By "complex structure" I am referring to 2x2 matrices which square to $-\mathrm{Id}_2$. I need to know those with integer entries and determinant equal to 1.
Thank you
What are all the complex structures on $\mathbb{R}^2$ which live inside $\mathrm{SL}_2(\mathbb{Z})$?
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$\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$– GH from MOCommented 17 mins ago
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The matrices in $\mathrm{SL}_2(\mathbb{Z})$ whose square is $-\mathrm{Id}_2$ are precisely the matrices in $\mathrm{SL}_2(\mathbb{Z})$ with characteristic polynomial $x^2+1$. So these are the matrices in $\mathrm{SL}_2(\mathbb{Z})$ whose trace is zero. By looking at the corresponding fixed points in the upper half-plane, it is not hard to show that the matrices in question form two conjugacy classes in $\mathrm{SL}_2(\mathbb{Z})$ with representative elements $$\begin{pmatrix}0&-1\\1&0\end{pmatrix}\qquad\text{and}\qquad\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$
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$\begingroup$ Thank you. As a follow up, can I ask if these conjugacy classes contain infinitely many distinct elements in $\mathrm{SL}_2(\mathbb{Z})$ ? Furthermore, would it be easy to determine all the roots (not just square) of $-\mathrm{Id}_2$ inside $\mathrm{SL}_2(\mathbb{Z})$ ? $\endgroup$– nayreelCommented 1 hour ago
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$\begingroup$ @nayreel Yes, both conjugacy classes contain infinitely many elements. The conjugacy class of $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ consists of the integral matrices $\begin{pmatrix}a&b\\c&-a\end{pmatrix}$ with $c>0$ and $a^2+bc=-1$. The conjugacy class of $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ consists of the integral matrices $\begin{pmatrix}a&b\\c&-a\end{pmatrix}$ with $c<0$ and $a^2+bc=-1$. I continue in the next remark. $\endgroup$ Commented 32 mins ago
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$\begingroup$ @nayreel To your second question: if an element of $\mathrm{SL}_2(\mathbb{Z})$ has finite order $n$, then $n\in\{1,2,3,4,6\}$. These orders do occur in $\mathrm{SL}_2(\mathbb{Z})$, and the corresponding finite order elements have a simple description as in the case $n=4$ discussed in my post. In particular, $-\mathrm{Id}_2$ only has square-roots and cube-roots in $\mathrm{SL}_2(\mathbb{Z})$. $\endgroup$ Commented 21 mins ago