= Solution
The <Lie algebra> of <SL2R> is the space of traceless real two-by-two <matrices>. For $A$ in this space, the <Cayley-Hamilton theorem> gives $A^2=cI$, where $c=-\det A$. Expanding the <matrix exponential>, if $c=s^2>0$ then
$$
\exp A=(\cosh s)I+\frac{\sinh s}{s}A,
\qquad \operatorname{tr}(\exp A)=2\cosh s\geq2.
$$
If $c=0$, then $\exp A=I+A$ and its <trace> is $2$. If $c=-s^2<0$, then
$$
\exp A=(\cos s)I+\frac{\sin s}{s}A,
\qquad \operatorname{tr}(\exp A)=2\cos s\in[-2,2].
$$
In every case $\operatorname{tr}(\exp A)\geq-2$. But
$$
g=\begin{pmatrix}-2&0\\0&-1/2\end{pmatrix}
$$
has determinant one and <trace> $-5/2$. Thus $g\in SL_2(\mathbb R)$ is not an exponential, and \b[the exponential map is not surjective].
Back to article page
