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 $Aut(V)$.

In this question (I apologize for the very same introduction) we established that $hol$ induces homomorphisms: $$ hol^k:\pi_{k+1}(X)\to \pi_{k}(Hol(\nabla)). $$

Therefore $\pi_{k}(Hol(\nabla))$ has to admit a subgroup which is isomorphic to a quotient of $\pi_{k+1}(X)$. As pointed out in the comments to the question, usually such image is trivial. But there is at least one case in which the image is not trivial, and it is linked to the Gauss-Bonnet theorem:

Consider $TS^2$, with the Levi-Civita connection. Then $Hol(\nabla)=SO(2)\cong S^1$.

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

To see this, consider a loop $L$ on $S^2$. The parallel transport around $L$ is $hol(L)\in SO(2)$, and it is given by: $$ hol(L)=\exp\left( \oint_L \omega \right) = \exp\left( \int_A R \right), $$

where $\omega $ is the connection form, $R$ is the Gaussian curvature, and both of them have values in the 1-dimensional Lie algebra $so(2)$ spanned by: $$ i\equiv \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right). $$

Since $SO(2)$ is Abelian, the formula above comes simply from Stokes's theorem.

For the unit sphere, $R=1$, so $hol(L)$ is a rotation by an angle $\alpha$ which is simply the area enclosed by the loop.

Consider now a continuous homotopy of loops $s\mapsto L_s, s\in [0,1]$ that generates $\pi_2(S^2)$, such that both $L_0$ and $L_1$ are trivial loops. Denote $\alpha_s$ the angles of rotation of $hol(L_s)$.

After suitable reparametrization, we can choose a homotopy $L_s$ for which $s$ is exactly the fraction of the area enclosed by the loop. In particular, for $s=1/2$, $\alpha_{1/2}=2\pi$, and for $s=1$, $\alpha_1=4\pi$.

From the point of view of $SO(2)$, $\alpha_s$ induces a double loop. So on $\pi_1(SO(2))$, $hol^2(1) = 2\in \Bbb{Z}$.

The factor 2 is not random, it is the Euler characteristic of the sphere. We know from the Gauss-Bonnet formula that what we obtained from the sphere generalizes, and that in particular for surfaces $A$: $$ \int_A R = 2\pi\chi(A). $$

For orientable Riemannian surfaces and $\nabla$ Levi-Civita, $Hol(\nabla)$ is always a subgroup of $SO(2)$, which has $\pi_1=\Bbb{Z}$. The subgroups of $\Bbb{Z}$ are in the form $n\Bbb{Z}$, for $n$ integer. So $Im(hol^2)=n\Bbb{Z}$ for some $n$, possibly $0$ in the trivial case. So:

For a general surface $A$, does it still hold that $Im(hol^2)=\chi(A)\Bbb{Z}$?

We have seen that for $S^2$ it holds. For the torus, the same result holds trivially, because $\pi_2$ is trivial and $\chi$ is zero.

Any counterexamples? Any proofs? Any generalizations for higher dimensions and orders?

Please also let me know if anything I wrote is not clear, or wrong.

Thank you.

  • $\begingroup$ Accidentally, I'm sorry for the downvoting, but when I hover over the down arrow, it shows "This question does not show any research effort...", and I truly believe that this is the case. $\endgroup$ Mar 27 '14 at 13:10

Considering your second display, you probably mean $\mathrm{hol}^1\colon\pi_2(A)\to\pi_1(S^1)$. In any case, for all closed surfaces $A$ except $S^2$ and $\mathbb{R}\mathrm{p}^2$, you have $\pi_k(A)=0$ for all $k\ge2$, so there's no hope. For proof, just consider the uniformization.

  • $\begingroup$ Oh right...should have thought about that! Thank you. $\endgroup$
    – geodude
    Mar 27 '14 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.