1
$\begingroup$

Let $X$ be a complex algebraic variety, and consider the presheaf

$U \mapsto H^i(U^{an}, \mathbb Z)$

in the Zariski topology. Is there a theorem that says this presheaf is already a sheaf, for certain values of $i$ and maybe under some assumptions on $X$?

It seems to be true for $i = 1$ and $X = \mathbb A^1$, but it fails for $i = 2$ and $X = \mathbb P^1$.

$\endgroup$
1
  • 2
    $\begingroup$ There seems to be absolutely no reason for this to be a sheaf, except maybe $i=0$ and $X$ irreducible. $\endgroup$ Mar 17 '14 at 19:04
5
$\begingroup$

Notice that the sheaf axiom is a sort of Mayer-Vietoris property. If $X=U\cup V$ for two opens $U,V$ then the sheaf axiom for some presheaf $F$ with values in an abelian category requires

$0\to F(X)\xrightarrow{res_U^X\oplus res_V^X} F(U)\oplus F(V) \xrightarrow{res_{U\cap V}^U-res_{U\cap V}^V} F(U\cap V)$

to be exact. The situation for singular cohomology is very different, there the Mayer-Vietoris-sequence is a long exact sequence. In particular will the short sequences for $H^i$ with $i>0$ will not be exact in general because no higher cohomology vanishes in general. The analogue for cohomology would propably be that the sequence of the cochain complexes

$0 \to S^\ast_{\{U,V\}}(X) \to S^\ast(U)\oplus S^\ast(V) \to S^\ast(U\cap V) \to 0$

is exact where $S^\ast_{\{U,V\}}(X)$ denotes the cochain complex that is generated by cochains supported either in $U$ or in $V$. This is homotopy equivalent to the full cochain complex $S^\ast(X)$, so there is a "sheaf axiom up to homotopy" somewhere in here.

On the other hand long exact sequences of (co)homology are exactly what makes an distinguished triangle in a derived category. Is there maybe a notion of a derived-category-valued sheaf?

$\endgroup$
4
  • 1
    $\begingroup$ The derived category has few limits, so it would be a bad idea to talk about sheaves with values in the derived category. Instead we look at presheaves of chain complexes satisfying a sheaf condition up-to-homotopy. $\endgroup$
    – Zhen Lin
    Mar 17 '14 at 19:34
  • $\begingroup$ Well along the lines of this answer one might try to require distinguished triangles instead of short exact sequences... $\endgroup$ Mar 17 '14 at 19:48
  • 1
    $\begingroup$ Minor correction: there is generally no $0$ at the end of the first short exact sequence. $\endgroup$ Mar 17 '14 at 19:58
  • $\begingroup$ @KeenanKidwell Oh you're right. I'll correct that. In that case the sheaf axiom (for two open sets anyway) is indeed satisfied by $H^0$. $\endgroup$ Mar 17 '14 at 20:07

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.