# Uniqueness of Equivariant Harmonic Map for Surface Group Representation

In section 1.2 of https://arxiv.org/pdf/1311.2919.pdf the following result is stated.

$$\textbf{Theorem}$$ (Labourie). Let $$S$$ be a closed Riemann surface of negative Euler characteristic, $$Γ$$ its fundamental group, $$(M, g)$$ a complete simply connected Riemannian manifold of sectional curvature $$≤ −1$$, and $$\rho$$ a representation of $$Γ$$ into $$\textrm{Isom}(M, g)$$. If $$\rho$$ does not fix a point in the boundary of $$M$$, or if $$\rho$$ fixes a geodesic in $$M$$, then there exists a $$ρ$$-equivariant harmonic map from $$\tilde{S}$$ to $$M$$. If $$\rho$$ does not fix a point in the boundary, this map is unique.

The statement above is a special case of Labourie's result in "Existence D'Applications Harmoniques Tordues à Valeurs Dans les Variétés à Courbure Négative" (my French is okay, so I could understand the paper). In his result, there is a technical condition on the geometry of $$M$$, and also $$\rho$$ is assumed to be reductive.

Let us assume for convenience that $$M=\mathbb{H}^n$$ (where I'm under the impression that the condition on $$M$$ is fulfilled).

$$\textbf{Question 1)}$$ How does "$$\rho$$ is reductive" correspond to not fixing a point on the boundary, or fixing a geodesic.

The above question is probably not hard, but I really don't know much about reductive groups past the basics.

Also, I cannot find a proof of the uniqueness statement in Labourie's paper. I did look in Corlette's paper, "Flat G-bundles with canonical metrics" but the language is quite different and I cannot see a proof of uniqueness there. So I ask

$$\textbf{Question 2)}$$ Why, under the hypothesis on $$\rho$$, is the map unique?

I have some ideas. The harmonic map descends to a harmonic map from $$S\to M/\rho$$. If $$M/\rho$$ is a manifold, then by a result of Sampson ("Some properties and applications of harmonic mappings," theorem 4), it is enough to show any two such maps are homotopic, when viewed as maps on $$S$$. I don't see why this should be the case.

I see an example of how you could not get uniqueness in the case that $$\rho$$ fixes a geodesic: take $$M$$ to be the disk with the Poincare metric and take $$\rho$$ to be a representation that commutes with the reflection across some geodesic. Then composing an equivariant harmonic map with this reflection gives a different harmonic map.

• Take a look at my answer here: math.stackexchange.com/questions/2548499/…. It contains all the ingredients that you need. – Moishe Kohan May 7 at 3:20
• Hi. I actually don't see how this addresses my first question. Please let me know if I'm missing something. As for the second question, maybe I'm being slow, but I don't understand why this helps. Sure, the method of Labourie (which is really a modified argument of Donaldson) uses the heat flow of Eels and Sampson, and so the limiting harmonic map is homotopic to a chosen initial map, but I can't quite finish an argument from there. – user470881 May 7 at 15:05
• My argument is a bit different and it uses geodesic interpolation instead of the heat flow, I will write more details when I have time. For now: A representation to $O(n,1)$ is reductive if the Zariski closure of the image is a reductive subgroup. The only non-reductive algebraic subgroups of $O(n,1)$ are ones which have precisely one fixed point in $S^{n-1}$: This is a nice exercise in hyperbolic geometry. Maybe Labourie also excludes the case of an invariant geodesic, but this is unnatural from the Lie theory viewpoint. – Moishe Kohan May 7 at 15:20
• Okay cool thanks a lot. Looking forward to hearing the details. – user470881 May 7 at 17:24