Why is Weak Hypercharge different for each fermion? I'm having trouble with Electro-Weak Theory. I understand that one wants to develop a gauge theory with $SU(2)_L \otimes U(1)_Y$. What buggs me is that the coupling with different fermions has different hypercharge. I understand that up-type and down-type particles in the $SU(2)_L$ doublet have $\frac{1}{2}$ and $-\frac{1}{2}$ weak isospin, that is related to different eigenvalues of the $SU(2)_L$ generators. But $U(1)_Y$ has only one generator and it's proportional to the identity! So I think that different fermion hypercharges are related to:
1) There is a different generator of U(1) for each fermion. (all of them are proportional to the identity but differ in a multiplying factor) This could be rephrased as: Each fermion transforms in a different equivalent representation of $U(1)$.
2) There's only one generator but each fermion has a different coupling constant to the weak hypercharge field. 
Neither of this sounds satisfactory... 
 A: Perhaps the right way for you to think about it is this: there is only one generator, and it is a big $N\times N$ matrix where $N$ is the number of all of the independent fermion degrees of freedom. To be exact, all of the generators of a symmetry are like this, but we are often able to ignore the full size of each matrix.
Any generator of a symmetry must specify an action of the symmetry on every degree of freedom. In general a symmetry can mix up all the different fermions. However, most of the symmetries of the Standard Model only mix up a few different groups of fermions. For example, if we write out all the fermions (just one generation) as $f = (u_L^r,u_L^g,u_L^b,d_L^r,d_L^g,d_L^b,\nu_L,e_L,u_R^r,u_R^g,u_R^b,d_R^r,d_R^g,d_R^b,e_R)^T$ then the generators of the color $SU(3)$ symmetries look like
$$T_{color}^a = \begin{pmatrix}\lambda^a_{3\times 3} &0 & \dots &\\
0 & \lambda^a_{3\times 3} & 0 & \dots \\ 0 & \dots & 0_{1\times 1} &\dots \\ 0 &\dots & & 0_{1\times 1} & \dots \\ 0 & \dots  & & &\lambda^a_{3\times 3} \\ 0 &\dots &&&& \lambda^a_{3\times 3}\\ 0 &\dots &&&&& 0_{1\times 1}\end{pmatrix} $$
where the $\lambda$'s are 3x3 Gell-Mann matrices that act on the subspaces $u_L^{\{r,g,b\}}$ etc. Likewise, the $SU_2(L)$ generators mix up the left-handed quarks and leptons in a specific way:
$$ T^i_{weak} = \begin{pmatrix}\Sigma^i_{6\times 6} & 0 &\dots  \\ 0 &\sigma^i_{2\times 2} & \dots \\ 0 &\dots & 0_{7\times 7}\end{pmatrix} $$
where $\sigma^i$ are the Pauli matrices and $\Sigma^i$ are a simple extension of the Pauli matrices, e.g.
$$\Sigma^2 = \begin{pmatrix}0_{3\times 3} & -iI_{3\times 3} \\ iI_{3\times 3} & 0_{3\times 3}\end{pmatrix}$$
Finally, the single $U(1)_Y$ generator is just a diagonal matrix:
$$T_Y =\begin{pmatrix}\frac{1}{6}I_{6\times 6} & 0 &\dots \\ 0 & -\frac{1}{2}I_{2\times 2} & \dots \\ 0 &\dots & \frac{2}{3} I_{3\times 3} &\dots \\ 0 &\dots && -\frac{1}{3}I_{3\times 3} & \dots \\ 0 &\dots &&& -1\end{pmatrix}$$
If you look at these for a while you can see that the matrices break down into blocks, and the  different blocks never mix with each other. The  $(u_L^r,u_L^g,u_L^b,d_L^r,d_L^g,d_L^b)^T$ fermions forms a 6x6 block. The $(\nu_L,e_L)$ part forms a 2x2 block, and so on. These subspaces are called irreducible subrepresentations of the symmetry.
To finally answer your question, the full $U(1)_Y$ generator is not proportional to the identity, but you can see that its action on each irreducible subrepresentation is proportional to the identity. Since irreducible subrepresentations never mix, we often refer to them separately and act like they have seperate generator matrices. This is not really the case, but it is much more convenient that writing out the huge matrices every time, and almost never causes confusion once you understand it.
The reason we can always split fields into irreducible subrepresentations like this is explained by representation theory, which is a beautiful mathematical subject that it is well worth learning. (And perhaps you are already trying to!)
