TITLE: Understanding what it means to be ''of general type''
QUESTION [9 upvotes]: 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?

REPLY [9 votes]: $X$ 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).