# Relation between Harmonic vector field and Harmonic 1-form

Definition 1: A unit vector field $X$ side to be harmonic if it is critical point for the following energy function $$E(X)=\frac{1}{2}\int_M\|dX\|^2dvol_g=\frac{m}{2}vol(M,g)+\int_M\|\nabla X\|^2dvol_g.$$

Definition 2: A 1-form $\omega$ side to be harmonic if it is in kernel of Laplace operator. i.e. $\Delta\omega=(d\delta+\delta d)\omega=0$.

Question: Is there relation between two above definition? Please give an simple example.

Update: I find some theorem in this topic:

Theorem 1. If $\omega$ is harmonic and $X$ is the dual vector field, we have that $\mathrm{div}X = 0$.

Theorem 2. If $X$ is a vector field on $(M,g)$ and $\omega(v) = g(X,v)$ is the dual 1-form, then $$\mathrm{div}X = −\delta\omega.$$

Thanks.

• It would help if you could provide more context. What is "the energy function" here? – Saal Hardali May 3 '16 at 18:54
• The duality from the metric between 1-forms and vector fields shows that your energy is the usual one on 1-forms for which the critical 1-forms are the harmonic ones (see any reference on Hodge theory). But the restriction to unit vector fields is a bit unusual, and doesn't give the same Euler-Lagrange equations, I imagine. – Ben McKay May 3 '16 at 19:35
• Note that $0 = \Delta_g(1) = \Delta_g (g(X,X)) = 2 g(X,\Delta_g X) + 2 g(\nabla X, \nabla X)$ you actually have that a unit harmonic vector field on a Riemannian manifold must be parallel. It seems very strange to require that $X$ is unit in the definition. – Willie Wong May 3 '16 at 19:46
• Harmonic forms are almost never unit length, as far as I am aware. It would appear to me that Definition 1 is a very special definition, largely unrelated to Definition 2. You can of course talk about harmonic vector fields (i.e. dual to harmonic forms) but this results in a different object than your "harmonic unit vector fields". – Ryan Budney May 3 '16 at 20:20
• I think that the OP is mixing up a couple of different things: The definition of harmonic for a 1-form is standard, but for unit vector fields, there is another notion of 'harmonic': Regard $S(M)$, the unit sphere bundle of $(M,g)$, as a Riemannian manifold in the natural way and then ask whether a unit vector field $X:M\to S(M)$ is harmonic as a mapping between two Riemannian manifolds. There is actually an additional subtlety, in that one can ask that $X$ be a critical point of the energy functional when one only varies $X$ through sections of $S(M)\to M$ (instead of through all maps). – Robert Bryant May 6 '16 at 9:11

The condition for a unit vector field $X$ on a Riemannian manifold $(M,g)$ to be harmonic is not the same as the condition that the dual $1$-form $X^\flat$ be harmonic. The point is that, for unit vector fields, one defines the energy as the energy of the map $X:M\to S(M)$ where $S(M)$ is the unit sphere bundle of $(M,g)$ endowed with the Sasaki metric and one says that a unit vector field is harmonic if it is a critical point of this energy. This is not the same as the energy of the $1$-form $X^\flat$ in general (though it can be sometimes, for example, if the metric is flat).
A simple example is to take $(M,g)$ to be $S^3=\mathrm{SU}(2)$ endowed with its natural bi-invariant metric. Then any unit left-invariant (or right-invariant) tangent vector field $X$ is harmonic in the above sense, but the dual $1$-form $\omega = X^\flat$ is not harmonic as a $1$-form because the only harmonic $1$-form on $S^3$ is the zero $1$-form. (Since $H^1(S^3) = 0$, this follows, for instance, from the Hodge Theorem.)