The Lie algebra can be introduced as the tangent space to the identity for . By an easy calculation it results that is three dimensional. Choose a basis as follows:
with commutation rules:
First we want to identify a typical irreducible representation for . We resume its properties in a theorem.
Theorem.For each positive integer there exists up to isomorphism a unique irreducible complex linear representation of on a complex vector space of dimension . Moreover, there are an integer and a basis of such that: 1. , 1. , with , 1. , with .
where for and is the action of on .
Proof. Let be a complex linear irreducible representation of . We show that satisfy 1. to 3. Let be an eigenvector for , say . Then are also eigen vectors because:
Since are distinct, these eigenvectors are indipendent. By finite dimensionality of we can find with
a. a. a.
Define . Then , by the same argument as above and so there is a minimum integer with . Then are indipendent and
a. a. a. with .
Now we show that . It is sufficient to show that is stable under . This results from:
with . We prove the calculation above by induction. Assume case for i. Then for i+1:
For ending the proof of uniqueness, we show that . It results:
due to the cyclic property of the trace. Note that here we intend and , being elements of . Then in the basis the trace is explicitely:
Then .
Remark that by 1. in the statement of the theorem the smallest of the eigenvalues is the negative of the greatest.
The proof of the existence is easily achieved by defining a on the generators of such that properties 1. 2. 3. are satisfied, and then by extending through linearity to whole . This is a representation for . To see irreducibility, let be a nonzero invariant subspace. Since is invariant under , is spanned by a subset of the basis . Applying on such a the action of several times, we can see that is in . Repeated application of then shows that . Hence the representation is irreducible. This completes the proof.
Since we built directly a representation that is irreducible in a very concrete way, we want a result that assure us that we can decompose a representation in irreducible ones, that we report without proof.
Theorem.Let be a complex linear representation of on a finite-dimensional vector space . Then V is completely reducible in the sense that there exist invariant subspaces of such that and such that the restriction of the representation to each is irreducible
The importance of studying is that a semi-simple Lie algebra contains several copies of . It follows that for a given representation of , there are several representantions of embedded in . This is shown shortly in Lie groups and algebras representations.
Revised on July 2, 2010 at 08:25:50
by
giuseppe_malavolta