It is known that there exists a fine moduli space for marked (nonalgebraic) K3 surfaces over $\mathbb{C}$. See for example the book by Barth, Hulek, Peters and Van de Ven, section VIII.12. Of course the marking here is necessary, otherwise the presence of automorphisms can be used to construct isotrivial non trivial families of K3.

Assume that we want to construct an algebraic analogue of this. Of course some structure has to be added; I am thinking of something like torsion points for elliptic curves.

Is there a way to add some structure and actually build a fine moduli space of (K3 + structure) which is defined over $\mathbb{Z}$?

Of course this will only parametrize algebraic K3 surfaces, so it will need to have infinitely many components. A second question, if we do not want to add structure, is

Is there a fine moduli stack of algebraic K3 surfaces which is algebraic (either in the DM or Artin sense)?

It is not difficult to produce K3 surfaces with a denumerable infinity of automorphisms, so I'm not really expecting the answer to be yes, but who knows.

Edit: since there seems to be some confusion in the answers, the point is that I'm asking for a FINE moduli space. I'm aware that one can consider moduli spaces of polarized K3, and this is why I wrote "Of course this will only parametrize algebraic K3 surfaces, so it will need to have infinitely many components."

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    $\begingroup$ You might want to polarize your K3, but I am not an expert $\endgroup$ Feb 20, 2010 at 21:25

3 Answers 3


I think you are looking for this


which is the same as:

Rizov, Jordan Moduli stacks of polarized $K3$ surfaces in mixed characteristic. Serdica Math. J. 32 (2006), no. 2-3, 131--178.

It builds on an earlier result:

Olsson, Martin C. Semistable degenerations and period spaces for polarized $K3$ surfaces. Duke Math. J. 125 (2004), no. 1, 121--203.

which builds on Friedman's Ph.D. (the algebraic proof of Piatetski-Shapiro and Shafarevich of global Torreli for K3's).

Enough history for one answer I guess.

  • $\begingroup$ Thank you, this paper seems what I was looking for. So it seems that the stack of primitively polarized K3 is DM, and if one adds a level structure, one gets even an algebraic space. I guess this answers both questions. $\endgroup$ Feb 21, 2010 at 23:43

This is in response to the second question: the stack of all K3 surfaces is not algebraic. If you're working with Artin's axioms, then the problem is that there exist formal deformations that are not effective, i.e., one can write down a compatible system of K3 surfaces $X_n \to \mathrm{Spec}(k[t]/(t^{n+1}))$ which do not come from an algebraic K3 surface over the power series ring, at least when the field $k$ is big enough. I learnt this fact from Jason Starr's paper www.math.sunysb.edu/~jstarr/papers/moduli4.pdf (see page 14 and 15).

Added later: As Scott mentions in the comments, this problem is specific to the non-polarised case. Any formal deformation of a pair (X,L) (where X is a projective variety, and L is line bundle giving a projective embedding) is automatically algebraic by formal GAGA.

  • $\begingroup$ Your answer complements in some way David's. Still I find some difficulties in reading Starr's paper, but I will have a closer look. How come this is not in contradiction with the paper cited by David which proves that the stack of primitively polarized K3 is DM? $\endgroup$ Feb 21, 2010 at 23:41
  • $\begingroup$ @Andrea: These deformations aren't polarized. $\endgroup$
    – S. Carnahan
    Feb 22, 2010 at 6:25
  • $\begingroup$ @Andera: I did not stress this because my answer came chronologically a day after Alexeevs, but the only way you can do anything in the algebraic context is first polarize. $\endgroup$ Feb 22, 2010 at 8:08

Why aren't you happy with the moduli space of polarized K3's, i.e. pairs $(X,L)$ where $L$ is an ample line bundle? This is standard, and this at least makes sense over any field or $\mathbb Z$. And the automorphism group of $(X,L)$ is finite, since it is discrete and algebraic.

If you insist on marking your K3s with an isomorphism $H^2(X,\mathbb Z)\to L_{K3}$ then yes, you need $H^2(X,\mathbb Z)$ for that, so you better work over $\mathbb C$.

I suppose one can look at $H^2(X_{et},\mathbb Z_l)$ instead... That would be a poor substitute, I think. Lattices over $\mathbb Z_l$ are much easier than lattices over $\mathbb Z$. Even if you add all prime $l$ and $\mathbb R$, these do not capture the isomorphism classes of lattices over $\mathbb Z$.

  • $\begingroup$ Is the moduli space of polarized K3 fine? $\endgroup$ Feb 21, 2010 at 10:42
  • $\begingroup$ No it is not, since the automorphism groups are not trivial; they are finite. Over $\mathcal C$, it can be written as the quotient $G\D$, $D$ a Hermitian symmetric domain, $G$ an arithmetic group. Choosing a finite index subgroup $G'$ of $G$, the space $G'\D$ may be considered to be a fine moduli space. $\endgroup$
    – VA.
    Feb 21, 2010 at 13:54
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    $\begingroup$ If you read my question, I was asking for a fine moduli space of K3 with some (algebraic) additional structure. Moreover, how do you construct moduli spaces of polarized K3 over $\mathbb{Z}$? I only know how to do it over $\mathbb{C}$. $\endgroup$ Feb 21, 2010 at 17:01

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