6
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ If $f$ and $g$ correspond to odd lattices, are you writing them as quadratic forms with half-integer coefficients? $\endgroup$
    – S. Carnahan
    Jul 19 '11 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 '11 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 '11 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 '11 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 '11 at 15:18
7
$\begingroup$

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).

$\endgroup$
3
  • $\begingroup$ 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 '11 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 '11 at 18:42
  • $\begingroup$ You're welcome! Good point about classically integral including both even and odd - I forgot about the diagonal :) $\endgroup$ Jul 19 '11 at 20:01

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.