# What is ample generator of a Picard group?

First a note of caution: I am a physicist with a rudimentary knowledge of algebraic geometry picked up here and there. So don't assume I know anything besides basic properties of sheaves and try to give as simple answers as possible. Also, if my questions don't make sense for any reason, try to point me in the right direction.

So, my questions:

1. I'd like to hear the definition of the ample generator of a Picard group. I know what a Picard group is but I am having a hard time finding an actual definition of its ample generator.

2. Why is this notion important and where is it used primarily?

3. If you could provide some simple examples to illustrate the notion for some special $X$ and ${\rm Pic}(X)$ that would be welcome.

Motivation: recently, Scott Carnahan provided a very nice answer about conformal blocks in CFT over at physics.SE. My problem is that the punchline of his example involves the notion of an ample generator of a Picard group of certain moduli space of $SU(2)$ bundles on the Riemann surface (which is the arena for the CFT).

-
Your difficulty is mostly semantical. There is no notion of "ample generator of a group". The Picard group is sometimes cyclic in which case it will have a generator, say g. Now -g is also a generator. Exactly one of these two will represent the class of an ample line bundle and this is the ample generator. – Felipe Voloch Dec 21 '10 at 1:36
@Felipe: yes, this is the picture I got from Henri's answer. But thanks for making that explicit. – Marek Dec 21 '10 at 9:01

First of all, the Picard group of a variety is not always monogen, so that the notion of "the ample generator" you are referring to surely concerns a restricted class of varieties.

Furthermore, an ample line bundle (or invertible sheaf) is a line bundle $L$ which satisfies any of the following properties :

1. For any coherent sheaf $\mathcal F$, the sheaf $\mathcal F \otimes L^{\otimes m}$ is generated by its global sections for every $m\geq m_0(\mathcal F)$.

2. The map $\Phi_{|L^{\otimes m}|} : X \to \mathbb P(H^0(X,L^{\otimes m})), x \mapsto [s_0(x): \ldots:s_N(x)]$ - where $(s_i)$ is a base of the space of global sections of $L^{\otimes m}$ - induces a embedding of $X$ in some projective space for every $m\geq m_0$.

3. If you are working over $\mathbb C$, which seems to be the case, this is also equivalent for $L$ to admit a smooth hermitian metric $h$ whose curvature $\Theta_h(L)$ is a (striclty) positive $(1,1)$-form.

Then I guess that the ample generator of some Picard group is in some case one generator which besides is ample.

For example, the most basic example consists in taking the ample line bundle $\mathcal O_{\mathbb P^n}(1)$ over $\mathbb P^n$, which generates $\mathrm{Pic}(\mathbb P^n)$.

Finally, if your question is : if the Picard group of a projective variety is monogen, then can we choose an ample generator? Then the answer is yes because for every $m\geq 1$, $L$ is ample iff $L^{\otimes m}$ is ample.

Edit : I forgot to mention that in this case, the unicity of "the" ample generator is clear : indeed, if $L$ and $L^{-1}$ admit non trivial sections (say $s$ and $t$) then $st$ is a non-zero section of $\mathcal O_X$ thus is constant, so that $s$ and $t$ are both non-vanishing sections, which implies that $L$ is trivial. You can apply this to $L^{\otimes m}$ to get the unicity property.

-
Thank you Henri, this is very enlightening (unfortunately I can't yet vote up). I'll spend some more time with your answer to grok all of it and then will update my question to make it more precise. – Marek Dec 20 '10 at 23:16

In terms of why these might be important, having an ample line bundle is equivalent to your object of study being projective. This is since, as in Henri's answer, ample implies that there in an embedding into $\mathbb{P}H^0(X,L^{\otimes m})$ for some $m$, while conversely pulling back $\mathcal{O}(1)$ from an embedding into projective space yields an ample line bundle. Being projective gives it many nice properties.

-