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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.
16
votes
Morphism between projective varieties
A morphism $X \to Y$ factorizes as an embedding $X \to X\times Y$ (the graph of $f$) followed by the projection $X\times Y \to Y$. The first is a closed embedding (and hence is projective) if $Y$ is s …
15
votes
Accepted
Does a resolution of a rational singularity have rationally connected fibers?
No. For instance the cone over an Enriques surface (with respect to any projective embedding) has rational singularity, but Enriques surface is not rationally connected.
15
votes
Accepted
Atiyah class for non-locally free sheaf
It is better to define the Atiyah class as an element of $Ext^1(E,E\otimes\Omega^1)$. Then it is defined for all coherent sheaves, and even for all objects of the derived category. The most convenient …
12
votes
A tale of two maps into a Grassmannian
I guess, when you say "in the sense of Hartshorne" you mean the projective spectrum of $\oplus S^kE$.
Yes, the morphisms are the same, and to see this just note that there is a natural (relative Veron …
12
votes
Accepted
Do non-projective K3 surfaces have rational curves?
Some of them do, and some don't.
Indeed, by global Torelli theorem, there is a K3 surface $X$ with $\mathrm{Pic}(X) = 0$. Such $X$ has no curves, in particular no rational curves.
On the other hand, t …
11
votes
Accepted
Stably trivial non-trivial vector bundles
Assume $E \oplus \mathcal{O} \cong \mathcal{O}^{\oplus n}$. Then, of course, $E \cong \mathcal{O}^{\oplus n}/\mathcal{O}$. On the other hand
$$
Hom(\mathcal{O}, \mathcal{O}^{\oplus n}) \cong \Gamma(X, …
11
votes
Accepted
Is the symmetric product of an abelian variety a CY variety?
When $\dim A = 1$, $S^nA$ is a $\mathbb{P}^{n-1}$-bundle over $A$, so its Kodaira dimension is $-\infty$.
When $\dim A = 2$, the minimal resolution of $S^nA$ is given by the Hilbert scheme $A^{[n]}$, …
9
votes
Accumulation of algebraic subvarieties: Near one subvariety there are many others (?)
Apart of the deformations of $X$ you can consider deformations of any subscheme supported on $X$ (that is of any subscheme $Y$ such that $Y_{red} = X$). Another choice is to consider deformations of a …
9
votes
Building all holomorphic vector bundles from the tangent bundle
I think it is a result of Verbitsky (probably proved in Coherent sheaves on general K3 surfaces and tori) that if $M$ is a very general (non-algebraic) K3 surface then the condition you want is satisf …
9
votes
Accepted
Families of Fano varieties over non-hyperbolic curves
Let $X = SO(10)/P_5 \subset P^{15}$ be the spinor variety. It is projectively self-dual and has codimension 5, so its generic linear section of codimension 5 is smooth, and, moreover, generic pencil o …
8
votes
Accepted
Calculating the decomposition of a vector bundle over rational curve
Let $(z,w) \mapsto (f_1(z,w),\dots,f_5(z,w)$, $\deg f_i = s$, be a map $P^1 \to P^4$ and $g(x_1,\dots,x_5)$, $\deg g = d$, be an equation of a hypersurface containing the image. Then the normal bundle …
8
votes
Given a smooth hyperplane section Y of a variety X there exists a Lefschetz pencil of hyperp...
A pencil of hyperplane sections of $X$ corresponds to a line in the dual projective space $\check{\mathbb{P}}^N$. It is a Lefschetz pencil if and only if the line is transverse to the projectively dua …
7
votes
Accepted
Does a projective variety have only finitely many associated Hilbert polynomials?
Yes for the first question, by Riemann--Roch.
No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).
7
votes
Accepted
How do we write a locally free resolution for...
In general, there is no straight way to write such a resolution (note by the way, that a resolution is in no way unique!). However, in some cases there is a distinguished resolution. For example, if $ …
7
votes
When are two resolutions of a coherent sheaf homotopic
No. The simplest example is given by the following two resolutions of the structure sheaf of a point $P \in \mathbb{P}^1$:
$$
0 \to \mathcal{O}_{\mathbb{P}^1}(-1)
\to \mathcal{O}_{\mathbb{P}^1}
\to \ …