$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\CO}[1]{\text{CO}(#1)}$ $\newcommand{\dist}{\operatorname{dist}}$ $\newcommand{\g}{\mathfrak{g}}$ $\newcommand{\h}{\mathfrak{h}}$ $\newcommand{\Volg}{\text{Vol}_\g}$ $\newcommand{\SO}[1]{\text{SO}(#1)}$

Let $\M,\N$ be compact oriented $d$-dimensional Riemannian manifolds with boundary. Given $x \in \M,y \in N$ we define $$\CO{\g_x,\h_y} =\{\lambda R : R \in \SO{\g_x,\h_y} \, | \, \lambda > 0\} $$

Consider the following functional $E:C^{\infty}(\M,\N) \to \R:$

$$E(f)= \int_\M \dist^p (df,\CO{\g,f^*h})\,\Volg, $$

which measures the deviation of $f$ from being conformal. (The distance on $\text{Hom}(T_p\M,T_{f(p)}\N)$ is the one induced by the metrics).

Define $F:=\{ f:\M \to \N \, | \,\, f \text{ is a smooth immersion}\}$.

Question: Suppose $ \inf_{f \in F}E(f)=0$. Is it true that $\M$ is conformally immersibly in $\N$? i.e. does there exist a smooth conformal immersion $\M \to \N$?

I assume $p >d$. (This might be necessary since when $p < \frac{d}{2}$ there are regularity issues, at least in the Euclidean case).

Comment:

The conformal group is not closed, $0$ belongs to its closure. On manifolds, it can happen that a sequence of conformal diffoemorphisms weakly converge to a constant map (which I do not consider conformal here).

Here is a classic example: Take $\M=\N=\mathbb{S}^n$. Consider the following one-parameter family of diffeomorphisms $\psi_{\lambda}:\mathbb{S}^n \to \mathbb{S}^n$, $\lambda >0$:

$\psi_{\lambda}$ is obtained by using the stereographic projection, then dilating by $\lambda$ and then projecting back. $(\psi_{\lambda})_{\lambda >0}$ is a family of conformal diffeomorphisms that weakly converge to the pole when $\lambda \to \infty$.

In essence, the question is whether something like this can happen between manifolds which are not conformally equivalent. (With maps which are asymptotically conformal in the sense defined above).

$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\CO}[1]{\text{CO}(#1)}$ $\newcommand{\dist}{\operatorname{dist}}$ $\newcommand{\g}{\mathfrak{g}}$ $\newcommand{\h}{\mathfrak{h}}$ $\newcommand{\Volg}{\text{Vol}_\g}$ $\newcommand{\SO}[1]{\text{SO}(#1)}$

Let $\M,\N$ be compact oriented $d$-dimensional Riemannian manifolds with boundary. Given $x \in \M,y \in N$, we define $$\CO{\g_x,\h_y} =\{\lambda R : R \in \SO{\g_x,\h_y} \, | \, \lambda > 0\} .$$

Consider the following functional $E:C^{\infty}(\M,\N) \to \R:$

$$E(f)= \int_\M \dist^p (df,\CO{\g,f^*h})\,\Volg, $$

which measures the deviation of $f$ from being conformal. (The distance on $\text{Hom}(T_p\M,T_{f(p)}\N)$ is the one induced by the metrics).

Define $F:=\{ f:\M \to \N \, | \,\, f \text{ is a smooth immersion}\}$.

Question: Suppose $ \inf_{f \in F}E(f)=0$. Is it true that $\M$ is conformally immersible in $\N$? i.e., does there exist a smooth conformal immersion $\M \to \N$?

I assume $p >d$. (This might be necessary, for when $p < \frac{d}{2}$, then there are regularity issues, at least in the Euclidean case).

Comment:

The conformal group is not closed, as $0$ belongs to its closure. On manifolds, it can happen that a sequence of conformal diffeomorphisms weakly converges to a constant map (which I do not consider conformal here).

Here is a classic example: Take $\M=\N=\mathbb{S}^n$. Consider the following one-parameter family of diffeomorphisms $\psi_{\lambda}:\mathbb{S}^n \to \mathbb{S}^n$, $\lambda >0$:

$\psi_{\lambda}$ is obtained by using the stereographic projection, then dilating by $\lambda$ and then projecting back. $(\psi_{\lambda})_{\lambda >0}$ is a family of conformal diffeomorphisms that weakly converge to the pole when $\lambda \to \infty$.

In essence, the question is whether something like this can happen between manifolds which are not conformally equivalent (with maps which are asymptotically conformal in the sense defined above).