MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's say we have two $n$-dimensional lattices $(V,b)$ and $(W,b_1)$ equipped with integral bilinear forms $b$ and $b_1$ respectively. Is there an implemented function in MAGMA that decides whether $(V,b)$ and $(W,b_1)$ are isometric? Equivalently given two symmetric $n \times n$ integer matrices $M$ and $N$, is there any function that decides if $T^{t}MT=N$ for some $T \in GL_{n}(Z)$. For positive definite $M$ and $N$ one can do it by defining LM:=LatticeWithGram(M) and LN:=LatticeWithGram(N) and then asking IsIsometric(LM,LN). Since the input of LatticeWithGram must be positive definite, the above does not work for indefinite matrices.

share|cite|improve this question
You should probably ask this is some MAGMA forum... – Mariano Suárez-Alvarez Nov 26 '09 at 12:16
Or e-mail Harris Nover...he knows all this stuff. – Ben Weiss Mar 20 '10 at 2:35

not an answer to the question, but: Checking isometry is much easier for indefinite forms; it's purely local, by strong approximation for the spin group. If interested search for "spinor genus."

share|cite|improve this answer
Yes by strong approximation the spinor genus and the isometry class are the same for indefinite forms( at least in dimension bigger than 2). The problem is that I don't know how to check whether the spinor genus of M and N is the same. If M is indefinite MAGMA does not accept LatticeWithGram(M). – Guillermo Mantilla Nov 27 '09 at 4:42
OK, if you really want to use MAGMA, you can use some trick like trying to replace M,N by p-adically close but definite forms M', N'; then use Magma's implemented functions to check if these are equivalent at p. (You do this at all primes p dividing the discriminant.) I'm not sure if Magma's implemented functions cover the "spinor" version of this, however. – moonface Nov 27 '09 at 7:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.