Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is there a deep mathematical reason for why bosons should be in the adjoint representation of the gauge group rather than any other representation?

share|improve this question

1 Answer 1

up vote 7 down vote accepted

I guess by "bosons" you're referring to gauge bosons?

If so then start with some matter field $ \psi(x)$ which transforms under the gauge group. For local gauge transformations the gauge group element $g$ is spacetime dependent $g(x)$, and the transformation is $$\psi(x) \longrightarrow \psi'(x) = g(x)\psi(x).$$

Derivatives would transform as

$$\partial_{\mu}\psi(x) \longrightarrow g(x)\partial_{\mu}\psi(x)+(\partial_{\mu}g(x))\psi(x),$$

i.e. inhomogeneously. We would like a gauge covariant derivative $D_{\mu}$ which transforms homogeneously as

$$D_{\mu}\psi(x) \longrightarrow g(x)D_{\mu}\psi(x).$$

To achieve this, we define

$$D_{\mu}\psi = \partial_{\mu}\psi - A_{\mu}\psi,$$ where $A_{\mu} = \mathbf{A}_{\mu} \cdot\boldsymbol{{\tau}}$ and $\boldsymbol{\tau}$ are the generators of the Lie algebra of the gauge group and $A_{\mu}$ is our bosonic gauge field. This introduction of gauge bosons via the derivative term is sometimes referred to as minimal coupling.

In order to achieve this, $A_{\mu}$ is forced to have the transformation law

$$A_{\mu} \longrightarrow A'_{\mu} = gA_{\mu}g^{-1} + (\partial_{\mu}g)g^{-1}.$$

Just looking at how the $A_{\mu}$ are transforming under the group action (the first term), we recognize the adjoint representation.

Of course, on the global stage, the fields $\psi$ can be interpreted as bundle sections and the gauge fields as bundle connections. $A_\mu$'s transformation law will be recognisable as a transformation of connection coefficients under the action of the bundle's structure group. A good reference is Nakahara, or this link.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.