Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Non-compactness of Lorentz Group ?

  1. Aug 17, 2011 #1
    Hello, I'm studying the Lorentz group and their properties... and I have some question for them..

    Peskin's text(p496) said that

    "we are primarily interested in Lie algebras that have finite-dimensional Hermitian representations, leading to finite-dimensional unitary representations of the corresponding Lie group.

    We will also assume that the number of generators is finite.

    Such Lie algebras are called compact, because these conditions imply that the Lie group is a finite-dimensional compact manifold."

    From this mentions of compactness,

    I've thought that Homogeneous Lorentz group is compact because the number of generators of it is just 10([itex]M_{\mu\nu},P_{\mu}[/itex] 3 rotations, 3 boosts, 4 translations)

    In Arfken and many other texts, However, they said that the Homogeneous Lorentz group is non-compact, because the limit of a sequence of rapidities going to infinity is no longer an element of the an element of the group. (Arfken, p278)

    From the same reason, translation is non-compact, too.
    (therefore, Poincare group is non-compact.)

    What's wrong with my assumption for compactness of the Lorentz group?

    I know that I have some misunderstanding for compactness and I need a help to figure it out...
    Last edited: Aug 17, 2011
  2. jcsd
  3. Aug 17, 2011 #2


    User Avatar
    Science Advisor
    Gold Member

    Peskin/Schroeder is sometimes (or better said often) pretty sloppy in the details. Of course, first of all we are interested in the Poincare group, if we like to do quantum theory in the relativistic realm, and the Poincare group for sure is not compact. The alone are not a compact Lie group, but neither is the Lorentz group (the same is true for the Galileo group, where also the translations and the Lorentz boosts are non-compact subgroups).

    A Lie group is called compact if there is a compact parameter set which maps out the whole group (more precisely there should be a finite number of such compact maps which cover the whole group, i.e., make an atlas). Indeed, if you look at the Lorentz group (more precisely: the orthochronous proper Lorentz group, which makes the subgroup of the full Lorentzgroup that is continuously connected with the group identity) is not compact, because to get boosts in the [itex]x[/itex] direction, you always need a non-compact parameter set to map this one-parameter subgroup. The natural choice is indeed rapdity:

    \cosh \eta & -\sinh \eta &0 & 0 \\
    -\sinh \eta & \cosh \eta &0 &0 \\
    0 & 0 & 1 &0 \\
    0& 0 & 0 & 1
    \end{pmatrix}, \quad \eta \in \mathbb{R}.[/tex]

    The first consequence of the Lorentz group not being compact is that there are no unitary finite-dimensional representations, except the trivial one, but there are unitary infinitely dimensional representations of the whole Poincare group, and these can be classified (Wigner 1939) and investigated whether they describe physical systems. Indeed, one finds representations describing massive and massless particles with integer and half-integer spin. This is the starting point for model building in high-energy physics (standard model etc.).

    You find a quite complete description of the mathematics in my qft manuscript,


    and of course in Weinberg's marvelous books (Quantum Theory of Fields). The unitary representations of the Poincare group are treated in the first chapters of volume 1. It's way better than Peskin/Schroeder in such fundamental issues!
  4. Aug 17, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper

    The set of all Lorentz transformations forms a non-compact topological space, because it contains the orthocronous proper transformations which form a non-compact tolopogical space by the following argument. The subset of all boosts in an arbitrary direction can be homemorphically mapped onto R, which is non-compact in the usual topology. But the set of all boosts in a direction is a subset of all restricted Lorentz transformations. So the full Lorentz group contains a subset which is homeomorphic to R. So it must be non-compact.
  5. Sep 29, 2011 #4

    I came to this same exact question this couple of days, not sure how useful it will be to post on a thread which seems inactive now, but still, quoting vanhees71

    "A Lie group is called compact if there is a compact parameter set which maps out the whole group"

    which is the rough definition of the compactness, then what about describing every Lorentz transformation with the parameter V (or V/c) of relative velocity of 2 inertial frames, which goes from 0 to c. then to every element of Lorentz group corresponds a number between 0 and c, so it seems to be compact, what's wrong here?
  6. Sep 29, 2011 #5

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Is c included?
  7. Sep 29, 2011 #6
    I guess I should make my point clearer.

    What is a Lorentz transformation - it's a transformation of coordinates from one intertial frame to another, which moves with a uniform velocity V relatively to the first frame.
    Different Lorentz transformations just correspond to different relative velocities V.

    The question is, the velocities are limited from above by c, so the velocities themselves form a compact set, to every V corresponds a Lorentz transformation - then why is the Lorentz group non-compact?
  8. Sep 29, 2011 #7

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    c is not included in the parameter space for the set of Lorentz transformations that you considered, i.e., the parameter space is the half-open, half-closed interval of real numbers [0 , c). By the Heine-Borel theorem, a subset of the real numbers is compact if and only if the subset is closed and bounded. [0 , c) is not a closed subset of the real numbers, and thus is not compact.
  9. Sep 29, 2011 #8
    ah, thanks, that explains it all
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook