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Using the fact that every matrix in $SL(2,\mathbb{R})$ is conjugate in $SL(2,\mathbb{R})$ to one of the following matrices:

$$ \left(\begin{array}{rr} a & 0\\ 0 & \frac{1}{a} \end{array}\right), \quad \left(\begin{array}{rr} 1 & t\\ 0 & 1 \end{array}\right), \quad \left(\begin{array}{rr} -1 & t\\ 0 & -1 \end{array}\right) \quad \mbox{y} \quad \left(\begin{array}{rr} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array}\right), $$ Show that the image of the exponential map $$\exp\: : \: \left\{ \left(\begin{array}{rr} x & y \\ z & -x \end{array}\right) \: : \: x,y,z \in \mathbb{R}\right\} \longrightarrow SL(2,\mathbb{R})$$ is \begin{equation}\left\{M\in SL(2,\mathbb{R}) \: : \: tr(M)>-2\right\} \cup \left\{-I\right\}. \qquad (*)\end{equation}

Remark: I have tried many things, for example, we know that $$\exp(PJP^{-1}) =P\exp(J)P^{-1}$$

We also know that the trace is invariant under cujugación, this is $$tr\left(\exp(PJP^{-1}) \right)=tr\left(\exp(J)\right).$$ Therefore, if $M \in \exp \left(SL(2,R)\right)$, then there is $A\in SL(2,R)$ such that $M=\exp(A)$, then, as $A$ is conjugate to one of above matrices, then $$tr(M)= \left\{ \begin{array}{} tr\left(\exp\left(\left(\begin{array}{rr} a & 0\\ 0 & \frac{1}{a} \end{array}\right) \right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr} a & 0\\ 0 & \frac{1}{a} \end{array}\right)\\ tr\left(\exp\left( \left(\begin{array}{rr} 1 & t\\ 0 & 1 \end{array}\right)\right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr} 1 & t\\ 0 & 1 \end{array}\right)\\ tr\left(\exp\left(\left(\begin{array}{rr} -1 & t\\ 0 & -1 \end{array}\right) \right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr} -1 & t\\ 0 & -1 \end{array}\right)\\ tr\left(\exp\left( \left(\begin{array}{rr} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array}\right) \right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array}\right)\\ \end{array} \right.$$ From the latter, the problem that I have is to show that all matris in $ \exp \left(SL(2,R)\right)$ is in the set $(*)$. The biggest problem I have when I want to show that all matris in set $(*)$ is in $ \exp \left(SL(2,R)\right)$.

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Notation: We will use the following notations.

$$ r(\theta) = \left ( \begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) $$

$$ n(x) = \left ( \begin{matrix} 0 & x \\ 0 & 0 \end{matrix} \right ) $$

$$ a(t) = \left ( \begin{matrix} t & 0 \\ 0 & -t \end{matrix} \right ) $$

where $\theta$, $x$, $t$ are real numbers.

We will use Iwasawa decomposition . It says that any matrix in $SL(2,\mathbb{R})$ is of the form $KAN$ where $K=$ a rotation matrix, $A$ is a diagonal matrix (with positive entries) and $N$ is a unipotent matrix.

We will show that we can find trace zero real matrices $(k,a,n)$ such that $e^{k} = K$, $e^{a} = A$ etc.

For any diagonal matrix $A$ we can take $a$ = diagonal matrix with entries $ln(a)$ and $ -ln(a)$ on the diagonal.
For any unipotent matrix $N(x)$ with $x$ on the off-diagonal matrix we can take $n(x)$.

Finally if $K(\theta)$ is the rotation matrix, which rotates the plane by $\theta$ then consider the matrix $r(\theta)$ above. The $n$-th product $r(\theta)^{n}$ will only have diagonal or off-diagonal elements depending on whether $n$ is odd or even. It is easy to see that $e^{r(\theta)} = K(\theta)$ when you substitute power-series expressions for $\cos(\theta)$ and $\sin(\theta)$.

The estimates on the eigenvalues follow easily.

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  • $\begingroup$ Be ideal is to consider the suggestion that I gave at the beginning of the problem statement. $\endgroup$ – Diego Fonseca Feb 18 '16 at 23:11

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