# Understanding what it means to be ''of general type''

I'm attempting to understand the Bombieri-Lang Conjecture:

If $X$ is a smooth projective variety of general type defined over a number field, then the set of rational points of $X$ is not dense.

I don't understand what it means for a variety to be ''of general type''. I know it's when the variety's Kodaira dimension is maximal, but this doesn't mean much to me. Is there an equivalent condition, or more intuitive way to visualise Kodaira dimension?

• Take a variety $X$. Think about the algebra generated by sections of tensor powers of the canonical bundle of $X$. Depending on $X$, this algebra might be the same as the algebra of sections of tensor powers of some line bundle on some, perhaps lower dimensional, variety $Y$. If not, $X$ has general type. So the canonical bundle of $X$ twists around wildly, feeling'' all of the dimensions of the variety $X$. – Ben McKay Jul 20 '18 at 10:47

## 2 Answers

To start understanding this, it's probably best to start with some examples.

First, the conjecture says that if a curve has Zariski dense rational points, then it is genus zero or one. This is known (Faltings).

Second, the conjecture, plus the Enriques-Kodaira classification, says that if a surface has Zariski dense rational points, then it is a rational surface, a ruled surface, an abelian surface, a K3 surface, an Enriques surface, an elliptic surface, a hyperelliptic surface, or a blow-up of one of these. A general type surface is simply one that is not any of those.

I don't know how many of these are possible to visualize but each of these has a much more definite structure and set of key properties than the class of all general type surfaces.

You could also try to understand general type surfaces through some positive examples like high-degree hypersurfaces, covers of products of two higher genus curves, and hyperbolic surfaces.

• Ah, I see. So the Bombieri-Lang Conjecture really just restricts the set of varieties with Zariski dense rational points to a well known list of possible suspects for that variety? – user221330 Jul 20 '18 at 10:21
• There is also a strong form of the Bombieri-Lang conjecture: If one removes the special subset (Zariski closure of the union of the images of all non-trivial rational maps from abelian varieties), there are only finitely many rational points. (See [Hindry-Silverman], F.5.2.3; section F.5 contains more things that might be of interest to you.) – user19475 Jul 20 '18 at 10:31
• In your third paragraph, are you assuming the surface is minimal? (I'm not sure if e.g. blowups of abelian surfaces or K3 surfaces are covered by your list, but I don't think they are.) – R. van Dobben de Bruyn Jul 20 '18 at 10:39
• @user221330 In low dimensions, yes. In higher dimensions, it becomes harder to classify them, or alternately one needs to classify them using broader categories. – Will Sawin Jul 20 '18 at 10:40
• @R.vanDobbendeBruyn Fixed. – Will Sawin Jul 20 '18 at 10:40

$X$ is of general type iff it is birational to its canonical model $X^c={\rm Proj}(\oplus _{m\geq 0}H^0(mK_X))$. Here $X^c$ has canonical singularities and $mK_{X^c}$ is a very ample Cartier divisor for some $m>0$. Thus there is an embedding $f:X^c\to \mathbb P ^N=|mK_{X^c}|$ and $\omega _{X^c}^{\otimes m}=\mathcal O _{\mathbb P ^N}(1)|_{X^c}$ under this embedding. This is a very useful characterization.

One of the main consequences/results about varieties of general type is that (up to birational isomorphism) they have good moduli spaces. Let $v = K_{X^c}^{\dim X}$ be the canonical volume which is given by the top self intersection of the canonical divisor on the canonical model. In any fixed dimension, these volumes belong to a discrete set and for any fixed dimension and fixed volume, canonical models are parametrized by a quasi-projective variety (see https://arxiv.org/abs/1503.02952 for some state of the art results, details and references). Eg if $\dim X =1$, then $v=2g-2$ and the corresponding moduli space has dimension $3g-3$.

An important feature is that by the easy addition theorem, $X$ can not be covered by by varieties not of general type, in particular by rational curves or abelian varieties (which tend to have many rational points).