Let $ G $ be a discrete group and $ X $ paracompact topological space. The classifying space $ BG $ classifies isomorphy classes principal bundles over $ X $ via

$$ [ X, BG] \cong PrinG(X)/Isom , [f] \to f^* \pi $$ where the brackets on the left denote the homotopy classes of maps $ X \to BG $ and $ \pi: EG \to BG $ the universal bundle. Since the classifying classifying space is only unique up to homotopy there are several ways to construct it. Arguebly in case of discrete group $ G $ there is a kind of canonical realization of $ BG $ as a $ \triangle $-complex whose $n$-simplices are given by $[g_0,g_1,...,g_n]$ glued together in the obvious way. (we are using here the notion of Delta complex in Hatcher's sense)

More precisely the universal bundle $EG$ carries structure of a $\triangle$-complex whose $n$-simplices are the ordered $(n + 1)$ tuples $[g_0, ... ,g_n]$ of elements of $G$. As quotient space the classifing space $BG=EG/G$ inherits structure $\triangle$-complex where a $n$-simplex $BG$ can be written uniquely in the form $[g_1 \vert g_2\vert ... \vert g_n]:= G \cdot [e_G, g_1, g_1g_2,..., g_1,.., g_n]$. For details see p 89 in Hatcher's book on algebraic topology.

Assume now $ X $ be a $ \triangle$-complex and $ F \to X $ be a principal $G$-bundle.

Question: Is it possible to construct combinatorically in explicit terms from this $G$-bundle $ F $ a simplicial map of $ \triangle $-complexes $X \to BG$ (i.e. by assigning to which $ n $-simplices of $ BG $ are mapped $n$-simplices of $ X$) which gives a kind of " natural" representantative of the homotopy class of $ X \to BG $ which corresponds via the correspondence above to isomorphism class of $F \to X$? In other words the question is how to construct combinatorically the map $ PrinG(X)/Isom \to [ X, BG] $ in other direction in case $ X $ combinatorically 'nice' ?

A short remark on the reason for dealing with $\triangle$-complexes intead of combinatorically more simpler simplicial complexes: Seemingly even though $ EG$ in Hatchers book is realised as simplicial complex, the classifying space $ BG $ in the way Hatcher constructed it as quotient of $ EG $carries only structure of a $\triangle$-complex because it involves boundary identifications which are not allowed for simplicial complexes (Here my reference is Greg Friedman's An Elementary illustrated Introduction to Simplicial Sets). It might be possible - I dont knot - that $ BG $ might be also endowed with structure of a cimplicial complex, but this would then differ from Hatcher's. So doubt if it's possible here to consider simplicial complexes first as 'simplified' case.

**Here are a few loose ideas what I tried so far: **

One idea how to manage it in the case $X$ is one dimensional, ie consists of vertices and $1$-simplices.

I think we can assume that every loop contains at least three vertices, if not, subdivide any $1$-simplex beeing part of the loop into three new $1$-simplices:

$\bullet$-------$ \bullet$ subdivide into $\bullet$----$\bullet$----$\bullet$----$\bullet$

(why we do it? To avoid troubles with transition functions on the intersections of trivializing cover; see later)

Now, we choose a cover $\{U_i\}_{i \in I}$ of $X$, over which the principal $G$-bundle $E \to X$ trivialize, ie $E \vert _{U_i} \cong U_i \times G$.

Now we make several assumptions on this cover which look rather realizable if $X$ is dimension one, but possibly fail in bigger dimension (that's why I think this approach only works in dimension one):

We choose a cover $\{U_i\}_{i \in X_0} $ indexed by the set $X_0= \{v_1, v_2,..., \} $ of vertices of $X$ and require that every member $U_i$ saisfies following properties:

  1. $v_i \in U_i$ and $v_j \not \in U_i$ for every other vertex $v_j \neq v_i$

  2. $U_i$ and $U_j$ have a non trivial intersection - which in that case is isomorphic to an open interval $ \cong (0,1)$ - if and only if vertices $v_i$ and $v_j$ are adjacent, ie form the boundary of a $1$-simplex $[v_i, v_j]$; in this case the intersection $U_i \cap U_j \cong (0,1) $ is completely contained in thee inner of this $1$-simplex $[v_i, v_j]$

(Attention: That's where we use the assumption that every loop in $X$ contains at least three vertices. Otherwise two unwanted things could happen: one thing that a $1$-simplex having a single vertex as it's boundary could be contained completely in a $U_i$ (we don't want it) and secoundly two or more different $1$-simplices having as boundary the same two vertices $v_i$ and $v_j$ - in that case the intersection $U_i \cap U_j$ would be not an open interval, but a union of open intervalls; one for each $1$-simplex. That would cause troubles with the contruction of the map $X \to BG$ later.)

Recall, that in an simplicial complex a higher simplex is completly determined by it's vertices; in a $\triangle$-complex there could be different higher simplices having same set of vertices, eg $1$-complex
$v_1$---$w$---$v_2$ after identifying vertices $v_1$ and $v_2$. And such behavior would cause problems when we try to to associate to a $1$-simplex of $X$ a $1$-simplex in $BG$.

Obviously every $U_i$ is contractible, since by construction it's isomorphic to a bouquet of intervalls $[1, 0)$ (one for each $1$-simplex having $v_i$ as one of it's boundary points), after identifying the left $1$'s with the vertex point $v_i$.

Now we construct the map $f:X \to BG$: the $\triangle$-complex of $BG$ in Hatcher's book has only one $vertex$ $[*]$, so we map every vertex of $X$ to it. What to do with the $1$-simplices of $X$?

Let $[v_i,v_j]$ a $1$-simplex with boundary points $v_i, v_j$. Then there exist a unique non trivial intersection of the covering pieces $U_i$, which is contained in $[v_i,v_j]$, obviously by construction $U_i \cap U_j \cong (0,1)$.

The transition function $g_{ij}: U_i \cap U_j \to G$ satisfying the patching isomorphism $U_i \cap U_j \times G \to U_i \cap U_j \times G, (u, g) \mapsto (u, g_{ij}(u) \cdot g)$ is determined up to homotopy and since $U_i \cap U_j$ contractible, $g_{ij}$ can be identified with an element in $G$. So we map the $1$-simplex $[v_i,v_j]$ to the $1$-simplex $[g_{ij}]$ of $BG$.


-does this construction give the $G$-bundle $E$ back? ie $E = f^*EG$? (Here we have to check that the transition function $g_{ij}: U_i \cap U_j \to G$ restricted to $U_i \cap U_j \cap f^{-1}(V_k \cap V_m) $ is given by $g_{ij}(u)= h_{km}(f(u)) $ where $V_k, V_m \subset BG$ are arbitrary two open subsets of $BG$ over which the universal bundle $\pi: EG \to BG$ trivializes, ie $EG \vert _{V_k}$ and $h_{km}: V_k \cap V_m \to G$ is the associated transition function. I not know how to check $g_{ij}(u)= h_{km}(f(u)) $. What is the transition function of $EG \to BG$ and is there a canonical choice of trivialiving cover $\{V_k\}_{k \in K}$ to $\pi: EG \to BG$?

Can this construction generalized to higher dimensional $X$? Observe, that for $X$ of dimension $1$ I have choosen a very special trivializing cover. I'm not sure if there is always possible to choose a cover satisfying the same conditions for higher dimensional $X$, eg a octahedron.

If the space $X$ becomes even more complicated (eg of dimension bigger that $1$ and it contain loops which are boundaries of higher simplices, my construction above seems to be irreparable.

Another naive idea was to discard all simplices of dimension $\le 2$ for the moment and perform the construction from above on the $1$-skeleton of $X$, but I doubt if it's possible to extend it to the map $X \to BG$ naturally, since in case of $\triangle$-complexes the higher dimensional simplces are almost never determined by vertices & $1$-simplices like that's the case for simplicial complexes.

  • $\begingroup$ @MarianoSuárez-Álvarez: I'm not completly sure if that's the best way to do it but lets try: we can regard a circle as as a triangle with removed inner. so it consists of three vertices $ v_i$ and three 1-simplices $(v_i, v_j)$. We assume that covered by three intervalls $ I_i $ over which the G-bundle trivializes and each of these intervalls contains exactly one vertex $ v_i$. The intersections of each two of these intervalls are intervalls too and are completely contained in a unique 1-simplex, namely $I_j \cap I_j \subset (v_i, v_j)$. $\endgroup$
    – user267839
    Dec 29, 2022 at 0:19
  • $\begingroup$ Each intersection of two such intervalls also determines a transition function, ie an element $ g_{ij} \in G$. Then we map the 1-simplex $(v_i, v_j)$ to $ [1 \vert g_{ij}]$, a 1-simplex of BG. Is the idea ok? $\endgroup$
    – user267839
    Dec 29, 2022 at 0:24
  • $\begingroup$ If that works, then this answers the case where X is 1-dimensional. I have no clue how to extend it to higher dimensional $\triangle$-complexes $\endgroup$
    – user267839
    Dec 29, 2022 at 0:27
  • $\begingroup$ If we try to mimic to same approach like for the circle, seemingly the resulting maps $ X \to BG$ would be completely determined only by what is going on 1-skeleta, and that sounds strange... $\endgroup$
    – user267839
    Dec 29, 2022 at 0:57
  • $\begingroup$ @MarianoSuárez-Álvarez: above I added a more precise idea how that could work for $X$ one dimensional $\triangle$-complex. hope it's not explained too confusingly. One point in your remark I not completely understand: you wrote that the bundle is trivial over $1$-simplices. If $X$ would be a simplicial complex, then I aggree with you, there every 1-simplex has two different vertices as boundaries, so it's homotopic to a line and therefore contractible. But in a $\triangle$-complex there could be some weird identifications performed on the boundary inside $X$, $\endgroup$
    – user267839
    Dec 29, 2022 at 23:17

1 Answer 1


It's been a long time since I studied these things, and principal bundles always baffled me (while vector bundles made more sense somehow), so maybe the following is nonsense. But I thought I'd give it a shot:

You might want to think about, say, circle bundles over the 2-sphere, so $X$ is $S^2$. There are $\mathbb Z$ of these, with the integer corresponding to the degree of the clutching-function on the equator.

If you represent $S^2$ as an octahedron, then you have to represent those uncountably many things by a map from a few triangles to BG, which you've represented as a finite simplicial complex. There are only finitely many such maps.

I feel pretty sure that you need to include subdivision as a possible step in building your explicit combinatorial map.

  • $\begingroup$ the octahedron example is great, I feel that it provides the right intuition what is going on there. Say we have a G-bundle and there is a covering $ \{ U_i \}_i$ of of the octahedron X such that over each $ U_i $ the bundle trivializes with additional mild assumption that each $ U_i $ contains exactly one vertex $ v_i $ of the octahedron. how can we we associate to the bundle canonically a map to BG? Naively I would ignore the 2-simplices and try the same idea as in the comments above using transitions functions along the intersection of two $ U_i $ but I'm not sure if it works here. $\endgroup$
    – user267839
    Dec 29, 2022 at 0:42
  • $\begingroup$ It should additionally be required that every intersection $ U_i \cap U_j $ which determine a transition function $ g_{ij} $ can be related to a unique 1-simplex of the octahedron in order to map this 1-simplex to the 1-simplex $ [1 \vert g_{ij}] $ of BG, but I think that for a 2-dimensional complex this might not always work like in case of a 1-dimensional complex, eg a circle $\endgroup$
    – user267839
    Dec 29, 2022 at 0:48
  • $\begingroup$ Or do you see another natural way how to associate a map $ X \to BG $ to a given principal G-bundle in the case X octahedron? $\endgroup$
    – user267839
    Dec 29, 2022 at 0:52
  • $\begingroup$ by the way: as far as I understand the motivation behind performing an extra subdivision correctly, the aim is that the bundle trivializes over every simplex of maximal dimension, right? $\endgroup$
    – user267839
    Dec 29, 2022 at 1:00
  • $\begingroup$ To your last question... yes. But if you want to wrap a circle (the equator of the octahedron) multiple times around some circle in BG , you need at least 3 edges for every "wrap", so having just four edges won't cut it. Hence the need for subdivision. To be honest, I can't quite make sense of your other 3 comments because I haven't had my first shot of caffeine, and I'm not going to be able to spend any more time on this for at least a week. But I'm glad it might have pushed you in the right direction. $\endgroup$ Dec 29, 2022 at 12:24

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