Picard Number One – Smooth Projective Varieties

ag.algebraic-geometry

Is there a classification theory of smooth projective varieties with Picard number 1?

By Lefschetz theorem, if $X$ is a complete intersection variety of dimension at least $3$, then the Picard number $\rho(X)=1$.

Can some one give an example of smooth projective variety with picard number 1 which is not a complete intersection? What kind of additional conditions, especially cohomological conditions, will force such a variety to be a complete intersection?

Best Answer

Picard number $1$ is the most frequent case among all varieties, so you cannot expect a classification. It's quite the opposite, you might stand a chance to classify those (within some class) that have Picard number larger than $1$. For instance a general $K3$ surface has Picard number $1$ and the locus of those with a given Picard number becomes smaller as the Picard number increases.

If the Picard number is larger than $1$, that usually means that the variety admits some non-trivial maps which gives you a handle on them or a starting point if you will. If the Picard number is $1$, it is hard to get any traction to get some way to study the object.

On the other hand that means that you can get lots of examples with Picard number $1$. Chances are, if you choose a variety at random it will have Picard number $1$. You can get lots of examples that are not complete intersections by the simple observation that for a complete intersection of dimension $d$, the middle cohomology groups of the structure sheaf vanish, that is, $H^i(X,\mathscr O_X)=0$ for $0<i<d$. This is actually another way to see that an abelian variety of dimension at least $2$ cannot be a complete intersection.

In particular, you can find lots of examples among surfaces. Surfaces with $H^1(X,\mathscr O_X)\neq 0$ are known as irregular, so any irregular surface with Picard number $1$ gives and example that you want. One way to ensure that the Picard number is $1$ is to make sure that $\mathrm{rk}\, H^2(X,\mathbb{Z})=1$. In other words, any surface with $q \ne 0$ and $b_2=1$ gives you an example.

Of course, you would want an explicit example. Unfortunately, I can't think of one at the moment, but I am fairly certain, that a general surface with $H^1(X,\mathscr O_X)\neq 0$ has Picard number one, so that should give you plenty of examples. In fact, one could argue that that's why I can't give an explicit one, because they are the general ones (and anything explicit is not general).

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