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There seems to be general opinion that, for positive integral quadratic forms in at least three variables, spinor genera in the same genus all have the same mass (not representation measures of some number, that is different, indeed some recent authors write of representation mass of numbers and it throws me off). Authors R. Scharlau and R. Schulze-Pillot have been loosely mentioned in this regard.

Does anyone know for sure, and in particular know a simple (published) reference for positive forms? Evidently there is an analogous formulation where it is true for indefinite forms, see The Hirzebruch-Mumford volume for the orthogonal group and applications, by Valery Gritsenko, Klaus Hulek, G.K. Sankaran,

I should add that Kneser (1961) is sometimes mentioned in this regard, but my take is that, while his methods can be used to reconstruct the result, he is not explicit about the mass.

Here is computer output on an example that appears in Benham, Earnest, Hsia, Hung (1990), formula (3.8), positive ternary forms, including one of the 29 known spinor regular forms that are not regular. The sextuple a b c d e f refers to the form $T(x,y,z) = a x^2 + b y^2 + c z^2 + d y z + e z x + f x y,$ with discriminant $\Delta = 4 a b c + d e f - a d^2 - b e^2 - c f^2.$


   ===Discriminant   343 ==Genus Size==  3
   343 = 7^3
 All forms in regular spinor genus represent      1
  Spinor genus misses        1     4    16    64   121   256
   484   529   841
       343:    2     7          8      7    1    0    auto 4   Level 196  irreg spin candidate 
--------------------------size 1 
  Spinor genus misses     no exceptions  
       343:    1     2         49      0    0    1    auto 8   Level 196
       343:    1     7         14      7    0    0    auto 8   Level 196
--------------------------size 2 
Disc    343
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What do you mean when you say "mass"? Could you give us a definition of the mass? Thanks. –  emiliocba Dec 28 '11 at 18:14
Yes, any positive form has a finite number of integral automorphs, those being integral matrices of determinant $\pm 1$ that preserve the quadratic form, so the collection is also called the orthogonal group of the form. For a genus, or, here, a spinor genus, the mass is the sum of the reciprocals of the automorph counts over all classes in the genus (spinor genus). In the example above, the mass of the genus is 1/4 + 1/8 + 1/8 = 1/2. See… –  Will Jagy Dec 28 '11 at 20:09
Isn't the mass a genus invariant? –  emiliocba Jan 3 '12 at 13:37
@emiliocba yes. In the same way, Siegel's weighted representation measure of a number can be calculated from purely local information. The mass is sometimes referred to as the measure of representing the 0 form. Meanwhile, for positive forms it is just a finite sum, and one may ask for the value of that finite sum on a subset of the genus, in particular one of the spinor genera when there are more than one of those. –  Will Jagy Jan 3 '12 at 21:15

2 Answers 2

up vote 3 down vote accepted

Better late than never:

In the adelic setup, the mass or measure of a lattice $L$ (the german word Maß translates to measure in English) is given as $\mu(O(V)\backslash O(V)O_{\mathbb A}(L))$, where $\mu$ is the (Tamagawa-normalized) Haar measure on the adelic orthogonal group $O_{\mathbb A}(V)$ of the underlying quadratic space $V$ and $O_{\mathbb A}(L)$ is the adelic isometry group of $L$; it is proportional to $\frac{1}{\vert O(L)\vert}$ in the definite case (here $O(L)$ is the global group of isometries of $L$, which is finite in the definite case).

If we sum that up over a set of representatives of the classes in the spinor genus of $L$ we obtain $\mu(O(V)\backslash O(V)O'_{\mathbb A}(V)O_{\mathbb A}(L))$, where $O'_{\mathbb A}(V)$ is the group of adelic orthogonal transformations of $V$ which have spinor norm $1$ at each place.

If we replace here the lattice $L$ by another lattice $\sigma L$ in the genus of $L$ (with $\sigma \in O_{\mathbb A}(V)$), we have

$$O(V)O'_{\mathbb A}(V)O_{\mathbb A}(\sigma L)=O(V)O'_{\mathbb A}(V)\sigma O_{\mathbb A}(L)\sigma^{-1}=O(V)O'_{\mathbb A}(V)O_{\mathbb A}(L),$$ since $O'_{\mathbb A}(V)$ contains all commutators.

So, the measures for the spinor genus of $L$ and for that of $\sigma L$ are the same.

The argument is implicit in Kneser's 1961 article and also in Weil's 1962 article, but unfortunately neither of these authors bothered to write down the statement explicitly.

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I'm late on that question ... I found it while looking at the posts related to this other question on spinor genera.

It seems to me that the answer should be yes, at least in dimension $d\geq 4$. Here is a justification.

Take a quadratic space $(V,q)$ over $\mathbf Q$, of dimension $d$, and $L$ an integral lattice on $V$. Let $\mathcal G$ be the genus of $L$, i.e. the set of lattices on $V$ that are equivalent to $L$ at each place. It decomposes in a certain number of spinor genera (the equivalence relation here is that two lattices are in a same spinor genus if there exists an isometry $g_p$ of spinor norm $\theta(g_p)=1$ between their completions at each place $p$). Let $S$ be the set of these spinor genera.

The group $\mathrm{O}(V\otimes \mathbf A_\mathbf Q)$ acts transitively on $S$. The stabilizer of the spinor genus of $L$ is generated by $\mathrm{O}'(V\otimes \mathbf A_\mathbf Q)$ and $\mathrm{O}(L\otimes \mathbf A_\mathbf Z)$. One deduces that $A$ is acted upon simply transitively by an elementary abelian $2$-group isomorphic to $\theta(\mathrm{O}(V\otimes \mathbf A_\mathbf Q))/\theta(\mathrm{O}(L\otimes \mathbf A_\mathbf Z)$.

If the map $\theta(\mathrm{O}(V))\to \theta(\mathrm{O}(V\otimes \mathbf A_\mathbf Q))/\theta(\mathrm{O}(L\otimes \mathbf A_\mathbf Z)$ happens to be onto, then there is a representant of each class in each spinor genus, so the masses are the same. By Hasse-Minkowski, this happens for example when $d\geq 4$.

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