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Nov 18, 2016 at 22:03 vote accept AndreaPaco
Nov 18, 2016 at 12:12 comment added Vincent There is actually one simple one of actual dimension 10: it is so(5) which is isomorphic to sp(4) (B_2 and C_2 in root system nomenclatura). If your Lie algebra actually turns out to be so(5) it has a representation in anti-symmetric 5 by 5 matrices, half as big as the ones in the adjoint representation. This is a common theme. The adjoint is never the smallest rep, except in the E_8 case.
Nov 18, 2016 at 12:09 comment added Vincent Also when the center is zero there is a finite list of Lie algebras your Lie algebra could be isomorphic to. By 'could' I mean it could not be isomorphic to any other so it must be in the list, only perhaps presented differently. The point is that a centerless Lie algebra is semisimple and a semisimple lie algebra is a direct sum of simple ones. If you look at the list of simple Lie algberas in wikipedia you can find all the ones with dimension $\leq 10$ and see which ones can add up to give dimension 10
Nov 18, 2016 at 12:00 history edited Vincent CC BY-SA 3.0
added last )
Nov 18, 2016 at 11:55 history edited Vincent CC BY-SA 3.0
more extra zeros
Nov 18, 2016 at 11:50 history edited Vincent CC BY-SA 3.0
Edited in description of adjoint representation
Nov 18, 2016 at 11:37 comment added AndreaPaco There are no linear combinations of operators of mine that commute with all the others. So the center of the Algebra is zero?! Can you please expand "things are very simple" and explain what I should do? As I said, I've not an algebraic / group-theoretic background, so I would appreciate a more detailed help. Thanks a lot in advance.
Nov 18, 2016 at 11:22 history edited Vincent CC BY-SA 3.0
removed superfluous word
Nov 18, 2016 at 11:12 history answered Vincent CC BY-SA 3.0