Let $M,N$ be smooth oriented $d$-dimensional Riemannian manifolds, $\, f:M \to N$ a smooth map. Let $\Omega^1(M,f^*TN)=\Gamma(T^*M \otimes f^*TN)$ be the space of $f^*TN$-valued one-forms.

Let $d$ be the covariant exterior derivative associated with the pullback connection of the Levi-Civita connection on $N$ (via $f$), and let $\delta$ be its adjoint.

Finally, define the section $Q(df) \in \Gamma(T^*M \otimes f^*TN)$ as the closest orientation-preserving isometry to $df$. (here "closest" is w.r.t the natural norm induced on $T^*M \otimes f^*TN$ by the metrics on $M,N$).

I am interested in mappings $f$ which satisfy:

$(1) \, \, f$ is an orientation-preserving immersion.

$(2) \, \, \delta (Q(df))=\delta(df)=0$.

(The motivation comes from studying the critical points of a distortion functional. The equations in $(2)$ are related to the associated Euler-Lagrange equations).

Recall that $\delta(df)=0$ is harmonicity, and that every isometry is harmonic. Thus, if $f$ is an orientation-preserving isometry it satisfies $(2)$.

Moreover, if $f$ is harmonic and there exist a constant $\lambda > 0$ such that $df$ is always $\lambda $ times an (orientation-preserving) isometry, then $Q(df)=\frac{1}{\lambda}df$, so again $(2)$ is satisfied. (In fact the harmonicity of $f$ follows from the other condition).

Question: Suppose $f:M \to N$ satisfies $(1),(2)$. Is it true that it is a "scaled isometry"? (i.e is it conformal with a "constant scaling factor")? Or at least conformal?


Here's a simple counterexample: Let $M=N=T^2$ (the standard torus, thought of as $\mathbb{R}^2/\mathbb{Z}^2$). Let $f:M\to N$ be the identity, and let the metrics on $M$ and $N$ be any two translation-invariant metrics on $T =\mathbb{R}^2/\mathbb{Z}^2$ (so, in particular, they don't need to be conformal). Then one easily sees that $f$ satisfies your equations, but if the two metrics aren't constant multiples of one another, $f$ won't be a 'scaled isometry'.


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