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I About Lie group product ([itex]U(1)\times U(1)[/itex] ex.)

  1. May 7, 2016 #1
    I recently got confused about Lie group products.
    Say, I have a group [itex]U(1)\times U(1)'[/itex]. Is this group reducible into two [itex]U(1)[/itex]'s, i.e. possible to resepent with a matrix [itex]\rho(U(1)\times U(1)')=\rho_{1}(U(1))\oplus\rho_{1}(U(1)')=e^{i\theta_{1}}\oplus e^{i\theta_{2}}=\begin{pmatrix}e^{i\theta_{1}} & 0 \\ 0 & e^{i\theta_{2}}\end{pmatrix}[/itex]? Can I say it's reducible, right? Because the way I see it, if the transformation is applied to a 2-dimensional vector, then the first (second) element is transformed by the first (second) [itex]U(1)[/itex] ([itex]U(1)'[/itex]), thus leaving us two invariant 1-dimensional subspaces under the group actions.

    Is it always possible to represent a group product as the direct sum of individual group representations? Or is it just an Abelian case? (IMHO, it seems so because the transformation [itex]SU(2)\times U(1)[/itex] on leptons isn't a [itex]3\times3[/itex] block-diagonal matrix (as one would expect, because fundamental rep. dimensions are 2+1 = 3) but a [itex]2\times 2[/itex] matrix).

    Thanks a lot

    edit: bonus question -- is [itex]2\times2[/itex] rep. of [itex]U(1)[/itex], [itex]\begin{pmatrix}e^{i\theta} & 0 \\ 0 & e^{i\theta}\end{pmatrix}[/itex] a reducible or irreducible representation?
     
    Last edited: May 7, 2016
  2. jcsd
  3. May 8, 2016 #2

    fresh_42

    Staff: Mentor

    You mix up terms here. We don't say a group is reducible in this context. You are talking about direct products of groups.
    Groups without proper normal subgroups are called simple, the others are simply not simple. Direct products cannot be simple because their factors are normal subgroups.

    What can be reducible are representations ##φ## of groups. A representation is a homomorphism of a group ##G## into the linear group ##GL(V)## of a vector space ##V##. ##φ## is called irreducible if there are no subspaces ##U⊆V## for which ##φ(G)(U) ⊆ U## holds beside ##U=\{0\}## and ##U=V##. Other representations are called reducible, i.e. there is a ##0 ⊂ U ⊂ V## with ##φ(G)(U) ⊆ U##.

    The confusion probably comes from the fact that the gauge groups you mentioned are themselves defined as subgroups of a general linear group in which case a representation comes in for free with its embedding. Your representation ##ρ## is correct. You may represent ##U(1) \times U(1)## in this way. (Just do me a favor and drop the ' on the second ##U(1)##. It's not a different one, but simply a second one.)

    Yes, ##ρ## is a reducible representation of ##U(1) \times U(1)## for the reason you said. Only your wording is a bit unfortunate.
    Formally I would say: ##ρ(U(1) \times U(1)) (\begin{pmatrix} 1 \\ 0 \end{pmatrix}) ⊆ ℂ \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} ## (and likewise for ##\begin{pmatrix} 0 \\ 1 \end{pmatrix}##) determine one dimensional invariant subspaces of ##ℂ^2## thus showing that ##ρ## is reducible.

    Yes. Two representations (not necessarily the same) of each one of the (normal) subgroups of a (direct) product of groups can be arranged via block matrices to a single representation of the whole group.

    No. it's not about Abelian, it's about the direct product where both factors are normal subgroups. Things change a bit for semidirect products.

    Define your representation before I can answer this! There is no natural way to do so.
    Edit: ... as long as you don't explicitly identify ##U(1)## with certain elements of ##ℂ##. In either case you should establish the homomorphism.
     
    Last edited: May 8, 2016
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