3
$\begingroup$

Given two Riemannian manifolds $M,N$, we say that $f:M \to N$ is harmonic if it is a critical point of the Dirichlet energy functional.

More precisely, this means that for every variation $f_t$ of $f$ with variation-field $v:=\frac{\partial f_t}{\partial t}|_{t=0}$ which is compactly supported in the interior of $M$, $\frac{d E(f_t)}{dt}|_{t=0}=0$.

Using the fact that this property of $f$ is equivalent to $f$ being a solution of a certain differential equation, one can deduce the following statement:

Claim: Suppose that for every point $p \in M$, there exist an open neighbourhood of $p$, $U_p\subseteq M$, such that $f|_{U_p}:U_p \to N$ is harmonic. Then $f$ is harmonic as a map $M \to N$.

Question: Is there a way to prove this claim without the passage $$\text{being critical} \to \text{satisfying E-L equation} \to \text{being critical}?$$

In other words, suppose you only know the "critical point definition" (and never heard of Euler-Lagrange equations). Is there a way to see directly that this property is local?

A naive idea is that given an arbitrary variation, we can somehow represent it as a finite number of compositions of "small variations" but I am not sure this makes any sense or really helpful.

For start, I am ready to assume $N=\mathbb{R}^n$ if it makes the problem easier.

$\endgroup$
  • $\begingroup$ Hint: A basis in the topological vector space $C^\infty(M)$ can be chosen to consist of functions whose support sets are as small as you like. Then use the fact that if $\delta E$ vanishes on a topological basis of the vector space $C^\infty(M)$, then $\delta E=0$. $\endgroup$ – Moishe Kohan Nov 25 '17 at 13:25
  • $\begingroup$ Thanks. Can you elaborate? I guess you refer to the case where the target is $N=\mathbb{R}$. What is the topology on $C^{\infty}(M)$ are you considering? $\endgroup$ – Asaf Shachar Dec 9 '17 at 15:10
  • $\begingroup$ I was only thinking about real-valued functions. $\endgroup$ – Moishe Kohan Dec 9 '17 at 22:34
  • $\begingroup$ I will write an answer in a day or so. $\endgroup$ – Moishe Kohan Dec 10 '17 at 2:19
2
$\begingroup$

Here is one way to solve this puzzle. Let ${\mathcal U}$ be the cover of $M$ by open subsets $U_p$ as in your question. Since $M$ is paracompact, without loss of generality, we may assume that this cover is locally finite. In particular, it admits a partition of unity $\{\eta_U\}_{U\in {\mathcal U}}$. The manifold $N$ admits an isometric embedding in some Euclidean space ${\mathbb R}^k$. Henceforth, I will consider $N$ as a submanifold of $E^k$ with the induced Riemannian metric. Then the energy of a map $$ f: M\to N $$ is the same as the energy of $f$ as a map $f: M\to {\mathbb R}^k$. One has to be careful here because $M$ is noncompact: We will have to take the energy of the restriction of $f$ to a relatively compact subset of $M$. To simplify the matters, let me simply assume that $M$ is compact, otherwise I have to repeatedly restrict to arbitrary relatively compact subsets.

For the purpose of variation of the energy functional we, of course, still have to restrict to maps $M\to N$. On the infinitesimal level, we will be computing $$ \delta E(f)(V) $$ where $V$ is a tangent vector field along $N$. It is convenient to regard $V$'s as defined on $M$ rather than on $N$, so technically speaking, they are sections of the pull-back of the tangent bundle: $f^*(TN)$. I will denote the space of such sections as ${\mathfrak X}_f(M,N)$.

Notice that I have no need to actually know the explicit form of the derivative of $E$. In particular, I do not even need to know what energy is, as long as $E$ is an integral $\int_M e(f)$, where $e(f)$ (which normally is the energy-density) is some (nonlinear) 1st order differential operator on maps $M\to {\mathbb R}^k$ (but I need to know that $E$ is differentiable in Gateaux's sense).

The condition that $f: M\to N$ is a critical point of $E$ is that for every tangent vector field $V$ along $N$ we have $$ \delta E(f)(V)=0. $$
Notice that $\delta E(f)(V)$ is linear in the variable $V$.

Now, suppose that I know only the vanishing property
$$ \delta E(f|_{U})(V)=0, \forall V\in {\mathfrak X}_f(U, N) $$
for each $U\in {\mathcal U}$. Given a vector field $V\in {\mathfrak X}_f(M, N)$, I define local vector fields $$ V_U:= \eta_U V $$ supported on $U$. Since $\{\eta_U\}$ is a partition of unity, we have $$ V= \sum_{U\in {\mathcal U}} V_U. $$ Then, by linearity, $$ \delta E(f)(V)= \sum_{U\in {\mathcal U}} \delta E(f|U)(V_U). $$ Each term in the right hand side of this equation vanishes since $f$ was a "local" critical point of $E$. Hence, $\delta E(f)(V)=0$ for every $V$ as above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.