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Group theory

  1. Mar 19, 2009 #1
    The fundamental representation of SU(N) has a basic form that allows you to deduce that there is a SU(N-1) subgroup. For example, in SU(3), the generators [tex]T_{1}[/tex], [tex]T_{2}[/tex], [tex]T_{3}[/tex] form an SU(2) subgroup.

    I'm reading a book right now that goes into the adjoint representation of SU(N) to show that SU(N) has subgroups [tex]SU(N_1)[/tex], [tex]SU(N_2)[/tex], [tex]SU(N_3)[/tex], ... U(1), where [tex]N_1+N_2+N_3+...=N[/tex]. For example, SU(5) has subgroups SU(3), SU(2), and U(1).

    My question is that since SU(N-1) is already subgroup, doesn't [tex]N_1=N-1[/tex], forcing [tex]N_2=1[/tex], and the rest of the N's zero?

    Also, for U(1) to be a subgroup, you have to find a linear combination of generators (call it the generator [tex]T_0[/tex]) such that the new structure constant has a zero whenever one of its indices is 0? If the structure constant changes, isn't this a new group, and not a subgroup?

    And why are the subgroups of SU(5) combined into the direct product group: SU(3)xSU(2)xU(1)?
  2. jcsd
  3. Mar 20, 2009 #2
    Seems that there is no SU(1) group.
  4. Mar 20, 2009 #3
    Actually SU(1) = SO(1) is the trivial group, i.e. it has a faithful representation consisting of the number 1 under multiplication.

    To the OP, maybe it would help if you state the theorem in the book i.e. I cannot tell whether it is a necessary or sufficient condition.
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