The Nahm Pole Solutions

Chapter IV: THE EXPANSIONS OF THE NAHM POLE SOLUTIONS TO

4.2 The Nahm Pole Solutions

In this section, we will summarize some basic properties of the Nahm pole solutions to the Kapustin-Witten equations, largely following [43], [30], [49].

The Setup of the Nahm Pole Solutions

LetY be a smooth 3-manifold with a Riemannian metric and we writeM :=Y×R⁺ and y be the coordinate ofR⁺. We choose the product metric over M and volume form VolY∧dy, where VolY is the volume form ofY. LetPbe anGbundle overM, whereG is a compact Lie group with Lie algebrag. Letg_P be the adjoint bundle, takeAbe a connection ofP, and letΦbe ag_P valued 1-form.

Given a principle embedding ρ :su(2) →g, for an integer a = 1,2,3 and a point x ∈Y, take{e_a}to be any unit orthogonal basis ofT_x^?Y, the cotangent bundle ofY and take {t_a}to be sections of the adjoint bundlegP lie in the image of ρwith the relation[t_a,t_b]=abct_c. We write the vierbein forme :=Í3

i=1t_ie_i, whereegives an endormorphism of the tangent bundleTY to the adjoint bundleg_P. The definition of the vierbein formedepends on the choice of ρ.

Definition 4.2.1. A solution (A,Φ) to(4.1) over M is a Nahm pole solution if for any point x ∈Y, there exist{e_a}, {t_a} as above such that when y → 0,(A,Φ)has the following expansions: A∼ O(y⁻¹⁺), Φ∼ y⁻¹e+O(y⁻¹⁺), for some constant > 0.

Now, we will introduce some basic terminology of the regularity of a function over manifold with boundary, largely follows from [41]. Let x® = (x₁,x₂,x₃) to be the local coordinates ofY. For any (gP-valued) differential form α, we say α is conormal if for any j ≥ 0 andk® = (k₁,k₂,k₃)with ki ≥ 0, sup|(y∂_y)^j∂_x®^®kα| ≤ C_j®k, where ∂_x^k_®^® = ∂_x^k₁¹∂_x^k₂²∂_x^k₃³ and the sup is taken over an open neighborhood of the boundary. We say αis polyhomogeneous if αis conormal and has an asymptotic expansionα ∼ Í

y^γj(logy)^pαj p( ®x). Here the exponentsγj lie in some discrete set E ⊂ C, called the index set of α, which has the properties that <γj → ∞ as j → ∞, the powers p of logy are all non-negative integers, and there are only finitely many log terms accompanying any given y^γj. The notation "∼" means sup|α−Í

j≤N y^γj(logy)^pαj p| ≤ y^<γN+1(logy)^q, and the corresponding statements must hold for the series obtained by differentiating any finite number of times.

Now, we will summarize the regularity theorem in [43]:

Theorem 4.2.2. ([43, Prop 5.3, Prop 5.9, Section 2.3)] For (A,Φ) a Nahm pole solution to the Kapustin-Witten equations, choose a smooth reference connectionA₀ and denoteΦ₀the leading term ofΦ, if(A,Φ)satisfies the gauge fixing equation

d^?_A⁴

0(A− A₀) −?₄[Φ₀, ?₄(Φ−Φ₀)]= 0,

where ?₄ is the 4-dimensional Hodge star operator, then (A,Φ) is polyhomoge- neous with the following expansions:

A∼ ω+ ⁺

∞

Õ

i=1 r_i

Õ

p=1

yⁱ(logy)^pa_i,p, Φ∼ y⁻¹e+ ⁺

∞

Õ

i=1 r_i

Õ

p=1

yⁱ(logy)^pb_i,p. (4.3) Here for eachi,riare finite positive integers andai,p, bi,pare 1-forms independent ofy coordinate and smooth inxdirection.

Remark. We write a := A− A₀, b := Φ−Φ₀, the statement that (A,Φ)is poly- homogeneous with the gauge fixing condition d^?_A⁴

0a−?₄[Φ, ?₄b] = 0 is proved in [43, Proposition 5.9] and it also works for many other gauge fixing conditions, for exampleA_y =0andd^?_A

0a−?[Φ, ?b]=0,or evend^?_A

0(A−A₀)= 0, whereA_yis the dy component of Aand?is the Hodge star operator ofY. It is straight forward to check that these two gauge fixing conditions will only bring integer expansions. The claim that(A,Φ)has the leading terms(ω+y(logy)^pa1,y⁻¹e+y(logy)^p)is proved in [43, Section 2.3].

For (A,Φ) a Nahm pole solution over Y × R⁺, under the temperal gauge, we can assume A doesn’t have dy component. We write Φ = φ+ φ_ydy. We have the following well-known vanishing claim for theφ_yterm, for a proof see [56, Page 36], [28, Corollary 4.7].

Proposition 4.2.3. For(A,Φ)a Nahm pole solution overY×R⁺, writeΦ= φ+φ_ydy, then iflim_y→+∞|φy|_C0 =0, thenφy =0.

It is straightforward to obtain the following:

Corollary 4.2.4. Let (A,Φ) be a Nahm pole solution convergences to a flat irre- ducibleG^Cconnection aty → ∞, thenΦdon’t havedycomponent.

As pointing out in [49], the reducible limit is important to considered, where φ_y term might appear. In our paper, we don’t make any assumption of the limit of the solution at y→ +∞.

Elementary Representation Theory

Now we will introduce some representation theory ofsu(2). Consider a principal embeddingρ:su(2) →g, we call the imagesu(2)tthe principal subalgebra. When ρ is a principal embedding, under the action of su(2)t, g decomposes as a direct sum of irreducible modules τσ with dimension 2σ +1, and we write g = ⊕_στσ.

Hereσare positive integers of which precisely one equals to 1, corresponding to the principle subalgebra. For simple Lie algebrag, the values ofσare precisely compute in [30]. For example, whenG= SU(N), the values ofσare 1,2,3,· · ·,N−1.

Under the action of su(2)t, the decomposition of g will automatically induce a decomposition of Ω_Y¹(g_P), which is the g_P-valued 1-from on Y. We write V_σ = Ω¹_Y(τσ), and then

Ω_Y¹(g_P)= ⊕_σVσ. (4.4)

We can define the projection mapP_σ :Ω_Y¹(g_P) →Vσ andP_σ : Ω_Y⁰(g_P) →Ω⁰(τσ), for a,b ∈ Ω_Y¹(g_P), we write a^σ := P_σa, b^σ := P_σb and φ^σ_y := P_σφ_y. The Clebsch-Gordan theorem will imply the following proposition:

Proposition 4.2.5. [35] Under the previous assumptions, for σ₁, σ₂ are positive integers, then

?a^σ1∧b^σ2 ∈ ⊕_σ^σ1₌⁺_|σ^σ2

1−σ₂|Vσ, ?[φ^σ_y¹,a^σ2] ∈ ⊕_σ^σ1₌⁺_|σ^σ2

1−σ₂|Vσ.

It is also important to understand the action of the vierbein form e. We define a linear operator:

L :V_σ →Vσ,

a→?[e,a], (4.5)

which obeys the following properties:

Proposition 4.2.6. [43, Section 2.3.2] (1) Lhas three eigenspacesV_σ⁻,V_σ⁰,V_σ⁺ with dimension 2σ − 1,2σ + 1,2σ + 3 and eigenvalues σ + 1,1,−σ. We can write Vσ =V_σ⁻⊕V_σ⁰⊕V_σ⁺.

(2) For 1-forma∈Vσ,[?e,a]=0impliesV_σ⁰component ofais zero.

We also denoteV^◦ := ⊕_σV_σ^◦, where◦ ∈ {+,−,0}. For an integer k, we can define the following operatorL^σ_k(a):= ka+?[e,a]and obtain

Corollary 4.2.7. L_k^σis an isomorphism fork , (σ+1),−1,−σ.

Letω be a connection onY, for the 3-dimensional differential operator?dω, which acts from the spaceΩ_Y¹(g_P)to itself, has the following properties:

Proposition 4.2.8. [30, Page 5]?dω : V_σ⁻ → V_σ⁰, V_σ⁰ → V_σ⁻ ⊕V_σ⁰ ⊕V_σ⁺, V_σ⁺ → V_σ⁰⊕V_σ⁺.

In addition, the vierbein form e will give an identification ofV₁⁰ and Ω⁰(τ₁). We define the operator

Γ:Ω¹(g_P) →Ω⁰(g_P),

Γ(a):=?[a, ?e], (4.6)

whereais a 1-form, then we have the following identities:

Proposition 4.2.9. [49, Appendix A] Fora ∈Vσ = Ω¹(τσ),b∈Ω⁰(τσ), we have:

(1)Γa = Γ(a)⁰, [e,Γa] = 2(a)⁰, Γ[e,b] = 2b, where (a)⁰ means the projection to theV_σ⁰part under the decompositionV_σ =V_σ⁺ ⊕V_σ⁰⊕V_σ⁰,

(2)[e,d_ω^?a]=2(?dωa)⁰,Γ(?dωa)= d^?_ωa, where(?dωa)⁰is theV_σ⁰part of?dωa.

With this preparation, we will state an algebraic lemma which will be heavily used in the following parts of the paper:

Lemma 4.2.10. Fora1,a2∈Vσ,b∈V_σ⁺,c1,c2∈Ω⁰(τσ)and a positive integerr, if they satisfy the following algebraic equations,

(σ+1)a₁=?[e,a₁] − [e,c₁], (σ+1)c1=−Γa₁,

r a₁+(σ+1)a₂=?[e,a₂]+[c₂,e]+?dωb, rc1+(σ+1)c2 =−Γa₂+d_ω^?b,

(4.7)

thena₁ =0andc₁=0.

Proof. Consider theV_σ⁺ part of the first equation, and we obtain (a₁)⁺ = 0, where (a₁)⁺ is theV_σ⁺ part of a₁. As −(σ +1)a₂+?[e,a₂] ∈ V⁺ ⊕ V⁰ and by Lemma 4.2.8,?dωb ∈V⁺ ⊕V⁰, consider theV_σ⁻ part of the third equation, and we obtain (a₁)⁻ =0.

The only situation left is theV⁰part. Consider theV⁰part of the first two equations, we obtain

σ(a₁)⁰ =−[e,c₁], c₁= − 1

σ+1Γa₁. (4.8)

Applying Proposition 4.2.9, we obtain σ(a₁)⁰ = _σ²₊₁(a₁)⁰. When σ , 1, then (a₁)⁰ =0.

Ifσ = 1, then the two equations in (4.8) are equivalent. To be explicit, we obtain (a1)⁰ = −[e,c1]orc1 = −¹

2Γa1. We consider theV_σ⁰ part of the last two equations,

we obtain

r(a₁)⁰+(a₂)⁰=[c₂,e]+(?d_ωb)⁰

rc1+2c₂ =−Γa₂+d_ω^?b. (4.9)

Using[e, ]acts on the second equation of (4.9), we obtain

−r(a1)⁰+2(a2)⁰= 2[c2,e]+2(?dωb)⁰.

Comparing the coefficients with the first equation of (4.9), we obtain(a1)⁰=0.

Lemma 4.2.11. Let a,Θ ∈ V_σ⁰, φ,Ξ ∈ Ω⁰(τσ), let λ be a real number such that λ,2or−1. If they satisfy

(λ−1)a =Θ− [e, φ], λφ=−Γa+Ξ, (4.10) then

a = 1

λ²−λ−2(λΘ− [e,Ξ]), φ = 1

λ²−λ−2((λ−1)Ξ−ΓΘ). (4.11) Specially, ifΘ=Ξ=0, thena =φ= 0.

Proof. Applying[e,]to the second equation, by Proposition 4.2.9, we compute λ[e, φ]=−[e,Γa]+ΓΞ=−2a+ΓΞ.

Combining with the first equation, we obtain a = _λ2−λ−2¹ (λΘ− [e,Ξ]). Acting the operator Γ to the first equation and combing with the second, we will obtain φ= _λ2−λ−2¹ ((λ−1)Ξ−ΓΘ). The rest follows immediately.