Led $D$ be a very ample divisor in $X$ projective variety.
I can't understand why the first Chern class $c_1(\mathscr{O}_X(D))$ equals the Poincaré dual of $D$, $\mathscr{P}(D)$
4
-
2$\begingroup$ This is a perfectly reasonable question, but not for this site. The best place for this question is math.stackexchange.com $\endgroup$– Michael JoyceCommented Nov 14, 2012 at 17:00
-
1$\begingroup$ Also if you post the question on Math stackexchange, you should state what definition of Chern class you are using. $\endgroup$– Michael JoyceCommented Nov 14, 2012 at 17:01
-
4$\begingroup$ Since this is a graduate level question, it does seem appropriate here. In a nutshell, since $D$ is very ample, you can reduce the problem to $X=\mathbb{P}^n$. Since $H^2(X)=\mathbb{Z}$ in this case, $D$ and $c_1(O(1))$ are multiples of each other. By restricting to a line, you can check the factor is $1$. I may flesh this out later, if I have time. $\endgroup$– Donu ArapuraCommented Nov 14, 2012 at 19:34
-
2$\begingroup$ We have the following short exact sequence $0\rightarrow \Omega ^1_X\rightarrow \Omega ^1_X(\log D)\rightarrow \mathcal{O}_D\rightarrow 0$ and since for the total Chern class we have from pervious short exact sequence $\ c(\Omega ^1_X(\log D))=c(\Omega ^1_X).c(\mathcal{O}_D)$ and since $c(\mathcal{O}_D)=(1-[D])^{-1}$ hence $c_p(\Omega ^1_X(\log D))=c_p(\Omega ^1_X)+c_{p-1}(\Omega ^1_X).[D]+\ldots +[D]^p$ hence $c_1(\Omega_X^1(\log D))= c_1(\Omega_X^1)+[D]$ $\endgroup$– user21574Commented Jun 5, 2017 at 21:09
Add a comment
|