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) $S^2$?

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    $\begingroup$ 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). $\endgroup$ Apr 19, 2020 at 0:40

1 Answer 1


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.

  • $\begingroup$ I think the result goes back to Wood, a discussion can be found here core.ac.uk/download/pdf/82593933.pdf. $\endgroup$ Apr 19, 2020 at 13:40
  • $\begingroup$ @AndySanders: Thanks for the reference! I knew it had been known a long time, but I had forgot where I learned it. $\endgroup$ Apr 19, 2020 at 14:01
  • $\begingroup$ What does this imply for the case of dimension $3$? $\endgroup$ Apr 19, 2020 at 15:42
  • $\begingroup$ @EduardoLonga: Do you mean dimension 3 for the range or the domain, or both? $\endgroup$ Apr 19, 2020 at 16:31
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    $\begingroup$ @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. $\endgroup$ Apr 19, 2020 at 17:47

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