Determinant line does not depend on the differential - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:26:55Z http://mathoverflow.net/feeds/question/81983 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81983/determinant-line-does-not-depend-on-the-differential Determinant line does not depend on the differential Pavel Safronov 2011-11-27T04:14:52Z 2011-11-27T04:34:08Z <p>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<br> $\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},\nabla_{X/S})$ is isomorphic to the determinant line bundle of the Dolbeault (Hodge)<br> cohomology $\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},0)$?</p> <p>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?</p>