Generating the derived category with line bundles

Question

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'.

Answer 1

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!).