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We have a map $f \in \mathcal{C}^{\infty}(M, N)$ with two manifolds $M$ and $N$ (with dimensions $m:\dim(M)$ and $n:=\dim(N)$). We define the graph $F: M \to M \times N$ by $F(p)=(p, f(p))$. I wish to prove:

1.) $e(F)=\frac{m}{2}+e(f)$, where $e$ is the energydensity.

2.) $f$ is harmonic iff $F$ is harmonic.

Thank you and best regards.

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1 Answer 1

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It is intended that $M\times N$ is endowed with the direct sum Riemann structure. In this case, for any smooth map $f:L\to M\times N$ it is true that $\frac{1}{2}|Df|^2=\frac{1}{2}|Df_1|^2+\frac{1}{2}|Df_2|^2$, whence (1) since the energy density of id is m/2. Also, $f$ is a local minimizer of the local energy integral $E(f)=E(f_1)+E(f_2)$ if and only of both $f_1$ and $f_2$ are, that implies (2).

rmk. Here of course "local minimizer" is a map f such that for any point p of the domain there exists a nbd U of p such that the integral of the density energy of f over U is minimum with respect to variations with compact support in U. This property is equivalent to harmonicity (and of course does not require that the total energy be finite).

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  • $\begingroup$ For 2) there is also a non-variational argument (I do not know that all harmonic maps are local minimisers): if $f=(f_1,f_2):M\to N_1\times N_2$ is a map of Riemannian manifolds with the target having product metric, then $f$ is harmonic iff each $f_i$ is harmonic (the tension field of $f$ has those of the $f_i$ as components). $\endgroup$ Jun 24, 2010 at 10:56
  • $\begingroup$ Sure, and maybe even simpler; nevertheless I preferred the variational definition as I feel it more natural (dunno why). I've added a rmk correspondingly. $\endgroup$ Jun 24, 2010 at 11:10

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