Let $G$ be a sufficiently nice topological or Lie group (e.g. compact), and let $H$ be a closed subgroup. This data determines a principal $H$ bundle $G \rightarrow G/H$ defined by the projection $g \mapsto gH$.


In this case, we have Haar measures on $G$ and $G/H$ which allow us to perform invariant integration. I am wondering how this can be generalized to more general principal bundles. 


Let $P$ be a principal $H$ bundle with automorphism group $\operatorname{Aut}(P)$. Is it possible to define an $\operatorname{Aut}(P)$-invariant measure on $P$, or otherwise define an invariant integral (e.g. using differential forms) of vector valued functions on P?


More specifically, I am interested in integrating over $P$ functions $f$ in the following space:
$$ M = \{ f : P \rightarrow V \; | \; f(ph) = \rho(h^{-1}) f(p) \}.
$$
Where $(\rho, V)$ is a representation of $H$. (Functions in $M$ are in one to one correspondence with sections of the associated vector bundle $P \times\_\rho V$)


The automorphism group $\operatorname{Aut}(P)$ acts on $M$ by $f(p) \mapsto f(a^{-1} p)$ for $a \in \operatorname{Aut}(P)$. If I'm not mistaken, we have for the bundle $G \rightarrow G/H$ that the Haar measure is invariant to this action. Hence my question is whether there is such a thing as "generalized Haar measure" on more general principal bundles.



Invariant integration on principal bundles
$\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\GL}{GL}$ $\DeclareMathOperator{\Aut}{Aut}$


You should start with a simple case


$$M=\bR^n,\;\; G=\GL\_m(\bR),\;\;P=G\times M. $$


In this case
$$
\Aut(P)=C^\infty\big(\; \bR^n,\;\GL\_k(\bR)\;\big).
$$
The action of gauge transformation $\gamma:\bR^n\to\GL\_m(\bR)$ on $P$ is given by
$$
\gamma\cdot(x,g)=(x,\gamma(x)\cdot g).
$$
For $g\in G$ denote by $\gamma\_g$ the constant gausge transformation $\gamma\_g(x)=g$, $\forall x\in\bR^n$.


Any top degree form $\omega^P$ on $P$ admits a canonical decomposition


$$
\omega^P= \rho(x,g)dV\_n\wedge dV\_G,
$$ 
where $dV\_G$ is a **left** invariant Haar form on $G$, $dV\_n$ denotes the Euclidean volume form on $\bR^n$ and $\rho$ is a function $P\to\bR$.


If $\omega^P$ is $\Aut(P)$ -invariant then from the equality $\gamma\_g^\*\omega^P=\omega^P$ we deduce that $\rho(x,g)$ is independent of $g$.


The exterior derivative $d^P$ on $P$ decomposes as a sum
$$
d^P= d^M+d^G.
$$
If we write 
$$
dV\_G =f(g)\bigwedge\_{i,j} dg\_{ij},\;\; g=(g\_{ij})\_{1\leq i,j\leq m},
$$
then for any $\gamma\in\Aut(P)$ we have
$$
\gamma^\*\omega= \rho(x)f(\gamma(x)g) \Bigg(d\_M\bigwedge \big(\gamma(x)g\big)\_{ij}+ d\_G\bigwedge d\_G\big(\gamma(x)g\big)\_{ij}\;\Bigg)\wedge dV\_n
$$
$$
=\rho(x)f(\gamma(x)g)\Bigg(d\_G\big(\gamma(x)g\big)\_{ij}\;\Bigg)\wedge dV\_n
$$
$$
=\omega^P\_{x,\gamma(x)g},
$$
where at the last step we used the left invariance of $dV\_G$.


Thus in this case there exist $\Aut(P)$-invariant volume forms on $P$.
$$
\omega^P\rho(x) dV\_n\wedge dV\_G.
$$ 
For a general bundle with general structure group $G$ the issue of existence of $\Aut(P)$-invariant forms localizes to the above situation and we conclude that such volume forms exist in general. 


**Remark.** The [Haar measure](https://en.wikipedia.org/wiki/Haar_measure) exists on *locally compact* groups. The group of gauge transformations is infinite dimensional and so it is not locally compact. 


