A well known theorem of Cartan states that every free homotopy class of closed paths in a compact Riemannian manifold is represented by a closed geodesic (theorem 2.2 of Do Carmo, chapter 12, for example). This means that every closed path is homotopic to a closed geodesic (through a homotopy that need not fix base points).

I am wondering if there is a higher dimensional generalization of this result. Let us define a free n-homotopy class of $M$ to be a set $L_n$ of continuous maps $S^n \to M$ such that if $f \in L_n$ and $g: S^n \to M$ is any continuous map which is homotopic to $f$ (no base points required) then $g \in L_n$.

Question 1: Can one use the geometry of $M$ to intelligently single out a preferred class in any free n-homotopy class?

The naive guess is that every free n-homotopy class has an area minimizing representative, but my intuition tells me that this is either not true or really hard to prove. This is because the proof in the 1-dimensional case develops the argument from certain continuity properties of arclength that tend to either fail or require tremendous care when generalized to area. But perhaps this idea abstracts the wrong property of geodesics; maybe one should instead look for representatives which extremize some intelligently chosen functional instead. I'm hoping someone has already thought about this and come up with a good answer.

Question 2: Can anything more be said in the presence of negative curvature?

In the 1-dimensional case, I believe that the closed geodesic guaranteed by Cartan is in fact unique if $M$ is compact with strictly negative curvature. If there is an affirmative answer to question 1, I would be curious to know if there is a corresponding uniqueness statement in negative curvature. And if question 1 doesn't seem to have a nice answer in general, maybe negative curvature helps.

Thanks in advance!

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    $\begingroup$ Regarding question 2, complete negatively curved manifolds are aspherical, so any spheroid is null-homotopic. Thus constant maps are prefered ones. :) The question becomes more meaningful if the domain is an arbitrary closed manifold. Then one can e.g. talk about harmonic representatives in a free homotopy class, and here negative curvature does help. $\endgroup$ Aug 11, 2010 at 17:55
  • $\begingroup$ You probably also mean for any pair in a free class to be freely homotopic to each other; otherwise you have defined what I would think of as a union of classes. $\endgroup$
    – Tracy Hall
    Aug 11, 2010 at 20:40
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    $\begingroup$ I don't have enough for an answer, but here's some interesting references: books.google.com/… atlas-conferences.com/c/a/w/j/13.htm arxiv.org/abs/math/0308090 $\endgroup$
    – Ian Agol
    Aug 11, 2010 at 22:40

1 Answer 1


A theorem of Sacks and Uhlenbeck tells us http://www.jstor.org/pss/1971131 that we are still in the good shape for free homotopy $2$-classes. Namely, any such class can be represented by a collection of minimal spheres joined by geodesic segments. I would guess that in the case when you consider $\pi_n(M^{n+1})$ sometimes there will be a minimal hypersurface that realises such a class, for example in the case $M^{n+1}=S^1\times S^n$.

But for a general situation I remember a "claim" of Gromov, that such an preferred class does not exist, in particular for the case of homotopy groups of spheres. Unfortunately, I don't remember a reference (in fact I might be wrong here, see the article of Gromov below). Though the complexity of the question can be traced in the article of Larry Guth, treating in particular the case of maps $S^3\to S^2$.

Isoperimetric inequalities and rational homotopy invariants


ADDED. In his new article Gromov says a bit about "good representatives" of maps in fixed homothopy classes $S^N\to S^n$. But this ounds more like a collection of questions and possible directions rather than some definite theorems (I guess it reflect the current state of these questions). This is contained in section 2,



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