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A simple question on representations and tensor products

  1. Jan 9, 2010 #1
    I have question, can someone please check whether my answer is correct or not:

    1)Let [tex]\pi_i[/tex] be representations of a group G on vector spaces Vi, i = 1, 2. Give a formula for the tensor product representation [tex]\pi_1 \otimes \pi_2[/tex] on [tex]V_1 \otimes V_2[/tex]

    Answer: [tex]\pi_1 V_1 \otimes \pi_2 V_2[/tex]

    2)Check that it obeys the representation property.

    Answer: A representation is a group homomorphism, ie it satisfies:

    [tex]\pi(g.h)= \pi(g) . \pi(h)[/tex]


    [\pi_1 V_1 \otimes \pi_2 V_2](g.h)
    =\pi_1 V_1 (gh) \otimes \pi_2 V_2 (gh)

    I am a little stuck here: we know that [tex]\pi_i[/tex] is a representation, can we also say that [tex]\pi_i V_i[/tex] is also a representation? If it is, we can use the homomorphism property and show that

    [tex] \pi_1 V_1 (gh) \otimes \pi_2 V_2 (gh)=\pi_1 V_1 (g)\pi_1 V_1 (h) \otimes \pi_2 V_2 (g) \pi_2 V_2 (h)=[\pi_1 V_1 \otimes \pi_2 V_2](g)[\pi_1 V_1 \otimes \pi_2 V_2](h)[/tex]

    which I think the question is trying to get at.
  2. jcsd
  3. Jan 9, 2010 #2
    If [tex]\pi_i[/tex] are representations, then they are functions [tex]G \to V_i[/tex], so how are they eating elements of V_i? You have things somewhat backwards here.
  4. Jan 9, 2010 #3
    I think it's because [tex]\pi_i \in End (V_i)[/tex]
  5. Jan 9, 2010 #4
    ah yes, excuse me...fair enough. You still have it a bit backwards though. [tex]\pi_i \not \in \operatorname{End}(V_i)[/tex], [tex]\pi_i(g) \in \operatorname{End}(V_i)[/tex] for all g in G. [tex]\pi_i : V \to \operatorname{End}(V_i)[/tex]. Pardon the initial confusion, its been a while since I though about this type of representation.
  6. Jan 9, 2010 #5


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    What do you mean by this? Answer: [tex]\pi_1 V_1 \otimes \pi_2 V_2[/tex]

    Aren't you supposed to find a function [itex]\pi:G\rightarrow\mbox{End}(V_1\otimes V_2)[/itex] and show that it's a representation? The first idea that occurs to me is


    where the right-hand side is defined by

    [tex]\pi_1(g)\otimes\pi_2(g)(x_1\otimes x_2)=\pi_1(g)(x_1)\otimes\pi_2(g)(x_2)[/itex]

    I haven't checked if it satisfies the requirements.

    (I think rochfor1's last LaTeX expression should be [itex]\pi_i : G \to \operatorname{End}(V_i)[/itex] ).
  7. Jan 9, 2010 #6
    Indeed it should.
  8. Jan 9, 2010 #7
    Hi Fredrik:

    Well, I was mindlessly reeling off the following in my notes:

    if [tex]A \in end(V)[/tex]

    and [tex]B \in end (W)[/tex]


    [tex]A \otimes B \in end(V \otimes W)[/tex], and this is defined via:

    [tex](A \otimes B)(V \otimes W)= (AV) \otimes (AW)[/tex]

    So I just assumed we can let [tex]A=\pi_1[/tex] and [tex]B=\pi_2[/tex],

    This clearly is silly because it would imply [tex] \pi_1: V_1 \rightarrow V_1[/tex] and [tex]\pi_2: V_2 \rightarrow V_2[/tex], where the maps infact should be [tex] \pi_1: G \rightarrow V_1[/tex] and [tex] \pi_2: G \rightarrow V_2[/tex] as you point out.

    I think you're first idea is correct.

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