1 b0 = b01yz + y c0(y, z), b1 = b10z + yc1(y, z).

2 If b01 = 0 then the first column is divisible by y . If b01 6= 0 then we can subtract b10/b01 times the first column from the second one to get that the second column is divisible by 2 y. As it does not influence the determinant, we get that y divides RP,Q(y, z). We can now describe when the intersection multiplicity is one

Proposition 6.3. Let C and D be projective curves in P2 and let p be any point in C ∩D. Then Ip(C,D) = 1 if and only if p is a non-singular point of both C and D and the tangent lines to C and D at p are distinct. Proof. We may assume that C and D have no common component, hence we may assume that p = [0 : 0 : 1] and conditions (a)-(c) of Theorem 6.1 are satisﬁed. By Lemma 6.2 we know that p is a non-singular point of C and D. Let C = {P = 0}, D = {Q = 0}. We 2 show that the resultant RP,Q(y, z) is divisible by y if and only if tangent lines to C and D at p coincide. Since RP,Q is a homogeneous polynomial in y and z divisible by y, the ﬁrst condition is equivalent to ∂R P,Q (0, 1) = 0. ∂y By (c), [1 : 0 : 0] does not lie on the tangent line ∂P ∂P ∂P x (0, 0, 1) + y (0, 0, 1) + z (0, 0, 1) = 0, ∂x ∂y ∂z

∂P hence ∂x (0, 0, 1) 6= 0. The implicit function theorem for P (x, y, 1) implies that there are open neighbourhoods U, V of 0 ∈ C and a holomorphic function λ1 : U → V such that

λ1(0) = 0 and if x ∈ V , y ∈ U then

P (x, y, 1) = 0 ⇔ x = λ1(y). Moreover, P (x, y, 1) = (x − λ1(y))l(x, y), where l(x, y) is a polynomial in x whose coeﬃcients are holomorphic functions of y. If we assume that the coeﬃcient P (1, 0, 0) of xn in P (x, y, z) is one, then

n Y P (x, y, 1) = (x − λi(y)). i=1

37 Similarly, if U and V are suﬃciently small, there exists µ1 : U → V such that

µ1(0) = 0,Q(x, y, 1) = 0, ⇔ x = µ1(y), for x ∈ V , y ∈ U. We can write

m Y Q(x, y, 1) = (x − µi(y)). i=1 The tangent lines to C and D at point p are deﬁned by

0 0 x = λ1(0)y, x = µ1(0)y. If y ∈ U then by Lemma 4.8,

RP,Q(y, 1) = (µ1(y) − λ1(y))S(y), where Y S(y) = (µi(y) − λj(y)). (i,j)6=(1,1) It is a product of holomorphic functions, hence it is holomorphic. Since λ1(0) = 0 = µ1(0), we have ∂R P,Q (0, 1) = (µ0 (0) − λ0 (0))S(0). ∂y 1 1

∂P As ∂x (0, 0, 1) 6= 0, polynomial P does not have repeated zeroes at p, similarly for Q. It shows that λi(0) 6= 0 and µi(0) 6= 0, for i > 1. Moreover, if λi(0) = µj(0) for some pair i, j, then [0 : 0 : 1] and [µj(0) : 0 : 1] = [λj(0) : 0 : 1] are distinct points of C ∩ D lying on ∂RP,Q a line y = 0 which contradicts (b). It shows that S(0) 6= 0. Thus ∂y (0, 1) = 0 if and 0 0 only if λ1(0) = µ1(0) if and only if tangent lines to C and D at p coincide. Corollary 6.4. Let C and D be projective curves in P2 of degrees n and m. Assume that every point p ∈ C ∩ D is a non-singular point of C and D and that the tangent lines to C and D at p are distinct. Then C ∩ D has exactly nm points. Theorem 6.5 (Implicit function theorem). Let A(z, w) be a polynomial with complex coeﬃcients. Suppose that ∂A A(z , w ) = 0 6= (z , w ). 0 0 ∂w 0 0 Then there is a holomorphic function f : U → V , where U and V are open neighbourhoods of z0 and w0 in C such that f(z0) = w0 and if z ∈ U, w ∈ V then A(z, w) = 0 if and only if f(z) = w. Moreover, on U × V

A(z, w) = (w − f(z))B(z, w), where B(z, w) is a polynomial in w whose coeﬃcients are holomorphic functions of z.

38 02.02. Homework II

The homework in due on Thursday the 16th of February. Please hand in your solutions before the lecture.

1. Prove that a non-singular cubic curve C in P2 has exactly 9 inﬂection points [Hint: use Remark 7.6 and Corollary 6.4].

2. Prove the following converse to Pascal’s theorem: if the intersections of the opposite sides of a hexagon lie on a straight line then the vertices lie on a conic. More 2 precisely, let p1, . . . , p6 be distinct points of P , no three of which lie on a line. If i 6= j let Lij be the line through pi and pj. Show that if three points of intersection L12 ∩ L45, L23 ∩ L56, L34 ∩ L16 are collinear then p1, . . . , p6 lie on a conic. 3. Let C be a non-singular cubic curve deﬁned by

x3 + y3 + z3 + λxyz = 0

where λ3 + 27 6= 0. Show that the points of inflection of C are the points of intersection of C with a different curve of the same form, and deduce that they satisfy x3 + y3 + z3 = 0 = xyz. Deduce that C has exactly nine points of inflection and that a line through any two of them meets C again at the third point of inflection.

39 Solutions to Homework II

1. Let C be a non-singular cubic curve in P2 defined by P ∈ C[x, y, z] and let D be a cubic curve defined by the Hessian HP ∈ C[x, y, z]. It follows from Corollary 6.4 that in order to prove that C has nine inflection points we need to show that if p ∈ C ∩ D then p is a non-singular point of D and the tangent lines to C and D at p are distinct. By Remark 7.6 we can assume that p = [0 : 1 : 0] and

P (x, y, z) = y2z − x(x − z)(x − λz) = y2z − x3 + (λ + 1)x2z − λxz2.

Then ∂P ∂P ∂P = −3x2 + 2(λ + 1)xz − λz2, = 2yz, = y2 + (λ + 1)x2 − 2λxz ∂x ∂y ∂z ∂P ∂P ∂P (0, 1, 0) = 0, (0, 1, 0) = 0, (0, 1, 0) = 1. ∂x ∂y ∂z We have ∂2P ∂2P ∂2P = −6x + 2(λ + 1)z, = 0, = 2(λ + 1)x − 2λz, ∂x2 ∂x∂y ∂x∂z ∂2P ∂2P ∂2P = 2z, = 2y, = −2λx. ∂y2 ∂y∂z ∂z2 hence D is given by a polynomial  −6x + 2(λ + 1)z 0 2(λ + 1)x − 2λz  HP (x, y, z) = det  0 2z 2y  = 2(λ + 1)x − 2λz 2y −2λx (24λ + 16λ(λ + 1))xz2 − (8λ(λ + 1) + 8λ2)z3 − 8(λ + 1)2x2z − 8(λ + 1)y2z + 24xy2.

We have ∂H P = (24λ + 16λ(λ + 1))z2 − 16(λ + 1)2xz + 24y, ∂x ∂H P (0, 1, 0) = 24, ∂x ∂H P = −16(λ + 1)yz + 24x, ∂y ∂H P (0, 1, 0) = 0, ∂y ∂H P = 2(24λ + 16λ(λ + 1))xz − 3(8λ(λ + 1) + 8λ2)z2 − 8(λ + 1)2x2 − 8(λ + 1)y2, ∂z ∂H P (0, 1, 0) = −8(λ + 1). ∂z It follows that the line z = 0

40 tangent to C at p is distinct from the line

24x − 8(λ + 1)z = 0

tangent to D at p. The statement follows.

2. Let Lij = {[x : y : z] | Pij(x, y, z) = 0} be lines through points pi, pj. Consider two cubics Q1 = {P12P34P56 = 0},Q2 = {P23P45P61 = 0}.

Cubics Q1 and Q2 intersect in

p1 = L12 ∩ L61, p2 = L12 ∩ L23, p3 = L34 ∩ L23,

p4 = L34 ∩ L45, p5 = L56 ∩ L45, p6 = L56 ∩ L61,

q1 = L12 ∩ L45, q2 = L23 ∩ L56, q3 = L24 ∩ L16.

If q1, q2, q3 lie on a line, i.e. a curve of degree one then, by Proposition 5.6 the remaining six points p1, . . . , p6 lie on a curve C of degree at most two. Since no three of the points p1, . . . , p6 lie on a line, the curve C has degree two, i.e. it is a conic.

3. Let P (x, y, z) = x3 + y3 + z3 + λxyz. Then ∂P ∂P ∂P = 3x2 + λyz, = 3y2 + λxz, = 3z2 + λxy, ∂x ∂y ∂z ∂2P ∂2P ∂2P = 6x, = λz, = λy, ∂x2 ∂x∂y ∂x∂z ∂2P ∂2P ∂2P = 6y, = λx, = 6z, ∂y2 ∂y∂z ∂z2 and  6x λz λy  HP (x, y, z) = det  λz 6y λx  = λy λx 6z 216xyz + λ3xyz + λ3xyz − 6λ2y3 − 6λ2x3 − 6λ2z3 = −6λ2(x3 + y3 + z3) + (216 + 2λ3)xyz.

If λ = 0 then inﬂection points of

C = {[x : y : z] | P (x, y, z) = 0}

satisfy

x3 + y3 + z3 = 0, 216xyz = 0.

41 If λ 6= 0, inﬂection points of C satisfy

108 + 2λ3 x3 + y3 + z3 + 3λxyz = 0, x3 + y3 + z3 − xyz = 0. 3λ2

108+2λ3 3 The above cubics coincide if and only if λ = − 3λ2 , if and only if 4λ = −108, i.e. λ3 = −27. By assumption λ3 6= −27, hence two cubic are diﬀerent. Note that the only pair (a, b) ∈ C2 such that a + λb = 0 and a + µb = 0 with µ 6= λ is the pair (a, b) = (0, 0). Thus, the inﬂection points need to satisfy

x3 + y3 + z3 = 0, xyz = 0.

2 πi Let ε3 = e 3 be the primitive third root of unity. Then the inﬂection points of C with coordinate x equal to zero are

[0 : 1 : −ε], [0 : 1 : −ε2], [0 : 1 : −1].

Similarly we can write down the inﬂection points with coordinate y or z equal to zero. To sum up C has nine inﬂection points

2 q1 = [0 : 1 : −ε], q2 = [0 : 1 : −ε ], q3 = [0 : 1 : −1], 2 q4 = [1 : 0 : −ε], q5 = [1 : 0 : −ε ], q6 = [1 : 0 : −1], 2 q7 = [1 : −ε : 0], q8 = [1 : −ε : 0], q9 = [1 : −1 : 0].

We have q1, q2, q3 ∈ {x = 0}, q4, q5, q6 ∈ {y = 0}, q7, q8, q9 ∈ {z = 0}, q1, q4, q9 ∈ {εx + εy + z = 0}, q2, q6, q7 ∈ {εx + y + εz = 0}, q3, q5, q8 ∈ {x + εy + εz = 0}, 2 2 q1, q5, q7 ∈ {ε x + εy + z = 0}, q2, q4, q8 ∈ {ε x + y + εz = 0}, q3, q6, q9 ∈ {x + y + z = 0}, q1, q6, q8 ∈ {x + εy + z = 0}, q2, q5, q9 ∈ {x + y + εz = 0}, q3, q4, q7 ∈ {εx + y + z = 0}.

It follows that a line through any pair of inﬂection points of C intersects C in another inﬂection point.

42 7 06.02. Lecture 7. Cubic curves: classiﬁcation and group structure

We show that every non-singular projective curve of degree greater than two has at least one and at most ﬁnitely many inﬂection points. As a corollary, we get that every smooth projective cubic curve is of the form

y2z = x(x − z)(x − λz).

Deﬁnition 7.1. Let P ∈ C[x, y, z] be a homogeneous polynomial. The Hessian HP of P is the polynomial deﬁed by   Pxx Pxy Pxz HP = det  Pyx Pyy Pyz  . Pzx Pzy Pzz

Definition 7.2. A non-singular point [a : b : c] of a projective curve C in P2 defined by P (x, y, z) is called a point of inflection (of flex) of C if

HP (a, b, c) = 0.

Note that if deg P = d then the second partial derivatives of P are homogeneous of degree d − 2. It follows that HP is homogeneous of degree 3d − 6, hence the inﬂection point is well-deﬁned. In order to prove

Lemma 7.3. Let C = {P (x, y, z) = 0} be an irreducible projective curve of degree d. Then every point of C is an inﬂection point if and only if d = 1.

we need to know that if deg P = d > 1 then   Pxx Pxy Px 2 2 z HP (x, y, z) = (d − 1) det  Pyx Pyy Py  . (5) Px Py dP/(d − 1)

Indeed, Euler’s relation gives

dP (x, y, z) = xPx(x, y, z) + yPy(x, y, z) + zPz(x, y, z),

(d − 1)Px(x, y, z) = xPxx(x, y, z) + yPyx(x, y, z) + zPzx(x, y, z),

(d − 1)Py(x, y, z) = xPxy(x, y, z) + yPyy(x, y, z) + zPzy(x, y, z),

(d − 1)Pz(x, y, z) = xPxz(x, y, z) + yPyz(x, y, z) + zPzz(x, y, z).

43 Then   Pxx Pxy Pxz z det  Pyx Pyy Pyz  = Pzx Pzy Pzz   Pxx Pxy Pxz det  Pyx Pyy Pyz  = xPxx + yPyx + zPzx xPxy + yPyy + zPzy xPxz + yPyz + zPzz   Pxx Pxy Pxz (d − 1) det  Pyx Pyy Pyz  Px Py Pz Applying the same to the columns of the above matrix we get the result. ∂P Now if ∂y is non-zero the equation P (x, y, 1) = 0 locally deﬁnes y as a holomorphic function of x (by the implicit function theorem). Then ∂P dy ∂P + = 0, ∂x dx ∂y ∂2P dy ∂2P dy ∂2P d2y ∂P + ( )2 + 2 + = 0, ∂x2 dx ∂y2 dx ∂x∂y dx2 ∂y

dy ∂P ∂P i.e. (as dx = − ∂x / ∂y )

 P P P  d2y xx xy x = (P )−3 det P P P dx2 y  yx yy y  Px Py 0 It follows that

2 d y HP (x, y, 1) 2 = 2 3 (6) dx (d − 1) (Py) Proof of Lemma 7.3. Since the second derivatives of a polynomial of degree 1 vanish, one implication is clear. Let now C be a curve whose every point is an inﬂection point and assume that d = deg C is greater than 1. By applying a suitable projective transformation we can assume that ∂P P (0, 0, 1) = 0 6= (0, 0, 1). ∂y The implicit function theorem tells us that there are open neighbourhoods of U and V of 0 ∈ C and g : U → V such that g(0) = 0 and, for x ∈ V , y ∈ U, P (x, y, 1) = 0 if and only if y = g(x). Formula (6) implies that g00 = 0, hence g(x) = λx. It follows that P (x, λx, 1) vanishes identically.

44 Then P (x, y, z) is divisible by y − λx. Indeed, let us ﬁrst consider an example

P (x, y, z) = ax2 + bxy + cy2 + dxz + ez2 + fyz.

Then P (x, λx, 1) = ax2 + bλx2 + cλ2x2 + dx + e + fλx = 0 if and only if e = 0, d = −fλ and a = −bλ − cλ2. It follows that

P (x, y, z) = −bλx2 + bxy − cλ2x2 + cy2 − fλxz + fyz = (y − λx)(bx + c(y + λx) + fz).

In general, if P = P a xi1 yi2 zi3 , we get i1+i2+i3=n i1,i2,i3

j aj,0,n−j + λaj−1,1,n−j + ... + λ a0,j,n−j = 0.

Hence, for any monomial we have:

i j n−j j−i i n−j − λ aj−i,i,n−jx z − aj−i,i,n−jx y z j−i n−j i i i = aj−i,i,n−jx z (y − λ x ) j−i n−j i−1 i−2 i−1 i−1 = aj−i,i,n−jx z (y − λx)(y + λxy + ... + λ x ).

Since C was assumed to be irreducible, it follows that C = {x − λy = 0} has degree one.

Proposition 7.4. Let C be a non-singular projective curve in P2 of degree d. (i) If d ≥ 2 then C has at most 3d(d − 2) points of inﬂection.

(ii) If d ≥ 3 then C has at least one point of inﬂection.

Proof. HP is homogeneous of degree 3d − 6, so provided it is not constant (when d > 2 this means not identically zero) it defines a projective curve in P2 (possibly with multiple components). By Lemma 7.3 we know that HP is not the zero polynomial. Statement (i) follows from the weak Bezout’s theorem 5.2 provided P and HP have no common factor. Since a non-singular curve is irreducible, Corollary 5.3, the common component would be the whole curve C. Thus, to finish the proof of (i) we need to show that the only curve whose every point is an inflection point has degree one. It is the statement of Lemma 7.3 below. To prove (ii) we note that curve C given by a polynomial P of degree d ≥ 3 intersects curve given by polynomial HP of degree greater than or equal to 1 in at least one point, see Theorem 5.1.

Corollary 7.5. Let C be a non-singular cubic curve in P2. Then C is equivalent under a projective transformation to the curve deﬁned by

y2z = x(x − z)(x − λz), for some λ ∈ C \{0, 1}.

45 Proof. By Proposition 7.4, C has a point of inﬂection. We may assume that [0 : 1 : 0] is the inﬂection point and that the tangent line to C at [0 : 1 : 0] is z = 0. Then C = {P (x, y, z) = 0} and ∂P ∂P P (0, 1, 0) = 0, (0, 1, 0) = 0, (0, 1, 0) = 0, ∂x ∂y ∂P H (0, 1, 0) = 0, (0, 1, 0) 6= 0, P ∂z as C is non-singular. Applying (5) with y and z exchanged, we get   Pxx Px Pxz 2 3 y HP (x, y, z) = 4 det  Px 2 PPz  Pzx Pz Pzz so   Pxx 0 Pxz 2 0 = 4 det  0 0 Pz  = −4Pz Pxx. Pzx Pz Pzz

It follows that Pxx(0, 1, 0) = 0. The above conditions imply that the coefficients in front 2 2 3 of monomials x y (Pxx = 0), xy (Px = 0) and y (Py = 0) in P must vanish. Hence, P can be written as P (x, y, z) = yz(αx + βy + γz) + ϕ(x, z), for some homogeneous polynomial ϕ of degree three. Moreover, ∂P β = (0, 1, 0) 6= 0. ∂z Projective transformation αx + γz [x : y : z] 7→ [x : y + : z] 2β maps C to a curve defined by βy2z + ψ(x, z) = 0, α2x2+2αγxz+γ2z2 where ψ(x, z) = ϕ(x, z) + z 4β2 . ψ(x, z) is a product of three linear factors. Moreover, as C is irreducible, ψ is not divisible by z, hence the coefficient of x3 in ψ is not zero. After a suitable diagonal transformation C is defined by y2z = (x − az)(x − bz)(x − cz). and a, b, c are distinct (otherwise C would be singular). Projective transformation x − az [x : y : z] 7→ [ , ηy, z], b − a where η2 = (b − a)3 puts C into form y2z = x(x − z)(x − λz), for some λ ∈ C \{0, 1}.

46 Remark 7.6. We have in fact proved that for any non-singular cubic curve C and an inﬂection point p there exists a projective transformation which maps C to the curve given by y2z = x(x − z)(x − λz) and p to the point [0 : 1 : 0].

Lemma 7.7. A line L in P2 meets a non-singular cubic C either (a) in three distinct points p, q, r each with intersection multiplicity one (i.e. L is not the tangent line to C at p, q and r),

(b) in two points, p with multiplicity one and q with intersection multiplicity two (i.e. L is the tangent line to C at q but not at p and q is not a point of inﬂection on C),

(c) in one point p with intersection multiplicity three (i.e. L is a tangent line to C at p and p is the inﬂection point).

Theorem 7.8. Given any non-singular projective cubic C in P2 and a point of inflection p0 on C there is a unique group structure on C such that p0 is the zero element and three points of C add up to zero if and only if they are the three points of intersection of C with some line in P2. Proof. To check uniqueness, not first that additive inverses are uniquely determined since −p0 = p0 and if p 6= p0 then −p is the third point on the intersection of C with the line through p and p0. Also if p and q are any points then p + q = −r, where r is the third point of intersection of C with the line through p and q or the tangent line to C at p = q. It remains to show that it is an additive group structure. Commutativity comes directly from the definition of the group structure. At this point we cannot show associativity. We just check that p + p0 = p. We have p + p0 = −r where r is the third point of the intersection of the line via p and p0 with C. This point is not p0 as any the line tangent to C to p0 does not intersect it in any other point, see Lemma 7.7(c). It follows that −r is the third point of the intersection with the line through r and p0 with C, which is of course p. By definition −p is the third point of the intersection of a line through p and p0 with C. Then the third point on the line through p and −p which lies on C is p0, hence p + (−p) = p0.

47 8 09.02. Lecture 8. Topology of non-singular plane 1 curves and branched covers of CP

As a subset of P2 a complex projective curve has a natural topology. In fact C is a sphere with g handles. This number g is called the genus of the curve. We shall show that we have a degree-genus formula 1 g = (d − 1)(d − 2). 2 Lemma 8.1. A complex projective line L in P2 is homeomorphic to the two-dimesnional sphere 2 2 2 2 2 S = {(u, v, w) ∈ R | u + v + w = 1}. Proof. By applying a projective transformation which is a homeomorphism by Lemma 2.8 we may assume that L = {z = 0}. We deﬁne ϕ: S2 → L via

ϕ(u, v, w) = [u + iv, 1 − w, 0]. its inverse is given by

2Re (xy) 2Im (xy) |x|2 − |y|2 ψ([x : y : 0]) = ( , , ). |x|2 + |y|2 |x|2 + |y|2 |x|2 + |y|2

To describe the approach we are taking to determine genus of a curve, we ﬁrst consider an example of a projective curve

C = {[x : y : z] | x3 + y3 + z3 = 3yz2}.

Its aﬃne part is given by x3 + y3 + 1 = 3y. Then

√ 1 √ 1 −(x3 + 1) + x6 + 2x3 − 3 3 −(x3 + 1) − x6 + 2x3 − 3 3 y = + 2 2 √ √ √ for an appropriate choice of the cube roots. If x∈ / S = {1, ω, ω, − 3 3, −ω 3 3, −ω 3 3} then 6 3 the value of y is unique (the set S is the set of roots of x + 2x − 3). If√ we cut C√along 3 3 straight√ line√ segments connecting√ points of S, i.e. if we remove [1, −ω 3], [−ω 3, ω], [ω, − 3 3], [− 3 3, ω] and [ω, −ω 3 3] then on the remaining open set D there are three functions f1, f2, f3 satisfying

3 3 fj(x) + x + 1 = 3fj(x).

If we add three points at the inﬁnity we can construct C by gluing three copies of D∪{∞}. We can read oﬀ the gluing from the behaviour of branches f1, f2, f3 when we move around points of S.

48 We will see that the map which assigns to (x, y) in the affine part of C the coordinate x extends to a map ϕ: C → P1 such that for all but finitely many points of P1, ϕ−1(p) has three elements. In general, we will show that any smooth projective curve admits a map ϕ: C → P1 which is a homeomorphism outside of finitely many points on C and P1 – called ramification points and the branch locus. On the next lecture we will construct a triangulation of P1 and the induced triangulation of C. They will tell us how to glue C from pieces of P1. It will allow us to relate the genus and degree of C. Note that curve C = {y2 = xz} admits a surjection ϕ: C → P1 defined ϕ[x : y : z] = [x : z] such that if [x : z] ∈ P1 then ϕ−1([x : z]) consists of exactly two points unless x = 0 or z = 0. Such a map ϕ: C → P1 is called a double cover of P1 branched over [0 : 1] and [1 : 0]. We can use ϕ to visualise C as two copies of P1 cut and glued together. We shall see that any non-singular projective curve C of degree d > 1 can be viewed as a branched cover of P1 in a similar way. Let C be a non-singular projective curve in P2 defined by a homogeneous polynomial P (x, y, z) of degree d > 1. By applying a suitable projective transformation we may assume that [0 : 1 : 0] ∈/ C. Then we have a well-defined continuous map ϕ: C → P1 given by ϕ[x : y : z] = [x : z].

Definition 8.2. The ramification index νϕ[a : b : c] of ϕ at a point [a : b : c] ∈ C is the order of zero of the polynomial P (a, y, c) at point y = b. Point [a : b : c] is a ramification point of ϕ if νϕ[a : b : c] > 1.

Remark 8.3. (i) νϕ[a : b : c] > 0 for any [a : b : c] ∈ C.

(ii) νf [a : b : c] > 1 if and only if ∂P P (a, b, c) = 0 = (a, b, c) = 0, ∂y i.e. if and only if [a : b : c] ∈ C and the tangent line to C at [a : b : c] contains the point [0 : 1 : 0].

(iii) νf [a : b : c] > 2 if and only if ∂P ∂2P P (a, b, c) = 0 = (a, b, c) = (a, b, c) = 0. ∂y ∂y2 This happens if and only if the tangent line to C at [a : b : c] contains [0 : 1 : 0] and [a : b : c] is the point of inﬂection on C. To prove this note that [a : b : c] 6= [0 : 1 : 0] ∂P so a 6= 0 or c 6= 0. Let us assume that c 6= 0. If P (a, b, c) = 0 = ∂y (a, b, c) then formula (5) from the last lecture gives  P P P  (d − 1)2 xx xy x (d − 1)2 H (a, b, c) = det P P 0 = − (P )2P . P c2  yx yy  c2 x yy Px 0 0

By Euler’s relation, if Px(a, b, c) = 0 then Pz(a, b, c) = 0 and the point would be singular. It follows that Px 6= 0, hence HP (a, b, c) = 0 if and only if Pyy(a, b, c) = 0.

49 Lemma 8.4. The inverse image ϕ−1([a : c]) of any [a : c] ∈ P1 under ϕ contains exactly X d − (νϕ(p) − 1) p∈ϕ−1([a:c]) points. In particular ϕ−1([a : c]) contains d points if and only of ϕ−1([a : c]) contains no ramiﬁcation points of ϕ.

Proof. A point of C lies in ϕ−1([a : c]) if and only if it is of the form [a : b : c] where P (a, b, c) = 0. By assumption P (0, 1, 0) 6= 0 and we may assume that P (0, 1, 0) = 1. Then P (a, y, c) is a monic polynomial of degree d:

Y mi P (a, y, c) = (y − bi) , 1≤i≤r where b1, . . . , br are distinct complex numbers and m1 + ... + mr = d. Thus

−1 ϕ ([a : c]) = {[a, bi, c] | 1 ≤ i ≤ r} and νϕ[a : bi : c] = mi. Then r r X X r = d − mi + r = d − (mi − 1). i=1 i=1

Note that the above proof does not require that C is non-singular.

Definition 8.5. Let R be the set of ramification points of ϕ. The image ϕ(R) of R under ϕ is the branch locus of ϕ and ϕ: C → P1 is a branch cover of P1. Lemma 8.6. (i) ϕ has at most d(d − 1) ramification points.

(ii) If νϕ[a : b : c] ≤ 2 for all [a : b : c] ∈ C then C has exactly d(d − 1) ramiﬁcation points.

Proof. Since C is non-singular it is irreducible, Corollary 5.3. By assumption [0 : 1 : 0] ∈/ C so the coefficient P (0, 1, 0) of yd in P is non-zero. Thus, the homogeneous polynomial ∂P ∂y (x, y, z) is homogeneous of degree d − 1. It cannot be divisible by P , so C and the ∂P curve D = { ∂y = 0} have no common component. It follows from Remark 8.3(ii) that any ramification point of ϕ lies on C ∩ D. By Theorem 5.2 C ∩ D ≤ d(d − 1) which finishes the proof of part (i). Now suppose that νϕ[a : b : c] ≤ 2, for all [a : b : c] ∈ C. By Corollary 6.4, in order to prove (ii) it suffices to check that for any p ∈ C ∩ D, p is a non-singular point of D and the tangent line to C and D at p are distinct. If not then p = [a : b : c] satisfies

P (a, b, c) = 0 = Py(a, b, c)

50 and the tangent vector to D

(Pxy(a, b, c),Pyy(a, b, c),Pzy(a, b, c)) is either zero or the scalar multiple of the tangent vector to C:

(Px(a, b, c),Py(a, b, c),Pz(a, b, c)).

It follows that Pyy(a, b, c) = 0, i.e. νϕ[a : b : c] ≥ 2. Lemma 8.7. By applying a suitable projective transformation to C we may assume that

νϕ[a : b : c] ≤ 2, for all [a : b : c] ∈ C.

Proof. We know that C has at most 3d − 6 points of inﬂection, see Proposition 7.4. Thus by applying a suitable projective transformation we may assume that [0 : 1 : 0] does not lie on C nor on any of the tangent lines to C at its inﬂection points. We conclude by Remark 8.3(iii).

51 10.02. Workshop II

2 3 2 1. Let p = [0 : 0 : 1]. Calculate Ip(x + 2yz, y + x z).

2. Prove Lemma 7.7: A line L in P2 meets a non-singular cubic C either (a) in three distinct points p, q, r each with intersection multiplicity one (i.e. L is not the tangent line to C at p, q and r), (b) in two points, p with multiplicity one and q with intersection multiplicity two (i.e. L is the tangent line to C at q but not at p and q is not a point of inﬂection on C), (c) in one point p with intersection multiplicity three (i.e. L is a tangent line to C at p and p is the inﬂection point).

Reduce to the case when L = {y = 0} and [1 : 0 : 0] ∈/ C. Describe points in C ∩ L. Calculate tangent line to C at p ∈ C ∩L and check when it coincides with L. Check that p is an inﬂection point if and only if |C ∩ L| = {p}.

3. Use Bezout’s theorem to show that if a projective curve C in P2 of degree d has strictly more than d/2 singular points all lying on a line L then L is a component of C.

2 4. Prove Pappus’ theorem: if L and M are two projective lines in P and p1, p2, p3 lie on L \ L ∩ M and q1, q2, q3 lie on M \ L ∩ M then if Lij is the line joining pi and qj the three points of intersection of the pairs of lines Lij and Lji are collinear.

52 Solutions to Workshop II

1. Since p does not lie on {z = 0}, we have

2 3 2 2 3 2 Ip(x + 2yz, y + x z) = Ip(z(x + 2yz), y + x z)

We use property (vi) to get

2 2 3 2 2 2 3 2 3 2 2 2 3 2 Ip(zx +2yz , y +x z) = Ip(zx +2yz −y −x z, y +x z) = Ip(y(2z −y ), y +x z).

Using (v) we get

2 2 3 2 3 2 2 2 3 2 Ip(y(2z − y ), y + x z) = Ip(y, y + x z) + Ip(2z − y , y + x z).

Since p does not lie on the curve {2z2 − y2 = 0}, the second intersection is zero. Using (vi) and (v) again, we get

3 2 2 Ip(y, y + x z) = Ip(y, x z) = 2Ip(y, x) + Ip(y, z) = 2 · 1 + 0 = 2.

2. Since C is irreducible, it does not contain L so we can assume that L = {y = 0} and that [1 : 0 : 0] ∈/ C. If

C = {[x : y : z] | P (x, y, z) = 0}

then C ∩ L = {[x : 0 : z] | P (x, 0, z) = 0}. By Lemma 1.7, P (x, 0, z) can be written as

P (x, 0, z) = µ(x − λ1z)(x − λ2z)(x − λ3z).

Then C ∩ L = {[λi : 0 : 1] | i = 1, 2, 3}.

The tangent line to L at [λi : 0 : 1] is ∂P ∂P ∂P L = { (λ , 0, 1)x + (λ , 0, 1)y + (λ , 0, 1)z = 0}. i ∂x i ∂y i ∂z i

∂P ∂P The line Li is L if ∂x (λi, 0, 1) = 0 = ∂z (λi, 0, 1). We know that ∂P ∂P λ (λ , 0, 1) + (λ , 0, 1) = 3P (λ , 0, 1) = 0. i ∂x i ∂z i i Hence, ∂P ∂P (λ , 0, 1) = 0 ⇔ (λ , 0, 1) = 0. ∂x i ∂z i

It follows that Li is L if and only if λi is a repeated root of the polynomial

P (x, 0, 1) = (x − λ1)(x − λ2)(x − λ3).

53 In other words, if P (x, 0, 1) has three pairwise different irreducible factors then L intersects C in three different points and is not tangent to C at any of these points. 3 If L is tangent to C at point [λi : 0 : 1] we need to check that P (x, 0, y) = (x − λi) if and only if [λi : 0 : 1] is the inflection point of C. From the lecture we know that   Pxx Pxy Px 2 2 z HP (x, y, z) = (d − 1) det  Pyx Pyy Py  . Px Py dP/(d − 1) It follows that   Pxx Pxy 0 2 HP (λi, 0, 1) = 4 det  Pyx Pyy Py  (λi, 0, 1) = −4(Py) Pxx(λi, 0, 1). 0 Py 0

∂P ∂P Since L is assumed to be tangent to C at [λi : 0 : 1], derivatives ∂x , ∂z vanish when ∂P evaluated at (λi, 0, 1). Since C is non-singular, it follows that ∂y (λi, 0, 1) 6= 0. It ∂2P shows that [λi : 0 : 1] is an inﬂection point of C if and only if ∂x2 (λi, 0, 1) = 0, i.e. 3 if and only if P (x, 0, z) = (x − λiz) . 3. Let C be curve of degree d and let L be a line which contains e singular points of P C. For any p ∈ C ∩ L we have Ip(C,L) ≥ 2. It follows that p∈C∩L Ip(C,L) ≥ 2e. P If e > |d|/2 then p∈C∩L Ip(C,L) > d. It follows from Bezout’s theorem that L is a component of C.

4. Let Lij be a line through pi and qj and let r1 = L12 ∩L21, r2 = L13L31, r3 = L23 ∩L32. Consider a line E through r1 and r2. We shall show that r3 ∈ E.

Let Σ = {p1, p2, p3, q1, q2, q3, r1, r2}. We show that the space of cubics which contain Σ is two-dimensional. A cubic is given by

P (x, y, z) = Ax3 + By3 + Cz3 + Dx2y + Ex2z + F xy2 + Gy2z + Hxz2 + Iyz2 + Jxyz,

i.e. the space V3 of cubics in 10 dimensional. We have the evaluation map given by Σ: 8 ψ : V3 → C , ψ(P ) = (P (p1),...,P (r2)). The space of cubics which contain Σ is the kernel of ψ. It is two dimensional if ψ is surjective, i.e. if for any point of Σ there exists a cubic which does not contain this point and contains the remaining 7 points of Σ. This is indeed the case. For p1 it is the cubic M ∪ L21 ∪ L31. Similarly, one can construct cubics for the remaining 7 points of Σ.

Let C1 be a cubic L12 ∪L23 ∪L31 and C2 the cubic L21 ∪L32 ∪L13. The corresponding polynomials P1 and P2 form a basis of cubics containing Σ. It follows that the cubic C = M ∪ L ∪ E corresponds to a polynomial P = λP1 + µP2. In particular, C1 ∩ C2 = Σ ∪ {r3} is contained in C which ﬁnishes the proof.

54 9 13.02. Lecture 9. Covering projections, triangulations, and Euler characteristic

Deﬁnition 9.1. A continuous map π : Y → X between topological spaces is a covering projection if each x ∈ X has an open neighbourhood U ⊂ X such that π−1(U) is a disjoint union of open subsets of Y each of which is mapped homeomorphically onto U by π. We have seen an example of a covering projection last time. Namely, let C be a projective algebraic curve of degree d which does not contain [0 : 1 : 0]. The map

1 ϕ: C → P , f([x : y : z]) = [x : z] is well-defined. If we denote by S ⊂ C the (finite!!) set of ramification points and by B = ϕ(S) the branch locus, ϕ induces a map π : C \ S → P1 \ B such that the preimage of every point has exactly d points. The same is true about a small open subset U of P1 \ B, π−1(U) is a disjoint union of d copies of U. Another example, which we shall use later is given by a lattice (i.e. an additive subgroup isomorphic to Z2)Λ ⊂ C. The natural projection π : C → C/Λ is a covering projection of a torus. Lemma 9.2. Let π : Y → X be a covering projection and f : [0, 1] → X a continuous map. Given y ∈ Y such that π(y) = f(0) there exists a unique map F : [0, 1] → Y such that F (0) = y and π ◦ F = f.

S −1 Proof. X has an open cover X = Ui such that π (Ui) is a disjoint union of open subsets of Y mapped homeomorphically onto Ui. We may assume that f(0) ∈ U0. Since [0, 1] is compact, there exist 0 = t0 < t1 < . . . < tn−1 < 1 = tn such that f[tj−1, tj] is contained in one of Ui’s. −1 Let π (U0) = V0 ∪ ... ∪ Vd. Fixed element y belongs to one of the Vk’s, without loss of generality we may assume y ∈ V0. Let ψ : U → V0 be a homeomorphism such that π ◦ ψ = IdU . Then F1 := ψ ◦ f|[0,t1] : [0, t1] → Y is a map such that F1(0) = y and

π ◦ F1 = f|[0,t1]. −1 Map F1 gives a point F1(t1) ∈ π (t1). Point t1 lies in some element of the open cover −1 0 0 of X, let us denote it by Ui1 . As before we have π (Ui1 ) = V0 ∪ ... ∪ Vd and without 0 0 0 loss of generality we may assume that F1(t1) ∈ V0 . A homeomorphism ψ : Ui1 → V0 such that π ◦ ψ0 = Id gives a map ψ0 ◦ f| :[t , t ] → Y . Since ψ0 ◦ f(t ) = F (t ), Ui1 [t1,t2] 1 2 1 1 1 0 maps ψ ◦ f|[t1,t2] and F1 can be glued to give F2 : [0, t2] → Y such that F2(0) = y and

π ◦ F2 = f|[0,t2]. Analogous argument gives maps F3,...,Fn. By deﬁnition Fn = F is the desired map [0, 1] → Y . Since at every step the construction of Fi was unique, map F is unique. We consider a closed simplex

2 ∆ = {(x, y) ∈ R | x ≥ 0, y ≥ 0, x + y ≤ 1}, its interior

0 2 ∆ = {(x, y) ∈ R | x ≥ 0, y ≥ 0, x + y < 1},

55 and boundary

∂∆ = ∆ \ ∆0.

Deﬁnition 9.3. A topological space X is simply connected if any continuous map g : ∂∆ → X can be extended to a continuous map g : ∆ → X. Lemma 9.2 can be generalised to Lemma 9.4. (i) Let π : Y → X be a covering projection and let f : A → X be a continuous map. Suppose that A is simply connected, path connected and locally path connected (i.e. every point in A has arbitrarily small path connected open neighbourhood in A). Then given any a ∈ A and y ∈ Y such that f(a) = π(y) there is a unique continuous map F : A → Y such that F (a) = y and π ◦ F = f.

(ii) If f is a homeomorphism onto its image then F is a homeomorphism onto a connected component of π−1f(A).

A non-singular complex algebraic curve is given cut out by a polynomial from P2, i.e. it has a complex dimension one. When considered over the field of real numbers, C is of dimension two, i.e. it is a surface. We shall introduce Riemann surfaces later, for now on let us just say that a surface X is a topological space such that every point x ∈ X has an open neighbourhood which is homeomorphic to an open subset of R2. Definition 9.5. Let X be a surface. A triangulation of X is the following data (a) A finite nonempty set V of points called vertices,

(b) A ﬁnite nonempty set E of continuous maps e: [0, 1] → X called edges,

(c) A ﬁnite nonempty set F of continuous maps f : ∆ → X called faces, satisfying (i) V = {e(0) | e ∈ E} ∪ {e(1) | e ∈ E}, i.e. vertices are the end points of edges;

(ii) If e ∈ E then the restriction of e to the open interval (0, 1) is a homeomorphism onto its image, and this image contains no points in V or in the image of any other e ∈ E; (iii) If f ∈ F then the restriction of f to ∆0 is a homeomorphism onto a connected component Kf of X \ Γ, where [ Γ = e([0, 1]). e∈E

If r : [0, 1] → [0, 1] and σi : [0, 1] → ∆ are deﬁned by

r(t) = 1 − t, σ1(t) = (t, 0), σ2(t) = (1 − t, t), σ3(t) = (0, 1 − t)

i then either f ◦ σi or f ◦ σi ◦ r is an edge ef ∈ E, for 1 ≤ i ≤ 3;

56 (iv) The mapping f 7→ Kf from F to the set of connected components of X \ Γ is a bijection;

+ + (v) for every e ∈ E there is exactly one face fe ∈ F such that e = fe ◦ σi, for some i − − and exactly one face fe ∈ F such that e = fe ◦ σi ◦ r, for some i. Note that the boundary of ∆ is oriented counter clock-wise. The Euler number χT (X) with respect to the triangulation T is

χT (X) = |V | − |E| + |F |.

By Lemma 8.1 a complex projective line in P2 is homemomorphic to a sphere. Thus it has a triangulation with three vertices, three edges and two faces. It follows that

1 χT (P ) = 2.

Theorem 9.6. χT (X) depends only on X, not on the choice of triangulation. In order to prove Theorem 9.6 we introduce complex vector spaces

T T T C0 (X),C1 (X),C2 (X) with bases V , E and F . Thus

T X C0 (X) = { λvv | λv ∈ C v ∈ V }. v∈V We also have linear maps

T T T T ∂2 T ∂1 T ∂0 C2 (X) −→ C1 (X) −→ C0 (X) −→ C

T X X ∂0 ( λvv) = λv, T X X ∂1 ( λee) = λe(e(1) − e(0)), T X X 1 2 3 ∂2 ( λf f) = λf (±ef ± ef ± ef ),

i i where the sign in front of ef is positive if and only if f ◦ σi = ef . We have

T T T ∂0 ∂1 (e) = ∂0 (e(1) − e(0)) = 0 and T T ∂1 ∂2 (f) = f(1, 0) − f(0, 0) + f(0, 1) − f(1, 0) + f(0, 0) − f(0, 1) = 0. It follows that

T T T T im ∂1 ⊂ ker ∂0 , im ∂2 ⊂ ker ∂1 .

57 Lemma 9.7. We have

T T ker ∂0 ker ∂1 χT (X) = dim( T ) − dim( T ) + k + 1 im ∂1 im ∂2 where k is the number of connected components of X.

Proof. We have

T T |V | = dim ker ∂0 + dim im ∂0 , T T |E| = dim ker ∂1 + dim im ∂1 , T T |F | = dim ker ∂2 + dim im ∂2 .

Then T T ker ∂0 ker ∂1 T T χT (C) = dim( T ) − dim( T ) + dim im ∂0 + dim ker ∂2 . im ∂1 im ∂2 T T Since V 6= ∅ map ∂0 is non-zero, hence surjective. It follows that dim im ∂0 = 1. It thus T suﬃces to show that dim ker ∂2 = k. By point (vi) of the deﬁnition of the triangulation, we have

T X X ∂ ( λ f) = (λ + − λ − )e. 2 f fe fe

It vanishes if and only if λ + = λ − for all e ∈ E. fe fe Let X1,...,Xk be connected components of X. Since ∆ is connected, f(∆) is 0 S contained in one of the Xis. Moreover, by (v), X = f∈F f(∆). Finally, f(∆) ∩ fe(∆) + − if and only if there exists e ∈ E such that {f, fe} = {fe , fe }. It follows that f(∆) and fe(∆) lie in the same connected component of X if and only if there is a sequence

f = f0, f1, . . . , fn = fe such that {f , f } = {f +, f −}. i i+1 ei ei P Therefore λ f satisﬁes λ + = λ − if and only if there exist µ , . . . , µ ∈ such that f fe fe 1 k C T λf = µi if and only if f(∆) ⊂ Xi. It follows that dim ker ∂2 = k which ﬁnishes the proof.

T T ker ∂0 ker ∂1 To prove Theorem 9.6 it suffices to check that dim( T ) and dim( T ) are im ∂1 im ∂2 independent of the triangulation. We introduce infinite dimensional spaces C0(X), C1(X) and C2(X). The basis of C0(X) are points of X, i.e. X C0(X) = { λx x | x ∈ X, λx ∈ C, λx = 0for all but finitely many x ∈ X}. x∈X

58 The basis of C1(X) are continuous maps [0, 1] → X and the basis of C2(X) are continuous maps ∆ → X. We have

∂2 ∂1 ∂0 C2(X) −→ C1(X) −→ C0(X) −→ C, (7)

T T T for ∂2, ∂1 and ∂0 deﬁned analogously as ∂2 , ∂1 , ∂0 . One checks that (7) is a complex. It is a part of singular cochain of X.

T T ker ∂0 ker ∂0 ker ∂1 ker ∂1 Proposition 9.8. T ' and T ' . im ∂1 im ∂1 im ∂2 im ∂2 The proof is based on Lemmas

T Lemma 9.9. If ξ ∈ C0(X) then there exists η ∈ C1(X) such that ξ − ∂1η ∈ C0 (X). T Lemma 9.10. If ξ ∈ C1(X) and ∂1ξ ∈ C0 (X) then there exists η ∈ C2(X) such that T ξ − ∂1η ∈ C1 (X). T Lemma 9.11. If ξ ∈ C2(X) and ∂2ξ ∈ C1 (X) then there exists η ∈ C2(X) such that T ∂2ξ = ∂2 η. Proof of Proposition 9.8. It suﬃces to show that for j = 0, 1,

T T T ker ∂j = im∂j+1 + ker ∂j , im∂j+1 ∩ ker ∂j = im∂j+1

T T as then inclusions ker ∂j ⊂ ker ∂j and im∂j ⊂ im∂j will give isomorphisms

T ker ∂j ker ∂j T ' . im∂j im∂j

T Indeed, any ξ ∈ ker ∂j can be then written as ζ + ∂jη for ζ ∈ ker ∂j , hence ζ and ξ are T T ker ∂j ker ∂j ker ∂j equivalent in . Moreover, the second isomorphism implies that ' T . im∂j+1 im∂j im∂j Inclusions

T T T T im∂j+1 + ker ∂j ⊂ ker ∂j, im∂j+1 ⊂ im∂j+1 ∩ ker ∂j are clear. T Let now ξ ∈ ker ∂j. There exists η ∈ Cj+1(X) such that ξ − ∂j+1η = ζ ∈ Cj (X). Moreover, T ∂j (ζ) = ∂j(ζ) = ∂jξ + ∂j∂j+1η = 0, T so ξ = ζ + ∂j+1η ∈ ker ∂j + im∂j+1. T Next, suppose that ζ ∈ ker ∂0 and ζ = ∂1ξ, for some ξ ∈ C1(X). There exists T η ∈ C2(X) such that ξ − ∂2η = χ ∈ C1 (X). Then

T ζ = ∂1ξ = ∂1χ = ∂1 χ,

T i.e. ζ ∈ im∂1 . T Finally, suppose that ζ ∈ ker ∂1 , ξ = ∂2(ξ) for some ξ ∈ C2(X). There exists η ∈ T T T C2 (X) such that ∂2 (η) = ∂2ξ = ζ, i.e. ζ ∈ im∂2 . Since the three objects above were arbitrary, we get the required isomorphisms.

59 10 16.02. Lecture 10. Triangulation of plane curves and the degree–genus formula

Last time we deﬁned a triangulation and its Euler number χT (X). We proved that χT (X) depends only on X, not on the triangulation. Today, we show

Theorem 10.1. Every nonsingular projective curve C in P2 has a triangulation. Moreover, if C is of degree d and r is a positive integer such that r ≥ d(d − 1) and r ≥ 3 then C has a triangulation with rd − d(d − 1) vertices, 3(r − 2)d edges and 2(r − 2)d faces.

Using map ϕ: C → P1 we will relate the Euler number of C to the Euler number of P1 and degree of C. As a corollary, we will obtain the degree-genus formula. First we relate the Euler number of a surface X to its genus.

Lemma 10.2. Let X be a sphere with g handles. Then its Euler number is 2 − 2g.

Proof. Picture!!!

Deﬁnition 10.3. The genus of a nonsingular projective curve is 1 g = (2 − χ), 2 where χ is the Euler number of C.

As a corollary of Theorem 10.1 we get

Corollary 10.4. The Euler number χ and genus g of a non-singular projective curve of degree d in P2 are given by 1 χ = rd − d2 + d − 3rd + 6d + 2rd − 4d = d(3 − d), g = (d − 1)(d − 2). 2 For the proof of Theorem 10.1 we need

1 Lemma 10.5. Let p1 . . . pr be any set of at least 3 distinct points in P . Then there is a 1 triangulation of P with p1, . . . , pr as vertices, 3r − 6 edges and 2r − 4 faces.

Proposition 10.6. Let C be a nonsingular projective curve in P2 not containing [0 : 1 : 0] and let ϕ: C → P1 be the branched cover deﬁned by ϕ([x : y : z]) = [x : z]. Suppose that (V,E,F ) is a triangulation of P1 such that V contains the branch locus of ϕ. Then there is a triangulation (V,e E,e Fe) of C such that

Ve = ϕ−1(V ), Ee = {e: [0, 1] → C | e is continuous ϕ ◦ e ∈ E}, Fe = {e: ∆ → C | fe is continuous ϕ ◦ fe ∈ F }.

60 Moreover, if νϕ(p) is the ramiﬁcation index of ϕ at p and d is the degree of C then X |Ve| = d|V | − (νf (p) − 1), p∈R |Ee| = d|E|, |Fe| = d|F |, where R ⊂ C is the set of ramiﬁcation points for ϕ.

It gives Riemann-Hurwitz formula

1 X χ(C) = dχ(P ) − (νϕ(p) − 1). p∈R

Proof of Theorem 10.1. Let P (x, y, z) be a homogeneous polynomial of degree d defining the curve C. By Lemma 8.7 after applying a suitable projective transformation we can assume that map 1 ϕ: C → P , ϕ([x : y : z]) = [x : z] is well-defined and the ramification index of ϕ at every point of C is less than or equal to two. Then, by Lemma 8.6 ϕ has exactly d(d − 1) ramification points. We denote the set of ramification points by R. By Lemma 10.5 if r ≥ 3 and r ≥ d(d − 1) then we can choose a triangulation (V,E,F ) of P1 such that ϕ(R) ⊂ V and |V | = r, |E| = 3r − 6, |F | = 2r − 4. Therefore, by Proposition 10.6, there is a triangulation (V,e E,e Fe) of C with X |Ve| = d|V | − (νϕ(p) − 1) = rd − d(d − 1), p∈R |Ee| = d|E| = 3(r − 2)d, |Fe| = d|F | = 2(r − 2)d.

Proof of Lemma 10.5. We proceed by induction on r. We know that P1 ' S2 so it has a triangulation with 3 vertices, 3 edges and 2 faces. Assume now that a triangulation with vertices p1, . . . , pr−1 exists. By inductive assumption it has 3r − 9 edges and 2r − 6 faces. Point pr might lie in an interior of a face or of an edge. In the ﬁrst case we consider the subdivision of the triangle. We add 3 edges and 2 faces. If pr lies on an edge e, we need to subdivide two faces adjacent to e. We add 3 edges and 2 faces. Hence in both cases we got a triangulation with r vertices, 3r − 6 edges and 2r − 4 faces which ﬁnishes the proof.

Sketch of proof of Proposition 10.6. Let f : ∆ → P1 be an element of F . If p ∈ C is a point such that ϕ(p) = f(t), for some t∈ / {(0, 0), (0, 1), (1, 0)}, then by Lemma 9.4 there exists a unique fe: ∆ → C such that fe(t) = p and ϕ◦fe = f. Since ϕ−1(f(t)) has d points, there are exactly d “faces over” fe.

61 1 Similarly, any e: [0, 1] → P can be lifted in d possible ways to e: [0, 1] → C. P By Lemma 8.4, Ve has d|V | − p∈R(νϕ(p) − 1) points. For a lift e of e, e(0) and e(1) lie in the fibers of ϕ over e(0) and e(1). As for any e, e(0) and e(1) are points of V , the end points of e are points of Ve. Similarly for the edges of fe ∈ Fe. It follows that conditions (i)-(iii) are satisfied. By construction the set G = S f(∆) contains C \ ϕ−1(V ). As ϕ−1(V ) is finite, G fe∈Fe e is dense. Moreover, G is a union of compact, hence closed sets. It follows that G = C and (iv) holds. The proof of (v) is technical and will be skipped.

62 16.02. Homework III

The homework in due on Thursday the 9th of March. Please hand in your solutions before the lecture.

T 1. Let (V,E,F ) be a triangulation of a path connected surface X and let C0 (X) be a complex vector space with basis V . Let further C0(X) be a complex vector space whose basis consists of all points of X and C1(X) a complex vector space with basis of all continuous maps [0, 1] → X. Let ∂1 : C1(X) → C0(X) be a linear map which to a basis element e: [0, 1] → X assigns e(1) − e(0) ∈ C0(X). Prove that for any T ξ ∈ C0(X) there exists η ∈ C1(X) such that ξ − ∂1η ∈ C0 (X).

2. Let ϕ: C → P1 be deﬁned by ϕ([x : y : z]) = [x : z]

where C ⊂ P2 is a non-singular projective curve not containing [0 : 1 : 0]. Show that if C has degree d > 1 then ϕ has at least one ramiﬁcation point. Show that if f = 1 then ϕ has no ramiﬁcation points and is a homeomorphism.

3. The theorem of isolated zeros in complex analysis states that for an open U ⊂ C and a non-constant holomorphic function f : U → C the set {u ∈ U | f(u) = 0} is totally disconnected. Use it to show that if f : R → S is a non-constant holomorphic map between connected Riemann surfaces then every x ∈ R has an open neighbourhood U ⊂ R such that f(y) 6= f(x) for all y ∈ U \{x}.

63 Solutions to Homework III

Pr 1. Let ξ = i=1 λixi and let v be a vertex of the triangulation. Since X is path connected, there exist paths connecting each of xi with v. Let ei : [0, 1] → X be such a path, i.e. assume that ei(0) = xi and ei(1) = v. Pr 0 Pr T Deﬁne η = i=1 λiei ∈ C1(X) and v = ( i=1 λi)v ∈ C0 (X). Then

r r r X X X 0 ξ − ∂1η = λixi − (λiei(0) − λiei(1)) = λiv = v , i=1 i=1 i=1

i.e. η is the required element of C1(X).

∂P 2. Ramification points on ϕ are the points [a : b : c] ∈ C such that ∂y (a, b, c) = 0. ∂P If d > 1 then deg ∂y ≥ 1, hence ramification points are the intersection points of ∂P C with D = {[x : y : z] | ∂y (x, y, z) = 0}. By Bezout’s theorem curves C and D intersect, hence ϕ has at least one ramification point. If C = {[x : y : z] | P (x, y, z) = 0} is of degree one, then for any [a : b : c] ∈ C the polynomial P (a, y, c) is of degree one, hence it cannot have multiple zeroes. By definition ϕ has no ramification points. Let now q be any point in P1. We know that ϕ−1(q) has precisely X 1 − (νϕ(p) − 1) = 1 p∈ϕ−1(q)

point. Thus, ϕ is surjective. Moreover, ϕ is injective as ϕ(p) = ϕ(p0) would imply that the preimage of ϕ(p) would have at least two points. It follows that ϕ is bijective. We know that ϕ is continuous. To prove that ϕ is a homeomorphism we need to check that ϕ−1 is continuous. If C = {αx + βy + γz = 0} then the condition [0 : 1 : 0] ∈/ C implies that β 6= 0. Thus −αx − γz ϕ−1([x : z]) = [x : : z] β 2 3 −αx−γz 1 is continuous, as map ψ : C → C , ψ(x, z) = (x, β , z) is continuous and P , P2 have quotient topology.

3. Let x ∈ R be any point, ϕα : Uα → Vα, ψβ : Wβ → Xβ atlases on R and S such that x ∈ Uα and f(x) ∈ Wβ. −1 −1 Function ψβ ◦ f ◦ ϕα : ϕα(Uα ∩ f Wβ) → Xβ is holomorphic, hence so is

−1 −1 0 g := ψβ ◦ f ◦ ϕα − ψβ ◦ f(x): ϕα(Uα ∩ f Wβ) → Xβ,

0 where Xβ = Xβ − ψβ ◦ f(x) is the image of Xβ under the homeomorphism C → C, z 7→ z − ψβ ◦ f(x).

64 0 Theorem on isolated zeros implies that ϕα(x) has an open neighbourhood V ⊂ Vα 0 such that g is not equal to zero on V \{ϕα(x)}. Since ϕα is a homeomorphism, 0 −1 0 U = ϕα (V ) is an open neighbourhood of x. 0 −1 −1 Let y be a point in U . If f(y) = f(x) then ψβ ◦f ◦ϕα (ϕα(y)) = ψβ ◦f ◦ϕα (ϕα(x)) 0 0 hence ϕα(y) ∈ V is a zero of g. The contradiction with the choice of V implies that f(x) 6= f(y).

65 11 27.02. Lecture 11. Holomorphic atlases, Riemann surfaces and nonsingular plane curves

We deﬁne Riemann surfaces and show that every non-singular projective curve is a Riemann surface.

Deﬁnition 11.1. A surface is a Hausdorﬀ topological space S which is locally homeomorphic to C (or equivalently to R2). It means that any x ∈ S has an open neighbourhood which is homeomorphic to an open subset in C. A homeomorphism ϕ: U → V between an open subset U of S and an open subset V of C is called a chart on S. An atlas Φ for S is a collection of charts

Φ = {ϕα : Uα → Vα | α ∈ A} such that [ S = Uα. α∈A

If ϕα : Uα → Vα and ϕβ : Uβ → Vβ are charts then

ϕα(Uα ∩ Uβ) is an open subset of C. If Φ is an atlas for S then the homeomorphisms

−1 ϕαβ = ϕαϕβ : ϕβ(Uα ∩ Uβ) → ϕα(Uα ∩ Uβ) are called transition functions. An atlas Φ is holomorphic if all of its transition functions are holomorphic. If U is an open subset then IdU is a holomorphic atlas on U.

Proposition 11.2. If C is a complex algebraic curve in C2 deﬁned by a polynomial P (x, y) then C \ Sing(C) has a holomorphic atlas.

∂P Proof. Suppose that (a, b) ∈ C and ∂y (a, b) 6= 0. The implicit function theorem tells us that there are open neighbourhoods V and W of a and b in C and a holomorphic function g : V → W such that g(a) = b and if x ∈ V , y ∈ W then

P (x, y) = 0 ⇔ y = g(x).

Since C \ Sing(C) is open in C, shrinking V and W if necessary, we may assume that

U = {(x, y) ∈ C | x ∈ V, y ∈ W } is an open neighbourhood of (a, b) in C \ Sing(C). Then ϕ: U → V

ϕ(x, y) = x

66 is a continuous function with a continuous inverse

x 7→ (x, g(x)).

∂P Similarly, if ∂x (a, b) 6= 0 there is an open neighbourhood U of (a, b) in C \ Sing(C) such that the map ψ : U → C, ψ(x, y) = y is a homeomorphism onto an open subset V of C with the inverse given by y 7→ (h(y), y).

Thus there is an atlas on C \ Sing(C) such that every chart is one of these two ϕ and ψ. The transition functions are either identity or

x 7→ (x, g(x)) 7→ g(x), y 7→ (h(y), y) 7→ h(y).

It follows that the atlas is holomorphic.

Proposition 11.3. If C = {P = 0} is a projective curve in P2 then C \ Sing(C) has a holomorphic atlas.

∂P Proof. Suppose that [a : b : c] ∈ C and that ∂y (a, b, c) 6= 0. By Euler’s relation ∂P ∂P ∂P a (a, b, c) + b (a, b, c) + c (a, b, c) = 0 ∂x ∂y ∂z so a = c = 0 implies a = b = c = 0. It follows that (a, c) 6= (0, 0). Assume that c 6= 0. Then ∂P ∂P (a/c, b/c, 1) = c−d−1 (a, b, c) 6= 0 ∂y ∂y where d = deg P . By implicit function theorem for P (x, y, 1) there exist open neighbourhoods V of a/c and W of b/c and holomorphic g : V → W such that for x ∈ V , y ∈ W , P (x, y, 1) = 0 ⇔ y = g(x). If V and W are small enough,

U = {[x : y : 1] ∈ C | x ∈ V, y ∈ W } is an open neighbourhood of [a : b : c] = [a/c : b/c : 1] in C \Sing(C). The map ϕ: U → V

ϕ([x : y : z]) = x/z has an inverse x 7→ [x : g(x) : 1]. Similarly, if a 6= 0 or the other partial derivatives of P at (a, b, c) do not vanish we can ﬁnd an open neighbourhood of [a : b : c] in C \ Sing(C) and a homeomorphism to an open subset of C given by z/x, y/z, z/y, x/y, y/x

67 and inverse

w 7→ [1 : g(w): w], [g(w): w : 1] [g(w) : 1 : w], [w : 1 : g(w)], [1 : w : g(w)].

Thus we get an atlas in which all the transition functions are of the form

w 7→ w, 1/w, g(w), 1/g(w), w/g(w), g(w)/w such that g is holomorphic and the denominator does not vanish on the set where the transition function is deﬁned. It follows that the atlas is holomorphic. When we have a holomorphic atlas, we can deﬁne a holomorphic function on a surface.

Deﬁnition 11.4. Let Φ = {ϕα : Uα → Vα | α ∈ A} be a holomorphic atlas on a surface S. A continuous map f : S → C is holomorphic with respect to Φ at x ∈ S if there is a −1 chart ϕα : Uα → Vα in Φ such that x ∈ Uα and f ◦ ϕα : Vα → C is holomorphic. Map ϕ is holomorphic if it is holomorphic with respect to Φ at every x ∈ S.

Because Φ is a holomorphic atlas, the above condition is equivalent to requiring that −1 for any Uα such that x ∈ Uα map f ◦ ϕα is holomorphic. It implies that

Lemma 11.5. A continuous function ϕ: S → C is holomorphic with respect to a −1 holomorphic atlas Φ if and only of f ◦ ϕα : Vα → C is holomorphic for every chart ϕα : Uα → Vα in Φ. Deﬁnition 11.6. Let S and T be surfaces with holomorphic atlases Φ and Ψ. A continuous map f : S → T is holomorphic with respect to Φ and Ψ if

−1 −1 −1 ψB ◦ f ◦ ϕα |ϕα(Uα∩f Wβ ) : ϕα(Uα ∩ f Wβ) → Yβ is holomorphic for every chart ϕα : Uα → Vα in Φ and every chart ψβ : Wβ → Yβ in Ψ.

−1 −1 Note that f (Wβ) is open because f is continuous, hence U ∩ f (Wβ) is open. Lemma 11.7. If f : S → T and g : T → R are holomorphic with respect to given holomorphic atlases Φ, Ψ, Θ on the surfaces S, T , R then g ◦ f : S → R is holomorphic with respect to atlases Φ and Θ.

Proof. Choose x ∈ S and charts ϕα : Uα → Vα, ψβ : Wβ → Yβ, θγ : Xγ → Zγ such that −1 x ∈ Uα, f(x) ∈ Vβ, gf(x) ∈ Xγ. We need to check that θγ ◦ g ◦ f ◦ ϕα is holomorphic at ϕα(x). It follows from the fact that locally we can write

−1 −1 −1 θγ ◦ g ◦ f ◦ ϕα = (θγ ◦ g ◦ ψβ ) ◦ (ψβ ◦ f ◦ ϕα ).

Deﬁnition 11.8. Two holomorphic atlases Φ, Ψ on a surface S are compatible if the identity map IdS : S → S is holomorphic both as map from S with atlas Φ to S with atlas Ψ and as a map from S with atlas Ψ to S with atlas Φ.

68 It follows from Lemma 11.7 that compatibility deﬁnes equivalence relation on the set of holomorphic atlases on S. The holomorphic atlases

{IdC : C → C} and

{h: U → V | U, V ⊂ C open, h is holomorphic with a holomorphic inverse} on C are compatible. They are not compatible with the atlas with the single chart

{ϕ: C → C}, ϕ(z) = z. Deﬁnition 11.9. A Riemann surface os a surface S together with an equivalence class H of holomorphic atlases on S.A holomorphic map f :(S, H) → (T, F) is a continuous map which is holomorphic with respect to any holomorphic atlas Φ ∈ H on S and Ψ ∈ F on T . Deﬁnition 11.10. Two Riemann surfaces S and T are biholomorphic if there is a holomorphic bijection f : S → T whose inverse is holomorphic. Examples

(a) Any open subset U of C with the holomorphic atlas {IdU : U → U} is a Riemann surface.

(b) If S is a Riemann surface with a holomorphic atlas

Φ = {ϕα : Uα → Vα | α ∈ A} and W ⊂ S is an open subset then

Φ|W = {fα|Uα∩W : Uα ∩ W → ϕα(Uα ∩ W ) | α ∈ A} is a holomorphic atlas on W .

(d) The Riemann sphere is P1 = C ∪ {∞} where z ∈ C is identiﬁed with [z : 1] ∈ P1 and ∞ = [0 : 1]. Let U = P1 \ {∞}, V = P1 \{0}. Deﬁne ϕ: U → C, ψ : V → C by ϕ([x : y]) = x/y, ψ([x : y]) = y/x.

It is a homolomorphic atlas with transition functions

−1 −1 ϕ ◦ ψ = ψ ◦ ϕ : C \{0} → C \{0} given by z 7→ 1/z.

69 If U is an open subset of C then a meromorphic function on U is f : U → C ∪ {∞} such that for any a ∈ U there exists ε > 0 such that f maps the punctured disc

{z ∈ C | 0 < |z − a| < ε} to C and admits a Laurent series expansion

X k f(z) = ck(z − a) . k≥−m

If this power series is identically zero then f(a) = 0. Otherwise we can assume that c 6= 0. Then −m   ∞ if m > 0, f(a) = c0 if m = 0,  0 if m = 0. Equivalently, we can write f(z) = (z − a)−mh(z) where h(z) is holomorphic and h(a) 6= 0. Thus f(z) is meromorphic in a neighbourhood of a if and only if 1/f(z) = (z − a)m/h(z) is holomorphic. This shows that a meromorphic function is a holomorphic function f : U → P1. Conversely, any holomorphic f : U → P1 such that f(U) ∈/ {∞} is a meromorphic function on U. If

n m p(z) = a0 + a1z + ... + anz , q(z) = b0 + b1z + . . . bmz are polynomials of degree n and m with no common factor there is a holomorphic map ϕ: P1 → P1 whose restriction to C is the rational function p(z)/q(z) and whose value at ∞ is   0 if m > n, f(∞) = an/bn if m = n,  ∞ if m < n. It follows that unless f is constant it must take value ∞ at least once.

Lemma 11.11. Any nonconstant holomorphic map f : P1 → P1 is rational hence takes the value ∞ at least once.

Proof. By definition if a meromorphic function has a pole at a, it is holomorphic on some neighbourhood of a. Since the restriction of f to C and map C → P1 defined by z 7→ f(1/z) are both meromorphic and P1 is compact, f can have only finitely many poles. Let us assume that they are at a1, . . . , ak.

70 In the neighbourhood of aj f has a Laurent expansion

X (j) n f(z) = cn (z − aj) . n≥−mj

We consider k −1 X X (j) n g(z) = cn (z − aj) . j=1 n=−mj It is a rational function and it extends to a holomorphic function g : P1 → P1 with g(∞) = 0. Map f − g : P1 → P1 has no poles in C so

X n f(z) − g(z) = cnz . n≥0

In the neighbourhood of ∞ with the local chart w = 1/z the diﬀerence f − g has Laurent expansion X −n f(w) − g(w) = cnw . n≥0

It is meromorphic if and only if there exists N such that cn = 0 for n > N. It follows PN n that f(z) = g(z) + n=0 cnz is a sum of rational functions, hence it is rational.

71 12 02.03. Lecture 12. Weierstrass’ elliptic function

Let W ⊂ C be open. A function f : W → C is holomorphic if its derivative f(z) − f(a) f 0(a) = lim z→a z − a exists at every a ∈ W . A function f is holomorphic on an open disc {z ∈ C | |z − a| < r} if and only if it can be expressed as a convergent power series

X n f(z) = cn(z − a) , |z − a| < r n≥0

1 (n) called the Taylor series of f about a. The coeﬃcients are cn = n! f (a). A meromorphic function on W is a function f : W → C ∪ {∞} such that f|W \f −1(∞) is holomorphic and f has a pole at every point a ∈ f −1(∞). It means that near a g(z) f(z) = , (z − a)m for some holomorphic function g such that g(a) 6= 0. Equivalently

X n f(z) = cn(z − a) . n≥−m

Then m is called the order or multiplicity of the pole. The coefficient c−1 is called the residue of the pole and is denoted by res {f(z), a}. f 0(z) The residue at a of f(z) is then −m. Similarly, if f has zero of order m in a then the f 0(z) residue of f(z) at a is m. Residues of meromorphic functions can be calculated by integrals over closed piecewise smooth contours γ, i.e. functions γ : [0, 1] → C such that γ(0) = γ(1), γ(t1) 6= γ(t2) if t1 6= t2 and {t1, t2}= 6 {0, 1} and γ is smooth except finitely many points. If γ :[c, d] → W is smooth, the integral of f over γ is by definition Z Z d f(z)dz = f(γ(t))γ0(t)dt. γ c Theorem 12.1 (Cauchy’s theorem). Let γ be a closed piecewise smooth contour in C and let f be a function which is holomorphic inside and on γ. Then Z f(z)dz = 0. γ Theorem 12.2 (Cauchy’s residue theorem). Let γ be a closed piecewise smooth contour in C and let ϕ be a meromorphic function inside and on γ with no poles on γ and poles a1, . . . , ar inside γ. Then r Z X f(z)dz = ±2πi res {f(z), aj}, γ j=1 where the sign depends on the orientation of γ.

72 There is an useful construction of holomorphic functions

Theorem 12.3. Let (fn : W → C)n≥1 be a sequence of holomorphic functions on an open subset W of C converging uniformly to a function f : W → C. Then f is holomorphic on 0 0 W and the derivatives fn converge uniformly to f on W . Recall that a sequence converges uniformly if for any ε > 0 there exists N such that for any n, m > N, |fn(z) − fm(z)| < ε for all z ∈ W . A direct application is Weierstrass M-test

Theorem 12.4 (Weierstrass M-test). Let (fn : W → C)n≥1 be a sequence of holomorphic functions on an open subset W of C. Suppose there exist positive real numbers Mn such that X Mn n≥1 converges and |fn(z)| ≤ Mn ∀z ∈ W. Then the series X fn(z) n≥1 converges uniformly on W to a holomorphic function f(z) such that

0 X 0 f (z) = fn(z). n≥1

Now we are ready to deﬁne Weierstrass elliptic ℘ function. Let ω1, ω2 be complex number linearly independent over R (i.e. ω1/ω2 ∈/ R). We deﬁne a lattice

2 Λ = {nω1 + mω2 | (n, m) ∈ Z } which is an additive subgroup of C isomorphic to Z2.

Proposition 12.5. There is a meromorphic function ℘(z) on C deﬁned by X ℘(z) = z−2 + ((z − ω)−2 − ω−2), ω∈Λ\{0} with derivative given by X ℘0(z) = −2(z − ω)−3. ω∈Λ

Proof. We want to show that function ℘ is meromorphic at any point of C. We will show that ℘(z) is a sum of a homolomorphic function and ﬁnitely many meromorphic functions. Note that any point lies in some disc {z ∈ C | |z| ≤ R}. Therefore it is enough to prove that, for any R there exists a ﬁnite subset Λ ⊂ Λ such that P ((z − ω)−2 − ω−2) R ω∈Λ\ΛR is holomorphic. Using Weierstrass M-test we will show that P ((z − ω)−2 − ω−2) ω∈Λ\ΛR converges uniformly on {z ∈ C | |z| ≤ R}.

73 We shall need to know that there exists δ such that

p 2 2 |xω1 + yω2| ≥ δ x + y for any x, y ∈ R. We prove it below. Let now ΛR = {ω ∈ Λ | |ω| < 2R}. The above inequality implies that

2 2 2 −2 ΛR ⊂ {nω1 + mω2 | n + m ≤ 4R δ } hence it is ﬁnite. 1 Let now z ∈ C be such that |z| ≤ R and let ω ∈ Λ\ΛR, i.e. |ω| > 2R. Then |z| ≤ 2 |ω|, 5 |ω| hence |z − 2ω| < |z| + 2|ω| ≤ 2 |ω| and |z − ω| > 2 . It follows that z(z − 2ω) 5R|ω|/2 |(z − ω)−2 − ω−2| = | | ≤ (z − ω)2ω2 |ω|4/4 3 −3 2 2 − 3 = 10R/|ω| ≤ 10Rδ (n + m ) 2 .

The result follows from the convergence of the series

X 2 2 − 3 X X 2 2 − 3 X 8 (n + m ) 2 = (n + m ) 2 ≤ ≤ ∞. k2 (n,m)6=(0,0) k≥1 max(|n|,|m|)=k k≥1

The second estimate comes from the fact that there are exactly 8k pairs (n, m) such 2 2 2 1 1 that max{|n|, |m|} = k. For each of them we have n + m ≥ k so n2+m2 ≤ k2 and 1 3 1 2 ( n2+m2 ) ≤ k3 . Lemma 12.6. There exists δ > 0 such that

p 2 2 |xω1 + yω2| ≥ δ x + y for all real numbers x, y.

Proof. Consider f : [0, 2π] → R, f(θ) = |(cos θ)ω1 + (sin θ)ω2|. Interval [0, 2π] is compact, hence f is bounded and attains its bounds. Moreover, f(θ) 6= 0 as ω1 and ω2 are linearly independent over R. It follows that f(θ) > δ, for some δ > 0. Since |λxω +λyω | = λ|xω +yω |, it follows that |xω +yω | = |(x, y)||x0ω +y0ω | ≥ p 1 2 1 2 1 2 1 2 δ x2 + y2 where (x0, y0) is the projection of point (x, y) 6= (0, 0) to the unit circle in R2 p and |(x, y)| is the distance x2 + y2 of the point (x, y) ∈ R2 to zero. Deﬁnition 12.7. ℘(z) is called the Weierstrass elliptic ℘-function associated to the lattice Λ.

Lemma 12.8. We have ℘(−z) = ℘(z) = ℘(z + ξ), for all z ∈ C and ξ ∈ Λ.

74 Proof. We have X ℘(−z) = z2 + ((z + ω)−2 − ω−2) Λ\0 and we can rearrange, by changing ω to −ω to get ℘(z) = ℘(−z). First we note that X ℘0(z + ξ) = −2 (z + ξ − ω)3. ω∈Λ 0 0 Since the series converges absolutely and since Λ = −Λξ, we have ℘ (z + ξ) = ℘ (z), i.e. 1 ℘(z + ξ) = ℘(z) + c(ξ). Putting z = − 2 ξ, we get 1 1 c(ξ) = ℘( ξ) − ℘(− ξ) = 0. 2 2

Deﬁnition 12.9. Functions f on C with the property that

f(z + ξ) = f(z), ∀ z ∈ C, ξ ∈ Λ or equivalently f(z + ω1) = f(z) = f(z + ω2), ∀z ∈ C are doubly periodic with period lattice Λ.

We use

Theorem 12.10 (Liouville’s theorem). Any bounded holomorphic function on C is constant.

to conclude that

Lemma 12.11. A doubly periodic holomorphic function f is constant.

Proof. Since f is holomorphic it is continuous, hence it is bounded on any compact subset of C such as P = {sω1 + tω2 | s, t ∈ [0, 1]}. Given any z ∈ C we can ﬁnd ξ ∈ Λ such that z + ξ ∈ P . It shows that f(z) = f(z + ξ) is bounded.

Lemma 12.12. 0 2 3 ℘ (z) = 4℘(z) − g2℘(z) − g3, where

X −4 X −6 g2 = g2(Λ) = 60 ω , g3 = g3(Λ) = 140 ω . ω∈Λ\{0} ω∈Λ\{0}

75 Proof. Function X ℘(z) − z−2 = ((z − ω)−2 − ω−2) ω∈Λ\{0} vanishes at 0 and restricts to a holomorphic function in an open neighbourhood of zero. Moreover it is even (℘(z) − z−2 = ℘(−z) − (−z)−2) so its odd derivatives vanish at zero. Then ℘(z) − z−2 = λz2 + µz4 + z6h(z), where h(z) is holomorphic in a neighbourhood of zero. Then

℘0(z) = −2z−3 + 2λz + 4µz3 + 6z5h(z) + z6h0(z).

The “meromorphic part” of ℘(z)3 is

z−6 + 3λz−2 + 3µ while the “meromorphic” part of ℘0(z)2 is

4z−6 − 8λz−2 − 16µ.

Then ℘0(z)2 − 4℘(z)3 = −20λz−2 − 28µ + zeh(z). It follows that k(z) = ℘0(z)2 − 4℘(z)3 + 20λ℘(z) + 28µ is a holomorphic function in an open neighbourhood of zero which vanishes at zero. Since ℘(z) and ℘0(z) are holomorphic on C \ Λ and doubly periodic, k(z) is holomorphic and doubly periodic. By Lemma 12.11 it is constant and equal to 0 = k(0). By deﬁnition, we have 2λ = ℘(2)(0) and 24µ = ℘(4)(0).

Proposition 12.13. The Weierstrass ℘-function

℘: C \ Λ → C is surjective. Also ℘(z) = ℘(w) if and only if w ∈ Λ ± z.

Proof. Consider c ∈ C and f(z) = ℘(z) − c. Let

P (a) = {a + sω1 + tω2 | s, t ∈ [0, 1]} be a parallelogram such that the boundary of P (a) does not pass through any zeroes or poles of f. If we denote by γ the boundary of P (a) then Cauchy’s residue theorem implies that 1 Z f 0(z) dz = Z − P 2πi γ f(z)

76 where Z is the number of zeroes and P the number of poles of f inside P (a). Since f is periodic, integral along opposite sides of the parallelogram cancel each other, hence

Z − P = 0.

We know that ℘z has exactly one pole of multiplicity two inside P (a), hence so does f. It follows that Z = 2, i.e. there exists ω0 such that f(ω0) = 0, i.e. ℘(ω0) = c. Since ℘(z) is even and doubly periodic we have

℘(z) = ℘(ω0) = c ∀ z ∈ Λ ± ω0.

There exists ω1 ∈ Λ − ω0 which lies in P (a). Then ω0 and ω1 are the two zeroes of f inside P (a). It remains to check that if ω1 = ω0 then f(z) has a double zero at ω0, i.e. 0 0 f (ω0) = ℘ (ω0) = 0 (we want to know that there is no other zero of f). 0 Equality ω1 = ω0 implies that Λ + ω0 = Λ − ω0. As ℘ (z) is odd and doubly periodic it follows that 0 0 0 ℘ (ω0) = −℘ (−ω0) = −℘ (ω0).

77 03.03. Workshop III

1. Show that there is a homeomorphism given by

[s : t] 7→ [st3 : s4 : t4]

from P1 onto a quartic curve in P2. Why doesn’t it contradict the degree-genus formula?

2. The identity theorem of complex analysis tells that if f : U → C and g : U → C are holomorphic functions from a connected open subset U of C to C, and if f(z) = g(z) for all z in some non-empty open subset W of U then f(z) = g(z) for all z ∈ U. Use this to show that if f, g : R → S are holomorphic functions between Riemann surfaces with R connected and if f(z) = g(z) for all z in some non-empty open subset W of R then f(z) = g(z) for all z ∈ R.

3. The open mapping theorem of complex analysis says that if f : U → C is a non- constant holomorphic function from a connected open subset U of C to C then f(U) is open in C. Use this to show that if f : R → S is a non-constant holomorphic map between Riemann surfaces and R is connected then f(R) is an open subset of S.

4. Let C be a cubic curve deﬁned as

2 3 2 3 C = {[x : y : z] | y z = 4x − g2xz − g3z }.

2 2 Show that C is non-singular if and only if g2 − 27g3 6= 0.

78 Solutions to Workshop III

1. Let C = {[x : y : z] | x4 = yz3} be a quartic curve in P2. Map ϕ: P1 → C, ϕ([s : t]) = [st3 : s4 : t4] is well-deﬁned as (st3)4 = s4 · (t3)4. Map ϕ is continuous since ψ : C2 → C3, ψ(s, t) = (st3, s4, t4) is continuous. Indeed, it follows that Π ◦ ψ : C2 \{(0, 0} → P2 is continuous. As P1 has the quotient topology and diagram

ϕ 1 / 2 PO PO Π Π ψ C2 \{(0, 0)} / C3 \{(0, 0, 0)}

commutes, it follows that ϕ considered as a map P1 → P2 is continuous. Hence it is continuous as a map from P1 to any subset of P2 containing ϕ(P1), in particular as a map P1 → C. To show that ϕ is a bijection, we construct its inverse

[ x : 1] if z 6= 0, ψ : C → 1, ψ([x : y : z]) = z P [1 : 0] if z = 0.

First, note that if [x : y : z] ∈ C and z = 0 then x = 0 hence [x : y : z] = [0 : 1 : 0]. Then we have ϕ ◦ ψ([0 : 1 : 0]) = ϕ([1 : 0]) = [0 : 1 : 0]. If z 6= 0 then equality yz3 = x4 implies that

x x x4 x yz3 x y ϕ ◦ ψ([x : y : z]) = ϕ([ : 1]) = [ : : 1] = [ : : 1] = [ : : 1] = [x : y : z]. z z z4 z z4 z z On the other hand, we have

ψ ◦ ϕ([1 : 0]) = ψ([0 : 1 : 0]) = [1 : 0], ψ ◦ ϕ([s : 1]) = ψ([s : s4 : 1]) = [s : 1],

hence ψ ◦ ϕ = Id and ϕ ◦ ψ = Id. Instead of showing that ϕ−1 is continuous we prove that any continuous bijection from a compact space to a Hausdorﬀ space is always a homeomorphism. First we show that for a continuous f : X → Y with X compact and Y Hausdorﬀ the image f(Z) ⊂ Y of any closed subset of X is closed. Indeed, let y ∈ Y \ f(Z). By assumption, for any z ∈ Z there exist open subsets Uz, Vz in Y such that f(z) ∈ Uz, S y ∈ Vz and Uz ∩ Vz = ∅. We thus have an open cover f(Z) ⊂ z∈Z Uz. Set Z is a closed subset of a compact space, hence it is compact and so is its image under Sn continuous map. It follows that there exist z1, . . . , zn such hat f(Z) ⊂ i=1 Uzi . Tn Then V := i=1 Vzi is open such that y ∈ V and Y ∩ f(Z) = ∅, i.e. f(Z) is closed. Assume now that f : X → Y as above is a continuous bijection. To prove that f −1 is continuous we need to check that f is open, i.e. the image of an open set U ⊂ X

79 is open. It follows immediately from the above argument as f(U) = Y \ f(X \ U) and f(X \ U) ⊂ Y is closed. 4 3 ∂P 3 ∂P 3 ∂P 2 If P (x, y, z) = x − yz then partial derivatives ∂x = 4x , ∂y = −z , ∂z = −3yz vanish at [0 : 1 : 0], hence C is singular and the degree-genus formula does not hold.

2. Let W ⊂ R be the union of all open subsets in R on which f(z) = g(z). By deﬁnition W is open and by assumption it is non-empty. To show that W = R we need to show that W is closed. Let z ∈ R \W. If f(z) 6= g(z) then, because S is Hausdorﬀ, there exist open neighbourhoods Wf of f(z) and Wg of g(z) such that Wf ∩ Wg = ∅. For any 0 −1 −1 0 0 −1 −1 z ∈ f (Wf ) ∩ g (Wg) we have f(z ) 6= g(z ) hence f (Wf ) ∩ g (Wg) is an open neighbourhood of z in R \W.

If f(z) = g(z) but z∈ / W let ϕα : Uα → Vα be a chart containing z and ψβ : Wβ → Xβ −1 −1 a chart containing f(z). Let U = f (Wβ) ∩ g (Wβ) be an open neighbourhood of −1 −1 z. If U ∩ W 6= ∅ then holomorphic maps ψβ ◦ f ◦ ϕα |ϕα(U) and ψβ ◦ g ◦ ϕα |ϕα(U) agree on an open subset ϕα(U ∩ W) of ϕα(U), hence by the identity theorem, they agree on ϕα(U). Then, by deﬁnition, z ∈ W which contradicts our choice. Thus U ∩ W = ∅, i.e. U is an open neighbourhood of z in R \W. It proves that W is closed. As the only open and closed subset of a connected space is the space itself, we conclude that W = R, i.e. f(z) = g(z) for all z ∈ R.

3. Let s be a point in f(R). We need to show that there exists an open W ⊂ S such that s ∈ W and W ⊂ f(R). Let r ∈ R be such that f(r) = s and let ϕα : Uα → Vα, ψβ : Wβ → Xβ be charts on R, respectively on S, such that r ∈ Uα and s ∈ Wβ. It follows from the previous exercise that f is non-constant on Uα and any of its open subsets (if it was constant on an open subset of R it would −1 −1 be constant on the whole R). Thus, map ψβ ◦ f ◦ ϕα : ϕα(Uα ∩ f Wβ) → Wβ is a non-constant holomorphic function. It follows that there exists an open subset 0 −1 −1 −1 0 W ⊂ ψβ ◦ f ◦ ϕα (ϕα(Uα ∩ f (Wβ))) which contains ψβ(s). Then ψβ (W ) is an −1 −1 open subset of f ◦ ϕα (ϕα(Uα ∩ f (Wβ))), in particular of f(R) which contains s. As s was arbitrary, it proves that f(R) is open.

2 3 2 3 4. Let us calculate singular points of C. For P (x, y, z) = y z − 4x + g2xz + g3z we have ∂P ∂P ∂P = −12x2 + g z2, = 2yz, = y2 + 2g xz + 3g z2, ∂x 2 ∂y ∂z 2 3

and [a : b : c] is a singular point of C if and only if

 2 2  −12a + g2c = 0, 2bc = 0, 2 2  b + 2g2ac + 3g3c = 0.

80 It follows from the second equation that either b = 0 or c = 0. If c = 0 then the ﬁrst equation implies that a = 0 and the third gives b = 0, i.e. [a : b : c] is not a point of P2.

If b = 0 then the third equation reads c(2g2a + 3g3c) = 0. As we have already seen 2 c = 0 does not give a point in P , hence we have 2g2a + 3g3c = 0. If g2 = 0 then vanishing of g3 would imply that [0 : 0 : 1] is a singular point of C. In other words, 3 2 if g2 = 0 and C is non-singular then g2 − 27g3 6= 0. 3g3 If g2 6= 0 we have a = − c. Then the ﬁrst equation gives 2g2

2 9g3 2 2 1 2 3 2 −12 2 c + g2c = 2 (−27g3 + g2)c = 0. 4g2 g2

3 2 If g2 − 27g3 6= 0 the above equality implies that c = 0, i.e. a = b = c = 0 and the curve C is non-singular. If, on the other hand, g3 − 27g2 = 0 then [ −3g3 : 0 : 1] is a 2 3 2g2 singular point of C.

81 13 06.03. Lecture 13. Elliptic curves as complex tori

Recall that to a lattice Λ ⊂ C we assigned Weierstrass ℘-function X ℘(z) = z−2 + ((z − ω)−2 − ω−2) ω∈Λ\{0} and proved that 0 2 3 ℘ (z) = 4℘(z) − g2℘(z) − g3 for g2 = g2(Λ), g3 = g3(Λ) as in Lemma 12.12.

2 Deﬁnition 13.1. Let CΛ be the projective curve in P deﬁned by the polynomial

2 3 2 3 QΛ(x, y, z) = y z − 4x + g2xz + g3z where g2 = g2(Λ) and g3 = g3(Λ) are deﬁned as in Lemma 12.12.

Lemma 13.2. The cubic curve CΛ is non-singular. Proof. Let 1 1 1 α = ℘( ω ), β = ℘( ω ), γ = ℘( (ω + ω )). 2 1 2 2 2 1 2 It follows from Proposition 12.13 that α, β and γ are distinct. We will show that

2 QΛ(x, y, z) = y z − 4(x − αz)(x − βz)(x − γz). Smoothness will follow. 3 We need to check that α, β and γ are roots of the polynomial 4x − g2x − g3. By 0 1 0 1 0 1 Lemma 12.12 it suﬃces to check that ℘ ( 2 ω1), ℘ ( 2 ω2) and ℘ ( 2 (ω1 + ω2)) are zero. 0 We know that ℘ (z) is an odd doubly periodic function with periods ω1 and ω2. Then 1 1 1 ℘0( ω ) = −℘0(− ω ) = −℘0( ω ). 2 1 2 1 2 1 1 1 Similarly for 2 ω2 and 2 (ω1 + ω2). Λ is an additive subgroup of C hence we can form the quotient

C/Λ = {Λ + z | z ∈ C}.

There is a canonical surjective map π : C → C/Λ, π(z) = z+Λ, which induces the quotient topology on C/Λ. Moreover, map π is open, i.e. the image of any open set is open. Indeed, if U ⊂ C is −1 S open then π π(U) = ω∈Λ ω + U is a union of open sets, hence it is open. By deﬁnition, π(U) ⊂ C/Λ is open. The restriction of π to parallelogram

P = {sω1 + tω2 | s, t ∈ [0, 1]} is surjective, hence C/Λ is an image of a compact set. Thus it is itself compact. Topologically C/Λ is a torus obtained by gluing opposite edges of P .

82 Lemma 13.3. There is a well-deﬁned map u: C/Λ → CΛ, [℘(z): ℘0(z) : 1] if z∈ / Λ, u(Λ + z) = [0 : 1 : 0] otherwise.

Proof. It follows from Lemma 12.8 that ℘(z) = ℘(z + Λ) and ℘0(z) = ℘0(z + Λ), thus ℘ and ℘0 are well-deﬁned on C/Λ. Lemma 12.12 implies that point [℘(z): ℘0(z) : 1] lies on CΛ for any z ∈ Z.

Proposition 13.4. Mapping u: C/Λ → CΛ is a homeomomorphism. Proof. We check that u is bijection, and that u and u−1 are continuous. First we prove that u is injective. Suppose that u(z) = u(w). Then ℘(z) = ℘(w), hence by Proposition 12.13 z ∈ Λ ± w. If z ∈ Λ − w then ℘0(z) = −℘0(w). Because u(z) = u(w) we know that ℘0(z) = ℘0(w), i.e. ℘0(z) = 0 = ℘0(w). As in the proof of 1 1 1 Lemma 13.2 it follows that ℘(w) = ℘( 2 ω1) or ℘(w) = ℘( 2 ω2) or ℘(w) = ℘( 2 (ω1 + ω2)). 1 Proposition 12.13 again implies that w ∈ 2 Λ, hence Λ + w = Λ − w, i.e. z ∈ Λ + w. Next we check that u is surjective. Let [a : b : c] be a point in CΛ. We can assume that [a : b : c] 6= [0 : 1 : 0]. It follows from the equation that if c = 0 then a = 0, so we can assume that c = 1. By Proposition 12.13 there is z such that ℘(z) = a. Then by Lemma 12.12 and the deﬁnition of CΛ, we have

℘0(z)2 = b2 so ℘(z) = ±b. Since ℘ is even and ℘0 is odd it follows that either u(z) or u(−z) is equal to [a : b : c]. Since C/Λ has quotient topology, to prove that u is continuous it is enough to check 0 that C → CΛ is continuous. As ℘(z) and ℘ (z) are holomorphic on C \ Λ, u can possibly be discontinuous on Λ. We can write ℘(z) = g(z)/z2 and ℘0(z) = h(z)/z3 for functions g and h holomorphic in the neighbourhood of 0 and such that g(z) 6= 0, h(z) 6= 0. Then in a neighbourhood of 0 u(z) = [zg(z): h(z): z3] which tends to [0 : 1 : 0]. Thus u is continuous at 0, hence at every ω ∈ Λ. We have shown that u: C/Λ → CΛ is a continuous bijection. Since C/Λ is compact and CΛ is Hausdorﬀ it follows that u is a homeomorphism.

Next, we want to show that u is holomorphic. First, we need to check that C/Λ is a Riemann surface. C/Λ has the quotient topology induced by the map π : C → C/Λ. It follows from Lemma 12.6 that there exists δ > 0 such that

|nω1 + mω2| ≥ δ.

Hence, if a ∈ C, the restriction on π to the open disc 1 U = {z ∈ | |z − a| < δ} a C 4 is a homeomorphism onto π(Ua). Indeed, the distance between z and z + Λ is at least δ.

83 Now, we check that the atlas is holomorphic. If π(Ua) ∩ π(Ub) 6= ∅, there is unique λ = nω1 + mω2 such that 1 |a − b + λ| < δ. 2 Then −1 −1 −1 (π|Ub ) ◦ π|Ua∩π π(Ub) : Ua ∩ π π(Ub) → Ub is given by the translation by λ. Therefore the charts

−1 ϕa = (π|Ua ) : π(Ua) → Ua form a holomorphic atlas. C/Λ is Hausdorﬀ because there is a homeomorphism u: C/Λ → CΛ and CΛ is Hausdorﬀ. If S is a Riemann surface then, by the property of quotient topology, map f : C/Λ → S is holomorphic if and only if π ◦ f : C → S is holomorphic. Proposition 13.5. The homeomorphism [℘(z): ℘0(z) : 1] if z∈ / Λ, u(Λ + z) = [0 : 1 : 0] otherwise. is holomorphic and so is its inverse.

Proof. Let us ﬁx some holomorphic charts ϕα : Uα → Vα on C/Λ and ψβ : Wβ → Yβ on CΛ such that w + Λ ∈ Uα and u(w + Λ) ∈ Wβ for some w ∈ C. We need to check that −1 −1 ψβ ◦ u ◦ ϕα : ϕα(Uα ∩ u (Wβ)) → Yβ is holomorphic. As we have just seen we might take ϕα to be the inverse of the projection map π : C → −1 C/Λ restricted to some suﬃciently small disc Vα ⊂ C. Then ϕα = π|Vα . If w∈ / Λ then u(Λ + w) = [℘(z), ℘(z0), 1] and as in the proof of the existence of a holomorphic atlas on a complex curve (see Proposition 11.3), we might assume that ψβ is given by [x : y : z] 7→ x/z or [x : y : z] 7→ y/z.

0 −1 Since both ℘(z) and ℘ (z) are holomorphic near w, so is ψβ ◦ u ◦ ϕα . If w ∈ Λ then u(Λ + w) = [0 : 1 : 0] and as in the proof of Proposition 11.3 we can ∂QΛ assume that ψβ([x : y : z]) = x/y because ∂z (0, 1, 0) 6= 0, where as before QΛ(x, y, z) = 2 3 2 3 y z − 4x + g2xz + g3z . We thus need to check that ℘(z)/℘0(z) for z∈ / Λ, ψ ◦ u ◦ ϕ−1(z) = β α 0 for z ∈ Λ is holomorphic at 0, hence at any point in Λ. g 0 h We have ℘(z) = z2 and ℘ (z) = z3 , where g and h are holomorphic and not equal to 0 zg(z) zero at zero. Thus, ℘(z)/℘ (z) = h(z) is holomorphic at zero. By the inverse function theorem, if f : U → V is a holomorphic bijection between open subsets of C then so is f −1. It follows that u−1 is holomorphic.

84 14 09.03. Lecture 14. Singular curves and their resolution of singularities

We shall consider a singular projective curve C ⊂ P2 and ﬁnd a compact Riemann surface Ce with a surjective continuous map π : Ce → C such that π−1(Sing(C)) is a ﬁnite set and π is a homeomorphism outside Sing(C) and its preimage. We call such π : Ce → C a resolution of singularities for C.

Definition 14.1. An ordered pair (f,g) of meromorphic functions defined on an open neighbourhood of 0 ∈ C is a pair if f is non-constant on any neighbourhood of 0 and the mapping t 7→ (f(t), g(t)) is one-to-one near 0. A parameter change is a holomorphic function ρ defined on an open neighbourhood 0 of 0 ∈ C such that ρ(0) = 0, ρ (0) 6= 0. Two pairs (f, g) and (f,e ge) are equivalent, (f, g) ∼ (f,e ge) if there is a parameter change ρ such that fe = f ◦ ρ, ge = g ◦ ρ. By the inverse function theorem it is an equivalence relation on the set of pairs. The equivalence class of a pair (f, g) is called a meromorphic element; we shall use notation hf, gi or hf(t), g(t)i for it.

We denote by M the set of meromorphic elements. We shall make M into a Riemann surface with infinitely many connected components which will provide resolutions for projective curves. We define open sets. Let (f, g) be a pair and r > 0 sufficiently small such that f and g are defined on the disc D(0, r) and t 7→ (f(t), g(t)) is one-to-one on D(0, r). Then (f(t + t0), g(t + t0)) is a pair for any t0 ∈ D(0, r) and we define

U(f, g, r) = {hf(t + t0), g(t + t0)i | t0 ∈ D(0, r)} ⊂ M.

Lemma 14.2. There is a topology on M such that any open set is a union of U(fi, gi, ri).

Proof. It suﬃces to show that any ﬁnite intersection of U(fi, gi, ri) is again open in the above sense, i.e. that for any

hf, gi ∈ U(f1, g1, r1) ∩ U(f2, g2, r2), there exists r > 0 such that

U(f, g, r) ⊂ U(f1, g1, r1) ∩ U(f2, g2, r2).

By deﬁnition

hf, gi = hf1(t1 + t), g1(t1 + t)i = hf2(t2 + t), g2(t2 + t)i, i.e. there exist ρ1, ρ2 such that

f1(t1 + ρ1(t)) = f(t) = f2(t2 + ρ2(t)), g1(t1 + ρ1(t)) = g(t) = g2(t2 + ρ2(t))

85 for all t is some open disc D(0, s). There exists a smaller disc D(0, r) ⊂ D(0, s) such that 0 0 ρ1 and ρ2 are holomorphic on D(0, r), ρ1, ρ2 do not vanish and

ρ1(D(0, r)) ⊂ D(0, r1 − |t1|), ρ2(D(0, r)) ⊂ D(0, r2 − |t2|).

For t0 ∈ D(0, r) function σ(t) = ρ1(t0 + t) − ρ1(t0) is a parameter change (it vanishes at zero and its diﬀerential does not), so

hf(t0 + t), g(t0 + t)i = hf1(t1 + ρ1(t0 + t)), g1(t1 + ρ1(t0 + t))i =

hf1(t1 + ρ1(t0) + σ(t)), g1(t1 + ρ1(t0) + σ(t))i = hf1(t1 + ρ1(t0) + t), g1(t1 + ρ1(t0) + t)i

(the last equality follows from the equivalence relation). Since |t1 + ρ1(t0)| < r1, we have hf(t0 + t), g(t0 + t)i ∈ U(f1, g1, r1). As t0 was arbitrary,

U(f, g, r) ⊂ U(f1, g1, r1).

Similarly one can show the other inclusion U(f, g, r) ⊂ U(f1, g1, r1). We need to deﬁne an atlas.

Deﬁnition 14.3. Let A be the set of all ordered triples (f, g, r) where (f, g) is a pair and r > 0 is such that f and g are deﬁned on D(0, r) and the mapping t 7→ (f(t), g(t)) is one-to-one on D(0, r).

Lemma 14.4. If (f, g, r) ∈ A then the map D(0, r) → U(f, g, r), t0 7→ hf(t0 +t), g(t0 +t)i is a homeomorphism.

Proof. The map is surjective by deﬁnition of U(f, g, r). If hf(t0 + t), g(t0 + t)i = hf(t1 + t), g(t1 + t)i then (f(t0), g(t0)) = (f(t1), g(t1)) and t0 = t1. It shows that the map is injective. From the deﬁnition of the topology on M it follows that the map is a homeomorphism.

Deﬁnition 14.5. If α = (f, g, r) ∈ A let Uα = U(f, g, r), Vα = D(0, r) and ϕα : Uα → Vα be the inverse of the homeomorphism of Lemma 14.4.

Proposition 14.6. M is a Riemann surface with the holomorphic atlas Φ = {ϕα : Uα → Vα | α ∈ A}. Proof. Direct calculation that the composition is holomorphic. The proof that M is Hausdorﬀ will be given later.

Functions ψ, χ: M → C ∪ {∞}, ψ(hf, gi) = f(0), χ(hf, gi) = g(0) are meromorphic because their compositions with inverses of ϕα are given by

−1 −1 ψ ◦ ϕα (t0) = f(t0), χ ◦ fα (t0) = g(t0).

86 Deﬁnition 14.7. If f is a meromorphic function deﬁned on a neighbourhood of a point a ∈ C∪{∞} then the Riemann surface of f is the connected component of M containing ha + t, f(a + t)i if a ∈ C, or ht−1, f(t−1)i, if a = ∞. This meromorphic element is called germ of f at a and denoted by [f]a.

P n Let (f, g) be a pair. If f is holomorphic at 0 then f(t) = n≥0 cnt and we have the multiplicity m of f(t) − c0 at 0, i.e. minimal m > 0 such that cm 6= 0. Then m f(t) = c0 + t h(t) and h(0) = cm 6= 0. We can ﬁnd m’th holomorphic root k(t) of h(t) and then ρ(t) = tk(t) m is such that f(t) = c0 + ρ(t) . Then

m −1 hf, gi = hc0 + t , g ◦ ρ (t)i.

If f has a pole at 0, then f(t) = t−mh(t) where h(t) is holomorphic near 0 and h(0) 6= 0. We can ﬁnd holomorphic m’th root of 1/h(t) and multiply it by t to get a parameter change ρ such that f(t) = ρ(t)−m. Then

hf, gi = ht−m, g ◦ ρ−1(t)i.

It shows that any element of M is of the form ha + tm, gi or ht−m, gi. If m = 1 it is a germ of a holomorphic function. If m > 1 it is called a branch point of the component of M to which it belongs.

Deﬁnition 14.8. Let P ∈ C[x, y, z] be a non-constant irreducible homogeneous polynomial of degree d not divisible by z. The Riemann surface SP of P is the open subset of M consisting of all those elements hf, gi ∈ M satisfying

P (f(t), g(t), 1) = 0 for all t in some neighbourhood of 0. If C = {[x : y : z] | P (x, y, z) = 0} ⊂ P2 we write Ce for SP and deﬁne π : Ce → C by π(hf, gi) = [f(0) : g(0) : 1]. if f and g are holomorphic near 0, and otherwise

π(hf, gi) = [fe(0) : ge(0) : 0]

n n where fe(t) = t f(t), ge(t) = t g(t) and n is the maximum of the multiplicites of the pole of f and g at 0.

Theorem 14.9. Ce is a compact Riemann surface, map π : Ce → C is continuous and surjective. If C is non-singular then π is a holomorphic bijection, and in general π−1(Sing(C)) is ﬁnite and

π : Ce \ π−1(Sing(C)) → C \ Sing(C) is a holomorphic bijection.

87 We sketch the proofs of some Lemmas needed in the proof.

Lemma 14.10. π is continuous and its restriction to Ce \ π−1(Sing(C)) is holomorphic.

Proof. π is continuous at any hf0, g0i with f0 and g0 holomorphic at 0 because it is the composition of the continuous projection Π: C3 \{0} → P2 with map Ce → C3 \{0}, hf, gi 7→ (ψ(hf, gi), χ(hf, gi), 1).

If f0 or g0 has a pole at 0 then, if the multiplicity of the pole of f0 is greater than or equal the multiplicity of the pole of g0, π is the composition of Π with χ(hf, gi) 1 hf, gi 7→ (1, , ), ψhf, gi ψhf, gi hence it is continuous, as both maps are holomorphic. To show that the map is holomorphic we consider any chart ϕα : Uα → Vα on M and −1 ψβ on C \ Sing(C). We know ϕα ; map ψβ is a quotient of homogeneous coordinates −1 which takes value in C. The composition ψβ ◦ π ◦ ϕα is then a quotient of f and g and 1, i.e. it is a meromorphic function which takes values in C (because ψβ takes values in C). Any such function is holomorphic which ﬁnishes the proof.

Lemma 14.11. The restriction of π to Ce \ π−1(Sing(C)) is a bijection.

∂P ∂P Proof. Let [a : b : c] be a non-singular point of C with c 6= 0. Then either ∂x or ∂y does ∂P not vanish at [a : b : c]. (If they both vanish, Euler’s relation implies that ∂z is zero and ∂P the point is singular). We can assume that ∂y 6= 0. The implicit function theorem implies existence of open neighbourhoods U and V of a and b in C and a holomorphic h: U → V such that P (x, y, 1) = 0 with x ∈ U and y ∈ V if and only if y = h(x). If π(hf, gi) = [a : b : 1] then g = h(f(t)) (it follows from the deﬁnition of π that f and g are holomorphic at 0). Since (f, g) is a pair and f is holomorphic at zero, ρ(t) = f(t)−a is a parameter change, hence

hf, gi = ha + ρ(t), h(a + ρ(t))i = ha + t, h(a + t)i so [a : b : c] has precisely one preimage. For c = 0 we consider polynomial P (x, 1, z) or P (1, y, z) for the implicit function theorem and proceed as above.

Lemma 14.12. π : Ce → C is surjective. Proof. We need to prove that the preimage of a singular point of C is non-empty. We can ∂P assume that [0 : 1 : 0] ∈/ C, so ∂y is not identically zero. Then Bezout’s theorem implies ∂P that there are ﬁnitely many point on C such that ∂y vanishes at them. Let [a : b : c] ∈ C be singular. First, assume c = 1. Then there exists ε > 0 such that ∂P if 0 < |a − x| < ε there is no y ∈ C such that [x : y : 1] ∈ C and ∂y (x, y, 1) = 0. Recall that we have a branched cover ϕ: C → P1, ϕ([x : y : z]) = ([x : z]). Let D+(a, ε) be an open disc with the line from a to a + ε removed, similarly D−(a, ε). As

88 D±(a, ε) is simply connected, ϕ restricts to a homeomorphism of connected component −1 1 of ϕ (D±(a, ε)) with D±(a, ε). We know that ϕ: C → P is a branch cover of degree d = deg C, so by inverting ϕ, we get 2d holomorphic homemorphisms on their images ± fj : D±(a, ε) → C. ± ± ± ± We have P (x, fj (x), 1) = 0 and fj (x) 6= fi (x) if i 6= j. Since C is compact, maps fj are bounded and as x → a the only possible value they can take is y s.t. [a : y : 1] ∈ C. ± Point y = b satisﬁes this property and lies in the closure of fj D±(0, ε). Shrinking the discs if necessary we can assume that b is the only such point, so f(x) → b as x → a. We have deﬁned 2d functions, the cover ϕ: C → P1 has d sheets. Renumbering, we + − can assume that fj agrees with fj on the lower half of D+(a, ε) ∩ D−(a, ε). It agrees − with fσ(j) on the upper half. We can further assume that there exists m such that i + 1 for i = 1, . . . , m − 1, σ(i) = 1 for i = m

± ± Then we can glue functions f1 , . . . , fm to a function + m 1 f (a + t ) if (2j − 2)π/m < arg(t) < 2jπ/m m j g : {t ∈ C | 0 < |t| < ε } → C, g(t) = − m fj (a + t ) if (2j − 1)π/m < arg(t) < (2j + 1)π/m. Since g is bounded and has b as a limit when t → 0, it can be extended to a holomorphic 1 m function g : D(0, ε m ) → C. We have P (a + t , g(t), 1) = 0 and the mapping t 7→ (a + m ± ± m t , g(t)) is injective since fj (x) 6= fi (x). Thus, ha + t , g(t)i ∈ M and we have π(ha + tm, g(t)i = [a : b : c]. 1 If c = 0 then a = 1 because [0 : 1 : 0] ∈/ C.There exists ε > 0 such that if |x| ≥ ε there ∂P is no y such that [x : y : 1] ∈ C and ∂y (x, y, 1) = 0. Then we take “cut” discs D±(∞, ε) and glue to ﬁnd an element ht−m, g(t)i ∈ M such that π(ht−m, g(t)i) = [1 : b : 0]. Remark 14.13. The permutation σ of the above proof can be decomposed into disjoint cycles σ = σ1 . . . σl, where σi is a cycle of length mi. Since σ was a permutation of d discs, m1 + ... + ml = d. Gluing as above, we can construct holomorphic functions gj such that

−1 mj ψ (a) = {ha + t , gj(t)i | 1 j l} π ϕ 1 mj for the map ψ : Ce −→ C −→ P . The multiplicity of 0 of ψ − a at ha + t , gj(t)i is mj, hence ψ takes value a ∈ P1 precisely d times, counting with multiplicities. Note that we have found m1 . . . , ml such that m1 + ... + ml = d and constructed mi g1 . . . , gl such that P (a + t , gi(t), 1) = 0. In other words, if x is close to a, for any mj’th 1 1 m m root (x − a) j of (x − a), y = gj((x − a) j ) is a solution to the equation P (x, y, 1) = 0. Finally, if |x − a| is very small positive, points [x : yj : 1] lie in diﬀerent sheets of the covering so d numbers 1 2πis/mj m yj = gj(e (x − a) j ) for 1 ≤ i ≤ l, 1 ≤ s ≤ mj are distinct. As deg P = d, it follows that

1 Y Y 2πis/mj m P (x, y, 1) = K (y − gj(e (x − a) j ))

1≤j≤l 1≤s≤mj for some K ∈ C∗.

89 We omit proofs that Ce is compact and C and Ce are connected. We sketch the proof of

Lemma 14.14. M is Hausdorﬀ.

Proof. Let hf, gi and hf,e gei be points in M which do not have disjoint open neighbourhoods. We want to show that they are equal. As we have seen, we can assume that f(t) = a + tm or f(t) = t−m and fe(t) = b + tn or fe(t) = b−n. For k large enough U(f, g, 1/k) and U(f,e g,e 1/k) are open neighbourhoods of hf, gi and hf,e gei hence by assumption they have a common element, i.e.

hf(sk + t), g(sk + t)i = hfe(tk + t), ge(tk + t)i, for sk, tk ∈ D(0, 1/k). In particular, f(sk) = fe(tk) and g(sk) = ge(tk). When k goes to m inﬁnity, sk and tk tend to zero, so f(0) = fe(0). It follows that either f(t) = a + t and n −m −n n m fe(t) = a + t or f(t) = t and fe(t) = t . Moreover, sk = tk , so either both are zero or both are non-zero. If for some k, sk = 0 = tk then the two meromorphic elements are equal and the proof ﬁnishes. So we assume that sk 6= 0 and tk 6= 0 for all k. Let σk be any any complex n’th root of sk. Then

m n m (σk) (sk) = n = 1 tk (tk)

m so (σk) is a complex n’th root of 1 for each k. Since there are only n complex n’th roots tk m of unity, there exists ω such that (σk) = ω for inﬁnitely many k. Then tk

n −1 m g(t ) = ge(ω t ) for t = tk and these values of k. The uniqueness theorem of complex analysis implies that n −1 m n −1 m g(t ) = ge(ω t ) in some neighbourhood of 0. We also have f(t ) = fe(ω t ) as both are equal to a + tmn or t−mn. Mappings t 7→ (f(t), g(t)) and t 7→ (fe(t), ge(t)) are one-to-one in a neighbourhood of n n −1 m −1 m zero, hence so are t 7→ (f(t ), g(t )) and t 7→ (fe(ω t ), ge(ω t )). We have shown that n −1 n −1 these mappings are the same so n = m and g(t ) = ge(ω t ), so g(t) = ge(ω t) for small t. Since also f(t) = fe(ω−1t), we have

(f(t), g(t)) = (fe(ρ(t)), ge(ρ(t))), for ρ(t) = ω−1t.

90 09.03. Homework IV

3 2 1. Let Λ be a lattice in C. Show that g2(Λ) − 27g3(Λ) 6= 0. 2. Find the ﬁrst three terms of the Puiseux expansion about [0 : 0 : 1] of the curve deﬁned by P (x, y, z) = x3 + y3 + 3xyz.

3. Calculate genus of the curve deﬁned by P (x, y, z) = x3 + y3 + 3xyz.

91 Solutions to Homework IV

2 3 2 1. We know that the cubic curve CΛ defined by polynomial y z − 4x + g2(Λ)xz + 3 2 g3(Λ)z is non-singular. In particular, its affine part D ⊂ P defined by polynomial 2 3 P (x, y) = y − 4x + g2(Λ)x + g3(Λ) is non-singular. ∂P Assume that (a, b) ∈ D is a singular point. Then ∂y (a, b) = 2b = 0, i.e. b = 3 0. Thus, a is a root of the polynomial P (x, 0) = −4x + g2(Λ)x + g3(Λ) and of ∂P 2 ∂ ∂x (0, x) = −12x + g2(Λ) = ∂x P (x, 0). Such a ∈ C exists if and only if P (x, 0) has multiple roots. In other words, if D is non-singular, then P (x, 0) has distinct roots λ1, λ2, λ3, i.e. P (x, 0) = −4(x − λ1)(x − λ2)(x − λ3). We have

2 − 4(x − λ1)(x − λ2)(x − λ3) = −4(x − (λ1 + λ2)x + λ1λ2)(x − λ3) 3 2 = −4x + 4(λ1 + λ2 + λ3)x − 4(λ1λ2 + λ1λ3 + λ2λ3)x + 4λ1λ2λ3.

Hence

λ1 + λ2 + λ3 = 0, 2 2 − 4(λ1λ2 + λ1λ3 + λ2λ3) = −4(λ1λ2 − (λ1 + λ2) ) = 4((λ1 + λ2) − λ1λ2) = g2(Λ),

4λ1λ2λ3 = −4λ1λ2(λ1 + λ2) = g3(Λ).

Polynomial P (x, 0) has distinct roots if and only if

2 2 2 2 2 2 0 6= (λ1 − λ2) (λ2 − λ3) (λ1 − λ3) = (λ1 − λ2) (λ1 + 2λ2) (2λ1 + λ2) 2 2 4 4 3 2 2 3 = (λ1 − 2λ1λ2 + λ2) (4λ1 + 4λ2 + 20λ1λ2 + 33λ1λ2 + 20λ1λ2) 6 6 5 5 4 2 2 4 3 3 = 4λ1 + 4λ2 + 12λ1λ2 + 12λ1λ2 − 3λ1λ2 − 3λ1λ2 − 26λ1λ2.

On the other hand, we have 1 (g (Λ)3 − 27g (Λ)2) = 4((λ + λ )2 − λ λ )3 − 27λ2λ2(λ + λ )2 16 2 3 1 2 1 2 1 2 1 2 6 4 2 2 2 3 3 = 4(λ1 + λ2) − 12λ1λ2(λ1 + λ2) − 15λ1λ2(λ1 + λ2) − 4λ1λ2 6 6 5 5 4 2 2 4 3 3 = 4λ1 + 4λ2 + 12λ1λ2 + 12λ1λ2 − 3λ1λ2 − 3λ1λ2 − 26λ1λ2.

3 2 Thus if the curve D is non-singular, then g2(Λ) − 27g3(Λ) 6= 0. 2. We have P (x, y, 1) = x3 + y3 + 3xy. The Newton polygon is a line segment joining 1 3 (0, 3) and (1, 1). Its equation is α + 2 β = 2 . Then

X α β 3 P (x, y, 1) = cαβx y + x . 1 3 α+ 2 β= 2

Function √ √ X β 3 f0(t) = cαβt = t + 3t = t(t + i 3)(t − i 3) 1 3 α+ 2 β= 2 √ has roots 0 and ±i 3.

92 √ √ 2 Let us consider the case ±i 3 ﬁrst. The ﬁrst substitution is x = x1, y = x1(±i 3+ y1) and √ √ 6 3 3 3 P (x1, y1, 1) = x1 + x1(±i 3 + y1) + 3x1(±i 3 + y1). Thus √ √ √ √ 3 3 2 3 3 2 P1(x1, y1) = x1 +y1 ±3i 3y1 −9y1 −3(±i 3)±3i 3+3y1 = x1 +y1 −6y1 ±3i 3y1.

Polynomial P1(x1, y1) has as the Newton polygon the line segment joining (0, 1) and (3, 0) with equation α + 3β = 3. Then √ X α β 3 2 P1(x1, y1) = dαβx1 y1 + y1 ± 3i 3y1 α+3β=3 and X β f1(t) = dαβt = 1 − 6t α+3β=3 1 has root 6 . We get substitutions 1 x = x , y = x3( + y ) 1 2 1 2 6 2 hence 1 1 √ 1 P (x , y ) = x3 + x9( + y )3 − 6x3( + y ) ± 3i 3x6( + y )2 1 2 2 2 2 6 2 2 6 2 2 6 2 and 1 1 1 √ 1 1 P (x , y ) = 1 + x6(y3 + y2 + y + ) − 1 − 6y ± 3i 3x3(y2 + y + ) 2 2 2 2 2 2 2 12 2 216 2 2 2 3 2 36 X α β = eαβx2 y2 . α+3β≥3

Polynomial √ X i 3 e tβ = −6t ± αβ 12 α+3β=3 √ i 3 has two roots (± 72 ). Then √ i 3 x = x , y = x3((± ) + y ). 2 3 2 3 72 3 Then, the expansion is √ √ √ √ 1 3 1 3 i 3 1 1 2 i 3 7 y = x 2 (±i 3 + x 2 ( + x 2 (± + ...))) = ±i 3x 2 + x ± x 2 + ... 6 72 6 72

93 2 If, on the other hand we take the root 0 of f0(t) then the ﬁrst substitution is x = x1, y = x1y1 and

6 3 3 3 P (x1, y1, 1) = x1 + x1y1 + 3x1y1 3 3 X α β P1(x1, y1) = x1 + y1 + 3y1 = gαβx y . α+3β≥3

Function X β f1(t) = gαβt = 3t + 1 α+3β=3 1 has one root t = − 3 . We get 1 x = x , y = x3(− + y ), 1 2 1 2 3 2

1 1 P (x , y ) = x3 + x9(y − )3 + 3x3(y − ) 1 2 2 2 2 2 3 2 2 3 1 1 P (x , y ) = 1 + x6(y3 − y2 + y − ) + 3y − 1 2 2 2 2 2 2 3 2 27 2 1 1 = x6y3 − x6y2 + x6y − x6 + 3y 2 2 2 2 3 2 2 27 2 2 X α β = hαβx y . α+6β≥6

Function X 1 f (t) = h tβ = 3t − 2 αβ 27 α+6β=6 1 6 1 has root 81 . Then x2 = x3, y2 = x3( 81 + y23) The Puiseux expansion is:

1 3 1 6 1 1 1 2 1 5 y = x 2 (x 2 (− + x 2 ( + ...))) = 0 · x 2 − x + x + ... 3 81 3 81

3. A point [a : b : c] is a singular point of a curve {x3 + y3 + 3xyz = 0} if

 2  3a + 3bc = 0, 3b2 + 3ac = 0,  ab = 0.

It follows from the last equation that either a = 0 or b = 0. If a = 0 then the second equation implies that b = 0. If b = 0 then the ﬁrst equation implies that a = 0. Hence p = [0 : 0 : 1] is the only singular point of C. It follows from Noether formula that (3 − 1)(3 − 2) g(C) = − δ(p). 2

94 Point [0 : 1 : 0] does not lie on C. To check whether it lies on tangent lines to inﬂection points of C we calculate the matrix of second derivatives

 6x 3z 3y   3z 6y 3x  3y 3x 0

Its determinant is 54(xyz − y3 − x3). It follows that inflection points of C satisfy 3 3 x + y = 0 and xyz = 0. There are three inflection points [1 : −1 : 0], [ε3 : −1 : 0] 2 ∂P 2 and [ε3 : −1 : 0]. Since ∂y (x, y, z) = 3y +3xz does not vanish at the above inflection points, [0 : 1 : 0] does not lie on any line tangent to the inflection point of C. It follows that 1 ∂P δ(p) = (I (P, ) − ν (p) + |π−1(p)|). 2 p ∂y ϕ We know that |π−1(p)| is the number of essentially distinct Puiseux expansions. From the previous exercise we learn that there are 3 Puiseux expansions: √ √ 1 1 2 i 3 7 y1(x) = i 3x 2 + x + x 2 + ..., 6 72√ √ 1 1 2 i 3 7 1 1 2 1 5 y (x) = −i 3x 2 + x − x 2 + . . . , y (x) = 0 · x 2 − x + x + ... 2 6 72 3 3 81 If we consider √ √ 1 i 3 g (t) = i 3t + t4 + t7 + ... 1 6 72 1 1 then y1(x) = g1(x 2 ) and y2(x) = g1(−x 2 ). Thus, these two expansions are not essentially distinct. It follows that |π−1(p)| = 2. 3 We have P (0, y, 1) = y so νϕ(p) = 3. ∂P 2 Finally, ∂y = 3y + 3xz and

3 3 2 4 3 3 2 3 2 2 Ip(x + y + 3xyz, y + xz) = Ip(z (x + y + 3xyz) − x z (y + xz), y + xz) 3 4 5 2 2 3 2 3 3 2 2 2 2 = Ip(y z + 3xyz − x y z , y + xz) = Ip(z (y z + 3xyz − x y ), y + xz) 2 2 2 2 2 2 2 2 2 = Ip(y(y z + 3xz − x y), y + xz) = Ip(y, y + xz) + Ip(y z + 3xz − x y, y + xz) 2 3 4 2 2 2 3 4 2 = Ip(y, xz) + Ip(y z + 3xz − x yz − 3y z − 3xz , y + xz) 2 2 2 3 2 2 2 2 = 1 + Ip(x yz + 2y z , y + xz) = 1 + Ip(x y + 2y z, y + xz) 2 2 2 2 3 2 2 2 2 = 2 + Ip(x + 2yz, y + xz) = 2 + Ip(x z + 2yz − xy z − x z , y + xz) 3 2 2 2 2 = 2 + Ip(2yz − xy z, y + xz) = 3 + Ip(2z − xy, y + xz) = 3.

It follows that 1 δ(p) = (3 − 3 + 2) = 1, g(C) = 1 − 1 = 0. 2

95 15 13.03. Lecture 15. Newton Polygons and Puiseux expansions

Today we investigate what a singular curve of degree d looks like in a neighbourhood of a singular point. We assume that C is given by P ∈ C[x, y, z] and that [0 : 0 : 1] ∈ C is singular. Newton’s idea was to think about equation P (x, y, 1) = 0 as an implicit equation for ∂P y as a function of x near 0. If ∂y (0, 0, 1) 6= 0 the implicit function theorem tells us that y is locally a holomorphic function in x, hence can be expanded as a power series in x. When the derivative vanishes we need to consider fractional powers of x. Let us suppose ﬁrst that P (x, y, 1) is a quasi-homogeneous polynomial:

X α β P (x, y, 1) = cαβx y . α+µβ=ν

Clearly, P (x, y, 1) is quasi-homogeneous if all (α, β) such that cαβ 6= 0 lie on one line in R2. Then putting y = txµ, we get

µ X α β µβ ν X β ν P (x, tx , 1) = cαβx t x = x ( cαβt ) = x f(t). α+µβ=ν α+µβ=ν

µ If t0 is such that f(t0) = 0 then y = t0x is the solution to the equation P (x, y, 1) = 0. In general, let X α β P (x, y, 1) = cαβx y α,β and deﬁne the carrier of P as

∆(P ) = {(α, β) ∈ Z | cαβ 6= 0}. Let x + µy = ν be a line which contains at least two points of ∆(P ) and such that α + µβ ≥ ν for all (α, β) ∈ ∆(P ). If y = txµ then

µ ν X β α+µβ P (x, tx , 1) = x f(t) + cαβt x . α+µβ>ν

µ For a root t0 of f(t) we can think of y = t0x as an approximate solution to P (x, y, 1) = 0. To show that this way we can construct a solution we ﬁrst deﬁne the Newton polygon of P :

Deﬁnition 15.1. If p, q ∈ R2 let [p, q] = {tp + (1 − t)q | t ∈ [0, 1]} be the line segment from p to q. Consider the convex subset of R2 consisting of those (x, y) ∈ R2 such that x ≥ a and y ≥ b for some (a, b) ∈ [δ1, δ2], δ1, δ2 ∈ ∆(P ). Its boundary consists of a vertical half-line and a horizontal half-line joined by a union of ﬁnitely many straight line segments. This union is the Newton polygon of P .

96 Because P (0, 0, 1) = 0, (0, 0) ∈/ ∆(P ). One can always choose coordinates such that P (x, y, z) is not divisible by x. If it is true then there is some (0, β) ∈ ∆(P ). β0 If ∆(P ) is a single point, it is (0, β0) and all terms of P (x, y, 1) are divisible by y , so P (x, y, 1) = yβ0 Q(x, y). and Q(0, 0) 6= 0. In this case the only solutions to P (x, y, 1) = 0 in the neighbourhood of 0 are given by y = 0. If ∆(P ) has more than one point, the steepest segment of the Newton polygon is the starting line of the procedure described above. Let (0, β0) be the upper endpoint of this 1 segment and let − be its slope. Then µ0 is a positive rational, µ0 p0 µ0 = . q0 We can write X α β P (x, y, 1) = cαβx y

α+µ0β≥ν0 where ν0 = µ0β0. Since there is at least one point (α, β) ∈ ∆(P ) other than (0, β0) which lies on this segment, polynomial

X β f0(t) = cαβt

α+µ0β=ν0 has a non-zero root, say t0. Then µ y0 = t0x0 gives us the ﬁrst approximate solution. Next we make the substitution

q0 µ p0 x = (x1) , y = x (t0 + y1) = x1 (t0 + y1) to get

q0 p0 X q0(α+µ0β) β q0ν0 P (x1 , x1 (t0 + y1), 1) = cαβx1 (t0 + y1) = x1 P1(x1, y1). α+µ0β≥ν0 Then X q0α+p0β−q0ν0 β P1(x1, y1) = cαβx1 (t0 + y1) q0α+p0β≥q0ν0 is a polynomial not divisible by x1. Remark 15.2. From the construction of Newton’s polygon we learn that yβ0 is the β1 smallest power of y appearing in P (x, y, 1). Let y1 be the smallest power of y1 appearing in P1(x1, y1). Then X β P1(0, y1) = (t0 + y1)

q0α+ρ0β=q0ν0 so β1 is the smallest integer such that cα,β 6= 0 and α + µ0β = ν0, for some α ≥ 0. Either β0 β1 < β0 or P1(0, y1) is a constant multiple of y1 . If this is the case then the degree β0 − 1 β0 β0−1 term of c0β0 (t0 + y1) must cancel with term of cαβ0−1(t0 + y1) . In particular, β0 − 1 must lie on the line α + µ0β = ν0. It follows that its slope µ0 needs to be integer, so q0 = 1.

97 We now repeat the whole process, replacing P (x, y, 1) by P (x1, y1) and continue p0 p1 indeﬁnitely. We obtain a sequence of positive rationals µ0 = , µ1 = ... and q0 q1 complex numbers t0, t1,... and successive “approximate solutions” (x, y), (x1, y1),... to the equation P (x, y, 1) = 0. We will have

1 1 q1 q2 µ0 µ1 x = x1 , x1 = x2 , y = x (t0 + y1), y1 = x1 (t1 + y2),.... We would like to conclude that

µ0 µ1 µ0 µ2 µ1 µ0 µ0 µ0+µ1/q0 µ0+µ1/q1+µ2/q0q1 y = t0x + t1x1 x + t2x2 x1 x + ... = t0x + t1x + t2x + ... is a genuine solution near (0, 0). This series is called a Puiseux expansion for the curve C = {[x : y : z] | P (x, y, z) = 0} near [0 : 0 : 1]. By the above remark, qi = 1 unless βi−1 > βi. Since β0 ≥ β1 ≥ β2 ... is a sequence of positive integers, qi = 1 for all but ﬁnitely many i. The product n of all qi is well-deﬁned 1 and the Puiseux expansion may be expressed as a formal power series in x n .

P r/n Theorem 15.3. Any Puiseux expansion y = r≥1 arx for the curve C near [0 : 0 : 1] is a power series in x1/n which converges for x suﬃciently close to 0 and satisﬁes

X r/n P (x, arx , 1) = 0. r≥1

Proof. From the last lecture, we know that there are m1 + ... + ml = d and holomorphic g1 . . . , gl such that

1 Y Y 2πis/mj m P (x, y, 1) = K (y − gj(e (x − a) j )).

1≤j≤l 1≤s≤mj

P (j) r We can expand each gj(t) = r≥0 ar t as a convergent power series near 0. If N is the lowest common multiple of m1, . . . , ml and n then series

2πis/mj 1/mj X (j) 2πirs/mj r/mj gj(e x ) = ar e x r≥0 and the Puiseux expansion can be regarded as elements of the ring C{x1/N } of formal power series. It is an integral domain, so if Q ∈ C{x1/N }[y] satisﬁes Q(c) = 0 for some 1/N c ∈ C{x } and Q(y) = k(y −c1) ... (y −cd) then c = cj for some j. Therefore, it suﬃces to check that as a formal power series

X r/n P (x, arx , 1) = 0 r≥1

P (j) 2πirs/mj r/mj as then the Puiseux expansion must coincide with one of the r≥0 ar e x , hence converge. 1/N PM r/n The exponent of the smallest power of x in P (x, r=1 arx , 1) is at least p0β0 + ... + pM βM . It tends to zero as M tends to inﬁnity, since each pj and βj are positive P r/n integers. This tell us that every coeﬃcient in the formal power series P (x, r≥1 arx , 1) is 0, so the power series is 0.

98 Recall that we have π : Ce → C and that

−1 mj π ([0 : 0 : 1]) = {ht , gj(t)i | 1 ≤ j ≤ l}.

Any Puiseux expansion of C is of the form

2πis/mj 1/mj y = gj(e x ) for j ∈ [1, l], s ∈ [1, mj]. We regard two Puiseux expansions essentially different if they −1 come from different functions gj. Then the number of points of π ([0 : 0 : 1]) is the number of essentially different Puiseux expansions of C near [0 : 0 : 1].

99 16 16.03. Lecture 16. The degree–genus formula for singular curves

We know that any irreducible projective curve C has a resolution of singularity π : Ce → C, that the map π is a homeomorphism when restricted to π−1(C \ Sing(C)) and that, for a singular p ∈ C, π−1(p) is the number of essentially diﬀerent Puiseux expansions of C in the neighbourhood of p. The curve Ce is smooth, hence its genus g is well-deﬁned. We shall call it the genus of C and relate to the degree of C. We shall assign to each singular p ∈ C a positive integer δ(p) and prove Noether’s formula 1 X g = (d − 1)(d − 2) − δ(p). 2 p∈Sing(C)

As before, assume that [0 : 1 : 0] ∈/ C and define ϕ: C → P1 via ϕ([x : y : z]) = [x : z]. π ϕ We regard ψ : Ce −→ C −→ P1 as a branched cover of P1 and define ∂P R = π−1{[a : b : c] ∈ C | (a, b, c) = 0} ∂y to be the set of ramification points of ψ and its image ψ(R) the branch locus of ψ.

Proposition 16.1. Given any triangulation (V,E,F ) of P1 such that ψ(R) is contained in the set of vertices V there is a triangulation (V,e E,e Fe) of Ce such that

Ve = π−1(V ), |Ee| = d|Ee|, |Fe| = d|F |. Proof. Since π−1(Sing(C)) ⊂ R and π| is a homeomorphism, the proof is a Ce\π−1(Sing(C)) straightforward modiﬁcation of the proof in the smooth case, see Proposition 10.6.

If p = [a : b : c] we deﬁne νϕ(p) to be the multiplicity of the polynomial P (a, y, c) at y = b.

P P −1 Lemma 16.2. |Ve| = d|V | − p∈π(R)(νϕ(p) − 1) + p∈Sing(C)(|π (p)| − 1). Recall that we proved Lemma 8.4 The inverse image ϕ−1([a : c]) of any [a : c] ∈ P1 under ϕ contains exactly X d − (νϕ(p) − 1) p∈ϕ−1([a:c]) points. In particular ϕ−1([a : c]) contains d points if and only of ϕ−1([a : c]) contains no ramiﬁcation points of ϕ.

Proof of Lemma 16.2. The inverse image of any q ∈ P1 under ϕ: C → P1 contains exactly P d − p∈ϕ−1(q)(νϕ(p) − 1). −1 If p∈ / π(R) then νϕ(p) = 1 and π(R) ⊂ ϕ (V ) so

−1 X |ϕ V | = d|V | − (νϕ(p) − 1). p∈π(R)

100 The set ϕ−1(V ) contains all singular points of C, hence

−1 −1 X X −1 |Ve| = |π ϕ V | = d|V | − (νϕ(p) − 1) + (|π (p)| − 1). p∈π(R) p∈Sing(C)

Recall that we proved that for a non-singular [a : b : c] ∈ C, we have νϕ[a : b : c] ≥ 2 if and only if P (x, y, z) = Py(x, y, z) = 0 and νϕ[a : b : c] > 2 if and only if P (a, b, c) = Py(a, b, c) = Pyy(a, b, c) = 0 if and only if [a : b : c] is the inﬂection point on C and the tangent line to C at [a : b : c] contains [0 : 1 : 0].

Lemma 16.3. Suppose that [0 : 1 : 0] does not lie on C or the tangent line to C at any of the ﬁnitely many p ∈ C \ Sing(C) which are inﬂection points. Then if p ∈ π(R) and p∈ / Sing(C) we have

∂P ν (p) = 2,I (P, ) = 1. ϕ p ∂y

Proof. Since [0 : 1 : 0] ∈/ C, the coefficient P (0, 1, 0) of yd in P (x, y, z) is non-zero, hence Py(x, y, z) is a polynomial of degree d − 1 which is not identically zero. Since P is irreducible, Py and P cannot have a common factor, hence C and D = {Py = 0} intersect in finitely many points. If p ∈ π(R) then by definition of R, p ∈ C ∩ D and νϕ(p) ≥ 2. Let us assume further that p ∈ C is a non-singular point. It follows from the description of the points with νϕ(q) > 2 that νϕ(p) ≤ 2, hence νϕ(p) = 2. To prove that the intersection index is equal to 1 we need to check that tangent lines are distinct (we know that p ∈ D is non-singular because Pyy(a, b, c) 6= 0). If they would coincide then

[Px(a, b, c): Py(a, b, c): Pz(a, b, c)] = [Pxy(a, b, c): Pyy(a, b, c),Pyz(a, b, c)] in particular Pyy(a, b, c) = 0 which contradicts the assumption on tangent lines. Corollary 16.4. If coordinates are chosen as above then

X ∂P −1 χ(Ce) = d(3 − d) + (Ip(P, ) − νϕ(p) + |π (p)|). ∂y p∈Sing(C)

Proof. By deﬁnition

X X −1 χ(Ce) = |Ve| − |Ee| + |Fe| = d(|V | − |E| + |F |) − (νϕ(p) − 1) + (|π (p)| − 1). p∈π(R) p∈Sing(C)

The sum |V | − |E| + |F | = χ(P1) = 2. By Lemma 16.3

X X ∂P (ν (p) − 1) = I (P, ). ϕ p ∂y p∈π(R)\Sing(C) p∈π(R)\Sing(C)

101 ∂P By construction, Sing(C) ⊂ π(R). Then, since the degree of C is d and D = { ∂y } is d − 1, Bezout’s theorem implies that

X ∂P X ∂P I (P, ) = d(d − 1) − I (P, ). p ∂y p ∂y p∈π(R)\Sing(C) p∈Sing(C)

Then we get

X ∂P X −1 χ(Ce) = 2d − Ip(P, ) + (|π (p)| − 1 − νϕ(p) + 1) ∂y p∈π(R)\Sing(C) p∈Sing(C) X ∂P = 2d − d(d − 1) + (I (P, ) + |π−1(p)| − ν (p)). p ∂y ϕ p∈Sing(C)

Deﬁnition 16.5. Let p be a singular point of the irreducible curve

2 C = {[x : y : z] ∈ P | P (x, y, z) = 0} and suppose that coordinates have been chosen so that [0 : 1 : 0] does not lie on C or on the tangent line to C at any of the inﬂection points. We deﬁne 1 ∂P δ(p) = (I (P, ) − ν (p) + |π−1(p)|). 2 p ∂y ϕ

Theorem 16.6. The genus g of an irreducible projective curve C of degree d in P2 is 1 X g = (d − 1)(d − 2) − δ(p). 2 p∈Sing(C)

Proof. Follows from Corollary 16.4 and χ(Ce) = 2 − 2g. It remains to check that δ(p) does not depend on the choice of coordinates. Extending the results about intersection multiplicity to holomorphic curves, one can check that

Lemma 16.7. We have ∂P X ∂P X I (P, ) = µ( (f, g, 1)), ν (p) = µ(f − f(0)), p ∂y ∂y ϕ hf,gi∈π−1(p) hf,gi∈π−1

∂P where µ( ∂y (f, g, 1)) is the multiplicity of zero or minus multiplicity of the pole of ∂P ∂y (f(t), g(t), 1) at t = 0. In the second formula, we omit f(0) if f has a pole at 0. Then we have

102 Lemma 16.8. Suppose that neither [0 : 1 : 0] or [α : β : γ] lie on C or the tangent line [α:β:γ] to any inﬂection point of C \ Sing(C). Suppose that p ∈ Sing(C) and let νϕ (p) be the smallest positive integer such that ∂ ∂ ∂ (α + β + γ )mP ∂x ∂y ∂z does not vanish at p. Then ∂ ∂ ∂ ∂P I (P, α + β + γ ) − ν[α:β:γ](p) = I (P, ) − ν (p). p ∂x ∂y ∂z ϕ p ∂y ϕ

Proof. Note that νϕ(p) was defined as the multiplicity of 0 of the polynomial P (a, y, c), dm i.e. minimal m such that dym P (a, y, c) 6= 0. It follows that if [α : β : γ] = [0 : 1 : 0] there is nothing to prove. Otherwise, we can find a projective transformation [x : y : z] 7→ [x0 : y0 : z0] which ∂P 0 0 0 ∂P 0 0 0 fixes [0 : 1 : 0], such that, up to scalar, ∂y (x , y , z ) = ∂y0 (x , y , z ) and which maps [α : β : γ] to [1 : 0 : 0]. Indeed, the first two conditions imply that the matrix is of the  a 0 b  form  c d e . Such a transformation does not change the intersection multiplicity, f 0 g hence without loss of generality we can assume that [α : β : γ] = [1 : 0 : 0]. For an element hf, gi ∈ Ce, we have P (f(t), g(t), 1) = 0 for all t near 0. Differentiating this equation gives ∂P ∂P f 0(t) (f(t), g(t), 1) + g0(t) (f(t), g(t), 1) = 0. ∂x ∂y We conclude that ∂P ∂P µ(f 0) + µ( (f, g, 1)) = µ(g0) + µ( (f, g, 1)). ∂x ∂y

Since µ(f 0) = µ(f − f(0)) − 1 and µ(g0) = µ(g − g(0)) − 1, summing the above equality over all hf, gi ∈ π−1(p), we get that

∂P ∂P ν[0:1:0](p) + I (P, ) = ν[1:0:0] + I (P, ). ϕ p ∂x ϕ p ∂y

103 17.03. Workshop IV

1. Calculate the ﬁrst two terms of the Puiseux expansion about [0 : 0 : 1] of the curve deﬁned by P (x, y, z) = y4z3 + 2x3y2z2 + 4x5yz + x6z + x7.

2. Calculate the genus of the curve deﬁned by P (x, y, z) = y2z − x3.

104 Solutions to workshop IV

1. The carrier of polynomial

P (x, y, 1) = y4 + 2x3y2 + 4x5y + x6 + x7

is ∆(P ) = {(0, 4), (3, 2), (5, 1), (6, 0), (7, 0)}. The Newton polygon is the line segment joining (0, 4) and (6, 0). Its equation is 3 α + 2 β = 6. Then

X α β P (x, y, 1) = cαβx y , 3 α+ 2 β≥6 X β 4 2 2 2 2 2 f0(t) = cαβt = t + 2t + 1 = (t + 1) = (t − i) (t + i) . 3 α+ 2 β=6

Polynomial f0(t) has roots ±i, hence the ﬁrst substitution is

2 3 x = x1, y = x1(±i + y1). We have

12 4 12 2 13 12 14 P (x1, y1, 1) = x1 (±i + y1) + 2x1 (±i + y1) + 4x1 (±i + y1) + x1 + x1 , 4 2 2 P1(x1, y1) = (±i + y1) + 2(±i + y1) + 4x1(±i + y1) + 1 + x1 4 3 2 2 2 = y1 ± 4iy1 − 6y1 − (±4iy1) + 1 + 2y1 ± 4iy1 − 2 ± 4ix1 + 4x1y1 + 1 + x1 4 3 2 2 = y1 ± 4iy1 − 4y1 ± 4ix1 + 4x1y1 + x1.

The carrier of polynomial P2 is

∆(P2) = {(0, 4), (0, 3), (0, 2), (1, 0), (1, 1), (2, 0)}.

1 Newton polygon is the line segment joining (0, 2) and (1, 0). Its equation is α+ 2 β = 1. Then

X α β P1(x1, y1) = dαβx y 1 α+ 2 β≥1 X 1√ 1√ f (t) = d tβ = −4t2 ± i = 4(t − ±i)(t + ±i). 1 αβ 2 2 1 α+ 2 β=1

1 √ Polynomial f1(t) has two roots ± 2 ±i. We have

1 1√ x = x 2 , y = x (± ±i + y ). 1 2 1 2 2 2 We get Puiseux expansion

3 1 1√ 3 1√ 7 y = x 2 (±i + x 4 (± ±i + ...)) = ±ix 2 ± ±ix 4 + ... 2 2

105 2. Let C = {y2z − x3 = 0} be a cubic curve. By Noether formula

1 X g(C) = (3 − 1)(3 − 2) − δ(p). 2 p∈Sing(C)

We need to ﬁnd singular points of C. If P (x, y, z) = y2z − x3 then

∂P ∂P ∂P = 3x2, = 2yz, = y2. ∂x ∂y ∂z

It follows that if [a : b : c] is a singular point then a = b = 0. Moreover [0 : 0 : 1] is a singular point of C. Let us ﬁnd inﬂection points of C. The matrix of second derivatives

 6x 0 0   0 2z 2y  0 2y 0

has determinant −24xy2. Hence, [a : b : c] is an inﬂection point if

a3 = b2c ab2 = 0 0 = b2c a = 0 a = 0 b = 0 b = 0 c = 0 b = 0 a = 0 a = 0 a = 0

The solutions are [0 : 0 : 1] and [0 : 1 : 0]. Since [0 : 0 : 1] is the singular point of C, [0 : 1 : 0] is the only inﬂection point. Tangent line to C at [0 : 1 : 0] is z = 0. Then [1 : 0 : 1] is a point which does not lie on C and on the tangent line to the inﬂection point of C. Then 1 ∂P δ(p) = (I (P, ) − ν[1:0:1](p) + |π−1(p)|), 2 p ∂(x + z) ϕ

[1:0:1] ∂ ∂ m where νϕ (p) is the smallest positive integer such that ( ∂x P + ∂z P ) does not vanish at p. First, let us calculate |π−1(p)|, i.e. number of essentially diﬀerent Puiseux expansions. We have

2 3 X α β P (x, y, 1) = y − x = cαβx y . 3 α+ 2 β=3

106 Polynomial X 2 f0(t) = = t − 1 = (t − 1)(t + 1) 3 α+ 2 β=3 has two roots ±1, hence

2 3 x = x1, y = x1(±1 + y1)

2 3 3 Note that P (x1, ±x1) = 0, hence P (x, y, 1) has ﬁnite Puiseux expansions y = ±x 2 . They are not essentially diﬀerent, hence

|π−1(p)| = 1.

Further ∂ ∂ ∂2 ∂2 ∂2 ( P + P )2 = P + 2 P + P = 6x ∂x ∂z ∂x2 ∂x∂z ∂z2 vanishes at p while

∂ ∂ ∂3 ∂3 ∂3 ∂3 ( P + P )3 = P + 3 P + 3 P + P = 6 ∂x ∂z ∂x3 ∂x2∂z ∂x∂z2 ∂z3 does note, hence [1:0:1] νϕ (p) = 3. Finally, ∂ ∂ P + P = 3x2 + y2 ∂x ∂z and

2 2 2 3 2 2 2 2 2 3 Ip(3x + y , y z − x ) = Ip(3x + y , x(3x + y ) + 3y z − 3x ) 2 2 2 2 2 2 2 = Ip(3x + y , y (x + 3z)) = 2Ip(3x + y , y) + Ip(3x + y , x + 3z)

= 4Ip(x, y) + 0 = 4

It follows that 1 δ(p) = (4 − 3 + 1) = 1, 2 g(C) = 1 − 1 = 0.

107 17 20.03. Lecture 17. Holomorphic diﬀerentials

2 We have associated to a lattice Λ in C a non-singular cubic curve CΛ ⊂ P . Now, we would like to understand how to a non-singular cubic curve assign a lattice in C. For this we need to know how to integrate a holomorphic differential along a piecewise smooth path in a Riemann surface. Definition 17.1. A piecewise-smooth path in a Riemann surface S is a continuous map γ :[a, b] → S such that if ϕ: U → V is a holomorphic chart and [c, d] ⊂ γ−1(U) then ϕ ◦ γ :[c, d] → V is a piecewise-smooth path in the open subset V of C. We say that γ is closed if γ(a) = γ(b). Recall that a meromorphic function on an open W ⊂ C can be interpreted as a holomorphic function W → P1 of Riemann surfaces. It motivates Definition 17.2. A meromorphic function on a Riemann surface S is a function f : S → P1 which is holomorphic in the sense of Riemann surfaces and not identically ∞ on any connected component of S. If S is a compact Riemann surface then every holomorphic function S → C is constant but there are lots of meromorphic functions (which is not easy to prove in general). Examples are Weiertrass ℘-function on a complex torus, and rational functions on non- singular projective curves. By a rational function on C\Sing(C), where C = {P (x, y, z) = 0} we mean a meromorphic function S(x, y, z) [x : y : z] 7→ T (x, y, z) where S and T are homogenous polynomials of the same degree ans T does not vanish identically at C. Does it make sense to differentiate a meromorphic function on a Riemann surface? If Φ = {ϕα : Uα → Vα | α ∈ A} is a holomorphic atlas on S then a meromorphic function g : S → P1 is determined by the collection of meromorphic functions

−1 1 {g ◦ ϕα : Vα → P | α ∈ A}. Conversely, a collection of meromorphic functions

1 {Gα : Vα → P | α ∈ A} deﬁnes a meromorphic function on S if and only if

Gα(ϕα(u)) = Gβ(ϕβ(u)), ∀u ∈ Uα ∩ Uβ.

−1 1 −1 0 1 We can diﬀerentiate g ◦ ϕα : Vα → P to get meromorphic functions (g ◦ ϕα ) : Vα → P −1 0 0 but they do not deﬁne a meromorphic function on S since (g ◦ ϕα ) and (g ◦ ϕβ) do not necessarily agree on Uα ∩ Uβ. Instead, the chain rule tells us that −1 0 −1 −1 0 −1 0 −1 0 (g ◦ ϕα ) (ϕα(u)) = ((g ◦ ϕβ )(ϕβ ◦ ϕα )) (ϕα(u)) = (g ◦ ϕβ ) (ϕβ(u))(ϕβ ◦ ϕα ) (ϕα(u)).

108 So if we differentiate gα we do not get a meromorphic function on S but an abstract −1 0 object called a meromorphic differential, denoted by dg which is glued from (g ◦ ϕα ) . We can then multiply a meromorphic differential dg by a meromorphic function f. We −1 −1 0 will say that f dg and fe dge are equal if the defining functions (f ◦ ϕα )(g ◦ ϕα ) and −1 −1 0 (fe◦ ϕα )(ge ◦ ϕα ) agree on every Vα. More precisely, we have Definition 17.3. A meromorphic differential on a Riemann surface S is an equivalence class of pairs (f, g) of meromorphic functions on S such that

(f, g) ∼ (f,e ge) if and only if for every holomorphic chart ϕ: U → V on S and every z ∈ V , we have

−1 −1 0 −1 −1 0 f ◦ ϕ (z)(g ◦ ϕ (z)) = fe◦ ϕ (z)(ge ◦ ϕ (z)) . Abstractly, we also have

Deﬁnition 17.4. Let {ϕα : Uα → Vα | α ∈ A} be a holomorphic atlas on a Riemann surface S. A meromorphic diﬀerential η on S is given by a collection

1 {ηα : Vα → P | α ∈ A} of meromorphic function such that if u ∈ Uα ∩ Uβ then

−1 0 ηα(ϕα(u)) = ηβ(ϕβ(u))(ϕβ ◦ ϕα ) (ϕα(u)).

Remark 17.5. If η and ζ are meromorphic differentials and ζ is not identically zero on any connected component of S then η/ζ defines a meromorphic function on S glued from ηα/ζα. It follows that η = fζ. Thus, to show that every abstract meromorphic differential is of the form f dg it suffices to show that there is at least one non-constant meromorphic function on every Riemann surface. This is beyond the scope of this lecture but we have seen examples on non-singular curves and complex tori.

Definition 17.6. The meromorphic differential f dg has a pole at a point p ∈ S is the meromorphic function (ϕ ◦ ϕ−1)(g ◦ ϕ−1)0 has a pole at ϕ(p) when ϕ: U → V is a holomorphic chart on an open neighbourhood U of p ∈ S. We call f dg holomorphic differential if it has no poles.

Deﬁnition 17.7. If f dg is a holomorphic diﬀerential on S then the integral of f dg along a piecewise-smooth path γ :[a, b] → S is

Z Z b f dg = f ◦ γ(t)(g ◦ γ)0(t)dt. γ a

We need to check that if (f, g) ∼ (f, g) then R f dg = R f dg, i.e. that e e γ γ e e

Z b Z b 0 0 f ◦ γ(t)(g ◦ γ) (t)dt = fe◦ γ(t)(ge ◦ γ) (t)dt. a a

109 We can ﬁnd a = a0 < a1 < . . . < ap = b and α1, . . . αp ∈ A such that γ([ai−1, ai]) ⊂ Uαi . Then p Z X Z ai f dg = f ◦ γ(t)(g ◦ γ)0(t)dt = γ i=1 ai−1 p X Z ai (f ◦ ϕ−1)(ϕ γ)(t)(g ◦ ϕ−1)0 ◦ (ϕ γ)(t)(ϕ ◦ γ)0(t)dt = αi αi αi αi αi i=1 ai−1 p X Z ai (f ◦ ϕ−1)(ϕ γ)(t)(g ◦ ϕ−1)0 ◦ (ϕ γ)(t)(ϕ ◦ γ)0(t)dt = e αi αi e αi αi αi i=1 ai−1 p Z ai Z X 0 fe◦ γ(t)(ge ◦ γ) (t)dt = fe dg.e i=1 ai−1 γ

R R 0 0 If the Riemann surface is C then γ f dg = γ f(z)g (z)dz is the integral of f(z)g (z) along γ in the usual sense of complex analysis. If g : S → C is a complex-valued holomorphic mapping on any Riemann surface S R then γ dg = g(γ(b)) − g(γ(a)). Definition 17.8. If ψ : S → R is a holomorphic mapping between Riemann surfaces S and R an if f dg is a holomorphic differential on R then we define a holomorphic differential ψ∗(f dg) on S by

ψ∗(f dg) = (f ◦ ψ) d(g ◦ ψ).

If γ :[a, b] → S is a piecewise-smooth path in S then

Z Z b Z ψ∗(f dg) = f ◦ ψ ◦ γ(t)(g ◦ ψ ◦ γ)0(t) = f dg. γ a ψ◦γ

Let C ⊂ P2 be an irreducible projective curve defined by P ∈ C[x, y, z]. An abelian R integral is an integral of the form γ f dg where f and g are rational functions on C \ Sing(C) and γ is a piecewise-smooth path not passing through any poles of f dg. We usually assume that C is not the line at infinity {z = 0} and we can take g to be g([x : y : z]) = x/z. If we work in inhomogeneous coordinates [x : y : 1], we write dx for dg. In affine coordinates f becomes a rational function R(x, y) in the usual sense and Z Z f dg = R(x, y)dx, γ γ where y is regarded as a multivalued function of x via the equation P (x, y, 1) = 0 which defines C in affine coordinates. Given a lattice Λ ∈ C we have defined a biholomorphism

u: C/Λ → CΛ, where 2 3 2 3 CΛ = {y z = 4x − g2(Λ)xz − g3(Λ)z }.

110 There is a meromorphic diﬀerential on CΛ given in inhomogeneous coordinates [x : y : 1] by y−1 dx. Let η = u∗(y−1 dx). Then η is a meromorphic diﬀerential on C/Λ. If π : C → C/Λ is the canonical projection then

π∗η = π∗u∗(y−1 dx) = (u ◦ π)∗(y−1dx) = (℘0)−1d℘ = (℘0)−1℘0 dz = dz. where z : C → C denotes the identity function. Since π is locally a holomorphic bijection with a holomorphic inverse and dz is a holomorphic differential on C, it follows that η has no poles. Since u is a holomorphic bijection with a holomorphic inverse it follows that y−1 dx has no poles. Now, choose any λ ∈ Λ and define γe: [0, 1] → C by γe(t) = tλ. If γ := π◦γe: [0, 1] → CΛ then γ(0) = Λ + 0 = Λ + λ = γ(1). By definition Z Z Z ∗ η = π η = dz = γe(1) − γe(0) = λ. γ γe γe On the other hand, if γ : [0, 1] → C/Λ is a piecewise-smooth closed path then we can find a continuous path γe: [0, 1] → C such that γ = π ◦ γe. It is locally smooth since π is a holomorphic bijection with holomorphic inverse. Moreover, π ◦ γe(a) = π ◦ γe(b), hence Z Z Z ∗ η = π η = dz = γe(b) − γe(a) ∈ Λ. γ γe γe Thus, we have proved R Proposition 17.9. Λ = { γ η | γ is a piecewise-smooth path in C/Λ}.

∗ −1 Since u: C/Λ → CΛ is a bijection with a holomorphic inverse and η = u (y dx), we get

R −1 Corollary 17.10. Λ = { γ y dx | γ is a piecewise-smooth path in CΛ}.

2 −1 This means that we can recover the lattice from CΛ ⊂ P . We can also describe u −1 in terms of integrals of the diﬀerential y dx on CΛ.

Proposition 17.11. The inverse of the holomorphic bijection u: C/Λ → CΛ is given by Z p u−1(p) = Λ + y−1 dx [0:1:0] where the integral is over any piecewise-smooth path γ in CΛ from [0 : 1 : 0] to p. Note that the definition of u−1(p) makes sense because any two paths from [0 : 1 : 0] to p differ by a closed path and a differential along a closed path gives an element in the lattice Λ.

111 Proof. Suppose that u(Λ + z) = p and that γ :[a, b] → CΛ is any piecewise-smooth path from [0 : 1 : 0] to p. Then u−1 ◦ γ is a path in C/Λ from Λ + 0 to Λ + z and hence Z Z Z Z y−1 dx = u∗(y−1dx) = η = π∗η = γ(b) − γ(a). −1 −1 e e γ u ◦γ u ◦γ γe

−1 −1 Then Λ + γe(b) = u (p) = Λ + z and Λ + γe(a) = u ([0 : 1 : 0]) = Λ + 0 so Z −1 Λ + y dx = Λ + γe(b) − γe(a) = Λ + z. γ

112 18 23.03. Lecture 18. Abel’s theorem

As an application of Bezout theorem we proved that any projective line intersects a smooth cubic in three points, counting with multiplicity. This allowed us to deﬁne an abelian group structure on a smooth cubic by the rule that three points add to zero if and only if they lie on an intersection of the cubic with a projective line. The neutral element could be chosen as any (out of nine) inﬂection point of C. Weierstrass ℘-function allowed us to view a non-singular cubic CΛ as C/Λ, for some lattice Λ ⊂ C. (In fact, one can show that any smooth cubic {y2z = x(x−z)(x−λz)} ⊂ P2 is biholomorphic to CΛ, for some lattice Λ.) Addition in C endows C/Λ with a structure of an abelian group. Today, we show that these two group structures coincide. In particular, we give the proof of associativity that was missing in Lecture 7. Recall that a line in P2 meets a non-singular projective cubic curve C in P2 either (a) in three distinct points p, q, r each with multiplicity one, or

(b) in two distinct points p with multiplicity one and q with multiplicity two (i.e. L is tangent to C at q but not at p and q is not an inﬂection point of C), or

(c) in one point p with multiplicity three (L is tangent to C at p and p is an inﬂection point of C). That means that we have

p + q + r = 0, p + 2q = 0, p + p + p = 0.

Since we choose one inflection point to be the neutral element 0, it follows from the last formula that the remaining inflection points are of order 3. There are precisely nine points of order 1 or 3 in C/Λ: j k Λ + ω + ω . 3 1 3 2 They can be arranged in a 3 × 3 array such that the three entries in a row, column or on a diagonal add up to Λ + 0 (recall Homework II). In order to state Abel’s theorem, recall that CΛ is defined by polynomial

2 3 2 3 QΛ(x, y, z) = y z − 4x + g2(Λ)xz + g3(Λ)z , map u: C/Λ → CΛ is deﬁned as [℘(z): ℘0(z) : 1] if z∈ / Λ, u(Λ + z) = [0 : 1 : 0] if z ∈ Λ, and its inverse Z p u−1(p) = y−1dx. [0:1:0]

113 Theorem 18.1 (Abel’s theorem for tori). If t, v, w ∈ C then t + v + w ∈ Λ

2 if and only if there is a line L in P whose intersection with CΛ consists of the points u(Λ + t), u(Λ + v), u(Λ + w) (allowing for multiplicities). Equivalently, if p, q, r ∈ CΛ then Z p Z q Z r Λ + y−1dx + y−1dx + y−1dx = Λ + 0 [0:1:0] [0:1:0] [0:1:0]

2 if and only if p, q, r are the points of intersection of CΛ with a line in P . Proof. First, we show that if L is a line intersecting C in p, q, r (allowing for multiplicities) then Z p Z q Z r Λ + y−1dx + y−1dx + y−1dx = Λ + 0. [0:1:0] [0:1:0] [0:1:0]

Case 1 Suppose that L is tangent line z = 0 to CΛ at the point of inﬂection [0 : 1 : 0]. Then p = q = r = [0 : 1 : 0].

Case 2 Suppose that L = {cy = bz}. Then L meets CΛ in three points

p1(b, c) = [a1 : b : c], p2(b, c) = [a2 : b : c], p3(b, c) = [a3 : b : c]

where a1, a2, a3 are roots of QΛ(x, b, c). Deﬁne

Z p1(b,c) Z p2(b,c) Z p3(b,c) 1 −1 −1 −1 µ: P → C/Λ, µ([b : c]) = Λ + y dx + y dx + y dx. [0:1:0] [0:1:0] [0:1:0]

We shall prove later that µ is holomorphic and that any holomorphic map from P1 → C is constant. 1 Since π : C → C/Λ is a covering projection, we can lift µ to a map µe: P → C. Since π is holomorphic and locally an isomorphism, µe is holomorphic, hence constant. It follows that µ(b, c) = µ(1, 0) = Λ + 0.

Case 3 Suppose that L is any line. Then

L = {sx + t(cy − bz) = 0}.

1 Fix b, c and let L ∩ CΛ = {q1(s, t), q2(s, t), q3(s, t)}. Deﬁne ν : P → C/Λ by

Z q1(s:t) Z q2(s,t) Z q3(s,t) ν([s : t]) = Λ + y−1dx + y−1dx + y−1dx. [0:1:0] [0:1:0] [0:1:0]

As in Case 2 ν is holomorphic, hence constant, i.e. ν([s : t]) = ν([0 : 1]) = Λ + 0.

114 It follows from the deﬁnition of u−1 that if t, v, w are complex numbers and u(Λ + t), 2 u(Λ + v), u(Λ + w) are the intersection points of CΛ with a line in P then t + v + w ∈ Λ. Let now t, v, w ∈ C be such that t + v + w ∈ Λ. Let p = u(Λ + t), q = u(Λ + v), r = u(Λ + w)

2 If p 6= q let L be a line in P through p and q. If p = q let L be the line tangent to CΛ at p. In both cases L meets CΛ in another point re. Then −1 −1 −1 −1 −1 −1 u (p) + u (q) + u (re) = Λ + 0 = Λ + t + v + w = u (p) + u (q) + u (r). It follows that r = re which ﬁnishes the proof. In the proof we have used Lemma 18.2. Map µ: P1 → C/Λ,

Z p1(b,c) Z p2(b,c) Z p3(b,c) µ([b : c]) = Λ + y−1dx + y−1dx + y−1dx [0:1:0] [0:1:0] [0:1:0] is holomorphic.

∂QΛ Proof. For all but ﬁnitely many b ∈ C the partial derivative ∂x (x, y, z) is non-zero at (a, b, 1) when QΛ(a, b, 1) = 0. For such a, b the polynomial Q(x, b, 1) has three distinct roots a1, a2, a3 and pi(b, 1) = [ai, b, 1]. By the implicit function theorem applied to QΛ(x, y, 1) there are open neighbourhoods U and V1, V2, V3 of b and a1, a2, a3 and holomorphic functions gi : U → Vi such that if x ∈ Vi and y ∈ U then QΛ(x, y, 1) = 0 ⇔ x = gi(y).

Hence, there are holomorphic maps U → CΛ given by

ψi(w) = [gi(w): w : 1].

We may choose V1, V2 and V3 to be disjoint. This means that for w ∈ U, g1(w), g2(w) and g3(w) are distinct roots of the polynomial QΛ(x, w, 1), so

pi(w, 1) = [gi(w): w : 1] = ψi(w).

Thus, if γ is a path in U from b to w then ψi ◦ γ is a path in Vi from pi(b, 1) to pi(w, 1) and Z Z g0(w) y−1dx = i dw. ψi◦γ γ w Then X Z w g0(y) µ[w : 1] = µ[b : 1] + i dy y 1≤i≤3 b 0 0 gi(y) We can ﬁnd Puiseux expansions to see that gi(0) = 0. It follows functions y are holmorphic on U, so their integrals from b to w are holomorphic functions of w near b. It follows that µ is holomorphic in a neighbourhood of [b : 1]. We have shown that µ is holomorphic except possibly at ﬁnitely many points of P1. Since a bounded continuous function U → C whose restriction to U \{a} is holomorphic, is holomorphic on U, we conclude that µ is holomorphic.

115 Lemma 18.3. Any holomorphic map f : P1 → C is constant.

Proof. We can regard f as a map P1 → P1 which never takes the value {∞}. We proved in Lemma 11.11 that any non-constant holomorphic map P1 → P1 is rational, hence takes value ∞ at least one.

116 19 27.03. Lecture 19. The Riemann-Roch theorem

We shall relate genus g of a curve to meromorphic differentials on the curve. Definition 19.1. A divisor D on C is a formal sum X D = npp p∈C such that np ∈ Z for every p ∈ C and np = 0 for all but finitely many p ∈ C. The degree of D is then X deg(D) = np p∈C

If np = 0 for p∈ / {p1, . . . , pl} then we also write

D = m1p1 + ... + mkpk.

We add and subtract divisors by adding and subtracting np’s. This way divisors on C form a C-vector space Div(C) and the degree defines a group homomorphism deg: Div(C) → Z. 0 If np ≥ 0 for all p, we write D ≥ 0 and say that D is effective. We write D ≥ D if D − D0 ≥ 0. Let f be a meromorphic function on C which is not identically zero. If p ∈ C we can −1 choose a holomorphic chart ϕα : Uα → Vα on C such that p ∈ Uα. Then f ◦ ϕα is a meromorphic function on Vα ⊂ C. If m is a positive integer we say that f has a pole or a −1 zero of multiplicity m at p if f ◦ ϕα has a pole or a zero of multiplicity m at ϕα(p). This is independent of the choice of a chart. Similarly, if ω = f dg is a meromorphic differential on C which is not identically zero we say that ω has a pole or a zero of multiplicity m at p if the meromorphic function −1 −1 0 (f ◦ ϕα )(g ◦ ϕα ) which represents ω has a pole or zero of multiplicity m at ϕα(p). Definition 19.2. The divisor of a meromorphic function f which is not identically zero is X (f) = np p where np = −m if f has a pole of multiplicity m at p, np = m if f has a zero of multiplicity m at p and np = 0 otherwise. Note that f (fg) = (f) + (g), = (f) − (g). g A divisor of a meromorphic function is called principal. Two divisors are linearly equivalent, D ∼ D0 if D − D0 is principal. If ω is a meromorphic differential we can analogously define (ω). The divisor of a meromorphic differential is called a canonical divisor and is denoted κ. We have noted that if η is another meromorphic differential then η = fω, hence (η) = (f) + (ω) ∼ (ω), i.e. the linear equivalence class of canonical divisor is well-defined.

117 Proposition 19.3. A principal divisor on C has degree zero; that is a meromorphic function on C which is not identically zero has the same number of zeroes and poles, counted with multiplicities.

dg Proof. Let g be a meromorphic function. We consider meromorphic differential g which has poles precisely at the points q1, . . . , qt where g has zeroes or poles. The same remains true if we compose g with some holomorphic chart. 2 We can choose coordinates in P such that [0 : 1 : 0] ∈/ C and 0, ϕ(q1), . . . , ϕ(qt) and ∞ are distinct and are not branch points of the map ϕ: C → P1 defined by [x : y : z] 7→ [x : z]. We can choose a triangulation (V,E,F ) of P1 such that every branch point of ϕ belongs to V and ϕ(q1), . . . , ϕ(qt) and 0 are in the interior of a face f0 while ∞ is in the interior of a different face f∞ ∈ F . There is a triangulation (V,e E,e Fe) of C given by

Ve = ϕ−1(V ), Ee = {e: [0, 1] → C continuous, s.t. ϕ ◦ e ∈ E}, Fe = {fe: ∆ → C continuous, s.t. ϕ ◦ fe ∈ F }.

By subdividing the triangulation if necessary we can assume that each face has at most one branch point among its vertices. This means that if fe ∈ Fe and f = ψ ◦ fe ∈ F then

ϕ: fe(∆) → f(∆) is a homeomorphism whose restriction to fe(∆ − {(0, 0), (0, 1), (1, 0)}) is the restriction of 1 a holomorphic chart if f 6= f∞. If f = f∞ we need to compose ϕ with the mapping z → z . The boundary γe of fe(∆) in C is the image of f ◦ σi, it is a closed piecewise-smooth path (recall that we chose an orientation on the boundary of ∆). Then

X Z dg = 0 γ g fe∈Fe e

−1 because if σi is the edge of some face f then σi is the edge of another face in the triangulation. R dg R (g◦ϕ−1)0(z) (g◦ϕ−1)0(z) On the other hand = −1 dz. The meromorphic function −1 has a γe g γ (g◦ϕ )(z) (g◦ϕ )(z) pole at a point a inside γ with residue ρ if and only if g ◦ ϕ−1 has either a zero at a with multiplicity ρ or a pole at a with multiplicity −ρ. Since ϕ| is a chart, the statement fe(∆o) is equivalent to g having a zero or a pole at ϕ−1(a). It follows from Cauchy’s theorem that Z (g ◦ ϕ−1)0(z) −1 dz = ±(Z(fe) − P (fe)) γ (g ◦ ϕ )(z) and the sign depends on whether γ is positively or negatively oriented in C. Since we −1 0 assumed that all poles of dg/g lie in ϕ (f0(∆ )), the sign is consistent, hence

X Z dg = ±(Z − P ). g ge fe∈Fe ,ϕ◦fe=f0

118 As the remaining faces of Fe contain no poles we get that X Z dg 0 = = ±(Z − P ). γ g fe∈Fe e

2 It remains to prove that we can choose a triangulation of P such that 0 and ϕ(qi) lie in an interior of one face.

Lemma 19.4. Let {p1, . . . , pr, q1, . . . , qs} be a set of r + s points in C with r ≥ 3. Then 1 there is a triangulation (V,E,F ) of P such that V = {p1, . . . , pr} and the set {q1, . . . , qs} is contained in the interior of a face. We may also assume that ∞ is in the interior of a diﬀerent face.

Proof. The proof is by induction on r. First suppose r = 3. We can assume that 0 does not belong to the set {p1, . . . , pr, q1 . . . , qs} and that arguments of the complex numbers p1, p2, p3, q1, . . . , qs are distinct. We can choose a real number R such that

R > max{|p1|, |p2|, |p3|, |q1|,..., |qs|} and we can reorder p1, p2, p3 so that

arg(p1) < arg(p2) < arg(p3).

For ε > 0 suﬃciently small we can ﬁnd a piecewise smooth path from p1 to R exp(ε + arg(p1)), then along the arc of the circle of radius R to R exp(−ε + arg(p2)) and then to 1 p2. Three paths like that give a triangulation of P with two faces, the interior, containing q1, . . . , qs, and the exterior, containing ∞. Now assume the inductive hypothesis that r > 3 and that we have a triangulation of 1 P with vertices p1, . . . , pr−1. If pr lies in the interior of the face containing q1, . . . , qs or on its boundary we can choose a subdivision of that face so that all q1, . . . , qs still lie in one face. In the remaining cases we proceed as in the proof that we have a triangulation of P1 with prescribed vertices, i.e. we choose arbitrary subdivisions.

Corollary 19.5. Two linearly equivalent divisors on C have the same degree. In particular, the degree of a canonical divisor is well-deﬁned.

Proposition 19.6. If κ is a canonical divisor on C then

deg κ = 2g − 2.

Proof. Since any two meromorphic differentials are linearly equivalent, it is enough to show that there is some meromorphic differential ω on C such that deg(ω) = 2g − 2. Let P (x, y, z) be a homogeneous polynomial defining C. We may assume that [0 : 1 : d ∂P 0] ∈/ C so that the coefficient P (0, 1, 0) of y in P is non-zero. Then ∂y is not identically

119 ∂P zero. Since C is irreducible, { ∂y = 0} does not have common components with C and ∂P by Bezout’s theorem there are ﬁnitely many points of C at which ∂y vanishes. Since [0 : 1 : 0] ∈/ C, applying a projective transformation of the form

x 7→ x, y 7→ y, z 7→ αx + z

∂P we can assume that if [a : b : c] ∈ C and ∂y = 0 then c 6= 0. ∂P Let now ω = d(x/z). Near points [a : b : c] ∈ C with c 6= 0 and ∂y (a, b, c) 6= 0 we can take x/z as a local holomorphic chart (because the implicit function theorem allow us to represent y as a holomorphic function of x). It follows that ω has no poles and no zeroes at such points. ∂P At a point [a : b : 0] ∈ C we have a 6= 0 (because [0 : 1 : 0] ∈/ C) and ∂y 6= 0 (by our choice of coordinates). Then v = z/x is a local holomorphic chart and

ω = d(1/v) = −v−2 dv

∂P has a pole of multiplicity 2. By assumption, ∂y (a, b, 0) 6= 0 whenever [a : b : 0] ∈ C so line {z = 0} is not tangent to C at any of the intersection points. It follows that the intersection multiplicity at each point of intersection is equal to one, hence by Bezout theorem, there are precisely d points of intersection. These points contribute to −2d of degree of ω. ∂P It remains to consider points with ∂y (a, b, c) = 0. These are precisely ramiﬁcation points of map ϕ: C → P1, ϕ([x : y : z]) = [x : z]. At these points c is non-zero, so ∂P ∂P ∂x (a, b, c) 6= 0. (If it was then Euler’s relation would give that c ∂z (a, b, c) = 0, hence ∂P ∂z (a, b, c) = 0 which cannot happen because C is non-singular). Therefore, u = y/z is a local chart and locally x/z is a holomorphic function f(u) of u satisfying

P (f(u), u, 1) = 0.

Diﬀerentiating we get

0 Pxf + Py = 0 0 2 0 (2) 0 0 2 0 (2) Pxx(f ) + Pxyf + Pxf + Pxyf + Pyy = Pxx(f ) + 2Pxyf + Pxf + Pyy = 0

0 If f (u0) = 0 then P f (2) = − yy . Px (k) In general, if f (u0) = 0, for k = 1, . . . , m − 1 then

∂m P (u , f(u ), 1) (m) ∂ym 0 0 f (u0) = − ∂P . ∂x (u0, f(u0), 1)

120 (m) It follows that the smallest positive integer m such that f (u0) 6= 0 is equal to the smallest positive integer m such that ∂mP (u , f(u ), 1) 6= 0. ∂ym 0 0 Since ω = d(f(u)) = f 0(u) du the multiplicity of zero of ω at a ramification point of ϕ is precisely one less than the ramification index at this point. We can assume that the coordinates were chosen is such a way that there are precisely d(d − 1) ramification points, each of index 2. Hence, we get d(d − 1) to the degree of ω. To sum up deg ω = d2 − d − 2d = d(d − 3). By degree-genus formula, deg ω = 2g − 2.

To state Riemann-Roch theorem we need one more definition P Definition 19.7. Let D = p∈C np p be a divisor on C. Then L(D) is a complex vector space L(D) = {(f) | (f) + D ≥ 0} ∪ 0. In other words a meromorphic function f on C belongs to L(D) if it is holomorphic outside of the points for which np > 0 and the order of the pole at such a point is at most np. Moreover, f has zero of order at least −np at every p ∈ C such that np < 0. We define l(D) = dim L(D).

Theorem 19.8 (Riemann-Roch). If D is any divisor on a non-singular projective curve C of genus g in P2 and κ is a canonical divisor on C then l(D) − l(κ − D) = deg(D) + 1 − g.

121 20 30.03. Lecture 20. The proof of the Riemann- Roch theorem

On the last lecture we deﬁned divisors, linear equivalence, degree of divisor and a vector space L(D). We calculated degree of the canonical divisor κ, i.e. divisor of any meromorphic diﬀerential on C, to be 2g − 2. We formulated the theorem Theorem 19.8 (Riemann-Roch). If D is any divisor on a non-singular projective curve C of genus g in P2 and κ is a canonical divisor on C then l(D) − l(κ − D) = deg(D) + 1 − g.

We have also proved that degree of a principal divisor (f) is 0. As a corollary, we get

Corollary 20.1. If deg(D) < 0 then l(D) = 0.

Proof. If f is meromorphic function which is not identically zero such that

(f) + D ≥ 0 then deg D = deg((f) + D) ≥ 0.

Lemma 20.2. If D ∼ D0 then l(D) = l(D0).

Proof. If D = D0 +(g) then f 7→ fg deﬁnes an isomorphism between L(D) and L(D0). Before proving Riemann-Roch theorem, we discuss some of its corollaries

Corollary 20.3. The genus of a non-singular projective curve C in P2 equals the dimension l(κ).

Proof. Riemann-Roch theorem for D = 0 tell us that

l(0) − l(κ) = 1 − g. l(0) is the dimension of the vector space of holmorphic functions. We shall see soon that any holomorphic function is constant, hence l(0) = 1. (More precisely, we will show that 0 ≤ l(0)−l(−p) ≤ 1. Together with l(−p) = 0 and the fact that there is a one dimensional space of constant functions we get l(0) = 1.) Let now X H = Ip(C,L)p p∈C 2 where Ip(C,L) is the intersection multiplicity at p of C with a line L in P , L = R(x, y, z) = 0. By Bezout’s theorem, H is a divisor of degree d on C. Then

deg(κ − mH) = deg κ − md

122 becomes negative if m is large enough. Thus, for m >> 0, l(κ − mH) = 0. On the other hand, if Q(x, y, z) is any homogeneous polynomial of degree m then Q(x, y, z) f = Rm(x, y, z) is a meromorphic function such that (f)+mH ≥ 0, i.e. f ∈ L(mH). Any two polynomials Q and Q0 deﬁne the same function on C if and only if their diﬀerence is divisible by P . Hence, if Ck[x, y, z] is the vector space of homogenous polynomials of degree k then

l(mH) ≥ dim Cm[x, y, z]/P (x, y, z)Cm−d[x, y, z] 1 1 1 = (m + 1)(m + 2) − (m − d + 1)(m − d + 2) = md + d(3 − d) = md + 1 − g. 2 2 2 We have shown that l(mH) − l(κ − mH) ≥ deg(mH) + 1 − g, when m is large enough. Riemann-Roch theorem tells us that there is equality, i.e. that any meromorphic function is of the form Q/Rm and in particular it is rational. 2 The same argument shows that if there are many lines L1,...,Lm in P and corresponding divisors Hj then if m is large enough every meromorphic function on C satisfying (f) + H1 + ... + Hm ≥ 0 is rational. On the other hand, every meromorphic function f satisﬁes

(f) + H1 + ... + Hm ≥ 0 for some lines L1,...,Lm (we just need to take lines through the poles). It proves Theorem 20.4. All meromorphic functions on a non-singular projective curve in P2 are rational. In order to prove Riemann-Roch, we need three lemmas

Lemma 20.5. Given any divisor D on C and any positive integer m0 there exists m ≥ m0 and points p1 . . . , pk of C (not necessarily distinct) such that

D + p1 + ... + pk ∼ mH. Proof. By adding points to D we can assume that D ≥ 0. For each of p ∈ C such that np > 0 we can choose a line through p whose points of intersections with C (allowing for multiple intersections) are (p) (p) (p) q1 = p, q2 , . . . , qd . Since the ratio of two linear homogeneous polynomials in x, y, z deﬁnes a meromorphic function on C, (p) (p) q1 + ... + qd ∼ H. Then X X (p) X mH ∼ np p + npqi = D + pj. np>0 2≤i≤d

123 Lemma 20.6. Let ω be a meromorphic diﬀerential on C with precisely one pole. Then this pole is not a simple pole (i.e. its multiplicity is at least 2).

Proof. Let us assume that ω = g dh has one simple pole at q. We choose ϕ: C → P1 a branch cover such that 0, ϕ(q) and ∞ are distinct and are not branch points of ϕ. As on the previous lecture, we chose a triangulation of P1 such that every branch point is a vertex, 0 and ϕ(q) belong to the same face f0 and ∞ to another face f∞. It induces a triangulation of C and, considering subdivision if necessary, we can assume that we have a triangulation (V,e E,e Fe) of C such that each face has at most one branch point among its vertices. It means that ϕ| is a holomorphic chart (we might have to compose it fe(∆) with z 7→ 1/z if ϕ ◦ fe = f∞. The boundary γe of every face in Fe is a closed path. Then Z Z ω = (g ◦ ϕ−1)(h ◦ ϕ−1)0(z)dz γe γ

0 −1 −1 0 where γ = ϕ ◦ γe. If q ∈ fe(∆ ) then f = f0 and (g ◦ ϕ )(h ◦ ϕ ) has a simple pole inside γ. Since a residue of a simple pole is always non-zero, we get by Cauchy’s theorem R ω 6= 0. By assumption q was the only pole of ω, hence the integral along boundary of γe any other face in Fe is zero. Thus, we get X Z ω 6= 0. γ fe∈Fe e

−1 On the other hand, in the above sum each edge σi occurs twice, once as σi, once as σi , hence the integrals cancel each other. It follows that

X Z ω = 0. γ fe∈Fe e The contradiction shows that our assumption that ω has one simple pole cannot be satisﬁed. Lemma 20.7. If D is a divisor on C, κ is a canonical divisor and p is any point of C then 0 ≤ l(D + p) − l(κ − D − p) − l(D) + l(κ − D) ≤ 1. P Proof. Suppose that D = q∈C nqq. Then L(D) = L(D + p) if there is no meromorphic function f such that (f) + D + p ≥ 0 and f has a pole at p of multiplicity exactly np +1 (if np ≥ 0) or a zero at p of multiplicity exactly −np − 1 (if np < 0). Otherwise, L(D) is a linear subspace of L(D + p) of codimension one (because if we subtract λf, for some λ 6= 0, we get an element of L(D)). In particular,

0 ≤ l(D + p) − l(D) ≤ 1, 0 ≤ l(κ − D) − l(κ − D − p) ≤ 1.

124 We need to show that we cannot have

l(D − p) − l(D) = 1 = l(κ − D) − l(κ − D − p).

If we had there would be functions f and g with

(f) + D + p ≥ 0, (g) + κ − D ≥ 0 with “equalities” at the point p. Then fgω is a meromorphic diﬀerential

(fgω) = (f) + (g) + κ ≥ −D − p + D = −p, i.e. a diﬀerential with a pole of order exactly one at p. By the previous Lemma, we know that such a meromorphic diﬀerential does not exist. Hence, either l(D + p) − l(D) = 0 or l(κ − D) − l(κ − D − p) = 0. Proof of Riemann-Roch theorem. First, we check that l(D) − l(κ − D) ≥ deg D − g + 1. We saw that there exists m0 such that if m ≥ m0 then

l(mH) − l(κ − mH) ≥ deg(mH) − g + 1.

We also know that D + p1 + ... + pk ∼ mH. Then deg(mH) = deg(D)+k, l(mH)−l(κ−mH) = l(D+p1 +...+pk)−l(κ−D−p1 −...−pk).

By the above Lemma

l(D + p1 + ... + pk) − l(κ − D − p1 − ... − pk) − l(D) + l(κ − D) ≤ k.

Then

l(D) − l(κ − D) ≥ l(D + p1 + ... + pk) − l(κ − D − p1 − ... − pk) − k = l(mH) − l(κ − mH) − k ≥ deg(mH) − g + 1 − k = deg(D) − g + 1.

The inequality applied to κ − D gives

l(κ − D) − l(D) ≥ deg(κ − D) − g + 1 = 2g − 2 − deg(D) − g + 1 = − deg(D) + g − 1.

Hence, we have equality. Riemann-Rich theorem gives us an alternative proof of the associativity of the abelian group structure on a smooth cubic curve in P2: Let p, q, r be points of C and let

a = p + q, b = a + r = (p + q) + r, c = q + r, d = p + c = p + (q + r).

Since p, q and −a are collinear, there is a homogeneous linear polynomial which vanishes at p, q and −a. Similarly, there is a linear polynomial which vanishes at a, −a and p0. The ratio of these polynomials deﬁnes a rational function ϕ with zeros at p and q and

125 poles at a and p0. By the same argument there is a rational function ψ with zeros at a and r and poles at b and p0. Then ϕψ is a meromorphic function with zeros at p, q, r and poles at b and p0 (with multiplicity two). Similarly, there is a meromorphic function with zeros at p, q, r and poles at d and p0 (with multiplicity two). If b 6= d the ratio of these functions is a meromorphic function with single zero at d and single pole at b. By the degree-genus formula, g = 1, so deg(κ) = 0 and κ − b < 0, i.e. l(κ − b) = 0. Then Riemann-Roch theorem gives

l(b) = deg(b) + 1 − g = 1 so the only meromorphic functions with at most simple poles at b are constant. It follows that b = d.

126 31.03. Workshop V

1. Show that given any nine points in P2 there is a cubic which contains them. 2. Let S be a compact connected Riemann surface of genus zero. Assuming that the Riemann-Roch theorem applies to S show that if D = p, for any p ∈ S, then l(D) = 2. Deduce that there exists a meromorphic function f on S with a simple pole at p and no other poles. (In fact f is a holomorphic bijection S → P1).

3. Show that a point p 6= [0 : 1 : 0] of the cubic curve CΛ associated with a lattice Λ ⊂ C has order two if and only if the tangent to CΛ at p passes through [0 : 1 : 0]. Show that the points of order two form a subgroup isomorphic to Z2 × Z2.

4. Use Abel’s theorem to show that if u, v, w ∈ C \ Λ and u, v, w are distinct modulo Λ then u + v + w ∈ Λ is and only if

 ℘(u) ℘0(u) 1  0 = det  ℘(v) ℘0(v) 1  . ℘(w) ℘0(w) 1

127 Solution to Workshop V

1. A general equation of a cubic is P (x, y, z) = Ax3 +By3 +Cz3 +Dx2y+Ex2z+F xy2 +Gy2z+Hxz2 +Iyz2 +Jxyz = 0

2 Let p1 = [a1 : b1 : c1], . . . , p9 = [a9 : b9 : c9 : d9] be arbitrary points in P . The cubic C = {P (x, y, z) = 0} contains them if and only if P (ai, bi, ci) = 0 for i = 1,..., 9. Thus, we get 9 linear equations on coeﬃcients A, . . . , J: P (ai, bi, ci) = 0. By rank-nullity theorem the space of solutions is at least one dimensional, hence there exists a non-zero solution (A0,...,J0). Then

3 3 3 2 2 2 2 2 2 {A0x +B0y +C0z +D0x y +E0x z +F0xy +G0y z +H0xz +I0yz +J0xyz = 0}

is a cubic which contains p1, . . . , p9. 2. We know that deg(κ) = 2g − 2 = −2. It follows that deg(κ − p) = −3 < 0 so l(κ − p) = 0. Then Riemann-Roch theorem gives l(p) − l(κ − p) = l(p) = deg(p) − g + 1 = 1 − 0 + 1 = 2.

Constant function f0 has a trivial divisor, hence (f0) + p > 0. Since L(0) $ L(p) there exists f which is an element of L(p) but not of L(0). It follows that (f) is not eﬀective, i.e. f has some poles. Since (f) + p ≥ 0, f can have only one pole of multiplicity one at p. Hence, f is the required meromorphic function.

3. By Abel’s theorem three point q, r and s on CΛ add up to zero if and only if they are collinear. We have 2q = −r where r is the second intersection point of the line tangent to C at q with C. It follows that 2p = 0 = −0 if and only if the tangent line to C at p contains 0 ∈ CΛ, i.e. the point [0 : 1 : 0]. 2 3 2 3 We know that CΛ is given by Q(x, y, z) = y z −4x −g2xz +g3z . A point [0 : 1 : 0] lies on the line tangent to CΛ at p = [a : b : c] if and only if ∂Q ∂Q ∂Q 0 · (a, b, c) + 1 · (a, b, c) + 0 · (a, b, c) = 0, ∂x ∂y ∂z

∂Q i.e. if and only if 0 · ∂y (a, b, c) = 2bc = 0.

It follows that points of order two lie on the intersection of CΛ with lines {z = 0} and {y = 0}. If z = 0 then x = 0 so the neutral element [0 : 1 : 0] is the 3 2 3 only point of intersection CΛ ∩ {z = 0}. If y = 0 then 4x + g2xz − g3z = 4(x − µ1z)(x − µ2z)(x − µ3z) = 0. Since curve CΛ is non-singular, all µi are pairwise distinct. It follows that CΛ ∩ {z = 0} = {p1, p2, p3}.

By assumption pi = −pi. Moreover, as p1, p2, p3 are collinear, p1 + p2 + p3 = 0. It follows that p3 = −p3 = p1 + p2. Then a map p1 7→ (1, 0), p2 7→ (0, 1), p3 7→ (1, 1), 0 7→ (0, 0) is an isomorphism of the subgroup of CΛ with points of order two and Z2 × Z2.

128 4. The sum u + v + w lies in Λ if and only if u + v + w = 0 in C/Λ. Riemann surface 0 C/Λ is biholomorphic to CΛ and the map is given by z 7→ u(z) = [℘(z): ℘ (z) : 1], for z∈ / Λ. Thus u + v + w ∈ Λ if and only if u(u) + u(v) + u(w) = 0 in CΛ, i.e. if and only if u(u), u(v) and u(w) are collinear. By assumption u(u), u(v) and u(w) are distinct. There exists a line αx + βy + γz = 0 which contains all three points if and only if  ℘(u) ℘0(u) 1   α   ℘(v) ℘0(v) 1   β  = 0 ℘(w) ℘0(w) 1 γ for some (α, β, γ) 6= (0, 0, 0). A vector (α, β, γ) like this exists if and only if  ℘(u) ℘0(u) 1  det  ℘(v) ℘0(v) 1  = 0 which ﬁnishes the proof. ℘(w) ℘0(w) 1

129 21 03.04. Lecture 21. Revision

Student should know

• What is an aﬃne and projective curve.

• What is a tangent line to a smooth point of a projective curve and an arbitrary point of an aﬃne curve.

• What is multiplicity of a point on a curve.

• What is projective space.

• That a projective curve is compact and Hausdorﬀ

• How to formulate Bezout theorem and its weak version

• What is intersection multiplicity and how to calculate it

• That a non-singular projective curve is irreducible

• When intersection multiplicity is one and when it is greater than one

• What is a Hessian of a polynomial and an inﬂection point of a curve

• How many inﬂection points can a curve of degree d have

• What is a general form of a non-singular cubic in P2 • How a line can intersect a non-singular cubic

• What is a group law on a non-singular cubic in P2 • What is a ramiﬁcation point and a branch locus

• How many ramiﬁcation points can there be

• What is a covering projection

• What is a triangulation

• What is Euler characteristic

• The degree–genus formula for smooth curves

• What is a Riemann surface

• That every C \ Sing(C) is a Riemann surface

• That a lattice in C gives a cubic curve • That an elliptic curve is biholomorphic to a complex tori

130 • That every curve has a resolution of singularities

• That the resolution of singularities is constructed by holomorphic elements

• How does a projective curve look like near singular point

• What is Newton polygon

• What is the algorithm for calculating Puiseux expansions

• When Puiseux expansions are essentially diﬀerent

• That there is a degree–genus formula for singular curves

• How to calculate genus of a singular curve

• What is a holomorphic diﬀerential and its integral

• That Λ is the lattice of integrals of η along closed paths

• What is Abel’s theorem

• What is a divisor, a principal divisor and divisor of a holomorphic diﬀerential

• What is degree of a divisor

• What is the formulation of Riemann-Roch theorem

131 06.04. Lecture 22. Mock Exam

Student should be able to

•

132 Exam

1. (a) Find singular points their multiplicity and tangent lines of the affine curve: y3 = x3 − x2. (b) Find singular points and inflections points of projective curve C defined by polynomial P (x, y, z) = y2z − x2(x + z). Find tangent lines to C at smooth inflection points. (c) Find the intersection points of C with tangent lines at smooth inflection points. [Similar to Workshop and Homework I and IV (for finding inflection points)]

2. (a) Use Bezout theorem to show that irreducible cubic curve cannot have more than one singular point. (b) Show that given any ﬁve points on P2 there is at least one conic containing them. (c) Use Bezout theorem to deduce that a projective curve of degree 4 in P2 with 4 singular points is reducible. [Hint: show that any conic containing the 4 singular points and any other point of C must have a component common with C]. [Similar to Workshop II]

3. (a) Show that the projective curve C = {y2z = x3} has a unique singular point p. (b) Show that a line through the singular point and point [t : s : 0] with s 6= 0 on the line {z = 0} intersects C at the singular point and [s2t : s3 : t3]. (c) Show that the map f : P1 → C, f[s : t] = [s2t, s3, t3] is a homeomorphism. Deduce that the degree-genus formula cannot be applied to singular curves. [Hint: recall that any continuous bijection from a compact space to a Hausdorﬀ space is always a homeomorphism] [Similar to Workshop III]

4. Recall that if p = [0 : 0 : 1] is a singular point of a curve C = {P (x, y, z) = 0} ⊂ P2 such that [0 : 1 : 0] doesn’t lie on C or the tangent line to C at any of non-singular 1 ∂P −1 inﬂection points of C then we deﬁne δ(p) = 2 (Ip(P, ∂y ) − νϕ(p) + |π (p)|), where ∂ m νϕ(p) is the minimal m such that ( ∂y P ) does not vanish at p and π : Ce → C is the resolution of singularities.

(a) Calculate the ﬁrst three terms of Puiseux expansion of C around the singular point. (b) Calculate the intersection multiplicity of C with the curve −4x2 − 2xz = 0 at the singular point of C. (c) Calculate the genus of the curve deﬁned by P (x, y, z) = y2z − x2(x + z).

[Similar to Workshop/Homework IV]

133 Solutions

1. (a) Let C ⊂ C2 be a curve deﬁned by polynomial P (x, y) = y3 − x3 − x2. Then ∂P ∂P = −3x2 − 2x = −x(3x − 2), = 3y2. ∂x ∂y A point [a : b] is a singular point of C if 3b2 = 0 and a(3a − 2) = 0. It follows 2 that b = 0 and either a = 0 or a = 3 . It remains to check whether points (0, 0) 2 and ( 3 , 0) lie on C. We have: 2 P (0, 0) = 0,P ( , 0) = −8/27 − 4/9, 3 hence p = (0, 0) is the only singular point of C. To calculate its multiplicity, we look at further diﬀerentials of P : ∂2P ∂2P ∂2P = −6x − 2, = 0, = 6y. ∂x2 ∂x∂y ∂y2

∂P Since ∂x (0, 0) 6= 0, point (0, 0) is a singular point of multiplicity two. Tangent lines are linear factors of the equation 1 ∂2P 1 ∂2P 1 ∂2P (0, 0)x2 + (0, 0)xy + (0, 0)y2 = −x2. 2! ∂x2 1!1! ∂x∂y 2! ∂y2 It follows that curve C has tangent line x = 0 at the singular point (0, 0). (b) Let now C ⊂ P2 be the projective curve defined by polynomial P (x, y, z) = y2z − x2(x + z). To calculate singular points, we first calculate derivatives ∂P ∂P ∂P = −3x2 − 2xz, = 2yz, = y2 − x2. ∂x ∂y ∂z Thus [a : b : c] is a singular point if and only if   a(−3a − 2c) = 0 bc = 0  a2 − b2 = 0. It follows from the second equation that either b = 0 or c = 0. In the first case b = 0, the third equation implies a = 0. Then the first equation holds, hence [0 : 0 : 1] is a singular point of C. If c = 0 then the first equation implies a = 0. Then the third equation holds if b = 0 which contradicts the assumption that [a : b : c] is a point in P2. It follows that [0 : 0 : 1] is the only singular point of C. To find inflection points of C we look at the matrix of second derivatives  −6x − 2z 0 −2x   0 2z 2y  −2x 2y 0

134 It determinant is −8x2z + 4y2(6x + 2z) = 8y2z + 24xy2 − 8x2z. A point [a : b : c] is an inﬂection point of C if

b2c = a3 + a2c b2c = −3ab2 + a2c b2c = a3 + a2c 0 = a3 + 3ab2 = a(a2 + 3b2) b2c = 0 b2c = a3 + a2c 0 = a a2 = −3b2 1 2 3 2 b = 0 c = 0 − 3 a c = a + a c a = 0 a = 0 a2 = −3b2 2 4 b = 0 c = 0 0 = a (a + 3 c) 2 1 2 a = 0 a = 0 b = − 3 a 3 b = 0 c = 0 a = 0 c = − 4 a 2 1 2 a = 0 a = 0 b = 0 b = − 3 a It follows that the inﬂection points are: √ √ i 3 3 i 3 3 [0 : 0 : 1], [0 : 1 : 0], [1 : : − ], [−1 : : ]. 3 4 3 4 Point [0 : 0 : 1] is singular. Tangent lines at smooth inﬂection points are:

z = 0, √ 3 2i 3 4 − x − y − z = 0, 2 √4 3 3 2i 3 4 − x + y − z = 0. 2 4 3 (c) Line z = 0 intersects C in points [a : b : c] such that

c = 0, −a3 = 0

It follows that [0 : 1 : 0] is the only intersection point. √ 3 i 3 4 Line − 2 x − 2 y − 3 z = 0 intersects C in points [a : b : c] such that √ 3 i 3 4 − 2 a − 2 b − 3 c = 0, b2c − a2(a + c) = 0 √ i 3 3 4 2 b = − 2 a − 3 c, b2c − a2(a + c) = 0 √ i 3 3 4 2 b = − 2 a − 3 c, 64 3 16 2 2 3 2 3 2 16 2 64 3 4 3 − 27 c − 3 ac − 3a c − a − a c = −(a + 4a c + 3 ac + 27 c ) = −(a + 3 c) = 0

135 √ i 3 3 It follows that [1 : 3 : − 4 ] is the only intersection point. √ 3 i 3 4 Finally, Line − 2 x + 2 y − 3 z = 0 intersects C in points [a : b : c] such that √ 3 i 3 4 − 2 a + 2 b − 3 c = 0, b2c − a2(a + c) = 0 √ i 3 3 4 2 64 2 16 2 2 b = 2 a + 3 c ⇒ b = − 27 c − 3 ac − 3a , b2c − a2(a + c) = 0 √ i 3 3 4 2 b = 2 a + 3 c, 64 3 16 2 2 3 2 3 2 16 2 64 3 4 3 − 27 c − 3 ac − 3a c − a − a c = −(a + 4a c + 3 ac + 27 c ) = −(a + 3 c) = 0

√ i 3 3 It follows that [1 : − 3 : − 4 ] is the only intersection point. 2 2. (a) Let C ⊂ P be a cubic and assume that p0, p1 are singular points of C. Let L be a line passing through p0 and p1. Since pi ∈ C is singular, Ip (C,L) ≥ 2. It P 1 follows that p∈C∩L Ip(C,L) ≥ 4. By Bezout theorem, if C and L had no common components then P p∈C∩L Ip(C,L) = 3. It follows that C and L have a common component. As L is irreducible, it is the common component of C. It shows that C is reducible. 2 (b) Let p0 = [a0 : b0 : c0], . . . , p4 = [a4 : b4 : c4] be arbitrary points in P . A conic Q is deﬁned by a polynomial

P (x, y, z) = Ax2 + By2 + Cz2 + Dxy + Exz + F yz.

Two polynomials deﬁne the same conic if they diﬀer by a multiplication by scalar, hence we can assume that A + B + C + D + E + F = 1. Conic Q contains points pi if and only if P (ai, bi, ci) = 0. Thus, we get a system of six linear equations in A, . . . , F . A + B + C + D + E + F = 1, P (ai, bi, ci) = 0

Such a system always has a solution (A0,...,F0) and the ﬁrst equation guarantees that it is not zero. Then, the conic

2 2 2 Q = {[x : y : z] | A0x + B0y + C0z + D0xy + E0xz + F0yz}

contains p0, . . . , p4.

(c) Let now C be a projective curve of degree 4 and assume that p1, . . . , p4 are singular points of C. Let p0 ∈ C be any point of C \{p1, . . . , p4}. By the above argument, there exists a conic Q which contains p0, . . . , p4. Then

4 4 X X X Ip(C,Q) ≥ Ipi (C,Q) + Ip0 (C,Q) ≥ 2 + 1 = 9 p∈C∩Q i=1 i=1

136 because if pi ∈ C is a singular point and p ∈ Q then Ip(C,Q) > 1. On the other hand, degree of C is 4 and degree of Q if 2. If C and Q had no common component, then by Bezout’s theorem X Ip(C,Q) = 2 · 4 = 8. p∈C∩Q

The contradiction implies that C and Q have common component, in particular C is reducible.

3. (a) Let P (x, y, z) = x3 − y2z. A point [a : b : c] is a singular point of curve C = {[x : y : z] ∈ P2 | P (x, y, z) = 0} if and only if ∂P ∂P ∂P = 3x2, = −2yz, = −y2 ∂x ∂y ∂z

vanish at [a : b : c], i.e. if

 2  3a = 0, 2bc = 0,  b2 = 0

The ﬁrst and the last equation imply that a = 0 = b. Then c is arbitrary, hence [0 : 0 : 1] is the unique singular point of C. (b) A line through [0 : 0 : 1] and [t : s : 0] is given by polynomial

Ls,t = {−sx + ty = 0}.

If s 6= 0, point [a : b : c] lies on the intersection of C with Ls,t if and only if

t a = s b, 2 t3 3 b c = s3 b t a = s b, 2 t3 b (c − s3 b) = 0 t a = 0, a = s b, t3 b = 0 c = s3 b

t t3 2 3 3 It follows that [ s : 1 : s3 ] = [ts : s : t ] is the intersection point of Lst with C which is diﬀerent than p. (c) Since map C2 → C3,(s, t) 7→ (s2t, s3, t3) is continuous, so is the map P1 → P2, [s : t] 7→ [s2t : s3 : t3]. As (s2t)3 = (s3)2t3, its image is contained in C, hence

1 2 3 3 ϕ: P → C, f([s : t]) = [s t : s : t ] is continuous.

137 Projective line P1 is compact and any complex projective curve is Hausdorﬀ, hence ϕ is a homeomorphism if it is a bijection. Let us construct its inverse. s s3 Since t = s2t , we put [y : x] if y 6= 0, ψ : C → 1, ψ([x : y : z]) = P [0 : 1] if y = 0. Then [s3 : s2t] = [s : t] if s 6= 0, ψ ◦ ϕ([s : t]) = ψ([s2t : s3 : t3]) = [0 : 1] if s = 0. i.e. ψϕ = Id. Let now [x : y : z] ∈ C. If y = 0 then x3 = 0, so [x : y : z] = [0 : 0 : 1]. Then

ψ ◦ ϕ[0 : 0 : 1] = ψ([1 : 0]) = [0 : 0 : 1].

If y 6= 0 then

ψ ◦ ϕ[x : y : z] = ψ([y : x]) = [y2x : y3 : x3] = [y2x : y3 : y2z] = [x : y : z],

i.e. ϕ ◦ ψ = Id. As discussed above, it follows that ϕ is a homeomorphism. In particular, the genus of C is the genus of P1, which is zero. On the other hand, C is of degree three, hence 1 (d − 1)(d − 2) = 1. 2 1 For smooth curves the degree-genus formula reads g(C) = 2 (d−1)(d−2). The example of {x3 = y2z} shows that the degree-genus formula does not hold for singular curves. 4. (a) Curve C is deﬁned by polynomial P (x, y, z) = y2z − x2(x + z). We know from the ﬁrst exercise that C has one singular point p = [0 : 0 : 1]. Then

2 3 2 X α β P (x, y, 1) = y − x − x = cαβx y α+β≥2 and X β 2 f0(t) = cαβt = t − 1 = (t − 1)(t + 1) α+β=2 has two roots ±1. The ﬁrst substitutions are

x = x1, y = x1(±1 + y1)

2 2 3 2 P (x1, y1, 1) = x1(±1 + y1) − x1 − x1, 2 2 X α β P1(x1, y1) = y1 ± 2y1 + 1 − x1 − 1 = y1 ± 2y1 − x1 = dαβx y . α+β≥1

138 Polynomial X β f1(t) = dαβt = ±2t − 1 α+β=1 1 has two roots ± 2 . The second substitutions are 1 x = x , y = x (± + y ), 1 2 1 2 2 2

1 1 P (x , y ) = x2(± + y )2 ± 2x (± + y ) − x , 1 2 2 2 2 2 2 2 2 2 1 1 X P (x , y ) = x (y2 ± y + ) + 1 ± y − 1 = x y2 ± x y + x ± y = e xαyβ. 2 2 2 2 2 2 4 2 2 2 2 2 4 2 2 αβ α+β≥1 Polynomial X 1 f (t) = e tβ = ±t + 2 αβ 4 α+β=1 1 has root −(± 4 ). Then 1 x = x , y = x (−(± + y )) 2 3 2 3 4 3 and the Puiseux expansions are 1 1 1 1 y (x) = x(1 + x( + x(− + ...))) = x + x2 − x3 + ... 1 2 4 2 4 1 1 1 1 y (x) = x(−1 + x(− + x(+ + ...))) = −x − x2 + x3 + ... 2 2 4 2 4 (b)

2 3 2 2 2 3 2 2 2 Ip(y z − x − x z, 2x + xz) = Ip(2y z − 2x − 2x z + x(2x + xz), 2x + xz) 2 2 2 2 = Ip(2y z, 2x + xz) = 2Ip(y, 2x + xz) + Ip(z, 2x + xz)

= 2Ip(y, x(x + z)) + 0 = 2Ip(y, x) + 2Ip(y, z) = 2 + 0 = 2.

(c) Curve C has one singular point, hence Noether formula reads 1 g(C) = (3 − 1)(3 − 2) − δ(p). 2 To calculate δ(p) we ﬁrst calculate |π−1(p)|, i.e. the number of essentially diﬀerent Puiseux expansions of C around p. From part (a) we know that there are two: 1 1 1 1 y (x) = x(1 + x( + x(− + ...))) = x + x2 − x3 + ... 1 2 4 2 4 1 1 1 1 y (x) = x(−1 + x(− + x(+ + ...))) = −x − x2 + x3 + ... 2 2 4 2 4

139 1 2 1 3 If g(t) = t + 2 t − 4 t + ..., then g(−εx) is not the second Puiseux expansion, for any root of unity ε. Indeed, we would necessarily have ε2 = 1 and g(−x) 6= y2(x). It follows that the two expansions are essentially diﬀerent, hence

|π−1(p)| = 2.

We learn from the first exercise that point [0 : 1 : 0] lies on the tangent line to C at the inflection point [0 : 1 : 0]. However, [1 : 0 : 1] is a point which does not lie on C and on tangent line to any smooth inflection point of C. [1:0:1] We calculate νϕ (p). We have

∂ ∂ ∂2 ∂2 ∂2 ( + )2P = P + 2 P + P = −6x − 2z − 2x ∂x ∂z ∂x2 ∂x∂y ∂y and it does not vanish at z. Hence

[1:0:1] νϕ (p) = 2.

∂ ∂ 2 2 2 Finally ( ∂x + ∂z )P = −3x − 2xz − x = −4x − 2xz. By part (b),

2 3 2 2 Ip(y z − x − x z, 2x + xz) = 2.

Then 1 δ(p) = (2 − 2 + 2) = 1 2 g(C) = 0.

140