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I would like to prove that the two algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb R)$ are not isomorphic as Lie algebras.

I started considering the following basis for $\mathfrak{su}(2)$: \begin{equation} \sigma_1=\frac{1}{2} \begin{pmatrix} i &0\\ 0 &-i\\ \end{pmatrix} \quad \sigma_2=\frac{1}{2} \begin{pmatrix} 0 &i\\ i &0\\ \end{pmatrix} \quad \sigma_3=\frac{1}{2} \begin{pmatrix} 0 &-1\\ 1 &0\\ \end{pmatrix} \end{equation} and for $\mathfrak{sl}(2,\mathbb R)$: \begin{equation} A_1=\frac{1}{2} \begin{pmatrix} 1 &0\\ 0 &-1\\ \end{pmatrix} \quad A_2=\frac{1}{2} \begin{pmatrix} 0 &1\\ 0 &0\\ \end{pmatrix} \quad A_3=\frac{1}{2} \begin{pmatrix} 0 &0\\ 1 &0\\ \end{pmatrix} \end{equation} It is possible to check that for the $\sigma$s the following commutation relation holds: \begin{equation} [\sigma_i,\sigma_j]=\epsilon_{ijk}\sigma_k. \end{equation} While if we consider: \begin{equation} [A_1,A_2]=A_2 \end{equation} Then it seems that I can't find any isomorphism between these 2 basis. Now I would like to generalize the argument to every basis, does anyone have any idea on how to proceed?

glS
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popoolmica
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2 Answers2

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Looking at a general basis will not help much, I guess, but what you have done can be completed: Note that the subspace $U = \def\<#1>{\left<#1\right>}\<A_1, A_2>$ of $\def\sl{\mathfrak{sl}(2, \mathbf R)}\sl$ generated by $A_1$ and $A_2$ is a subalgebra, as $[A_1, A_2] = A_2 \in U$. So $\sl$ has a two-dimensional subalgebra. We will prove that $\def\su{\mathfrak{su}(2,\mathbf C)}\su$ does not have any. Define $\phi \colon \su \to \mathbf R^3$ by $\phi(\sigma_i) = e_i$, where $e_i$ denotes the $i$-th standard unit vector. Then we have $$ \phi([\sigma_i, \sigma_j]) = \phi(\epsilon_{ijk}\sigma_k)=\epsilon_{ijk}e_k = e_i \times e_j = \phi(\sigma_i) \times \phi(\sigma_j) $$ where $\epsilon_{ijk}$ is the Levi-Civita symbol and $\times$ denotes the usual cross product. Hence $$ \phi([A, B]) = \phi(A) \times \phi(B),\qquad A,B \in \su $$ by linearity which implies that $\phi$ is an isomorphism of algebras of $\su$ with $(\mathbf R^3, \times)$. Now let $A, B \in \su$ be linearly independent, then $\phi(A)$ and $\phi(B)$ span a two-dimensional linear subspace $V$ of $\mathbf R^3$, hence $\phi(A) \times \phi(B)$ is a non-zero vector orthogonal to $V$; that is, $\phi([A,B]) \not\in V$, giving - as $\phi$ is invertible - $[A,B]\not\in \phi^{-1}[V] = \<A,B>$. So $A, B$ do not generate a two-dimensional subalgebra. As $A$ and $B$ were arbitrary, there isn't any.

Therefore $\su$ and $\sl$ aren't isomorphic.

Community
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martini
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There are several ways to see that the two Lie algebras are not isomorphic. First, $\mathfrak{sl}(2)$ has a $2$-dimensional subalgebra $\langle A_1,A_2\rangle$, but $\mathfrak{su}(2)$ has not - see above. Secondly, the Killing forms are not equivalent, hence the Lie algebras cannot be isomorphic, see this duplicate:

The Killing forms of $\mathfrak{su}(n)$ and $\mathfrak{sl}(n,\Bbb R)$ are not isomorphic (as real Lie algebras)

Furthermore, this question has appeared already here:

Showing the Lie Algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$ are not isomorphic.

Dietrich Burde
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