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For a Matrix Lie group, the exponential map of a matrix Lie group is the convergent matrix exponentialIts domain is the Lie algebra $\mathfrak{g}=T_{I}G$, and $e^{tX}\in G$ for every real $t$ when $X\in \mathfrak{g}$. Because multiples of the same matrix commute, $e^{sX}{e}^{tX}=e^{(s+t)X}$ and $(e^{tX}{)}^{-1}=e^{-tX}$. Thus take $I=\mathbb{R}$ to obtain the image of a one-parameter subgroup.

Here is the precise classification of nontrivial one-parameter subgroups. For $X\ne 0$, the smooth group homomorphism ${\gamma }_X:t\mapsto e^{tX}$ has a closed kernel $K\subset \mathbb{R}$. A closed subgroup of $\mathbb{R}$ is zero, $\tau \mathbb{Z}$ for some $\tau >0$, or all of $\mathbb{R}$. Indeed, if positive elements have infimum zero, their integer multiples approximate every real number, so closedness makes the subgroup all of $\mathbb{R}$; otherwise the infimum is attained and is its least positive generator. The last possibility is excluded by ${\gamma }_X^{\prime }(0)=X\ne 0$. Give the image the quotient topology and smooth structure from $\mathbb{R}/K$. Its inclusion in $G$ is an injective immersion, since ${\gamma }_X^{\prime }(t)=e^{tX}X$ never vanishes. Consequently it is a Lie subgroup, withIn the periodic case $[0,\tau )$ is also a parameter interval covering the image, with endpoints understood modulo $\tau$.

Two qualifications matter. If $X=0$, the image is the trivial group, a third possibility omitted by the wording “two”. Also, a general one-parameter subgroup need not be closed or embedded: an irrational winding $t\mapsto \mathrm{diag}(R(t),R(\sqrt{2}t))$ in a two-torus is an example, where $R(t)$ is a planar rotation. The classification uses the intrinsic immersed Lie subgroup structure, rather than assuming that the matrix subspace topology is the topology of $\mathbb{R}$.

For the real special linear group, differentiating $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$ at the identity shows that its Lie algebra consists of traceless matrices. Conversely $\det e^{tX}=e^{t\mathrm{tr}X}=1$ for a traceless matrix. HenceChoose the standard basis of the real sl2 Lie algebra,Their squares have the required values. Direct matrix multiplication gives the Lie bracketsWriting $[L_a,L_b]=f_{ab}{}^c{L}_c$, these specify all the nonzero Lie algebra structure constants: $f_{0+}{}^{+}=2$, $f_{0-}{}^{-}=-2$, $f_{+-}{}^0=1$, together with their negatives on reversing the lower indices.

The three requested exponentials areThus the first two images are isomorphic to the additive real group, while the last is $SO(2)$, the circle group. The PDF gives $L_{+}-L_{-}$ here; the local TeX's $L_{-}-L_{-}$ is a transcription error.

For a general generator, put $\Delta ={\alpha }_0^2+{\alpha }_{+}{\alpha }_{-}$. Direct multiplication gives $X^2=\Delta I_2$. If $\Delta >0$, the eigenvalues $\pm \sqrt{\Delta }$ give an unbounded exponential image. If $\Delta =0$ but $X\ne 0$, it is a nilpotent linear map with exponential $I_2+tX$, again unbounded. If $\Delta =-{\omega }^2<0$, the power series instead giveswith least positive period $2\pi /\omega$. Its image is a continuous image of a circle and is compact. Equivalently, $X/\omega$ is a real complex structure and is similar over $\mathbb{R}$ to the rotation generator. Therefore the exact criterion for compact one-parameter subgroups of SL2R isFor a nonzero generator only the strict inequality remains.