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As I understand it, this comes down to calculating the slope of the expanding eigenvector of each matrix... but I am having trouble with the details. I feel that the fact that we have identified every matrix with its negation means that this should be true, but again, I have not been able to make anything precise. Can anyone help with this?

user672836
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  • Many somewhat related recent threads 1,2,3. Linking them here to enhance connectivity for present and future readers. – Jyrki Lahtonen Aug 16 '22 at 19:08
  • I find this question unclear, because the elements of ${\rm PSL}(2,{\mathbb Z})$ are not matrices. – Derek Holt Aug 16 '22 at 19:22
  • @DerekHolt They aren't matrices, but they are equivalence classes of matrices, and presumably one should be able to think about many questions in $PSL(2,\mathbb Z)$ by lifting things to $SL(2,\mathbb Z)$. When OP talks about eigenvalues, it seems clear to me that he is thinking about eigenvalues of a representative of the equivalence class. I would still agree that the question is unclear (slope of the expanding eigenvectors?), but not exceptionally so. – Aaron Aug 16 '22 at 19:48
  • @JyrkiLahtonen (or anybody) I had asked question 2 that you linked--just wondering, are the examples given there of matrices not conjugate to their transposes also valid in $\text{PSL}(2, \mathbb{Z})$? As in, are there elements of $\text{PSL}(2, \mathbb{Z})$ not conjugate to their transpose? – user672836 Aug 16 '22 at 20:38

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They are not. Consider $$g=\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\operatorname{PSL}_2(\mathbb Z).$$ Every conjugate of $g$ is (the projection of a matrix) of the form $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\begin{pmatrix}1-ac&a^2\\-c^2&1+ac\end{pmatrix}$$ for some $(a,b,c,d)\in\mathbb Z^4$ with $ad-bc=1$. If this is to be $g^{-1}$, we must have $$\begin{pmatrix}1-ac&a^2\\-c^2&1+ac\end{pmatrix}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$$ (these matrices cannot be negatives of each other since they both have trace $2$), which implies $a^2=-1$. This has no solutions over $\mathbb Z$.

Carl Schildkraut
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  • Thank you, can't believe I missed this!! I guess I would need something different than inverses then for a general property like this... sorry to ask like this, but do you know if the examples given there of matrices not conjugate to their transposes in $\text{SL}(2, \mathbb{Z})$ are also valid in $\text{PSL}(2, \mathbb{Z})$? As in, are there elements of $\text{PSL}(2, \mathbb{Z})$ not conjugate to their transpose? – user672836 Aug 16 '22 at 20:40
  • @user672836 I believe the same counterexample works; you get $c^2=-1$ instead of $a^2=-1$. (For transpose inverse, i.e. $(g^{-1})^T$, I'm not sure, but I'd guess you can cook up a counterexample somehow.) – Carl Schildkraut Aug 16 '22 at 20:44