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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?

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  • $\begingroup$ Is the goal to get a Lie algebra isomorphic to $(\Bbb{R}^3,\times)$, i.e. the Lie product in $\Bbb{R}^3$ would be the cross product of vectors? Indeed, you left out the r.h.s. of the constraint equation from the definition of $\mathfrak{sp}(2,\Bbb{R})$. You could copy/paste it from your other post to the question body. Logically it belongs there IMO. $\endgroup$ Mar 22, 2018 at 18:05

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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.

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    $\begingroup$ This is not an answer. $\endgroup$ Mar 22, 2018 at 17:50
  • $\begingroup$ Can you comment on the answer now? $\endgroup$
    – The Bosco
    Mar 22, 2018 at 17:57
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    $\begingroup$ It is not an answer. Why don' you add this to the question, and delete this? $\endgroup$ Mar 22, 2018 at 17:57
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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, $$\mathfrak{sl}(2,\mathbb{R})\otimes\mathbb{C}\cong\mathfrak{sl}(2,\mathbb{C})\cong\mathfrak{so}(3)\otimes\mathbb{C}.$$

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