Let $A$ be a matrix in $SL(n,\mathbb Z)$, let $a_{ij}$ denote the element in the $i$th row and $j$th column. Suppose $|tr(A)|\leq n$. Is it true that $A$ is conjugate to a matrix $B$ such that $|b_{ij}| \leq 1$ for all $i,j= 1, \ldots, n$?

Or what are the necessary and sufficient conditions for a matrix in $SL(n, \mathbb Z)$ to be conjugate to a matrix with all entries equal to $0$, $1$ or $-1$? It is obvious that not all matrices are of this form, for example $[5, 19; 1, 4]$ is in $SL(n,\mathbb Z)$ but can't be a matrix satisfying the property since trace is preserved under conjugation. I have found some families of matrices that satisfy it but can't come up with a criterion for this question.

Also any references on the subject would be appreciated.

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    $\begingroup$ For more on integer conjugacy classes in $SL(n,\mathbb{Z})$ see Multidimensional Gauss reduction. $\endgroup$ – Dietrich Burde Mar 31 '16 at 20:14
  • $\begingroup$ Thank you for the source, I will have a look at it. $\endgroup$ – Z. L. Mar 31 '16 at 21:12

Assuming that by "conjugate" you mean "conjugate within $\operatorname{SL}(n,\mathbb Z)$", this is not true. Consider the matrix

$$ \pmatrix{1&x\\0&1}\in\operatorname{SL}(2,\mathbb Z) $$

with $x\in\mathbb Z$ and with trace $2\le2$. We have

$$ \pmatrix{d&-b\\-c&a}\pmatrix{1&x\\0&1}\pmatrix{a&b\\c&d}=\pmatrix{1+cdx&d^2x\\-c^2x&1-cdx}\;. $$

For $|x|\gt1$, the condition can only be fulfilled if $c=d=0$, contradicting $ad-bc=1$.

  • $\begingroup$ Ok, I upvoted because that answers my first question. Do you know anything about the second? Maybe some geometrical criterion, such the matrix coming from the action of a diffeomorphism on the integral homology of a manifold? $\endgroup$ – Z. L. Mar 31 '16 at 21:24
  • $\begingroup$ @Z.L.: I don't, unfortunately. $\endgroup$ – joriki Mar 31 '16 at 21:27
  • $\begingroup$ Do you think that question applies for MO? $\endgroup$ – Z. L. Mar 31 '16 at 21:27
  • $\begingroup$ @Z.L.: I'm not the best person to ask that; I don't spend much time on MO. I suspect that it would help if you could motivate why you think that such a criterion should exist. $\endgroup$ – joriki Mar 31 '16 at 21:29
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    $\begingroup$ @Z.L. Definitely it would help to know why matrices with entries $0,1,-1$ are interesting here. Without motivation it appears a bit random, e.g., we also could ask the same question for alternating sign matrices, or for matrices with entries only prime numbers. $\endgroup$ – Dietrich Burde Apr 1 '16 at 8:13

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