2
$\begingroup$

What is the definition of relative differential forms of a family $\pi: X \to B$ of (nodal) curves, where $B$ is the base space.

$\endgroup$
3
  • 2
    $\begingroup$ hi quanting, welcome to mathoverflow! $\endgroup$ Jan 4, 2011 at 5:14
  • $\begingroup$ You can find a definition of relative differentials in almost any book on algebraic geometry. Differential forms are just exterior powers of the sheaf of differentials. $\endgroup$
    – S. Carnahan
    Jan 4, 2011 at 5:33
  • $\begingroup$ hi shizhuo,it is nice to see you here --zhaoquanting $\endgroup$ Jan 4, 2011 at 6:47

1 Answer 1

5
$\begingroup$

I learnt the following from the paper On the relative de Rham sequence (MathSciNet JSTOR) by Nick Buchdahl. It is more general than the OP situation, but should easily specialise to it.


Let $f:X \to B$ be a smooth map between manifolds. Then the sheaf $\Omega^1_f$ of germs of relative 1-forms on $X$ is defined by the following exact sequence $$ f^*\Omega^1_B \to \Omega^1_X \to \Omega^1_f \to 0 $$

Then the relative differential $k$-forms are defined by $$ \Omega^k_f = \Lambda^k \Omega^1_f $$ as usual. Differentiating along the fibres gives a differential $d_f: \Omega^k_f \to \Omega^{k+1}_f$. and $(\Omega^\bullet_f, d_f)$ is called the relative de Rham complex.

When $f$ is a submersion, the relative de Rham complex gives a resolution $$ 0 \to f^{-1}\mathcal{E}_B \to \Omega^\bullet_f $$ of the sheaf of functions of $X$ which are constant on the fibres.

$\endgroup$
3
  • $\begingroup$ Again JSTOR is selling something which is in free access: ams.org/journals/proc/1983-087-02/S0002-9939-1983-0681850-3/… $\endgroup$ Jan 4, 2011 at 8:15
  • $\begingroup$ My apologies. It was not my intention to peddle JSTOR's wares, I simply followed the "Article" link in MathSciNet. Interesting that MathSciNet, owned by the AMS, should link to JSTOR instead of to its own journals... $\endgroup$ Jan 4, 2011 at 9:01
  • $\begingroup$ Dear José, I was definitely not criticizing you! On the contrary, I am quite grateful to you for letting me discover this paper. $\endgroup$ Jan 4, 2011 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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