Good covers on complex algebraic varieties with normal crossings singularities - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:02:29Z http://mathoverflow.net/feeds/question/101611 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101611/good-covers-on-complex-algebraic-varieties-with-normal-crossings-singularities Good covers on complex algebraic varieties with normal crossings singularities Kevin Kordek 2012-07-07T23:24:02Z 2012-07-09T06:46:23Z <p>Let $X$ be a topological space. A good cover on $X$ is an open cover such that all finite non-empty intersections are contractible. It is a theorem of Hironaka that (complex) algebraic sets admit triangulations. As a consequence, all algebraic sets admit a good cover - simply take as an open cover the collection of open stars of the vertices in the triangulation. In case $X$ is an algebraic curve with nodes, there is a way of constructing an open cover geometrically. </p> <p>Consider the resolution of singularities $\pi:\widetilde{X}\rightarrow X$. This is a compact Riemann surface obtained by blowing up the singular points of $X$. The singular locus $S$ of $X$ is finite and $\pi^{-1}(s)$ consists of two points when $s\in S$. If $E = \pi^{-1}(S)$, then $\pi|_{\widetilde{X}\backslash E}: \widetilde{X}\backslash E\rightarrow X\backslash S$ is a biholomorphism. Note that a node is an example of a normal crossings singularity - one which is locally isomorphic to a union of coordinate hyperplanes. In fact, nodes are the only examples of normal crossings singularities on curves. </p> <p>Since $\widetilde{X}$ is a smooth (paracompact) manifold, it admits a good cover by differential geometry (choose a metric on $\widetilde{X}$, then cover it by sufficiently geodesically convex balls). There exists a finite cover $(U_j)_{j\in J}$, consisting of convex balls, for which each point of $E$ is contained in exactly one $U_j$. We can assume that the $U_j$ which intersect $E$ are mutually disjoint. Each $\pi(U_j)$ is homeomorphic to an open disc $\mathbb{D}\subset \mathbb{C}$. However, $\pi(U_j)$ is open in $X$ if and only if it does not contain a singular point. Define a subset $K\subset J$ where $j\in K$ if and only if $\pi(U_j)\cap S = \emptyset$ . To get an open neighborhood of a singular point $s\in S$, define $W_s = \pi(U_{j_1})\cup \pi(U_{j_2})$ where the $U_{j_k}$ each contain exactly one point of $\pi^{-1}(s)$. $W_s$ is contractible because it is homeomorphic to a wedge sum $\mathbb{D}\vee \mathbb{D}$, where the two discs are joined at the origin. Then $(W_s)_{s\in S}\cup (\pi(U_k))_{k\in K}$ is a good cover for $X$. </p> <p>My question is: </p> <blockquote> <p>Does this construction generalize to varieties with normal crossings singularities? More precisely, does there exist a good (?) open cover $(U_{\alpha})$ of $\widetilde{X}$ such that finite unions of the $\pi(U_{\alpha})$ can be used to build a good cover for $X$ ? If so, is there an analogously simple description of the open sets which happen to contain singular points? </p> </blockquote>