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Results tagged with complex-geometry
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user 40297
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.
5
votes
Accepted
semiample of canonical bundle in a smooth family (Campana's proof)
Let $f:X\rightarrow \Delta $ be your family. $\ (*)$ implies that $f_*K_{X/\Delta }^{N}$ is a vector bundle on $\Delta $, with fiber $H^0(X_t, K_{X_t}^N)$ at $t\in\Delta$. The canonical homomorphism $ …
- 38k
7
votes
Accepted
Roots of line bundles in a family
The locus you consider is either empty, or equal to $B$.
This can be seen as follows. Line bundles on a fiber $F$ of $\pi $ are parameterized by $H^1(F,\mathscr{O}^*_F)$. This group fits into an exact …
- 38k
7
votes
Fixed-point free holomorphic involutions
For $n\geq 4$, a smooth hypersurface $X\subset\mathbb{P}^n$ never admits a fixed point free involution. By the Lefschetz theorem $\operatorname{Pic}(X) $ is cyclic, so any automorphism of $X$ preserv …
- 38k
7
votes
Accepted
Difference between stabilizer and automorphism group of subvariety of an abelian variety
They have absolutely no reason to be equal. Consider the case where $A$ is the Jacobian of a genus 2 curve $C$, and $X=C$ embedded in $A$ by $x\mapsto [x]-[p]$ for some fixed point $p\in C$. Then $X$ …
- 38k
6
votes
Accepted
Projective variety of general type such that $S^m \Omega_X^1$ is globally generated
If $X$ is a surface it is true. In general, a smooth projective variety with $S^m\Omega ^1_X$ globally generated does not contain any smooth rational curve $C$. Indeed $\Omega ^1_C$ is a quotient of …
- 38k
1
vote
Accepted
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 …
- 38k
8
votes
Accepted
Deformation invariance of Chern classes
This is actually true for all Chern classes, but you must first say how you identify $H^*(X,\mathbb{Z})$ and $H^*(X_t,\mathbb{Z})$. There is no problem for small deformations, that is if $B$ is a ball …
- 38k
4
votes
Accepted
Short exact sequence of trivial holomorphic line bundles not splitting
For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by the surjection $p:L\oplus L\rightarrow L$ must be nonzero on one of the summands, say the first one; hen …
- 38k
7
votes
Accepted
Branched covers of the sphere branched over few points
Let me post my comment as an answer. Take a Weierstrass point on $X$, that is, a point $P$ for which there exists a meromorphic function $f$ with a pole of order $k\leq g$ at $P$ (there always exist …
- 38k
10
votes
Accepted
Existence of holomorphic retraction
No, this is actually very rare. Indeed the existence of such a retraction implies that the exact sequence $$0\rightarrow T_X\rightarrow T_{M|X}\rightarrow N_{X/M}\rightarrow 0$$ splits. In particular, …
- 38k
3
votes
Accepted
Fixed locus of a Kahler $S^1$-action
It doesn't give a trivialization (think of the trivial action...), but indeed a splitting. For $y\in F_{\alpha }$, $S^1$ acts on $T_y(M)$; denote by $t_y$ the action of an element $t\in S^{1}$. The …
- 38k
4
votes
Accepted
Cup Product with Ample Line Bundles
No. If $\alpha $ is of type $(n-2,n)$, its product with any class of type $(1,1)$ is zero, but $\alpha $ is not necessarily zero (you can take for instance $\alpha = c_1(L)^{n-2}[\omega ]$, where $L$ …
- 38k
9
votes
Accepted
Holomorphic structures for line bundles over projective manifolds
The group of isomorphism classes of complex line bundles on $M$ is isomorphic to $H^2(M,\mathbb{Z})$, via the map $L\mapsto c_1(L)$. The analogous group $\operatorname{Pic}(M) $ for holomorphic line b …
- 38k
7
votes
Accepted
Maps between grassmannians with inclusion property
I think there is no holomorphic such map. Consider the incidence variety $Z=\{(p,\ell)\in \mathbb{P}^3\times \mathbb{G}(2,4)\,|\, x\in\ell\} $. The projection $p:Z\rightarrow \mathbb{P}^{3}$ is a $\m …
- 38k
3
votes
Stable vector bundle and Hitchin map
There is one important case where $H$ is surjective, namely when $E$ is very stable. This means, by definition, that any nilpotent homomorphism $E\rightarrow E\otimes K$ is zero; or, equivalently, tha …
- 38k