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Scalar as one dimensional representation of SO(3)

  1. Apr 1, 2015 #1
    Hi to all the readers of the forum.
    I cannot figure out the following thing.

    I know that a representation of a group G on a vector spaceV s a homomorphism from G to GL(V).
    I know that a scalar (in Galileian Physics) is something that is invariant under rotation.

    How can I reconcile this with the following sentence:
    "an ordinary scalar belongs to the one-dimensional representation of the SO(3) group".
    (It is taken from Griffiths' Introduction to elementary particles, but it is written in all introductory books on Particle Physics, as you will know)

    What is the one-dimensional representation in question? Is it the function constantly equal to 1?
    Even in that case, I am not sure I would understand.

    Please, can someone elaborate on that? I cannot see which is the homomorphism involved!

    Thank you for any help :-)
     
  2. jcsd
  3. Apr 1, 2015 #2

    Orodruin

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    Yes, this is indeed the case.

    If g is a homomorphism, it must hold that g(ab) = g(a)g(b). For the mapping g(a) = 1 for all a, we have that g(ab) = 1 = 12 = g(a)g(b) so it is a homomorphism (it is a homomorphism for all groups.
     
  4. Apr 1, 2015 #3
    Thanks! So far so good.
    Maybe my next questions betrays a more fundamental problem in my comprehension of the subject.

    To each group element we associate the number 1. Okay.
    So what does it mean to say that tha modulus of a vector, say (1,1,0) which is 2, is a scalar? I just know that this is invariant if I rotate the vector!

    What does it mean to say that the modulus belongs to a unidimensional representation of SO(3)?
    Is the previous sentence precise/true/correct?
     
  5. Apr 1, 2015 #4

    Orodruin

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    It is sqrt(2) ... but anyway.

    A scalar is a number which is invariant under the group action. In general, if a quantity transforms according to the n-dimensional representation g, then it is represented by a column matrix v (with n entries), which transforms according to
    $$v\to g(a) v$$
    under the transformation ##a##. In the case with the trivial representation, a scalar is a number ##s## which transforms as ##s \to g(a)s = 1s = s##.
     
  6. Apr 1, 2015 #5
    Perfect!
    Last thing:
    Is this a matter of definition/convention (isomorphisms between Rn and the groups of matrices, etc. ) or is there a "reason" for that?
     
  7. Apr 2, 2015 #6
    Maybe it is simply the definition that the dimension of a representation is the dimension of the vector space on which it acts?
     
  8. Apr 2, 2015 #7

    Orodruin

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    That would be by definition, yes.
     
  9. Apr 2, 2015 #8
    Thanks!
     
    Last edited: Apr 2, 2015
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