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As real Lie algebras, both are three-dimensional. The basis of $su(2)$ is

$$ \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right), \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right),\left( \begin{matrix} 0 & i \\ i & 0 \end{matrix} \right) . $$

The basis of $sl(2;R)$ is

$$ \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right), \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right),\left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) . $$

But how to prove that there is no isomorphism mapping between the two algebras?

John
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1 Answers1

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The first of these has no two-dimensional subalgebras, whereas the second does. See $\mathfrak{su}(2)$ not isomorphic to $\mathfrak{sl}(2,\mathbb R)$ for a fuller explanation.

Community
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David Towers
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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. - From Review – SchrodingersCat Dec 28 '15 at 11:18
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    Why does it not answer the question? Any isomorphism would map a two-dimensional subalgebra to a two-dimensional subalgebra. – David Towers Dec 28 '15 at 11:44
  • The language was a copy-paste, don't get disheartened at that. The point is: This better suits to be a comment than an answer since this does not have a proper explanation or links to backup. You might say this requires no explanation, its totally straightforward ; but the general trend shows people asking such questions often get confused by $1$ line answers. No offence intended to the OP. – SchrodingersCat Dec 28 '15 at 11:56
  • Thank you for the clarification. As you guessed I posted it as an answer as I couldn't post it as a comment. I've since seen that a fuller explanation is given in http://math.stackexchange.com/questions/1466917. – David Towers Dec 28 '15 at 12:06
  • You're welcome. – SchrodingersCat Dec 28 '15 at 12:07