Show $\phi$ is a isomorphism as a lie algebra homomorphism Show $\phi$ is a isomorphism as a lie algebra homomorphism
$\phi: \textbf{su}_2  \bigotimes_{\mathbb{R}} \mathbb{C}\rightarrow sl_2(\mathbb{C})$ and $\phi: a(I  \bigotimes 1)+b(J  \bigotimes 1)+c(K  \bigotimes 1) \rightarrow (ai+b)X+(ai-b)Y+ciH$

Where {$I, J, K$} is a basis of $\textbf{su}_2$ and $a, b, c \in \mathbb{C}$ and {$X, Y, H$} is a basis of $\textbf{sl}_2\mathbb{C}$

$\textbf{What I know:}$ Requirements for $\phi$ to be a lie-algebra isomorphism:


*

*lie-bracket must be preserved $\phi([X,Y])=[\phi(X), \phi(Y)]$

*$\phi$ must be 1:1 and onto

*I believe that since $\phi$ is a lie algebra homomorphism, then we must just show $\phi$ is a vector-space isomorphism



My problem is I can't apply these facts to the given map; I just can't get my head around it. Any thanks would be very much appreciated
 A: $\phi  $ is indeed a vector space isomorphism as it takes $\{(I\otimes1) +i(J\otimes1) \space ,(I\otimes1) -i(J\otimes1)\space , (K\otimes1)\}$ to $\{2iX\space,2iY \space ,iH\}$ which is a basis(!) of $\mathfrak{sl}_2(\mathbb{C})$. So $\phi \\ $ is onto and since dimensions are same on both sides, $\phi$ is 1-1 also.
Now the bracket in $\mathfrak{su}_2 \otimes_\mathbb{R}\mathbb{C}$ is given on typical generators by $ [  M\otimes x\space,N\otimes y ] =[M\space,N] \otimes xy$ and  $\phi $ is not actually a Lie algebra homomorphism when your $I,J,K,X,Y,H $ are given by your comment for the following reason.
 $\hspace{30cm}$ We can check that $\phi(I\otimes1)=iX+iY,\phi(J\otimes1)=X-Y,\phi (K\otimes1)=iH,[I,J]=2K$ .  $\hspace{30cm}$Now $  \phi[I\otimes1\space ,J\otimes1]=\phi([I,J]\otimes1)=\phi(2K\otimes1)=2\phi(K\otimes1)=2iH$  $\hspace{30cm}$ But$[\phi(I\otimes1),\phi(J\otimes1)]=[iX+iY,X-Y]=2i[Y,X]=-2iH \neq 2iH$ $\hspace{30cm}$  Now if you replace $J\space$ by $-J$ one can easily check(on generators) in a similar fashion as above that $\phi$ is indeed a Lie algebra homo(iso)morphism.
