Finding an isomorphism $\mathfrak{sp}(2) \to \mathbb{R}^3$ I am studying Lie groups and Lie algebras and I am trying to find an isomorphism between the groups/algebras and $\mathbb{R}^n$.
For $Sp(2,\mathbb{R})$, using the standard skew-symmetric matrix, and the condition
$$\begin{pmatrix}a & c \\ b & d \end{pmatrix} \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} \begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$
I got that the matrices of $Sp(2,\mathbb{R})$ are of the form 
$$\begin{pmatrix}a & b \\ c & \frac{1+bc}{a}\end{pmatrix}$$
I am trying to come up with a similar thing for the Lie algebra $\mathfrak{sp}(2,\mathbb{R})$ but I end up with
$$\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} \begin{pmatrix}a & b  \\ c & d \end{pmatrix} + \begin{pmatrix}a & c \\ b & d\end{pmatrix} \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$
$$= \begin{pmatrix}0 & a + d \\ -a-d & 0\end{pmatrix}$$ which does not have dimension $0$. According, to https://en.wikipedia.org/wiki/Table_of_Lie_groups this Lie algebra has dimension $n(2n+1)$ which is $3$ in this case, but I am not obtaining matrices isomorphic to $\mathbb{R}^3$. What am I doing wrong?
 A: I just noticed something really dumb. I forgot to use the constraint
$$\begin{pmatrix}0 & a + d \\ -a -d & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}$$
Which gives me $a = -d$.
Therefore, the matrices for $\mathfrak{sp}(2)$ are those of the form 
$$\begin{pmatrix}a & b \\ c & -a\end{pmatrix}$$
Correct me if I'm wrong please.
A: Note that $\mathfrak{sp}(2)$ is different from $\mathfrak{sp}(2,\mathbb{R})$. You are talking about the latter. The former consists of $2\times 2$ quaternionic matrices. (In fact, $\mathfrak{sp}(1)\cong\mathfrak{su}(2)\cong\mathfrak{so}(3)$ as lie algebras.)
Note that the condition that $A\in\mathrm{Sp}(2,\mathbb{R})$ turns out to be $\det A=1$, so in fact we have an equality (not just an isomorphsim, an equality) $\mathrm{Sp}(2,\mathbb{R})=\mathrm{SL}(2,\mathbb{R})$.
I assume by $\mathbb{R}^3$ you mean $(\mathbb{R}^3,\times)$, which is isomorphic to $\mathfrak{so}(3)$. I also assume you are talking about real lie algebras. In this case, you will not find an isomorphism  $\mathfrak{sp}(2,\mathbb{R})\cong(\mathbb{R}^3,\times)$, because $\mathfrak{sl}(2,\mathbb{R})\not\cong\mathfrak{so}(3)$. They are not isomorphic as real lie algebras. On the other hand, if you're talking about their complexification, then they are isomorphic complex lie algebras, $$\mathfrak{sl}(2,\mathbb{R})\otimes\mathbb{C}\cong\mathfrak{sl}(2,\mathbb{C})\cong\mathfrak{so}(3)\otimes\mathbb{C}.$$
