Projector operator in Representation theory I am reading some introductory stuff on Representation theory applied to physics and I am a bit confused about some things. The book I use is Lie Algebra in Particle Physics by Georgi (you can find it here) and the part where I got confused is part 1.16 where he finds the normal modes for 3 particles connected by springs, forming a triangle. So I want to understand how all this works.
As far as I understood, you find the symmetry of the system and choose the proper dimensions for the representation (in this case 6D and S3). Then you get the characters of the matrices in this representation and calculate the projectors onto the space of irreducible representations. Now I am a bit lost. How do one gets the normal modes (or in general the eigenvectors associated to a certain irreducible representation, from the projector operators). For the projection operators associated to 1D representation I think I get it, as they are just the product between the column and row vector, that span the space of that representation. But for projector onto higher dimensional representations I am quite lost.
Moreover, I am not sure I understand what do I need to do when the same representation appears multiple times. Why do we have just a projection operator for both? The 6D vectors (in this case) forming the space on which each of them acts is different. So I would really appreciate if someone can give me a more clear explanation of all this. Thank you!
 A: The first step is to obtain any one vector in every irrep and this is what the projectors can do for you.  If the irrep is 1-dimensional this is enough but as you point out, when the irrep has greater dimension the character projector is not enough to complete the calculation.  
If you have one state in the irrep of dimension greater than 1, you need to act on this state with the remaining elements of the group to generate all the vectors in the space, and construct an orthonormal basis from these vectors.  It is essential that you act with group operators as they are guaranteed not to take you outside of your irrep.
Now to get eigenvectors you need to choose operators to diagonalize.  If the irrep is 1-d that's automatic since all operators act by multiplication.  For the $2$-dimension irrep of $S_3$ you will need to select a subset of mutually commuting operators to get eigenvectors since $S_3$ is not abelian.  The commuting operators are usually taken as the identity and $P_{12}$: they have distinct spectrum so getting the common eigenvectors is enough.  (This part is extremely not clear from the discussion in Georgi; I would have to read the whole thing to get a handle on the notation.)
The projection operator depends on the irreps alone, and not on their multiplicities.  As a result, the projection operator will give you one basis state in one irrep, but it's not always enough to get all the basis states, especially if there are multiple irreps.  It is usually better to proceed by diagonalization, selecting once more a set of commuting operators.  When you have multiple copies of an irrep you will have degenerate eigenvalues; the degenerate subspace is of the same dimension as the multiplicity of the eigenvalues since the representation is equivalent to a unitary representation.  As with all problem with degeneracies, there is no unique way of selecting eigenvectors.  One choice is to pick a first vector to be the one obtained by the projector, but that's not always a good choice: you might want solution that exhibits special physics features of your problem.  
Anyways, having constructed one eigenvector in each irrep, and making sure these initial vectors are orthogonal, you can then proceed and construct the others in the same irrep one irrep at the time by acting on the initial vector of that irrep using the group elements, thus guaranteeing that you do not get out of the irrep.  
When it comes to the permutation group there are natural subgroup chains and "tricks" using class operators: the good reference for this is the text by Chen, Ping and Wang.  There is also the method of Young symmetrizers that will work for any irrep; details can be found in the text of Wu-Ki Tung.
