$\newcommand{\M}{\mathcal{M}}$
$\newcommand{\N}{\mathcal{N}}$
$\newcommand{\Hom}{\operatorname{Hom}}$
$\newcommand{\tr}{\operatorname{tr}}$
$\newcommand{\TM}{\operatorname{T\M}}$
$\newcommand{\TN}{\operatorname{T\N}}$
$\newcommand{\sAverage}[1]{\langle#1\rangle} $
$\newcommand{\IP}[2]{\sAverage{#1,#2}}$
$\newcommand{\Cof}{\text{Cof}}$


Lately I derived the following equation (details are at the end):



> 
> **Theorem:**
>  Let $(\M,g),(\N,h)$ be smooth $d$-dimensional oriented Riemannian manifolds, 
> 
> 
> Let $\,f:\M \to \N\,$ be a smooth conformal map. Then
> 
> 
> $$ \det df \cdot \delta(df) + (\frac{2}{d}-1) \tr\_{g}\big(df \otimes d \det(df) \big) =0. \tag{1} $$
> 
> 
> 


($\delta$ is the adjoint of the pullback connection on $f^\*{\TN}$. The equation is an equality of sections of $f^\*{\TN}$.)


*Equation $(1)$ has a few immediate (well-known) corollaries:*


1. In dimension $d=2$, conformal mappings are harmonic.
2. Homotheties are harmonic (in every dimension).
3. In dimension $d > 2$, conformal harmonic maps are homotheties (assuming $\M$ is connected): 
Let $e\_i$ be an orthonormal frame for $\TM$. Then $$ 0=\tr\_{g} \big( df \otimes d \det(df) \big)= \sum\_{i=1}^d d \det(df)(e\_i) \cdot df(e\_i) .$$
Since $f$ is an immersion, $df(e\_i)$ is a basis of $f^\*\TN$.


(Summary: A conformal map is harmonic if and only if $d=2$ or it is a homothety).


**Questions:**


**1.** Is equation $(1)$ known? (Perhaps in a Euclidean setting? Or in coordinates?)


**2.** Is equation $(1)$ an Euler-Lagrange's equation of some functional?




---


*Sketch of derivation:*


My colleagues and I showed the following: 



> 
> **Theorem(2):**
>  Let $(\M,g),(\N,h)$ be smooth $d$-dimensional oriented Riemannian manifolds, 
> 
> 
> Let $\,f:\M \to \N\,$ be an **arbitrary** smooth map. Then
> 
> 
> $$ \delta (\Cof df)=0. \tag{2}$$ 
>  where $\Cof df:= (-1)^{d-1} \star\_{f^\*TN}^{d-1} (\wedge^{d-1} df) \star\_{TM}^1.$ (It's a section of $T^\*\M \otimes f^\*{\TN}$).
> 
> 
> 


(For a proof see Proposition 3.4 [here](https://arxiv.org/pdf/1701.08892v1.pdf)). 


Now, if $f$ is conformal, then
$$ \Cof df=(\det df)^{1-\frac{2}{d}} df. \tag{3}$$


Plug $(3)$ into $(2)$ (the rest is Leibniz rule + simplification).



Is this expression for the Laplacian of conformal maps between Riemannian manifolds known?
Well, *for conformal maps* equation $(1)$ is merely $d$-harmonicity in disguise:)


The equation is
$$
\delta\big((\det df)^{1-\frac{2}{d}} df\big)=0. \tag{1}
$$


Since for conformal maps, $\det df=\|df\|^d$ up to a constant, we equivalently get


$$
\delta\big(\|df\|^{d-2} df\big)=0. \tag{2}
$$


(Which is trivial of course, since conformal maps are source symmetries of the $d$-energy).


The funny part is that while equation $(2)$ is an Euler-Lagrange's equation, equation $(1)$ is **not** (for $d \neq 2$). 


The moral here is that if we start from an E-L equation, write it differently for specific class of maps (the conformal maps in this example) then the resulted equation may not be an E-L equation.


*Let's prove equation $(1)$ is not E-L:*


Indeed, if it were an E-L equation (of an isometrically invariant smooth functional), this would imply the existence of a smooth map $ h:M\_d \to \mathbb{R}$, such that $dh\_X(V)=\langle (\det X)^r X,V\rangle $.


This is not possible since $d^2h\neq 0$:


$d^2h=\sum\_{ij}(\det X)^rx\_{ij}dx^{ij} \Rightarrow d^h=\sum\_{ijks} \frac{\partial((\det X)^rx\_{ij}) }{\partial x\_{sk}} dx^{sk} \wedge dx^{ij}=r \sum\_{ijks} x\_{ij} \frac{\partial \det X }{\partial x\_{sk}} dx^{sk} \wedge dx^{ij}$ which is zero iff 
$$x\_{ij} \frac{\partial \det X }{\partial x\_{sk}} =x\_{sk} \frac{\partial \det X }{\partial x\_{ij}} , $$ which is false. (The RHS does not depend on $x\_{ij}$ while the LHS is dependent of it). 


