A map or a vector field $g: \mathbb{R}^n \to \mathbb{R}^n $ is called a harmonic map if all its components are harmonic functions.

Motivated by conversations on this questions we ask:

Is the flow of a Harmonic vector field on $\mathbb{R}^n$, a harmonic function?

More precisely, assume that every component of the vector field $x'=f(x)$ is a harmonic map. Is it true to say that for every $t$, $\phi_t$ is a Harmonic map? Note that the converse is true. Namely, a vector field is Harmonic if its flow is harmonic.

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    $\begingroup$ No. Try some simple cases for which you can solve the differential equations explicitly. $\endgroup$ May 18, 2017 at 21:04
  • $\begingroup$ @MichaelRenardy yes thank you $(y^2_x^2)\partial_x$ is a counter example.but the converse is true, i think. This is a motivation to modify this post to have a true version. $\endgroup$ May 19, 2017 at 7:26
  • $\begingroup$ As stated, the answer to your question is no for a trivial reason. Any diffeomorphism $\phi_0$ can be extended to a flow along a vector field $V$. $\endgroup$
    – Deane Yang
    May 19, 2017 at 22:24

1 Answer 1


There is a "known true version", but it is basically what you have already as the converse.

Defn Let $(M,g)$ and $(N,h)$ be Riemannian manifolds, and let $\phi: M\to N$ be a smooth mapping. The tension field of $\phi$ is defined to be $$ \tau(\phi) = \mathrm{trace}_g D \mathrm{d}\phi $$ where $D$ is the pull-back of the Levi-Civita connection on $N$ to $M$ via $\phi$.

Defn A mapping $\phi:M\to N$ is said to be "harmonic" if $\tau(\phi) \equiv 0$.

Rmk In the case $M$ and $N$ are Euclidean spaces, this reduces to the components of mapping being harmonic functions.

Defn Given a Riemannian manifold $(M,g)$, a vector field is said to be 1-harmonic-Killing if, denoting by $\varphi_t$ its corresponding one parameter family of diffeomorphisms, that $$ \frac{\mathrm{d}}{\mathrm{d}t} \tau(\varphi_t) \Big|_{t = 0} = 0 $$

Rmk $\tau(\varphi_0) = \tau(\mathrm{id}) = 0$. So the definition of 1-harmonic-Killing asks that the flow to be linearly harmonic.

Thm A vector field $V$ is 1-harmonic-Killing if and only if $$ \triangle X^\flat = \mathrm{Ric}(X,\cdot) \tag{1}$$ where $\triangle$ is the Hodge-Laplacian and $\mathrm{Ric}$ is the Ricci tensor.

Rmk In the Euclidean case, Ricci curvature vanishes, and (1) reduces to what you state as "harmonic vector field".

This result can be found in: Dodson, Trinidad Pérez, and Vázquez-Abal; Stepanov and Shandra; and Nouhaud. The first reference also contains examples showing that there exists 1-harmonic-Killing vector fields whose flow is not always harmonic.

Notice however, Remark 3.1 and Proposition 3.1 in the first-cited paper above, which combine to say that on a compact Riemannian manifold with non-positive Ricci curvature, a vector field $V$ is 1-harmonic-Killing if and only if $V$ is a parallel vector field. In particular, this implies that on such manifolds any 1-harmonic-Killing vector field will give rise to a harmonic flow. While this statement gives a positive answer to your question, we should note that in this setting $V$ is also necessarily Killing and hence the flow generated is a one parameter family of isometries, and hence the result can also be viewed as a rigidity statement.

A geometrically interesting fact is this:

Let $V$ be a vector field and $\varphi_t$ its local one parameter family of isometries. In the above we asked about the property of $\varphi_t$ being harmonic. If we replace the word harmonic by

  • isometric; we get Killing vector fields;
  • conformal isometric: we get conformal Killing vector fields;
  • totally geodesic: we get affine Killing vector fields.

In all three of the above situations, having the "1-" version hold (meaning that the condition holds for $\varphi_t$ to first order as $t \to 0$) implies that the condition in fact holds (to infinite order). So in the literature we do not distinguish between 1-Killing vector fields and Killing vector fields.

One reason behind this distinction is the fact that compositions of harmonic maps need not be harmonic, in contrast to the other three cases mentioned above.

  • $\begingroup$ Thank you very much for your very helpful answer. I learn a lot from your answer. $\endgroup$ May 20, 2017 at 5:46

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