Proof of the transversality lemma

Chapter VI: Moduli spaces of harmonic surfaces

6.5 Proof of the transversality lemma

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

Here,G ( ¤𝜈)is the derivative of the tension field in the𝜈¤-direction:

G ( ¤𝜈) = 𝑑 𝑑 𝑡

|_𝑡=0𝜏(𝜇, 𝜈+𝑡𝜈, 𝑓¤ _𝜇,𝜈). Accordingly,𝑉 is called a harmonic variation. It follows that

𝑑Θ(0,0,0,0,0,𝜈¤)= (𝑉(𝑟), 𝑉(𝑠), 𝑉(𝑝), 𝑉(𝑞)).

Suppose Θ(𝑟 , 𝑠, 𝑝, 𝑞, 𝜇, 𝜈) ∈ 𝐿. To simplify notation, we rename the points as 𝑧1= 𝑝,𝑧

2=𝑟,𝑤

1 =𝑞,𝑤

2=𝑠. The proof of Lemma 6.4.10 is another contradiction argument. Suppose the lemma is incorrect. Then, there are four vectors 𝑍_𝑖 ∈ F𝑧_𝑖, 𝑊_𝑖 ∈F𝑤𝑖, not all of them zero, such that

• 𝑍_𝑖and𝑊_𝑖are either zero or normal to the surface 𝑓(Σ)at the points 𝑓(𝑧_𝑖)and 𝑓(𝑤_𝑖) respectively and

• for every𝑉 such thatJ𝑉 =G ( ¤𝜈), the following holds:

∑︁2

𝑖=1

(⟨𝑉(𝑧_𝑖), 𝑍_𝑖⟩ + ⟨𝑉(𝑤_𝑖), 𝑊_𝑖⟩) =0. (6.8)

We now invoke reproducing formulas for the zeroth derivative. We showed the existence of reproducing kernels in Section 6.3. Adding up the four zeroth or- der reproducing kernels associated to the points 𝑧_𝑖, 𝑤_𝑖, we find a section 𝑋 : Σ\{𝑧

1, 𝑧

2, 𝑤

1, 𝑤

2} →Fwith maximum regularity and such that

∑︁2

𝑖=1

(⟨𝑊(𝑧_𝑖), 𝑍_𝑖⟩ + ⟨𝑊(𝑤_𝑖), 𝑊_𝑖⟩) =

∫

Σ

⟨J𝑊 , 𝑋⟩𝑑𝐴

for all𝑊 ∈ Γ(F). We also record here thatJ𝑋(𝑝) =0 for every 𝑝 ≠ 𝑧

1, 𝑧

2, 𝑤

1, 𝑤

2

and 𝑋 ∈ 𝐿^𝑝(F) for every 𝑝 ≥ 1. 𝑋 is not identically equal to zero as we can certainly find sections𝑊 ∈ Γ(F) such that the left-hand side above is not zero. On the other hand, from (6.8) we conclude that

∫

Σ

⟨J𝑉 , 𝑋⟩𝑑𝐴 =0 for every harmonic variation𝑉.

Stepping back for a moment, if 𝜈¤ has support near 𝑓(𝑧

1), then the associated J𝑉 is supported near all preimages of 𝑓(𝑧

1). The kernel 𝑋 may have singularities at 𝑧1 and𝑧

2, while 𝑋 is smooth at the other preimages of 𝑓(𝑧

1). The tangent planes

𝑑 𝑓(𝑇_𝑧

1Σ)and𝑑 𝑓(𝑇_𝑧

2Σ)are either tangential or span a𝑘-plane for𝑘 =3 or 4, and we find it convenient to treat the cases separately. In both cases, it is possible to choose

¤

𝜈 so that G ( ¤𝜈) is negligible at 𝑧

2 but not so at 𝑧

1. In the tangential case, we use the argument from [Mar18, Section 7]. This is where the super-regular condition comes into play (and this is the only place it does). In this way, we can eliminate the singularity of 𝑋 at𝑧

1. Repeating the procedure, but interchanging the roles of 𝑧1and𝑧

2, we’re able to show that 𝑋is a global Jacobi field, which means 𝑋 ≡0.

The time derivative of the tension field

We compute⟨J𝑉 , 𝑋⟩in coordinates for a general variation𝜈¤. We let(𝑥

1, 𝑥

2)and(𝑢

1, . . . , 𝑢_𝑛)denote local coordinates near𝑧

1 ∈Ωand 𝑓(𝑧

1) ∈ 𝑀 such that𝑧

1= (0,0), 𝑓(𝑧

1) = (0, . . . ,0). Near𝑧

1, the reproducing kernel 𝑋 can be expressed as a linear combination of the sections 𝜕 𝑓^𝜕𝑗

= 𝑓^{∗ 𝜕}

𝜕 𝑢𝑗, 𝑗 =1, . . . , 𝑛. We let 𝑋^𝑗 denote the real valued functions onΩsuch that

𝑋_𝑘 =

𝑛

∑︁

𝑗=1

𝑋^𝑗

𝜕

𝜕 𝑓_𝑗 .

In local coordinates onΣ(not necessarily holomorphic), the tension field𝜏is given by

𝜏^𝛾 =𝜏^𝛾(𝑓 , 𝜇, 𝜈) = 𝜇^{𝑖 𝑗}

𝜕²𝑓^𝛾

𝜕 𝑥_𝑖𝜕 𝑥_𝑗

−^𝜇Γ_{𝑖 𝑗}^𝑘 𝜕 𝑓^𝛾

𝜕 𝑥_𝑘 +^𝜈Γ^𝛾

𝛼 𝛽(𝑓)𝜕 𝑓^𝛼

𝜕 𝑥_𝑖

𝜕 𝑓^𝛽

𝜕 𝑥_𝑗

,

where𝛾 =1, . . . , 𝑛, and we’re using the Einstein summation convention. Here𝜇^{𝑖 𝑗} are the components of the inverse of the metric tensor 𝜇. Let𝜈¤ be a variation of 𝜈 and set 𝜈_𝑡 = 𝜈+𝑡𝜈¤. Recall we have defined G ( ¤𝜈) = _{𝜕 𝑡}^𝜕𝜏(𝑓_𝜇,𝜈, 𝜇, 𝜈_𝑡) |𝑡=0. Since 𝜏(𝑓 , 𝜇, 𝜈) =0, we see

⟨G ( ¤𝜈), 𝑋⟩= 𝑑 𝑑 𝑡

|_𝑡=0⟨𝜏(𝑓 , 𝜇, 𝜈_𝑡), 𝑋⟩.

The only term that does not die upon taking the derivative is the term involving

𝜈Γ^𝛾

𝛼 𝛽(𝑓). Thus,

⟨G ( ¤𝜈), 𝑋⟩= 𝑑 𝑑 𝑡

|𝑡=0

∑︁

𝛼, 𝛽

𝜈_{𝛼 𝛽}𝜇^{𝑖 𝑗 𝜈𝑡}Γ_{𝛾 𝛿}^𝛼 (𝑓)𝜕 𝑓^𝛾

𝜕 𝑥_𝑖

𝜕 𝑓^𝛿

𝜕 𝑥_𝑗

𝑋^𝛽. (6.9)

Set

𝜈𝑡

Γ𝛾 ,𝛼 𝛽 = 1 2

(𝜈^𝑡

𝛼𝛾 , 𝛽+𝜈^𝑡

𝛾 𝛽,𝛼−𝜈^𝑡

𝛼 𝛽,𝛾), where𝜈^𝑡

𝛼 𝛽,𝛿 =

𝜕 𝜈^𝑡

𝛼 𝛽

𝜕 𝑢_𝛿 and𝜈

𝛾 𝛿

𝑡 denote inverse components of𝜈^𝑡. Under this notation, the Christoffel symbols are computed by the well-known formula

𝜈𝑡

Γ^𝛾

𝛼 𝛽=∑︁

𝛿

𝜈

𝛾 𝛿

𝑡 ·^𝜈𝑡Γ𝛿,𝛼 𝛽.

Inserting back into (6.9) yields

⟨G ( ¤𝜈), 𝑋⟩ = 𝑑 𝑑 𝑡

|𝑡=0

∑︁

𝛼

𝜇^{𝑖 𝑗 𝜈𝑡}Γ𝛾 𝛿,𝛼(𝑓)𝜕 𝑓^𝛾

𝜕 𝑥_𝑖

𝜕 𝑓^𝛿

𝜕 𝑥_𝑗

𝑋^𝛼 =∑︁

𝛾

𝜇^{𝑖 𝑗}Γ¤𝛼 𝛽,𝛾

𝜕 𝑓^𝛼

𝜕 𝑥_𝑖

𝜕 𝑓^𝛽

𝜕 𝑥_𝑗 𝑋^𝛾.

(6.10) Here we are using the notation

Γ¤_{𝛼 𝛽}^𝛾 =lim

𝑡→0

𝜕^𝜈𝑡Γ^𝛾

𝛼 𝛽

𝜕 𝑡 . We also record that

Γ¤𝛾 ,𝛼 𝛽 = 1 2

( ¤𝜈_{𝛼𝛾 , 𝛽}+ ¤𝜈_{𝛾 𝛽,𝛼}− ¤𝜈_{𝛼 𝛽,𝛾}). Tangential harmonic disks

LetΩbe a small neighbourhood of𝑧

1such that 𝑓 :Ω → 𝑀 is an embedding. We let (𝑥

1, 𝑥

2) be conformal coordinates near 𝑧

1 and (𝑢

1, . . . , 𝑢_𝑛) normal coordinates centered at 𝑓(𝑧

1) ∈ 𝑀 such that

• 𝑓(𝑧

1) =(0, . . . ,0),

• the (regular) surface 𝑓(Ω) is tangent to the plane𝑃= {𝑢

3 =· · ·=𝑢_𝑛=0}at 𝑓(𝑧

1), and 𝑓(𝑥_𝑖) =𝑢_𝑖for𝑖=1,2, and

• 𝜈_{𝑗 𝑘} =𝜈_{𝑗 𝑘} =0 and𝜈_{𝑗 𝑗} =1 for 𝑘 =1,2 and 𝑗 =3, . . . , 𝑛when restricted to 𝑃 at 𝑓(𝑧

1).

Note that, as observed in Section 6.4, the set 𝑓⁻1(𝑓(𝑧

1))is finite. Set 𝑓⁻¹(𝑓(𝑧

1)) ={𝑧

1, 𝑧

2, . . . , 𝑧_𝑚}.

For𝜖 ∈ (0,1) small enough, we let 𝐷(𝜖) denote the disk of radius𝜖 in the plane 𝑃, and let 𝐷_𝜖 be the ball of radius𝜖 in the (𝑢

1, . . . , 𝑢_𝑛)-coordinates centered at 0.

Since𝑧_𝑘 ∈Σ^{𝑟 𝑒𝑔}(𝑓), we may choose𝜖 so that 𝑓⁻¹(𝐷_𝜖)=

𝑚

Ø

𝑘=1

Ω𝑘,

where Ω𝑘 = Ω𝑘(𝜖) is the corresponding neighbourhood of 𝑧_𝑘. If we choose a variation 𝜈¤ with support in 𝐷_𝜖, then the induced variation of the pullback metric 𝑓^∗𝜈 is supported in 𝑓⁻¹(𝐷_𝜖). If J𝑉 = G ( ¤𝜈), we will see that this implies J𝑉 is supported there as well, and we obtain

∫

Σ

⟨J𝑉 , 𝑋⟩𝑑𝐴=

𝑚

∑︁

𝑘=1

∫

Ω𝑘

⟨J𝑉 , 𝑋⟩𝑑𝐴=0. (6.11)

Our proof of transversality ofΘinvolves analyzing each integral in the sum above.

We split into cases: (i) the harmonic surfaces 𝑓(Ω₁) and 𝑓(Ω₂) are tangential at 𝑓(𝑧

1) and (ii) they are not tangential. In each case, we pick a different variation of the target metric to find our contradiction.

We first treat case (i). For𝜖 > 0 small enough, 1 ≲ |𝑑 𝑓| ≲ 1

on eachΩ𝑘. Here | · |is the operator norm. Since the surface 𝑓(Ω𝑘)is regular and proper, it follows that

𝜖²≲

∫

Ω𝑘

𝑑𝐴 ≲ 𝜖² (6.12)

for all𝑘.

We now specify our variation. We let𝜑_{𝛼 𝛽} =𝜑_𝛽𝛼 denote a set of real numbers such that𝜑_{𝛼 𝛽}=0 if at least one𝛼, 𝛽is greater than two. Let𝜒be a non-negative function of (𝑢

1, . . . , 𝑢_𝑛) with support in 𝐷

2, equal to 1 on 𝐷

1/2, and such that it has total integral 1 with respect to the induced Euclidean area form on 𝑃. These conditions guarantee that, restricted to this plane, 𝜒^𝜖(𝑢) = 𝜖⁻²𝜒(𝑢/𝜖) converges in the sense of distributions to the Dirac delta function as 𝜖 → 0. 𝜒^𝜖 has compact support in 𝐷(2𝜖), and the product𝜒^𝜖𝜑_{𝛼 𝛽}is equal to𝜑_{𝛼 𝛽}on𝐷(𝜖/2). Define𝜈¤ =𝜈¤(𝜖) by

¤

𝜈_{𝛼 𝛽}(𝑢) =

𝑛

∑︁

𝑗=3

−2𝑢_𝑗𝜒^𝜖(𝑢)𝜑_{𝛼 𝛽}.

We suppress the𝜖 from our notation wherever possible. Referring back to (6.10), we are interested in the variation ofΓ𝛼 𝛽,𝛾. For𝛾 ≥ 3,

Γ¤𝛼 𝛽,𝛾 =−1 2

¤

𝜈_{𝛼 𝛽,𝛾} =𝜑_{𝛼 𝛽}𝜒^𝜖(𝑢) +𝑢_𝛾𝜑_{𝛼 𝛽}𝜒^𝜖

𝛾(𝑢) and hence, on(−𝜖/2, 𝜖/2), this is

Γ¤𝛼 𝛽,𝛾 =𝜈¤_{𝛼 𝛽,𝛾} =𝜑_{𝛼 𝛽}𝜒^𝜖(𝑢). For𝛾 =1,2,

| ¤Γ𝛼 𝛽,𝛾|≲ max

𝛼 𝛽,𝛿

| ¤𝜈_{𝛼 𝛽,𝛿}|≲ 𝜖|∇𝜒^𝜖| ≲ 𝜖⁻². In any case, we always have a𝑂(𝜖⁻²)bound.

The local coordinates (𝑥¹

1, 𝑥¹

2) near𝑧

1satisfy

𝜕 𝑓^𝛼

𝜕 𝑥1 𝑖

(𝑧) =𝛿_{𝑖 𝛼}.

Since 𝑓(Ω₂) is tangent to{𝑢

3 = 0} at 𝑧

2, we can choose coordinates (𝑥²

1, 𝑥²

2) near 𝑧2such that

𝜕 𝑓^𝛼

𝜕 𝑥²

𝑖

(𝑧) =𝛿_{𝑖 𝛼}+𝑂(𝜖). (6.13) Note that, by our restrictions on 𝜇, 𝜇is no longer conformal in these coordinates.

Remark 6.5.1. When the two harmonic disks are equal, the𝑂(𝜖)term is identically zero.

Inserting these expressions into (6.9) gives, near𝑧

1,

⟨G ( ¤𝜈), 𝑋⟩=∑︁

𝛾

𝜇^{𝑖 𝑗}Γ¤𝑖 𝑗 ,𝛾𝑋^𝛾 (6.14)

and near𝑧

2,

⟨G ( ¤𝜈), 𝑋⟩=∑︁

𝛾

𝜇^{𝑖 𝑗}Γ¤𝛼 𝛽,𝛾(𝛿_{𝑖 𝛼}𝛿_{𝑗 𝛽}+𝑂(𝜖))𝑋^𝛾 =∑︁

𝛾

𝜇^{𝑖 𝑗}Γ¤𝑖 𝑗 ,𝛾𝑋^𝛾+𝑂(𝜖) ∑︁

𝛼, 𝛽,𝛾

Γ¤𝛼 𝛽,𝛾𝑋^𝛾. (6.15) Incompatible asymptotics

The reproducing kernel 𝑋 is regular near each point 𝑧_𝑘 when 𝑘 > 2. Trivially,

|J𝑉|≲ 1 near 𝑧_𝑘. Recalling (6.12), we deduce

∫

Ω𝑘

⟨J𝑉 , 𝑋⟩𝑑𝐴 ≲ 𝜖⁻²

for 𝑘 > 1,2. For 𝑘 = 1,2, it may be the case that 𝑋 has a singularity near𝑧

1or𝑧

2

(or both). We computed this singularity in Proposition 6.3.7, the result being that in a trivialization near𝑧

1,

𝑋(𝑧) = 1 2𝜋

log 1

|𝑧|

𝑍_𝑘 +𝐵

1(𝑧), where𝐵

1(𝑧) is a𝐶^0,𝛼 local section ofFnear𝑧

1, and𝑍_𝑘 ∈F𝑧𝑘 is the vector normal to the surface 𝑓(Σ)at 𝑓(𝑧

1), defined above. Our coordinate is not conformal around 𝑧2.

Here we are considering the zero vector to be normal. Recall we are assuming that for𝑘 =1,2, the patches 𝑓(Ω𝑘)are tangent to the plane𝑃at 𝑓(𝑧_𝑘). 𝑍_𝑘 is normal to 𝑃because𝜈

1𝛾 =𝜈

2𝛾 =0 for𝛾 ≥ 3. Incorporating these asymptotics into (6.14) and (6.15), we isolate that at𝑧

1,

⟨G ( ¤𝜈), 𝑋⟩=∑︁

𝛾≥3

𝜇^{𝑖 𝑗}Γ¤𝑖 𝑗 ,𝛾𝑋^𝛾+ ∑︁

𝛿=1,2

𝜇^{𝑖 𝑗}Γ¤𝑖 𝑗 ,𝛿𝑋^𝛿 =∑︁

𝛾≥3

𝜇^{𝑖 𝑗}Γ¤𝑖 𝑗 ,𝛾𝑋^𝛾+𝑂(𝜖⁻²) (6.16)

and at𝑧

2,

⟨G ( ¤𝜈), 𝑋⟩=∑︁

𝛾≥3

𝜇^{𝑖 𝑗}Γ¤𝑖 𝑗 ,𝛾𝑋^𝛾(1+𝑂(𝜖)) +𝑂(𝜖⁻²). We now use the fact that the restrictions of the metric 𝜇at the points𝑧

1and𝑧

2are not conformal to each other via 𝑓. By the choice of local coordinates, this means the matrix 𝜇^{𝑖 𝑗}(𝑧

1) is not a multiple of the matrix 𝜇^{𝑖 𝑗}(𝑧

2), where both matrices are found by trivializing the pullback bundle Fover 𝑓⁻1(𝐷_𝜖) using a trivialization of 𝐷_𝜖. Furthermore, the two spaces of 2×2 matrices orthogonal to𝜇^{𝑖 𝑗}(𝑧

1)and𝜇^{𝑖 𝑗}(𝑧

2) respectively (with respect to the Frobenius inner product) do not coincide. Thus, we can choose𝜑_{𝑖 𝑗} uniformly bounded above and such that

∑︁2

𝑖, 𝑗=1

𝜇^{𝑖 𝑗}(𝑧

1)𝜑_{𝑖 𝑗}(𝑓(𝑧

1))=1 and

∑︁2 𝑖, 𝑗=1

𝜇^{𝑖 𝑗}(𝑧

2)𝜑_{𝑖 𝑗}(𝑓(𝑧

2)) =0. Taylor expanding (6.16), we see that near𝑧

1,

⟨G ( ¤𝜈), 𝑋⟩=∑︁

𝛾≥3

𝜖⁻²𝜒(𝑧/𝜖) 1 2𝜋

log 1

|𝑥|

𝑍

𝛾

1(𝑥) +𝐵^𝛾(𝑥)

+𝑂(𝜖⁻²)

=∑︁

𝛾≥3

𝜖⁻²log|𝑥|⁻¹𝜒(𝑧/𝜖) 2𝜋

𝑍

𝛾

1(0) +𝑂(𝜖⁻²). As for𝑧

2,

∑︁2 𝑖, 𝑗=1

𝜇^{𝑖 𝑗}(𝑧

2)𝜑_{𝑖 𝑗} ≲ 𝜖 follows by Taylor expansion, and therefore

⟨G ( ¤𝜈), 𝑋⟩=∑︁

𝛾≥3

𝜇^{𝑖 𝑗}(𝜑_{𝑖 𝑗}𝜒^𝜖(𝑢) +𝑢_𝛾𝜑_{𝑖 𝑗}𝜒^𝜖

𝛾(𝑢))𝑋^𝛾(1+𝑂(𝜖)) +𝑂(𝜖⁻²)

≲ ∑︁

𝛾≥3

𝜖(𝜒^𝜖(𝑢) +𝑢_𝛾𝜒^𝜖

𝛾(𝑢)) (1+𝑂(𝜖)) log 1

|𝑥|

𝑍

𝛾

2(𝑥) +𝐵^𝛾(𝑥)

+𝑂(𝜖⁻²)

≲ ∑︁

𝛾≥3

𝜖⁻¹log𝜖⁻¹|𝑍

𝛾

2(0) | +𝑂(𝜖⁻²). Taking integrals yields

∫

Ω₁

⟨J𝑉 , 𝑋⟩𝑑𝐴= 𝜖⁻2

2𝜋

∑︁

𝛾≥3

∫

Ω₁

log 1

|𝑥|

𝜒(𝑧/𝜖)𝑍

𝛾

1(0)𝑑𝐴(𝑥¹

1, 𝑥¹

2) +𝑂(1)

∫

Ω₂

⟨J𝑉 , 𝑋⟩𝑑𝐴 ≲ 𝜖log𝜖⁻¹|𝑍

2(0) | +𝑂(1) ≲ 1,

and replacing back into (6.11) gives

∫

Σ

⟨J𝑉 , 𝑋⟩𝑑𝐴= 𝜖⁻² 2𝜋

∑︁

𝛾≥3

∫

Ω₁

log 1

|𝑥|

𝜒(𝑧/𝜖)𝑍

𝛾

1(0)𝑑𝐴(𝑥¹

1, 𝑥¹

2) +𝑂(1). Our standing assumption is that for all 𝜖 > 0, the left-hand side is equal to 0.

Therefore,

∑︁

𝛾≥3

|𝑍

𝛾 1(0) | 1

2𝜋

∫

Ω₁

log 1

|𝑥|

𝜒(𝑧/𝜖)𝑑𝐴(𝑥¹

1, 𝑥¹

2)

=

∑︁

𝛾≥3

∫

Ω₁

1 2𝜋

∫

Ω₁

log 1

|𝑥|

𝜒(𝑧/𝜖)𝑍

𝛾

1(0)𝑑𝐴(𝑥¹

1, 𝑥¹

2) ≲ 1.

In coordinates, Ω₁ is contains a ball of radius 𝜖/2 with respect to the Euclidean metric, and in such a ball 𝜒(𝑧/𝜖) =1. Thus,

𝜖⁻² 2𝜋

∫

Ω₁

𝜒(𝑧/𝜖)log 1

|𝑥|

𝑑𝐴(𝑥¹

1, 𝑥¹

2) ≳ 𝜖⁻²

∫

Ω₁

log 1

|𝑥|

𝑑𝑥¹

1𝑑𝑥¹

2 ≳ log(𝜖⁻¹). If there exists𝛾 ≥ 3 such that 𝑍

𝛾

1(0) ≠0, this implies log𝜖⁻¹ ≲ 1, which is nonsensical. This forces 𝑍

𝛾

1(0) =0 for all 𝛾 ≥ 3. Furthermore, recalling that 𝑍

1 is normal to 𝑓(Ω₁) at 𝑧

1, we must have that 𝑍

1(𝑧

1) = 0 identically. This proves the following lemma.

Lemma 6.5.2. Suppose 𝑓(Ω₁)and 𝑓(Ω₂)are tangential at 𝑓(𝑧

1). Then𝑋extends smoothly over𝑧

1.

Non-tangential harmonic disks

We have essentially proved transversalityΘ, if we assume the images of the harmonic map are tangential at 𝑧

1, 𝑧

2, and at 𝑤

1, 𝑤

2. In this subsection, we consider other intersections. Namely, we prove the following.

Lemma 6.5.3. Suppose 𝑓(Ω₁) and 𝑓(Ω₂) are not tangential at 𝑓(𝑧

1). Then 𝑋 extends smoothly over𝑧

1.

Equipped with this lemma, we can prove transversality with ease.

Proof of transversality ofΘ. AssumeΘis not transverse, so that we have the section 𝑋 as in the work above. Applying Lemma 6.5.2 or Lemma 6.5.3, depending on the