4
$\begingroup$

The determinant line bundle of a coherent sheaf $\mathcal{F}$ on an $n$-dimensional (smooth) analytic space is defined as \begin{equation} \det \mathcal{F} := \bigotimes_i^n (\det \mathcal{E}_i)^{⊗ (-1)^i} \end{equation} where $\mathcal{E}_\bullet \to F$ is a locally free resolution of $\mathcal{F}$ (which we can take to have length at most $n$). It can be shown that this is independent of the resolution taken, and that if $\mathcal{F}$ is torsion-free then \begin{equation} \det \mathcal{F} \cong \left(\bigwedge^{\operatorname{rk} \mathcal{F}} \mathcal{F}\right)^{**} \end{equation} . Since $\bigwedge^k$ and double duals are both functorial we see that a morphis $\mathcal{F} \to \mathcal{F}$ of *torsion-free shaves of the same rank induces a morphism between their determinant line bundles. Can we say the same thing about any two coherent sheaves of the same rank?

My thoughts so far: It seems like the obvious way to do this would be to take free resolutions $\mathcal{E}_\bullet, \mathcal{E}'_\bullet$ of $\mathcal{F}, \mathcal{F}'$, which gives a map $f_\bullet: \mathcal{E}_\bullet \to \mathcal{E}'_\bullet$ and to just take the alternating tensor product of the maps induced by $f_i$ from $\det \mathcal{E}_i \to \det \mathcal{E}'_i$. However those maps don't exist because $\det$ is only functorial on vector bundles of the same rank?

$\endgroup$

1 Answer 1

10
$\begingroup$

No.

Working on projective space, consider a composition $$ \mathcal O \to \mathcal O \oplus \mathcal O/\mathcal O(-1) \to \mathcal O $$ where $\mathcal O/\mathcal O(-1)$ is the constant sheaf on a hyperplane, and the maps are just defined to be the identity on $\mathcal O$ and $0$ on $\mathcal O/\mathcal O(-1)$.

Since the composition is the identity, the induced map on determinants of the composition is the identity. By functoriality, the composition of the induced map on determinants must be the identity.

But taking determinants, we get $$\mathcal O \to \mathcal O(1) \to \mathcal O$$ and since every map $\mathcal O(1) \to \mathcal O$ vanishes, there is no composition of maps $\mathcal O \to \mathcal O(1)$ and $\mathcal O(1) \to \mathcal O$ that gives the identity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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