Assume that $(M,g)$ is a Riemannian manifold. A vector field $X$ on $M$ is called a harmonic vector field if the corresponding $1$-form $\alpha$ with $\alpha(Y)= \langle X,Y \rangle_g$ is a harmonic $1$-form.

Motivated by this conversations we ask the following question:

Assume that $X$ is a vector field on $M$. Assume that for every $t$ the flow $\phi_t$ of the vector field is a harmonic map. Does this imply that the vector field is a harmonic vector field?

  • 3
    $\begingroup$ It appears to me that this can be resolved by a straightforward calculation. Did you try that already? $X$ satisfies the linearized harmonic map equation. Either that matches or there is an extra curvature term. $\endgroup$
    – Deane Yang
    May 19 '17 at 13:36
  • $\begingroup$ @DeaneYang I just tried for Euclidean structure. $\endgroup$ May 19 '17 at 13:40
  • $\begingroup$ May you elaborate your comment? $\endgroup$ May 19 '17 at 13:41
  • $\begingroup$ Differentiate $\Delta \phi_t = 0$ with respect to $t$. Compare it to the equator for a harmonic vector field. $\endgroup$
    – Deane Yang
    May 19 '17 at 14:26
  • 1
    $\begingroup$ My answer to your other question already contains the answer. The answer is that the corresponding one form solves $\triangle \alpha = \mathrm{Ric}(X,\cdot)$ (with an extra curvature term as @DeaneYang suspects), and so is guaranteed harmonic only in the case $M$ is Ricci flat. $\endgroup$ May 19 '17 at 21:11

Well, right away, you can see that the answer is 'no', in general. Consider the round $n$-sphere $S^n$ with its standard metric. When $n>1$, it has no nonzero harmonic $1$-forms, but it has nontrivial Killing vector fields since it is a homogeneous Riemannian manifold. Since the flow $\phi_t$ of such a Killing field $X$ is a harmonic map of $S^n$ to itself, this example shows that the flow of $X$ can be harmonic even though $X$ is not.

  • $\begingroup$ Thank you very much for your interesting answer. $\endgroup$ May 20 '17 at 5:48

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