MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've come across a comment in a paper that hints at a relation between the set of (flat) connections for a principal bundle and the homotopy group of the base of the bundle. Can anyone tell me what the exact correspondence is here? Is the homotopy space some kind of moduli space for the flat connections?

share|cite|improve this question
To a flat principal connection you can assign its holonomy group. In general, I believe that the answer to your question is an explicit bijective correspondence between moduli space of flat connections modulo gauge equivalence and the set $\mathrm{Hom}(\pi_1(B),G)/G$, where $B$ is the base manifold, $G$ is the principal group and the action of $G$ on homomorphisms is via conjugation on their image. – Vít Tuček Mar 8 '11 at 19:30
Given a flat connection for $E\to B$ the parallel transport along two different curves starting and ending at the same points is the same. This is essentially due to Stokes' theorem and the flatness of the connection. Therefore, given a smooth loop $\gamma$ one can parallel transport along $\gamma$ to get an element of $G$. When you quotient out by gauge equivalence on one side, you quotient out by the conjugation action on the right side. – Somnath Basu Mar 8 '11 at 22:10
@Somnath Basu: You missed the word homotopic in your first sentence. If the two curves are not homotopic, parallel transport can detect it. – José Figueroa-O'Farrill Mar 8 '11 at 22:58
To clarify RoBoT's answer a tiny bit. Flat connections on a principal G-bundle P modulo gauge equivalence are in bijection (by taking the holonomy) with a union of some of the components of $\Hom(\pi_1(B),G)/conj$. To get all components you need to consider all (isomorphism classes of) principal G-bunldes over $B$. Also, with some care one can show this identification preserves more structure, i.e. is a homeo, or a (real or complex) analytic isomorphism, etc. – Paul Mar 8 '11 at 23:34
@Robot and pAuL: Why don't you guys formulate your answers as answers, as opposed to comments? – André Henriques Mar 9 '11 at 16:52
up vote 1 down vote accepted

The comments have addressed relating fundamental group and a flat connection. Something can be said about the moduli space of flat connections. Goldman, in, considers the moduli space of flat connections over a surface genus $>1$. He defines a symplectic form and relates is to the free homotopy classes of closed curves. Karshon, in, and Abbaspour and Zeinalian, in, discuss some generalizations.

Here is a brief description of Goldman's work. Let $S$ be a surface of genus $g>1$. The moduli space of flat connections over $S$ has singularities, but at manifold points the tangent space can be identified with the first cohomology of $S$ with coefficients in a flat bundle, $H^1(S; \text{flat bundle})$ (defined in the paper). A two form is defined by taking the cup product of two elements in $H^1(S; \text{flat bundle})$ and evaluating the product on $[S]$. This form is closed and non-degenerate- a symplectic form. Then the smooth functions on the moduli space of flat connections, denoted $C^\infty(\text{moduli space})$, with the Poisson bracket is a Lie algebra.

Let $\hat \pi$ be the conjugacy classes of elements in $\pi_1(S)$- i.e. the free homotopy classes of closed curves in $S$ and let $\mathbb{Z} \hat \pi$ be the free abelian group generated by the classes. The Goldman bracket is a Lie bracket on $\mathbb{Z} \hat \pi$ defined by choosing representatives, taking the free homotopy class represented by the composition of the curves at intersection points, and taking the sum of all such classes. This is well-defined and satisfies the Jacobi identity.

A closed curve in $S$ defines a function on the moduli space of flat connections defined by taking the trace of the holonomy of the curve with respect to a connection. Since the connections are flat, the function is independent of choice of representative in $\mathbb{Z} \hat \pi$. Goldman shows that this map $$\mathbb{Z} \hat \pi \rightarrow C^\infty(\text{moduli space}) $$ is a map of Lie algebras.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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