Is gauge group infinite dimensional Lie group?

Question

Let $P$ be a smooth principal bundle on a manifold $M$ with a structure Lie group $G$. Then we define the gauge group $\mathcal{G}$ by the automorphism group of $P$, that is, the group of diffeomorphisms on $P$ which preserve the fiber and the action of $G$. This pdf says that the gauge group is a Hilbert Lie group. Is it true? And what is its Lie algebra?

Answer 1

the reason why a gauge field is in the Lie algebra of a gauge group $G$ is that we have to cancel out the term which comes from the kinetic term by acting gauge transformation. For simplification I want to use the $F_{\mu \nu}$. I know this transforms as $$ F_{\mu \nu} \rightarrow g F_{\mu \nu} g^{-1} $$ and it's called adjoint representation $(g \in G)$. The field strength $F_{\mu \nu}$ is a (local representation of) a Lie-algebra valued two-form. In components, it is often written as $F_{\mu \nu}^a$, where the $\mu \nu$ are the space-time indices (making it a two-form) and a is the Lie-algebra index. As such, it is indeed in the Lie-algebra.

However, it is not any Lie-algebra valued two form. Instead, it is the (local representation of) the covariant derivative of a special one-form, the connection one-form (whose local representation is usually denoted by $A_\mu$ ). This necessarily transforms in the adjoint representation. An intuitive reasoning for this can be found in this question (1). This is the mathematical viewpoint. The physical viewpoint is that the adjoint rep gives the "correct" transformation, simply because of the way we have defined the field strength in nonabelian gauge theories. Specifically, the field strength is defined as (this definition comes from the differential geometry picture sketched above): $$ F_{\mu \nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-i\left[A_\nu, A_\mu\right] $$

And if you perform a gauge transformation of this object (as is done e.g. here (2), you find that it transforms in the adjoint representation.