# What does “higher monodromy” tell us about a principal bundle

Let $P \to X$ be a principal $G-$bundle and let $f: X \to BG$ be its classifying map. As I understand there's some way to associate a monodromy representation $\pi_1(X) \to G$ to it. I know how to construct such a presentation out of a principal connection on $P$ by taking an appropriate quotient of the holonomy representation but I failed in the past weeks to find a source that defines a purely topological construction of this representation.

Question 1: Where can I find the definition and some elementary topological treatment of the monodromy representation associated to a principal bundle?

Extending the classifying map $f$ to a fiber sequence gives:

$$\cdots\to \Omega G \to \Omega E_f \to \Omega X\to G \to E_f \to X \to BG$$

Where $E_f=\{(x,\gamma)\in X \times Y^I:\gamma(1)=f(x)\}$.

Question 2: How is the map $\Omega X \to G$ above related to the usual notion of monodromy represnetation (which i'm unfamiliar with) of the the principal bundle $P\to X?$ How does it relate to the holonomy representation of a given connection?

My naive attempt continues by taking the fundamental group of the sequence above to get the following long exact sequence of groups:

$$\cdots \to \pi_2(G) \to \pi_2(E_f) \to \pi_2(X)\to \pi_1(G) \to \pi_1(E_f) \to \pi_1(X) \to \pi_0(G)$$

Since the procedure was very natural I imagine the above sequence must yield some important information about the principal bundle at hand. For example ff $X$ is weakly contractible the above gives another motivation for why every fiber bundle on $X$ is trivial.

Question 3: To what extent does the above "higher monodromy sequence" determine the principal bundle $P \to X$?

• Any model of $BG$ satisfies $\pi_1BG=\pi_0 G$. Also, I don't thing you're going to get much more than a homomorphism $\pi_1X\rightarrow\pi_0 G$ out of the bundle. – Fernando Muro Dec 1 '15 at 11:20
• @FernandoMuro Why do you think that? Can you point me to a resource where the monodromy representation of a bundle is constructed? – Saal Hardali Dec 1 '15 at 11:22
• Because the bundle is just determined by a map $X\rightarrow BG$. I don't know of any reference, I'm kind of far from this stuff. Maybe I'm totally wrong. – Fernando Muro Dec 1 '15 at 11:31

There is a monodromy of sorts for any topological bundle (or even fibration) $\pi \colon E \to X$ with fiber $F$. Let $\gamma \colon [0,1] \to X$ be a path and set $F = \pi^{-1}(\gamma(0))$. The homotopy lifting property applied to the square $$\begin{array}{ccc} F & \hookrightarrow & E \\ \downarrow & & \downarrow \\ F \times [0,1] & \xrightarrow{ \gamma \circ \pi_2 } & X \end{array}$$ yields a map $h \colon F \times [0,1] \to E$ with $h(x,0)=x$ and covering the homotopy $\gamma \circ \pi_2$. If $\gamma$ is a loop, then you can regard the map $m \colon F \to F$ defined by $x \mapsto h(x,1)$ as a monodromy map. The map $m$ is not well defined as it depends on a choice of lift, but any two choices of lift are fiber homotopic (again by the homotopy lifting property). You can do the procedure above for all loops at the same time, by applying homotopy lifting to the square $$\begin{array}{ccc} F \times \Omega X & \xrightarrow{i} & E \\ \downarrow & & \downarrow \\ F \times \Omega X \times [0,1] & \xrightarrow{ j } & X \end{array}$$ with $i(x,\gamma)=x$ and $j(x,\gamma,t)=\gamma(t)$ to get a "monodromy representation" $$m \colon \Omega X \times F \to F$$ Again this is not unique, but it is "homotopy unique". More precisely $m$ can be extended to an $A_\infty$-action of $\Omega X$ on $F$ (this means an action up to homotopy, together with the data of the homotopy and in fact higher homotopies) which is essentially equivalent to the data of the fibration $E \to X$. In the case of a principal bundle you get an $A_\infty$ map $m \colon \Omega X \to G$ (which is loops on the classifying map $X \to BG$).

Your "higher monodromy" is just the effect of the map $m \colon\Omega X \to G$ (the connecting map in the fiber sequence) on homotopy groups and as such it is a pale reflection of the map itself (although it does contain some information about it). In general it is pretty far from determining the bundle.

To address your second question, any choice of connection (even of Ehreshman connection on a not necessarily principal fiber bundle) will give you a specific way to lift (smooth) paths and hence a specific choice of the monodromy map $m$ above. The topological story described above is, in a sense, what all these geometric monodromies have in common.

At the moment I can't really give you references for the claims above but I will try to get back to it when I can.

• whis was very helpful, thanks. One thing still bugs me though. Suppose I choose a connection for the bundle classified by $f$. This connection gives a holonomy representation $\Omega X \to Aut(P) \subset G$ into the gauge group of the bundle. What's the relation, if there's any at all, between this map and map $\Omega f : \Omega X \to G$? (I know the second map is unique only upto homotopy but stil...) – Saal Hardali Dec 1 '15 at 21:43
• I am a bit confused by your $Aut(P)$. I think the holonomy representation will just be a homomorphism $\phi \colon \Omega X \to G$. Then the answer to your question is that $\phi$ is homotopic to $\Omega f$ (in fact homotopic as an $A_\infty$ map). – Gustavo Granja Dec 2 '15 at 0:05
• This is exactly what I wanted thanks.The part with $Aut(P)$ was me incorrectly involving the global gauge group in the discussion. Thanks! Where would you recommend to read about all this homotopical algebraic perspective on fibre bundles? – Saal Hardali Dec 2 '15 at 0:10
• No problem. Regarding the homotopy lifting property, etc, I would recommend Husemoller - "Fibre bundles". Steenrod's book, "The Topology of Fibre bundles" is even better but it is a bit old fashioned. I am afraid I don't know a reference for the $A_\infty$ stuff. My first guess would be May's "Classifying spaces and fibrations" (available on his webpage) but a cursory search doesn't confirm my guess... I will try to look for a reference. – Gustavo Granja Dec 2 '15 at 0:28

You won't be able to construct a representation $\pi_1(X)\rightarrow G$ unless the bundle is flat i.e it is endowed with a connection whose curvature form vanishes identically. In this case the "monodromy" is just the holonomy of the connection. In the general case you have the so-called infinitesimal holonomy computed around the contractible loops who does not always vanishes for example if $P\rightarrow X$ is the bundle of frames associated to a differentiable metric whose curvature does not vanish.

• So there's no such thing as "monodromy" for a topological fiber bundle? (i.e. one that doesn't admit a smooth structure). – Saal Hardali Dec 1 '15 at 12:51
• What does "flat" mean if your bundle has no smooth structure? – Saal Hardali Dec 1 '15 at 12:53
• Thanks! What's the name of the "theory" that concerns itself with the kind of "topological monodromy" i'm talking about? Just so I could find my way in the literature – Saal Hardali Dec 1 '15 at 13:05
• @Saal: you can use the fact that a map $X \to BG_{\delta}$ is (fixing a basepoint of $X$ and assuming it's connected) the same thing as a map $\pi_1(X) \to G$, then appeal to the classification of flat vector bundles in the connection sense. The problem is that the flat connection associated to a given monodromy representation is not literally unique, so it's unclear how one would go about producing a canonical such connection. – Qiaochu Yuan Dec 1 '15 at 21:59
• I should say, more precisely, that either a free homotopy class of such maps is the same as a conjugacy class of maps $\pi_1(X) \to G$, or that a based homotopy class of such maps is the same as a map $\pi_1(X) \to G$ on the nose. – Qiaochu Yuan Dec 1 '15 at 22:07

Question 3: To what extent does the above "higher monodromy sequence" determine the principal bundle $P \to X$?

In general, not at all. There is no loss of generality in replacing $BG$ with an arbitrary connected space $Y$ (every such space is some $BG$, morally for $G = \Omega Y$), so your question is to what extent the homotopy class of a (pointed) map $f : X \to Y$ is determined by the long exact sequence in homotopy for the fiber sequence

$$F \to X \to Y$$

where $F$ is the homotopy fiber. What can happen in general is that $X$ and $Y$ have homotopy groups in degrees that are far apart from each other, so that this sequence tells you nothing about the homotopy class of $f$.

For example, if $X = BG$ for some discrete group $G$ and $Y = B^3 A$ for some discrete abelian group $A$, then all you learn is that $F$ is a space with $\pi_1(F) \cong G, \pi_2(F) \cong A$, and other homotopy groups trivial, and all of the maps in the long exact sequence are either isomorphisms or zero, regardless of $f$. But of course the group cohomology $H^3(BG, A)$ is interesting in general.

• I've just discovered (during reading an $n$-lab thread) that $EG$ is a fibrant replacement for $*$ in the pullback diagram of the principal bundle. So the homotopy fiber of the classifying map is the principal bundle. Am I understanding this correctly? – Saal Hardali Jan 2 '16 at 23:22
• @Saal: yes, that's right. – Qiaochu Yuan Jan 3 '16 at 5:51

You could look at

P.R. Heath, Groupoid operations and fibre homotopy equivalences, Math Z. 130 (1973) 207-233.

There is also a II, and a set of lecture notes by him on the area.