SU(3) defining representation (3) decomposition under SU(2) x U(1) subgroup.

Karatechop250
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I have been reading Georgi "Lie Algebras in Particle Physics" and on page 183 he mentions how that the SU(3) defining representation decomposes into an SU(2) doublet with hyperchage (1/3) and singlet with hypercharge (-2/3). I am confused on how he knows this. I apologize if this is not the right place to put such a question.
 
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Are you familiar with how to split a representation into irreducible representations?
 
Yes you decompose it into block diagonal form. Or another way you can do it is do tensor products of the fundamental rep and decompose it using young tableaux methods.
 
So, when you break SU(3), the defining representation of SU(3) is not an irreducible representation of SU(2)xU(1) but is naturally still a representation (just map any element in the subgroup to the element it would have been mapped to in the SU(3) defining representation). Decomposing this representation into irreducible representations should give you the desired result.

Hint: If it helps you think about it, pick a basis such that the SU(2) is generated by ##\lambda_1, \lambda_2, \lambda_3## and the U(1) by ##\lambda_8## (where ##\lambda_i## are Gell-Mann matrices).
 
Is there anyway you could go into a little more detail. And how did you know in the first place to pick such an SU(2)xU(1) basis. Also do you know of any good reading material besides Georgi's book which I could look at that would include this.
 
Well, if you are familiar with the Pauli matrices of SU(2), it is fairly straightforward to see that those four matrices will generate a subgroup which is SU(2)xU(1). The first three are just the Pauli matrices in the upper left block and the last is the only remaining generator which commutes with those as it is proportional to unity in this block.

Tinkham's book Group Theory and Quantum Mechanics seems fairly popular (although I must admit not having read it) and is published by Dover and so is available at a very reasonable price.
 
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