# #(cardinality of irreps Lie algebras) > #(irreps of ssociative algebras). Proof?

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

$A\cong\oplus_{i=0}^{N}M_{n_i}(\mathbb{C})$

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

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.
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
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
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