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A local $U(1)$ transformation is given by \begin{equation} f(x) = e^{i\epsilon(x)} \qquad \text{with} \qquad \epsilon(x) \in C^\infty \, . \end{equation}

Why do we require that the functions in the exponent are infinitely differentiable functions? What would go wrong without this restriction?

Qmechanic
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jak
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  • Related: https://physics.stackexchange.com/q/1324/2451 and links therein. – Qmechanic Oct 10 '18 at 18:35
  • In physics, one require the functions to be $C^\infty$ just to mean that they are regular enought to not cause any troubles. – Kevin De Notariis Oct 10 '18 at 18:41
  • @KevinDeNotariis thanks! However, my question is specifically what would go wrong if the functions that parametrize gauge transformations wouldn't be $C^\infty$. – jak Oct 10 '18 at 18:52
  • If you perform a Gauge Transformation you change the Bundle on which the Theory lives. If the transformation would not be $C^{\infty}$, the Bundles where not Isomorphic (as Principle Fibre Bundles), hence they would not describe the same Theory. This is not completly true however, since one usually just says ''smooth'' meaning that we can differentiate as many times as we want to (not necesarily $C^{\infty}$), and then one would require the Gauge Transformations to be ''smooth'' as well. – Creo Oct 10 '18 at 22:39

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In the Standard Model, Gauge Symmetries are made local to introduce interactions. For example in the free Dirac Lagrangian: $$\mathcal{L}_D=\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ it is easy to verify that under a $U(1)$ transformation, it is invariant. By making $U(1)$ local, an interaction is brought into play (namely, the interaction with the electromagnetic photon).

In a more complicated theory, namely in the Glashow-Weinberg-Salam model of the electro-weak interaction, the group that is made local is the $SU(2)_L\times U(1)_Y$.

This procedure leads to a new Lagrangian in which there appear lots of new terms. After lots of rearrangements and new definitions of the fields, there appear 1 Complex massive Field (The $W^\pm$ i.e. the charged bosons mediating weak interactions), 1 real massive field (The $Z_0$ i.e. the neutral bosons mediating weak interactions) and 1 real massless field (The $A^\gamma$ i.e. the photon).

These fields are made up of parts which where the old fields introduced in the process of making local the gauge symmetry, and other parts which are the derivatives of the local phase of the local gauge transformations.

In other words for a gauge transformation of $SU(2)_L\times U(1)_Y$: $$\phi(x)\to \phi'(x)=e^{\frac{i}{2}\vec{\sigma}\cdot\vec{\alpha}(x)}e^{\frac{i}{2}\beta(x)}$$ At the infinitesimal level, we identify: $$\vec{\alpha}(x)=-\frac{\vec{\theta}(x)}{v}\quad\quad\quad\beta(x)=\frac{\theta_3(x)}{v}$$ where $v$ characterized the vacuum choosen, and these $\vec{\theta}$(x) are Goldstone Bosons. The Fields $W^\pm,Z_0,A^\gamma$ are defined in function of these Goldstone Bosons (to be precise are defined in function of the derivatives of these $\theta_i(x)$). And since Goldstone Bosons are defined in function of the local parameters of the local gauge transformation, if the local gauge transformation's parameters happen to be hill defined, then our interacting bosons will be hill defined and all the particle physics will be meaningless.

Kevin De Notariis
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