Principal Bundle and Cocycle

Let $G$ be a Lie Group and X a smooth manifold. Let $ G Bund(X)$ be the category of $G$-Principal Bundles. Objects are maps $\pi: P \rightarrow X$ where $P$ is a right $G$-space such that the local triviality is satisfied and maps $f: \pi_1 \rightarrow \pi_2$ are $G$-morphisms $f: P_1 \rightarrow P_2$ such that $\pi_2\circ f = \pi_1$.

A standard result is that there is a bijection between the first Chech Cohomology group $\check{H}^1(X, G)$ and isomorphism classes of $G$-Principal Bundles. To see that, given a $G$-cocycle $\{{g_{\alpha\beta}}\}$ over an open cover $\{U_{\alpha}\}$ of $X$, one can form the space $P = \bigcup_{\alpha}(\{\alpha\}\times U_\alpha\times G)$ and quotient it by $(\alpha, x, g) \sim (\beta, y, h) \Leftrightarrow (x = y) \wedge (h = g_{\beta\alpha}(x)\cdot g)$.

That being said, my question is do we have an equivalence of categories (groupoids here) $$C \simeq G Bund(X)$$ for a category $C$ that is described in term of $G$-cocycles. I know that there is such an equivalence given a classifying space $BG$ and in the principal bundle article of the ncatlab they are talking about an equivalence:

$$\mathbf{H}(X, \mathbf{B}G) \stackrel{\simeq}{\to} G Bund(X)$$.

But can we state that with a category $C$ whose objects are $G$-cocycles or cohomologous classes $\omega\in \check{H}^1(X, G)$?

Also, I would be glad if someone explain the equivalence found in the ncatlab article but concretely in our case (not in the abstract context of ncatlab). I don't figure out wether this abstract construction is more related to cocycles or to classifying spaces.

Thanks, Paul.

Solution 1:

I've found my answer. Let $\check{Z}^1(X, G)$ be the following category:

Objects: $(\{U_\alpha\}, \{\Phi_{\alpha\beta}\})$ where $\{U_\alpha\}$ is an open cover of $X$ and $\Phi_{\alpha\beta}: U_\alpha\cap U_\beta\longrightarrow G$ are smooth functions such that: $$\forall\alpha,\beta,\gamma: \Phi_{\alpha\beta}\cdot \Phi_{\beta\gamma} = \Phi_{\alpha\gamma}$$ So objects are cocycles.

Morphisms: $(\{U_\alpha\}, \{\Phi_{\alpha\beta}\})\stackrel{t}\longrightarrow (\{V_i\}, \{\varphi_{ij}\})$ is a collection $t = \{t_{i\alpha}\}_{i\alpha}$ of smooth functions $t_{i\alpha}: U_\alpha\cap V_i\longrightarrow G$ such that: $$\forall \alpha, \beta, i, j: \varphi_{ji}\cdot t_{i\alpha}\cdot\Phi_{\alpha\beta} = t_{j\beta}$$ Composition: Let $(\{U_\alpha\}, \{\Phi_{\alpha\beta}\})\stackrel{t}\longrightarrow (\{V_i\}, \{\varphi_{ij}\})$ and $(\{V_i\}, \{\varphi_{ij}\})\stackrel{u}\longrightarrow (\{W_a\}, \{\theta_{ab}\})$ be two morphisms. Then $(\{U_\alpha\}, \{\Phi_{\alpha\beta}\})\stackrel{v}\longrightarrow (\{W_a\}, \{\theta_{ab}\})$ is defined by $v_{a\alpha} = u_{ai}\cdot t_{i\alpha}$ on $U_\alpha\cap W_a\cap V_i$. Note that it is well defined because $\forall i, j: u_{ai}\cdot t_{i\alpha} = u_{aj}\cdot \varphi_{ji}\cdot t_{i\alpha}= u_{aj}\cdot t_{j\alpha}$

So now define the functor $K: \check{Z}^1(X, G)\longrightarrow GBund(X)$ as follow. For $\{\Phi_{\alpha\beta}\} \in \check{Z}^1(X, G)$, let $P = \bigcup_{\alpha}(\{\alpha\}\times U_\alpha\times G)/\sim$ as explained in my question. Then $K(\{\Phi_{\alpha\beta}\})$ is the $G$-principal bundle $\pi:P\longrightarrow X$. Given an other cocyle $\{\varphi_{ij}\}$ and a map $\{\Phi_{\alpha\beta}\}\stackrel{f}{\longrightarrow}\{\varphi_{ij}\}$ one can check that there is an unique map $K(\{\Phi_{\alpha\beta}\})\stackrel{K(f)}{\longrightarrow}K(\{\varphi_{ij}\})$ such that $\varphi_i\circ K(f)\circ \Phi^{-1}_\alpha = f_{i\alpha}$. Hence $K$ is full and faithful. Also, for all $P\in GBund(X)$ there exists $\{\Phi_{\alpha\beta}\}$ such that $P \simeq K(\{\Phi_{\alpha\beta}\})$. Note that here you need to choose a $G$-atlas $\{\Phi_\alpha\}$ for $P$ and take $\{\Phi_{\alpha\beta}\} = \{\Phi_\alpha\circ\Phi^{-1}_\beta\}$. So $K$ is dense.

Because $K$ is full, faithfull and dense, there exists a quasi-inverse for $K$ and so we have proved $$GBund(X)\simeq\check{Z}^1(X, G)$$ Note that the quasi-inverse depend of the choice of a $G$-atlas for each $G$-Principal Bundle. Note also that taking the connected component of the categories we have: $$GBund(X)_0\simeq\check{H}^1(X, G)$$ Finally, this construction works also for $(G, \lambda)$-Bundle where $\lambda: G\times F\longrightarrow F$ is a faithfull action of $G$ on a typical fiber $F$.

I haven't wrote all the details so I hope it's clear enough.