8
$\begingroup$

Let $X$ be a smooth complex projective algebraic variety and $E$ a line bundle on $X$. It is a classical result that if $E$ carries an integrable connection, then the first Chern class $c_1(E)$ vanishes. I am interesting in the following variant: let $D$ be a normal crossing divisor and $\nabla: E \to E \otimes \Omega^1_X(\log D)$ an integrable connection with logarithmic singularities along $D$. Denote $$ \mathrm{Res}_{D_i} \nabla $$ the residue of $\nabla$ at an irreducible component $D_i$ of $D$. As $E$ has rank one, it can be identified with a complex number.

Proposition. One has $c_1(E)=-\sum_i \mathrm{Res}_{D_i} \nabla \cdot [D_i]$

Could anybody explain me how to prove such a result?

$\endgroup$

2 Answers 2

3
$\begingroup$

More generally, for a vector bundle $E$ consider a short exact sequence $$0\rightarrow End(E)\otimes\Omega^1\rightarrow End(E)\otimes \Omega^1(D)\rightarrow End(E)\otimes \mathcal{O}_D\rightarrow 0,$$ where the rightmost map is the residue. Then the residue of a connection gives a well-defined class in $H^0(X, End(E)\otimes\mathcal{O}_D)$ and its image under the connecting homomorphism $H^0(X, End(E)\otimes\mathcal{O}_D)\rightarrow H^1(X, End(E)\otimes\Omega^1)$ is minus the Atiyah class $a(E)$, i.e. the class of the Atiyah sequence $$0\rightarrow End(E)\rightarrow \mathcal{A}_E\rightarrow T_X\rightarrow 0,$$ where $\mathcal{A}_E$ is the bundle of first-order differential operators on $E$.

One can compute Chern classes by $$c_k(E) = \frac{(-1)^k}{k!} tr(a(E)^{\wedge k}).$$ For instance, you get the formula for the first Chern class of a line bundle you wrote above.

The very first claim is an easy exercise in Cech cohomology and can be found in Appendix B in H. Esnault, E. Viehweg, Logarithmic De Rham complexes and vanishing theorems, Invent. Math. 86, 161-194.

$\endgroup$
0
4
$\begingroup$

There exist a paper due to Makoto Ohtsuki about a Residue Formula for Chern Classes Associated with Logarithmic Connections:

http://projecteuclid.org/download/pdf_1/euclid.tjm/1270215030

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.