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

  • $\begingroup$ You are not taking any basis here, you are taking the canonical generators. $\endgroup$ – Captain Lama Mar 16 '16 at 1:11
  • $\begingroup$ @CaptainLama could you explain canonical generators please? $\endgroup$ – thinker Mar 16 '16 at 1:14
  • $\begingroup$ Are $I$, $J$, $K$, $X$, $Y$ and $H$ not explicited in your exercise ? $\endgroup$ – Captain Lama Mar 16 '16 at 1:16
  • $\begingroup$ @CaptainLama no, they are not explicitly given $\endgroup$ – thinker Mar 16 '16 at 1:21
  • $\begingroup$ They are probably somewhere in the book ; otherwise, your result can't be true (you can't just take any basis...). Most likely, $X$, $Y$ and $H$ are what is called $e$, $f$ and $h$ here en.wikipedia.org/wiki/Special_linear_Lie_algebra and $I$, $J$ and $K$ are what is called $u_1$, $u_2$ and $u_3$ here en.wikipedia.org/wiki/Special_unitary_group#n_.3D_2 $\endgroup$ – Captain Lama Mar 16 '16 at 1:31

$\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.


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