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Let $V={\mathbb{R}}^4$ have symplectic basis $(e_1,f_1,e_2,f_2)$. Fix $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$ and define a symmetric bilinear form $Q$ on ${\Lambda }^{2}V$ by $u\wedge v=Q(u,v)\mathrm{vol}$. The pairings between complementary basis bivectors split this six-dimensional space into three hyperbolic planes, so $Q$ has signature $(3,3)$.

The symplectic group preserves volume and the bivector $\Omega =e_1\wedge f_1+e_2\wedge f_2$, obtained from the inverse symplectic form, up to its fixed sign convention. Since $Q(\Omega ,\Omega )=2$, its orthogonal complement $W={\Omega }^{\perp }$ has dimension five and signature $(2,3)$. On $W$ use $-Q$, which has signature $(3,2)$. This constructs the exterior-square double cover of SO0(3,2).

If an element acts trivially on $W$, it also fixes $\Omega$, so its action on the entire exterior square is the identity. Every decomposable bivector then shows that its two-plane is preserved. A line is the intersection of two such planes; hence every line is preserved, and the matrix is scalar. Its exterior-square action forces its scalar square to equal one, giving precisely $\pm I$.

The derivative is injective by the same infinitesimal argument: zero action on ${\Lambda }^{2}V$ forces an infinitesimal scalar, whose induced scalar there is twice its value and therefore zero. The symplectic Lie algebra has dimension $2^2+2(2\cdot 3/2)=10$, from blocks $\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with symmetric $B,C$. The orthogonal Lie algebra in five dimensions also has dimension ten. Thus the image is open. Real symplectic polar decomposition has connected compact factor $U(2)$ and a contractible positive symplectic factor, so $Sp(4,\mathbb{R})$ is connected and its image is exactly the identity component:As in the preceding noncompact cases, this does not identify the quotient with the disconnected full $SO(3,2)$.