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Problem. Given a matrix $A\in\mathrm{GL}(n,\mathbb{Z})$ such that $A^k=1$ for some $k\geq 1$, is there a matrix $g\in\mathrm{GL}(n,\mathbb{Z})$ such that $gAg^{-1}$ has only $0$, $1$, and $-1$ as possible entries?

Edit: after the remark by Mark Sapir that it is a famous open problem (which for me was already sufficient as an answer), I changed my question into the following ones, so now maybe it is more suitable for staying on MO without being closed.

What is known about this problem? Which other parts of mathematics is it connected to?

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    $\begingroup$ A possible counterexample is the $48\times 48$ companion matrix for the cyclotomic polynomial $\Phi_{105}(x)$, since this polynomial has two coefficients equal to $-2$. $\endgroup$
    – Richard Stanley
    Commented Nov 25, 2019 at 2:40
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    $\begingroup$ @MarkSapir I think it would not be optimal to close this. It would be better to record as an actual answer that it is a famous open problem. $\endgroup$
    – Neil Strickland
    Commented Nov 25, 2019 at 11:36
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    $\begingroup$ @Mark, yes, it can be proved that in order for the cyclotomic polynomial $\Phi_n$ to have a coefficient exceeding $1$ in absolute value, $n$ must have three or more distinct odd prime factors, and $105=3\times5\times7$ is the smallest such $n$. $\endgroup$
    – Gerry Myerson
    Commented Nov 25, 2019 at 11:46
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    $\begingroup$ @AntonyQuass: I also do not see any strong connection. I heard about the OP problem long ago, and the only approach I heard about is what in Richard Stanley's comment. If a 0,1 -1 matrix has finite order, its char polynomial should divide $x^k-1$ for some $k$. That ia a connection with cyclotomic polynomials. The conjugacy problem for $SL_n(Z)$ is not trivial and the conjugacy to 0,1,-1matrix must be even harder. $\endgroup$
    – user6976
    Commented Nov 25, 2019 at 15:56
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    $\begingroup$ Also, if it is a famous open problem, it'd be nice to have at least one reference to where it's been discussed. I've found it not so easy to search for... @MarkSapir do you know one? $\endgroup$
    – Joshua Grochow
    Commented Nov 25, 2019 at 16:02

2 Answers 2

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I found a proof here for $n=4$:

Yang, Qingjie, Conjugacy classes of torsion in (\mathrm{GL}_N(\mathbb Z)), Electron. J. Linear Algebra 30, 478-493 (2015). ZBL1329.15063. MR3414308

See the discussion in the last paragraph on p. 482 for the case that the characteristic polynomial is irreducible, and Theorem 1.7 for the reducible case.

On the other hand, I suppose it's possible that the number of conjugacy classes of finite-order elements in $GL_n(\mathbb{Z})$ could grow faster than the number of $0,\pm1$ matrices intersected with $GL_n(\mathbb{Z})$. One can get a lower bound on the number of conjugacy classes of finite-order elements in $GL_n(\mathbb{Z})$ by counting the number which are block-diagonalizable with irreducible blocks. This should correspond to a sum over decompositions of $n$ into $\varphi(m)$ by $|Cl(\mathbb{Z}[e^{2\pi i/m}])|$, a sum over class numbers, since one obtains a conjugacy class of element of $GL_{\varphi(m)}(\mathbb{Z})$ of order $m$ for every ideal class in $\mathbb{Z}[e^{2\pi i/m}]$. I have no intuition though for the growth of this function, especially since the class numbers of cyclotomic fields behave erratically.

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For the record, the case $n=3$ of the problem can also be easily deduced by the lists presented in Tahara, On the finite subgroups of GL(3,Z)

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