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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Aut{Aut}$Fixed $n \geq 2$, given $K \in \GL(n,Z)$. One can view $K$ is a Gram matrix of Lattice. I also imposed that $K$ is symmetric i.e $K^{T}=K$. We can define the automorphism group of $K$, $\Aut(K)=\{ W \in \GL(n,Z)\mathrel|W^TKW=K \}$. One can see a similar question at matrix congruence and smith normal form.

I know one can use Magma or Sage to compute the generator of $\Aut(K)$ if $K$ is positive definite (now $\Aut(K)$ is finite). But now I want to ask is there a way to compute some element of $\Aut(K)$? What I understand is that if $K$ is indefinite, then $\Aut(K)$ may be infinite. So I understand why both software can not compute the whole group. But what I just want to know: Is there a way at least compute to some of the elements in $\Aut(K)$?

To be more precise, given $K$ is indefinite, I know that $\pm I \in \Aut(K)$. But I want to know some other elements (I do not need to know all $\Aut(K)$). Where can I get such an algorithm or software? I have searched a lot but I still can not find such one.

Another question is that if I restricted some element of $W$ or some of the element $W$ satisfy some relation, can I check whether such $W \in Aut(K)$? I want to check whether such solution exists?

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  • $\begingroup$ for a binary with dicriminant positive but not a square, there is an explicit recipe once we solve $x^2 - \Delta y^2 = 4.$ I like treatment in Buell, Binary Quadratic Forms. I wrote my own software for dimension 3 and 4. For indefinite, I just tell it an upper bound on the entries (absolute value). Not hard. I also allow different quadratic forms, so, given symmetric integer $G,H$ I program $kH=P^t G P$ where $k$ is an integer constant I choose. Useful for many things. $\endgroup$
    – Will Jagy
    Commented Jan 27, 2021 at 18:21
  • $\begingroup$ see mathoverflow.net/questions/141284/… $\endgroup$
    – Will Jagy
    Commented Jan 27, 2021 at 18:46
  • $\begingroup$ @WillJagy, thanks. But I may need to solve that for the 6 by 6 matrix K. I will check the post. $\endgroup$
    – en kuo
    Commented Jan 27, 2021 at 20:08

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