0
$\begingroup$

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?

$\endgroup$
5
  • 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
12
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you very much for your interesting answer. $\endgroup$ May 20 '17 at 5:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.