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Structure constants of su(2) and so(3)

  1. Aug 2, 2008 #1
    SU(2) and SO(3) "have the same Lie algebra".

    While I understand that their corresponding lie algebras su(3) and so(2) have the same commutator relations

    [tex] \mbox{SO(3)}: \left[ \tau^i, \tau^j\right] = \iota \varepsilon_{ijk} \tau^k[/tex]

    [tex] \mbox{SU(2)}: \left[ \frac{\sigma^i}{2}, \frac{\sigma^j}{2}\right] = \iota\varepsilon_{ijk} \frac{\sigma^k}{2}[/tex]

    so that the structure constants of each are identical, and as each Lie Algebra is uniquely defined by it structure constants, both algebras are identical.

    The elements being operated on are, clearly, very different. If I were to distinguish the two groups in a discussion, I cannot speak of Algebra as the Algebra is purely the rule of composition between elements of the group(not the elements themselves)?. The Algebras are truly identical.

    To distinguish, I would say

    "Both Groups obey the same Lie Algebra, but whose infinitesimal generators are different"

    yes?

    Thanks

    edit: I did search the forum, but could only find much more general questions about Lie Algebras so I decided to make a new topic.
     
  2. jcsd
  3. Aug 2, 2008 #2

    CompuChip

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    The algebra is the set of rules that determines the operation. In this case, the operation is commutation and by linearity of that operation, it is uniquely determined once given on the basis elements:
    [tex]\left[ \sum_i a_i \tau^i, \sum_j b_j \tau^j \right] = \sum_{i, j} a_i b_j c^{ijk} \tau^k[/tex]
    where a, b are coefficients (real or complex), [itex]\tau[/itex] are the basis elements and c are the structure constants; i and j index the basis elements.

    Now a Lie group, L, is a group with elements of the form [itex]g(a_1, a_2, \cdots) = \exp\left( \sum_i a_i \tau^i \right)[/itex], where the exponential is defined by the power series:
    [tex]g(a_1, a_2, \cdots) = \sum_n \frac{1}{n!} \left( \sum_i a_i \tau^i \right)^n[/tex].
    The [itex]\tau[/itex] are called the generators of the group; though the group itself is usually infinite, in practice the number of generators for such a group is often finite (but it can be infinite of course).

    Finally, there is an object called a presentation. For a group, this is a map [itex]\phi: G \to GL_n[/itex] which assigns to each group element g an nxn matrix which preserves the group structure, e.g. [itex]\phi(g \circ h) = \phi(g) \phi(h)[/itex] (group composition on the left, matrix multiplication on the right). One group can have many representations, for many values of n. (Some are more interesting than others). So there is now a distinction between the elements of a group and the matrix representation of those elements. You can similarly imagine linking matrices to elements of the algebra. Because matrix multiplication is linear, it suffices to assign to each generator [itex]\tau^i[/itex] a matrix [itex]M^i[/itex] and then for each element [itex]\sum_i a_i \tau^i[/itex] you have the matrix [itex]\sum_i a_i M^i[/itex]. To be a representation of the algebra, these matrices must be chosen such that their commutation relations are the same as those of the algebra elements. Again, there can be several sets of matrices which satisfy these commutation relations, and they are all equally valid representations of the algebra. The matrices can however look very differently. Of course, once you have such a representation, you can make a representation for the corresponding Lie group by just exponentiating the matrices. Note that you can also build as many as you want. One trivial trick, you can always use, is to take your favorite representation [itex]M_i[/itex] and make a new one with matrices N defined by
    [tex]N_i = \begin{pmatrix} M_i & 0 & 0 & \cdots & 0 \\ 0 & M_i & 0 & & \vdots \\ 0 & 0 & \ddots & & 0 \\ 0 & 0 & 0 & \cdots & M_i \end{pmatrix} [/tex].

    What probably confuses you as well, is this structure of creating Lie-algebras. As I just described, what we formally do is fix the algebra - which is the thing we want to work with. Then we find a convenient representation and do our calculations there, because it's easier for both us and computers to calculate with matrices. In practice, however, one often does it the other way around. We have some set of matrices (for example, rotations in three dimensions) and notices that this is a Lie-group. One can then isolate the generators and write down the generators and the commutation relations. On one hand the matrices have been used to define the algebra. On the hand, this original set of matrices is already a representation of this group. It may not be the only one, but it is usually the one which is most intuitive to us.

    Hope that clears some of the confusion.
     
  4. Aug 3, 2008 #3
    The usual 'equal' and 'isomorphic' misunderstanding. Clearly, they are not equal, since thet are different, but equally clearly they are isomorphic.
     
  5. Aug 3, 2008 #4
    Yeah, the second post makes sense. I decided not to reply to the first one till I digested it entirely.

    I can see how they're isomorphic (I can't formalise right now but by analogy with everyday manifolds).

    You see, my lecturer contradicted himself. He says that su(2) and so(3) were the same algebra. And then, almost directly afterwards, Gives the Theorem:

    "To every Lie algebra there corresponds a unique simply connected Lie
    group."

    Taking the su(2) algabra which is supposedly the "same as" so(3) so that the "one" algebra corresponds to both SO(3) and SU(2). Which would contradict the theorem. However, if they are realised to be simply isomorphic then there is no difficulty.

    Have I picked you up correctly?

    CompuChip: Thank you. I will reply to your post as soon as I have made my best efforts to understand it.
     
  6. Aug 3, 2008 #5
    There is no contradiction there: SO(3) isn't simply connected. It's fundamental group is C_2 (group with 2 elements), and has SU(2) as its simply connected cover. These facts are illustrated quite nicely with 'the soup bowl trick' and quarternions.
     
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