All Questions
8 questions
4
votes
1
answer
296
views
Fundamental group of the smooth locus of a normal algebraic surface is a quotient of that of a Zariski open subset
Let $X$ be a normal algebraic surface (over $\mathbb{C}$) and $Y$ its smooth locus, i.e., the complement of the singularities of $X$. Suppose $Z\subset Y$ is a Zariski open subset of $X$. Then is it ...
4
votes
1
answer
218
views
Topological interpretation of the canonical cover of a logarithmic Enriques surface
A normal projective surface $Z$ with at worst quotient singularities is called a logarithmic (log) Enriques surface if its canonical Weil divisor $K_Z$ is numerically equivalent to zero, and $H^1(Z,\...
9
votes
1
answer
284
views
Fundamental group of a smoothing of a complex surface
Let $X_0$ be a compact complex algebraic surface with an isolated singularity and let $X_t$ be a smoothing of $X_0$ over the disc. How can we compute the fundamental group of $X_t$ say in terms of the ...
7
votes
1
answer
2k
views
Relating the holomorphic Euler characteristic of a family of algebraic varieties to properties of the base and fibers
Let $f : X\rightarrow Y$ be a proper flat morphism (of schemes) with connected fibers over a smooth projective curve $Y$ over $\mathbb{C}$. Let $X_{y_0}$ denote a smooth fiber over $y_0\in Y$.
If $f$ ...
4
votes
1
answer
639
views
Intersection form in Algebraic Geometry/Topology
Let $S$ be a smooth complex projective surface. We let define an intersection form $(-)\cdot(-)$ on $\mathsf{Pic}(S)$ by setting $$D\cdot D':=\mathcal{O}_S(D)\cdot\mathcal{O}_S(D')$$ where the ...
5
votes
1
answer
461
views
Homeomorphism between del Pezzo surfaces
Let $X$ and $Y$ be smooth del Pezzo surfaces of the same degree $K_X^2=K_Y^2$.
Are the sets $X(\mathbb{C})$ and $Y(\mathbb{C})$ homeomorphic, or at least homotopy equivalent?
23
votes
1
answer
718
views
Del Pezzo surfaces and homotopy groups of spheres
A (complex) del Pezzo surface is a smooth projective complex surface with ample anticanonical line bundle. Such surface has a degree defined as the self intersection of the canonical divisor. It is ...
8
votes
1
answer
652
views
How do branched coverings of complex surfaces "fit" with branched coverings of curves?
Since I'm used to working with algebraic $\pi_1$'s, which don't work well with surfaces, I find myself lacking geometric intuition when I attempt to do these types of purely geometric arguments. I'm ...