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?

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    $\begingroup$ For example, Grassmannian $Gr(k,n)$ is in general not complete intersection of any projective spaces, but is smooth projective and has Picard number $1$. $\endgroup$ Feb 26, 2013 at 6:24
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    $\begingroup$ Thanks a lot for the comment. Are there other types? $\endgroup$
    – Fei YE
    Feb 26, 2013 at 8:29
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    $\begingroup$ There are many. Not only most Grassmannians, but most flag varieties (of semisimple groups) with Picard number one are not complete intersections. A still bigger series of examples is as follows. Consider an arbitrary smooth projective variety $X\subset\mathbb P^n$ with Picard number one. Even if it is a complete intersection in $\mathbb P^n$, its $m$-fold embedding, for $m\gg0$, is not a complete intersection anymore (proof: canonical class of a c.i. must be a multiple of the class of hyperplane section). $\endgroup$ Feb 26, 2013 at 8:54
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    $\begingroup$ As for classification, in general no. Sometimes if you fix some additional geometric conditions, say smooth Fano threefold, then there is a classification of those with Picard number 1. But for other geometric conditions, say Calabi-Yau threefolds, then (last time I checked) there is not yet a classification of those with Picard number 1. $\endgroup$
    – Chris Brav
    Feb 26, 2013 at 11:25

2 Answers 2


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).


Every abelian variety $X$ of given dimension $g>1$ over the field $\mathbb{C}$ of complex numbers is not a complete intersection, since it is not simply connected. On the other hand, a ``generic" $g$-dimensional $X$ has endomorphism ring $\mathbb{Z}$ and therefore Picard number 1 (A. Weil).

Explicit examples of such $X$ are provided by jacobians $J(C_f)$ of hyperelliptic curves $C_f:y^2=f(x)$ where $f(x)$ a polynomial of degree $n=2g+1$ or $2g+2$ without multiple roots that enjoys the following properties. There is a subfield $K$ of $C$ such that $f(x)$ is an irreducible polynomial in $K[x]$ and its Galois group over $K$ is either the full symmetric group $S_n$ or the alternating group $A_n$. (See arXiv math/9909052 .)

For example, if $K$ is the field $\mathbb{Q}$ of rational numbers then $f(x)=x^n-x-1$ is irreducible over $\mathbb{Q}$ (Selmer) and its Galois group is $S_n$ (Serre). This implies that if $n\ge 5$ then the jacobian $J$ of $y^2=x^n-x-1$ has Picard number 1 (and is not a complete intersection).


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