Let $X$ be an algebraic variety (separated quasi-compact scheme of finite type) over a field $k$.

One of the possible definitions of an ample line bundle goes as follows:

Def 1: A line bundle $\mathcal{L}$ on $X$ is said to be ample iff some tensor power of it $\mathcal{L}^{\otimes k}$ admits $n+1$-generating sections (for some $n$) s.t. the associated morphism $X \to \mathbb{P}^n$ is a closed embedding (then $\mathcal{L}^{\otimes k}$ is said to be very ample).

I always found this definition rather subtle and mysterious. The classical story goes through showing that this definition is equivalent to the following one (which is manifestly much more useful in practice and much less easy to check):

Def 2: A line bundle $\mathcal{L}$ on $X$ is said to be ample if for every coherent sheaf $\mathcal{F}$ there exists some $n>0$ (depending on $\mathcal{F}$) such that for all $m>n$ the sheaf $\mathcal{L}^{\otimes m} \otimes_{\mathcal{O}_X} \mathcal{F}$ is generated by global sections (i.e. is a quotient of a trivial vector bundle).

The proof I know of this equivalence is subtle and goes through a reduction argument to the projective case and using serre vanishing (notice that may be why the relation between $k$ in the first definition and $n$('s) in the second is highly indirect).

Let $QCoh(X)$ denote the derived (stable $\infty$-)category of sheaves of quasi-coherent $\mathcal{O}_X$-modules with symmetric monoidal structure given by $\otimes_{\mathcal{O}_X}$.

Given this structure we can easily to detect (shifted) line bundles inside $QCoh(X)$ as those are given by the $\otimes$-invertible objects. Here's the question:

Questions: Given a (shifted-)line bundle $\mathcal{L}$ in $QCoh(X)$ can we... (increasing level of difficulty).

  1. Detect whether $\mathcal{L}$ is ample (in the classical sense above) without "leaving" the derived category $QCoh(X)$? (using $\otimes$-structure).

  2. Detect whether $\mathcal{L}$ is ample by considering $QCoh(X)$ without the $\otimes$-structure, but remembering the action of $Pic(X)$ (the $\infty$-picard groupoid of line bundles) on it.

  3. Detect whether $\mathcal{L}$ is very ample without using the $\otimes$-structure at all?. ֿ

  • $\begingroup$ Usually, $\operatorname{QCoh}(X)$ refers to the category of quasi-coherent sheaves on $X$, not its associated derived category. Which of the two are you interested in? $\endgroup$ Nov 13, 2017 at 1:00
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    $\begingroup$ he means DQCoh(X), or maybe some infinity version of that. The question can be answered by cheating, as the tensor structure allows you to recover X itself (and therefore any ample line bundle on it). But I don't know how to define ampleness intrinsically. $\endgroup$ Nov 13, 2017 at 3:06
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    $\begingroup$ Well I would say "mysterious" depends on point of view. From another (maybe old-fashioned) point of view, describing all possible maps to the projective space is what the line bundles are for. $\endgroup$ Nov 13, 2017 at 5:32
  • $\begingroup$ @მამუკაჯიბლაძე Of course I agree, still the fact that these two classical definitions agree has always been a miracle for me. I can't think of a proof which doesn't reduce to some form of computation (e.g. serre vanishing). $\endgroup$ Nov 13, 2017 at 7:57
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    $\begingroup$ For any line bundle $\mathcal{L}$, the functor $-\otimes\mathcal{L}$ is a self-equivalence of the derived category that commutes with the action of $\text{Pic}(X)$, so I don't think you can distinguish different line bundles without using the $\otimes$-structure, or only using the $\text{Pic}(X)$-action. $\endgroup$ Nov 16, 2017 at 19:07

1 Answer 1


Yes, to be ample amounts to be a generator of the derived category taking into account $\otimes$-powers, specifically an invertible sheaf $\mathcal{L}$ is ample if and only if the family $$ \{\mathcal{L}^{\otimes t}[n]\,/\, n\in \mathbb{Z}, t \geq 0\} $$ generartes de derived category of quasi-coherent sheaves on the scheme. This is explained in greater detail in Neeman's JAMS 1996 "The Grothendieck Duality Theorem via Bousfield’s Techniques and Brown Representability" in the general setting of divisorial schemes.

I guess this answers question 1 at least.

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    $\begingroup$ I don't see how this can be true. In what way does $\mathcal O(1)$ on $\mathbb P^n$ generate the derived category, that $\mathcal O(-1)$ does not? Both of them do not generate on their own (even with shifts) but do if you allow tensor powers. $\endgroup$
    – Will Sawin
    Nov 12, 2017 at 23:06
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    $\begingroup$ @Leo Alonso: Adding tensor powers does not help, since powers of an anti-ample line bundle $L$ generate the whole derived category. This follows easily by using the pullback of the (twisted) Koszul complex from a projective space; it allows to express $L^{i+1}$ by means of $L^i$, $L^{i-1}$, \dots, $L^{i-n}$, where $n = \dim X$. $\endgroup$
    – Sasha
    Nov 13, 2017 at 11:32
  • $\begingroup$ @Sasha What I mean is to generate the unbounded derived category $\mathbf{D}(Qco(X))$ as a triangulated category with coproducts, not $\mathbf{D}^b(Coh(X))$ as a triangulated category. The key point is that for every $\mathcal{F} \in Qco(X)$ a coproduct of negative tensors of $\mathcal{L}$ surjects onto $\mathcal{F}$, see SGA 6 p. 169. I guess this is the sign that distinguishes $\mathcal{L}$ from $\mathcal{L}^{-1}$. $\endgroup$
    – Leo Alonso
    Nov 13, 2017 at 12:07
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    $\begingroup$ @LeoAlonso: I understand, but what I say just means that all positive powers of $L$ are contained in the subcategory generated by negative powers. And the other way round, all negative powers are contained in the subcategory generated by positive powers. $\endgroup$
    – Sasha
    Nov 13, 2017 at 12:30
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    $\begingroup$ If we are given the tensor structure then we have the unit, and Hom from unit distinguishes ample and anti-ample, no? $\endgroup$ Dec 15, 2017 at 16:21

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