4
$\begingroup$

Let $X$ be a (reduced) affine variety with two irreducible components, $X_1$ and $X_2$, and let $\mathcal{E}$ be a torsion-free coherent sheaf on $X$. Denote the pullbacks of $\mathcal{E}$ to the closed subvarieties $X_1$, $X_2$, and $X_1 \cap X_2$ by $\mathcal{E}_1$, $\mathcal{E}_2$, and $\mathcal{E}_{12}$, respectively. Then there is a short exact sequence in the category of coherent sheaves on $X$

$$0 \to \mathcal{E} \to \mathcal{E}_1 \oplus \mathcal{E}_2 \to \mathcal{E}_{12} \to 0$$

which gives rise to an "inclusion-exclusion" formula in the Grothendieck group of coherent sheaves on $X$

$$[\mathcal{E}] = [\mathcal{E}_1] + [\mathcal{E}_2] - [\mathcal{E}_{12}]$$

If we try to generalize this formula to affine varieties with more than two irreducible components, the induction gets stuck: the pullback of a torsion-free sheaf to a union of components might not remain torsion-free. Are there natural conditions on $\mathcal{E}$ (or perhaps the components of $X$) that guarantee a formula like this holds? Would it help if I told you that everything is equivariant (for a reductive group $G$) and each component of $X$ contains an open dense $G$-orbit?

$\endgroup$
5
$\begingroup$

Everything is fine for locally free sheaves. Also, you can do the same for arbitrary objects of the derived category, which are perfect complexes (i.e., admit a finite locally free resolution); in this case, of course, you need to restrict in the derived sense (i.e., restrict the resolution).

Another way to say the same is the following. First, you can apply this construction to the structure sheaf of $X$ (by the way, the affinness assumption is absolutely irrelevant here). What you get is a complex, whose terms are sums of structure sheaves of intersections of components, and which is quasiisomorphic to $O_X$. Now you can tensor it with any sheaf on $X$ (or with any complex on $X$). You will get a spectral sequence, that will converge to the original object. However, unless the object is a perfect complex, the spectral sequence will be infinite, and will not give a direct expression for the class in the Grothendieck group.

$\endgroup$

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.