13
$\begingroup$

$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\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).

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).

$\endgroup$
  • 2
    $\begingroup$ A couple of quick comments: 1) If you pull back the metric on $\mathcal{N}$ back to $\mathcal{M}$, then $f$ becomes effectively the identity map and your equations become equations for the conformal metric. Those probably can be found in the literature. However, I always prefer writing such equations the way you have. 2) You can probably prove a regularity result. Work in this direction, I believe, can be found in papers discussing quasiconformal maps. A random paper that comes to mind is one by Donaldson and Sullivan. $\endgroup$ – Deane Yang Apr 28 '17 at 17:20
1
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.