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Let $F$ be a number field with ring of integers $\mathfrak{o}$. Let $(V,Q)$ be a quadratic space of dimension $n$ over $F$, and let $L$ be a free lattice in $V$ (i.e. $L\cong\mathfrak{o}^n$). If the class number of $F$ is odd, then the genus of $L$ consists of free lattices, because the squares of their Steinitz classes are equal to the class of the volume of $L$, i.e. $1$. Also, in this case, the genus of $L$ consists of a single spinor genus.

Question 1. When is it true that the genus of $L$ consists of free lattices?

Question 2. When is it true that the spinor genus of $L$ consists of free lattices?

The questions above and the ones below are motivated by comparing the mass formula in its original form proved by Siegel, and in its adelic form developed by Tamagawa, Weil, etc. Assume $F$ is totally real and $Q$ is totally positive definite. The formula in its original form concerns the average number of representations of an integer $a\in\mathfrak{o}$ by the classes of quadratic forms over $\mathfrak{o}$ in the genus of $Q$. The theorems of Hasse-Minkowski and Witt allow one to reformulate this average as one over a genus of classes of representations $(x,M)$, where $x\in V$ is fixed with $Q(x)=a$, and $M$ is any free lattice in the genus of $L$ that contains $x$. In the adelic form it is no longer required that $M$ be free.

Question 3. Is there a simple reason why the classical and adelic formulations give the same answer? It seems that Siegel averages over fewer representations $(x,M)$ than Tamagawa, Weil.

Question 4. Is there a classical notion of the spinor genus of $Q$?

I would appreciate any thoughts, results, or pointers to the literature.

Added 1. It seems that the answer to Questions 1-2 are "always true" and this also gives the answer to Questions 3-4: "there is really no difference between the classical and adelic notions in the present setting". Namely, by this paper (based on a method of Kneser from 1957) each class in the genus of $L$ can be reached from the class of $L$ by successively taking $\mathfrak{p}$-neighbors for two suitable prime ideals $\mathfrak{p}$ (which can be chosen to lie outside any prescribed finite set of prime ideals), and $\mathfrak{p}$-neighbors always have the same Steinitz class. I suspect that this conclusion can be reached more directly, but I am no expert, so I am still awaiting thoughts, results, or pointers to the literature.

Added 2. Inspired by Rainer Schulze-Pillot's answer below, here is a slightly simpler way to show that the Steinitz class of a lattice $L$ is determined by the volume of $L$, hence also by the genus of $L$. We can write $L$ as $$ L=\mathfrak{a}_1x_1+\dots+\mathfrak{a}_nx_n, $$ where $(x_i)$ is a suitable basis of $V$, and the $\mathfrak{a}_i$'s are fractional ideals in $F$. By definition, the volume of $L$ is the fractional ideal $$ \mathfrak{v}L = \mathfrak{a}_1^2\dots \mathfrak{a}_n^2\cdot d(x_1,\dots,x_r), $$ where $d(x_1,\dots,x_r)$ is the discriminant of the basis $(x_i)$. The latter equals the discriminant of $V$ modulo $(F^\times)^2$, hence $\mathfrak{vL}$ really determines $\mathfrak{a}_1\dots \mathfrak{a}_n$ modulo $F^\times$ (because the multiplicative group of fractional ideals is torsion free), which equals the Steinitz class of $L$.

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As in your "Added": indeed, the "classical" and "adelic" do count the same thing(s), although the translation between the two may be unobvious or non-trivial in some cases. For example, Strong Approximation fails in general for orthogonal groups... but, as Kneser investigated, etc., this failure is not relevant to this matter, although it might give a false impression of a discrepancy. –  paul garrett Dec 30 '12 at 16:53

1 Answer 1

up vote 9 down vote accepted

Question 1: This is not really a question of genus theory but of elementary divisor theory. If $F$ is the quotient field of the Dedekind domain ${\mathfrak o}$ any two lattices $L_1, L_2$ of full rank on the $F$-vector space $V$ of dimension $n$ can be written as $L_1={\mathfrak a}_1 x_1+ \dots + {\mathfrak a}_n x_n, L_2={\mathfrak b}_1{\mathfrak a}_1x_1+ \ldots + {\mathfrak b}_n{\mathfrak a}_nx_n$ with a basis $x_1,\ldots,x_n$ of $V$. Then $L_1,L_2$ have the same Steinitz class if the product of the elementary divisors ${\mathfrak b}_i$ is principal.

Now, if we have a non degenerate quadratic form on the space $V$ and $L_1, L_2$, considered as quadratic modules with this form on them, are in the same genus, it is clear that the product of the elementary divisors is ${\mathfrak o}$ (consider the local discriminants), so all lattices in the same genus have the same Steinitz class, in particular they are either all free or all non free.

Question 2: Is obsolete after Question 1.

Question 3 is nevertheless of interest: It appears not to be widely known that Siegel treated what we would now call the case of a non free lattice with a non-degenerate quadratic form. Of course, in his language of matrices this takes a different shape: In "Über die analytische Theorie der quadratischen Formen III" (Gesammelte Abhandlungen, vol. 1., p. 469) he treats symmetric $(n+1)\times (n+1)$ matrices of rank $n$ which are not integrally equivalent to a non-singular $n \times n$-matrix. He writes:... die hierzu nötige Verallgemeinerung ... führt zu ziemlich umfangreichen Hilfsbetrachtungen" (the necessary generalizations lead to rather extensive auxiliary considerations).

Question 4: If I remember right this is in Watson's book on integral quadratic forms (which is on my bookshelf in my office, but not here at home). The equivalence of both notions has been shown in an unpublished diploma thesis of a student of Martin Kneser in the 1970s. Since the notion of spinor genus is not much older than the adelic treatment I don't think Watson's definition has ever been used seriously.

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Thank you for the nice answer and welcome to the community! –  GH from MO Dec 30 '12 at 20:32
    
Regarding Question 1: By the same argument, lattices of the same volume are isomorphic as $\mathfrak{o}$-modules. –  GH from MO Dec 31 '12 at 9:00
    
I added a slightly simpler argument under "Added 2" of the original post. –  GH from MO Dec 31 '12 at 11:28
    
I would say that the argument is different but not simpler - my argument deals with lattices as such, regardless of any quadratic form that may be around. –  Rainer Schulze-Pillot Dec 31 '12 at 11:48
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I just edited my answer to make clear where I think of lattices with quadratic form and where the quadratic form plays no role. –  Rainer Schulze-Pillot Dec 31 '12 at 12:22

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