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Harmonic vector fields are critical points of Dirichlet energy function on the set of all unit vector fields on $M$, which is defined as follows:

$$E(X):=\frac{1}{2}\int_M\|dX\|^2\mathrm{dVol_g}\qquad X: (M,g)\to (TM,g_s), \|X\|=1.$$

Where it's Euler-Lagrange equation satisfies the following: $$\bar{\Delta}X=\lambda X,\quad \lambda\in C^\infty(M)$$ where $\bar{\Delta}$ denote rough Laplacian.

$p$-Harmonic vector field: If we use $p$-energy function instead of Dirichlet energy function: $$E_p(X):=\frac{1}{p}\int_M\|\mathrm{d}X\|^p\mathrm{dVol_g},\qquad X: (M,g)\to (TM,g_s),$$ The Euler-Lagrange equation of $E_p(.)$ in all vector fields (not in unit case) is as follows (for function case see this paper): $$\|dX\|^{p-2}\left(\tau(X) + dX\left(\mathrm{grad}(\log \|dX\|^{p-2})\right)\right) =0.$$ where $\tau(X)=\mathrm{div}(dX)$ called tension field of $X$.

My questions are :

Question 1: What is the Euler-Lagrange of $E_p(.)$ on the set of all unit vector fields on $M$?

Question 2: Can one hope to get statements which is generalized and new (not same to exactly) 2-harmonic (harmonic vector field) case?

Thanks.

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  • $\begingroup$ Isn't Question 1 just Lagrange multipliers? Also, what do you mean by a tension field of $X$? Is it not just $\overline\Delta X$? How do you understand the term involving $dX$? Is $dX$ supposed to be the antisymmetrized derivative for just the covariant derivative of $X$ w.r.t. $g$? $\endgroup$
    – Willie Wong
    Commented Jul 7, 2017 at 13:14
  • $\begingroup$ Tension field is just an equivalent equation to Euler-Lagrange equation and here I don't know what is it exactly. $dX: TM\to TTM$ is derivation of vector field $X$. $\endgroup$
    – C.F.G
    Commented Jul 7, 2017 at 17:35
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    $\begingroup$ Just to check my understanding, is it the case that $\| dX\|^2 = n + \|\nabla X\|^2$? Imposing that you are keeping variations within the sphere bundle, and denoting by the infinitesimal variation $V$, you have that $\langle X, V\rangle = 0$, which means that $\| d (X + V)\|^2 = n + \|\nabla (X+V)\|^2$ which to linear order gives $n + \|\nabla X\|^2 + 2 \langle \nabla X, \nabla V\rangle$. This would imply that for the $p$ case you have the ELE being $\endgroup$
    – Willie Wong
    Commented Jul 8, 2017 at 2:36
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    $\begingroup$ $$ \mathrm{div}_g \cdot \left[ (n + \|\nabla X\|^2)^{p/2 - 1} \nabla X \right] = \lambda X $$ since the integral of the inner product of the left hand side against any vector field orthogonal to $X$ must be 0. The constant $\lambda$ can also be computed using that $X$ is of unit norm as $$ \lambda = - (n + |\nabla X\|^2)^{p/2 - 1} \|\nabla X\|^2 $$ (These are all predicate in my assumption that $\|dX\|^2 = n + \|\nabla X\|^2$ is correct.) $\endgroup$
    – Willie Wong
    Commented Jul 8, 2017 at 2:43
  • $\begingroup$ Yes, I think your assumption $\|dX\|^2 = n + \|\nabla X\|^2$ is correct. $\endgroup$
    – C.F.G
    Commented Jul 8, 2017 at 8:16

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