Groups in QFT

  1. Qhy does SO(n) have the same number of dimensions of O(n), whereas SU(n) reduces the dimensions of U(n)? Isn't the constraint the same for both cases, i.e. detM=1?
  2. jcsd
  3. If I haven't misunderstood something myself:

    The constraint is the same, sure, but the other is complex and the other real:

    If U∈U(n), |det(U)|=1 (it preserves the norm) and det(U) is in ℂ, so it's of the form e for some θ∈ℝ -- there is one free parameter "in the determinant". Further requiring U∈SU(n) means that det(U) is exactly one, so one "degree of freedom" (one free parameter) is lost and θ=0 exactly. So dim(SU(n))=dim(U(n))-1.

    If O∈O(n), |det(O)|=1 still, but since det(O) is now in ℝ, there is a discrete set of two possible values instead of a free parameter: Either det(O)=1 or det(O)=-1. So, requiring O∈SO(n) just picks one of these two options, that det(O)=1, but it doesn't remove any actual free parameters, so dim(SO(n))=dim(O(n)).

    (That's obviously not a proof, of course, but merely something that made me understand the reasoning behind it.)
  4. Thank you, I had just gotten to a similar conclusion myself. Also, I thought about it topologically: in the orthogonal case, O(n) is formed of two disjoint sets, (i) detR=1 and (ii) detR=-1, of which only one is a subgroup {i.e. SO(n), with detR=1, as it contains the identity}. Therefore removing the bit (ii) detR=-1 doesn't "get rid" of any subgroup, but only of a coset in O(n). In U(n) nothing of this happens, as it is compact and connected (which I think reflects on the fact that we have a continuous set of possible values, as opposed to the discrete one in O(n)).

    I don't enough topology or group theory to be sure about this, but this is the way I thought about it. Can somebody tell me if I'm wrong?
  5. DarMM

    DarMM 331
    Science Advisor

    It seems right to me. ##U(n)## is a connected ##n^2 - 1##-dimensional manifold, you can always create a coordinate system where one of the coordinates is ##\theta## the logarithm of the determinant. Hence, ##SU(n)## is simply the ##\theta = 0## submanifold, and so has one dimension less.

    In the ##O(n)## case we simply have two manifolds of the same dimension, one indexed with ##+1## and the other indexed with ##-1##. ##SO(n)## is simply one of these manifolds.
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