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](https://mathoverflow.net/questions/269940/vector-field-with-holomorphic-flow) 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. 



The flow of Harmonic vector fields
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". 




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This result can be found in: [Dodson, Trinidad Pérez, and Vázquez-Abal](http://www.ams.org/mathscinet-getitem?mr=2016229); [Stepanov and Shandra](http://www.ams.org/mathscinet-getitem?mr=1996772); and [Nouhaud](http://www.ams.org/mathscinet-getitem?mr=290289). 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. 




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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. 


