$\mbox{PSL}(2,\mathbb{R})$ and $\mbox{Sp}(2,\mathbb{R})$

Question

I am wondering if somebody could point me to some kind relationship between groups $\mbox{PSL}(2,\mathbb{R})$ and $\mbox{Sp}(2,\mathbb{R})$. I know very little about Lie Algebras and group representation theory. The question is motivated by my study of the flow of a system of variational equation of a Hamiltonian system and the flow of its projectivization.

Answer 1

The relationship is $\mathrm{Sp}(2,\mathbb{R})=\mathrm{SL}(2,\mathbb{R})$. (This is equality, not just isomorphism.)

Here's why. The unique symplectic form on $\mathbb{R}^2$ up to scaling is given by

$$\begin{array}{ll} \omega\left(\begin{bmatrix}a \\ b \end{bmatrix},\begin{bmatrix} c \\ d \end{bmatrix}\right) & =\left\langle \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}a \\ b \end{bmatrix},\begin{bmatrix} c \\ d \end{bmatrix} \right\rangle \\ & =\left\langle \begin{bmatrix} -b \\ a \end{bmatrix},\begin{bmatrix} c \\ d \end{bmatrix} \right\rangle \\ & =ad-bc \\ & =\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} \end{array} $$

Then, the condition that $A\in\mathrm{GL}(2,\mathbb{R})$ preserves the symplectic form $\omega$, which may be written out in full as $\omega(Au,Av)=\omega(u,v)$, is equivalent to $\det [Au~Av]=\det [u~v]$, but since the left side is given by $\det(A)\det[u~v]$, the condition is equivalent to $\det(A)=1$, i.e. $A\in\mathrm{SL}(2,\mathbb{R})$.