[Riemann-Roch's Theorem and Applications IV] First steps to the classification of curves

In this post, we will give some results related to the classification of irreducible projective smooth curves of small genus, and make a careful study on hyperelliptic curves. In this note, a curve means a projective smooth irreducible curve.

Curves of genus 0 and 1. We will first begin with a curve $X$ of genus $0$. By Riemann-Roch's theorem, for any divisor of the form $p\in Div(X)$, we have $l(p)=2$. This yields $p$ is a very ample divisor. Hence, the induced map from $X\to \mathbb{P}^1$ is an isomorphism. We have another way to look at this: because $l(p)=2$, there exists $f\in L(p)$ a non-constant rational function, i.e. $f$ has simple pole at $p$. Hence, $f$ defines a map from $X$ to $\mathbb{P}^1$ sending $p\mapsto \infty$. Hence, $f$ is a map of degree 1, and hence, an isomorphism.

Let $X$ be a smooth projective curve of genus 1. We will prove that in fact $X$ is a smooth cubic curve in $\mathbb{P}^2$. By using Riemann-Roch, one can see $l(np)=n$, for all $n\ge 1$. In particular, $(1,\phi)$ is the basis for $L(2p)$, $(1,\phi,\psi)$ is the basis for $L(3p)$, $(1,\phi,\psi,\phi^2)$ is the basis for $L(4p)$, $(1,\phi,\psi,\phi^2,\phi\psi)$ is the basis for $L(5p)$. From this, one can see that $(1,\phi,\psi,\phi^2,\phi\psi, \psi^2,\phi^3)$ for $L(6p)$. This yields the non-trivial relation between 7 functions:

$$a_0+a_1\phi+a_2\psi+a_3\phi^2+a_4\phi\psi+a_5\psi^2+a_6\phi^3$$

This yields

$$a_5\psi^2 + a_4\phi\psi + a_2\psi =a_6\phi^3 + a_3\phi^2 + a_1\phi+a_0$$

By change of variable, one can assume that $a_5=a_6=1$. And $(\phi,\psi)$ defines a regular map between $X$ and a cubic curve $E$. By more careful analysis, the cubic curve $E$ defined about (in terms of $\phi,\psi$) is smooth, and the map from $X$ to $E$ is an isomorphism (For reference, see Milne's "Elliptic Curves").

Hyperelliptic curves and curves of genus 2 and 3. We begin with the following

Definition 1. Let $X$ be a smooth projective irreducible curve with $g_X\ge 1$. Then $X$ is a hyperelliptic curve if there exists a degree 2 morphism $X\to \mathbb{P}^1$.

We then prove that

Proposition 2. Any curve $X$ of genus 2 is hyperelliptic.

Proof. By Riemann-Roch, we have $\deg(K)=2$, and $l(K)=2$, where $K$ is the canonical divisor of $X$. That means, there exists $f\in K(K)$, and $f$ is non-constant, and $div(f)+K\ge 2$ with $\deg(div(f)+K)=\deg(K)=2$. And hence, because $div(f)+K\sim K$, we can assume without loss of generality that $K$ is of the form $p+q$.

We will prove next that $f$ has simple pole at $P,Q$. Because $f$ is non-constant, it must have poles somewhere. And its possible poles is simple at $P$ and $Q$. If $f$ has only simple pole at $P$, then we know that $f$ will define a map to $\mathbb{P}^1$, and $\deg(f)=1$ (since $\#f^{-1}(\infty)=1$), and $f$ defines an isomorphism from a curve of genus 2 to a curve of genus 0, a contradiction. Hence, $f$ must have simple pole at $P,Q$. Hence, $div(f)=P+Q-R-S$. And one can see that $f$ defines a degree two map from $X$ to $\mathbb{P}^1$, and $X$ is a hyperelliptic curve. (Q.E.D)

It is natural to ask: with given $g\ge 0$, there exists a curve of genus $g$?

Proposition 3. There exists hyperelliptic curve of every genus $g\ge 1$.

Proof. Let us consider the affine curve $X'$ defined by $y^2=\prod_{i=1}^{2g+1}(x-a_i)$, where $a_i\in k$ are distinct values. It can be seen that $X'$ is a smooth affine curve, but for $g\ge 2$, its compactification is not smooth, because the point of infinity is a singular point. We can now assume for simplicity that $k=\mathbb{C}$, then $X'\to \mathbb{P}^1\setminus\{a_1,...,a_{2g+1},\infty\}$ defines an unbranched 2-sheeted holomorphic map. Now, using a theorem for Riemann surfaces, there exists a compact Riemann surface $X$ with a branched covering $f: X\to \mathbb{P}^1$ that extend the map $X'\to \mathbb{P}^1\setminus\{a_1,...,a_{2g+1}, \infty\}$. And because $2g+1$ is odd, the map $f$ has critical values at $a_1,...,a_{2g+1},\infty$, and hence, we have $2g+2$ branched points of $X$ with multiplicity 2 for each. Now, we use the Riemann-Hurwitz's formula: $2g_X-2=2g+2-4$. This implies $g_X=g$.

It remains to show that any compact Riemann surface is an algebraic curve. But it is not a trivial fact. We can sketch the proof here. First, the Riemann-Roch's theorem can be applied for compact Riemann surfaces, and using it, we can deduce that there is a non-constant meromorphic function on $X$. This function defines a branched covering map from $X$ to $\mathbb{P}^1$, and it is a finite map, since $X$ is compact. And then, one can deduce that the meromorphic function field of $X$, denoted $M(X)$ is an extension of $M(\mathbb{P}^1)$ of degree $n$, and hence, there exists an irreducible polynomial $F(w)$ of degree $n$ such that $M(\mathbb{P}^1)[w]/(F(w))\cong M(X)$. Now, it yields $X$ is biholomorphic with the Riemann surfaces defined by $F(w)\in M(\mathbb{P}^1)[w]$, i.e. all coefficients of $F$ are meromorphic functions on $\mathbb{P}^1$. But meromorphic functions on $\mathbb{P}^1$ are just quotient of polynomials. This yields $X$ is in fact defined by an algebraic equation, and $X$ is an algebraic curve. (Q.E.D)

There is an useful criterion that help us to detect hyperelliptic curves via divisors.

Proposition 4. Let $X$ be a curve of genus $g\ge 1$, then $X$ is a hyperelliptic curve iff there exists a divisor $D$ on $X$ such that $\deg D=2, l(D)=2$.

Proof. We already saw in our previous results that the statement holds for $g=1,2$. Assume that $X$ is a hyperelliptic curve of genus $g\ge 3$, i.e. there exists a map $\phi: X\to \mathbb{P}^1$ and $\deg \phi=2$. This yields $\phi^*: k(\mathbb{P}^1)\to k(X)$ is a field embedding of degree 2. We know that there exists a non-constant rational function $f$ on $\mathbb{P}^1$, such that $div(f)=p-q$, where $p,q\in\mathbb{P}^1$, i.e. $q$ is the only simple pole of $f$. This yields $\phi^*f$ have exactly two poles $q_1,q_2\in X$, which lie on the fiber of $q$ via $\phi$. Now, let us consider $D:=q_1+q_2\in Div(X)$. It can be seen that $\deg(D)=2$, and $l(D)\ge 2$. If $l(D)=3$, then $D$ is a very ample divisor, since $l(D-p_1-p_2)\le 1$, for all $p_1,p_2\in X$. And hence, $X$ can be embedded into $\mathbb{P}^2$ via $D$. Let $F$ be the homogeneous equation for $X$, then due to the genus degree formula, $\deg f=d$, such that $g=\frac{(d-1)(d-2)}{2}$, and because $g\ge 3, d\ge 4$. But then, the function field $k(X)$ is an extension field of $k(\mathbb{P}^1)$ of degree 4. It is a contradiction. Hence, $D$ is not a very ample divisor, and $l(D)=2$. The converse is already done by our earlier arguments. (Q.E.D)

This result will lead to the classification of curves of genus $3$. Let $X$ be a curve of genus $3$. It can be seen that the canonical divisor $K$ has degree $3$. We will prove that it is actually base-point free.

Proposition 5. Let $X$ be a curve of genus $g\ge 1$. Then the canonical divisor $K$ is base point free.

Proof. It can be seen that $l(K)=g$, and $K$ is base point free iff $l(K-p)=l(K)-1$, for all point $p\in X$. By Riemann-Roch's theorem, we have

$$l(p)-l(K-p)=2-g$$
And it is sufficient for us to prove that $l(p)=1$, for all point $p\in X$. We have $1\le l(p)\le 2$. If $l(p)=2$, then it is a very ample divisor, and one can embedded $X$ into $\mathbb{P}^1$, a contradiction. So, $l(p)=1$ for all point $p\in X$, and hence $K$ is a base point free divisor (Q.E.D)

If $K$ is a base point free divisor, we then have a canonical map $\phi: X\to \mathbb{P}^{g-1}$ defined by $K$. In the case $g=3$, we will have a canonical map $\phi: X\to \mathbb{P}^2$. If furthermore, $K$ is base point free, then the map above is actually an embedding. And in this case $X$ is given by a polynomial of degree 4, by the genus degree formula.

Otherwise, $K$ is not a base point free divisor, and there exists two points $p,q$ such that $l(K-p-q)=l(K)-1$, i.e. there exists a divisor $D$ of degree 2, and $l(D)=2$. This yields by our previous proposition that $X$ is a hyperelliptic curve. In conclusion, we have

Proposition 6. Let $X$ be a curve of genus 3. Then either $X$ is hyperelliptic, or $X$ is a plane curve, given by a polynomial of degree 4.

Maps from hyperelliptic curve to $\mathbb{P}^1$. We can now deduce that if $X$ is a hyperelliptic curve, there exists only one map from $X$ to $\mathbb{P}^1$, by recalling some knowledge about morphisms to projective space.

We recall that from a curve $X$ to $\mathbb{P}^n$ is uniquely characterized by $n+1$-tuples of meromorphic function $\lambda f_0,...,\lambda f_n$ on $X$, where $\lambda\in k(X)^\times$, and $f_0,..,f_n$ do not vanish simultaneously at any point $p\in X$, and if we define a divisor $D\in Div(X)$, where $-D(p)$ is the minimum of $ord_p(f_i)$, then it can be seen that $D$ (up to linearly equivalent) is independent to the choice of $f_i$. One can see that in fact $f_i\in L(D)$. And hence, the vector space spanned by $(f_0,...,f_n)$ is a subspace of $L(D)$.

In the case $X$ is a hyperelliptic curve, one has the map $\phi$ from $X\to\mathbb{P}^1$, which is characterized by 2-tuples of meromorphic functions $f_0,f_1$ on $X$. The condition that $f_i$ do not vanish simultaneously at any point $p\in X$ implies that $f_0,f_1$ are linearly independent. And hence, $(f_0,f_1)$. In an open subset of $X$, where $f_0$ does not vanish, we have $\phi$ is given by $(1:f_1/f_0)$. Let us denote $f:=f_1/f_0$. Because $\phi$ is of degree 2, $f$ is not defined at exactly 2 points $q_1,q_2$, which are the fiber of $\infty\in \mathbb{P}^1$. Hence, the map $\phi:X\to\mathbb{P}^1$ is exactly $p\mapsto (1:f(p))$, for $p\ne q_i$ and $q_i\mapsto \infty$. It can be seen that the divisor $D$ associated to $\phi$ defined as above in this case is $q_1+q_2$, which has degree 2, and due to Lemma 4, $l(D)=2$. This yields by our previous argument that $(1,f)$ is the basis for $L(D)$. In conclusion, we have

Proposition 7. There exists only one map of degree 2 from a hyperelliptic curve to $\mathbb{P}^1$.

We now conclude this section by considering the canonical morphism from a hyperelliptic curve $X\to\mathbb{P}^{g-1}$ defined by the canonical divisor. It is quite surprising: $X$ will be mapped to the rational normal curve in $\mathbb{P}^{g-1}$.

Lemma 8. The canonical divisor on a hyperelliptic curve $X$ is of the form $(g-1)p+(g-1)q$, for some points $p,q\in X$.

Proof. We begin with a divisor $D=p+q$ on $X$ such that $l(D)=2$. This yields $(1,\phi)$ is the basis for $L(D)$, where $\phi$ is a non-constant meromorphic function. This yields $(1,\phi,...\phi^{g-1})$ are linearly independent in $L((g-1)D)$. Hence, $l((g-1)D)\ge g$. By Riemann-Roch, because $\deg (g-1)D=\deg K=2g-2$, one has $l((g-1)D)=l(K-(g-1)D)+g-1$. But we know that for any divisor $E$ with $\deg(E)=0$, then $l(E)\le 1$, and $l(E)=1$ iff $E$ is a principal divisor. This yields $l((g-1)D)=g$, and hence, $K\sim (g-1)D$. (Q.E.D)

Using this, we have

Proposition 9. The canonical morphism from a hyperelliptic curve $X$ maps $X$ into a rational normal curve in $\mathbb{P}^{g-1}$, and it is a map of degree 2.

Proof. One can see by the previous lemma that there exists $D=p+q\in Div(X)$, and $l(D)=2$, and $K=(g-1)D$. This yields, if $(1,\phi)$ is a basis for $l(D)$, we have $(1,\phi,...,\phi^{g-1})$ is a basis for $K$. From this, the canonical morphism is defined by $p\mapsto (1:\phi(p):...:\phi^{g-1}(p))$, and it is exactly the equation for the rational normal curve in $\mathbb{P}^{g-1}$. Because rational normal curves are isomorphic to $\mathbb{P}^1$, this induces a morphism from $X$ to $\mathbb{P}^1$, given by $p\mapsto (1:\phi(p))$. And it is exactly the map in Proposition 7, i.e. it is a degree 2 map. Hence, the canonical map from $X$ to the rational normal curve is of degree 2. (Q.E.D)

[Elliptic Curves I] A Note on Isogenies

This short note is taken from the book of Washington "Elliptic Curves: Number Theory and Cryptography", and Milne "Elliptic Curves".

1. Brief review on Weierstrass $\wp$-function. We first begin with a lattice $L:=\mathbb{Z}\omega_1+\mathbb{z}\omega_2$, where $\omega_1,\omega_2\in \mathbb{C}$ are linearly independent over $\mathbb{R}$. The fundamental domain $\mathcal{F}$ of $L$ is the set $\mathcal{F}:=\{a\omega_1+b\omega_2|a,b\in [0,1)\} = \mathbb{C}/L$. One can consider $\mathcal{F}$ as a group by the following addition law: $z_1+z_2 = z_1+z_2\mod L$. The fundamental domain $\mathcal{F}$ is called a torus associated to $L$. The Weierstrass $\wp$-function associated to $L$ is defined as follow:

$$\wp(z) := \frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}[\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}]$$

It can be seen that $\wp(z)$ is a non-constant meromorphic even function on $\mathbb{C}$. More clearly, poles of $\wp(z)$ is exactly the lattice points of $L$, and $\wp(z)=\wp(-z)$. If we take the derivative of $\wp(z)$, we get 

$$\wp'(z) = -2\sum_{\omega\in L}\frac{1}{(z-\omega)^3}$$

We can see that $\wp'(z)$ is an odd function, and both $\wp(z)$ and $\wp'(z)$ are doubly periodic, i.e. for all $\omega\in L$, we have $\wp(z+\omega)=\wp(z)$, and $\wp'(z+\omega) = \wp'(z)$. We next define the $k$-th Eisentein's series associated to $L$ as follows.

$$G_k=\sum_{\omega\in L\setminus \{0\}}\frac{1}{\omega^k}$$

Around a point $\omega$ in $L$, we can can see

$$\wp(z) = \frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}\omega^{-2}[\frac{1}{(z/\omega-1)^2}-1]$$

Using the geometric series, we get

$$\wp(z) = \frac{1}{z^2} + \sum_{\omega\in L\setminus \{0\}}\omega^{-2}(\sum_{n\ge 1}(z/\omega)^n)^2=\frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}\sum_{n\ge 1}(n+1)\frac{z^n}{\omega^{n+2}}=$$

$$=\frac{1}{z^2}+\sum_{n\ge 1}(n+1)z^n\sum_{\omega\in L\setminus \{0\}}\frac{1}{\omega^{n+2}}=\frac{1}{z^2}+\sum_{n\ge 1}G_{n+2}(n+1)z^n$$

Because $\wp(z)$ is an even function, we get

$$\wp(z) = \frac{1}{z^2}+\sum_{n\ge 1}G_{2n+2}(2n+1)z^{2n}$$

Using this identity, we can obtain the power series expansion for $\wp(z)$ as follows

$$\wp'(z) = \frac{-2}{z^3}+\sum_{n\ge 1}G_{2n+2}(2n+1)(2n)z^{2n-1}$$

and by direct computation, we have

$$f(z):=\wp'(z)^2 - 4\wp(z)^3 +  60G_4\wp(z) + 140G_6 = c_1z+c_2z^2 + ...$$

That means $f(z)$ is a holomorphic function, this yields $f(z)$ is bounded in the fundamental domain $\mathcal{F}$,  and it is doubly periodic (since $\wp(z)$ and $\wp'(z)$ are). Now, it follows that $f(z)$ is an entire funcion and bounded. By Louville's theorem, $f(z)=f(0)$ is a constant function. Because $f(0)=0$, we have $f(z)\equiv  0$. This yields a following important identity:

$$\wp'(z)^2 = 4\wp(z)^3 - g_2\wp(z) - g_3$$

where $g_2 = 60G_4, g_3 = 140 g_6$. That means, $(\wp(z),\wp'(z))$ is a point on a curve $ (E): y^2= x^3 - g_2x - g_3$. This curve is non-singular, and for any $(x,y)$ on the curve, there exists $z$ such that $(x,y)=(\wp(z),\wp'(z))$. Moreover, one obtain the following important bijection $\mathbb{C}/L \ to (E)$ by sending $z\ne 0 \mapsto (\wp(z):\wp'(z):1)$ and $0\mapsto (0:1:0)$. This gives a natural group law on $(E)$ induced from the group law on $\mathbb{C}/L$. For any $z\ne 0 \in \mathbb{C}/L$, because $z+0=0+z=z$, we can define 

$$(\wp(z):\wp'(z):1) + (0:1:0) = (\wp(z):\wp'(z):1)$$

If $z=0$, we have $(0:1:0)+ (0:1:0) = (0:1:0)$. If $z_1,z_2\in \mathbb{C}/L\setminus\{0\}$, and $z_1\ne z_2$, then $z_1+z_2$ correspond to the point $(\wp(z_1+z_2):\wp'(z_1+z_2):1)$ on the curve. So, we can define

$$(\wp(z_1):\wp'(z_1):1)+(\wp(z_2):\wp'(z_2):1):=(\wp(z_1+z_2):\wp'(z_1+z_2):1)$$

Similarly, when $z_1 = -z_2 in \mathbb{C}/L$, and both are non-zero, because $z_1+z_2=0$, we can define

$$(\wp(z_1), \wp'(z_1):1)+(\wp(z_2):\wp'(z_2):1)=(0:1:0)$$

This gives $(E)$ the natural group structure, and $(E)\cong \mathbb{C}/L$ as abelian groups.

NOTE. We are familiar with the fact that the group law on an elliptic curve comes from the intersection of curves with lines. Yes, it is true also in this context. Let us consider the line goes through two (affine) points $(\wp(z_1), \wp'(z_1))$ and $(\wp(z_2),\wp'(z_2))$ with equation: $y-ax-b=0$, it will cut the curve at the third point: $(\wp(-z_1-z_2), \wp'(-z_1-z_2))$.

To prove this fact, let $f(z) = \wp'(z) - a\wp(z) - b$ (correspond to our line: $y-ax-b$), then we know that $f(z)=0$ at two points $z_1,z_2 in \mathbb{C}/L$, and it has unique pole in this domain (at 0) with multiplicity 3 (since $\wp'(z)$ has). We now use a theorem from complex analysis, if $f$ is a meromorphic map from $\mathbb{C}/L$ to $\mathbb{C}$, and it has zero and poles at $z_i$ with multiplicity $n_i$, then $\sum_{i}n_i=0$, and $\sum_{i}n_iz_i=0$. From this, we can see $f$ has another zero $z$ such that $z_1+z_2+z+0=0$, i.e. $z=-z_1-z_2$. 

This will yields the third intersection point between the line and our curve is $(\wp(-z_1-z_2), \wp'(-z_1-z_2))$. So, taking the inverse point, we get $(\wp(-z_1-z_2), -\wp'(-z_1-z_2))$. Because $\wp(z)$ is an even function, and $\wp'(z)$ is an odd function, we get $((\wp(-z_1-z_2), -\wp'(-z_1-z_2))=(\wp(z_1+z_2),\wp'(z_1+z_2))$. And this is identical with our familiar definition on group law.

2. Complex theory of isogenies. Let $L_1, L_2$ be lattices in $\mathbb{C}$ with fundamental domain $E_1:=\mathbb{C}/L_2,E_2:=\mathbb{C}/L_2$, which are considered as elliptic curves defined over $\mathbb{C}$. Let $\alpha\in \mathbb{C}$ such that $\alpha L_1\subset L_2$, we define the isogeny $[\alpha]: E_1\to E_2$ by sending $z\mod L_1$ to $\alpha z\mod L_2$. This is a well-defined map, and is a homomorphism of groups.

If $\alpha\ne 0$, it can be seen that $\alpha L_1$ is a sublattice of $L_2$, i.e. a subgroup of rank 2 of $L_2$. This will yields the index $[L_2:\alpha L_1]$ is finite. In the case $\alpha\ne 0$, the degree of $[\alpha]$ is defined by the index $[L_2:\alpha L_1]$. If $\alpha=0$, we define the degree of $[\alpha]$ as 0.

Assume that $\alpha\ne 0$ and $n$ is the degree of the map $\alpha$, we have $n L_2\subset \alpha L_1$. This implies $\hat{\alpha} L_2\subset L_1$, where $\hat{\alpha} := (n/\alpha)$ and it will yields the isogeny $[\hat{\alpha}]$ from $E_2$ to $E_1$, which is called the dual of $[\alpha]$. We will prove that $\deg [\alpha] = \deg [\hat{\alpha}]$.

In fact, let $\{\omega_1,\omega_2\}$ is the basis for $L_1$, and $\{\omega_3,\omega_4\}$ is the basis for $L_2$. Because $\alpha L_1\subset L_2$, one can represent $\alpha \omega_1 = a\omega_3 + b\omega_4$, $\alpha \omega_2=c\omega_3 + d\omega_4$, with $a,b,c,d\in \mathbb{Z}$. Let us denote $A$ the $2\times 2$ matrix with the first row is $a,b$ and second row is $c,d$. Then the degree of $[\alpha]$ is $|\det A| = n$ Also, since $n/\alpha L_2\subset L_1$, we can also represent $n \omega_3 = e (\alpha\omega_1) + f (\alpha\omega_2) = (ea+fc)\omega_3 + (eb+fd)\omega_4$, and $n \omega_3 = g(\alpha \omega_1) + g(\alpha\omega_2) = (ga+hc)\omega_3 + (gb+hd)\omega_4$. If we denote $B$ the matrix with the first row $e,f$ and the second row $g, h$, then $\deg([\hat{\alpha}])=|\det B|$. Furthermore, it can be seen that $BA$ is the matrix send $(\omega_3, \omega_4)$ to $(n\omega_3, n\omega_4)$, i.e. $|\det B||\det A|=n^2$. This yields $\deg [\hat{\alpha}]=\deg \alpha$.

One can also easily deduce that $[\hat{\hat{\alpha}}]=\alpha$, and $[\alpha]\circ [\hat{\alpha}] = [\deg \alpha]$ is an isogeny from $E_1$ to $E_1$. Also, $[\hat{\alpha}]\circ [\alpha]=[\deg \alpha]=[\deg \hat{\alpha}]$ is an isogeny from $E_2$ to $E_2$. The kernel of $[\alpha]$, it is exactly $z\in E_1$ such that $\alpha z\in L_2$. 

Furthermore, we can deduce the complex version of Velu's formula.

Proposition 2.1. Let $G\subset E_1$ is a finite subgroup. Then there exists a lattice $L_2$ and an isogeny from $E_1\to E_2$ such that its kernel is $G$.

Proof. Due to the correspondence theorem of groups, there exists $L_2\subset \mathbb{C}$ such that $G=L_2/L_1$. If we denote $\#G= n$, then it can be seen that $L_1\subset L_2\subset (1/n)L_1$, i.e. $L_2$ is also a lattice. The map $E_1\to E_2$ sending $z\mod L_1$ to $z\mod L_2$ has the kernel $G$. (Q.E.D)

We can also obtain the complex version of the following statement

Proposition 2.2. Let $f$ be a holomorphic map between $E_1$ and $E_2$, then $f(z)$ is of the form $\alpha z+\beta$ for some $\alpha,\beta \in \mathbb{C}$, i.e. $f$ is a composition of an isogeny and a translation map.

Proof. We will use a bit theory of Riemann surfaces, because $\mathbb{C}$ is a universal covering of $E_1, E_2$, with projection maps $\pi_1, \pi_2$. The holomorphic map between $E_1$ and $E_2$ will induce the holomorphic map from $\tilde{f}: \mathbb{C}\to \mathbb{C}$ such that $\tilde{f}(z\mod L_1)=\tilde{f}(z)\mod L_2$, i.e. it makes the diagram commutes (the reader should draw it, I cannot draw it here). 

From this, for any $\omega\in L_1$, we have 

$$\tilde{f}(z+\omega) \equiv \tilde{f}(z+\omega)=\tilde{f}(z)\equiv \tilde{f}(z)\mod L_2$$

If we let $g(z) = \tilde{f}(z+\omega) - \tilde{f}(z)$, then $g(z)$ is a holomorphic function, and $g(z)$ takes values only on $L_2$, which is a discrete subset of $\mathbb{C}$. Hence, $g(z)$ is a constant function, then $0=g'(z) = \tilde{f}'(z +\omega)-\tilde{f}'(z)$, i.e. $\tilde{f}'(z) = \tilde{f}'(z+\omega)$, i.e. $\tilde{f}'$ is a bounded entire function. By Louville's theorem again, $\tilde{f}'(z)$ is a constant function. This yields, $\tilde{f}(z) = \alpha z + \beta$, for some $\alpha,\beta \in \mathbb{C}$. (Q.E.D)

3. Brief review on divisors. We will recall something about intersection numbers and divisors. Let $E$ be any projective curve in $\mathbb{P}^2$, the formal sum $D:=\sum_{P\in E}n_P [P]$, where $n_P\in \mathbb{Z}$ and $n_P\ne 0$ at finitely many points $P\in E$, is called a divisor on $E$. The set of all divisors on $E$ is denoted $Div(E)$. It can be seen that $Div(E)$ has the abelian group structure induced from $\mathbb{Z}$. The degree of $D$, denoted by $\deg D$ is defined as $\sum_{P\in E}n_P$.

If $E, F$ are projective curves in $\mathbb{P}^2$, without common irreducible components. it can be seen that $E\cap F$ is a finite set (since $E,F$ is of dimension 1, their intersection is of dimension 0, and hence, discrete and finite). At a point $P\in E$, we can define the intersection multiplicity (with $F$) at $P$, which is denoted $I_P(E,F)$, it is a non-negative integer, and is positive if $P\in E\cap F$. Furthermore, it has the following properties: $I_P(E,F)=I_P(F,E)$ and $I_P(E,FG) = I_P(E,F)+I_P(E,G)$, where we identify $FG$ with the zeros of $fg$, with $F, G$ are given by $f=0, g = 0$ resp.  From this, one obtains the Bezout's theorem:

$$\deg E \deg F = \sum_{P\in E\cap F} I_P(E,F)$$

If $E$ is a curve, and $f:=f_1/f_2$ is a rational function, where $f_1,f_2\in k[x_0,x_1,x_2]\setminus \{0\}$ are homogeneous polynomial of the same degree, such that $F_i$ (the curve defined by $f_i=0$) has no common component with $E$. We define

$$div(f) := \sum_{P\in E\cap F_1}I_P(E,F_1)[P] - \sum_{Q\in E\cap F_2}I_P(E,F_2)[Q]$$

Because $\deg F_1=\deg F_2$, due to Bezout's theorem, we get $\deg (div(f))=0$. If $D\in Div(E)$, and there exists $f=f_1/f_2$, where $f_1,f_2\in k[x_0,x_1,x_2]\setminus \{0\}$ are homogeneous polynomial of the same degree, such that $F_i$ (the curve defined by $f_i=0$) has no common component with $E$, such that $D = div(f)$, then $D$ is called a principal divisor. Due to the properties of the intersection numbers, we have $div(fg)=div(f)+div(g)$, i.e. the set of all principal divisors form a subgroup of $Div(E)$. We denote this group $Prin(E)$. If $D_1, D_2\in Div(E)$, and $D_1-D_2\in Prin(E)$, we denote $D_1\sim D_2$. It is obvious to see that $\sim$ is an equivalent relation.

It follows from above that any principal divisor has degree 0. Hence, $Prin(E)\subset Div^0(E)$, which is the subgroup of $Div(E)$ containing all degree zero divisors.The quotient group $Div^0(E)/Prin(E)$ is denoted $Pic^0(E)$.

If $E$ is an elliptic curve (we now view $E$ a projective curve), and $L$ is a line. There is 3 possibilities:

1. $L$ cuts $E$ at three distinct points
2. $L$ cuts $E$ at two distinct points, and the intersection multiplicity at a point is $2$, i.e. $L$ is a tangent line of $E$ at this point.
3. $L$ cuts $E$ only at 1 point, and the intersection multiplicity at this point is $3$. In this case, $L$ is also a tangent line of the curve at this point. For example, if we consider the Weierstrass form of $(E): y^2z=x^3+Axz^2+Bz^3$, then the line at infinity $(L): z=0$ cuts the curve only at $(0:1:0)$. Hence, $I_{(0:1:0)}((E), L)=3$. Another example is $(E): y^2z+yz^2=x^3$. The line $(L): y=0$ cuts the curve only at $(0:0:1)$, and $I_{(0:0:1)}(E, L)=3$.

Let $P,Q$ are two distinct points on $E$, the line $L_1$ given by equation $f_1=0$ connecting $P,Q$ will cut the curve at the third point $R$. If this line is the tangent line at $P$ (or $Q$), then $R$ can be considered as $P$ (or $Q$, resp.). If we take the line $L_2$ connecting $R$ and $-R$, given by equation $f_2=0$ then it will cut the curve at the third point $\infty$. So, in this case, by definition,

$$div(f_1/f_2)=[P]+[Q]+[R]-([R]+[-R]+[\infty])= [P]+[Q] - [-R] - [\infty]$$

This implies $[P]+[Q]\sim [-R]+[\infty]$. If $P\equiv Q$, we take the tangent line $L_1$ of $E$ at $P$, this will cut the curve at $R$ ($R\equiv P$ in the case the tangent line cuts the curve only at 1 point), we again take the line connecting $R$ and $-R$, this will cut the curve at $\infty$. We again get $div(f_1/f_2)=2[P] - [-R] - [\infty]$, i.e. $2[P]\sim [-R]+[\infty]$.

What we can see from this is that we can define the group law on $E$ as $P + Q = -R$. It is identical with our familiar definition for the group law on an elliptic curve, and it is more accurate, because in the case of tangent line of order 3, we are tricked by our geometric intuition. And we do not have to prove the associativity law, because it is natural induced from the group law on $Pic^0(E)$. Furthermore, one obtains the bijection (and hence, a group isomorphism) between $E$ and $Pic^0(E)$ given by $P\mapsto [P]-[\infty]$.

4. A quick look on algebraic theory of isogenies. Let $E_1,E_2$ are two elliptic curves over an algebraically closed field $k$. An isogeny is both a non-constant morphism and a group homomorphism from $E_1$ to $E_2$, i.e. if we look at affine parts of $E_1$ and $E_2$, an isogeny a rational map between $E_1$ and $E_2$ that sends $\infty$ to $\infty$. Let $\alpha: E_1\to E_2$ be an isogeny, it will induce the injective field homomorphism $\alpha^*: k(E_2)\to k(E_1)$. The field extension $k(E_1)$ of $\alpha^*(k(E_2))$ is finite, and the degree of $\alpha$ is defined as the degree of the field extension. If the field extension is separable, $\alpha$ is called separable isogeny. Otherwise, $\alpha$ is called inseparable.

 It follows from Chapter II of Washington's book that:

1. An isogeny is always surjective (We can use a little bit of algebraic geometry to deduce this fact: $E_1$ is projective variety, and hence, is complete, and $\alpha(E_1)$ is closed and irreducible in $E_2$ (since $E_1$ is irreducible), i.e. $\alpha(E_1)=E_2$, since $E_2$ is also irreducible and of dimension 1, and $\alpha$ is non-constant).

2. The kernel of an isogeny $\alpha$ is always finite, and $\# \ker \alpha =\deg \alpha$ if $\alpha$ is separable, i.e. $\deg \alpha$ is actually the number of points in a fiber $\alpha^{-1}(\infty)$. It can be proved that in the case $\alpha$ is an separable isogeny, the fiber of $\alpha$ at every point is equal, and it is $\deg \alpha$.

3. If $\alpha$ is non-separable, then $\deg \alpha >\#ker \alpha$. An example is the Frobenius endomorphism, its degree is $q$ (in the case $E$ are defined over $\mathbb{F}_q$), and it is bijection from $E(\overline{\mathbb{F}_q})$ to $E(\overline{\mathbb{F}_q})$, hence, the its kernel is just $\infty$.

So from 2 and 3, we can see that the kernel of an isogeny is always finite.

4. If $\alpha: E_1\to E_2$ is a morphism, then it induces the push-forward map $\alpha_*$ from $Div(E_1)\to Div(E_2)$ defined by $\alpha_*(\sum_{P\in E} n_P [P])=\sum_{P\in E} n_P [\alpha(P)]$. This can be seen that $\alpha_*$ is a group homomorphism. If furthermore, $\alpha(\infty)=\infty$, then it can be proved $\alpha_*$ maps principal divisors to principal divisors, i.e. the induced map $\alpha_*: Pic^0(E)\to Pic^0(E)$ is well-defined.

Using the isomorphism between $E$ and $Pic^0(E)$, we can prove a beautiful

Proposition 4.1. Let $E_1,E_2$ as above, and $\alpha:E_1\to E_2$ is a non-constant morphism, and $\alpha(\infty)=\infty$, then $\alpha$ is an isogeny.

Proof. It is sufficient to prove that $\alpha$ is a group homomorphism. From the isomorphism $\phi_i$ between $E_i$ and $Pic^0(E_i)$, which sends $P\to [P]-[\infty]$, and Remark 4 above, we have

$$\phi_2^{-1}\circ \alpha_*\circ \phi_1(P) = \phi_2^{-1}\circ \alpha_* ([P]-[\infty])=\phi_2^{-1}([\alpha(P)]-[\infty])=\alpha(P)$$

Because $\phi_i$ and $\alpha_*$ are group homomorphism, we have $\alpha$ is also a group homomorphism. (Q.E.D)

In section 3, we give the proof of the Velu's theorem for complex elliptic curves. It is the original version of Velu's theorem.

Proposition 4.2. Let $E_1$ be an elliptic curve defined over $k$, and $G\subset E_1$ a finite subgroup. Then there exists the curve $E_2$ and an isogeny $\alpha: E_1\to E_2$ such that $\ker(\alpha)=G$ and $E_2, \alpha$ can be computed explicitly via $G$ and the equation of $E_1$.

The existence of dual isogenies (for the separable case) can be deduced by the proposition above and the following suprising

Lemma 4.3. Let $E_1, E_2, E_3$ be three elliptic curves, with $\alpha_2: E_1\to E_2$ and $\alpha_3: E_1\to E_3$ are separable isogenies, and $\ker \alpha_2=\ker \alpha_3$. Then $E_2\cong E_3$, via the isomorphism $\beta$ such that $\alpha_3=\beta\circ\alpha_2$.

We now deduce the existence of dual isogeny for separable case (note that this also holds for the non-separable case).

Proposition 4.4. Let $\alpha: E_1\to E_2$ be a separable isogeny, then there exists an isogeny $\hat{\alpha}:E_2\to E_1$ such that $\hat{\alpha}\circ \alpha= [\deg \alpha]$.

Proof. Let $n:=\deg \alpha$, we will prove this proposition in the case $char(k)$ does not divide $n$.  Since $\alpha$ is separable, we have $\deg\alpha=\#\ker\alpha$. It follows $n=\#\ker\alpha$, and hence, $\ker\alpha\subset E_1[n]$, where $E_1[n]$ is the $n$-torsion subgroup of $E_1$. It can be seen then $\alpha(E_1[n])\cong E_1[n]/\ker\alpha$, i.e. $\#\alpha(E_1[n])=n$, since $E_1[n]\cong \mathbb{Z}_n\oplus \mathbb{Z}_n$ in the case $char(k)\nmid n$.

Due to Velu's formula, there exists a curve $E_3$ with an isogeny $\alpha_1$ from $E_2$ to $E_3$ such that $\ker\alpha_1 = \alpha(E_1[n])$. Hence, the composition map $\alpha_1\circ \alpha$ form $E_1$ to $E_3$ has the kernel $E_1[n]$. This is also the kernel of the map $[n]: E_1\to E_1$ sending $P$ to $nP$. Due to Lemma 4.3, we have an isomorphism $\beta$ between $E_3$ and $E_1$, such that $\beta\circ \alpha_1\circ \alpha=[n]$. Now, we just let $\hat{\alpha}:=\beta\circ \alpha_1$ (Q.E.D) 

[Vector Bundles II] Differential Forms and the Theorems of de Rham and Dolbeault on Riemann Surfaces

We will first begin with differential forms in Riemann surfaces then use the Poincare and Dolbeault's lemmas to obtain the fine resolutions for the locally constant sheaf $\mathbb{C}$, and $\Omega^1$. These help us easily deduce the theorems of de Rham for $d$-operator and of Dolbeault for $\overline{\partial}$-operator.

This note will become very long if we continue our discussion about differential forms on higher dimensional differentiable manifolds, and the generalizations of the two theorems above ( mainly based on the book of K. Kodaira "Complex Manifolds and Deformation of Complex Structures"). So, it will be mentioned later in our Part III (Our old Part III about Serre's duality should be changed to Part IV).

1. Differential forms on Riemann surfaces.

We will give a quick survey in this section. For further reference, one can take a look on the book of R. Miranda "Algebraic Curves and Riemann Surfaces".

Let $V_1\subset\mathbb{C}$ be an open subset with coordinate $z$, then a holomorphic 1-form on $V_1$ is of the form $\omega_1:=f(z)dz$, where $f$ is a holomorphic function on $U$. Assume that $V_2\subset \mathbb{C}$ is another subset with $V_1\cap V_2\ne \emptyset$, and $\omega_2:=g(w)dw$ be a holomorphic 1-form on $V_2$, then $\omega_1$ and $\omega_2$ is called compatible with each other if $f(z)dz=g(w)dw$ on $V_1\cap V_2$, i.e. in this case, we have $f(z)dz=g(w)dw$. This is equivalent to $f(z)\frac{\partial z}{\partial w}=g(w).$

Let $M$ be a 1-dimensional complex manifold (i.e. a Riemann surface), with charts $\phi_j: U_j\to V_j$. Then a holomorphic 1-form on $M$ is given by a collection of $\{\omega_j\}$, where $\omega_j$ is holomorphic 1-form on $V_j$, which are compatible with each other. Holomorphic $1$-form on $M$ forms a sheaf, which is often denoted by $\Omega^1$.

As we know in complex analysis. the Cauchy-Riemann (C-R) equation gives us the criterion when is a differentiable function $f(z)=f(x+iy)=u(x,y)+iv(z,y)$ on an open subset $U\subset\mathbb{C}$ is holomorphic. If we denote
$$f_x=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}, f_y=\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}$$
Then C-R's criterion lets us know that $f$ is holomorphic iff $f_x+if_y=0$. Using this, we will investigate the differential 1-form on a Riemann surface $M$.

A differential 1-form in an open subset $U\subset \mathbb{C}$ is given by the form $f(z)dx+g(z)dy$, where $f,g$ are differentiable functions on $U$, and $z=x+iy$. We see $\bar{z}=x-iy$, and $x=\frac{1}{2}(z+\bar{z}), y=\frac{1}{2i}(z-\bar{z})$. Then
$$\frac{\partial x}{\partial z}=\frac{\partial x}{\partial \bar{z}}=\frac{1}{2},\frac{\partial y}{\partial z}=-\frac{\partial y}{\partial \bar{z}}=\frac{1}{2i}$$
Hence, we can represent a differential 1-form on $U$ as $f(z,\bar{z})dz+g(z,\bar{z})d\bar{z}$. Then for any differentiable function $f$, we have
$$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial z}=\frac{1}{2}(f_x-if_y), \frac{\partial f}{\partial \bar{z}}=\frac{1}{2}(f_x+if_y)$$
Due to the C-R equation, one can see $f$ is holomorphic iff $\frac{\partial f}{\partial \bar{z}}=0$. For any differentiable function $f$ on $U$, we define three important operators
i. $\partial f=\frac{\partial f}{\partial z}$
ii. $\bar{\partial} f=\frac{\partial f}{\partial\bar{ z}}$
iii. $df = \partial f+\bar{\partial }f$

Remark 1.1. Due to what we have explained, a differentiable function $f$ on $U$ is holomorphic iff $\bar{\partial }f=0$.

From the definition of differentialbe 1-form on an open subset of $U$, we can generalize this to define differential 1-form on a Riemann surfaces $M$, which is a collection of differential 1-form on charts with compatible condition.

For a differential $1$-form $\omega$ on $M$, we say it is of type $(0,1)$ if it is locally represented by $g(z,\bar{z})d\bar{z}$. And it is of type $(1,0)$ if it is locally represented by $f(z,\bar{z})dz$. Because these types do not change under the compatible condition, they are well-defined definition. From this, we have four kinds of sheaves on a Riemann surface $M$
i. $\mathscr{E}$; the sheaf of differentiable functions on $M$.
ii. $\mathscr{E}^{1,0}$: the sheaf of differential 1-form on $M$ of type $(1,0)$.
iii. $\mathscr{E}^{0,1}$: the sheaf of differential 1-form on $M$ f type $(0,1)$.
iv. $\mathscr{E}^1$: the sheaf of differential $1$-form on $M$.

The natural question is that how the sheaf of type $(1,1)$ looks like? For this, we need to introduce the notions of differential $2$-form. A differential $2$-form on $M$ locally is represented by $fdz\wedge d\bar{z}$, where $f$ is a differentiable function on a chart of $M$, with $dz\wedge d\bar{z}=-d\bar{z}\wedge dz$, $dz\wedge dz=d\bar{z}\wedge d\bar{z}=0$.

Notation 1.2. We denote $\mathscr{E}^2$ the sheaf of differential $2$-form on $M$ (i.e. it is the sheaf of type $(1,1)$ on M). Also, because a Riemann surface is a 1-dimensional complex manifold, i.e. a two dimensional real manifold, for all $n\ge 3$, $\mathscr{E}^n$ is the actually the zero sheaf.

From a differential 1-form $\omega:=fdz+gd\bar{z}$, we can obtain a differential $2$-form by applying to $\omega$ one of three kinds of operators we have just defined above
i. $\partial \omega:= \frac{\partial g}{\partial z}dz\wedge d\bar{z}$
ii. $\bar{\partial}\omega:=\frac{\partial f}{\partial \bar{z}}d\bar{z}\wedge dz=-\frac{\partial f}{\partial \bar{z}}dz\wedge d\bar{z}$
iii. $d\omega:=\partial \omega +\bar{\partial}\omega=(\frac{\partial g}{\partial z}-\frac{\partial f}{\partial \bar{z}})dz\wedge d\bar{z}$

Remark 1.3. By Remark 1.1, $\bar{\partial}\omega=0$ iff $\omega$ is a holomorphic $1$-form.

Now, let $\alpha$ be a differential $2$-form (resp. $1$-form) on $M$, we say $\alpha$ is $d$-exact (resp. $\partial$-exact,$\bar{\partial}$-exact) if there exists a differential 1-form $\omega$ (resp. a differential function $\omega$) on $M$ such that $d\omega=\alpha$ (resp. $\partial\omega=\alpha, \bar{\partial}\omega=\alpha$). And a $1$-form $\omega$ is $d$-closed if $d\omega=0$ (similarly for $\partial$-closed, $\bar{\partial}$-closed). And one can see that any exact $1$-form is closed.

2. The lemmas of Poincare and Dolbeault and their consequences for short exact sequences of sheaves.

We will state without proof for the two lemmas, and deduce their consequences.

Proposition 2.1 (Poincare's Lemma). Let $\omega$ be a differential $1$-form on a neighbor hood of $p$ on a Riemann surface $M$ with $d\omega=0$, then there exists a neighborhood $U$ at $p$ and a differentiable function $f$ defined on $U$ such that $df=\omega$ on $U$.

If we denote $\mathscr{K}:=\ker: \mathscr{E}^1\xrightarrow{d} \mathscr{E}^2$ the kernel sheaf of the operator $d$ from $\mathscr{E}^1\to \mathscr{E}^2$. Then by the Poincare's lemma, the following exact sequence is exact of sheaves
$$\mathscr{E}\xrightarrow{d}\mathscr{K}\to 0$$
i.e. the sheaf map $d$ is onto. This is equivalent to say the $Im(d)=\ker:\mathscr{E}^1\xrightarrow{d} \mathscr{E}^2$, and hence, we obtain the following exact sequence\
$$\mathscr{E}\xrightarrow{d} \mathscr{E}^1\xrightarrow{d} \mathscr{E}^2\to 0$$
For the kernel of the first map, one can see that a $df=0$ locally at $p$ iff $f$ is locally constant function at $p$. If we denote $\mathbb{C}$-the locally constant sheaf on $M$, then we obtain the following exact sequence of sheaves
$$0\to \mathbb{C}\to \mathscr{E}\xrightarrow{d} \mathscr{E}^1\xrightarrow{d} \mathscr{E}^2\to 0$$
This exact sequence will play an important role in the proof of the de Rham's theorem.

Proposition 2.2 (Dolbeault's Lemma). Let $\omega$ be a differential 1-form defined on a neighborhood of a point $p$ on a Riemann surface $M$ of type $(0,1)$. Then on some neighborhoods $U$ of $p\in M$, there exists a differentiable function $f$ defined on $U$ such that $\bar{\partial }f=\omega$

The theorem is equivalent to the sheaf map $\mathscr{E}\xrightarrow{\bar{\partial}}\mathscr{E}^{0,1}$ is onto. And due to Remark 1.1. we have the following short exact sequences of sheaves
$$0\to \mathscr{O}\to \mathscr{E}\xrightarrow{\bar{\partial}} \mathscr{E}^{0,1}\to 0$$
Also, if we view $\mathscr{E}^2$ as $\mathscr{E}^{1,1}$, then the sheaf map $\bar{\partial }$ will map $\mathscr{E}^{1,0}$ to $\mathscr{E}^{1,1}$. And due to the Doulbeault's lemma, $\mathscr{E}^{1,0}\xrightarrow{\bar{\partial }}\mathscr{E}^{1,1}$ is onto. And for any differential form $\omega$ of type $(1,0)$, we have $\bar{\partial }\omega =0$ iff $\omega$ is the holomorphic $1$-form (Remark 1.3). We finally obtain the following short exact sequence of sheaves
$$0\to \Omega^1\to \mathscr{E}^{1,0}\xrightarrow{\bar{\partial}}\mathscr{E}^{1,1}\to 0$$
The two later short exact sequences play an important role in the proof of the Dolbeault's theorem.

3. Fine resolution of sheaves and the theorem of de Rham.

Definition 3.1. Let $\mathscr{F}$ be a sheaf on a differentiable manifold $M$. Then $\mathscr{F}$ is a fine sheaf if for any locally finite open covering $\{U_j\}$ of an open subset $U\subset M$, there is a family of homomorphism $h_j: \mathscr{F}(U_j)\to \mathscr{F}(U)$ such that:
i. $Supp(h_j)\subset U_j$
ii. $\sum_{j}h_j(s|_{U_j})=s$ for all $s\in \mathscr{F}(U)$

Theorem 3.9 in the book of Kodaira states that if $\mathscr{F}$ is a fine sheaf on $M$, then $H^q(M,\mathscr{F})=0$, for all $q\ge 1$. And in example 3.2, he shows that $\mathscr{E}, \mathscr{E}^{p,q}$ are fine sheaves, where $\mathscr{E}^{p,q}$ is the sheaf of differential $(p,q)$-form on $M$ (In the case $M$ is a Riemann surface, $(p,q)$ can be $(0,1),(1,0), (1,1)$).

Now, if $\mathscr{F}^0,...,\mathscr{F}^n$ be sheaves on a differentiable manifold $M$, where $\mathscr{F}^1,...,\mathscr{F}^n$ is fine, and there exists an exact sequence of sheaves
$$0\to \mathscr{F}^0\to \mathscr{F}^1\to...\to \mathscr{F}^n\to 0$$
then this exact sequence is called the fine resolution of $\mathscr{F}^0$. Back to our case, where $M$ is a Riemann surface, we have the fine resolution for the locally constant sheaf $\mathbb{C}$
$$0\to \mathbb{C}\to \mathscr{E}\xrightarrow{d} \mathscr{E}^1\xrightarrow{d}\mathscr{E}^2\to 0$$
This induces the following short exact sequences
$$0\to \mathbb{C}\to \mathscr{E}\xrightarrow{d} d\mathscr{E}\to 0$$
And
$$0\to d\mathscr{E}\to \mathscr{E}^1\to \mathscr{E}^2\to 0$$
The first short exact sequence yields the following long exact sequence (note that $\mathscr{E}$ is a fine sheaf)
$$0\to \mathbb{C}\to H^0(M,\mathscr{E})\xrightarrow{d} H^0(M,d\mathscr{E})\to H^1(M,\mathbb{C})\to 0\to H^1(M,d\mathscr{E})\to H^2(M,\mathbb{C})\to 0...$$
In particular, due to the ker-coker exact sequence, we have
$$H^1(M,\mathbb{C})\cong H^0(M,d\mathscr{E})/dH^0(M,\mathscr{E})$$
One can see that $H^0(M,d\mathscr{E})$ consists all $d$-closed $1$-form on $M$, and $dH^0(M,d\mathscr{E})$ consists of $d$-exact $1$-form on $M$. And hence, $H^0(M,\mathbb{C})$ measures how large the difference of these objects are. It is called the $0$-th de Rham's cohomology group. And also from the long exact sequence, for $n\ge 1$, we have $H^{n+1}(M,\mathbb{C})\cong H^n(M,d\mathscr{E})$.

Now, the second short exact sequence induces the following long exact sequence
$$0\to H^0(M,d\mathscr{E})\to H^0(M,\mathscr{E}^1)\xrightarrow{d} H^0(M,\mathscr{E}^2)\to H^1(M,d\mathscr{E})\to 0$$
And due to the ker-coker exact sequence, we have $H^1(M,d\mathscr{E})\cong H^0(M,\mathscr{E}^2)/dH^0(M,\mathscr{E}^1)$, and $H^n(M,\mathscr{E})$ vanishes for $n>1$. But due to the Poincare's lemma, we have $\mathscr{E}^2=d\mathscr{E}^1$, i.e. $H^1(M,d\mathscr{E})\cong H^0(M,d\mathscr{E}^1)/dH^0(M,\mathscr{E}^1)$, which is called the $1$-st de Rham cohomology group of $M$. We can see that, this quotient, again, is actually measure how large the difference $\{\text{closed 1-form on M}\}/\{\text{exact 1-form on M}\}$.

Combining things together, we have $H^1(M,\mathbb{C})\cong H^0(M,d\mathscr{E}^0)/dH^0(M,\mathscr{E}^0)$, and $H^2(M,\mathbb{C})\cong H^0(M,d\mathscr{E}^1)/dH^0(M,\mathscr{E}^1)$. And an important consequence is that the $k$-th de Rham cohomology group is independent on the differentiable structure of $M$, it is just dependent on the topological structure of $M$.

Theorem 3.2 (de Rham). Let $M$ be a differentiable manifold. We denote $\mathscr{E}^p$ the sheaf of differential $p$-forms on $M$. Then for all $n\ge 1$, we have 
$$H^n(M,\mathbb{C})\cong H^0(M,d\mathscr{E}^{p-1})/dH^0(M,\mathscr{E}^{p-1})$$
So, to compute the $k$-th de Rham's cohomology group, we need to compute $H^{k+1}(M,\mathbb{C})$. For example, if $M$ is a Riemann surface, and $M$ is simply connected, then $H^n(M,\mathbb{C})=0$ for $n\ge 1$, and in this case, the following de Rham's complex is exact.
$$0\to \mathscr{E}(M)\xrightarrow{d} \mathscr{E}^1(M)\xrightarrow{d} \mathscr{E}^2(M)\to 0$$

4. The theorem of Dolbeault.

Recall that in Section 2, we obtain the fine resolution for $\mathscr{O}$ and $\Omega^1$. We first consider the fine resolution for $\mathscr{O}$,
$$0\to \mathscr{O}\to \mathscr{E}\xrightarrow{\bar{\partial}} \mathscr{E}^{0,1}\to 0$$
It then induces the following long exact sequence
$$0\to \mathscr{O}(M)\to \mathscr{E}(M)\xrightarrow{\bar{\partial}} \mathscr{E}^{0,1}(M)\to H^1(M,\mathscr{O})\to 0$$
Due to the ker-coker exact sequence, we have
$$H^1(M,\mathscr{O})\cong coker(\bar{\partial})=\mathscr{E}^{0,1}(M)/\bar{\partial }\mathscr{E}(M)$$
One can see that $\mathscr{E}^{0,1}(M)$ consists of all $\bar{\partial }-closed $ $(0,1)$-form, and $\bar{\partial }\mathscr{E}(M)$ consists of all exact $(0,1)$-form of $M$. Hence $H^1(M,\mathscr{O})$ measure how far for a $\bar{\partial }$-closed $(0,1)$-form to be exact. If we consider $\mathscr{O}$ as $\Omega^0$, and $\mathscr{E}(M)$ as $\mathscr{E}^{0,0}(M)$, we then have
$$H^1(M,\Omega^0)\cong \mathscr{E}^{0,1}(M)/\bar{\partial }\mathscr{E}^{0,0}(M)$$
Now, let us investigate the fine resolution of $\Omega^1$. Recall that we have the following exact sequence of sheaves
$$0\to \Omega^1\to \mathscr{E}^{1,0}\xrightarrow{\bar{\partial}}\mathscr{E}^{1,1}\to 0$$
This induces the long exact sequence below
$$0\to \Omega^1(M)\to \mathscr{E}^{1,0}(M)\xrightarrow{\bar{\partial }}\mathscr{E}^{1,1}(M)\to H^1(M,\Omega^1)\to 0$$
Again, due to the ker-coker exact sequence, we have
$$H^1(M,\Omega^1)\cong \mathscr{E}^{1,1}(M)/\bar{\partial }\mathscr{E}^{1,0}(M)$$
We finally obtain the theorem of Dolbeault.

Theorem 4.1 (Dolbeault). Let $M$ be a differentiable manifold, and $\mathscr{E}^{p,q}$ the sheaf of $(p,q)$-form on $M$. Then we have $H^q(M,\mathscr{E}^p)\cong H^0(M,\bar{\partial }\mathscr{E}^{p,q})/\bar{\partial }H^0(M,\mathscr{E}^{p,q-1})$.

The differential form on higher dimensional complex manifolds as well as the two theorems in this note will be discussed in our next part.

What do you know about square roots?

In this note, we will make fun with square root functions on $\mathbb{C}$. Let $z\in\mathbb{C}$ be a complex number, we can represent $z=re^{i\theta}$. The square root of $z$ is $f(z):=r^{1/2}e^{i\theta/2}$. We now move $z$ along the circle centered at $0$ with radius $r$. Let $z_1$ be any point on this circle, then $z_1=re^{i\theta_1}$, and $f(z_1)=r^{1/2}e^{i\theta_1/2}$. When $\theta_1=\theta+2\pi$, we have $z_1$ is exactly $z$, but then $f(z_1)=r^{1/2}e^{i(\theta+2\pi)/2}\ne f(z)$. What does this means? It means that $f$ is not a well-defined function. However, if we choose $z\ne 0$, and choose $r$ such that the circle centered at $z$, with radius $r$ does not contain $0$, then by the same process, moving $z$ around this circle, we will get $f(z)$ in this case is defined! So what is wrong here?

The reason for the problem we get is that $0$ is the branch point of the square root function, and if we remove $0$, i.e. consider the map $f:\mathbb{C}^*\to \mathbb{C}^*$ sending $re^{i\theta}$ to $r^{1/2}e^{i\theta/2}$. It is a well-defined holomorphic map. We will prove that such $f$ cannot be extended to $\mathbb{C}$.

Proposition 1. There exists no square root function, which is continuous on $\mathbb{C}$.

Proof. Assume that such function $f$ exists, then actually $f$ is the inverse of the map $g:\mathbb{C}\to \mathbb{C}$ sending $z$ to $z^2$. Due to the inverse function theorem, $f'(z)\ne 0$ iff $z\ne 0$, and hence, for any $z\ne 0$, there exists $U_z, U_{z^2}$ are neighborhoods of $z$ and $z^2$, respectively, then $f|U_z=U_{z^2}$, and $U_z$ is homeomorphic with $U_{z^2}$, and the inverse map $f$ of $g$ is holomorphic in $\mathbb{C}^*$. If $f$ can be extended to $0$, by the assumption, $f$ is continuous at $0$, and now, by the Riemann's removable singularities theorem, we can extend $f$ to $\mathbb{C}$. But then $z=f(z)^2$, taking the derivative both sides at $z=0$, one gets $1=2f(0)f'(0)=0$, a contradiction. Hence, the square root function cannot be extended to the whole complex plane. (Q.E.D)

     This reflects partly our desire, when we want to extend the domain of the holomorphic function to a larger domain. It is a very important process, for example, look at my previous note about the Riemann's zeta function, and see how its analytic continuation play a key role in proving the existence of infinite prime numbers, or computing some integrals. But in this note, we just focus on the simpler function, the square root. 

     Before doing this, we represent a common method to analytically continue a convergent power series $f_0$ on an open disk $D(z_0,r)$, where $D(z_0,r)$ is the open disk centered at $z_0$ with radius $r$. Pick any $z_1$ in this disk, we represent the series by changing the center, from $z_0$ to $z_1$. We then get a new series $f_1$, centered at $z_1$ with the radius of convergent is at least $r-|z_1-z_0|$ (draw a picture, then you can see this). But if we are lucky, the radius of convergence $r'$ of a new series $f_1$ is bigger than $r-|z_1-z_0|$, then we can analytic continue $f$ from a domain $D_0:=D(z_0,r)$ to a new domain $D_1:=D(z_1,r')$. And by doing that as far as possible, we obtain a chain $(f_0,D_0),...,(f_n, D_n)$, with $f_i|{D_i\cap D_j}=f_j|{D_j\cap D_i}$, i.e. they are the same on the intersection. One often do this on a path from $z_0$ to $z_n$, and $z_i$ are in this path. 

But we still hope that there exists a function $f$ on $D:=\cup_{i=1}^n D_i$ such that $f$ is the analytic continuation of all $f_i$, i.e. $f$ is a holomorphic map, and $f|D_i=f_i$. Also, another question arises, if we analytically continue $(f_0, D_0)$ along a path $\gamma_0$ from $z_0$ to $z_n$ to obtain $(f_n, D_n)$, and then analytically continue $(f_0, D_0)$ along another path $\gamma_1$ from $z_0$ to $z_n$ to obtain $(f_n', D_n')$, then at a small neighborhood $U\subset D_n\cap D_n'$ of $z_n$, can we have $f_n|U=f'_n|U$? If this happens, by the famous identity theorem, one can conclude that $f_n$ and $f'_n$ is the same in $D_n\cap D_n'$. The answer of these questions are known under the name Monodromy Theorem.

Theorem 2. Let $G\subset \mathbb{C}$ be a region, $P,Q$ are two points of $G$. Let $f_0$ be a analytic function in $D_0:=D(P,r)\subset G$. Let $\gamma_0, \gamma_1$ is two path from $P$ to $Q$, such that $(f_0, D_0)$ can be analytically continued via $\gamma_0$ to $(f, D)$, where $D\subset G$ is an open neighborhood of $Q$, and $f$ is holomorphic in $D$.  Also, $(f_0,D_0)$ can be analytically continued via $\gamma_1$ to $(D',f')$, where $D'\subset G$ is an open neighborhood of $Q$, and $f'$ is holomorphic in $D'$. Then $f|D\cap D'=f'|D\cap D'$ if $\gamma_0$ is homotopic with $\gamma_1$ in $G$. In the case $G$ is simply connected, and $(f_0,D_0)$ admits the analytic continuation along any path from $P$ to any point $D\in G$, then there exists a unique $F$ holomorphic in $G$ such that $F|D_0=f_0$.

As an example, we will do analytic continuation and then detect why the condition of simply connected domain is necessary via the square root function. For any $\alpha\in\mathbb{R}$, we can define 

$${\alpha\choose k}:=\frac{\alpha(\alpha-1)...(\alpha-(k-1))}{k!}$$

And one can easily that this extends our familiar ${n\choose k}$, where $n$ is a natural number. Via this, we can define the binomial series for $z^{1/2}$ centered at $1$ as follow. One can see

$$f(z):=z^{1/2}=(z-1+1)^{1/2}=\sum_{k\ge 0}{1/2\choose k}(z-1)^k$$

And by the ratio test, this series has radius of convergence $1$. That means, for any $z_1\in D(1,1)$, where $D(1,1)$ is the open disk center at $1$ with radius 1, we have

$$f(z_1)=\sum_{k\ge 0}{1/2\choose k}(z_1-1)$$

Now, we want to represent the series with center $z_1$, and compute its radius of convergence. We first define if $z_1\in\mathbb{C}, z_1\ne 0$, and $z_1=re^{i\theta}$, then $\sqrt{z_1}:=re^{i\theta/2}$, i.e. $\sqrt{z_1}=f(z_1)$ .  From this, if $z_1\in D(1,1)$, then for any point $z\in D(z_1, 1-|z_1-1|)\subset D(1,1)$, we have $\frac{z}{z_1}<1$, and we can represent the series $\sqrt(z)$ centered at $z_1$ as follows.

$$\sqrt{z}=\sqrt{z_1}\sqrt{\frac{z}{z_1}}=\sqrt{z_1}\sum_{k\ge 0}{1/2\choose k}(\frac{z}{z_1}-1)^k=\sqrt{z_1}\sum_{k\ge 0}{1/2\choose k}z_1^{-k}(z-z_1)^k$$

By ratio test again, this series has radius of convergence $1$. In particular, if we choose $z_1=e^{i\pi/4}$, which lies in $D(1,1)$, and lies on $S^1$, we have

$$\sqrt{e^{i\pi/4}}=e^{i\pi/8}\sum_{k\ge 0}{1/2\choose k}e^{-k\pi i/4}(z-e^{i\pi/4})^k$$

And because this series has radius of convergence 1, and $i\in D_1:=D(e^{\pi i/4},1)$ but $i\notin D(1,1)$, we now successfully analytic continue $f$ from the domain $D_0 := D(1,1)$ to the domain $D(e^{\pi i/4},1)$. Continue this process, with the point $i\in D(e^{\pi i/4},1)$ we obtain the new domain $D_2:=D(i,1)$. And respectively, our domain will be $D_3:=D(e^{3\pi i/4},1)$, $D_4:=D(-1,1)$, $D_5:=D(e^{5\pi i/4},1)$, $D_6:=D(e^{6\pi i/4, 1})$, $D_7:=D(e^{7\pi i/4},1)$. And we finally obtain again $D_8:=D(1,1)$ by doing analytic continuation on $D(e^{7\pi i/4},1)$. Let us denote $f_0:=f$, and for $i>0$, $f_i$ is the analytic continuation of $f_{i-1}$ to $D_i$. What we have just done is that we are doing analytic continuation along the circle centered at 0, with radius 1.

This finishes our example for doing analytic continuation by using power series along a path. Now, we turn to the failure of the Monodromy Theorem in this case. When we back to $(f_8, D_8)$ from $(f_0, D_0)$, we hope that $f_8(z) = f_0(z)$ for all $z\in D_0$. Unfortunately, by using the same series for computing $e^{i\pi/4}$, we have

$$f_8(z)=e^{\pi i/2}\sum_{k\ge 0}(e^{\pi i})^k{1/2\choose k}(z-e^{\pi i})^k=i\sum_{k\ge 0}(-1)^k{1/2\choose k}(z-(-1))^k=$$

$$=i\sum_{k\ge 0}{1/2\choose k}(-z-1)^k=i\sqrt{-z}=i\sqrt{-1}\sqrt{z}=-\sqrt{z}=-f_0(z)$$

So, what is wrong here? Actually, the reason is that $\mathbb{C}^*$ is not simply connected (can you prove that? As an exercise, you may wanna prove that $\pi_1(\mathbb{C}^*)\cong \mathbb{Z}$, i.e. the fundamental group is not trivial). Hence, the Monodromy Theorem does not hold in this case.