# Lie Subgroups of $SL(2,\mathbb{R})$

I'm wondering about the Lie subgroups of $$SL(2,\mathbb{R})$$. It's Lie algebra is the algebra of real traceless matrices and has basis elements $$L_0 = \left( \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right), L_+ = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right), L_- = \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right).$$

I think I can use the exponential map to find lie subgroups associated with each of these basis vectors. $$G= Exp(tX)$$, but I'm struggling to see how to do this.

• Would Mathematics be a better home for this question? – Qmechanic May 28 '19 at 13:03

You can use the standard expansion of exponential to write, $$G = e^{tX} = \mathbb 1+tX+\frac12 t^2 X^2+\frac1{3!} t^3 X^3+\cdots$$ Note that $$L_0^2=\mathbb1$$, $$L_+^2=0$$ and $$L_-^2=0$$. Therefore, $$\begin{split}e^{t L_0} &= Cosh[t] \ \mathbb1 + Sinh[t] L_0 = \begin{pmatrix}e^{-t}&0\\0&e^t\end{pmatrix}\\ e^{t L_+} &= \mathbb 1+ t L_+ = \begin{pmatrix}1&t\\0&1\end{pmatrix}\\ e^{t L_-} &= \mathbb 1+ t L_- = \begin{pmatrix}1&0\\t&1\end{pmatrix}\end{split}$$ In fact, using these three matrices you can generate any element of $$\mathbb{SL}(2,\mathbb R)$$ by multiplication.