Let $L$ be a Lie algebra on $\mathbb{R}$. We consider $L_{\mathbb{C}}:= L \otimes_{\mathbb{R}} \mathbb{C}$ with bracket operation $$ [x \otimes z, y \otimes w] = [x,y] \otimes zw $$ far all $x,y \in L$ and $z,w \in \mathbb{C}$. We have that $L_{\mathbb{C}}$ is a Lie algebra. If $L= \mathbb{R}^{3}$ and for $x,y \in L$ we define $[x,y]:= x \wedge y$ (where $\wedge$ denotes the usual vectorial product). We have that $(L, \wedge)$ is a Lie algebra. I have to prove that $L \simeq \mathfrak{sl}(2)$. In order to do this I'd like to prove that $L \simeq \mathfrak{so}(3,\mathbb{R})$. Than, because $\mathfrak{so}(3,\mathbb{R}) \otimes \mathbb{C} \simeq \mathfrak{sl}(2)$ and $\mathfrak{sl}(2)$, up to isomorphism, is the unique $3$-dimetional semisimple algebra, I complete my proof. So my questions are: 1) How to prove that $(\mathbb{R}^{3}, \wedge) \simeq \mathfrak{so}(3, \mathbb{R})$ ? 2) Why $\mathfrak{so}(3,\mathbb{R}) \otimes \mathbb{C} \simeq \mathfrak{sl}(2)$ ?
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The claim in your title is actually wrong: the isomorphism only holds for the complexified algebras. – Mariano Suárez-Álvarez Feb 20 '13 at 01:06
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Also, it is not true that there is a unique $3$-dimensional algebra! – Mariano Suárez-Álvarez Feb 20 '13 at 01:07
1 Answers
For (1):
Let $$i:=(1,0,0),\quad j:=(0,1,0)\text{ and } k:=(0,0,1)\in\mathbb R^3$$ and $$ A:=\left(\begin{array}{ccc}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right),\quad B:=\left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{array}\right) \text{ and } C:=\left(\begin{array}{ccc}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{array}\right) \in \mathfrak{so}(3,\mathbb R). $$ So, $$i\wedge j = k,\quad i\wedge k = -j\text{ and }j\wedge k = i \in \mathbb R^3$$ and $$[A,B] = C,\quad [A,C] = -B\text{ and }[B,C]=A\in \mathfrak{so}(3,\mathbb R).$$
Then the linear application $\mathbb R^3\to\mathfrak{so}(3,\mathbb R)$ given by $$\begin{array}{rcl} i & \to & A \\ j & \to & B \\ k & \to & C \end{array}$$ is an isomorphism of Lie algebras.
For (2):
Let $$ X:=\left(\begin{array}{cc}0 & i \\ i & 0\end{array}\right),\quad Y:=\left(\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right)\text{ and } Z:=\left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right)\in\mathfrak{sl}(2,\mathbb C) $$ and $$ X'=A\otimes 2,\quad Y'=B\otimes 2\text{ and } Z'=C\otimes 2\in\mathfrak{so}(3,\mathbb R)\otimes\mathbb C, $$ for $A$, $B$ and $C\in\mathfrak{so}(3,\mathbb R)$ as in item (1). So, $$[X,Y] = 2Z,\quad [X,Z] = -2Y\text{ and }[Y,Z]=2X\in \mathfrak{sl}(2,\mathbb C)$$ and $$[X',Y'] = 2Z',\quad [X',Z'] = -2Y'\text{ and }[Y',Z']=2X'\in \mathfrak{so}(3,\mathbb R)\otimes\mathbb C.$$
Then, the linear application $\mathfrak{sl}(2,\mathbb C)\to\mathfrak{so}(3,\mathbb R)\otimes\mathbb C$ given by $$\begin{array}{rcl} X & \to & X' \\ Y & \to & Y' \\ Z & \to & Z' \end{array}$$ is an isomorphism of Lie algebras.
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