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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 definitions? Please give a 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 $\mathrm{div}X = 0$.

<p><strong>Theorem 2.</strong> If <span class="math-container">$X$</span> is a vector field on <span class="math-container">$(M,g)$</span> and <span class="math-container">$\omega(v) = g(X,v)$</span> is the dual 1-form, then
<span class="math-container">$$\mathrm{div}X = −\delta\omega.$$</span></p>

Thanks.

YCor
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C.F.G
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    It would help if you could provide more context. What is "the energy function" here? – Saal Hardali May 03 '16 at 18:54
  • Ok. the question is updated. – C.F.G May 03 '16 at 19:08
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    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 03 '16 at 19:35
  • @BenMcKay: maybe not exactly a reference on Hodge theory; the OP may be slightly confused by Weitzenbock if we bring the form Laplacian into the picture. – Willie Wong May 03 '16 at 19:44
  • Thanks Ben McKay. But in every reference i can't find this relation. – C.F.G May 03 '16 at 19:44
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    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 03 '16 at 19:46
  • @Willie Wong thanks for your Hint. what is your means of harmonic? Def. 1 or Def. 2? – C.F.G May 03 '16 at 19:54
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    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 03 '16 at 20:20
  • Perhaps you don't mean to have the word "unit" in your definition of a harmonic vector field? – Deane Yang May 03 '16 at 20:59
  • I meant Harmonic in the sense of in the kernel of the Laplace-Beltrami operator, which is the sense given by both of your definitions. – Willie Wong May 03 '16 at 21:33
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    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 06 '16 at 09:11
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    Many thanks R. Bryant. I still did not get the answer my question. is this true: if $X:M\to S(M)$ is Harmonic in the sense of a mapping between two Riemannian manifolds if and only if $\omega= X^\flat$ is Harmonic as 1-form.? – C.F.G May 06 '16 at 18:45
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    @C.F.G: Oh. The answer to that is 'no'. For example, if one takes a left-invariant unit vector field $X$ on $M^3=\mathrm{SU}(2)\simeq S^3$ endowed with its bi-invariant metric as a Lie group, then $X:M\to S(M)$ is harmonic as a map between Riemannian manifolds (and as a section of $S(M)$ as well), but $X^\flat$ is not a harmonic $1$-form on $M$, since the only harmonic $1$-form on $M$ is the one that vanishes identically. – Robert Bryant May 07 '16 at 20:31

1 Answers1

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The two notions are related, but they are not the same.

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.)

There are several good sources for study of this notion of harmonicity of unit vector fields. There is a whole book, Harmonic Vector Fields: Variational Principles and Differential Geometry, by S. Dragomir and Domenico Perrone (Elsevier, 2012), but there are also articles that you may find useful: For example, see the survey article Volume, energy and generalized energy of unit vector fields on Berger spheres. Stability of Hopf vector fields by Olga Gil-Medrano and Ana Hurtado (http://www.ugr.es/~ahurtado/PDF/correcciones.pdf) and the references therein.

Robert Bryant
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