# Is every finite-order unimodular matrix conjugate to a $0,1,-1$ matrix?

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.

• 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 $-2$. – Richard Stanley Nov 25 at 2:40
• @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. – Neil Strickland Nov 25 at 11:36
• @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 $n$. – Gerry Myerson Nov 25 at 11:46
• @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. – Mark Sapir Nov 25 at 15:56
• 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? – Joshua Grochow Nov 25 at 16:02

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