Somewhere injective harmonic maps

Chapter VI: Moduli spaces of harmonic surfaces

6.4 Somewhere injective harmonic maps

construction, we can choose𝜇_𝑛and𝑆_𝑛to vary nicely with𝑧for each𝑛, and then we get the correct regularity in the limit.

Observe

𝜕

𝜕 𝑧

⟨𝑉(𝑧), 𝑈(𝑧)⟩ =⟨∇𝑧𝑉(𝑧), 𝑈(𝑧)⟩ + ⟨𝑉(𝑧),∇𝑧𝑈(𝑧)⟩

in Ω^′. In terms of our integrals, differentiating under the integral via dominated convergence, we get

𝜕

𝜕 𝑧

∫

Σ

⟨J𝑉(𝜁), 𝑋(𝑧, 𝜁)⟩ =

∫

Σ

⟨J𝑉(𝜁),∇𝑧𝑋(𝑧, 𝜁)⟩ = ⟨∇𝑧𝑉(𝑧), 𝑈(𝑧)⟩+⟨𝑉(𝑧),∇𝑧𝑈(𝑧)⟩. Setting𝑧 =0, we find that the first order kernel is given by∇𝑧𝑋(0, 𝜁). From this we deduce the following.

Proposition 6.3.7. In the complex coordinate 𝑧, the reproducing kernel is of the form

𝑋(𝑧) = 1 𝜋 𝑧

𝑈(𝑝) +𝐵(𝑧) (6.7)

where𝐵(𝑧) is a𝐶^0,𝛼 local section ofEnear 𝑝, for any𝛼∈ (0,1).

Lemma 6.4.2. Suppose there is a pair of disks Ω₁,Ω₂ ⊂ Σ and a conformal diffeomorphismℎ:Ω₁→Ω₂such that 𝑓 ◦ℎ = 𝑓 onΩ₁. Then the Riemann surface Σis exceptional.

Proof. According to [Sag21a, Theorem 1.1], ifℎ :Ω₁→ Ω₂is a holomorphic map between open subsets ofΣsuch that 𝑓◦ℎ= 𝑓, then 𝑓 factors through a holomorphic branched covering map onto a surface Σ₀. If Σ₀ has genus less than 2, then it is either a sphere or a torus. In both cases, the subgroup

𝑓_∗(𝜋

1(Σ)) < 𝜋

1(𝑀)

is abelian, which contradicts our assumption that the homotopy classfis admissible.

Ifℎis anti-holomorphic, the result follows from Theorem 1.1 and the discussion in

Section 4 of [Sag21a]. □

This next result is well understood and one can find details in [Mar18, Appendix B].

Proposition 6.4.3. We let 𝜇 ∈ 𝔐^′(Σ) ifΣ𝜇 is not exceptional. 𝔐^′(Σ) is an open, dense, and connected subset of𝔐(Σ).

For ease of notation, we write 𝔐 = 𝔐^′(Σ) ×𝔐(𝑀) instead of 𝔐(Σ) ×𝔐(𝑀) throughout the rest of the chapter.

Super-regular points

Denote by 𝐴(𝑓) the set of 𝑝 ∈ Σ such that 𝑓⁻1(𝑓(𝑝)) ⊂ Σ^{𝑟 𝑒𝑔}(𝑓). Given metrics (𝜇, 𝜈)and𝑝, 𝑞 ∈ 𝐴(𝑓), we say that the inner products𝜇(𝑝)and𝜇(𝑞)are conformal to each other via 𝑓 if the tangent planes𝑑 𝑓(𝑇_𝑝Σ)and𝑑 𝑓(𝑇_𝑞Σ)agree in𝑇_𝑓(𝑝)𝑀, and if the push forwards 𝑓_∗𝜇(𝑝) and 𝑓_∗𝜇(𝑞)are collinear.

Definition 6.4.4. Given a map 𝑓, a point 𝑝 ∈Σis said to be super-regular if

• 𝑝 ∈ 𝐴(𝑓)and

• if 𝑓(𝑝) = 𝑓(𝑞), then𝜇(𝑝)and 𝜇(𝑞) are not conformal to each other via 𝑓. We denote the set of super-regular points for a map 𝑓 by Σ^SR(𝑓). We define SR ⊂ Σ×𝔐 by(𝑝, 𝜇, 𝜈) ∈ SRif𝑝 ∈ Σ^SR(𝑓_𝜇,𝜈).

Proposition 6.4.5. Continuing to exclude the exceptional metrics from𝔐, the set SR is open inΣ×𝔐 andΣ^SR(𝑓)is open and dense inΣ.

We first treat 𝐴(𝑓) on its own. It is due to Sampson [Sam78, Theorem 3] that the set of regular points of an admissible harmonic map is open and dense.

Lemma 6.4.6. 𝐴(𝑓) is open and dense inΣ.

Proof. Openness is obvious. As for density, suppose on the contrary that there is an open setΩ ⊂ Σon which 𝑓 is regular but no point is in𝐴(𝑓). By shrinkingΩwe may assume 𝑓|_Ωis an embedding. We then find a small tubular neighbourhood𝑁 ⊂ 𝑀 of the submanifold 𝑓(Ω), in which the nearest point projection 𝜋 : 𝑁 → 𝑓(Ω) is well-defined. The set𝑆 = 𝑓⁻¹(𝑁) ⊂Σ is then an open submanifold ofΣ.

Let𝑔 = 𝜋 ◦ 𝑓 : 𝑆 → 𝑓(Ω). If 𝑦 ∈ 𝑆 is a singular point of 𝑓, then it is a singular point of 𝑔. By assumption, for each𝑢 ∈ 𝑓(Ω), the set 𝑓⁻¹(𝑢) contains a singular point of𝑔. Thus, each point in 𝑓(Ω) is the image of a singular point 𝑦 ∈ 𝑆 of the

map𝑔. This contradicts Sard’s theorem. □

Proof of Proposition 6.4.5. It is clear that bothΣ^SR(𝑓)andSRare open. It remains to prove Σ^SR(𝑓) is dense. Note that the set 𝑓⁻1(𝑓(𝑥)) is finite provided 𝑥 ∈ 𝐴. Indeed, if |𝑓⁻1(𝑓(𝑥)) | = ∞, then the closed set 𝑓⁻1(𝑓(𝑥)) has an accumulation point, at which the rank of𝑑 𝑓 is necessarily strictly less than two (as 𝑓 cannot be an embedding near that point).

From the previous lemma, we are left to show that the conformality condition holds on a dense subset of 𝐴(𝑓). Arguing by contradiction, suppose that on an open subsetΩ ⊂ 𝐴 we have that for every 𝑝 ∈ Ω there exists 𝑞 ∈ 𝑓⁻1(𝑓(𝑝)) such that 𝜇(𝑝) and 𝜇(𝑞) are conformal to each other via 𝑓. Given 𝑝 ∈ Ω, we have a finite number of disks 𝐷

1, . . . , 𝐷_𝑛with centers 𝑝_𝑖 such that 𝑓(𝑝) = 𝑓(𝑝_𝑖)and with 𝜇(𝑝) and𝜇(𝑝_𝑖)conformal via 𝑓. We also assume 𝑓 is a regular embedding on𝐷_𝑖andΩ. Let𝐶_𝑖 ⊂ 𝐷_𝑖 be the closed set of points𝑥 ∈ 𝐷_𝑖 with the property that there exists 𝑦 ∈ Ωwith 𝑓(𝑥) = 𝑓(𝑦) and such that 𝜇(𝑥) is conformal to𝜇(𝑦) via 𝑓. We claim that for at least one𝑖,𝐶_𝑖has non-empty interior. If not, then

C =∪𝑖𝑓(𝐶_𝑖) ∩ 𝑓(Ω)

has empty interior, for it is a finite union of closed nowhere dense sets. Choosing a sequence(𝑝_𝑛)^∞

𝑛=1 ⊂ Ω\(𝑓⁻1(C) ∩Ω)converging to𝑝, we can find another sequence (𝑞_𝑛)_𝑛=^∞

1 ⊂ Σ\(∪𝑖𝐷_𝑖) with 𝑓(𝑝_𝑛) = 𝑓(𝑞_𝑛) and 𝜇(𝑝_𝑛) and 𝜇(𝑞_𝑛) are conformal via 𝑓. Passing to a subsequence, the 𝑞_𝑛 converge to some point 𝑞 ∈ Σ\(∪𝑖𝐷_𝑖) such

that 𝑓(𝑝) = 𝑓(𝑞) and 𝜇(𝑝) and 𝜇(𝑞) are conformal via 𝑓. This contradicts our construction of the𝐷_𝑖, and so the claim is proved.

Relabelling so that 𝑓(Ω) and 𝑓(𝐷

1) intersect with non-empty interior, we can find open setsΩ₁ ⊂ ΩandΩ₂ ⊂ 𝐷

1as well as a diffeomorphismℎ:Ω₁ →Ω₂such that 𝑓 ◦ℎ = 𝑓 onΩ₁. The metrics 𝜇andℎ^∗𝜇are pointwise conformally equivalent on Ω, and thus ℎis a conformal map. This contradicts Lemma 6.4.2. □ The mapΘ

Denote byJ the subset of nowhere injective maps.

Lemma 6.4.7. J is closed.

Proof. If a somewhere injective map 𝑓 (which need not be harmonic) has an injective point at 𝑝, meaning 𝑓⁻1(𝑓(𝑝)) = {𝑝}, then there is an open set containing 𝑝 that consists only of injective points. Indeed, choose a diskΩaround 𝑝 on which 𝑓 is regular and 𝑓|_Ω is injective. If the claim fails, we can find𝑝_𝑛 → 𝑝and 𝑞_𝑛 ∈ Σ\Ω such that 𝑓(𝑝_𝑛) = 𝑓(𝑞_𝑛). By compactness, the𝑞_𝑛subconverge to a point𝑞at which

𝑓(𝑝) = 𝑓(𝑞), a contradiction.

So, suppose ( (𝜇_𝑛, 𝜈_𝑛))^∞

𝑛=1 ⊂ J converges to (𝜇, 𝜈), and 𝑓 = 𝑓_𝜇,𝜈 is somewhere injective with injective point 𝑝. There is a disk Ω around 𝑝 such that 𝑓_𝜇

𝑛,𝜈_𝑛 is injective onΩ. Thus, there exists 𝑝_𝑛 ∈ Σ\Ωsuch that 𝑓_𝜇

𝑛,𝜈_𝑛(𝑝_𝑛) = 𝑓_𝜇

𝑛,𝜈_𝑛(𝑝), and again we find a contradiction by extracting an accumulation point𝑞 ≠ 𝑝. □ For the remainder of Sections 6.4 and 6.5, let us replace 𝔐 with the complement of the set of pairs with exceptional metrics onΣ. We hope this harmless change of notation does not cause any confusion.

Let𝑃, 𝑄 ⊂ Σbe two disjoint open embedded disks inΣ. For𝛿 >0 we let D (𝑃, 𝑄 , 𝛿)={(𝜇, 𝜈) ∈𝔐 :𝑑_𝜈(𝑓(𝑃), 𝑓(𝑄)) > 𝛿, 𝑃, 𝑄 ⊂ Σ^SR(𝑓)}.

It follows from Proposition 6.4.5 that D (𝑃, 𝑄 , 𝛿) is an open subset of 𝔐. By Proposition 6.4.5 we also know that the setΣ^SR(𝑓_𝜇,𝜈) is dense in Σ, and therefore non-empty. Thus, each pair (𝜇, 𝜈) ∈ 𝔐 is contained inD (𝑃, 𝑄 , 𝛿) for some disks 𝑃, 𝑄and𝛿 >0.

Lemma 6.4.8. Let 𝐴 ⊂ 𝔐 be a subset. Suppose every pair (𝜇, 𝜈) ∈ 𝔐 has a neighbourhood D ⊂ 𝔐such that D\𝐴is open, dense, and connected in D. Then 𝔐\𝐴is open, dense, and connected in𝔐.

The proof is trivial point-set topology and left to the reader. Thus, toward Theorem 6A it suffices to prove that everyD\J is connected, whereDranges over connected components of D (𝑃, 𝑄 , 𝛿). Henceforward we work on a single such component D. SetΣ² = Σ×Σ, 𝑀² = 𝑀 × 𝑀, andY = Σ²× (𝑃×𝑄 × D). Define the map Θ:Y →𝑀²by

Θ(𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈)= (𝑓(𝑟), 𝑓(𝑠), 𝑓(𝑝), 𝑓(𝑞))

where we abbreviate 𝑓 = 𝑓_𝜇,𝜈. As we have noted earlier, the map (𝜇, 𝜈) ↦→ 𝑓_𝜇,𝜈 is 𝐶^𝑘 and the evaluation map has the same regularity as 𝑓. We deduceΘis𝐶^𝑚, where 𝑚 =min{𝑘 , 𝑛+1}.

Let𝐿 be the diagonal

𝐿 ={( (𝑢, 𝑣),(𝑢, 𝑣)) ∈ 𝑀²×𝑀²}. The significance ofΘand 𝐿is contained in the fact that

𝜋⁻¹(J ) ⊂ Θ⁻¹(𝐿),

where𝜋 :Y → Dis the projection onto the last factor. Indeed, suppose(𝜇, 𝜈) ∈ J. Then for each pair of points (𝑝, 𝑞) ∈ 𝑃 ×𝑄 there exists (𝑟 , 𝑠) ∉ 𝑃×𝑄 such that

𝑓(𝑝) = 𝑓(𝑟) and 𝑓(𝑞) = 𝑓(𝑠). ThusΘ(𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈) ∈ 𝐿.

Remark 6.4.9. 𝜋⁻1(J )also contains the set𝐿_𝑃,𝑄×𝔐, where𝐿_𝑃,𝑄 is the diagonal of𝑃×𝑄. This set has codimension 4 and will not play a role in any of our analysis.

Proof of Theorem 6A

Assuming the transversality lemma below, we prove Theorem 6A.

Lemma 6.4.10. Let(𝜇, 𝜈)be a pair of metrics with 𝜇not exceptional. Then for all (𝑟 , 𝑠, 𝑝, 𝑞) ∈ Y such thatΘ(𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈) ∈ 𝐿,Θis a submersion at that point. In particular,Θis transverse to 𝐿at such points.

Via Lemma 6.4.10, we can shrink D (𝑃, 𝑄 , 𝛿) so that Θ is transverse to 𝐿 on all of Y × D. Beginning the proof of Theorem 6A, it is enough to show it is dense and connected. If (𝜇, 𝜈) yields a somewhere injective harmonic map, then by openness there is nothing to do. According to Lemma 6.4.2, we can also dismiss pairs(𝜇, 𝜈) such that𝜇is exceptional. Henceforth fix(𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈)such that𝜇is non-exceptional, and 𝑓 = 𝑓_𝜇,𝜈is nowhere injective. Define𝜃_{𝑝,𝑞, 𝜇,𝜈} :Σ² → 𝑀2×𝑀2

by

𝜃_{𝑝,𝑞, 𝜇,𝜈}(𝑟 , 𝑠) = Θ(𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈).

The map 𝜃_{𝑝,𝑞, 𝜇,𝜈} is a direct sum of evaluations of 𝑓_𝜇,𝜈, and hence 𝐶^𝑛+1. The evaluation map is just Θ, and hence it is 𝐶^𝑚, with 𝑚 as above. Therefore, the association

(𝑝, 𝑞, 𝜇, 𝜈) ↦→ 𝜃_{𝑝,𝑞, 𝜇,𝜈}

defines a 𝐶^𝑚 representation. 𝐿 is of codimension 2 dim𝑀 ≥ 6, while Σ² has dimension 4, so the Parametric Transversality Theorem ensures that for a generic (𝑝, 𝑞, 𝜇, 𝜈) ∈ 𝑃×𝑄× D the corresponding map 𝜃_{𝑝,𝑞, 𝜇,𝜈} is transverse to 𝐿. From dimensional considerations,𝜃_{𝑝,𝑞, 𝜇,𝜈}(Σ²)is therefore disjoint from𝐿. It follows that (𝜇, 𝜈) does not belong toD ∩ J. This is true for generic(𝑝, 𝑞, 𝜇, 𝜈), and therefore a generic pair (𝜇, 𝜈) does not live in D ∩ J. This establishes that J is nowhere dense.

Recalling the projection 𝜋 : Y → D, since 𝜋(Θ⁻¹(𝐿)) may not be a manifold, we cannot conclude immediately from transversality thatD ∩ J is connected. We argue directly, using transversality theorems.

Let𝛾 : [0,1] → D be a path whose endpoints lie inD \ J. We show that 𝛾 can be perturbed, while keeping the endpoints fixed, to lie entirely in𝔐\ J. First we partition [0,1] into sufficienltly small intervals, each of whose images under 𝛾 is contained in a sufficiently small subset ofD that fits into a single (convex) chart in the model Banach space for𝔐.

Suppose (𝜇_𝑖, 𝜈_𝑖) ∈ D, 𝑖 = 0,1, are contained in such a chart. We show that one can perturb (𝜇

1, 𝜈

1) so that the straight line connecting (𝜇

0, 𝜈

0) and the perturbed (𝜇

1, 𝜈

1)is contained in D \ J. Let𝑈 ⊂ 𝑃×𝑄× D be a small neighbourhood of (𝜇

1, 𝜈

1)and consider the𝐶^𝑚 map

𝜂: [0,1] ×Σ²×𝑈 → 𝑀²× 𝑀² given by

𝜂(𝑡 , 𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈) = Θ(𝑟 , 𝑠, 𝑝, 𝑞, 𝑡(𝜇

0, 𝜈

0) + (1−𝑡) (𝜇, 𝜈)).

𝜂 arises as the evaluation map for the𝐶^𝑚 representation given by (𝑝, 𝑞, 𝜇, 𝜈) ↦→

𝜌_{𝑝,𝑞, 𝜇,𝜈}, where

𝜌_{𝑝,𝑞, 𝜇,𝜈}(𝑡 , 𝑟 , 𝑠)=𝜂(𝑡 , 𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈).

The representation does take values in𝐶^𝑚 because it may be realized as the compo- sition

(𝑡 , 𝑟 , 𝑠) ↦→ (𝑡(𝜇

0, 𝜈

0) + (1−𝑡) (𝜇, 𝜈), 𝑟 , 𝑠) ↦→ (𝑓_𝑡, 𝑟 , 𝑠) ↦→ (𝑓_𝑡(𝑝), 𝑓_𝑡(𝑞), 𝑓_𝑡(𝑟), 𝑓_𝑡(𝑠)),

where 𝑓_𝑡 = 𝑓_𝑡(𝜇

0,𝜈

0)+(1−𝑡) (𝜇,𝜈). Using 𝑚 ≥ 1 we appeal to the Parametric Transver- sality Theorem to find that for a generic point (𝑝, 𝑞, 𝜇, 𝜈) ∈ 𝑈, the map 𝜌_{𝑝,𝑞, 𝜇,𝜈} is transverse to 𝐿. Counting dimensions, we see that𝛿( [0,1] ×Σ²) does not intersect 𝐿. This implies that the path

𝜋(𝜂(𝑡 , 𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈)) is contained inD \ J and connects (𝜇

0, 𝜈

0)and (𝜇, 𝜈). SinceD\ J is open, once (𝜇, 𝜈) is close enough, we can connect it to (𝜇

1, 𝜈

1) via a straight line in a model chart contained inD\ J.

Returning to the path𝛾, we do the above procedure over all of the coordinate charts, which perturbs 𝛾 to a new path contained entirely in D\ J and connecting the endpoints. This completes the proof.

Remark 6.4.11. A slightly simpler transversality argument is possible when𝑀has dimension≥ 4. We leave this for the reader to understand on their own. The proof of the analogue of Lemma 6.4.10 is essentially the same.

6.5 Proof of the transversality lemma