Let $G$ be a (topological) group whose identity element $e_G$ is a nondegenerated basepoint (e.g. if $G$ is a Lie group). Then that's a known fact that there is for every 'nice' enough topological space $X$ (eg paracompact sp, or more elementary a CW complex) a natural bijection

$$ [X, BG] \cong \mathcal{P}_G(X) $$

between homotopy classes of cont maps $X \to BG$ and isomorphism classes of of principal $G$-bundles over base $X$. Explicitely this is given by associating to a homotopy class of a map $f: X \to BG$ the pullback bundle $f^*EG$ of the universal principal $G$-bundle $u:EG \to BG$.

Now like the trivial bundle the universal bundle $EG$ is the second 'canonical' (up to isomorphism) $G$-bundle and besides it's universal property I'm trying to understand if it's possible to deduce it's universal property only from study of it's $1$-cocycle represenation. Let's recall that we can over every nice enough base space $X$ encode the complete information about a $G$-bundle $E \to X$ with fiber $F$ (therefore $G \subset Aut(F)$; in our case later it will be $G$ itself since we are interested on principal bundles) in terms of a $1$-cocycle. Let $\mathcal{U}= \{U_i \}_i$ covering of $X$ over which $E$ trivializes, then

(1) a colection of homeomorphisms $\psi_i: U_i \times F \to p^{-1}(U_i)$ compatible with projection $p$

(2) the $1$-cycle (ie the set of transition functions) $g_{ij}: U_j \cap U_i \to G$ such that for every $(x, f) \in (U_j \cap U_i) \times F$ we have

$$ \psi_j^{-1} \circ \psi_i (x,f) = (x, (g_{ji}(x))(f))$$

(3) the transition functions satisfy $g_{ii} = e_G$ (constant function), $g_{ji}= g_{ij}^{-1}$ and $g_{kj} \cdot g_{ji}= g_{ki}$

Moreover two $G$-bundles $E_1, E_2$ over same base $X$ are isomorphic if there exist fine enough covering $\{U_i \}_i$ of $X$ such that there eist a family $g_i: U_i \to G$ with

$$ g^2_{ij} = g_i \cdot g^1_{ij} \cdot g_j^{-1} $$

where $\{g^1_{ij}\}$ and $\{g^2_{ij}\}$ are $1$-cycles of $E_1$ and $E_2$.

Question: Is it known which $1$-cocycle $\{g^{EG}_{ij}\}$ is naturally associated to the universal bundle $u:EG \to BG$? Is there any canonical choice of the covering $\{U_i \}_i$ of $BG$ known over which this $1$-cocycle can be in most transparent way be declared? Is it known at least if we work with a finite group? The model of $EG$ and $BG$ I have here in mind is that one by Milnor explaned in detail for example here.

If we work with this model for $EG \to BG$ can it's $1$-cocycle be writen down? It must be something canonical but I don't know how to determine it. My motivation is to understand if it's possible to understand only by study the structure of it's $1$-cocycle that this $G$-bundle is in certain way the most 'flexible' one in the sense such that every other can be derived up to iso by the pullback of it, as the universal property says. So speaking informally every 'twist' of $G$-bundle should be already 'known' in some way already to the universal bundle, such it 'knows' all possible twists can happen only by taking pullbacks. (Ok, I admit, that the last formulation is a bit vague, but that's the core of my motivation: I understand 'formally' the universal property of the universal bundle, but have no intuition on it's intrinsical geometry why and why exacly this geometry is the most 'flexible' to allow to recover all over $G$-bundles by taking pullbacks of it.


For any $x \in (0,1)$, I will define an open set $U_x$ of $BG$.

The way we can define it is by defining an open set $\tilde{U}_x$ of $EG$, stable under the action of $G$, with $|G|$ connected components that are permuted by the action of $G$. Then any one of these components will project to an open subset of $BG$.

The set $\tilde{U}_x$ in each simplex of $E_G$ by the condition that there exists some $i$ such that $\sum_{j< i} t_j < x$ and $\sum_{j \leq i } t_j > x$. Because these are strict inequalities, this is an open subset of each simplex. Because they are unaffected by inserting a new $i$ with $t_i=0$, this remains open in $EG$. Because the tuple $g_1,\dots, g_n$ is not used in the definition, it is certainly stable under the action of $G$.

The connected components associated to $g\in G$ consists of those points where $g_i=g$ for the unique $i$ satisfying $\sum_{j< i} t_j < x$ and $\sum_{j \leq i } t_j > x$. This is a component because the function $g_i$ is locally constant, because, on each simplex, the unique value of $i$ satisfying these inequalities is locally constant - it only changes at the points where $\sum_{j < i} t_i =x$ or $\sum_{j\leq i} t_i = x$, which are not in the set.

The open sets $U_x$ cover $BG$ (it suffices to take any infinite set of $x$) since at each point, such an $i$ exists for all but finitely many $x$.

The cocycle is defined as follows: Given a point in $BG$, expressed as a tuple $t_1,\dots, t_n$ of nonnegative reals summing to one and a tuple $g_1,\dots, g_n \in G$ modulo the left action of $G$, that lies in $U_x \cap U_y$, let $i_x$ be the unique $i$ such that $\sum_{j < i} t_i <x$ and $\sum_{j\leq i} t_i > x$, and define $i_y$ similarly, and then set $$\psi_{xy} ( ((t_1,\dots, t_n), (g_1,\dots, g_n)) = g_{i_x}^{-1} g_{i_y} .$$

This cocycle is flexible in the sense that for any distinct tuple $x_1,\dots, x_n$ of open sets $U_{x_1},\dots, U_{x_n}$, for any tuple of $\psi_{ij} \in G$ for $1\leq i,j\leq n$ such that $\psi_{ik} = \psi_{ij} \psi_{jk}$ for all $i,j,k$, there is a unique component of $U_{x_1} \cap \dots \cap U_{x_n}$ such that $\psi_{x_i x_j} $ restricted to this component is $\psi_{ij}$.

Indeed, such a tuple must have the form $\psi_{ij} = c_i^{-1} c_j$ for some tuple $c_i \in G$ for $i$ from $1$ to $n$, and then the component where $g_{i_{x_j}} = c_j$ for all $j$ does the trick.

But possibly this is not actually what you want to do. Probably you want to use a different cocycle presentation of $G$-bundles, the one where you fix a simplicial complex structure on the base, and present a cocycle as a tuple consisting of one element of $G$ for each edge satisfying one relation for each triangle.

For this one, the key point is that given a simplicial complex with such a cocycle, you can map it to BG by mapping each edge to the corresponding edge of $BG$, and then there will be a unique way of extending to triangles and then to higher simplices.

  • $\begingroup$ Thank you for your answer. One detail still confuses me. You define the transition function $\psi_{xy}: U_x \cap U_y \to G$ as $\psi_{xy} ( ((t_1,\dots, t_n), (g_1,\dots, g_n)) = g_{i_x}^{-1} g_{i_y}$. But isn't this family of transition functions defined in that way conjugated via coboundary family' $\{ \psi_{U_x}\}$ with $\psi_{U_x}(((t_1,\dots, t_n), (g_1,\dots, g_n)):= g_{i_x}$ to the transtion family of constant functions $\{g_{UV}(x)= e_G \}$ which represent the trivial $G$-bundle $BG \times G$? But EG isnt trivial. Or do I confuse some impotant aspect in your construction ? $\endgroup$ Apr 8 at 23:25
  • $\begingroup$ @IsaktheXI It's not my construction - it's the construction of $BG$. The $g_1,\dots, g_n$ are only well-defined up to the action of $G$, so $g_{i_x}$ is not a well-defined function. It is on $EG$, which proves that this bundle is trivial on $EG$. $\endgroup$
    – Will Sawin
    Apr 8 at 23:40
  • $\begingroup$ On your last 'But possibly...'. I think that this last paragraph in your answer hits exactly my motivation behind study of these cycle representations. Do I understand you correctly that if we reduce our considerations to spaces which are simplicial complexes then all $G$-bundles over it can be always uniquely encoded in these say 'discrete' $1$-cocycle versions associating to every edge ${ij}$ (satisfying the relation you mentioned) between two vertices $a_i$ and $a_j$ a $g_{ij} \in G$? $\endgroup$ Apr 9 at 23:50
  • $\begingroup$ Intuitively every vertex would then represent a disc (=local trivialization) around it over which the bundle trivializes and every edge (+relation) the intersection of such two discs? $\endgroup$ Apr 9 at 23:50
  • 1
    $\begingroup$ @IsaktheXI Such $i$ are unique since if $i,i'$ both satisfy these inequalities with (wlog) $i<i'$ then $x< \sum_{j \leq i} t_j \leq \sum_{j <i'} t_j < x$, a contradiction. $\endgroup$
    – Will Sawin
    Aug 8 at 13:44

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