# Does the holonomy map define a homomorphism $\pi_k(X)\to\pi_{k-1}(Hol(\nabla))$?

Consider a vector bundle $V\to E\to X$ with fiber $V$, with structure group $G$, and $X$ path-connected. Consider a connection $\nabla$ on $E$. Then for any loop $L$ in $X$, based at $p$, we have a mapping: $$hol: L\mapsto hol(L)\in Aut(V),$$

the holonomy map, which gives us the (linear) transformation of vectors after parallel transport around the loop $L$. Its image $Hol(\nabla)$ (dropping the reference to the base point) is a subgroup of $Aut(V)$.

If $X$ is simply connected, then it is known that $Hol(\nabla)$ is a path-connected Lie subgroup of $Aut(V)$. (See for example "Riemannian Holonomy Groups and Calibrated Geometry" by D. Joyce, chapter 2.)

This can be seen in the following way: a continuous homotopy $L_s, s\in[0,1]$ between two loops $L_0$ and $L_1$ defines a continuous curve in $Hol(\nabla)$, given by: $$\gamma:s\mapsto \gamma_s= hol(L_s).$$

Therefore, homotopic loops have path-connected images in $Hol(\nabla)$.

My question is: how further can we go?

Between points in $Hol(\nabla)$, having a continuous curve that joins them is an equivalence relation. If we denote by $\pi_0(Hol(\nabla))$ the equivalence classes of such relation, we have a well defined map $hol^1:\pi_1(X)\to \pi_0(Hol(\nabla))$. In fact, if two loops are homotopic, their images lie in the same class.

$\pi_0$ has no group structure defined (would it be possible to define it?), so our map $hol^1$ is not a morphism of groups.

Now, a 2-dimensional homotopy $S^2\to X$ can be thought of a homotopy $L_s$ of loops, like before, but where both $L_1$ and $L_0$ are trivial (while for generic $s$, $L_s$ may not be trivial). The curve $s\mapsto hol(L_s)$ now defines a closed loop in $Hol(\nabla)$, based at the identity. A homotopy between two 2-dimensional homotopies in $X$ defines a homotopy of loops in $Hol(\nabla)$. In particular, we can define a mapping $hol^2:\pi_2(X)\to \pi_1(Hol(\nabla))$. It is well defined, because as we saw $hol$ preserves homotopy of different orders.

So in general, I would expect that $hol$ defines mappings: $$hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla)).$$

Moreover, for $k>0$, $\pi_k$ are groups, and since composing homotopies of order $k$ in $X$ corresponds to composing homotopies of order $k-1$ in $Hol(\nabla)$ (and the image of trivial homotopies are trivial), the maps $hol^k$ are morphisms at least for $k>1$. So I would say:

The holonomy map defines morphisms $hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla))$.

This would imply:

A bundle $V\to E\to X$ with holonomy $Hol(\nabla)$ can exist only if every $\pi_k(Hol(\nabla))$ admits as a subgroup a homomorphic image (i.e. a quotient) of $\pi_{k+1}(M)$.

Is this reasoning correct? Is it helpful? Has it been done anywhere?

EDIT: For $TS^2$, the tangent bundle to the 2-sphere, and the Levi-Civita connection, the Gauss-Bonnet theorem can be stated in this (beautiful!) way: since $Hol(\nabla)=SO(2)\cong S^1$,

The morphism $hol^2: \pi_2(S^2) \to \pi_1(S^1)$ induces an isomorphism $\Bbb{Z}\to 2\Bbb{Z}$.

Does this result generalize? Thank you!

• I think the implication provides no real restriction -- at least in the way it is stated, since the quotient of $\pi_{k+1}(M)$ contained in $\pi_k(Hol(\nabla))$ might be the trivial group. – Ulrich Pennig Mar 20 '14 at 9:34
• I think what you are saying is that the holonomy defines a continuous map of the loop space $\Omega X$ onto $Hol(\nabla)$, so there are induced homomorphisms $\pi _k(\Omega X)=\pi _{k+1}(X)\rightarrow \pi _k(Hol(\nabla))$. As Ulrich points out, I am afraid that doesn't give you any restriction on the possible holonomy groups. – abx Mar 20 '14 at 9:47
• @Ulrich Right. Except in some interesting cases, like the sphere (see the edit). Is the rest of the reasoning true? Where has it been done? – geodude Mar 20 '14 at 10:06
• @abx: nice! Another remark: the map from $\Omega X$ to $Hol(\nabla)$ is a "group homomorphism"; usually, this imposes conditions on the possible maps. – Konrad Waldorf Mar 20 '14 at 11:46

I can try to answer to myself, using the very helpful comments you wrote. (You are all still free to write an answer yourselves!)

Notice: For convenience, here I will denote by $hol^k$ what in the question I denoted by $hol^{k+1}$. So we look for group morphisms $hol^k:\pi_{k+1}(X)\to \pi_k(Hol(\nabla))$.

First of all, there is a very natural answer to:

Is it possible to define a group structure on $\pi_0(Hol(\nabla))$?

Yes. Since $Hol_0(\nabla)$ is a normal subgroup of $Hol(\nabla)$, we simply define: $$\pi_0(Hol(\nabla)) := \frac{Hol(\nabla)}{Hol_0(\nabla)}.$$

Holonomies of loops of the same homotopy class differ by an element of $Hol_0(\nabla)$. This is easy to see, if loops $L,L'$ are homotopic, take their composition with reversed $L$: $L'L^{-1}$ is homotopic to the trivial loop. Therefore $hol^0$ maps homotopic loops into the same element of $\pi_0(Hol(\nabla))$.

It is naturally a morphism, because if $L,L'$ are loops: $$hol(LL') = hol(L)hol(L'),$$

where the composition on the left member is loop concatenation, and the composition in the right member is group multiplication on $Hol(\nabla)$. The quotients $\pi_1(X)$ and $\pi_0(Hol(\nabla))$ inherit the multiplication law, which for $\pi_1(X)$ becomes a group law, and therefore we have a morphism. (Of course, trivial loops are mapped into the identity.)

Does $hol$ define morphisms $\pi_{k+1}(X)\to\pi_k(hol(\nabla))$?

Yes. As abx pointed out, $hol:\Omega X\to Hol(\nabla)$ defines maps: $$hol^k:\pi_k(\Omega X)\to \pi_k(Hol(\nabla)),$$

where $\Omega X$ is the loop space of $X$, and $\pi_k(\Omega X) = \pi_{k+1}(X)$.

Does this impose restrictions on the bundles and connections?

Hardly, as was pointed out by abx, Ulrich Pennig, and Konrad Waldorf, because the image of $hol^k$ can be trivial.

This, nevertheless, and the remark on $TS^2$, raises the following question:

What happens when $Im(hol^k)$ is not trivial? What properties does it have?

Or, following the example of $TS^2$:

Are characteristic classes linked to the "size" of $Im(hol^k)$?

When I gather enough details of the problem, I will ask a new question, and post the link here.

EDIT: I asked the question, here it is. EDIT 2: It doesn't generalize in the way I (wrongly) thought. I'll ask more when I find something actually interesting.