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Adjoint representation of Lorentz group

  1. Feb 4, 2016 #1

    There are some posts about the reps of SO, but I'm confused about some physical understanding of this.

    We define types of fields depending on how they transform under a Lorentz transformation, i.e. which representation of SO(3,1) they carry.
    The scalar carries the trivial rep, and lives in a 1-dim vector space.
    The vector carries a rep generated by 4x4 matrices, and lives in a 4-dim vector space, i.e. it is a 4-column.
    Weyl spinors and Dirac spinors carry a irreducible 2-dim rep and a reducible 4-dim rep respectively.

    I don't quite understand how to think of tensors. In particular, two things:

    1. What objects carry the adjoint representation? This is generated by 6x6 matrices, so should act on 6-columns?

    2. I read that a rank-2 tensor can be thought of something that transforms under the tensor product of two 4x4 matrices, i.e. a 16x16 matrix. Hence it should be a 16-column? How can I reconcile with writing a tensor as usual as a 4x4 matrix?

    Thanks a lot!
  2. jcsd
  3. Feb 4, 2016 #2


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    A general vector space need not be represented by column vectors. Several matrix spaces are vector spaces as well. You can represent those with a column vector by simply listing the coefficients of some basis, but you do not have to.

    Also note that the tensor representation is not irreducible. You can decompose it into several different irreps.
  4. Feb 4, 2016 #3
    Thanks Orodruin.

    So, group-theoretically, what am I doing when I am transforming a tensor g_\mu\nu with two Lambda matrices? I am decomposing the tensor product vector space into two spaces carrying the vector representation?

    Also, any insight on what fields carry the adjoint rep?
  5. Feb 4, 2016 #4


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    No, generally the tensor product does not decompose into two vector representations. The prime example of this is looking at the tensor product of two copies of the fundamental representation of SU(2), which decomposes into a singlet and a triplet.

    For the Lorentz group, the rank 2 tensor representation does not even have the correct number of dimensions to decompose into two vectors.
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