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.

My question is :

why if the support of F is of codimension greater or equal to 2 is the determinant of F trivial ?

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.

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    $\begingroup$ The best reference for these kinds of questions is the following article. MR0437541 (55 #10465) Reviewed Knudsen, Finn Faye; Mumford, David The projectivity of the moduli space of stable curves. I. Preliminaries on "det'' and "Div''. Math. Scand. 39 (1976), no. 1, 19–55. 14H10 (14F05 14C05) $\endgroup$ – Jason Starr Aug 7 '12 at 19:22

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.


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).

  • $\begingroup$ I want to show that the determinant bundle is algebraically trivial, what can not be seen in cohomology. Even if one works over \mathbb{C}, I don't know how to prove that it is topologically trivial. I don't understand why the first Chern class should live in the cohomology with support in Supp(F). $\endgroup$ – user25309 Aug 7 '12 at 18:43
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    $\begingroup$ $c_1(F) = 0$ does not imply that $F$ is trivial: Consider an elliptic curve $E$. Set $X = E \times E$. Let $g : (x,y) \mapsto (x+1/2,-y)$ be an involution w/o fixed points on $X$. Then the canonical bundle of the quotient $Z = X / \langle g \rangle$ has zero first Chern class, but is not trivial since it has no non-zero sections. $\endgroup$ – Gunnar Þór Magnússon Aug 7 '12 at 18:46

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