# Every homotopy class contains at least a harmonic representative

Let $$(M^3,g)$$ be a closed, connected and oriented Riemannian $$3$$-manifold. A circle-valued map $$v : M \to S^1$$ is harmonic iff the gradient $$1$$-form $$\omega_v = v^* d\theta \in \Omega_1(M)$$ is harmonic in the Hodge sense: $$d \omega_v = 0$$ and $$\delta \omega_v = 0$$. It can be seen that this happens precisely when $$v$$ minimizes the Dirichlet energy in its homotopy class $$[v] \in [M:S^1]$$. Thus, by Hodge theory, each homotopy class of a circle-valued map contains a harmonic representative.

My question is whether every homotopy class of $$S^2$$-valued maps contains a harmonic representative. More precisely: given $$u : M \to S^2$$ a smooth map, does there exist a harmonic map $$u_0 : M \to S^2$$ such that $$u$$ is smooth and homotopic to $$u$$?

A parallel question: if $$u_0 : M \to S^2$$ is any harmonic map, can we say that $$u_0^* \sigma \in \Omega_2(M)$$ is a harmonic $$2$$-form, where $$\sigma$$ is the area form of (the round) $$S^2$$?

• No, the theory for targets with some positive curvature is very complicated. For example, there is no harmonic map from the 2-torus to the 2-sphere of degree 1 (for any choice of metrics). Apr 19, 2020 at 0:40

As Andy says, the answer is 'no': It is known that there is no harmonic map of degree $$1$$ from the torus to the $$2$$-sphere. I forget who first observed this. (Amended after Andy's comment: It's originally due to J. C. Wood in the early 1970s, see Andy's comment for the exact reference.)

If I have time, I can put in the argument, but the essential outline of the argument is this:

There are two kinds of harmonic maps from the torus to the $$2$$-sphere. Those that are conformal and those that are not.

If it is conformal, then, up to reversing the orientation on the torus, it is a holomorphic map, and it is well-known that a non-constant holomorphic map from the torus to the $$2$$-sphere has degree at least 2. (In fact, there is such a holomorphic map of any degree $$d\ge 2$$.)

If it is not conformal, then a simple calculation shows that the degree of the mapping is zero. (Essentially, one produces an explicit $$1$$-form on the torus whose differential is the pullback of the area form on the $$2$$-sphere.)

Thus, there is no harmonic map of degree 1 from the torus to the $$2$$-sphere.

• I think the result goes back to Wood, a discussion can be found here core.ac.uk/download/pdf/82593933.pdf. Apr 19, 2020 at 13:40
• @AndySanders: Thanks for the reference! I knew it had been known a long time, but I had forgot where I learned it. Apr 19, 2020 at 14:01
• What does this imply for the case of dimension $3$? Apr 19, 2020 at 15:42
• @EduardoLonga: Do you mean dimension 3 for the range or the domain, or both? Apr 19, 2020 at 16:31
• @EduardoLonga: I don't know an example off the top of my head, but I suspect that existence of a harmonic mapping in a given homotopy class of maps $f:M^3\to S^2$ fails in many cases, just because the regularity theory for a nonlinear PDE gets harder as the dimension of the domain goes up. Maybe Andy knows something more specific about this. A good place to start is the work of Schoen and Uhlenbeck on regularity of harmonic maps. Apr 19, 2020 at 17:47