Proving the existence of an integral quadratic form

Question

Theorem 11 (Conway & Sloane, Sphere Packings, Lattices and Groups, 3rd Edition, pp 383, Ch 15). If a system of putative $p$-adic symbols for each $p$ satisfies the determinant, oddity, and $p$-adic existence conditions, then there exists an integral quadratic form with these $p$-adic symbols.

The book does not provide any references or pointers on where to find a proof. We checked the obvious reference i.e., B. W. Jones, Cassels, O'Meara, and could not find the relevant theorem. Unfortunately, we find it far from obvious and hence are unable to prove it ourselves.

We have two quick questions.

1. Can someone please recommend a reference where a proof appears ?
2. Is the Theorem if and only if ?

Thank you for your time.

Answer 1

The Conway–Sloane method is getting popular, partly because they gave the version of The Mass Formula that everyone uses. Some students of Gabriele Nebe, at Aachen, have begun doing calculations with it, see Lorch and Kirschmer - Single-Class Genera of Positive Integral Lattices. Meanwhile, Daniel Allcock taught a course on this last year and all the homework was in this C–S slang. Now, Allcock's adviser was Borcherds, and Borcherds' adviser was Conway. You can't tell me that's a coincidence.

Meanwhile, given a description of a genus by discriminant, signature, and a representative in each of the $p$-adic integers, why is the genus not empty? It seems magic sometimes. For positive ternary forms, the careful article is Levels of positive definite ternary quadratic forms by Larry Lehman (1992), I put a pdf at ternary. I wrote an article using that, it said there was a genus, I looked for it in my huge computer list, there they were. Amazing.

I think the most reassuring thing should be Siegel's weighted representation. Along with the mass formula, Siegel also showed how to calculate the average number of times a given integer was represented by a genus. For positive forms, the actual number of representations by an individual form is just divided by the number of integer automorphs of the form. For indefinite forms, the automorphism groups are infinite, so instead, the count is taken of distinct orbits. Here is the magic part: if an integer is $p$-adically eligible for each prime (including $\infty,$ which Conway has taken to calling $-1$), then the average number of representations, over the genus, is nonzero. Meaning the genus is not empty.

If you have not looked at it, Watson's book gives the beginnings of this argument. Remember that we create genera of binary forms by taking an eligible prime $p,$ solving $ \beta^2 \equiv \Delta \pmod {4p},$ finally arriving at the form $$ \left\langle p, \beta, \frac{\beta^2 - \Delta}{4p} \right\rangle. $$

Answer 2

The existence of integral quadratic forms with prescribed “local” data (i.e., essentially the genus) has been also proved by Nikulin. For a proof of Nikulin's existence theorem, see for example Miranda and Morrison - Embeddings of Integral Quadratic Forms, Theorem 5.2.

On the other hand, Conway and Sloane do provide some references – perhaps they are helpful. They write at the beginning of section 7: "No proofs will be offered. For $p\neq 2$ several accounts are readily available (e.g. Cassels, [Cas3]), and all cases are handled by O'Meara [O'Me 1], … etc.".

Answer 3

The theorem is actually presented in full details in Cassels, but it is a bit hard to equate to Conway. First, note that a genus symbol as in Conway-Sloane directly describes a sum of Jordan factors. Then apply Cassels's theorem about the existence of forms in genera given the right local factors, and the fact that the determinant and Hilbert reciprocity are the only conditions on existence of a rational form locally isomorphic to given local forms.