Consider a smooth map $f : M \to N$ between two Riemannian manifolds $(M,g)$ and $(N,h)$. I would like to think of the tension field of $f$ and the harmonicity of $f$ in terms of centers of mass.
I wish to know if the following formula holds for the tension field of $f$ at a point $x\in M$, for $r>0$: $$\tau(f)_x = C_r \int_{\Sigma_r} \exp_{f(x)}^{-1} \left(f(y)\right)\,d\sigma(y)$$ where
- $\Sigma_r$ is the sphere of radius $r$ centered at $x$ in $M$. In other words $\Sigma_r = \exp_x(S_r)$ where $S_r$ is the sphere of radius $r$ in $T_xM$.
- $d\sigma$ is the volume element on $\Sigma_r$ for the induced metric from $g$. Or maybe instead $d\sigma$ is the pushforward by $\exp_x$ of the volume element on $S_r$.
- $C_r$ is some constant depending on $r$, maybe something like $\dfrac{\dim M}{r^2 \mathrm{vol}_{d\sigma}(\Sigma_r)}$.
It's possible that such a formula for the tension field would not be true, but only a good approximation when $r$ is small, I don't know.
Allow me to quickly recall what a center of mass is in a Riemannian manifold $(N,h)$. Let $A$ be a subset of $N$ such that $A$ is contained in some geodesically convex ball and let $\mu$ be a Borel probability measure on $A$ (a "distribution of mass"). Then the the center of mass of $(A, \mu)$ is the unique minimizer of the function $\psi : x \in N \mapsto \int_A d(x,y)^2 d\mu(y)$. The center of mass is characterized by the fact that it is the unique zero of the gradient vector field $\mathrm{grad}\, \psi = \int_A \exp_x^{-1}(y) d\mu(y)$. This can be generalized a tiny bit: if $(A, \mu)$ is a probability space and $f : A \to N$ is a measurable map whose image is contained in some geodesically convex ball, then the center of mass of $f$ is defined as the unique minimizer of $\psi : x \in N \mapsto \int_A d(x,f(y))^2 \, d\mu(y)$.
So, if the formula for the tension field above was true, it would mean that the harmonicity of $f$ is characterized by the property that at each point $x\in M$, $f(x)$ is the center of mass of $f$ restricted to a sphere around $x$ (for the appropriate measure). I should say, provided the radius of the sphere is small enough that the ball it bounds is convex.
Strangely I could not find anything of the sort searching the web. Can someone please help me understand this?
Allow me to add one thought although the question is already long, apologies. I understand that since the tension field can be described as the trace of the second fundamental form of $f$, and since the mean value of a symmetric bilinear map on a sphere is its trace (divided by dimension), then the following formula for the tension field is true: $$\tau(f)_x = C_r \int_{S_r} \frac{\nabla_h^2}{dt^2}_{|t=0} \left(f(\exp_x(tu))\right)\,d\sigma(u)$$ where I have abusively written $\frac{\nabla_h^2}{dt^2}$ for the operator ``acceleration of a curve'' in $(N,h)$. I was hoping I could derive the formula for the tension field that I want from this formula, but maybe this would be "too easy"? (I seem to remember that mean value properties of harmonic maps $\mathbb{R}^n \to \mathbb{R}$ are already kind of hard to show).
