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How can I show that $\mathfrak{so}(3, \mathbb{R}) \otimes \mathbb{C} \simeq \mathfrak{sl}(2)$?

Is there a way that doesn't involve systems of roots?

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One approach is to use generators and relations. The generators for $\mathfrak{sl}_2$ are given by $e,f,h$ subject to the relations $[e,f]=h, [h,e]=2e$ and $[h,f]=-2f$. For $\mathfrak{so}_3$, you can take a basis of antisymmetric matrices with $A=E_{1,2}-E_{2,1}$, $B=E_{1,3}-E_{3,1}$ and $C=E_{2,3}-E_{3,2}$. The relations are $[A,B]=-C$, $[B,C]=-A$ and $[A,C]=B$. We'd like to see that these two Lie algebras are isomorphic upon complexification. Set $e=B+iA, f=B-iA$ and $h=2iC$, and check that the three $\mathfrak{sl}_2$ relations are satisfied. (I found this transformation by playing around with different combinations of $A$ and $B$.)

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By the way there is a missing $\otimes \mathbb C$ on the right of your equation. –  Grumpy Parsnip Feb 20 '13 at 19:31
    
My $\mathfrak{sl}(2)$ is $\mathfrak{sl}(2, \mathbb{C})$... –  ArthurStuart Feb 20 '13 at 23:25

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