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Representations of the lorentz group

  1. Nov 5, 2008 #1
    I'm very very very confused and extremely thick.

    If [itex] \Lambda_i [/itex] is some element of the Lorentz group and [itex] \Lambda_j [/itex] is another, different element of the group then under multiplication....

    [itex] \Lambda_i \Lambda_j [/itex] is also an element of the Lorentz group, say

    [itex] \Lambda_i \Lambda_j =c_{ij}^k\Lambda_k[/itex]

    where [itex] c_{ij}^k [/itex] has value 1 for one unique combination of i,j and k and 0 for the others and with a sum over k.

    Now I appreciate the i,j and k should be continous but for the moment assume they are discrete because it's easier. i,j and k run over all the integers because there are infinitely many elements of the group.

    How the hell do I find the [itex] c_{ij}^k [/itex]? I have absolutely no idea. My knowledge of rep theory is **** poor.
  2. jcsd
  3. Nov 5, 2008 #2


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    You seem to have omitted something crucially important. In particular

    the Lorentz group doesn't have an addition operation. So, on the surface at least, this makes no sense....
  4. Nov 5, 2008 #3
    No no, that's ok because the [itex]C_{ij}^k [/itex] is one for only one combination of i,j and k, and 0 for all others

    So infact there is no summation.

    The reason I have done this that way is because of what I intend to do with the C's afterwards.

    This was my supervisor's idea and he made that explicitly clear, that it's very tempting to think of this as summation over the group, when infact it isn't

    Hope that's clear. Thanks in advance for any help.
  5. Nov 5, 2008 #4


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    Okay, so c is really nothing more than the multiplication table? Or more precisely, if you view the graph of the multiplication operator as a subset of [itex]G^2 \times G[/itex], c is its characteristic function.

    If so, I don't understand what you're asking.
  6. Nov 5, 2008 #5
    Yes, I guess so.

    I should say I have tried to think about this in simpler terms for SU(2) where


    sigma are the pauli matrices but I am confused there too.
    Last edited by a moderator: Nov 6, 2008
  7. Nov 6, 2008 #6


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    Based on your notation, it looks like you want to talk about not the Lorentz group itself, but the Lie algebra of the Lorentz group. This is the space of infinitesimal transformations, ie, transformations [itex]\Lambda^\mu_\nu[/itex] of the form:

    [tex] \Lambda^\mu_\nu = \delta^\mu_\nu + \epsilon \omega^\mu_\nu [/tex]

    where [itex]\epsilon[/itex] is thought of as a fixed, very small number. Studying these infinitesimal transformations tells you a lot about the group, and is often easier than studying the group itself. The [itex]\omega^\mu_\nu[/itex] can be added and multiplied by scalars, so form a vector space, and if we pick a basis [itex]\omega_i[/itex], we can write (this is the standard notation):

    [tex] [\omega_i, \omega_j ] = f_{ij}^k \omega_k [/tex]

    where the LHS is what is called the commutator: it is roughly the difference between the element you get by applying [itex]\omega_i[/itex] then [itex]\omega_j[/itex] and the element you get by applying them in the reverse order. These dozen or so constants [itex]f_{ij}^k[/itex] actually tell you everything there is to know about the group (up to some large scale topological things).

    Anyway, I don't know if this is actually what you're talking about, but since the notation is so similar, maybe this is what your supervisor meant. Because it makes very little sense for you to write things that way if the [itex]\Lambda_i[/itex] really are (non-infinitesimal) Lorentz transformations. For one thing, there are uncountably many of those, so you can't index them by the integers.
  8. Nov 6, 2008 #7
    Ok thanks status x, don't worry about it, that's not what I meant at all.
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