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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.
12
votes
To differently gluing of two Riemann surfaces with boundary we get different surfaces
Here is a particularly simple example. We will take a cylinder and glue its ends together in two different ways. The cylinder will be
$$C:=\{ z: 0 \leq \mathrm{Im}(z) \leq 1 \} / \mathbb{Z},$$
where …
12
votes
Problem in Rick Miranda: finding genus of a Projective curve
Here is the most algebraic way I can see to compute this. Let $Q_1$ and $Q_2$ be two quadratic polynomials in four variables. Let $R$ be the graded ring $k[x_1, x_2, x_3, x_4]/(Q_1, Q_2)$. Let $V_d$ b …
9
votes
If Spec Z is like a Riemann surface, what's the analogue of integration along a contour?
I think that you have understood the analogy correctly, and you have pinpointed one of its weaknesses. Although number fields are like one dimensional functional fields in many ways, one of the differ …
3
votes
Accepted
Image of boundary circle under map from punctured elliptic curve to ℂ
I don't know if this is the kind of answer which will satisfy. Write $f$ for the function on $\mathbb{C} - \Lambda$ and $z$ for the coordinate on $\mathbb{C}$. Write $D^{\ast}$ for the punctured disc. …
3
votes
"Physical" construction of nonconstant meromorphic functions on compact Riemann surfaces?
The answers to this previous question are relevant. To fill in some gaps, that question is about building functions on a disc with specified Laplacian. As Tim Perutz says, you want to generalize that …
10
votes
Is this lattice in the Tate module of an elliptic curve, coming from complex-analytic unifor...
Any construction along these lines is going to run into an obstruction pointed out by Serre. Consider the elliptic curve $E = \{ y^2 = x^3+x \}$ over $\mathbb{Z}[i]$, and let $p$ be a prime which is $ …
12
votes
Automorphisms of genus 6 surfaces
In case you don't know the general context: There is a curve of genus $g$ with endomorphism group contained in $G$ if and only if $G$ can be generated by elements $g_1$, $g_2$, ..., $g_k$ with orders …
3
votes
A question about Abel-Jacobi map
Every fiber of $A$ is a projective space. If $V = H^0(X, \mathcal{O}(D))$, then $A^{-1}(A(D))$ is naturally identified with $\mathbb{P}(V)$. A divisor $E$ in $A^{-1}(A(D))$ is the zero locus of a nonz …
3
votes
Is there an algorithm to compute efficiently the dessin d'enfant from a Belyi pair?
I'd look into numerical homotopy software, such as Bertini or PHCPack. Numerical homotopy software attempts to solve problems of the following sort: Suppose we have a family of polynomial equations $f …
16
votes
Accepted
Analogy between the nodal cubic curve $y^2=x^3+x^2$ and the ring $\mathbb{Z}[\sqrt{-3}]$?
I think this will be a needlessly confusing example. In algebraic geometry over an algebraically closed field, there are two basic examples of nonnormal curves: the node and the cusp. Explicit equatio …
5
votes
Dolbeault cohomology
I like "Hodge Theory and Complex Algebraic Geometry" by Voisin. The focus is on the Kahler case, but the early explanations of Dolbeaut cohomology are for all complex manifolds.
7
votes
$\partial \bar{\partial}$ on a riemann surface
In a comment above, marco asks whether this is true for larger $n$: That is to say, $M$ a complex $n$-fold, $R$ a totally real sub-real-$n$-fold and $\alpha$ a $(1,1)$-form on $R$. The answer is no fo …
6
votes
Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?
The answer is no. This is a corrected version of Nicolast's comment.
Let $E$ be an elliptic curve, let $f: E \to E$ be an endomorphism and let $H_1(f) : H_1(E) \to H_1(E)$ be the induced map on $H_1$. …
5
votes
Accepted
Finite orbits on an elliptic curve with two generic involutions
No. Letting $\sigma$ and $\tau$ denote the two involutions, $\sigma \circ \tau$ is a translation by an element of $\mathrm{Pic}^0(C)$. In general, this translation will not be torsion, so its orbit th …
8
votes
Accepted
Deep/precise relationship between two approaches to FLT for polynomials, $n = 3$
This answer is basically a longer version of Felipe Voloch's, but maybe it will be useful. Both proofs take a class in $H^1(E)$, pull it back to $H^1(\mathbb{P}^1)$ and note that $H^1(\mathbb{P}^1)$ i …