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Simple Q about direct product representation of a group

  1. Oct 23, 2011 #1
    (At least, I think it's simple.)

    Disclaimer: I'm approaching this subject from the vantage point of a chemist, so be careful with how much lingo/jargon/rigor you lay on me :redface:

    The claim is that if you have two representations of a group, [itex]\Gamma_1[/itex] and [itex]\Gamma_2[/itex], with bases [itex]\{ f_i \}[/itex] and [itex]\{ g_j \}[/itex], then the set of products [itex]\{f_i g_j\}[/itex] is a basis for the direct product representation. Fine. But in the process of showing this, the following assertion is made: If [itex]\hat R[/itex] is an element of the group, then

    [tex]
    \hat R(f_i g_j) = \hat R(f_i) \hat R(g_j).
    [/tex]

    ...huh? How does that work? How do I know that if I apply a group operation to a product of basis elements, then it's just the product of the operation applied to the individual basis elements? I missed something there, because (a priori) that move is just as suspect as saying that [itex]x(yz) = (xy)(xz) = x^2yz[/itex] for [itex]x,y,z\in \mathbb R[/itex].
     
  2. jcsd
  3. Oct 23, 2011 #2

    lavinia

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    Gold Member

    If each representation is thought of a a set of matrices then the product representation is just the matrices of each representation joined into a diagonal block. Multiplication preserves these blocks.
     
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