## February 15, 2022

### Questions About the Néron–Severi Group

#### Posted by John Baez

A friend of mine with good intuitions sometimes says things without proof, and sometimes I want to know why — or even whether — these things are true.

Here are some examples from algebraic geometry.

First some background, just to see if I understand the basics, and maybe make this post interesting to other people who are just starting to learn algebraic geometry.

If you have a smooth complex projective variety $X$, the set of isomorphism classes of holomorphic line bundles on $X$ is called the Picard group of $X$. This has a topology — to indicated this it’s sometimes called the Picard scheme — where each connected component consists of different holomorphic line bundles that are isomorphic as topological line bundles. So, the identity component consists of different ways to give the trivial topological line bundle a holomorphic structure.

If you mod out the Picard scheme by its identity component you get the Néron–Severi group of $X$, or $NS(X)$ for short. Elements of this group —- in other words, the group of connected components of the Picard scheme — are isomorphism classes of topological line bundles over $X$ that actually admit a holomorphic structure. (Not all of them do, in general.) And this group is isomorphic to $\mathbb{Z}^n$ for some number $n$ called the Picard number of $X$.

(As a Star Trek fan all this terminology never ceases to amuse me. I just wish that Néron had been called something like Dukat.)

Now, my friend and I have been looking at the case where $X$ is a 2-dimensional abelian variety, also called an abelian surface. Then he claims that $NS(X)$ can be seen as a discrete subgroup of the Doubeault cohomology group $H^{1,1}(X)$. How generally is this true? I just need it for an abelian surface right now.

For $X$ an abelian surface, $H^{1,1}(X) \cong \mathbb{C}^4$. And he claims $NS(X) \cong \mathbb{Z}^4$. Is that right?

If so, we can form the real vector subspace $V \subset H^{1,1}(X)$ spanned by $NS(X)$, and $NS(X)$ will be a lattice in this.

Next, he claims something that I’m interpreting like this: the intersection pairing on $H^{1,1}(X)$ gives $V$ a symmetric bilinear form $B$ of signature $+++-$. Is that right?

If so, we can think of $NS(X)$ as a lattice in Minkowski spacetime.

Next, he claims that the ample line bundles — that is, those line bundles $L$ such that sufficiently high tensor powers $L^{\otimes p}$ have enough holomorphic sections to separate points — give precisely the elements of the Néron–Severi group that lie in the future cone of the Minkowski spacetime $V$. Here I’m borrowing some more terminology from physics: I mean that they lie in one of the two components of the set

$\{L \in NS(X) : B(L,L) > 0 \}$

Is that right?

I also have some independent evidence to support another guess of my own: the ‘principal polarizations’ of $X$ correspond to elements $L$ of the Néron–Severi group that not only lie in the future cone but have

$B(L,L) = 2$

Is that right?

Posted at February 15, 2022 8:16 PM UTC

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### Questions About the NéronSeveri Group

Those claims are correct. I know that you are looking for arguments that make sense for beginners, not just keywords. Nonetheless, there are some keywords. The claim that the Neron-Severi group is a discrete subgroup of $H^{1,1}$ follows from the long exact sequence of cohomology associated to the short exact sequence called the “exponential sequence”. The claim about the signature of $B$ on $V$ is called the “Hodge index theorem”. The claim about the positive cone is part of the Nakai-Moishezon Criterion (probably it can also be proved in other ways).

Posted by: Jason Starr on February 15, 2022 11:31 PM | Permalink | Reply to this

### Re: Questions About the Néron–Severi Group

Great! Thanks!

I think I can figure out the exponential sequence business.

I had not known the keyword Hodge index theorem, so that’s really useful: I see now that the signature is best thought of as having one plus and the rest minuses, and that it works this way for any surface.

I hadn’t quite keyed into the keyword ‘Nakai–Moishezon Criterion’, though I’d seen people talk about it. Anyone who wants a blog article explaining it can go here.

I’m still curious about my guess concerning a characterization of principal polarizations. And if anyone wants to say more about why any of the above results are true, or say anything interesting about this set of ideas, I’m all ears.

Posted by: John Baez on February 15, 2022 11:57 PM | Permalink | Reply to this

### Asvin

Isn’t the Neron Severi group rank generically 1 and 4 only if you are a product of isogenous elliptic curves?

Posted by: Asvin G on February 16, 2022 4:30 AM | Permalink | Reply to this

### Re: Néron–Severi group

I saw someone say that somewhere, but I didn’t trust them. This would completely change my mental picture, so if it’s true please someone confirm it — ideally with a reference or explanation!

Also: what’s the rank of the Néron–Severi group of an abelian variety of dimension $d$, generically?

Posted by: John Baez on February 18, 2022 8:54 PM | Permalink | Reply to this

### Re: Néron–Severi group

[BL]=Complex Abelian Varieties by Birkenhake and Lange.

The Néron-Severi group can be identified with the intersection of $H^{1,1}(X)$ and $H^2(X,\mathbb{Z})$ inside $H^2(X, \mathbb{C})$ and the embedding of $H^{1,1}(X)\cong \mathbb{C}^4$ into $H^2(X,\mathbb{C})\cong \mathbb{C}^6$ given by Hodge decomposition for an abelian variety $X=V/\Gamma$ varies with $\Gamma$.

An abelian variety of dimension $g$ and Néron-Severi group of rank $g^2$ is isogeneous to the $g$-th power of an elliptic curve with complex multiplication [BL, Exercise 5.10].

A general abelian variety has Néron-Severi group isomorphic to $\mathbb{Z}$, see [BL, Exercise 10.7.b] for the case of surfaces.

If an abelian surface is isogeneous to the product of elliptic curves $E_1$ and $E_2$ then we have three possibilities: $NS(X) = \mathbb{Z}^2$ if $E_1$ is not isogenous to $E_2$; $NS(X) = \mathbb{Z}^3$ if $E_1$ is isomogenous to $E_2$ and $End(E_1) = \mathbb{Z}$; and $NS(X) = \mathbb{Z}^4$ if $E_1$ is isogeneous to $E_2$ and $End(E_1) \neq \mathbb{Z}$ (i.e. $E_1$ has complex multiplication).

Posted by: Jorge Vitório Pereira on August 2, 2022 2:41 PM | Permalink | Reply to this

### Re: Questions About the Néron–Severi Group

You might find Chris Peters’ notes useful — they’re pretty succinct, but give a good overview of all of the things you’re asking about (plus they explicitly use the phrase “light cone”!) http://www.numdam.org/book-part/CIF_1995__23__A7_0/

Posted by: Tim Hosgood on February 16, 2022 1:12 AM | Permalink | Reply to this

### Re: Questions About the Néron–Severi Group

Thanks, Tim! I’ll check it out.

Posted by: John Baez on February 17, 2022 8:40 PM | Permalink | Reply to this

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