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Suppose we take the "even" indefinite lattice from page 50 in Serre A Course in Arithmetic (1973) $$ U \; = \; \left( \begin{array}{cc} 0 & 1 \\\ 1 & 0 \end{array} \right),$$ called $H$ in pages 189-191 of Larry J. Gerstein Basic Quadratic Forms.

What I cannot find in any detail is a proof of this arithmetic statement in SPLAG 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.

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 SPLAG, 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 The Leech Lattice and 1990 Lattices Like the Leech Lattice, but I don't see the SPLAG claim in an explicit manner.

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.

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  • $\begingroup$ If $f$ and $g$ correspond to odd lattices, are you writing them as quadratic forms with half-integer coefficients? $\endgroup$
    – S. Carnahan ♦
    Jul 19, 2011 at 3:32
  • $\begingroup$ My understanding is that an even lattice has all inner products integral and all norms even, so that includes the hyperbolic plane above, which I would prefer to write as $h(x,y) = 2 x y.$ Now, as far as I can make out, the sum of $k$ squares would be an odd lattice, as all inner products are integral but many vectors have odd norm (such as 1). So, while I would write a perfectly good ternary forms as $f(x,y,z) = x^2 + y^2 + z^2 + y z + z x + x y,$ in order to get integral inner products we need to double it, giving an even lattice. Annoying to me. $\endgroup$
    – Will Jagy
    Jul 19, 2011 at 3:42
  • $\begingroup$ In short, I do not know why Conway mixes terminology. If I am to consider the genus of my ternary above, by adding in variables $u,v,$ is he talking about $f(x,y,z) + 2 u v$ or $f(x,y,z) + u v?$ I do not know. $\endgroup$
    – Will Jagy
    Jul 19, 2011 at 3:46
  • $\begingroup$ Oh, most of the root lattices are written as even. The most important single item in the present project is $E_8,$ let me find a link, math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/E8.html if you look at the Gram matrix it is integral with all even numbers on the diagonal. So it is "even," and half of it is still an "integer-valued" quadratic form. $\endgroup$
    – Will Jagy
    Jul 19, 2011 at 3:51
  • $\begingroup$ I see. Somehow, I got used to the convention that one starts with quadratic forms as fundamental objects, and derives even bilinear forms by the rule $B(v,w) = Q(v+w)-Q(v)-Q(w)$. This would suggest that the hyperbolic plane is written as $xy$, your ternary quadratic form produces an even lattice, and the odd bilinear form giving the sum of $k$ squares is derived from $\frac12 \sum_{i=1}^k x_i^2$. It looks like you (and perhaps many others) are viewing integral bilinear forms as fundamental. At any rate, doubling doesn't lose any information with respect to integral equivalence or genera. $\endgroup$
    – S. Carnahan ♦
    Jul 19, 2011 at 15:18

1 Answer 1

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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 ...)".

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.

Finally, since the forms are indefinite of dimension at least $3$, the spinor genus consists of a single class.

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).

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  • $\begingroup$ Thank you so much for a proof out of Cassels, which is approachable. I should say, although it may not matter, that "classically integral" includes both "odd" and "even" lattices, see the first page in chapter 26 of SPLAG for $I_{n,1}$ and $II_{n,1}.$ The odd ones are like $f(x,y) = x^2 + y^2,$ some vectors have odd norm. The even ones are like $g(x,y) = x^2 + x y + y^2,$ to get "integral" we must double to $2 x^2 + 2 x y + 2 y^2,$ and now vector norms are even. So the root lattice $E_8$ and the Leech lattice are "even" in the same sense, as polynomials it is possible to divide through by 2. $\endgroup$
    – Will Jagy
    Jul 19, 2011 at 18:11
  • $\begingroup$ Excellent. Cassels, page 111, for prime $p=2$ he says $x_1^2 + 2 x_2 x_2$ is "properly primitive" (and I believe "odd lattice") while $ 2 x_1^2 + 2 x_1 x_2$ is "improperly primitive" (and I believe "even lattice"). Then we have Lemma 4.2 on page 120 for unary cancellation in $\mathbb Z_2,$ Corollary on page 122 for improperly primitive. Next, odd $p$... $\endgroup$
    – Will Jagy
    Jul 19, 2011 at 18:42
  • $\begingroup$ You're welcome! Good point about classically integral including both even and odd - I forgot about the diagonal :) $\endgroup$
    – Abhinav Kumar
    Jul 19, 2011 at 20:01

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