3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $ad-bc=1$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.