$\mathfrak{sp}(4,\mathbb{R})$ isomorphic to $\mathfrak{so}(5,\mathbb{R})$. How to show this isomorphism?
In this link (proposition 4) there's a prescription on how to build a proof for the complex case, how to use this for the real part, and how to show that this is indeed a Lie algebra homomorphism, i.e that the Lie bracket is invariant under this mapping?
 A: These algebras are not isomorphic, since $\mathfrak{so}(5,\mathbb R)$ is a compact real form of its complexification while $\mathfrak{sp}(4,\mathbb R)$ is a split real form of its complexification. 
The basic source of isomorphism of that type is the representation of $\mathfrak{sp}(4,\mathbb R)$ on $\Lambda^2\mathbb R^4$. This has dimension $6$, but there is an invariant subspace $\Lambda^2_0\mathbb R^4$ of dimension $5$, the kernel of the symplectic form preserved by $\mathfrak{sp}(4,\mathbb R)$. Moreover, the wedge product defined a symmetric bilinear form on $\Lambda^2\mathbb R^4$ with values in $\Lambda^4\mathbb R^4\cong\mathbb R$, which is invariant under the action of $\mathfrak{sp}(4,\mathbb R)$. One easily verifies that the restriction of this to $\Lambda^2_0\mathbb R^4$ is non-degenerate but not positive definite (indeed it has split signature $(2,3)$). Hence the representation of $\mathfrak{sp}(4,\mathbb R)$ on $\Lambda^2_0\mathbb R^4$ actually defines a homomorphism to $\mathfrak{so}(2,3)$. This is non-zero and hence injective since $\mathfrak{sp}(4,\mathbb R)$ is simple and counting dimensions, one sees that it has to be surjective, too. Hence $\mathfrak{sp}(4,\mathbb R)$ is isomorphic to $\mathfrak{so}(2,3)$ rather than to $\mathfrak{so}(5)$. 
