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Orthogonality of inner product of generators

  1. Jul 20, 2015 #1
    Hi, this is a rather mathematical question. The inner product between generators of a Lie algebra is commonly defined as [itex] \mathrm{Tr}[T^a T^b]=k \delta^{ab} [/itex]. However, I don't understand why this trace is orthogonal, i.e. why the trace of a multiplication of two different generators is always zero.
     
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  3. Jul 20, 2015 #2
    This is a choice we make for convenience. The generators form a basis for vector space that is the Lie algebra, and it is convenient to choose an orthogonal basis. We can always choose a basis that is orthogonal.
     
  4. Jul 20, 2015 #3

    Orodruin

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    The only thing you should worry about is whether the trace tr(AB), where A and B are elements of the Lie algebra, is an inner product or not. If you have an inner product, you can always select an orthogonal basis.
     
  5. Jul 20, 2015 #4

    vanhees71

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    This is the great thing with semi-simple compact Lie groups. Their generators can always be chosen such as they are "orthogonal" in the sense you wrote. This implies that the Lie group, as a differentiable manifold (with the group operations providing differentiable mappings), has a "natural" metric (Riemann-space) structure and you can thus easily derive things like the invariant measure for integrations over the group (Haar measure), using Weyl's unitarity trick (proving that all finite-dimensional representations are equivalent to a unitary one) etc. Last but not least, all semi-simple compact Lie groups are identified ("Cartan catalogue"). For details, see Weinberg, Quantum Theory of Fields, Vol. 2.
     
  6. Jul 20, 2015 #5
    I see, that makes sense. Thank you all for answering
     
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