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A basis for $sl(2,\mathbb{C})$ is given as $\{e = \left(\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right), \; f = \left(\begin{matrix}0 & 0 \\ 1 & 0\end{matrix}\right), \; h = \left(\begin{matrix}1 & 0 \\ 0 & -1\end{matrix}\right)\}$ with relations $[e,f]=h$, $[f,h]=2f$ and $[h,e]=2e$.

A basis for $so(3,\mathbb{C})$ is given as $\{x = \begin{pmatrix}0&0&0 \\ 0&0&-1 \\ 0&1&0 \end{pmatrix}, \; y = \begin{pmatrix}0&0&1 \\ 0&0&0 \\ -1&0&0 \end{pmatrix}, \; z = \begin{pmatrix}0&-1&0 \\ 1&0&0 \\ 0&0&0 \end{pmatrix} \}$ with relations $[x,y]=z$, $[y,z]=x$ and $[z,x]=y$.

My thought are, that I somehow need to map these basis elements in the right way st. two basis have the same relations, and then construct a map from the mapping of the basis elements.

My first idea was $e\mapsto 2x$, $f\mapsto y$ and $h\mapsto 2z$, but it doesn't seem to work out, when expecting the map together with the lie bracket. I've tried some different maps, but without any succes.

KJA
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    You need to use non-real coefficients, since over $\Bbb R$ these two Lie algebras are not isomorphic. So far, you didn't. You can find an isomorphism on this site. – Dietrich Burde Feb 18 '20 at 23:17
  • Are there any specific method to find these coefficients of the map? So far, I've just tried to make some guesses from the lie bracket relations, and then checked if it worked (however no luck so far). – KJA Feb 18 '20 at 23:25
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    Yes, there are specific methods. The easiest one is to write an operator equation for a potential isomorphism $f$ and first to compare traces of operators to get "linear conditions" (in general you have polynomial equations in the coefficients of $f$). – Dietrich Burde Feb 19 '20 at 09:45
  • @DietrichBurde Could we also prove this isomorphism through Dynkin diagrams, using that $B_1 = A_1$? – cip Jan 01 '21 at 16:00

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