Let $M,N$ be oriented $d$-dimensional Riemannian manifolds, $M$ compact*, and let $f:M \to N$ be smooth.

Consider the Dirichlet energy functional: $E_{M,N}(f)=\int_M \|df\|^2 \operatorname{Vol}_g$. ($\operatorname{Vol}_g$ is the Riemannian volume form of $M$).

Fix another $d$-dimensional Riemannian manifold $W$. It's easy to see that for any isometric immersion $\phi:N \to W$,

$$ E_{M,N}(f)=E_{M,W}(\phi \circ f) \, \text{ for any } \, f:M \to N. \tag{1}$$

Let us call a map $\phi$ which sarisfies $(1)$ a symmetry of the Dirichlet's integral.

In this terminology, every isometric immersion is a symmetry.

Question: Is every symmetry an isometric immersion?

Does anything change if restrict the symmetries to be invertible maps? (or diffeomorphisms)? Does this notion of symmetry has a name somewhere in the literature?

I guess the answer is positive.

*We can relax the compactness assumption, but then we need to restrict the discussion to maps which are constant outside a compact domain.


$(1)$ The "pointwise analog" is quite trivial:

If $B \in M_d$, and for every $A \in M_d$, $ \|BA\|=\|A\|$ (where $\| \|$ is the Euclidean norm), then $B$ is an orthogonal matrix.

The challenge is that the notion of a Dirichlet's-symmetry is a global one, while the concept of isometric immersion is local.

I think the rough idea should be to choose maps $f$ which are very "localized" (are constant outside small balls). However, this does not seem trivial, since the differential would have to pass from a given linear map to zero, so its norm would vary. (Think of a bump function which goes quickly from one to zero, you can't make the integral of the derivative small).

$(2)$ The answer could a-priori depend on the manifolds. Even the case where $M$ is an Euclidean ball in $\mathbb{R}^d$, $N=W=\mathbb{R}^d$, the answer does not seem to be trivial (see the previous comment).


1 Answer 1


The answer is yes, every symmetry must be an isometric immersion. This is explained here.


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