Timeline for Transversal intersection in the moving lemma
Current License: CC BY-SA 3.0
6 events
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Jul 1, 2015 at 17:44 | comment | added | user115940 | Thanks Mohan. What if $A$ and $B$ are both assumed to be smooth? This is actually the case. | |
Jul 1, 2015 at 15:50 | comment | added | Mohan | If you only move $A$, this is impossible. For example, take $X$ to be a projective space, $A$ a hypersurface and $B$ to be a subvariety with positive dimensional singular locus. Whatever $A$ is moved to, it is still a hypersurface and will meet $B$ in some singular points. | |
Jul 1, 2015 at 14:32 | comment | added | Jason Starr | Hmm, actually my last suggestion won't work. The determinantal locus can be made smooth away from the locus where the $\mathcal{O}_A$-module homomorphism drops rank (further). However, it will typically be singular on that locus. | |
Jul 1, 2015 at 14:17 | comment | added | Jason Starr | This would be simpler if you allowed to replace $A$ by $N\cdot A$ for $N$ a sufficiently positive and divisible integer. Use the isomorphism of $K$-theory and Chow theory, i.e., write $A$ as a polynomial in Chern classes $c_r$ of locally free sheaves $\mathcal{E}$. Using finite differences, $c_r(\mathcal{E})$ can be expressed in terms of Chern classes of $\mathcal{E}(d)$ for $d$ large. Thus, you need only find global sections of $\mathcal{E}(d)|_B$ whose determinantal locus is transversal (i.e., a Bertini-type theorem). Finally, extend those to global sections of $\mathcal{E}(d)$. | |
Jul 1, 2015 at 9:33 | review | First posts | |||
Jul 1, 2015 at 9:42 | |||||
Jul 1, 2015 at 9:19 | history | asked | user115940 | CC BY-SA 3.0 |