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According to "The Geometry of Four-Manifolds" by Donaldson and Kronheimer, indefinite unimodular forms are classified by their rank, signature and type. This is the Hasse-Minkowski classification of indefinite forms, they say.

However, this seems to be a bit of a folklore theorem, as I cannot find a single citation for it; all of my searches for any permutation of "hasse minkowski indefinite quadratic classification" yield instead a different theorem, namely one about solving the equation $Q(x) = r$ over $\mathbb{Q}$ for a given quadratic form $Q$.

Is it simply the case that the integral classification (which essentially states, for the case of an even form, that it is a sum of $E_8$ lattices and Hyperbolics) is an easy consequence of this other theorem? I'm not familiar enough with quadratic forms to see how this should be so.

If it isn't an easy consequence, is there a reference for the integral classification that I just haven't found?

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Google is good, but sometimes you have to actually read the book. Don-Kron cite two references in the bibliographic notes at the end of ch. 1: Serre's "Course in Arithmetic" and Milnor-Husemoller's "Symm. Bilinear Forms", as in Allan Edmonds's answer. Serre's book is on my shelf and contains a complete proof; probably M-H does too. –  Tim Perutz Mar 5 '10 at 22:03
    
Perhaps part of the problem is that the classification result you want -- for indefinite unimodular quadratic forms over $\mathbb{Z}$ -- is not part of the Hasse-Minkowski theory, which considers quadratic forms over $\mathbb{Q}$ and its completions. I think the result in question is simply more recent (certainly than Minkowski), but I don't have a reference on hand to back this up. –  Pete L. Clark Mar 6 '10 at 3:06
    
If that's the case, then why is it cited as the Hasse Minkowski classification in Donaldson-Kronheimer? –  Simon Rose Mar 6 '10 at 7:36
    
"Cited" as in they give a reference to paper of Hasse and/or Minkowski? Or just called the "Hasse-Minkowski classification"? I suspect the latter, and then I couldn't tell you why they say that. It is certainly true that you use H-M to prove this result, and that the proof is simpler than the derivation of H-M itself, so in some sense it's a consequence. I'm just saying that a quadratic form theorist -- rather than a geometric topologist -- would probably not say it this way, which explains your search engine difficulties. –  Pete L. Clark Mar 6 '10 at 14:04
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2 Answers

up vote 4 down vote accepted

For a survey of the topic check out Milnor and Husemoller, Symmetric Bilinear Forms, Springer, 1973, II.3. They cite Borevich-Shafarevich (Number Theory), O'Meara (Introduction to Quadratic Forms), and Serre (Cours d'Arithmétique).

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This seems to be what I'm looking for. Thanks. –  Simon Rose Mar 5 '10 at 22:10
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You want Theorem 6 in Section 2.2 of Chapter 5 of Serre. –  Qiaochu Yuan Mar 5 '10 at 23:09
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For integral forms, I like Rational Quadratic Forms by J. W. S. Cassels, it is available as a Dover reprint. But some of your choice of language suggests you might prefer Conway and Sloane, Sphere Packings, Lattices, and Groups. I get useful results from Integral Quadratic Forms by G. L. Watson. Then there is The Arithmetic Theory of Quadratic Forms by B. W. Jones. Any of these is easier than O'Meara. But perhaps you will learn all you need from Serre and Milnor and Husemoller, as mentioned above.

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