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$M$ is a Riemannian manifold with metric $g$ and we have a map $F: M \to T^{\*}M$ with $F(p)=(p,f(p))$ with a 1-form $f$. On $T^{*}M$ we use the Sasaki-metric.

How can I prove or it is wrong?:

$F$ is harmonic iff $f$ is harmonic.

Thank you and best regards.

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What is the difference between f and F? – Michael Bächtold Jun 24 '10 at 12:47
Why don't you sit down and actually compute, relative to a local coordinate system, what the tension-field/energy-density is, and what the local coordinate expressions for harmonicity is for the map $F$ and for the one-form $f$? – Willie Wong Jun 24 '10 at 12:52
When I compute this, I have to inverse the following metric $g^{\prime}_{ij}=g_{ij}+\left( \frac{\partial f^{k}}{\partial x^{i}}\delta_{kn}-f^{t}\delta_{tk}\Gamma_{in}^{k} \right)g^{nM} \left( \frac{\partial f^{a}}{\partial x^{j}}\delta_{aM}-f^{a}\delta_{ab}\Gamma_{jM}^{b} \right).$ Can you help me to get the inverse metric? – Differentialgeometer Jun 24 '10 at 13:01
Oh, now I understood the question. – Michael Bächtold Jun 24 '10 at 13:06
What's the Sasaki metric on the cotangent bundle? I'm not familiar with this. (This looks like a straightforward formal computation, and it is unlikely that you need to find an explicit formula for the inverse of $g'_{ij}$ to do the calculation. Just call it $g'^{ij}$ and plunge ahead.) – Deane Yang Jun 24 '10 at 13:17

It is perhaps easier to look at the the harmonicity as Euler Lagrange equations for some action functional. The energy density of the map $F$ can be computed to be $$e_F = N + |\nabla f|^2$$ directly using the definition of the Sasaki metric. $N$ is the dimension of the manifold. So the Euler-Lagrange equation gives that the equation satisfied by $f$ for $F$ to be a harmonic map is $$ \triangle_g f = 0$$ On the other hand, for $f$ to be a harmonic one-form, you need $$ (d\delta + \delta d) f = 0 $$ The Weitzenbock formula tells us that the two equations differ by a term coming from the Riemann curvature of $(M,g)$. So no, in general the two expressions are not equal.

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