To edify my understanding of fiber bundles with structure groups, I was currently trying to reconcile two classifications (in a particular case). For simplicity, I'm taking the base to be $S^1$ and the group $G$ to be discrete. Initially, I was getting a different answer from each. Update: The mistake there has been pointed out to me, but I'm still trying to resolve some cofusion.

We can classify principal $G$-bundles as homotopy classes of maps $S^1\rightarrow BG$. In other words, we want to take free homotopy classes of loops in $BG$, which is the same thing as conjugacy classes in $\pi_1(BG,\text{pt})$. Using $\Omega BG\simeq G$, we get $\pi_1(BG,\text{pt})\cong G$ and thus principal $G$-bundles over $S^1$ correspond to conjugacy classes in $G$.

(Initially, I forgot to take the conjugacy action when looking at homotopy classes of maps $S^1\rightarrow BG$, so I falsely assumed that principal bundles corresponded bijectively with elements of $G$. This had the same effect as working in a category of pointed spaces, similarly to how the classification of connected covering spaces involves subgroups of $\pi_1$ or their conjugacy classes, depending on whether or not our spaces are pointed.)

I see two ways of being more explicit about this correspondence, i.e. not going through the classifying space. The first is just a helpful thought process. The second is where my remaining confusion enters.

  1. We can look at transition functions. If we consider a fiber bundle over $S^1$ with fiber a $G$-torsor, it will be trivializable over the upper and lower semicircles and therefore given by a transition map on their intersection, which is just two elements $g_L,g_R\in G$. Two different pairs $(g_L,g_R)$ and $(h_L,h_R)$ define the same bundle if and only if they differ by some change of trivialization on the upper and lower semicircles, i.e. there exist $r_U,r_D\in G$ such that $$h_L=r_U^{-1}g_Lr_D\quad\text{and}\quad h_R=r_U^{-1}g_Rr_D.$$ This allows us to assume that every bundle is defined by a pair $(e,g)$, where $e\in G$ is the identity. But now two bundles defined by $(e,g)$ and $(e,h)$ will be isomorphic if and only if $h=r^{-1}gr$ for some $r\in G$ (we must have $r_U=r_D$ to keep $e$ in the first coordinate). This implies that the isomorphism classes of bundles correspond to conjugacy classes in $G$.
  2. In both formulations, the bundle corresponding to some $g\in G$ is just the mapping torus of left-multiplication by $G$. Explicitly, we take $G\times [0,1]$ and identify $(h,1)$ with $(gh,0)$ for any $h\in G$. Then the map $(x,t)\mapsto e^{2\pi it}$ clearly descends to the quotient and gives a covering of the circle. The $G$-action on the principal bundle is now just multiplication in the $G$-coordinate.

What still confuses me is the following: it seems like we can recover an explicit element $g\in G$ from the principal $G$-bundle, by taking the monodromy on a fiber. Explicitly, let $\pi:E\rightarrow S^1$ be a principal $G$-bundle, pick some $x\in \pi^{-1}(1)$ and let $\gamma:[0,1]\rightarrow E$ be the unique path with $\gamma(0)=x$ and $\pi\circ\gamma(t)=e^{2\pi i t}$. Then we get $\gamma(1)=gx$ for some $g\in G$ and this bundle is the mapping torus of left-multiplication by $G$. Initially, I thought that perhaps picking a different $x$ would lead to a conjugation of $g$, but in trying it out, I am getting the same $g$ before and after. (Note: this was wrong.)

In particular, I tried constructing the bundle corresponding to $(12)\in S_3$. As a covering space, this is just three copies of the connected double cover of $S^1$. But once we include the $G$-action, I get a bit confused. For now acting by $(12)$ corresponds to monodromy and therefore preserves each connected component of the total space. But acting by $(23)$ interchanges connected components! This makes it seem like the bundle cannot be isomorphic to one where the action of $(23)$ corresponds to monodromy. Did I compute this example wrong, or what is going on here? (Note: this was wrong.)

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    $\begingroup$ Unpointed homotopy classes of maps $S^1\to BG$ correspond to conjugacy classes of elements of $G$, not to elements. $\endgroup$ – Gregory Arone Feb 5 at 6:42

Okay, I figured out my mistake! I was trying to use left-actions everywhere, but we need some left and some right. This comes up when constructed associated bundles: if you have a left $G$-space $F$, then you put it together with a right principal $G$-bundle $E\rightarrow X$ to get the total space $E\times_G F$ of the associated bundle. The issue here isn't exactly that, since we're only dealing with principal bundles, but it's in a similar vein.

Essentially, I have to be more careful about my uses of left and right. This way, we can see both what goes wrong (i.e. left) and what goes right:

Wrong version: Let $E$ be the mapping torus of left-multiplication by $g\in G$. This is a quotient of $G\times [0,1]$ and it seems like we might be able to act on the left by $G$. But if we try, we'll get $$(gh,0)=(h,1)=h(e,1)=h(g,0)=(hg,0).$$ This implies that $gh=hg$, which won't always be the case when $G$ is non-abelian. (If $G$ is abelian, then there is no issue of elements versus conjugacy classes: they are the same thing.) So it might seem like monodromy corresponds to acting on the left by $g$, but there is no left action at all!

Right version: Again, let $E$ be the mapping torus of left-multiplication by $g\in G$. Since we are taking a quotient via left-multiplication, we still have a well-defined right action $(x,t)h=(xh,t).$ Fix some point $(x,0)$, we can look at the monodromy and find a unique $h\in G$ with $(xh,0)=(x,1)$. If we pick a different $(y,0)$ to start with, then there is a unique $r\in G$ with $y=xr$, so we get $$(yr^{-1}hr,0)=(xhr,0)=(xh,0)r=(x,1)r=(xr,1)=(y,1).$$ Therefore, this element $h\in G$ defined by monodromy is actually only defined up to conjugacy.

If we were instead working in the category of pointed spaces, then we would have a fixed choice of point in the fiber over $1\in S^1$ (this is equivalent to having a fixed trivialization over the basepoint). Then the monodromy of this point would indeed give a well-defined element of $G$. Passing back to the case without basepoints, we see that we lose some data about the fiber over the basepoint. This is the same as allowing some new isomorphisms, which correspond to change of basepoint in the fiber: these changes of basepoint in a $G$-torsor are exactly the same as inner automorphisms of $G$, which is another way to see the conjugacy coming into play.


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