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Conformal Group

  1. Sep 27, 2013 #1
    Hi. I'm reading about the compactification of Minkowski Space, and there is a subject that is keeping me awake. They say that the group of conformal transformations is isomorphic to the group of pseudoorthogonal transformations with determinant equal to 1. I don't know how this happen and it would of great help if someone could tell me why this happen.
     
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  3. Sep 27, 2013 #2

    strangerep

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    More accurately, the group SO(4,2) is a double cover of the conformal group of Minkowski space. The Lie algebras of these two groups coincide, as can be seen by direct evaluation.
     
  4. Sep 28, 2013 #3
    First of all, thank you so much for answer me. Second, I'm sorry, but I'm not familiar with Lie algebra. I don't know if it is possible to solve this in a more elemental way or maybe you can tell me about a book where I can find this. Again, thanks.
     
    Last edited: Sep 28, 2013
  5. Sep 28, 2013 #4

    Bill_K

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    Consider a 3-space with orthogonal coordinates T, X, S and signature + - -. Consider the rotations in this 3-space, which form the group SO(2,1), and their actions on the unit hyperboloid T2 - X2 - S2 = 1.

    Parametrize points on this hyperboloid with coordinates x, s whose relation to T, X, S is

    T - X = ex
    T + X = s2 ex + e-x
    S2 = s2 e2x

    (By substitution, one can verify that for any values of x and s we have T2 - X2 - S2 ≡ 1, which lies on the hyperboloid.)

    Now consider the action of SO(2,1) on x and s. One of the rotations, a boost in the T,X plane, amounts to a translation in x and a rescaling of s:

    x → x + c
    s → s e-c

    Another, a rotation about the null direction T - X, is a translation of s,

    s → s + c

    Ok, now generalize this to any number of dimensions. Replace the original S2 everywhere by Ʃ εi Si2 where εi = ±1. (This is the line element in an n-dimensional space with signature determined by the choice of εi's.) We are now looking at the unit hyperboloid T2 - X2 - Ʃ εi Si2 = 1. The parameters x and si

    T - X = ex
    T + X = Ʃ si2 ex + e-x
    Ʃ Si2 = Ʃ si2 e2x

    and the transformations

    x → x + c
    si → si e-c

    and

    si → si + ci

    Thus we have a rescaling of the si's and a translation of the si's, which, together with rotations among the si's, produce the conformal group in n dimensions.

    This shows explicitly that the transformations in the rotation group SO(n+1, m+1) map to those in the conformal group C(n,m).
     
    Last edited: Sep 28, 2013
  6. Sep 28, 2013 #5

    strangerep

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    Hmm. You said you were "reading about the compactification of Minkowski Space", but not knowing anything about Lie group theory is a serious handicap for most subjects in physics.

    You could try the book by Greiner & Muller: "QM - Symmetries". It contains a lot of introductory material on Lie groups (though of course it's not about compactification of Minkowski Space).

    Nice explanation. I hope the OP can appreciate it eventually. (BTW, since it's been ages since I studied this stuff closely, where does the double-cover arise?)
     
    Last edited: Sep 28, 2013
  7. Sep 28, 2013 #6
  8. Sep 30, 2013 #7
    Ok, let me be completely honest with you guys. I'm reading the book "Conformal differential geometry and its generalizations", of Akivis. In this book the author don't use anything what you are using. He starts with the euclidean case (n dimensions) and then he generalize to the pseudo euclidean case. Since he want to study the conformal aspects of this space, he introduces the point at infinity, son R^n (euclidean space) becomes C^n (conformal space). Next, he use "polyspherical coordinates", which are points in a (n+1)-projective space, P^(n+1), and gets an inner product wich helps him to identify every point in C^n with a point on a quadric in P^(n+1).This identification is knows as "Darboux mapping". He then says that every conformal transformation is equivalent, up to a factor c, with a linear transformation that leaves the quadric invariant. This linear transformations is of order n+2. Since, if n is odd, you can take an appropriate value of c to make 1 the determinat of the linear transformation, he concludes that the conformal group is isomorphic to SO(2+n,1)

    And I am like... "What?", Why?. If the linear transformations are of order 2+n, why it is isomorphic to a matrix of SO(2+n,1)?

    So, please you guys. I don't know anything about Lie algebra and I have weeks thinking on this. Thank you.
     
  9. Sep 30, 2013 #8

    Bill_K

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    This quadric is the "unit hyperboloid" that I referred to above in #4.

    I didn't mention anything about Lie algebras.
     
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