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Let $\mathsf{Prin}_G$ be the category of (right) $G$-principal bundles, with a morphism from the bundle $p: P \to M$ to the bundle $p': P' \to M'$ being a pair of arrows $\chi: P \to P'$ and $\bar{\chi}: M \to M'$ such that $\chi$ is $G$-equivariant and the obvious diagram commutes. We only need to specify the morphism $\chi: P \to P'$ since the corresponding morphism down on the base is uniquely determined by $\chi$.

Let $\mathsf{Space}_G$ be the category of (left) $G$-spaces, with a morphism from $S$ to $S'$ being a $G$-equivariant arrow $f: S \to S'$.

Let $\mathsf{Bund}$ be the category of fibre bundles, with morphisms from $E \to M$, $E' \to M'$ being a pair of arrows such that we get a commutative square.

Then associated bundle assignment $$\mathsf{Prin}_G \times \mathsf{Space}_G \to \mathsf{Bund}$$ sending a pair of morphisms $\chi: P \to P'$ and $f: S \to S'$ to the morphism
$$\chi \times_G f: P \times_G S \to P' \times_G S'$$ given by $$(\chi \times_G f) [u, s] = [\chi(u), f(s)]$$ is a bifunctor.

Now suppose we fix a principal $G$-bundle $P \to M$. Then this gives us an associated bundle functor $$P[-] := P \times_G (-): \mathsf{Space}_G \to \mathsf{Bund}(M),$$ where $\mathsf{Bund}(M)$ is the category of fibre bundles over $M$, with morphisms covering the identity on $M$.

Does this functor have any left or right adjoints? Do we need to restrict the target category to get an adjunction? What if instead we fix the other argument (the $G$-space) and let the principal bundle vary?

To provide a bit of motivation, I have read that for any fixed principal bundle the associated bundle functor from representations to vector bundles is exact, which leads me to guess that it might have both a left and a right adjoint.

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  • $\begingroup$ This might depend on the niceness of the underlying notion of spaces. If there exists an internal $\mathrm{Hom}$ in the category of fibre bundles over $M$ (with respect to $\times_ M$), then the right-adjoint should be given by $\mathrm{Hom}(P,–)$. $\endgroup$
    – Ben
    Commented Jan 10 at 12:35
  • $\begingroup$ @Ben The Associated bundle construction is not an endofunctor $\endgroup$
    – ಠ_ಠ
    Commented Jan 13 at 7:02
  • $\begingroup$ But $\mathop{Hom}(P,–)$ will lift to the category of $G$-bundles over $M$ and when I did some scribbling it seemed to me that post-composing with the forgetful functor to $G$-spaces should work as well, but I'm not completely sure. $\endgroup$
    – Ben
    Commented Jan 13 at 17:58
  • $\begingroup$ Okay, sorry, you still have to take global sections of that and I don't know what's the 'correct' topology then... $\endgroup$
    – Ben
    Commented Jan 15 at 10:23

2 Answers 2

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I have at least managed to construct a left adjoint to the associated bundle functor $$P [-]: \mathsf{Space}_G \to \mathsf{Bund}(M),$$ Given by the "fibre product" functor $$ P \times_M (-): \mathsf{Bund}(M) \to \mathsf{Space}_G,$$ where the $G$ action is the obvious one $(u_x, v_x).g = (u_x.g, v_x)$. The counit is given by the $G$-equivariant maps $$\varepsilon_S: P \times_M (P[S]) \to S$$ induced by the unique isomorphism $P \times_M (P[S]) \to P \times S$ (since both are pullbacks in the same pullback square) composed with the projection $P \times S \to S$. Explicitly, $\varepsilon_S(u_x, v_x) = q_{u_x}^{-1}(v_x)$, where $q_{u_x}: \{u_x\} \times S \to P[S]_x$ is the isomorphism given by restricting the quotient map $q: P \times S \to P[S]$. In the special case $S=G$, we find that the counit component $\varepsilon_G: P \times_M P[S] = P \times_M P \to G$ is just the usual division map for a principal bundle.

The unit map, however, seems a bit more elusive to describe, and I have not yet found a right adjoint to the associated bundle functor, if it exists.

EDIT: I've found the unit map. Will write a short paper this year with the proof.

As a nice application of this adjunction, we obtain the following classical result. Since the global sections functor $\Gamma: \mathsf{Bund}(M) \to \mathsf{Sets}$ is representable by the trivial bundle $\text{id}_M: M \to M$, we have a natural bijection $$\text{Hom}_{\mathsf{Space}_G}(P, S) \cong \Gamma(M, P[S])$$ which identifies the sections of the associated bundle $P[S] \to M$ with $G$-equivariant maps $P \to S$.

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$\newcommand\mhy{{\operatorname{-}}}$Here is a conditional answer concerning the existence of a right-adjoint, depending on the existence of two things:

  1. A right-adjoint $[P,-]$ to the endo-functor $-\times_M P$ on $\mathsf{Bund}(M)$ (think 'internal Hom')
  2. A right-adjoint $\Gamma\colon G\mhy\mathsf{Bund}(M)\to G\mhy\mathsf{Space}$ of the functor $\underline{(\cdot)}\colon G\mhy\mathsf{Space}\to G\mhy\mathsf{Bund}(M)$ from the category of $G$-(left-)spaces to that of $G$-(left-)bundles (with fibre-preserving actions) taking a $G$-space $S$ to the trivial $S$-bundle $\underline S=S\times M$ over $M$ (with the obvious $G$-action). (Think 'global sections'.)

(That is, we should be working in a sufficiently convenient setting. Depending on the concrete setting, it could easily be argued that the below is unreasonably cluttered. It could very well be sufficient to 'just' topologise the set of bundle-morphisms $P\to X$ in a sufficiently nice way, but I've always been ignorant of the different methods of how to do that, so, sorry for that and bear with me, please.)

First, I will construct a lift of $[P,-]$ along the forgetful functor $G\mhy\mathsf{Bund}(G)\to G\mhy\mathsf{Space}$, i.e., a natural $G$-action on $[P,X]$ for each $X$, which is going to have the property that $\hom_{\mathsf{Bund}(M)}(P[S],X)\cong \hom_{G\mhy\mathsf{Bund}(M)}(\underline{S},[P,X])$, naturally in $X$. By assumption 2, we conclude that $\Gamma\circ[P,-]$ is the desired right-adjoint.

The counit of the adjunction defining $[P,-]$ is called the evaluation morphism, $\varepsilon_X\colon [P,X]\times_M P\to X$. Therefore, a bundle-morphism $\psi\colon Y\to [P,X]$ is uniquely determined after evaluation, i.e., by the composite morphism $Y\times P\xrightarrow{\psi\times 1_P}[P,X]\times_M P\xrightarrow{\varepsilon_X} X.$

In this vein, the $G$-action on $[P,X]$ may be defined as follows: Let $\rho_P\colon \underline G\times_M P\to P$ be the left-action, $(g,p)\mapsto p{.}g^{-1}$. Then the morphism $\rho_{[P,X]}\colon\underline G\times_M[P,X]\to[P,X]$ shall correspond to the composition $$\begin{align}\underline G \times_M[P,X]\times_M P&\xrightarrow{\mathop{swap}\times 1_P}[P,X]\times_M\underline G \times_M P\\&\xrightarrow{1_{[P,X]}\times\rho_P}[P,X]\times_M P\\&\xrightarrow{\epsilon_X}X.\end{align}$$ (On elements, this maps $(g,f,p)\mapsto \varepsilon_X(f,p{.}g^{-1})$.) This is clearly natural in $X$.

Here comes the key observation. Let's have a closer look at the defining property of $\rho_{[P,X]}$. It is the unique morphism $\underline G\times_M [P,X]\to[X,P]$ such that after evaluating at $P$ we get back the morphism specified above, i.e., $\varepsilon_X\circ(\rho_{[P,X]}\times 1_P)=\varepsilon_X\circ(1_{[P,X]}\times\rho_P)\circ (\mathop{swap}\times 1_P)$; on elements: $\varepsilon_X(g{.}f,p)=\varepsilon_X(f,p{.}g^{-1})$. Does look familiar to the definition of $P\times_G-$, doesn't it?

A bundle morphism $\psi\colon \underline S\to [P,X]$ induces, via the adjunction, a corresponding morphism $\psi'\colon S\times P=\underline S\times_M P\to X$, $(s,p)\mapsto \varepsilon_X(\psi(s),p)$. Moreover, the observation above implies that $\psi$ is $G$-equivariant if and only if $\psi'$ satisfies $$(\ast)\quad\psi'(g{.}s,p)=\psi'(s,p{.}g^{-1})$$ for all $s\in S$, $g\in G$ and $p\in P$, which, on the other hand, is exactly the condition for $\psi'$ to descend to a bundle morphism $P\times_G S\to X$ along the (swapped) quotient projection $S\times P\to P\times_GS$!

Putting all the pieces together, we get the desired adjunction as the composition of natural equivalences

$$\begin{align} \hom(P\times_G S,X) &\cong \{\psi'\in\hom_{\mathsf{Bund}(M)}(S\times P,X)\mid (\ast)\}\\ &\cong\{\psi\in\hom_{\mathsf{Bund}(M)}(\underline S,[X,P]) \mid\psi\text{ is }G\text{-equivariant} \}\\ &=\hom_{G\mhy\mathsf{Bund}(M)}(\underline S,[P,X])\\ &\cong\hom_{G\mhy\mathsf{Space}}(S,\Gamma([P,X])).\end{align}$$

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