Can L Be Isomorphic to sl(2,C)?

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ElDavidas
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Homework Statement



Take

[tex]L = \left(\begin{array}{ccc}0 & -a & -b \\b & c & 0 \\a & 0 & -c\end{array}\right)[/tex]

where a,b,c are complex numbers.

Homework Equations



I find that a basis for the above Lie Algebra is

[tex]e_1 = \left(\begin{array}{ccc}0 & -1 & 0 \\0 & 0 & 0 \\1 & 0 & 0 \end{array}\right)[/tex]

[tex]e_2 = \left(\begin{array}{ccc}0 & 0 & -1 \\1 & 0 & 0 \\0 & 0 & 0 \end{array}\right)[/tex]

[tex]e_3 = \left(\begin{array}{ccc}0 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & -1 \end{array}\right)[/tex]

I then calculate all the products [itex][e_i,e_j][/itex] and see that L is non-abelian and simple

The Attempt at a Solution



The question then asks show L is isomorphic to sl(2,C). I have found [itex]e,f,h \in L[/itex] such that [itex][h,e] = 2e, [h,f] = -2f, [e,f] = h[/itex]
where,

[tex]h = \left(\begin{array}{ccc}0 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & -2 \end{array}\right)[/tex]

[tex]e = \left(\begin{array}{ccc}0 & 0 & -\sqrt{2} \\ \sqrt{2} & 0 & 0 \\0 & 0 & 0 \end{array}\right)[/tex]

[tex]f = \left(\begin{array}{ccc}0 & -\sqrt{2} & 0 \\0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \end{array}\right)[/tex]

if I haven't made any mistakes. I see L is homomorphic to sl(2,C) but how do I show it's an isomorphism? (i.e show the injection and surjection). I know the definitions for an injection map and a surjection map but don't know how to apply it in this case.

Thanks in advance for any help.
 
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You've just written down a sub-algebra isomorphic to sl_2, and clearly it is all of the space (just by dimension arguments). There is nothing more to show.

You haven't actually written down a map so you can't apply the notion of injection or surjection. If you want to put in a map - there is an obvious one - then it is trivially an injection (and a surjection).