# An element of $SL(2,\mathbb{R})$

Find the relationship between an elliptic element of $SL(2,\mathbb{R})$ and rotation..

An element $A$ of $SL(2,\mathbb{R})$ is called an elliptic element if $|\text{tr}(A)|<2$

As $|\text{tr}(A)|<2$ so the characteristics equation of A does not have any real roots...so it has no real eigen value..but I am unable to go ahead...

• Who says this is unclear? $SL(2, \Bbb{R})$ is the group of invertible $2\times 2$ real matrices. – apnorton Aug 28 '14 at 22:50

Let $g$ be your elliptic element. Since the eigenvalues are complex conjugate, $g$ will have an eigenvector $\binom z1\in\Bbb C^2$ with the imaginary part of $z$ positive.

It can be shown that you can always find some $h\in{\rm SL}_2(\Bbb R)$ such that $h\binom i1=c\binom z1$ for some $c\in\Bbb R$.

Thus $h^{-1}gh$ will have $\binom i1$ as eigenvector.

The final step is to show that the elements of ${\rm SL}_2(\Bbb R)$ having $\binom i1$ as an eigenvector are precisely the elements in the subgroup ${\rm SO}_2(\Bbb R)$, i.e. the rotations.

I leave the details as an exercise.

• That can be easily obtained just writing it down what it means. – Andrea Mori Aug 29 '14 at 13:18
• Suppose that $\left(\begin{array}{cc}a & b \\c & d\end{array}\right)\binom i1=\lambda\binom i1$ for some $\lambda\in\Bbb C$. Then $\frac{ai+b}{ci+d}=\frac{\lambda i}\lambda=i$. This yelds $a=d$ and $b=-c$. Combining this with $ad-bc=1$ you get $a^2+b^2=1$ so that $a=\cos t$, $b=\sin t$ for some $t\in \Bbb R$. But then your original matrix is rotation by $t$. – Andrea Mori Aug 30 '14 at 10:55

Here is a second proof. Let $\theta\in\Bbb R\setminus\pi\Bbb Z$ be such that $\mathrm{Tr}(A)=2\cos(\theta)$. The characteristic polynomial of $A$ equals $\chi_A(X)=X^2-\mathrm{Tr}(A)X+\det(A)=(X-e^{i\theta})(X-e^{-i\theta})=\chi_{R_{\theta}}(X)$ where $$R_{\theta}=\begin{pmatrix} \cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta) \end{pmatrix}$$ Since $\theta\notin\pi\Bbb Z$, $e^{i\theta}\neq e^{-i\theta}$ and $A$ and $R_{\theta}$ are diagonalzable (over $\Bbb C$) and conjugate (as complex matrices) to $$\begin{pmatrix} e^{i\theta}&0\\0&e^{-i\theta} \end{pmatrix}$$ We conclude by invoking the general fact that two real matrices that are conjugate as complex matrices are conjugate as real matrices.