Definitions and Schurs Lemma

In this section we will use the idea of invariant subspaces of a representation (see Definition 5.1) to define irreducible representations. Then we will prove Schur s lemma, which tells us that irreducible representations are indeed good building blocks. 

For some representations, the largest and smallest subspaces are the only invariant ones. Consider, for example, the natural representation of the group G = 50(3) on the three-dimensional vector space C . Suppose W is an invariant subspace with at least one nonzero element. We will show that W = C-. In other words, we will show that only itself (all) and the trivial subspace 0 (nothing) are invariant subspaces of this representation. It will suffice to show that the vector (1, 0, 0) lies in W, since W would then have to contain both 

To show that (1, 0, 0) lies in W, consider any nonzero vector w e W. Note that w e C, and it might not be pure real or pure imaginary. Define 

So if V = 0 then, by the argument above, we have W =. Onthe other hand, if V 7 0, then again by Exercise 4.11 we can choose a rotation M e SO(3) such that Mv = (ryOyOy for some nonzero real number r. Thus the invariant subspace VT contains the vector Mw = a + b, cY for some real numbers a, b and c. It follows that the subspace W also contains the vector 

Note that because r is nonzero, so is 2a + 2ir. So in this case as well we have (1, 0, 0) e VE and hence, as argued above, W — C. This shows that the only nonzero invariant subspace of the representation is the whole space 

---