Lie algebra isomorphism between $\mathfrak{sl}(2,{\bf C})$ and $\mathfrak{so}(3,\Bbb C)$

Question

I think that this is an exercise. I can not find a solution.

We can define Lie bracket multiplication on $\mathbb{C}^3$ : $$ x\wedge y $$ where $x=(x_1,x_2, x_3)$, $y= (y_1,y_2,y_3)$, and $\wedge $ is the wedge product we know.

Consider the Lie algebra $\mathfrak {sl}(2,\mathbb{C})= \{ X\in M_2(\mathbb{C}) \mid\ {\rm Trace} (X) =0\}$ and $$ e= \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right),\ f= \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right),\ h= \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right). $$ Note that $$ [e,f]=h,\ [e,h]=-2e,\ [f,h]=2f.$$

Here, the problem is to find an explicit isomorphism between $\mathfrak {sl}(2,\mathbb{ C})$ and $\mathbb{C}^3$.

Thank you in advance.

Answer 1

Some hints: you have introduced a basis $\{e,f,h\}$ for the Lie algebra $sl(2,\mathbb C)$; all you need is a suitable basis $\{e_i\}$ in $\mathbb C^{3}$ and an isomorphism $\phi:\mathbb C^{3}\rightarrow sl(2,\mathbb C)$ of vector spaces s.t. $\phi(e_i\wedge e_j)=[\phi(e_i),\phi(e_j)]$. Can you find such basis? Try with the simplest one... Then you should define the isomorphism $\phi$ simply as $\phi(e_i)=...$ (choose the right element of the basis for $sl(2,\mathbb C)$: you need to preserve compatibility with brackets) and extend it $\mathbb C$-linearly.

Answer 2

Here is a description of the isomorphism $\mathfrak{sl}_2(\mathbb C)\cong\mathfrak{so}_3(\mathbb C)$ which is perhaps more enlightening:

In general given a semisimple Lie algebra $\mathfrak g$ we have the adjoint representation $\mathrm{Ad}\colon\mathfrak g\to\mathfrak{gl}(\mathfrak g)$, sending $X$ to $\mathrm{ad}\ X\colon Y\mapsto [XY]$. In fact, since $[\mathfrak g\mathfrak g]=\mathfrak g$, the homomorphism $\mathrm{Ad}$ lands in $\mathfrak{sl}(\mathfrak g)$. Moreover, $\mathrm{Ad}$ is injective since its kernel is $\mathfrak z(\mathfrak g)=0$.

Note that $\mathfrak{g}$ comes with a non-degenerate pairing (the Killing form) $K\colon\mathfrak g\times\mathfrak g\to\mathbb C$, given by $K(X,Y):=\mathrm{tr}(\mathrm{ad}\ X\circ\mathrm{ad}\ Y)$. Then we see that $K([XY],Z)=K(X,[YZ])$, so $\mathrm{ad}\ X$ is orthogonal with respect to the pairing $K(-,-)$. Thus, the adjoint representation is in fact an injective homomorphism $\mathfrak g\hookrightarrow\mathfrak{so}(\mathfrak g)$.

When $\mathfrak g=\mathfrak{sl}_2(\mathbb C)$, this gives an isomorphism, since $\dim\mathfrak g=\dim\mathfrak{so}(\mathfrak g)=3$.

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Note: This argument does not work over $\mathbb R$, since not all non-degenerate bilinear pairings $V\times V\to\mathbb R$ on a $\mathbb R$-vector space $V$ are equivalent to each other. For instance, $(x,y)*(z,w):=xz+yw$ and $(x,y)*'(z,w):=xz-yw$ are not equivalent. Bilinear forms are classified by its signature $(p,q)$, corresponding to the bilinear form on $\mathbb R^{p+q}$ given by the symmetric matrix $\begin{pmatrix}1_{p\times p}\\&-1_{q\times q}\end{pmatrix}$. Correspondingly, we have Lie algebras $\mathfrak{so}_{p,q}$, which only become $\mathfrak{so}_{p+q}$ over $\mathbb C$.

One can check that the Killing form $K(-,-)$ on $\mathfrak{sl}_2(\mathbb R)$ has signature $(2,1)$, so $\mathfrak{sl}_2(\mathbb R)\cong\mathfrak{so}_{2,1}(\mathbb R)$, not $\mathfrak{so}_3(\mathbb R)$.