MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $h$ be a polynomial. Then results of several authors (including Chow, Grothendieck, Matsusaka, Mumford, Kollar and Viehweg) imply that the moduli space of polarized varieties with Hilbert polynomial $h$ is of finite type (even quasi-projective).

This means, in particular, that the moduli space has only finitely many connected components, and therefore, that there are only finitely many polarized varieties with Hilbert polynomial $h$ up to deformation.

Proving that the moduli space is of finite type is formally more difficult than the latter statement, I believe.

Question. Is there an "easy" proof for the finiteness of the number of deformation classes of polarized varieties with Hilbert polynomial $h$?

share|cite|improve this question
Why does your list of names omit Grothendieck? His finite type result for Hilbert schemes (combined with stratification thereof, depending on what you mean by the word "variety") seems extremely relevant, and such "boundedness" was the key difficulty in his original construction. – user36938 Aug 27 '13 at 12:51
There is in fact the more precise finiteness theorem of Chow, which states that the set of all closed subvarieties of $\mathbb{P}^n$ of a bounded degree is contained in only finitely many deformation types. Its proof for curves is particularly easy. – Vesselin Dimitrov Aug 27 '13 at 13:42
You say that, at least formally, proving finiteness up to deformation might be easier than proving "boundedness". In fact, "local finite typeness" is considerably simpler than proving boundedness. So, in fact, quasi-compactness (i.e., finiteness up to deformation) really is the difficult part of the proof. – Jason Starr Aug 27 '13 at 13:47
up vote 6 down vote accepted

I believe the two things you are relating:

The finiteness of the

  • number of deformation types
  • number of components of the moduli space

of polarized varieties are essentially equivalent problems.

The way moduli spaces are usually constructed is that first one finds a projective space that contains all the objects in the class in question embedded by the given polarization (or possibly a fixed power). This requires a "Matsusaka's Big Theorem"-type result.

Next one considers the locus in the appropriate Hilbert scheme that parametrizes these objects. Then one needs various results, for instance that this locus is locally closed. This is usually difficult, but has little to do with finiteness.

Now, the deformation types correspond to the components of this locus. The reason this is not yet a moduli space is that the same object may be embedded several times depending on the choice of sections of the appropriate line bundle. So, the moduli space (if it exists) is constructed as the quotient of this by the action of the appropriate PGL (and of course one needs a theorem that says that it exists). Since this is an algebraic group, the quotient will be of finite type exactly when this locus in the Hilbert scheme is.

So, as Jason already said in a comment, to prove finiteness of deformation types, the difficult thing to prove is that there is some kind of boundedness. Being locally of finite type often follows directly from the same property of the Hilbert scheme (see comment of user36938), although I have to add that there are some situations when this last statement is not entirely true and in that case boundedness is really hard to prove. (Think of how $\mathbb Z$ is inside something of finite type over $\mathbb C$, say $\mathbb C$, but itself is not of finite type (over $\mathbb C$).)

So, I think the answer to your question seems to be "no".

Added: I believe that those proofs you mention that these moduli spaces are of finite type actually prove the finiteness of deformation types. The main difference in difficulty is the existence. That's pretty tricky in general for the moduli space.

share|cite|improve this answer
Thank you for your answer. – Doedan Aug 29 '13 at 17:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.