8
$\begingroup$

$\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\TM}{\operatorname{TM}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{T^*\M}$

Let $\M,\N$ be $d$-dimensional oriented Riemannian manifolds. Let $f:\M \to \N$ be smooth, and let $\delta=d^*$ be the adjoint of the exterior derivative.

Let $1 \le k \le d$. Consider the following two properties $f$ can have:

  1. $\delta^{\N} \omega=0 \Rightarrow \delta^{\M}(f^*\omega)=0$ for every $k$-form $ \omega \in \Omega^k(\N)$.

$\,\,\,\,$ 2.$\,\delta_{\nabla^{\Lambda_k(f^*{\TN})}} (\bigwedge^k df)=0$.

Question: Does property 1 implies property 2?

Here $df \in \Omega^1(\M,f^*{\TN})$, and $\bigwedge^k df\in \Omega^k\big(\M,\Lambda_k(f^*{\TN})\big)$ is the induced map. $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} $ is the adjoint of $d_{\nabla^{\Lambda_k(f^*{\TN})}}$. (where $\nabla^{\Lambda_k(f^*{\TN})}$ is the induced connection on $\Lambda_k(f^*{\TN})$ by the Levi-Civita connection on $\N$).

Note that for one-forms ($k=1$), the second property is exactly harmonicity of $f$.

Edit 1: In the Euclidean case, the answer is positive:

Let $\M=\N=\R^d$ (endowed with the standard metrics), property $(1)$ implies property $(2)$. Indeed, let $\omega=dx^{i_1} \wedge dx^{i_2} \wedge \dots \wedge dx^{i_k}$. Then property $(1)$ implies

$$ \delta(df^{i_1} \wedge df^{i_2} \wedge \dots \wedge df^{i_k})=0. \tag{3}$$

For a given increasing multi-index $I=(i_1,\dots,i_k)$, write $df^I=df^{i_1} \wedge df^{i_2} \wedge \dots \wedge df^{i_k}$, and $e_I=e_{i_1} \wedge e_{i_2} \wedge \dots \wedge e_{i_k}$. Then a direct calculation (which is valid only in the Euclidean context) shows that $$ \delta_{\nabla^{\Lambda_k(f^*T\R^d)}} (\bigwedge^k df)=\sum_I \delta (df^I) \otimes e_I $$

This is an equality of elements in $\Omega^{k-1}(\R^d,\Lambda_k(\R^d))$.

Edit 2: property $(2)$ does not imply property $(1)$.

Let $\M=\N=\R^2$ (with the standard metrics). Define $f(x,y)=(x,x)$. $f$ is affine, so $\delta(df)=0$.

Let $k=1$ and set $\omega=ydx$. Then $$ \delta(\omega)=\star d \star ydx=\star d ( y dy)=\star 0=0,$$ but

$$ \delta(f^*\omega)=\delta(x dx) =\star d \star xdx=\star d ( x dy)=\star (dx \wedge dy)=1 \neq 0.$$

Improve this question
$\endgroup$
3
  • $\begingroup$ Even for a local diffeomorphism and $k=1,$ your formula isn't correct for non-conformal maps - try a coordinate calculation for a map between Euclidean spaces. The term $f^*\delta \omega$ should be replaced with something like $\operatorname{tr}_{g_{\mathcal M}} (f^* \nabla \omega).$ $\endgroup$
    – Anthony Carapetis
    Commented Mar 5, 2018 at 3:34
  • 1
    $\begingroup$ Note that your property (1) implies (since exterior differentiation commutes with pull-back) that if $\omega$ is harmonic on $\mathcal{N}$, its pull-back is also harmonic on $\mathcal{M}$. When $k = 1$ this means that $f$ is a harmonic morphism, and there exists a characterization of the form "$f$ is harmonic morphism iff $f$ is harmonic map + some other conditions". See e.g. global.oup.com/academic/product/… or numdam.org/article/AIF_1978__28_2_107_0.pdf $\endgroup$
    – Willie Wong
    Commented Mar 7, 2018 at 14:42
  • $\begingroup$ @WillieWong Thanks! This is really interesting. We are left with the question of what happens when $k>1$. $\endgroup$
    – Asaf Shachar
    Commented Mar 7, 2018 at 15:37

0

Reset to default

You must log in to answer this question.