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I know that irreducible representations of associative $*$-algebras are fairly restricted: any $*$-algebra $A$ is isomorphic to a finite sum of simple algebras


What's the cardinality of the irreps of a Lie algebra?

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It depends on what you mean by irrep... and it depends on the Lie algebra. Usually, though, there are many, many irreps. – Mariano Suárez-Alvarez Jun 5 '12 at 15:34
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The smallest non-zero Lie algebra of all is the one-dimensional Lie algebra.

If you only look at finite dimensional representations, its irreps are in bijection with pairs $(\lambda,n)$ with $\lambda\in\mathbb C$ and $n\in\mathbb N$: this fact is an immediate consequence of the theorem of Jordan Canonical forms.

If you want to consider infinite dimensional representations, then things are much more complicated. A representation of this algebra is roughly the same as an endomorphism of a vector space, so in a sense the study of its representions is the same thing as operator theory: whole books have been written about this.

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So, given a Lie algebra, you cannot form a canonical representation, can you? (unlike the $*$-algebra case). – c.p. Jun 5 '12 at 17:21
What do you mean by canonical representation? – Mariano Suárez-Alvarez Jun 5 '12 at 22:57
by canonical I meant, that if I have a $*$-algebra, I can get an canonical representation in the sum of all its irreps. I'm not sure of having used the right word though. Can you shed some light on the bijection you mentioned? – c.p. Jun 5 '12 at 23:00
I don't understand what you mean by «if I have a ∗-algebra, I can get an canonical representation in the sum of all its irreps». – Mariano Suárez-Alvarez Jun 6 '12 at 16:45
Since $A\cong\oplus_{i=0}^{N}M_{n_i}(\mathbb{C})$ for any fininte dimensional $*$-algebra $A$, and any matrix algebra direct summand has a unique irrep, to wit $\mathbf{n}=\mathbb{C}^n$ acted on by the left, I get an "obvious" representation for $A$ in the sum $V:=\oplus_i^N \mathbb{n}_{i}$. It seems to me pretty canonical-but, as I told you, I am not sure if that is the term I should use. The point is that I cannot do the same for semisimple Lie algebras, for their classification is quite more complicated, can I?Sorry to ask again, but could you shed some light on the bijection you mentioned? – c.p. Jun 6 '12 at 17:58

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