Do harmonic maps into spaces of negative curvature locally minimize energy?

$\newcommand{\N}{\mathcal{N}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\g}{\mathfrak{g}}$ $\newcommand{\g}{\mathfrak{h}}$ $\newcommand{\IP}{\left\langle #1,#2 \right\rangle}$ $\newcommand{\Volg}{\operatorname{Vol}_\g}$ Let $\M,\N$ be oriented Riemannian manifolds, and suppose the sectional curvature of $\N$ is negative. (Suppose also $\M$ is compact and connected). Let $\phi:\M \to \N$ be harmonic map.

Is it true that $\phi$ is a local minimizer for the Dirichlet energy? i.e let $\phi_t$ be a smooth variation of $\phi$. Is it true that $E(\phi) \le E(\phi_t)$ for sufficiently small $t>0$?

(The sufficiently small interval $I$ of the "good" $t$'s can depend on the variation of course).

Let us restrict the discussion for variations $\phi_t$ whose variation fields $\left. \frac{\partial\phi_t}{\partial t} \right|_{t=0}$ are not identically zero.

I proved that if the rank of $d\phi$ is everywhere $\ge 2$, then $\phi$ is a local minimizer. (See my proof below).

Also, if $\text{rank }d\phi=0$ everywhere, i.e $\phi$ is constant, then $\phi$ is also a local minimizer (it's a global minimizer...).

So, the question remains: Suppose $\phi$ is a non-constant harmonic map which has at least one point of degree less than $2$. Is $\phi$ a local minimizer?

Proof that $\text{rank }d\phi \ge 2 \Rightarrow$ $\phi$ is locally minimizing:

We first note that $$\left. \frac{\partial^2}{\partial t^2} E(\phi_{t}) \right|_{t=0}=H_{\phi}(V,V),$$ where $H_{\phi}$ is the hessian of the energy functional at $\phi$, and $V=\left. \frac{\partial\phi_t}{\partial t} \right|_{t=0}$ is the corresponding variation field.

From the second variation formula, we obtain

$$\begin{split} H(E)_{\phi}(V,V)&= \int_{\M} \IP{J_{\phi}(V)}{V} \Volg \\ &=\int_{\M} -\sum_i \IP{R^{T\N}(V,d\phi(e_i))d\phi(e_i)}{V}+\IP{d_{\nabla^{\phi^*(T\N)}}V}{d_{\nabla^{\phi^*(T\N)}}V} \Volg \\ & \ge -\sum_i \int_{\M} \IP{R^{T\N}(V,d\phi(e_i))d\phi(e_i)}{V}. \end{split}$$

By our assumption (non-zero variation field), there exist $p \in \M$ such that $V_p \neq 0$. If $V_p,d\phi_p(e_i(p))$ are linearly dependent, then since $V_p \neq 0$, $d\phi_p(e_i(p)) \in \text{span} \{V_p\}$. Since we assumed $\text{rank }(d\phi_p) \ge 2$, not all the $d\phi_p(e_i(p))$ are linearly dependent of $V_p$; Thus, there exist an $1 \le i \le d$ where they are independent, hence $\IP{R^{T\N}(V,d\phi(e_i))d\phi(e_i)}{V} < 0$ by the curvature assumption.

Recall that the sectional curvature of the plane spanned by $x,y \in T_q\N$ is $$K(x,y)=\frac{\IP{R^{T\N}(x,y)y}{x} }{|x \wedge y|^2}.$$

• I do not understand what's the issue here: If on a small open subset $U\subset M$ you have $rank df\le 1$, $E(f|U)=0$, and, hence, under small variations of $f$ supported on $U$ the energy is minimized by $f$. The alternative is that $rank(df)\le 1$ on a measure zero subset, so it should be irrelevant for your purposes. – Moishe Kohan Dec 3 '17 at 21:26
• I am not sure what you are saying exactly. If there exist an open subset where $\text{rank} df =0$ (I think you accidentally wrote $\text{rank} df \le 1$) then as you say, $f$ is minimizing w.r.t small variations. But I asked whether or not $f$ is minimizing w.r.t all variations, not just "small" ones. – Asaf Shachar Dec 4 '17 at 6:05
• Why are you trying to avoid the low-rank case? If $d\phi(e_i)$ and $V$ are dependent then $R(V, d\phi(e_i))=0$, so you can still conclude $H_\phi(V,V) \ge 0$, no? Seems to me this would only be an issue if you were trying to establish some kind of strong stability $H_\phi(V,V) > 0.$ – Anthony Carapetis Dec 5 '17 at 12:23
• @AnthonyCarapetis Well, I am trying to show $\phi$ is a local minimizer. This is not always the case if the second derivative is merely zero instead of strictly positive, right? Even in the finite dimensional case, $f(x)=x^3$ has a critical point at $x=0$, and $f''(0)=0$, but $x=0$ is not a minimizer. Am I missing something? – Asaf Shachar Dec 5 '17 at 18:54
• Right, sure. This approach of just showing the second derivative is strictly positive at a critical point cannot work in general - you should expect that even when every harmonic map is a minimizer, there could be a "centre manifold". For example if $\mathcal N$ has a nontrivial Killing field you can push $\phi$ around by its flow to generate a smooth family of minimizers. – Anthony Carapetis Dec 5 '17 at 22:55

I will assume that the domain manifold $M$ is compact, otherwise, one needs to be much more careful with the definition of energy and which variations are admissible. However, the same argument works if $M$ is noncompact, but you have to restrict to variations which are trivial outside of a compact subset of $M$. It also works if $N$ is a locally CAT(0) space (the theory was worked out by Jost, Schoen and Korevaar in this setting). But, instead of working with the maps to $N$, you lift to a map between the unuversal covers and work equivariantly.

The key is the following convexity property of energy of maps to manifolds of nonpositive curvature:

Let $f_0, f_1: M\to N$ be smooth and let $f_t$ be a geodesic interpolation between these maps, i.e. for each $x\in M$ the curve $f_t(x)$ is a parameterized geodesic in $N$. Then the energy function $E(f_t)$ is convex as a function of $t$. This convexity property is already present in Eells-Sampson's 1964 paper on harmonic maps (the formula is more-or-less the one that you wrote): $$\frac{d^2}{dt^2} E(f_t)= 2\int_M \left( \sum_i \left[|\nabla^N_{e_i} V|^2 -K_N(V, (f_t)_*(e_i)) + \langle \nabla^N_{e_i} \nabla^N_{\frac {\partial}{\partial t}} V, (f_t)_*(e_i)\rangle \right] \right)dt.$$ Here $\{e_i(x)\}$ is an orthonormal frame at $T_x(M)$, $V$ is the variational vector field of $f_t$ which we pull-back to $M\times [0,1]$, $\nabla^N$ is the pull-back of the L-C connection on $TN$. The key is that $$\nabla^N_{\frac {\partial}{\partial t}} V= 0$$ since $f_t(x), t\in [0,1]$, is geodesic. Hence, the integrand is $\ge 0$.

Then, if $f_0$ is harmonic, for every $f_1$ near $f_0$ we can construct a geodesic interpolation $f_t$ (using compactness of $M$ to ensure that the injectivity radius along $f_0(M)$ is bounded below) and obtain that $\frac{d}{dt}E(f_t)=0$ at $t=0$.

Notice that I am not using the variation $\phi_t$ that you are considering, only $\phi_0=f_0, \phi_1=f_1$.

Convexity of $E(f_t)$ then will imply that $E(f_0)\le E(f_t)$ for all $t$. qed

• Thanks. Just to be clear- are you saying that the claim "harmonic maps are local minimizers" holds only for "geodesic variations" and not for all kinds of variations? Or does your proof demonstrates the more general claim? – Asaf Shachar Dec 5 '17 at 19:51
• @AsafShachar: The variation was arbitrary: You take an arbitrary smooth $f_1$ sufficiently close to $f_0$. But then I connect them by a different homotopy than the one you gave me, since I like my homotopy better. In the end, local minimality means that $E(f_0)\le E(f_1)$. – Moishe Kohan Dec 5 '17 at 20:46
• Thanks again for this answer. By the way, I have only seen (in the Eells-Sampson's 1964 paper and also elsewhere) the convexity statement for $E(f_t)$ when $f_t$ is satisfying the heat flow. I did not see this "geodesic convexity" version you stated here. (convexity of $E(f_t)$ along geodesic interpolations). Do you have a reference for that? (I guess you referred to their paper "Harmonic mappings of Riemannian manifolds", right?). – Asaf Shachar Jan 2 '18 at 10:29
• Also, I think you have a normalization typo in your formula; there is a missing factor of the area of the parallelogram spanned by $V,(f_t)_*e_i$ (in the second variation formula only the curvature appears directly). This is very important however. – Asaf Shachar Jan 2 '18 at 10:29