# An example for a stable harmonic map which is not a local minimizer

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).

• Hint: Such examples already exist among geodesics $f: S^1\to N$. – Moishe Kohan Jan 25 '18 at 9:46
• Hmmm... I guess you mean to take geodesics on some compact surface with negative curvature? Can you say what surface you had in mind? – Asaf Shachar Jan 25 '18 at 10:18
• 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). – Moishe Kohan Jan 25 '18 at 16:49

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.
• 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). – Asaf Shachar Jan 28 '18 at 8:09
• @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. – Anthony Carapetis Jan 28 '18 at 11:00
• 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. – Moishe Kohan Jan 28 '18 at 11:45
• @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?). – Asaf Shachar Jan 30 '18 at 6:44