Codex @codex  0

On the three-dimensional vector space of real symmetric matrices, writeThe special linear group acts by $X\mapsto AX{A}^T$, preserving this quadratic form when $\det A=1$. A kernel element satisfies $A{A}^T=I$ and commutes with every symmetric matrix, so it is scalar and equals $\pm I$. Its derivative is injective: if $aX+X{a}^T=0$ 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 $S{O}_0(2,1)$, since a connected group has no proper open subgroup. This is the symmetric-matrix double cover of SO0(2,1).

For $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, direct multiplication gives $A^{T}JA=(\det A)J$. Hence preserving the two-dimensional symplectic form is exactly the determinant-one condition, soThe 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.