20
$\begingroup$

The following lemma is useful and well-known:

LEMMA If $L^{\pm 1}$ is ample on proper scheme over a field $k$, then some number of powers $\mathcal{O},L,...,L^{m}$ generate the unbounded derived category of quasi-coherent sheaves $D(X)$ (or split generate the subcategory of perfect complexes).

QUESTION: What about a converse? Suppose that I know some number of powers of $L$ generate $D(X)$. Then can I conclude that $L^{\pm 1}$ is ample?

The best I can do so far is see that the restriction of $L$ to any integral curve $C$ in $X$ has non-zero degree. (Since by adjunction $\mathcal{O},L,...,L^{m}$ generates $D(C)$, but if $L$ had degree $0$ on $C$, there would be something orthogonal $\mathcal{O},L,...,L^{m}$, for instance a generic line bundle of degree $g-1$ having no cohomology.)

Something I don't know yet: does the degree of $L$ must have the same sign on all curves? This would be useful for numerical tests of ampleness.

Note: I think that one doesn't need properness in the above lemma, but I am willing to assume it to get a converse. It makes life easier when restricting to closed subschemes.

Note 2: When saying a collection of objects generates a triangulated category with all coproducts, like $D(X)$, one usually means that you take the smallest triangulated subcategory closed under all coproducts and containing the the collection. Once you have all coproducts, then idempotents automatically split, by a standard argument called, I think, the Eilenberg swindle. If you are working with a smaller triangulated category having only finite coproducts, like perfect complexes on a scheme, then the smallest triangulated subcategory containing a collection might not be 'thick', in the sense that some idempotents might not split, so in this case one usually adds in the missing summands. To emphasize this, some people speak of 'split generation'.

$\endgroup$
15
  • 1
    $\begingroup$ I think you mean to say that these powers split-generate the derived category (otherwise, all varieties would have finite rank $K_0$ which fails already for curves). $\endgroup$ – Mohammed Abouzaid Aug 17 '10 at 14:51
  • 1
    $\begingroup$ Can you give a reference for the lemma? Thanks! $\endgroup$ – Fei YE Aug 17 '10 at 15:29
  • 1
    $\begingroup$ One reference would be arXiv:0804.1163, Theorem 4. If you are willing to assume $X$ proper, then you can give a shorter argument by replacing $L$ with some power that is both ample and globally generated, use this to get a finite morphism to some projective space, and then pull-back the usual generator $ \mathcal{O}\oplus \mathcal{O}(1)\oplus \cdots \mathcal{O}(n)$ from there. $\endgroup$ – Chris Brav Aug 17 '10 at 19:34
  • $\begingroup$ What are your assumptions on $X$? At the very least you will need irreducible... $\endgroup$ – Arend Bayer Aug 18 '10 at 18:10
  • $\begingroup$ I am happy to assume $X$ is irreducible and even integral, but why do you think the former is necessary? $\endgroup$ – Chris Brav Aug 19 '10 at 7:13
2
$\begingroup$

I think you can construct a (many) counter-examples as follows. Let $X$ be the blow-up of $\mathbb{P}^2$ along a point $p_0 \in \mathbb{P}^2$. I denote by $E$ the exceptional divisor. Let $\pi : X \longrightarrow \mathbb{P}^2$ be the blow-up map and let $L = \pi^* \mathcal{O}_{\mathbb{P}^2}(1) \otimes \mathcal{O}_{X}(E)$.

The line bundle $L$ is not ample because its restriction to $E$ has negative degree. The dual of $L$ is not ample because its retsriction to any curve in $X$ disjoint from $E$ has negative degree.

Now, I claim that $L^{-2},L^{-1},\mathcal{O}_X$ generates $D^b(X)$. This can be proved as follows. I denote by $A$ the full sucategory of $D^b(X)$, closed under taking direct summands, which is generated by $L^{-2}, L^{-1}, \mathcal{O}_X$. Let $F \in D^b(X)$ such that $\mathrm{Ext}^k(a, F) = 0$ for all $a \in A$. I want to prove that $F = 0$ in $D^b(X)$.

Let $x \in X \backslash E$. There exists a line $l \subset \mathbb{P}^2$ through $\pi(x)$, disjoint from $p_0$ so that there is a section $s$ of $L$ whose vanishing locus is exactly $\pi^{-1}(l) \cup E$. Hence, we have an exact sequence:

$$ 0 \rightarrow L^{-1} \stackrel{s}\longrightarrow \mathcal{O}_X \rightarrow \mathcal{O}_{\pi^{-1}(l)} \oplus \mathcal{O}_E \rightarrow 0.$$

We deduce that $\mathcal{O}_E$ and $\mathcal{O}_{\pi^{-1}(l)}$ are in $A$. Twisting the above exact sequence by $L^{-1}$, we get that $\mathcal{O}_{\pi^{-1}(l)} \otimes \pi^* \mathcal{O}_{\mathbb{P}^2}(-1)$ and $\mathcal{O}_{E}(-E)$ are also in $A$.

Now, we have an exact sequence:

$$0 \rightarrow \mathcal{O}_{\pi^{-1}(l)} \otimes \pi^* \mathcal{O}_{\mathbb{P}^2}(-1) \rightarrow \mathcal{O}_{\pi^{-1}(l)} \rightarrow \mathcal{O}_x \rightarrow 0.$$

As a consequence, we have $\mathrm{Ext}^k(\mathcal{O}_x, F) = 0$ for all $k\in \mathbb{Z}$. In particular, the support of $F$ does not contain $x$.

This is true for all $x \in X \backslash E$ so that the reduced support of $F$ is included in $E$.

Let $y \in E$. We have an exact sequence on $E$:

$$0 \rightarrow \mathcal{O}_E(E) \rightarrow \mathcal{O}_E \rightarrow \mathcal{O}_y \rightarrow 0.$$

Twisting by $\mathcal{O}_{E}(-E)$, we get an exact sequence on $E$:

$$0 \rightarrow \mathcal{O}_E \rightarrow \mathcal{O}_E(-E) \rightarrow \mathcal{O}_y \rightarrow 0.$$

Pushibg forward by $\mathrm{R} i_*$, we get an exact squence on $X$:

$$0 \rightarrow \mathcal{O}_E \rightarrow \mathcal{O}_E(-E) \rightarrow \mathcal{O}_y \rightarrow 0.$$

Since we know that $\mathcal{O}_E$ and $\mathcal{O}_E(-E)$ are in $A$, we deduce that $\mathrm{Ext}^k(\mathcal{O}_y, F) = 0$ for all $k\in \mathbb{Z}$. But this is true for all $y \in E$. We conclude that the support of $F$ is zero and that $F \simeq 0$ in $D^b(X)$.

Of course it seems difficult to find any condition on $X$ to prevent such a counter-example to appear. Indeed, we'll get a counter-example similar to the one above a soon as we get a birational map $X \rightarrow Y$ which contracts a curve ($Y$ may well be singular!).

$\endgroup$
7
  • $\begingroup$ what is the scheme-theoretic support of a complex? $\endgroup$ – Yosemite Sam Jan 12 '17 at 4:37
  • $\begingroup$ @YosemiteSam : union of scheme-theoretic support of its cohomology sheaves. $\endgroup$ – Libli Jan 12 '17 at 8:38
  • $\begingroup$ by scheme-theoretic union you mean taking the intersection of the ideal sheaves? And how do you show that F = i_* G for some G? (or do you have a reference for it?) I think one might be able to use induction on the length of the complex, but I'm not sure I have a rigorous argument. $\endgroup$ – Yosemite Sam Jan 12 '17 at 13:36
  • $\begingroup$ Well I thought that the following was kinf of standard. If $F$ is quasi-isomorphic to a sheaf, then its true. Assume that it is true when there are at most $m$ non vanishing cohomology sheaves for $F$. Not let $F'$ with $m+1$ non-vanishing cohomology sheaves. There is an exact triangle $F'' \rightarrow \mathcal{H}^{i_0}(F') \rightarrow F'$ where $i_0$ is the smallest integer such that $\mathcal{H}^{i}(F') \neq 0$. $\endgroup$ – Libli Jan 12 '17 at 18:54
  • $\begingroup$ Furthermore, we have $\mathcal{H}^{i}(F'') \simeq \mathcal{H}^{i}(F)$ for all $i>i_0$. By recursion, we know that $F'' = i_*G''$ and $\mathcal{H}^{i_0}(F') = i_* H$ for $H, G'' \in D^b(mE)$. Furthermore, by adjunction, the map $F'' \rightarrow \mathcal{H}^{i_0}(F')$ lifts to a map $G'' \rightarrow H$. Denote by $G'$ the cokernel of this map. Then, the five lemma shows that $F' \simeq i_* G'$ in $D^b(mE)$. $\endgroup$ – Libli Jan 12 '17 at 19:01

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.