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Su(2), so(3) and their representations

  1. Feb 28, 2015 #1
    I try to understand the statement "Every representation of SO(3) is also a representation of SU(2)".

    Does that mean that all the matrices of an integer-spin rep of SU(2) are identical to the matrices of the corresponding spin rep of SO(3)?

    Say, the j=1 rep of SU(2) has three 3x3 matrices, so has the j=1 rep of SO(3). Are the matrices for SU(2) identical to those for SO(3)? But their are different groups, so that can't be right.

    Or are the matrices identical only for the Lie algebras su(2), so(3)?


    THANK YOU
     
  2. jcsd
  3. Mar 1, 2015 #2

    Fredrik

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    There's a group homomorphism ##R:SU(2)\to SO(3)##. So if ##\Pi:SO(3)\to GL(V)## is a representation of ##SO(3)##, then ##\Pi\circ R## is a representation of ##SU(2)##.

    The map ##R## isn't an isomorphism. It fails to be injective, but has the property that ##R(-\lambda)=R(\lambda)## for all ##\lambda\in SU(2)##.

    This is a sketch of how to find the map ##R##:

    Every complex 2×2 hermitian matrix ##x## can be written in the form
    $$x=\begin{pmatrix}x_3 & x_1-ix_2\\ x_1+ix_2 & -x_3\end{pmatrix} =\sum_{i=1}^3 x_i \sigma_3.$$ The ##\sigma_i## are the Pauli spin matrices. The set ##H## of all such matrices is a 3-dimensional vector space over ##\mathbb R## (not ##\mathbb C##). We can define an inner product on ##H## by ##\langle x,y\rangle=\operatorname{Tr}(x^*y)## for all ##x,y\in H##. The map ##x\mapsto(x_1,x_2,x_3)## is an isometric isomorphism from ##H## to ##\mathbb R^3##.

    For all ##\lambda\in SU(2)##, the map ##R_\lambda : H\to H## defined by ##R_\lambda(x)=\lambda x\lambda^*## for all ##x\in H##, is a linear isometry. (If you want to prove this, note that ##\det x=-(x_3)^2-(x_1)^2-(x_2)^2##). Since ##H## is isometrically isomorphic to ##\mathbb R^3##, this means that we can identify ##R_\lambda## with an element of ##SO(3)##. It's easy to show that the map ##\lambda\mapsto R_\lambda## is a homomorphism. This is the map I'm denoting by ##R##.
     
    Last edited: Mar 1, 2015
  4. Mar 2, 2015 #3

    mathwonk

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    I am not too familiar with this subject, but I think you can imagine of this map roughly, i.e. topologically, by noting that elements of SU(2), being complex invertible transformations of complex 2 space, induce an isomorphism of the set of complex lines through the origin, i.e. an isomorphism of the projective line to itself. But this projective complex "line" is homeomorphic to the 2-sphere (it equals the complex numbers plus a point at infinity). This much is true for all elements of GL(2,C), but the ones in SU(2) also preserve the usual metric on the sphere.

    There seems to be another analogue, but more complicated, taking SL(4,C) onto SO(6,C). This is defined by choosing a skew symmetric form on complex 4 space, which, in every basis, determines a skew 4x4 matrix. Changing bases, i.e. acting by GL(4,C), transforms these matrices into each other, i.e. defines an action on the 6 dimensional space of such matrices. Those basis changes coming from SL(4,C) and not just GL(4,C) preserve the "Pfaffian" (square root of the determinant) of the skew matrix, hence preserve the rank 6 quadric defined by that determinant in 6 space, and this seems to make its action belong to SO(6,C).

    I think both these maps are double covers, i.e. 2:1 onto.
     
    Last edited: Mar 2, 2015
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