Which matrices of the special linear group are exponentials?

Idea. The trace of $e^X$ is $2\cos\omega$ or $2$ or $2\cosh\omega$, according to the sign of $\det X$.

Cayley-Hamilton. For $X$ with $\operatorname{tr}X=0$: $X^2-(\operatorname{tr}X)X+(\det X)I=0$, so $X^2=-(\det X)I$.

Inclusion $\subset$. Let $X\in\mathfrak{sl}_2$.
- *$\det X=\omega^2>0$.* Then $X^2=-\omega^2I$, so $X^{2k}=(-1)^k\omega^{2k}I$ and $X^{2k+1}=(-1)^k\omega^{2k}X$. Summing the series,
$$e^X=\cos\omega\,I+\frac{\sin\omega}\omega\,X,\qquad\operatorname{tr}e^X=2\cos\omega\in[-2,2].$$
The value $-2$ occurs iff $\cos\omega=-1$, which forces $\sin\omega=0$ and then $e^X=-I$. Otherwise $\operatorname{tr}e^X>-2$.
- *$\det X=0$.* Then $X^2=0$ and $e^X=I+X$, with trace $2>-2$.
- *$\det X=-\omega^2<0$.* Then $X^2=\omega^2I$ and $e^X=\cosh\omega\,I+\frac{\sinh\omega}\omega X$, with trace $2\cosh\omega>2$.

Also $\det e^X=e^{\operatorname{tr}X}=1$, so $e^X\in SL_2$. Thus every exponential is in the right-hand set.

Inclusion $\supset$. Let $g\in SL_2(\mathbb{R})$ with $\operatorname{tr}g>-2$; its characteristic polynomial is $\lambda^2-(\operatorname{tr}g)\lambda+1$.
- *$|\operatorname{tr}g|<2$:* eigenvalues $e^{\pm i\theta}$ with $\theta\in(0,\pi)$ (non-real, conjugate). Then $g=PR_{\pm\theta}P^{-1}$ for some $P\in GL_2(\mathbb{R})$ (real canonical form), and $g=\exp(\pm\theta PJP^{-1})$ with $\theta PJP^{-1}$ traceless.
- *$\operatorname{tr}g>2$:* distinct positive eigenvalues $\lambda,\lambda^{-1}$. Then $g=P\,\mathrm{diag}(\lambda,\lambda^{-1})P^{-1}=\exp\bigl(P\,\mathrm{diag}(\ln\lambda,-\ln\lambda)P^{-1}\bigr)$.
- *$\operatorname{tr}g=2$:* the only eigenvalue is $1$ (double). Either $g=I=e^0$, or $g=P(I+N)P^{-1}$ with $N=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ (Jordan form, with the scaling absorbed into $P$), and $g=\exp(PNP^{-1})$ with $PNP^{-1}$ traceless.
- *$g=-I$:* $-I=e^{\pi J}$ with $\operatorname{tr}J=0$.

Not onto. The matrix $\begin{pmatrix}-1&1\\0&-1\end{pmatrix}\in SL_2(\mathbb{R})$ has trace $-2$ and is not $-I$, so it is not an exponential. Likewise $\mathrm{diag}(-2,-\frac12)$ (trace $-\frac52<-2$).

Remark. The failure of surjectivity reflects the topology: $SL_2(\mathbb{R})\simeq SO(2)\simeq S^1$ is not simply connected, and the image of $\exp$ misses exactly the elements with negative eigenvalues other than $-I$ (the hyperbolic ones with trace $<-2$ and the non-trivial parabolic ones with trace $-2$). Compare with the problem on a matrix of positive determinant that is not an exponential.