U(2) charge commutator being zero

[kuecken]

Hi,
For SU(2) I can have that all Noether charges commute with one of the charges as one of the generators of the Lie algebra is the identity.
Can somebody explain me how this is related to the properties of SU(2)?
Charges can be considered to be generators of the transformation. So if this commutation [Qi,Qj]=0 what does this mean?
Many thanks.

 

[ChrisVer]

and yet, I thought that you can show that in a given basis the generators of SU(2) were the Pauli matrices ... (?)

 

[kuecken]

sorry i meant U(2) from the title!

 

[samalkhaiat]

sorry i meant U(2) from the title!

The $U(2)$ algebra is given by $$[Q_{a} , Q_{b}] = i \epsilon_{4 a b c} Q_{c}, \ \ (a,b,c) = 1, 2, \cdots , 4 .$$ If we call $Q_{4} = B$ and $Q_{i} = T_{i}$ for $i = 1, 2, 3$, then the above algebra is equivalent to $$[T_{i} , T_{j}] = i \epsilon_{i j k} T_{k} , \ \ \ [T_{i} , B] = 0 .$$ Clearly, the $T_{i}$’s generate the Lie algebra of $SU(2)$ which commutes with the Lie algebra of $U(1)$ generated by the $B$. This simply means that we can write $U(2)$ as direct product of the two commuting groups $SU(2) \times U(1)$. This, in turn, means that particles in an $SU(2)$ multiplet (say the fundamental representation $( p , n )$ ) mix under $SU(2)$ transformations but do not do so under the $U(1)$ transformations. In other words, $p$ and $n$ have different $SU(2)$ charges $T_{i}$(in particular $T_{3}(p) + T_{3}(n) = 0$) but the same $U(1)$ charge, $B(p) = B(n)$. If, $p$ and $n$ represent the proton and neutron fields, then the $U(1)$ charge $B$ represents the Baryon (Fermion) number.

Sam