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Dear all,

once again my question is all about $SL_2(\mathbb{Z})$ and $SL_2(\mathbb{Q})$ ! Which elements in $M \in SL_2(\mathbb{Q})$ can you write in the following form:

$M= NBN^{-1}$

with $N \in GL_2(\mathbb{Q})$ and $B \in SL_2(\mathbb{Z})$?

It is surely nor all of $SL_2(\mathbb{Q})$ (look at traces), but I do not have any guess which matrices I get!

Thank you very much again! Karl

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  • $\begingroup$ Thank you for your comment! I will edit my post at once! $\endgroup$
    – Karl
    Commented Oct 26, 2010 at 9:58

1 Answer 1

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You can do this if and only if the trace of $M$ is an integer. By the theory of the rational canonical form if matrices $A$ and $B$ over $\mathbb{Q}$ have the same characteristic polynomial and neither has a repeated eigenvalue they are conjugate by a matrix over $\mathbb{Q}$. This almost does it, save for some fiddling about when the eignvalue of $M$ is repeated.

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  • $\begingroup$ What if you work over an arbitrary number field instead of $\mathbb{Q}$? Does this argument still hold? $\endgroup$
    – Karl
    Commented Oct 26, 2010 at 10:10
  • $\begingroup$ Yes, rational normal forms (and the statement) work over any field... $\endgroup$
    – Bugs Bunny
    Commented Oct 26, 2010 at 10:37
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    $\begingroup$ You also need to know there is a matrix in SL(2,Z) with the same characteristic polynomial as M's. The original question starts only with the data of one matrix M in SL(2,Q). One possibility is to write down something explicit: if t = tr(M) is in Z then the char. poly. of M is x^2 - tx + 1 and the matrix [t-1 1|t-2 1] is in SL(2,Z) and has the same char. polynomial. This goes through over number fields too. $\endgroup$
    – KConrad
    Commented Oct 26, 2010 at 13:26

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