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

1. Jul 27, 2014

### center o bass

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. Jul 27, 2014

### Ben Niehoff

Under a gauge transformation, we have

$$F_{\mu\nu} \to g F_{\mu\nu} g^{-1}$$
and hence

$$F^{\mu\nu} F_{\mu\nu} \to g F^{\mu\nu} F_{\mu\nu} g^{-1}$$
so yes, gauge invariance of the action requires using the bi-invariant metric.

3. Jul 27, 2014

### samalkhaiat

See page 12, sec. 4 of the PDF in
https://www.physicsforums.com/showpost.php?p=4238251&postcount=1

Sam