Showing that the exponential map $\mathrm{exp}:\mathfrak{sl}(2,\mathbb{R})\to\mathrm{SL}(2,\mathbb{R})$ is not surjective I am having a difficult time showing that the exponential map $\mathrm{exp}: \mathfrak{sl}(2, \mathbb{R}) \rightarrow \mathrm{SL}(2, \mathbb{R})$ is not surjective.  I have, however,  worked out that $\mathfrak{sl}(2, \mathbb{R})$ is given by $\{A \in M(n, \mathbb{R}) \mid \mathrm{trace}(A) = 0\}$.
I will be appreciative of any help.  Thank you
 A: Can you find a matrix $X\in\mathfrak{sl}(2,\mathbb R)$ such that $\exp X=\left(\begin{smallmatrix}-1&1\\0&-1\end{smallmatrix}\right)$? 
Suppose we have such a matrix $X$. Since $\exp X$ is not diagonalizable, $X$ is also not diagonalizable, and it must have a double eigenvalue. Since it has trace zero, that eigenvalue is necessarily zero. Now, if $\lambda$ is an eigenvalue of $X$, $\exp(\lambda)$ is then an eigenvalue of $\left(\begin{smallmatrix}-1&1\\0&-1\end{smallmatrix}\right)$, so that $\lambda=\pi i+2\pi i k$ for some $k\in\mathbb Z$. Yet this cannot be zero!
A: The same answer as Mariano's, but with a different explanation:
Let $J=\pmatrix{-1&1\\ 0&-1}$. We claim that there does not exist a real matrix $A$ (traceless or not) such that $\exp(A)=J$. Suppose the contrary. Then $JA=(A+A^2+\frac{A^3}{2!}+\ldots)=AJ$. Yet, every matrix $A$ that commutes with $J$ must be of the form $\pmatrix{a&b\\ 0&a}$(note 1). Hence $J$ has a repeated eigenvalue $e^a>0$, which is a contradiction.
Note 1: In general, the matrices that commute with a Jordan block with a nonzero eigenvalue are precisely those upper triangular circulant matrices Toeplitz matrices.
