I'm trying to build some intuition for a very particular definition of the notions global and local gauge symmetries. The definition goes as follows and appears, for example, in "Quantum Field Theory - A Modern Perspective" by V. P. Nair:

  • The group of gauge transformations $\mathcal{G}$ is a group of smooth functions on spacetime that takes values in $G$, where $G$ is the gauge group.
  • The group of local gauge transformations $\mathcal{G}_\star$ consists of all transformations where the parametrizing functions have compact support. This implies \begin{align} \mathcal{G}_\star &= \big \{ \text{ set of all } g(x) \text{ such that } g \to 1 \text{ as } |x| \to \infty \big \} \end{align}
  • The group of global gauge transformations is given by $ G/G_\star$.

The crucial point is now that transformations in $\mathcal{G}_\star$ are redundancies, while $ \mathcal{G}/ \mathcal{G}_\star$ are the physical symmetries of the system. The difference between the two is that $\mathcal{G}_\star$ only change something within a finite region, while the transformations in $ \mathcal{G}/ \mathcal{G}_\star$ have an effect all the way to the boundary at infinity.

Is there any intuitive example that motivates this distinction? In other words, why are transformations that only change something within a finite region redundancies, while real symmetries have an effect all the way to infinity?


Physical systems are described by differential equations plus suitable boundary conditions. (Only when we combine differential equations with proper boundary conditions, we can expect unique solutions.)

For example, we can impose $$\phi(x) \to \phi_0 \quad \text{as} \quad |x| \to \infty \, .$$

Now, states connected by a redundancy are to be identified. In contrast, a symmetry connects physically distinct states which happen to have the same properties. However, the crucial point is that states connected by symmetry transformations are, in principle, distinguishable.

This means especially that only global transformations are allowed to change our boundary conditions since different boundary conditions correspond to physically distinct states. Therefore, redundancies must preserve our boundary conditions which implies

$$ g(x) \to 1 \quad \text{as} \quad |x| \to \infty \, . $$

This answer is basically just a summary of the comments by @Prahar here.


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