2
$\begingroup$

I am looking for an example of a harmonic mapping between compact Riemannian manifolds, which is stable but not a local minimizer of the Dirichlet energy.

A harmonic map $f:M \to N$ is said to be stable if the hessian of the energy at $f$ is non-negative.

(This is mentioned in "Two reports on harmonic maps", by Eells and Lemaire, without a proof or a reference).

$\endgroup$
  • $\begingroup$ Hint: Such examples already exist among geodesics $f: S^1\to N$. $\endgroup$ – Moishe Kohan Jan 25 '18 at 9:46
  • $\begingroup$ Hmmm... I guess you mean to take geodesics on some compact surface with negative curvature? Can you say what surface you had in mind? $\endgroup$ – Asaf Shachar Jan 25 '18 at 10:18
  • 1
    $\begingroup$ Actually, a surface of mixed curvature, such as the surface of revolution which would be (locally) the graph of, say, $y=x^3+1$ (rotate about the x-axis). $\endgroup$ – Moishe Kohan Jan 25 '18 at 16:49
3
$\begingroup$

Take the surface of revolution $M\subset R^3$ obtained by rotating the curve $$ x^{2n} + y^{2n}=1 $$ ($n\ge 2$) around the $x$-axis. On the surface $M$ take the closed curve $C$ $$ y^2+z^2=1 $$ parameterized by its arclength. This curve will give you a stable minimal map $S^1\to M$ which is not a local minimum of the energy functional.

$\endgroup$
  • $\begingroup$ Thanks. Just to make sure I understand: The reason your geodesic is stable is because of this inequality, right? I am also not sure why is it not a local minima. (The length of the curve $x^{2n} + y^{2n}=1$ is bigger than the length of a meridian, which is the length of the original curve you rotated). $\endgroup$ – Asaf Shachar Jan 28 '18 at 8:09
  • 1
    $\begingroup$ @AsafShachar: another description of $C$ is simply $x = 0$. Nearby curves $x = \epsilon$ are circles of radius $(1-\epsilon^{2n})^{1/2n},$ which are strictly shorter; but the amount by which they are shorter is $o(\epsilon^2),$ explaining the failure of the Jacobi operator to detect them. $\endgroup$ – Anthony Carapetis Jan 28 '18 at 11:00
  • $\begingroup$ There are several ways to see that $C$ is a geodesic. One is that it is the fixed point set of an isometric reflection $M\to M$ (restriction of the reflection in the yz-plane). You can also see this by realizing that along the circle $C$ the surface $M$ has contact of order $\ge 4$ with the cylinder $y^2+z^2=1$ in $R^3$. This fact proves that $C$ is a geodesic on $M$ and that it is stable. $\endgroup$ – Moishe Kohan Jan 28 '18 at 11:45
  • $\begingroup$ @AnthonyCarapetis Thanks. Just to make sure: The paths $x=\epsilon$ are not geodesics, right? (I think I got confused, since I tried to compare $x=0$ to other nearby geodesics, but this is not required, right?). $\endgroup$ – Asaf Shachar Jan 30 '18 at 6:44
  • 2
    $\begingroup$ Right, they're not geodesics - whenever we talk about locally minimizing geodesics (harmonic maps, minimal surfaces, etc.) we mean minimizing amongst all nearby curves, not just nearby geodesics. $\endgroup$ – Anthony Carapetis Jan 30 '18 at 9:39

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.