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Results tagged with complex-geometry
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user 18060
Complex geometry is the study of complex manifolds, complex algebraic varieties, complex analytic spaces, and, by extension, of almost complex structures. It is a part of differential geometry, algebraic geometry and analytic geometry.
4
votes
Complex manifold mapping to a Riemann surface it contains
As requested, an example where there is no retract:
Take $E$ an elliptic curve, and $C$ a smooth curve of genus $\geq 2$ in $E \times E$. Take $D$ a ramified double cover of $C$, and let $M$ be the qu …
- 149k
1
vote
Semistability and normal crossing divisors
I don't know what you mean by "the same time". Given any smooth surface any any divisor, we can always blow up enough times to make that divisor normal crossings. (I think this is discussed in Hartsho …
- 149k
1
vote
Ramification divisor and degenerate locus of jacobian
Given an effective divisor $D$, how do we associate a subscheme to it. If it is an irreducible divisor, then it corresponds to an ideal $I$ that is locally principal, and the subscheme corresponding t …
- 149k
3
votes
Accepted
First Chern class of canonical bundle ?
Yes. This is true for every vector bundle. By functorialuty, it is sufficient to check on just the infinite Grassmanian. But its integral cohomology is torsion-free, so Chern-Weil works.
- 149k
2
votes
Accepted
Criteria for a coherent sheaf pushing forward from the universal cover
Let $E$ be the elliptic curve. Let $E_1$ and $E_2$ be two different double covers of $E$, with $E = E_1 /x_1$ and $E=E_2/x_2$ for two-torsion points $x_1,x_2$.
Let $M$ be the minimal resolution of …
- 149k
7
votes
Accepted
How to properly verify that $E\times E'$ has no non-trivial effective divisors with Kodaira ...
If $D$ has Iitaka dimension zero, then $\dim H^0 ( n E) =1$ for all $n$, because if it were any larger than $\dim H^0(k ne) \geq k+1$.
If we have an automorphism $\sigma$ with $[ n \sigma (D)] = [n D …
- 149k
7
votes
Accepted
Existence of a real structure on the tangent bundle of a complex manifold
For a vector bundle to have real structure, it must be isomorphic to its complex conjugate, hence it's $i$th Chern class must be equal to its own negation (i.e. $2$-torsion) for all odd $i$. So $2c_1, …
- 149k
2
votes
When is bijective map between closed point of varieties a morphism?
It's important to distinguish two things. One is what it takes to uniquely specify a morphism, and the other is what it takes to construct a morphism.
When we say a morphism is uniquely specified by t …
- 149k
6
votes
Accepted
isotrivial elliptic fibration and kodaira's table of singular fiber
I think one ususally uses isotrivial in this case to mean that the smooth fibers are all isomorphic. Then it is certainly the case that there can be singular fibers.
Which singular fibers can appear …
- 149k
4
votes
Accepted
A simple question about a statement of Kähler Manifold and Moishezon Manifold
The answer probably depends a lot on how exactly you define a Kähler manifold as there are multiple equivalent definitions. In any case, the definition of a positive line bundle is that it has a metri …
- 149k
3
votes
Holomorphic maps into a symmetric product of Riemann surface
As SashaP already pointed out, this is false. In fact it fails very badly: The space of non-constant holomorphic maps from $X$ to $\operatorname{Sym}^2 Y$ can already contain infinitely many connected …
- 149k
10
votes
Accepted
Status of Hodge conjecture over subrings of $\mathbb{C}$
The $k$-Hodge conjecture is false for any $k$ containing an irrational real $\alpha$.
Indeed we may clearly assume $\alpha$ is negative. Consider the elliptic curve $E = \mathbb C / \langle 1, \sqrt{ …
3
votes
Accepted
Embeddings of of quotient singularities
Yes. By Hilbert's theorem, the ring of $G$-invariant functions on $\mathbb C^n$ is finitely generated. Say the number of generators is $N$, then this gives a $G$-invariant holomorphic map to $\mathbb …
- 149k
6
votes
Moduli Spaces of Higher Dimensional Complex Tori
A complex torus of dimension $d$ can be written as a quotient $\mathbb C^d/\mathbb Z^{2d}$. Thus it is determined by a map $\mathbb Z^{2d} \to \mathbb C^{d} \cong \mathbb R^{2d}$. We can specify this …
- 149k
2
votes
Accepted
Is the composition of a finite branched cover and a non-isotrivial Riemann surface bundle st...
If $E'$ is isotrivial, then the fibers $E'\to B$ are all isomorphic to a single fiber, so the Jacobians of every fiber are isomorphic to a single Jacobian abelian variety $A$. Then the Jacobian of eve …
- 149k