Let $M$ be a closed hyperbolic $3$-manifold, and $e \in H^2(M)$ an integral cohomology class which is the first Chern class of a $Spin^c$ structure on $M$. Suppose there is a solution to the monopole equations on $M$ in this $Spin^c$ class with respect to the hyperbolic metric on $M$. Does it follow that $e$ is a monopole class? More generally, is it true that as a metric on a $3$-manifold evolves by (nonsingular) Ricci flow, pairs of solutions to the monopole equations can disappear, but new ones never appear?

(Added:) One reason to be interested is that if the answer is "yes", then one can deduce that a given class is a monopole class directly from geometry. One can then hope to use this information to show that certain families of classes (eg. on families of manifolds obtained by hyperbolic Dehn surgery on some fixed cusped manifold) are all monopole classes. More abstractly, if one has a "natural" PDE on a manifold which depends on the metric, then one should try to understand how/whether solutions to that PDE evolve under "natural" flows on the space of metrics. A similar (more geometric) question might be: are minimal surfaces destroyed by Ricci flow but never created? Eg. if a hyperbolic $3$-manifold contains an embedded minimal surface, is there an isotopic minimal surface in every other metric on the manifold? The reason to speculate about a connection between the monopole equations and Ricci flow is the key role of scalar curvature in both cases: via the Weitzenbock formula for the Dirac operator on the one hand; and via Hamilton's monotonicity formula for the infimum of scalar curvature under (rescaled) Ricci flow on the other.

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    $\begingroup$ Spin^c structures are only an affine space over H^2 M. How does e determine a spin^c class without a choice of basepoint in the spin^c structures? $\endgroup$ Nov 12, 2009 at 2:21
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    $\begingroup$ A Spin^c structure on a 3-manifold determines (amongst other things) a U(2) bundle, which has a first Chern class, which is a 2-dimensional cohomology class. $\endgroup$ Nov 12, 2009 at 2:57
  • $\begingroup$ because $U(2)=Spin^c(3)$ $\endgroup$
    – Ben Webster
    Nov 12, 2009 at 15:12

1 Answer 1


Ian Agol, in his 03/04/03 blog states (claims?) that minimal surfaces propagate backwards under the Ricci flow. Hmmm. Ok, in 05/19/03 he gives more details, using CMC collars. I confess that I can't follow the argument in detail. Also, it's not obvious that this would suffice to answer:

Eg. if a hyperbolic 3-manifold contains an embedded minimal surface, is there an isotopic minimal surface in every other metric on the manifold?

ie backwards propagation from the limit is harder than what Agol is claiming?

  • $\begingroup$ This observation certainly seems relevant. $\endgroup$ Nov 12, 2009 at 21:12

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