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

In summary: Yang-Mill's theory and whether using a non-bi-invariant inner product would destroy its gauge invariance. The action is often written as$$ S= \int \frac{1}{4}\text{Tr} (F^{\mu \nu} F_{\mu \nu})d^4 x = \int d^4 x\frac{1}{4} F^{k \mu \nu} F_{k \mu \nu}$$The trace is taken with respect to the inner product ##\delta_{kl}##, which is ad-invariant under gauge transformations. However, implementing a more general inner product ##g_{ij}## with the action
  • #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##?
 
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  • #2
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.
 
  • #3
center o bass said:
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##?

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

Sam
 

1. What is a bi-invariant metric and why is it necessary for Yang-Mills theory?

A bi-invariant metric is a type of metric that is invariant under both left and right translations in a Lie group. In Yang-Mills theory, this metric is necessary because it allows for a consistent and meaningful way to define distances and angles between different points on the Lie group, which is essential for the theory's calculations and predictions.

2. How does a bi-invariant metric affect the behavior of Yang-Mills fields?

A bi-invariant metric plays a crucial role in the behavior of Yang-Mills fields. It allows for the formulation of gauge-invariant quantities, which are essential for understanding the interactions between particles and fields. Additionally, it helps to determine the strength of these interactions and how they change over space and time.

3. Can Yang-Mills theory be formulated without a bi-invariant metric?

No, a bi-invariant metric is a fundamental aspect of Yang-Mills theory and cannot be removed from its formulation. It is necessary for defining gauge-invariant quantities and determining the behavior of Yang-Mills fields.

4. How does the choice of bi-invariant metric affect the predictions of Yang-Mills theory?

The choice of bi-invariant metric does not affect the predictions of Yang-Mills theory. The theory is independent of the specific choice of metric and only relies on the fact that it is bi-invariant. However, different metrics may be useful for specific calculations or interpretations of the theory.

5. Are there any other theories that require a bi-invariant metric?

Yes, bi-invariant metrics are necessary for many other gauge theories, including general relativity and gauge theories for other fundamental forces such as the weak and strong nuclear forces. They are also used in other areas of mathematics, such as differential geometry and Lie theory.

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