MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\pi:X\rightarrow S$ be a proper morphism of schemes over $\mathbf{C}$ and $\mathcal{E}$ a vector bundle on $X$ with a relative flat connection $\nabla_{X/S}$. Is it true that the determinant line bundle of the de Rham cohomology
$\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},\nabla_{X/S})$ is isomorphic to the determinant line bundle of the Dolbeault (Hodge)
cohomology $\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},0)$?

Heuristically, the determinant line bundle behaves like the Euler characteristic, which does not depend on the differential. What is a reference for such a statement?

share|cite|improve this question
added empty space to counter erroneous rendering – Sándor Kovács Nov 27 '11 at 4:34
I think this is simply a consequence of the (relative, with coefficients) Hodge to de Rham spectral sequence $E_1^{p,q}=R^q \pi_*(\Omega_{X/S}^p\otimes{\cal E})\Longrightarrow R^{p+q}\pi_*(\Omega_{X/S}^\cdot\otimes{\cal E})$. In any spectral sequence $E_1^{p,q}\Longrightarrow E^{p+q}$ in an abelian category, you have $\sum_{p+q=k}\chi(E_r^{p,q})=\chi(E^{p+q})$, if $\chi$ is a map from the abelian category to an abelian group, which is additive for exact sequences. The map ${\rm det}(\cdot)$ has this property by Knudsen-Mumford. – Damian Rössler Nov 27 '11 at 7:56
...sorry, replace $\chi(E^{p+q})$ by $\chi(E^k)$ in the above comment. The map $det(\cdot)$ is viewed as having values in the Picard group of line bundles. – Damian Rössler Nov 27 '11 at 8:04
Thanks! I would accept this if it were an answer. – Pavel Safronov Nov 27 '11 at 15:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.