1-dimensional pure gauge theory

Just to add something about Gauss' law to the excellent previous answer: The Hamiltonian of a pure gauge theory, as initially derived from the Lagrangean, typically has the structure $H=E^2 +B^2 - A_0 \nabla \cdot E$, where $E$ and $B$ are the electric and magnetic fields, and $A_0 $ is the temporal component of the gauge field. In addition, the Lagrangean contains no time derivative of $A_0 $, and therefore there is no conjugate momentum for $A_0 $ - it is not a dynamical degree of freedom. Instead, it acts as a Lagrange multiplier enforcing $\nabla \cdot E =0$. That is Gauss' law, a constraint on the remaining dynamical fields.

There are distinct ways of dealing with this. One avenue is to use the gauge freedom to choose $A_0 =0$. Then the non-dynamical degree of freedom is gone, but one has lost the knowledge about the Gauss law constraint. In $A_0 =0$ gauge, one must impose the constraint in addition to the dynamical equations generated by the Hamiltonian.

Another avenue is to choose a different gauge, but then the formalism still contains $A_0 $. The "equation of motion" for $A_0 $ is, again, not dynamical, but a constraint on $A_0 $, which one can then solve to eliminate $A_0 $ (this, e.g., generates the Coulomb interaction). In this manner of speaking, Gauss' law doesn't play quite such a central role - one thinks of $A_0 $ being constrained, as opposed to the other fields being constrained.

Either way, one eliminates two components of the gauge field: One by choice of gauge, one by constraint. Only the remaining components are dynamical degrees of freedom. Thus, light comes with two polarizations, even though one starts out with a 4-component field $A_{\mu } $.

A gauge theory in mathematical terms (as I understand it) is a field theory whose fields include a gauge field: a $G$-bundle with connection. In this case, $G$ is a finite group, so there are no non-trivial connections. Thus a gauge field is simply a $G$-bundle.

1) Here, by pure gauge theory, I think the authors mean that this is a theory whose fields are precisely $G$-bundles and nothing more (in particular, no "matter" fields).

2,3) The path-integral is one approach to quantizing a field theory. It says, in particular, we should compute the partition function of the theory by a suitably weighted integral over a space of fields of the theory. The weight is prescribed by the action, a certain function on the space of fields.

Fields in this theory are $G$-bundles. The collection of $G$ bundles on a manifold $M$ (spacetime) can naturally be organized in to a groupoid, whose objects are $G$-bundles and morphisms are isomorphisms of $G$-bundles. For example, the groupoid of $G$-bundles on the circle $S^1$ may naturally be identified with the quotient groupoid $G//G$ (where $G$ acts on itself by conjugation). This identification assigns to a $G$-bundle on $S^1$ its holonomy (which is well-defined up to conjugation).

The (exponentiated) action functional is given by the abelian character $\lambda:G \to T$. This assigns to a $G$-bundle on $S^1$ with holonomy $g$, the function $\lambda(g)$. Thus the partition function on $S^1$ is given by an ''integral'' over the groupoid $G//G$ weighted by the function $\lambda$. This really just means the sum of $\lambda(g)$ as $g$ ranges over $G$, and one must divide by $|G|$ to get the right notion of volume (we are integrating over $G//G$ not $G$). You can check that one gets the stated results.

There is a similar idea happening for the $0$-manifold $pt_+$. In general QFT, one expects to assign to a codimension 1 submanifold a vector space (the "Hilbert space" of the theory). Roughly speaking one might think of elements of this vector space as being functions on the space of fields - or better, sections of a line bundle (which again is associated to the action).

In this case, the function $\lambda$ is actually the same thing as a line bundle on the space of fields on a point (which is the groupoid $pt//G$). The space of sections of this line bundle is either 1-dimensional if the bundle is trivial ($\lambda =1$) or $0$-dimensional if the bundle is non-trivial ($\lambda \neq 1$).

I wish I had something reasonable to add about the Gauss Law. Perhaps someone more knowledgeable could weigh in?