An example of a matrix in $\mathrm{SL}(4, \mathbb{Z})$ with the following properties

Question

I am looking for a matrix $M\in \mathrm{SL}(4, \mathbb{Z})$, with all eigenvalues equal to $1$, and with the following properties:

Write $M=\begin{bmatrix} A_1&A_2\\ A_3&A_4 \end{bmatrix}, $ where the $A_i$ are $2$ by $2$ sumbatrix of $M$.

Let $d_i$ be the dot product of two rows of $A_i$, i.e. if $A_i = \begin{bmatrix} a&b\\ c&d \end{bmatrix}$, then $d_i = ac +bd$.

Let $a_i = \mathrm{det}(A_i) - d_i$.

For example if $A_1 =\begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}$, then $d_1 = 11$ and $a_1 = - 2 - 11 = -13$.

Consider the matrix $A = \begin{bmatrix} a_1&a_2\\ a_3&a_4 \end{bmatrix}$, I would like to find $M$ such that $A$ is in $\mathrm{GL}(2, \mathbb{Z})$, and has one eigenvalue with absolute value not equal to $1$.

The matrices I have tried so far: since $M$ needs to have eigenvalues all equal to $1$, I tried the matrices consisting of just Jordan blocks, unfortunately, they didn't work. I also tried the matrix in the form $M=\begin{bmatrix} A_1&A_2\\ 0&A_4 \end{bmatrix}, $ where $A_1$ and $A_4$ have all eigenvalues all equal to $1$, but in this case $A$ will have all eigenvalues with absolute values $1$. I am a bit stuck as I can't think of other $4$ by $4$ matrices that have all eigenvalues equal to $1$.

Any idea to construct such matrices will be really appreciated.

Alright, I will try it with some $A \in SL_4(\Bbb Z)$ with $16$ variables and solve the equations (yes, might be hard). — Dietrich Burde, Jun 21 '22 at 11:37
@DietrichBurde I've been running a Groebner basis for this problem for the last hour. It hasn't finished, but the memory has been stable so far. — Michael Burr, Jun 21 '22 at 11:48
What happens if you change $\mathbb{Z}$ to $\mathbb{R}$ or $\mathbb{C}$, can you find such a matrix? — Michael Burr, Jun 21 '22 at 11:48
ghc1997, I have found an integral solution. Please check my computation. — Dietrich Burde, Jun 22 '22 at 08:51

Dietrich Burde · Accepted Answer · 2022-06-22T08:51:06.563

I have no difficulties to come up with several types of matrices $A\in SL_4(\Bbb Z)$ having all eigenvalues equal to $1$. For example, take $$ M=\begin{pmatrix} 0 & 0 & -1 & 0 \cr 1 & 0 & 0 & 1 \cr 1 & 0 & 0 & 2 \cr 0 & 1 & -3 & 4 \end{pmatrix}. $$ Then the matrix of determinants minus dot products is $$ A=\begin{pmatrix} 0 & -1 \cr 1 & 2 \end{pmatrix}\in SL_2(\Bbb Z). $$ However, this one has also only eigenvalues $1$.

The polynomial system for finding $M$ is indeed solvable with Groebner, however adding the condition $\det(A)=\pm 1$ is far too much for Groebner in my case. So one has to choose a certain form to reduce the complexity.

Edit: I found solutions over $\Bbb Q$, namely $$ M=\begin{pmatrix} 0 & \frac{3}{2} & -1 & 0 \cr 0 & 0 & \frac{1}{3} & 0 \cr 2 & -3 & 3 & 0 \cr -5\eta-3 & \eta & 0 & 1 \end{pmatrix} \in SL_4(\Bbb Q). $$ for all $\eta\in \Bbb Q$. Then the matrix $A$ is given by $$ A=\begin{pmatrix} 0 & \frac{1}{3} \cr -3 & 3 \end{pmatrix}\in SL_2(\Bbb Q). $$ and has characteristic polynomial $t^2-3t+1$.

Edit: Finally, by the same procedure, I found infinitely many integral solutions:

$$ M=\begin{pmatrix} 0 & 2 & -1 & 0 \cr 1 & 0 & 1 & 0 \cr 2 & -3 & 3 & 0 \cr -5\eta-7 & \eta & 0 & 1 \end{pmatrix} \in SL_4(\Bbb Z). $$ Then we obtain $$ A=\begin{pmatrix} -2 & 1 \cr -7 & 3 \end{pmatrix}\in SL_2(\Bbb Z), $$ which has characteristic polynomial $t^2-t+1$.

You are welcome. Did you confirm it? Is it related to automorphism groups of your previous question? — Dietrich Burde, Jun 22 '22 at 10:05
It is indeed correct. Yes, I am trying to look for some elements in the automorphism groups. — ghc1997, Jun 22 '22 at 11:37
If we let $a_i = \mathrm{det}(A_i) $, would such matrix exist? don't worry if this is not obvious, I might ask another question if I can't find one myself. — ghc1997, Jun 22 '22 at 12:19

Michael Burr · Answer 2 · 2022-06-21T14:15:59.470

More of a long comment over a full answer.

If this question does not have a solution in $\mathbb{C}$, then it cannot have a solution in $\mathbb{Z}$. Therefore, I wrote the following Macaulay2 code to study the problem over $\mathbb{C}$:

R = QQ[r_0..r_3,s_0..s_3,t_0..t_3,u_0..u_3]
S = R[lambda]
M = matrix{{r_0,r_1,s_0,s_1},
           {r_2,r_3,s_2,s_3},
           {t_0,t_1,u_0,u_1},
           {t_2,t_3,u_2,u_3}}
L = lambda*id_(S^4)
D = det (M-L)
C = coefficients D
-- dot products of the matrices
d1 = r_0r_2+r_1r_3
d2 = s_0s_2+s_1s_3
d3 = t_0t_2+t_1t_3
d4 = u_0u_2+u_1u_3
-- determinants of the matrices
e1 = r_0r_3-r_1r_2
e2 = s_0s_3-s_1s_2
e3 = t_0t_3-t_1t_2
e4 = u_0u_3-u_1u_2
-- values of a's
a1 = e1 - d1
a2 = e2 - d2
a3 = e3 - d3
a4 = e4 - d4
T = QQ[first entries vars R|{w},MonomialOrder=>Eliminate 16]
-- Conditions to be eigenvalues 1.
f1 = sub((C_1)(1,0)-(-1)^1*binomial(4,1),T)
f2 = sub((C_1)(2,0)-(-1)^2binomial(4,2),T)
f3 = sub((C_1)_(3,0)-(-1)^3binomial(4,3),T)
f4 = sub((C_1)_(4,0)-(-1)^4*binomial(4,4),T)
-- w is value of trace of A
I = ideal(f1,f2,f3,f4,w-sub(a1+a3,T),1-sub(a1a4-a2a3,T))
G = gb I
selectInSubring(1,G)
J = ideal(f1,f2,f3,f4,w-sub(a1+a3,T),1+sub(a1a4-a2a3,T))
H = gb J
selectInSubring(1,H)

The code first computes the coefficients of the characteristic polynomial for M and the polynomials $f1,\dots,f4$ force those coefficients to be the coefficients of $(\lambda-1)^4$. The $d1,\dots,d4$ are the dot products and the $e1,\dots,e4$ are the determinants of the $A_i$'s. Then, the $a$'s are calculated as you described.

Finally, we make a new ring where $w$ is the value of the trace of $A$. We make an ideal whose elements restrict the coefficients of the characteristic polynomial of $M$ to be correct, the determinant of $A$ to be $1$, and set $w$ to be the trace of $A$. If we eliminate all variables except for $w$, then we get the possible values for the trace.

If the trace has only finitely many values, then one can hope that these cases can be studied individually.

I gave up on the computation, the memory usage was gettinglarge. — Michael Burr, Jun 21 '22 at 17:00

An example of a matrix in $\mathrm{SL}(4, \mathbb{Z})$ with the following properties

2 Answers2

Linked