Fixed $\sigma_x=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)$ and $K=\left( \begin{array}{ccc} 3 & 32 & -64 \\ 1 & 32 & -32 \\ -2 & -32 & 64 \\ \end{array} \right),$

I want to find $n$ such that, $K_n=\sigma_x \oplus \sigma_x \oplus \cdots\oplus\sigma_x \oplus K$. and the corresponding $W \in GL(2n+3,\mathbb{Z})$ such that $W^{T}K_n^{-1}W=K_n^{-1},$ the only constraints for $W$ the downright corner should be $\left( \begin{array}{cc} 8 & -5 \\ 5 & 13 \\ \end{array} \right).$ For example, if $n=1,$ I manage to find a $W$ looks like $W= \begin{bmatrix} a_1 & a_2 & a_3 & a_4 & a_5 \\ b_1 & b_2 & b_3 & b_4 & b_5 \\ c_1 & c_2 & c_3 & c_4 & c_5 \\ d_1 & d_2 & d_3 & 8 & -5 \\ e_1 & e_2 & e_3 & 5 & 13 \end{bmatrix} $ such that: \begin{equation} W^{T}K_1^{-1}W=K_1^{-1}. \end{equation} I try to use Mathematica and find no solutions. Then I try $n=2$ and I still can not find the solution. But I don't know how to prove there is no solution for $W^{T}K_n^{-1}W=K_n^{-1}.$ for any $n$. I believe there should be $1$.

My comment about this problem:

1 I can not find much literature about the diophantine equation solution of integer quadratic form (at least for $4 \times 4$). But my imagination is that when $n$ grows, there should be more freedom and more variables. So one should be able to find a solution or at least prove the solution exists.

2 Since $K_n \oplus \sigma_x$ is an indefinite matrix so that there is no computational system to find that automorphism group of $K_n$. See related question here. An indefinite matrix will lead to an infinite automorphism group so the capacity of the automorphism group should be large enough to capture such $W$.

3 Notice that the smith normal form for $=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 32 & 0 \\ 0 & 0 & 32 \\ \end{array} \right). $ So for any $K_n$ it always has two nontrivial components $32$. So this is why there is a constraint for $2 \times 2$ corner for $W$.

My research has a problem related to this type of question given $K$ and $W$ (restrict $W$ for the right down corner and claim that one can find large enough $K$ (enlarge $K$ and $W$) with the same and nontrivial Smith normal ($\neq 1$) form and satisfy $W^{T}K_1^{-1}W=K_1^{-1}.$ )

Any comments and results are very welcome, thanks a lot.

  • $\begingroup$ Let's use simple English: what do you already know, and what are you looking for? For instance, you say that "My claim is that I should be able to find large enough $n$ such that $K_n=\sigma_x \oplus \sigma_x \oplus \dots \oplus \sigma_x \oplus K$", but a bit further you seem to look for a matrix $W$. I'm confused. $\endgroup$
    – Alex M.
    Oct 21, 2021 at 19:48
  • $\begingroup$ OK, I just want to find $n$ and corresponds $W$. I change the text, sorry for the confusion $\endgroup$
    – En-Jui Kuo
    Oct 21, 2021 at 20:03
  • 1
    $\begingroup$ I take it that $n$ is meant to be the number of copies of $\sigma_x$ in the definition of $K_n$. But then I'm confused when you write that you managed to find $W$ for $n=1$, but Mathematica found no solutions. Also, it's not clear to me why this particular $K$ is interesting, nor why the insistence on that particular $2\times2$ lower right corner. $\endgroup$ Oct 21, 2021 at 22:00
  • $\begingroup$ n is the number of pauli x, this is correct $\endgroup$
    – En-Jui Kuo
    Oct 22, 2021 at 2:04


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