= Solution
On the three-dimensional vector space of real symmetric matrices, write
$$
X=\begin{pmatrix}t+x&y\\y&t-x\end{pmatrix},\qquad -\det X=x^2+y^2-t^2.
$$
The <special linear group> acts by $X\mapsto AXA^T$, preserving this <quadratic form> when $\det A=1$. A kernel element satisfies $AA^T=I$ and commutes with every symmetric matrix, so it is scalar and equals $\pm I$. Its derivative is injective: if $aX+Xa^T=0$ for every symmetric $X$, taking $X=I$ makes $a$ skew-symmetric, and commuting with diagonal matrices then makes $a=0$.
Both real <Lie group> dimensions are three. The derivative is consequently an isomorphism, so the image is an open subgroup. The source $SL(2,\mathbb R)$ is connected: polar decomposition deforms it onto $SO(2)$, with the positive symmetric determinant-one factor contractible. Its image is thus the full identity component $SO_0(2,1)$, since a connected group has no proper open subgroup. This is the <symmetric-matrix double cover of SO0(2,1)>.
For $J=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$, direct multiplication gives $A^TJA=(\det A)J$. Hence preserving the two-dimensional <symplectic form> is exactly the determinant-one condition, so
$$
\boxed{SO_0(2,1)\cong SL(2,\mathbb R)/\{\pm I\}\cong Sp(2,\mathbb R)/\{\pm I\}.}
$$
The subscript zero is necessary. The full $SO(2,1)$ also contains a component reversing time orientation, which cannot be the image of the connected source.
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