- #1
center o bass
- 560
- 2
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##?
$$ S= \int \frac{1}{4}\text{Tr} (F^{\mu \nu} F_{\mu \nu})d^4 x = \int d^4 x\frac{1}{4} F^{k \mu \nu} F_{k \mu \nu}$$
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##?