Notes of Yves Benoist’s Southampton lecture 29-03-2017

Harmonic quasisometric maps

Joint with Dominique Hulin.

1. The result

We deal with Hadamard manifolds, i.e. complete simply connected nonpositively curved Riemannian manifolds. Say it is pinched if sectional curvature is between two negative constants.

Harmonic maps are critical points of the energy functional {\frac{1}{2}\int |Df|^2}. They satisfy the second order PDE trace{(D^2 f)=0}.

Theorem 1 Let {X} and {Y} be pinched Hadamard manifolds with dim{(X)\geq 2}. Every quasi-isometric embedding {X\rightarrow Y} is within bounded distance of a unique harmonic map.

If dim{(X)=1}, the theorem fails. Indeed, a bi-Lipschitz parametrization of a geodesic is a quasi-isometry which is not within bounded distance of the harmonic geodesic whose parametrization has constant speed.

2. History and first steps

Motivated by quasi-isometric rigidity of symmetric spaces, Schoen-Li-Wang conjectured this fact in 1995 for rank one symmetric spaces. Markovic-Lemm proved it for {X=Y=H^n_{\mathbb R}}, in a series of 3 papers.

Note that if curvature is not pinched, one gets into trouble. Indeed, let {X=H^2_{\mathbb R}} and {Y=X\times{\mathbb R}}. The map

\displaystyle  \begin{array}{rcl}  x\mapsto f(x)=(x,d(x_0,x)) \end{array}

is a quasi-isometric embedding. Every harmonic map {X\rightarrow X\times{\mathbb R}} is a pair {(h_1,h_2)} where {h_2} is a harmonic function, asymptotic to {d(x_0,\cdot)}. Hence {h_2} achieves a minimum, contradiction. Note that every higher rank symmetric space contains isometric copies of {X\times{\mathbb R}}, hence the counterexample is ubiquitous.

Yet, the Dirichlet problem has a unique solution on every bounded subset of {X}. One sees that, when restricting to a ball {B(x_0,R)} in {X} and solving the Dirichlet problem, the solution is {h_2\equiv R}, which is far away from {f}.

Thus the key point is to get control of the size of harmonic maps. The first step is the fact that the distance between two harmonic maps is a subharmonic function. Hence it must be large somewhere. This applies to contant maps. The second is an apriori estimate, due to Cheng, of the size of the derivative at the center of a unit ball in terms of the diameter of the image of this ball.

3. Proof of existence

One can assume that {f} is smooth with bounded first and second derivatives. Solve the Dirichlet problem on {B(x_0,R)}, get {h_R}.

Proposition 2 There exists {M>0} such that for all {R>0},

\displaystyle  \begin{array}{rcl}  \sup_{x\in B(x_0,R)}d(f(x),h_R(x))\leq M. \end{array}

Once this is done, with Cheng’s estimate, a subsequence converges to a global harmonic map within distance {M} of {f}.

3.1. First, a boundary estimate

Inspired by J. Jost.

Lemma 3

\displaystyle  \begin{array}{rcl}  d(f(x),h_R(x))\leq M_0(R-d(x_0,x)). \end{array}

Indeed, fix {y\in Y} and set

\displaystyle  \begin{array}{rcl}  G(x)=d(y,h_R(x))-d(y,f(x))-M_0(R-d(x_0,x)). \end{array}

As we have seen, {x\mapsto d(y,h_R(x))} is subharmonic, {x\mapsto d(y,f(x))} has bounded second derivatives, {x\mapsto d(x_0,x)} has large positive Laplacian, due to pinched curvature. So for {M_0} large enough, {G} is subharmonic, vanishes on the boundary, so it is nonpositive everywhere.

3.2. The interior estimate

The argument is geometric. Choose {x} that maximizes {d(f(x),h_R(x))=\rho}. By contradiction, assume that {\rho\gg\ell} are very large, and focus on the images of {S(x,\ell)}. Let {z\in S(x,\ell)}. Viewed from {y=f(x)}, consider angles {\theta_1} (resp. {\theta_2}) of {f(z)} and {h_R(z)} (resp. {h_R(z)} and {h_R(x)}). Let

\displaystyle  \begin{array}{rcl}  U=\{z\in S(x,\ell)\,;\,d(y,h_R(z))\geq \rho-\frac{\ell}{2C}\}. \end{array}

Lemma 4

  1. For all {z\in U}, {\theta_1} is small,

    \displaystyle  \begin{array}{rcl}  \theta_1\leq 4 e^{aC/2}e^{-al/2C}, \end{array}

    by hyperbolic trigonometry.

  2. For all {z\in U}, {\theta_2} is small,

    \displaystyle  \begin{array}{rcl}  \theta_2\leq 100 e^{aC/2}e^{-al/2C}, \end{array}

    by Cheng’s estimate played against subharmonicity of distance to {y}.

  3. {U} is not small. Its measure (on the sphere) is bounded below by {1/3C^2}.

From the Lemma, the angle {\theta_0} of {f(z)} and {h_R(x)} is not small, since {f} is quasi-isometric and its boundary value is a quasi-symmetric homeomorphism. This contradicts

\displaystyle  \begin{array}{rcl}  \theta_0\leq\theta_1+\theta_2. \end{array}


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