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For $k$ a positive integer, let us consider the rank 3 ternary indefinite lattice $L=L_k$ with quadratic form $$6kx^2-2(y^2+yz+z^2).$$ Its discriminant group has length $2$.

Question. Is this lattice unique in its genus?

Theorem 21 Chapter 15 of the book "Sphere packing, Lattices and Groups" by Conway and Sloane states that, in order to be not unique, one should have that $4\cdot 18k$ is divisible by $t^3$ for some non-square natural number $t=0$ or $1\, \operatorname{mod}\, 4$. But I would like a general result and to know exactly what is going on for any $k$.

Perhaps it is too much asking; to see one example of $k$ such that the lattice $L_k$ is not unique in its genus would also be interesting.

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  • $\begingroup$ why these particular forms? $\endgroup$
    – Will Jagy
    Commented Apr 9, 2021 at 17:29
  • $\begingroup$ oh, well. From Watson's little book, the forms $x^2 +xy-y^2 +25 z^2$ and $5x^2+ 5xy-5y^2 + z^2$ are in the same genus but are distinct. Page 116. $\endgroup$
    – Will Jagy
    Commented Apr 9, 2021 at 20:38
  • $\begingroup$ same for the positive forms $x^2 + xy + y^2 + 9 z^2$ and $x^2 + 3 y^2 + 3yz + 3z^2,$ same genus, each alone in its spinor genus so both are regular in the sense of Dickson. zakuski.utsa.edu/~jagy/papers/Mathematika_1997.pdf $\endgroup$
    – Will Jagy
    Commented Apr 9, 2021 at 22:49
  • $\begingroup$ @WillJagy I am interested by these forms because they have an order 3 automorphism preserving them. Thank you very much for the examples ; do you know examples that are indefinite ? $\endgroup$
    – X77 Math19
    Commented Apr 10, 2021 at 5:33
  • $\begingroup$ the example with 5 and 25 is indefinite forms. Your (indefinite) pattern does not seem to be cooperating as far as producing more than one class in a genus. $\endgroup$
    – Will Jagy
    Commented Apr 10, 2021 at 15:45

1 Answer 1

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The answer is in Rick Miranda and David R. Morrison, Embeddings of Integral QuadraticForms https://web.math.ucsb.edu/~drm/manuscripts/eiqf.pdf Chapter VIII Theorem 7.5 (2) and the following Lemmas

Namely, for any $k$ the quadratic form is

  • 2-regular (Lemma 7.7 (1))
  • 3-semiregular (Lemma 7.6 (2))
  • p-regular for all $p\neq 2,3$ (Lemma 7.6 (1))

If you have a finite number of examples in mind you can use sage. I implemented spinor genera for the recent version of sageMath following Conway Sloane's description in SPLAG.

for k in range(1,100): 
    D = matrix(ZZ,3,3,[2,1,0,1,2,0,0,0,-3*2*k]) 
    rep = Genus(D).representatives() 
    len(rep)
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  • $\begingroup$ Thank you much Simon! I will study that paper! $\endgroup$
    – Xavier49
    Commented May 4, 2021 at 17:58

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