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By Weyl's theorem every finite-dimensional representation of $\mathfrak{sl}_2(\mathbb{C})$ is completely reducible, because $\mathfrak{sl}_2(\mathbb{C})$ is a (semi) simple Lie algebra. It seems there should be a more elementary proof that is suited for $\mathfrak{sl}_2(\mathbb{C})$ itself, instead of making use of its abstract structural property (namely its semisimplicity). Is there anyone who happens to know such a proof? Thanks!

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    $\begingroup$ I've asked something related, math.stackexchange.com/questions/71578/… $\endgroup$ – Ehsan M. Kermani Nov 17 '11 at 6:30
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    $\begingroup$ Your first attempt should be to read the proof of the abstract fact, but applying it to your concrete situation. In this case the proof is by the unitarian trick, which derives the complete reducibility from that of a compact subgroup, here $SU(2,\mathbf C)$. You probably cannot do better than that; just because a proof is abstract does not mean it isn't a great idea in concrete situations. $\endgroup$ – Marc van Leeuwen Nov 17 '11 at 9:18
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A more elementary proof for the case of $\mathfrak{sl}_2(\mathbb{C})$ is given in [Erdmann-Wildon: Introduction to Lie algebras, Exercises 8.6 and 9.15, Solutions of which can be found in Appendix E].

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  • $\begingroup$ Could you give some idea on how the simplified proof goes? $\endgroup$ – Syang Chen Jun 12 '15 at 17:52

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