Notation: $GL(n, \mathbb{Z})$ to be the set of $n \times n$ invertible matrix, and $M_{m \times n}(\mathbb{Z})$ be the set of $m \times n$ matrix with integer entries.

I have the following conjecture: Given $n \geq 1$ and two matrices $W_4$ and $S_2$ satisfy \begin{equation} W_4 \in GL(n, \mathbb{Z}), S_2 \in GL(n, \mathbb{Q}) \end{equation} Moreover $S_2^{-1} \in GL(n, \mathbb{Z}).$

then there is $S_1 \in PSL(m, \mathbb{Z})$ and $W_1 \in GL(m,\mathbb{Z}), W_2 \in M_{m \times n}(\mathbb{Z}), W_3 \in M_{n \times m}(\mathbb{Z}) $ such that \begin{equation} \left( \begin{array}{cc} W_1 & W_2 \\ W_3 & W_4 \end{array} \right)^{T}.\left( \begin{array}{cc} S_1 & 0 \\ 0 & S_2 \end{array} \right).\left( \begin{array}{cc} W_1 & W_2 \\ W_3 & W_4 \end{array} \right)=\left( \begin{array}{cc} S_1 & 0 \\ 0 & S_2. \end{array} \right) \end{equation} I also use $W=\left( \begin{array}{cc} W_1 & W_2 \\ W_3 & W_4 \end{array} \right).$ below to shorten the notations

One nontrivial example Maybe it is a little abstract, let us see an example, given $S_2=\left( \begin{array}{cc} 0 & \frac{1}{8} \\ \frac{1}{8} & 0 \end{array} \right)$ and $W_4=\left( \begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array} \right).$ One can choose the following $W$ and $S_1=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$ \begin{equation} W=\left( \begin{array}{cccc} 3 & 0 & 0 & 8 \\ 0 & 3 & -8 & 0 \\ 0 & -1 & 3 & 0 \\ 1 & 0 & 0 & 3 \\ \end{array} \right) \end{equation} One can check the following is True. \begin{equation} W^{T}.\left( \begin{array}{cccc} 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 8 \\ 0 & 0 & 8 & 0 \\ \end{array} \right).W=\left( \begin{array}{cccc} 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 8 \\ 0 & 0 & 8 & 0 \\ \end{array} \right) \end{equation}

My conjecture is that always works for any $W_4$ and $ S_2.$ In other words, even though sometimes, we can not have $W_4^{T}S_2W_4=S_2,$ but we can provide extra space for our $W_4$ to embed it into larger integer matrix $W \in PSL(n+m.\mathbb{Z})$ so that if we focus on the subspace part $S_2$. We still get some kinds of preservation. I also require $det(S_1)=1.$ One can ask the same question by adding constraints that $W_4$ and $S_2$ are symmetric. But I believe the conjectures are always true if we allow $m$ to be arbitrarily large. I have searched the literature. However, except for using computers, I do not have any systematic way to tackle this conjecture.

In terms of integer lattice language, we can define an integer lattice $S_2^{-1}$ and I want to claim that there is another unimodular $det(S_1)=1.$ lattice $S_1$ such that one can direct sum two lattice $S_1+S_2$. Then that combined lattice has an automorphism group such that its corner (or focus on $S_2$ part) can realize that $W_4$.

Special case if $W_4 \in Aut(S_2)=\{x|x \in GL(n, \mathbb{Z}), x^{T}S_2x=S_2. \}$ see related. Then the conjecture is true by choosing $S_1$ and all $W_1,W_2. W_3$ to be empty. The original equation already holds. This is the simplest case. One can further require $S_2$ to be positive definite.

One Suspicion I suspect that when the difference of $W_4^{T}S_2W_4-S_2=x \in GL(n, \mathbb{Z}).$ The conjecture should be easily proved.

Simplest nontrivial case: However, for the following example, $S_2=\frac{1}{8}, W_4=3.$ I can not find any solutions. Simplest nontrivial case: Even I required $W_4, S_2$ are just integers. In this case, $S_2=\frac{1}{p}, p \geq 2.$ ansd suppose $W_4^2=1(mod p).$ I can not prove it.

Any suggestions or ideas are really welcome. Feel free to stronger the condition and prove any case except for the trivial one are really welcome.