12
$\begingroup$

I have a question about indefinite lattices.

QUESTION: Let $\Lambda\times\Lambda\rightarrow {\Bbb Z}$ be a lattice, that is, ${\Bbb Z}^n$ with a non-degenerate integer quadratic form, not necessarily unimodular, and $G:=O(\Lambda)$ the group of (integer) isometries. Denote the set of all vectors $v\in\Lambda$ such that $v^2=r$ by $S_r$. I think that it is true (under some additional assumptions on rank) that $G$ acts on $S_r$ with finitely many orbits, but I don't know a good reference.

In this paper we have an argument proving this for $r=0$: http://arxiv.org/abs/1208.4626 (Theorem 3.6), when the rank of a lattice is $\geq 7$.

For unimodular lattices I think there is just one orbit ("Eichler's theorem"), probably for rank $\geq 5$.

I would be very grateful for any reference to this result in bigger generality, with arbitrary $r$ and without unnecessary rank restrictions.

The question comes from complex geometry: if $M$ is a hyperkahler manifold, there is a canonical non-unimodular integer quadratic form in $H^2(M)$, and its automorphisms are identified (up to finite index) with the mapping class group of $M$. Various geometric questions about $M$ are translated into lattice-theoretic questions about this lattice.

| cite | improve this question | | | | |
$\endgroup$
6
$\begingroup$

It is in Kneser's book Quadratische Formen. For each r, there are only finitely many classes of representations of r by the lattice.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I did not find this theorem in Kneser (my German is bad, but I can read a bit, and I tried to leaf through this book). Could you be so kind to tell me where in the book I can find it? $\endgroup$ – Misha Verbitsky Oct 24 '13 at 11:11
  • $\begingroup$ It should be Satz 30.2 on page 126. $\endgroup$ – user41811 Oct 24 '13 at 17:14
1
$\begingroup$

I think you are describing Siegel's construction for counting the representations of a number by a genus of quadratic forms. For positive forms, you count the number of representations by each class in the genus, but divide each one by the number of integer automorphs (isometries) of the particular form. In the end, you divide by the mass of the genus, the sum of the reciprocals of the automorph counts. For indefinite forms, all of that goes sideways, instead you count the orbits that you are describing, and this is finite, and not dependent on dimension. I will see if I can find a good description...

Found it, you want Schulze-Pillot_2004 at TERNARY, especially Siegel's Main Theorem on pages 305-306. Apparently this is also in Kitaoka's book.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ well, I tried.. $\endgroup$ – Will Jagy Oct 23 '13 at 23:45
  • $\begingroup$ The problem with Siegel's Main Theorem at page 304-305 is that it starts with "definite lattice", in the end there is two lines explaining what happens for indefinite lattices, but it's not immediately clear how this implies finiteness of the number of orbits. I hope the Kneser's book is more direct. $\endgroup$ – Misha Verbitsky Oct 24 '13 at 8:58
1
$\begingroup$

There's a simple criterion for equivalence of two vectors in a lattice if the lattice contains two copies of the hyperbolic plane: you look at the length of the vector and its image in the discriminant group of the lattice. Gritsenko, Hulek and Sankaran mention it here http://arxiv.org/abs/0810.1614 but the result goes back to Eichler (they refer to it as the Eichler criterion). If you're working with the Beauville form, then all known Hyperkahlers satisfy the hyperbolic condition.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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