# construct map $X \to BG$ associated to principal $G$ bundle $F \to X$ explicitly combinatorically

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 $$BG$$.

Problems:

-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.

• @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 $I_j \cap I_j \subset (v_i, v_j)$. Dec 29, 2022 at 0:19
• 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? Dec 29, 2022 at 0:24
• 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 Dec 29, 2022 at 0:27
• 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... Dec 29, 2022 at 0:57
• @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 $X$, Dec 29, 2022 at 23:17

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.

• 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. Dec 29, 2022 at 0:42
• 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 Dec 29, 2022 at 0:48
• 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? Dec 29, 2022 at 0:52
• 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? Dec 29, 2022 at 1:00
• 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. Dec 29, 2022 at 12:24