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Riemann surfaces(Riemannian surfaces) is one dimensional complex manifold. For questions about classical examples in complex analysis, complex geometry, surface topology.

5 votes
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Existence of a holomorphic map between Riemann surfaces

I think this is what Picard says, translated in modern algebraic geometry language: You have a curve $X$ with a map $\pi :X\rightarrow \mathbb{P}^1$. Assume for simplicity that for each $z\in \mathbb{ …
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3 votes
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Jacobians of twisted coverings

They are not isogeneous in general. For an explicit example, take the case $n=0$ and $C$ hyperelliptic, so that we have a double covering $C\rightarrow \mathbb{P}^1$ branched along a subset $B$ of $\m …
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2 votes

Finding an algebraic equation given divisors

If I understand correctly your question (?), you want a smooth curve $C$ of genus 5 and 3 holomorphic forms $\omega _i$ with the divisors you have written down, satisfying $\ (\omega_1^2 - \omega_2^2 …
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4 votes

polynomial branched cover of the sphere with specified monodromy

Easy in your example : the covering curve must be $\Bbb{P}^1$ (by Hurwitz formula), with an action of $(\Bbb{Z}/2)^2$. Up to conjugacy there is no choice for such an action, it must be given by the in …
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15 votes

Area of a smooth complex projective curve

A much more general result is given by Mumford, in Projective varieties I, Theorem 5.22: the volume of any $r$-dimensional smooth projective variety in $\Bbb{P}^N$ (the area in your case) is its deg …
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5 votes

Rational functions on hyperelliptic Riemann surface

Yes (the answer was given, then deleted, by Francesco Polizzi). If $D$ and $D'$ are the divisors of zeroes (resp. poles) of a rational function, the linear system $|D|$ has dimension $r\geq 1$ and is …
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14 votes
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Quotients of curves of genus $4$ by a free $\mathbb{Z}/ 3 \mathbb{Z}$-action

Yes. Start from a genus 2 curve $C_2$, and choose a point of order 3 in $JC_2$, giving rise to an étale $\mathbb{Z}/3$-covering $C_4\rightarrow C_2$. Then $C_4$ cannot be hyperelliptic: a $g^1_2$ on …
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7 votes
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Image of the map induced on homology by a covering

No, this is already false if $\pi $ is a Galois covering (i.e. $Y\cong X/G$): the index is the order of the abelianized group $G_{ab}$. Indeed from the exact sequence $$\pi _1(X)\rightarrow \pi _1(Y)\ …
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10 votes
Accepted

Stable extensions by line bundles on Riemann surfaces

This never happens. Pick a point $p\in X$; the exact sequence $0\rightarrow L^{2}\rightarrow L^{2}(p)\rightarrow \mathbb{C}_p\rightarrow 0$ gives rise to an exact sequence $0\rightarrow \mathbb{C}\x …
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1 vote
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operations on matrices preserving the property of being the Riemann matrix of a surface

No, this is not true. If $C$ is a general curve of genus $\geq 4$ with period matrix $\Omega$ and $p$ is a prime, $p\Omega$ is not the period matrix of a curve. This is proved (in an equivalent, more …
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1 vote

Any no-zero homomorphism of holomorphic vector bundles over a compact Riemann surface factor...

Let $V_2$ be the image of $f:V\rightarrow W$; it is a subsheaf of $W$, hence locally free. Let $W_2$ be the quotient of $W/V_2$ by its torsion subsheaf, and let $W_1$ be the kernel of the projection …
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3 votes
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Symplectic representation of modular group

For your last question: the map from the hyperelliptic modular group to $\operatorname{Sp}(2g,\mathbb{Z}) $ is not surjective as soon as $g\geq 3$. This was proved by V. Arnold, A remark on the branc …
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4 votes
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Extended Abel-Jacobi theorem

This is true. Given two points $a,b\in X$ with $u(a)$ and $u(b)\neq 0$, one can choose a path from $a$ to $b$ and a determination of $\log u$ along that path, so that $\exp\int^b_a d\log u=u(b)/u(a)$. …
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4 votes

Paths $tg_1+(1-t)g_0$ in the moduli space of Riemann surfaces

Answering your PS: as you point out, the complex structure $J_t$ is given in each coordinate chart by a matrix which depends real-analytically on $t$. Now you can find a neighborhood $U$ of $[0,1]$ i …
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6 votes
Accepted

Moduli of stable bundles - analytic approach

$\mathcal{M}$ is what is called a coarse moduli space. In concrete terms, this means the following: 1) As a set, $\mathcal{M}$ can be viewed as the set of isomorphism classes of stable bundles (of r …
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