0
$\begingroup$

Let $\Lambda$ be an integral lattice in some definite ternary quadratic space $(V,Q)$ over $\mathbb{Q}$.

Consider the usual class set $\text{Cl}(\Lambda) = O(V)\backslash\text{Gen}(\Lambda)$, i.e. the finite set characterising global isometry classes of lattices in the genus of $\Lambda$.

Choose representative lattices $\Lambda_1, \Lambda_2, ..., \Lambda_h$ for the classes in $\text{Cl}(\Lambda)$. Are there (simple?) conditions on $\Lambda$ that guarantee the equality $|\text{Aut}(\Lambda_i)| = |\text{Aut}(\Lambda_j)|$ for all $1\leq i < j\leq h$?

$\endgroup$
3

1 Answer 1

1
$\begingroup$

I put lots of references at http://zakuski.math.utsa.edu/~kap/

I've got an early version working. At first I thought it would be just class number one or two.

The six coefficients $a,b,c,d,e,f$ refer to form $$ q(x,y,z) = a x^2 + b y^2 + c z^2 + d yz + e zx + f xy $$ The discriminant is $$ 4abc + def - a d^2 - b e^2 - c f^2$$

Ummm: the set contains all class number ones. Regular forms have discriminants divisible only by primes $2, 3, 5, 7, 11, 13, 17, 23.$ Makes sense to ask whether all these are so restricted(no). Also, is this set finite? I still think so.

These genera come from Alexander Schiemann's table 2

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

       495 :     1     2         71      1    1    1  auto  4  genus: 6 form: 1
       495 :     1     7         18      3    0    0  auto  4  genus: 6 form: 2
       495 :     1    10         13      5    0    0  auto  4  genus: 6 form: 3
       495 :     1    11         13      8    0    1  auto  4  genus: 6 form: 4
       495 :     2     7          9      0    0    1  auto  4  genus: 6 form: 5
       495 :     4     4          8     -1    1    1  auto  4  genus: 6 form: 6
 WOW 495 = 3^2 * 5 * 11 formcount 6

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

       285 :     1     2         41      1    1    1  auto  4  genus: 3 form: 1
       285 :     1     7         11      4    1    0  auto  4  genus: 3 form: 2
       285 :     4     4          6      3    3    3  auto  4  genus: 3 form: 3
 WOW 285 = 3 * 5 * 19 formcount 3

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

       438 :     1    11         13     11    1    1  auto  4  genus: 3 form: 1
       438 :     2     5         11      1    0    0  auto  4  genus: 3 form: 2
       438 :     5     5          7      5    5    4  auto  4  genus: 3 form: 3
 WOW 438 = 2 * 3 * 73 formcount 3

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

       876 :     1    10         22      2    0    0  auto  4  genus: 1 form: 1
       876 :     2     7         17      2    0    2  auto  4  genus: 1 form: 2
       876 :     3     7         11      4    0    0  auto  4  genus: 1 form: 3
 WOW 876 = 2^2 * 3 * 73 formcount 3

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

The last two genera are strongly linked; each form in one of them represents double a form in the other genus. With Hessian matrices $G,H$ we have integral $P$ of appropriate determinant such that $P^T G P = 2 H$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

       582 :     1     5         31      2    1    1  auto  4  genus: 3 form: 1
       582 :     1    13         13      9    0    1  auto  4  genus: 3 form: 2
       582 :     2     5         15      3    0    0  auto  4  genus: 3 form: 3
       582 :     5     5          7      1    1    4  auto  4  genus: 3 form: 4
 WOW 582 = 2 * 3 * 97 formcount 4

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

       625 :     2     2         42     -1    1    1  auto  4  genus: 2 form: 1
       625 :     2     5         17      5    1    0  auto  4  genus: 2 form: 2
       625 :     3     3         18      1    1    1  auto  4  genus: 2 form: 3
       625 :     3     5         12      5    2    0  auto  4  genus: 2 form: 4
       625 :     5     7          7      6    0    5  auto  4  genus: 2 form: 5
 WOW 625 = 5^4 formcount 5

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Here is a nice one. The genus has a regular form as well as a spinor regular form. There is a bit of an extra chance for your condition when the genus splits into spinor genera with equal numbers of classes, by the mass formula.

     27648 :     1    48        144      0    0    0  auto  8  genus: 1 form: 1
     27648 :     4    48         49     48    4    0  auto  8  genus: 1 form: 2
     27648 :     9    16         48      0    0    0  auto  8  genus: 1 form: 3
     27648 :    16    25         25     14   16   16  auto  8  genus: 1 form: 4
 WOW 27648 = 2^10 * 3^3 formcount 4

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

       846 :     1     5         45      3    0    1  auto  4  genus: 5 form: 1
       846 :     1    10         23      8    1    0  auto  4  genus: 5 form: 2
       846 :     1    11         22     10    0    1  auto  4  genus: 5 form: 3
       846 :     1    13         17      5    1    0  auto  4  genus: 5 form: 4
       846 :     5     5         10      4    4    1  auto  4  genus: 5 form: 5
       846 :     5     7          7      5    1    1  auto  4  genus: 5 form: 6
 WOW 846 = 2 * 3^2 * 47 formcount 6

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Interesting, the 846 sits above this 94. Kap had all sorts of maps from a genus to another with discriminant ratio a square. I'll need to look up some things.

        94 :     1     3          9      2    1    1  auto  4  genus: 1 form: 1
        94 :     1     5          5      1    0    1  auto  4  genus: 1 form: 2
 WOW 94 = 2 * 47 formcount 2

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .