3
$\begingroup$

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 $\mathfrak{so}(3,\mathbb{R}) \otimes \mathbb{C} \simeq \mathfrak{sl}(2)$ ?

$\endgroup$
  • $\begingroup$ The claim in your title is actually wrong: the isomorphism only holds for the complexified algebras. $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '13 at 1:06
  • $\begingroup$ Also, it is not true that there is a unique $3$-dimensional algebra! $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '13 at 1:07
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.