triviality of determinant sheaf - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:46:07Z http://mathoverflow.net/feeds/question/104213 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104213/triviality-of-determinant-sheaf triviality of determinant sheaf unknown (google) 2012-08-07T17:19:08Z 2012-08-07T19:45:15Z <p>On a smooth algebraic variety X, every coherent sheaf F has a finite resolution by locally free sheaves. Using such resolution, we can define the determinant of F, det F, which is a line bundle on X.</p> <p>My question is :</p> <p>why if the support of F is of codimension greater or equal to 2 is the determinant of F trivial ?</p> <p>It is mentionned without proof on the book "The geometry of moduli spaces of sheaves", D. Huybrechts, M. Lehn. I have verified this result on some explicit examples for which I know some explicit locally free resolutions but I don't see how to do the general case.</p> http://mathoverflow.net/questions/104213/triviality-of-determinant-sheaf/104217#104217 Answer by Youloush for triviality of determinant sheaf Youloush 2012-08-07T18:19:35Z 2012-08-07T18:19:35Z <p>Just an idea, using the first Chern class which should live in the cohomology with support in Supp(\$F\$), you should then get that \$c_1(F) = 0\$ which makes \$F\$ trivial since it's a line bundle (perhaps modulo linear equivalence). </p> http://mathoverflow.net/questions/104213/triviality-of-determinant-sheaf/104227#104227 Answer by Will Sawin for triviality of determinant sheaf Will Sawin 2012-08-07T19:45:15Z 2012-08-07T19:45:15Z <p>Outside the support of \$F\$, the resolution is an exact sequence, so the alternating tensor product of the determinants is trivial. On a smooth scheme, a line bundle trivial outside a codimension \$2\$ subset is trivial.</p>