Necessity of bi-invariant metric for Yang-Mill's theory.

  1. The action for Yang-Mill's theory is often written as
    $$ S= \int \frac{1}{4}\text{Tr} (F^{\mu \nu} F_{\mu \nu})d^4 x = \int d^4 x\frac{1}{4} F^{k \mu \nu} F_{k \mu \nu}$$
    where latin indices are indicies in the lie algebra, and the trace is taken with respect to the inner product ##\delta_{kl}##. The great thing about this inner product is that it is ad-invariant. I.e. when a gauge transformation is performed ##F\to g F g^{-1}## and
    $$\text{Tr} (F^{\mu \nu} F_{\mu \nu}) \to \text{Tr} (g F^{\mu \nu} F_{\mu \nu}g^{-1}) = \text{Tr} (F^{\mu \nu} F_{\mu \nu})$$.

    We could imagine implementing more general inner products ##g_{ij}## with the action
    $$ S= \int \frac{1}{4} g_{kl} F^{k \mu \nu}, F_{\mu \nu}^l d^4 x$$
    that is in general not bi-invariant. Am I wrong that this would destroy the gauge invariance of yang-mills theory? I.e. that gauge theory requires a bi-invariant inner product on ##\mathfrak g##?
  2. jcsd
  3. Ben Niehoff

    Ben Niehoff 1,766
    Science Advisor
    Gold Member

    Under a gauge transformation, we have

    [tex]F_{\mu\nu} \to g F_{\mu\nu} g^{-1}[/tex]
    and hence

    [tex]F^{\mu\nu} F_{\mu\nu} \to g F^{\mu\nu} F_{\mu\nu} g^{-1}[/tex]
    so yes, gauge invariance of the action requires using the bi-invariant metric.
  4. samalkhaiat

    samalkhaiat 1,194
    Science Advisor

    See page 12, sec. 4 of the PDF in

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