Invariant integration on principal bundles

Question

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.

Answer 1

$\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\GL}{GL}$ $\DeclareMathOperator{\Aut}{Aut}$

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 exists on locally compact groups. The group of gauge transformations is infinite dimensional and so it is not locally compact.