# Invariant integration on principal bundles

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.

• First, you need to clarify on which space you want to construct this measure. Next, recall that the group of gauge transformations is infinite dimensional. On which space is this group supposed to act? There are many possible choices. – Liviu Nicolaescu Dec 18 '18 at 20:51
• Thanks for your comment. I have added some information to my question: I am interested in integrating vector-valued functions on P, as well as integrating sections of vector bundles associated to $P$, i.e. $P \times_\rho V$ for some representation $\rho$ of $H$. Let me know if you need more information (I'm new to this stuff). Thanks! – John von N. Dec 19 '18 at 17:58
• You want to integrate these over what space? On which space is the $Aut(P)$-invariant measure defined? – Liviu Nicolaescu Dec 19 '18 at 20:30
• Sorry if I wasn't clear. I've edited the question again. I'm looking for an $\operatorname{Aut}(P)$-invariant measure on $P$ itself. – John von N. Dec 20 '18 at 9:44

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

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

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.

• I think you forgot the target of your Haar measure link. There are concepts of Haar-type measures for more exotic groups than locally compact ones, but they tend also to take more exotic values, and so maybe aren't appropriate here. – LSpice Dec 20 '18 at 12:23
• Thank you! I will study your answer. If I understand correctly you consider here the group of automorphisms that fix the base space ("B-automorphisms"). What happens if we include more general ones? E.g. the one given by translation in $\mathbb{R}^n$ or left translation on $G$, where $G$ is viewed as a principal $H$ bundle? I guess an invariant measure still exists then? – John von N. Dec 20 '18 at 16:48
• Also, if you have any references to textbooks or papers where this material is covered, I would appreciate it a lot. – John von N. Dec 20 '18 at 16:52
• Give me a concrete example that interests you and I will attempt to give you a precise answer. – Liviu Nicolaescu Dec 20 '18 at 18:21
• Let me ask next, once you have such an invariant measure what would you use it for? – Liviu Nicolaescu Dec 20 '18 at 19:56