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

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up vote 8 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.

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Why do we still say A is in the adjoint representation if there is that second term? – Lagrangian May 27 '15 at 14:27

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