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

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1 Answer 1

up vote 6 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 $\psi(x)->\psi'(x) = g(x)\psi(x).$

Derivatives would transform as

$\partial_{\mu}\psi(x)->g(x)\partial_{\mu}\psi(x)+(\partial_{\mu}g(x))\psi(x)$

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

$D_{\mu}\psi(x)->g(x)D_{\mu}\psi(x)$

To achieve this, we define

$D_{\mu}\psi = \partial_{\mu}\psi - A_{\mu}\psi$, where $A_{\mu} = \mathbf{A_{\mu}}.\boldsymbol{{\tau}}$ 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

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

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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