# How does gauge symmetry constrain the dynamics of a field's physical degrees of freedom?

My rough understanding of gauge theory is that some of a field's degrees of freedom (d.o.f.) may turn out to be "non-physical" due to local symmetries. But does gauge symmetry constrain the dynamics of the remaining "physical" d.o.f. at all? Suppose we take a field with very complicated dynamics, and then artificially add some "non-physical"/"fake" d.o.f. to obtain a field with gauge symmetry. As far as I can tell, this new field will still have complicated dynamics, since the gauge symmetry doesn't say anything about the physical d.o.f.. Therefore I don't understand what's so constraining about gauge symmetry, and feel I must be missing something.

To make this more concrete, consider the example given by Wikipedia of scalar $$O(n)$$ gauge theory. In this case, the field $$\Phi$$ takes values in $$\mathbb{R}^n$$, and there is local $$O(n)$$ symmetry: that is, at a given point $$x$$, the field values $$\Phi(x)$$ and $$O\Psi(x)$$ are physically equivalent for any $$O\in O(n)$$. It therefore seems that the only physical d.o.f. is $$|\Phi|^2$$, since this is $$O(n)$$-invariant. Can $$|\Phi|^2$$ behave however it likes? If so, why do we bother thinking about an $$n$$-component field $$\Phi$$ with gauge symmetry when we could just think about a single scalar field $$|\Phi|^2$$ with no gauge symmetry?

I'll focus on the scalar $$O(n)$$ gauge theory that was mentioned in the question. The set of observables in the gauged version of the model is not a subset of the observables in the ungauged version. Gauging the $$O(n)$$ symmetry does eliminate some observables, but it also introduces others.
Observables in the gauged version are required to be invariant under the combined transformation \begin{align} \Phi(x) &\to O(x)\Phi(x) \tag{1} \\ A_\mu(x) &\to O(x)A_\mu(x)O^{-1}(x) + iO(x)\frac{\partial}{\partial x^\mu} O^{-1}(x) \tag{2} \end{align} where $$A$$ is the gauge field. Most observables are not invariant under either (1) or (2) individually. The observable $$|\Phi|^2$$ that was mentioned in the question is invariant under both (1) and (2) individually, but most observables are not. The operator $$\sum_{j,k}\Phi_j^*(x)U_{jk}(C)\Phi_k(y) \tag{3}$$ qualifies as an observable (it is gauge invariant), where $$U(C)$$ is the unitary operator constructed from the gauge field by taking a path-ordered integral of $$\sim\exp(iA(s))$$ along some specified contour $$C$$ from $$s=y$$ to $$s=x$$. If $$C$$ is a closed contour, then the operator $$\text{trace}\,U(C) \tag{4}$$ also qualifies as an observable (gauge invariant). This is called a Wilson loop observable.
By the way, this is all more clear in lattice QFT, which is also the only known way to really define this theory. I mean, it's the only known way to define it nonperturbatively. In any other part of physics, the adjective "nonperturbatively" goes without saying whenever the word "define" is used, but the QFT culture is a little weird in this respect, for both historical and technical reasons. Ref 1 uses a generous dose of lattice QFT: the word "lattice" occurs on $$42$$ of the paper's $$176$$ pages.