Lorentzian characterization of genus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:52:57Z http://mathoverflow.net/feeds/question/70666 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus Lorentzian characterization of genus Will Jagy 2011-07-18T20:21:40Z 2011-07-19T15:58:54Z <p>Suppose we take the "even" indefinite lattice from page 50 in Serre <em>A Course in Arithmetic</em> (1973) $$U \; = \;<br> \left( \begin{array}{cc} 0 &amp; 1 \\ 1 &amp; 0<br> \end{array} \right),$$ called $H$ in pages 189-191 of Larry J. Gerstein <em>Basic Quadratic Forms</em>.</p> <p>What I cannot find in any detail is a proof of this arithmetic statement in <em>SPLAG</em> by Conway and Sloane, page 378 in the first edition(1988), anyway chapter 15 section 7, that quadratic forms $f,g$ are in the same genus if and only if $f \oplus H$ and $g \oplus H$ are integrally equivalent. Then they say this follows from properties of the spinor genus, presumably including Eichler's theorem that indefinite rank at least 3 means spinor genus and class coincide. Also, if f and g do not correspond to "even lattices," I'm not entirely sure what is being claimed. Oh, I absolutely cannot assume $f,g$ are in any way "unimodular." Very popular, that unimodular. Matter of taste, though. I'm not sure it matters, but my $f,g$ are going to be positive, which is surely the difficult case here.</p> <p>Everybody with whom I have discussed this regards this as either obvious or, essentially, an axiom. I would very much like a reference for this, plus an explanation of what is meant if $f,g$ correspond to "odd" lattices. For example, it would be wonderful if somewhere this claim and the words Theorem or Proposition or Lemma happened in the same sentence. I think I am making progress on the other bits I need, essentially ch. 26,27 in <em>SPLAG</em>, but this claim has me snowed, or perhaps buffaloed, thrown, stumped. As far as books that I own, I do not see the claim being discussed in Jones, Watson, O'Meara, Serre, Cassels, Kitaoka, Ebeling, Gerstein. I stopped by the office of R. Borcherds and discussed related matters for a while, the relevant articles are 1985 <em>The Leech Lattice</em> and 1990 <em>Lattices Like the Leech Lattice</em>, but I don't see the SPLAG claim in an explicit manner. </p> <p>EDIT... Sexy application: the Leech lattice and all the Niemeier lattices are in the same genus. Pointed out in an MO comment by Noam Elkies, who knows things.</p> http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus/70748#70748 Answer by Abhinav Kumar for Lorentzian characterization of genus Abhinav Kumar 2011-07-19T15:58:54Z 2011-07-19T15:58:54Z <p>A good reference for this assertion is Cassels's "Rational Quadratic Forms", though you have to dig a bit. Let me see if I can outline the proof. First, I think Conway and Sloane assume $f$ and $g$ are classical integral (i.e. correspond to even lattices). In my copy of SPLAG, at the end of subsection 2.1 of that chapter, they say "so in this book we call $f$ an integral form if and only if its matrix coefficients are integers (i.e. if and only if it is classically integral ...)".</p> <p>Now suppose $f$ and $g$ are in the same genus. Then so are $f\oplus H$ and $g \oplus H$. Next, we want to show they're in the same spinor genus. This follows from the Corollary of Lemma 3.6 of Chapter 11 of Cassels: "If we show $U_p \subset \theta(\Lambda_p)$ for all $p$, then the genus of $\Lambda$ consists of a single spinor genus". Here $\Lambda = f \oplus U$, where I'm identifying the form and the lattice by a bit of abuse of notation. Since $\theta(\Lambda_p) \supset \theta(H_p)$ (see a few sentences below the corollary), and $\theta(H_p) \supset U_p$ by Lemmas 3.7 and 3.8, we've proved that the genus consists of a single spinor genus.</p> <p>Finally, since the forms are indefinite of dimension at least $3$, the spinor genus consists of a single class.</p> <p>To go back is the easier direction (I think): if $f \oplus U$ is equivalent to $g \oplus U$, then they are equivalent over $\mathbb{Z}_p$ for every $p$. Then an analogue of Witt cancellation will do the job (see Chapter 8 of Cassels).</p>