# Diophantine equation about the automorphism group of lattice by constraints

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: $$$$W^{T}K_1^{-1}W=K_1^{-1}.$$$$ I try to use Mathematica and find no solutions. Then I try $$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$$.

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.

• 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. Oct 21, 2021 at 19:48
• OK, I just want to find $n$ and corresponds $W$. I change the text, sorry for the confusion Oct 21, 2021 at 20:03
• 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. Oct 21, 2021 at 22:00
• n is the number of pauli x, this is correct Oct 22, 2021 at 2:04