# All the symmetries of the Dirichlet energy are conformal

It seems to be "folklore" knowledge that all the (source) symmetries of the $d$-Dirichlet energy are conformal maps.

Specifically, I have found this nice proof for the following claim:

Proposition: Let $M,N$ be oriented $n$-dimensional Riemannian manifolds, let $f:M \to N$ be a (local) diffeomorphism. Suppose that for any $h \in C^{\infty}(N)$, $$E_N^p(h)=E_M^p(h \circ f)$$ where $E^p$ is the $p$-energy: $E_N^p(h)=\int_N \|dh\|^p \operatorname{Vol}_N$. (we assume $p\ge 1$).

Then,

If $p \neq n$, $\,f$ is an isometry. If $p = n, \,f$ is conformal.

Are there any other proofs around? While the proof I mentioned above is nice, there are some "magical choices" involved, and I wonder if there is a simpler proof (with less computations).

Edit: (The pointwise problem is easy-the challenge is to pass from global to local)

Using michelle's comment, let's try to build an elementary proof:

Suppose $f:M \to N$ is an orientation-preserving diffeomorphism:

We want $$\int_M \|dh \circ df\|^p \operatorname{Vol}_M=\int_N \|dh\|^p \operatorname{Vol}_N=\int_M f^*(\|dh \|^p \operatorname{Vol}_N)=\int_M (\|dh \|^p \circ f ) f^*\operatorname{Vol}_N=\int_M (\|dh \|^p \circ f ) \det df \operatorname{Vol}_M. \tag{1}$$

So, $$\int_M \|dh \circ df\|^p \operatorname{Vol}_M=\int_M (\|dh \|^p \circ f ) \det df \operatorname{Vol}_M \, \, \text{for every} \, h \in C^{\infty}(N).$$

Since $h$ can be arbitrary one can hope to deduce equality of the integrands, i.e

$$\|dh \circ df\|^p = (\|dh \|^p \circ f ) \det df \, \, \text{for every} \, h \in C^{\infty}(N). \tag{2}$$

The main challenge: Passing from equation $(1)$ to equation $(2)$; Something like selecting $h$ to be very "localized" and using a Lebesgue differentiation-type theorem might work. We can try selecting $h$ which are quickly decaying, and are zero outside a small set. But then at the "transition phase" their differential will have a non-negligible integral.

(Note that if we assume that $f$ restricted to arbitrary small open subsets preserve the energy, then the proof is trivial.)

Writing equation $(2)$ explicitly, in a given point $x \in M$, we get

$$\|dh_{f(x)} \circ df_x\|^p = (\|dh_{f(x)} \|^p ) \det df_x$$

Since $dh_{f(x)}$ can be chosen at will, this means (thinking now in matrix terms, writing $df_x=A \in M_n,dh_{f(x)}=B \in \mathbb{R}^n$) that

$$\|BA\|^p=\|B\|^p\det A \tag{3} \, \, \text{for every} \, B \in \mathbb{R}^n.$$

Taking $B=e_i$, we get $$\|A_{i\rightarrow}\|^p=\det A, \tag{4}$$ i.e the norms of all rows are identical.

Taking $B=e_i+e_j$, we get $\|A_{i\rightarrow}+A_{j\rightarrow}\|^p=(\sqrt{2})^p\det A=(\sqrt{2}\|A_{i\rightarrow}\|)^p$. This implies $A_{i\rightarrow},A_{j\rightarrow}$ are orthogonal; Indeed

$$2\|A_{i\rightarrow}\|^2=\|A_{i\rightarrow}+A_{j\rightarrow}\|^2=2\|A_{i\rightarrow}\|^2+2\langle A_{i\rightarrow}, A_{j\rightarrow} \rangle.$$

So the rows of $A$ are orthogonal and have the same norm, thus $A$ is conformal. Suppose $A \in \lambda \text{SO}(n)$, and plug this into equation $(3)$:

$$\|B\|^p\lambda^p=\|B\|^p\lambda^n \Rightarrow \lambda^p=\lambda^n.$$

So, if $p \neq n$, $\lambda=1$ and $A$ is an isometry. If $p=n$, we can only conclude $A$ is conformal.

• Perhaps it will be easier to prove only conformal maps preserve harmonicity. (While composing at the source; If we compose at the target then affine maps are also symmetries of the E-L equations). – Asaf Shachar Aug 23 '17 at 20:02
• Locally, in coordinates, it's a linear algebra statement: the preservation of $p$-energy means the derivative matrix $A=df$ satisfies $\|AB\|^p = \|B\|^p|\det A|$ for every $B$, where the norm is the Frobenius norm. This leads to $A$ being orthogonal unless $p=n$ when scaling is also allowed. It takes work to make a proof for manifolds out of this, but the idea is there, and isn't magic. – user357151 Aug 23 '17 at 21:27
• Thanks. I agree that finding the right scaling is not magical. (The "magical choices" I was referring to were part of the proof I mentioned). Your idea seems very nice, however I really do not see how to construct a rigorous proof based on it (though I hope it can be done). Note that the "preservation of energy condition" I defined was a global one, not local. Of course you can apply it for test functions which are quickly decaying, and are zero outside a small set. But then at the "transition phase" - – Asaf Shachar Aug 23 '17 at 23:29
• @Michelle (cont) - say from $1$ to $0$, their differential will have a finite non-zero integral which cannot be made arbitrarily small, so you can't just ignore the cut-off... – Asaf Shachar Aug 23 '17 at 23:29