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where FY stands for the curvature of a connection over Y . Now use inequali- ties (18) and (16).
Lemma 4.3 There exists a constant C1 > 0 such that for any τ > 0 and any (p, q)–monopole S over (0,τ) Y one has × ϑ 2 2(ϑ(S ) ϑ(S )) + C2τ p 2 . |∇ S| ≤ 0 − τ 1 k k∞ Z[0,τ]×Y Recall that by convention a (p, q)–monopole over (0,τ) Y is actually a con- × figuration over ( 1 ,τ + 1 ) Y , so the lemma makes sense. − 2 2 × Proof We may assume S is in temporal gauge. Then τ ϑ(S ) ϑ(S )= ∂ ϑ(S ) dt τ − 0 t t Z0 = ϑ , ϑ + E dt h∇ S −∇ S Si Z[0,τ]×Y ϑ ( E ϑ ), ≤ k∇ Sk2 k S k2 − k∇ Sk2 where the norms on the last line are taken over [0,τ] Y . If a, b, x are real × numbers satisfying x2 bx a 0 then − − ≤ x2 2x2 2bx + b2 2a + b2. ≤ − ≤ Putting this together we obtain ϑ 2 2(ϑ(S ) ϑ(S )) + E 2, k∇ Sk2 ≤ 0 − τ k Sk2 and the lemma follows from the estimate (16).
Lemma 4.4 For all C > 0 there exists an ǫ> 0 with the following significance. Let τ 4, p P with p τ 1/2 ǫ, and let S = (A, Φ) be a (p, q)–monopole ≥ ∈ k k∞ ≤ over (0,τ) Y satisfying Φ C . Then at least one of the following two × k k∞ ≤ statements muste hold: (i) ∂ ϑ(S ) 0 for 2 t τ 2, t t ≤ ≤ ≤ − (ii) ϑ(S ) <ϑ(S ) for 0 t 1, τ 1 t τ . t2 t1 ≤ 1 ≤ − ≤ 2 ≤
Proof Given C > 0, suppose that for n = 1, 2,... there exist τn 4, pn P 1/2 ≥ ∈ with p τn 1/n, and a (p , q )–monopole S = (A , Φ ) over (0,τ ) Y k nk∞ ≤ n n n n n n × satisfying Φ C such that (i) is violated at some point t = t and (ii)e k nk∞ ≤ n also does not hold. By Lemma 4.3 the last assumption implies 1/2 ϑ 2 C p τ C /n. k∇ Sn kL ([1,τn−1]×Y ) ≤ 1k nk∞ n ≤ 1
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For s R let : R Y R Y be translation by s: ∈ Ts × → × (t,y) = (t + s,y). Ts Given p> 2 then by Lemmas 4.2 and 4.1 we can find un : ( 1, 1) Y U(1) p − × →p in L such that a subsequence of u ( ∗ (S )) converges weakly in L over 2,loc n Ttn n 1 ( 1 , 1 ) Y to an Lp solution S′ to the equations (13) with µ = π∗η. Then − 2 2 × 1 2 ϑ ′ = 0. After modifying the gauge transformations we can even arrange that ∇ S S′ is smooth and in temporal gauge, in which case there is a critical point α of ϑ such that S′(t) α. After relabelling the subsequence above consecutively ≡ we then have h (t ) h ′ (0) Ξ. Sn n → S 6∈ Since Ξ is closed, h (t ) Ξ for n sufficiently large. Hence, ∂ ϑ(S (t)) = Sn n 6∈ t|tn n ϑ 2 0, which is a contradiction. −k∇ Sn(tn)k ≤
Proposition 4.2 For any constant C > 0 there exist C′,δ > 0 such that if S = (A, Φ) is any (p, q)–monopole over ( 2, T +4) Y where T 2, p δ, − × ≥ k k∞ ≤ and Φ C , then for 1 t T 1 one has k k∞ ≤ ≤ ≤ − 2 ′ 2 ϑS 2 sup ϑ(S−r) inf ϑ(ST +r) + C p ∞. |∇ | ≤ − 0≤r≤4 k k Z[t−1,t+1]×Y 0≤r≤1
Proof Choose ǫ > 0 such that the conclusion of Lemma 4.4 holds (with this constant C ), and set δ = ǫ/√6. We construct a sequence t0,...,tm of real numbers, for some m 1, with the following properties: ≥ (i) 1 t 0 and T t T + 4, − ≤ 0 ≤ ≤ m ≤ (ii) For i = 1,...,m one has 1 t t 5 and ϑ(S ) ϑ(S ). ≤ i − i−1 ≤ ti ≤ ti−1 The lemma will then follow from Lemma 4.3. The ti ’s will be constructed ′ ′ inductively, and this will involve an auxiliary sequence t0,...,tm+1 . Set t−1 = ′ t0 = 0. ′ ′ Now suppose ti−1,ti have been constructed for 0 i j . If tj T then we ′ ≤ ≤ ′ ≥ set tj = tj and m = j , and the construction is finished. If tj < T then we ′ define tj,tj+1 as follows: If ∂ ϑ(S ) 0 for all t [t′ ,t′ + 2] set t = t′ and t′ = t′ + 2; otherwise set t t ≤ ∈ j j j j j+1 j t = t′ 1 and t′ = t′ + 4. j j − j+1 j Then (i) and (ii) are satisfied, by Lemma 4.4.
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Proposition 4.3 For all C > 0 there exists a δ > 0 such that if S = (A, Φ) is any (p, q)–monopole in temporal gauge over ( 1, 1) Y such that p 1 δ, − × k kC ≤ Φ C , and ϑ C then the following holds: Either ∂ ϑ(S ) < 0, k k∞ ≤ k∇ Sk2 ≤ t|0 t or there is a critical point α such that S = α for t 1 . t | |≤ 2 Proof First observe that if S is any C1 configuration over R Y then × t2 ϑ(S ) ϑ(S )= ϑ , ∂ S dy dt, t2 − t1 h∇ St t ti Zt1 ZY 1 hence ϑ(St) is a C function of t whose derivative can be expressed in terms of the L2 gradient of ϑ as usual.
Now suppose there is a C > 0 and for n = 1, 2,... a pn P and a (pn, qn)– ∈ 1 monopole S = (A , Φ ) over ( 1, 1) Y such that p 1 , Φ C , n n n − × k nkC ≤ n k nk∞ ≤ ϑ C , and ∂ ϑ(S (t)) 0. Let p> 4 and 0 <ǫ< 1e. After passing to k∇ Sn k2 ≤ t|0 n ≥ 2 a subsequence and relabelling consecutively we can find un : ( 1, 1) Y U(1) p p − × → 1 in L3,loc such that Sn = un(Sn) converges weakly in L2 , and strongly in C , over ( 1 ǫ, 1 + ǫ) Y to a smooth solution S′ of (13) with µ = π∗η. We − 2 − 2 × 2 may arrange that S′eis in temporal gauge. Then
′ 0 ∂t 0ϑ(Sn(t)) = ϑ , ∂t 0Sn(t) ∂t 0ϑ(S ). ≤ | h∇ Sn(0) | i→ | t ZY But S′ is a genuine monopole, so ∂ ϑe(S′) = ϑ ′ 2 . It also follows that e t t e St 2 1 1 −k∇ k ϑ ′ = 0, hence ϑ ′ = 0 in ( ǫ, + ǫ) Y by unique continuation as ∇ S (0) ∇ S − 2 − 2 × in [15, Appendix]. Since h h ′ uniformly in [ ǫ,ǫ], and h ′ const Ξ, Sn → S − S ≡ 6∈ the function h maps [ ǫ,ǫ] into the complement of Ξ when n is sufficiently Sn − large. In that case, S restricts to a genuine monopole on [ ǫ,ǫ] Y , and the n − × assumption ∂ ϑ(S (t)) 0 implies that ϑ = 0 on [ ǫ,ǫ] Y . Since this t|0 n ≥ ∇ Sn − × holds for any ǫ (0, 1 ), the proposition follows. ∈ 2 We say a (p, q)–monopole S over R Y has finite energy if inf ϑ(S ) > . +× t>0 t −∞ A monopole over a 4–manifold with tubular ends is said to have finite energy if it has finite energy over each end.
Proposition 4.4 Let C, δ be given such that the conclusion of Proposition 4.3 holds. If S = (A, Φ) is any finite energy (p, q)–monopole over R Y with + × p 1 δ, q 1, Φ C , and k kC ≤ ≡ k k∞ ≤ sup ϑS L2((t−1,t+1)×Y ) C t≥1 k∇ k ≤ then the following hold:
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(i) There is a t > 0 such that S restricts to a genuine monopole on (t, ) Y , ∞ × (ii) [S ] converges in to some critical point as t . t BY →∞
Proof Let p > 4. If tn is any sequence with tn as n then by { } p → ∞ → ∞ Lemmas 4.1 and 4.2 there exist u L (R Y ; U(1)) such that a subsequence n ∈ 3,loc × of u ( ∗ S) converges weakly in Lp (hence strongly in C1 ) over compact subsets n Ttn 2 of R Y to a smooth (p, q)–monopole S′ in temporal gauge. Proposition 4.3 × guarantees that ∂tϑ(St) 0 for t 1, so the finite energy assumption implies ′ ≤ ≥ that ϑ(St) is constant. By Proposition 4.3 there is a critical point α such that ′ St = α for all t. This implies (i) by choice of the set Ξ (see Subsection 3.3). Part (ii) follows by a continuity argument from the facts that contains only BY finitely many critical points, and the topology on defined by the L2 –metric BY is weaker than the usual topology.
The following corollary of Lemma 3.1 shows that elements of the moduli spaces defined in Subsection 3.4 have finite energy.
Lemma 4.5 Let S be a configuration over R+ Y and α a critical point of p × ϑ such that S α L for some p 2. Then − ∈ 1 ≥ ϑ(S ) ϑ(α) as t . t → →∞
4.3 Neck-stretching I
This subsection contains the crucial step in the proof of Theorem 1.4 assuming (B1), namely what should be thought of as a global energy bound.
Lemma 4.6 Let X be as in Subsection 1.3 and set Z = X:1 . We identify Y = ∂Z . Let µ ,µ Ω2(Z), where dµ = 0. Set η = µ and µ = µ + µ . 1 2 ∈ 1 1|Y 1 2 Let Ao be a spin connection over Z , and let the Chern–Simons–Dirac functional ϑ over Y be defined in terms of the reference connection B = A . Then for η o o|Y all configurations S = (A, Φ) over Z which satisfy the monopole equations (13) one has
2ϑ (S )+ Φ 2 + Fˆ + iµ 2 η |Y |∇A | | A 1| ZZ 2 CVol(Z) 1+ Φ 2 + F + µ + µ + s , ≤ k k∞ k A o k∞ k 1k∞ k 2k∞ k k∞ for some universal constant C , where s is the scalar curvature of Z .
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The upper bound given here is not optimal but suffices for our purposes.
′ ′ Proof Set F = FˆA + iµ1 and define F similarly. Set B = A Y . Without A Ao | the assumption dµ1 = 0 we have
F ′ 2 = (2 (F ′ )+ 2 + F ′ F ′ ) | A| | A | A ∧ A ZZ ZZ = 2 Q(Φ) iµ+ 2 + F ′ F ′ 2idµ (A A ) | − 2 | Ao ∧ Ao − 1 ∧ − o ZZ + (Fˆ + Fˆ + 2iη) (B B ). B Bo ∧ − o ZY Without loss of generality we may assume A is in temporal gauge over the collar ι([0, 1] Y ). By the Weitzenb¨ock formula we have × s 0= D2 Φ= ∗ Φ+ Fˆ+ + . A ∇A∇A A 4 This gives
Φ 2 = ∗ Φ, Φ + ∂ Φ, Φ |∇A | h∇A∇A i h t i ZZ ZZ ZY 1 s = Φ 4 Φ 2 + iµ+Φ, Φ + ∂ Φ, Φ . −2| | − 4| | h i h B i ZZ ZY
Consider now the situation of Subsection 1.4. If (B1) holds then we can find a closed 2–form µ on X whose restriction to R ( Y ) is the pull-back of η , 1 + × ± j j and whose restriction to R Y ′ is the pull-back of η′ . From Lemma 4.6 we + × j j deduce:
Proposition 4.5 For every constant C < there exists a constant C < 1 ∞ 2 ∞ with the following significance. Suppose we are given
τ,C < and an r–tuple T such that τ T for each j , • 0 ∞ ≤ j real numbers τ ± , 1 j r and τ ′ , 1 j r′ satisfying 0 T τ ± τ • j ≤ ≤ j ≤ ≤ ≤ j − j ≤ and 0 τ ′ τ . ≤ j ≤ Let Z be the result of deleting from X(T ) all the necks ( τ −,τ +) Y , 1 j r − j j × j ≤ ≤ and all the ends (τ ′, ) Y ′ , 1 j r′ . Then for any configuration S = j ∞ × j ≤ ≤ (A, Φ) representing an element of a moduli space M(X(T ); ~α′; µ;~p,~p′) where
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′ r ϑ(α′ ) > C and p , p′ ,µ all have L∞ norm 5 Local compactness II
This section, which is logically independent from Section 4, provides the local compactness results needed for the proof of Theorem 1.4 assuming (B2). While the main result of this section, Proposition 5.5, is essentially concerned with local convergence of monopoles, the arguments will, in contrast to those of Section 4, be of a global nature. In particular, function spaces over manifolds with tubular ends will play a central role.
5.1 Hodge theory for the operator d∗ + d+ − In this subsection we will study the kernel (in certain function spaces) of the elliptic operator = d∗ + d+ :Ω1(X) Ω0(X) Ω+(X), (25) D − → ⊕ where X is an oriented Riemannian 4–manifold with tubular ends. The no- tation ker( ) will refer to the kernel of in the space Ω1(X) of all smooth D D 1–forms, where X will be understood from the context. The results of this subsection complement those of [10]. We begin with the case of a half-infinite cylinder X = R Y , where Y is any + × closed, oriented, connected Riemannian 3–manifold. Under the isomorphisms
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(14) there is the identification = ∂ + P over R Y , where P is the D ∂t + × self-adjoint elliptic operator 0 d∗ P = − d d − ∗ acting on sections of Λ0(Y ) Λ1(Y ) (cf (22)). Since P 2 is the Hodge Laplacian, ⊕ ker(P )= H0(Y ) H1(Y ). ⊕ Let h be a maximal orthonormal set of eigenvectors of P , say P h = λ h . { ν } ν ν ν Given a smooth 1–form a over R Y we can express it as a = f h , + × ν ν ν where f : R R. If a = 0 then f (t) = c e−λν t for some constant c . If ν + → D ν ν P ν in addition a Lp for some p 1 then f Lp for all ν , hence f 0 when ∈ ≥ ν ∈ ν ≡ λ 0. Elliptic inequalities for then show that a decays exponentially, or ν ≤ D more precisely, ( ja) β e−δt | ∇ (t,y)|≤ j for (t,y) R Y and j 0, where β is a constant and δ the smallest ∈ + × ≥ j positive eigenvalue of P . Now let σ > 0 be a small constant and a ker( ) Lp,−σ . Arguing as above ∈ D ∩ we find that a = b + c dt + π∗ψ, (26) where b is an exponentially decaying form, c a constant, π : R Y Y , and + × → ψ Ω1(Y ) harmonic. ∈ We now turn to the case when X is an oriented, connected Riemannian 4– manifold with tubular end R Y (so X R Y is compact). Let Y ,...,Y + × \ + × 1 r be the connected components of Y and set s Y ′ = Y , Y ′′ = Y Y ′, j \ j[=1 where 0 s r. Let σ > 0 be a small constant and κ: X R a smooth ≤ ≤ → function such that σt on R Y ′ κ = − + × σt on R Y ′′, ( + × where t is the R+ coordinate. Our main goal in this subsection is to describe ker( ) Lp,κ . D ∩ We claim that all elements a ker( ) Lp,κ are closed. To see this, note first ∈ D ∩ that the decomposition (26) shows that a is pointwise bounded, and da decays
eometry & opology, Volume 9 (2005) G T 42 Kim A Frøyshov exponentially over the ends. Applying the proof of [11, Proposition 1.1.19] to X = X (T, ) Y we get :T \ ∞ × ( d+a 2 d−a 2)= da da = d(a da) | | − | | ∧ ∧ ZX:T ZX:T ZX:T = a da 0 as T . ∧ → →∞ Z∂X:T Since d+a = 0, we conclude that da = 0. Fix τ 0 and for any a Ω1(X) and j = 1, . . . , r set ≥ ∈ R a = a. (27) j ∗ Z{τ}×Yj Recall that d∗ = d on 1–forms, so if d∗a = 0 then R a is independent of − ∗ ∗ j τ . Therefore, if a ker( ) Lp,κ then R a =0 for j>s, hence ∈ D ∩ j s Rja = a = d a = 0. ∂X:τ ∗ X:τ ∗ Xj=1 Z Z Set Ξ= (z ,...,z ) Rs : z = 0 . { 1 s ∈ j } Xj Proposition 5.1 In the situation above the map α: ker( ) Lp,κ ker(H1(X) H1(Y ′′)) Ξ, D ∩ → → ⊕ a ([a], (R a, . . . , R a)) 7→ 1 s is an isomorphism.
Proof We first prove α is injective. Suppose α(a) = 0. Then a = df for some function f on X . From the decomposition (26) we see that a decays exponentially over the ends. Hence f is bounded, in which case
0= f d∗a = a 2. | | ZX ZX This shows α is injective.
1 Next we prove α is surjective. Suppose b Ω (X), db = 0, [b Y ′′ ] = 0, 1 ∈ | and (z1,...,zs) Ξ. Let ψ Ω (Y ) be the harmonic form representing 1 ∈ ∈ [b Y ] H (Y ). Then | ∈ ∗ b R = π ψ + df, | +×Y
eometry & opology, Volume 9 (2005) G T Monopoles over 4–manifolds containing long necks, I 43 for some f : R Y R. Choose a smooth function ρ: X R which vanishes + × → → in a neighbourhood of X and satisfies ρ 1 on [τ, ) Y . Set z = 0 for :0 ≡ ∞ × j j>s and let z be the function on Y with z Vol(Y )−1z . Define |Yj ≡ j j b = b + d(ρ(tz f)). − Then over [τ, ) Y we have b = π∗ψ + z dt, so d∗b = 0 in this region, and ∞ × e d∗b =e b = z =e 0. − ∗ − ZX Z∂X:τ ZY Letκ ¯ : X R be a smoothe function whiche agrees with κ outside a compact → | | p,κ¯ set. By Proposition 2.3 we can find a smooth ξ : X R such that dξ L → ∈ 1 and d∗(b + dξ) = 0. Set a = b + dξ. Then (d + d∗)a =0 and α(a) = ([b], (z ,...,z )). e 1 s
The followinge proposition is essentially [10, Proposition 3.14] and is included here only for completeness.
Proposition 5.2 If b1(Y ) = 0 and s = 0 then the operator : Lp,κ Lp,κ (28) D 1 → has index b (X)+ b (X) b+(X). − 0 1 − 2
Proof By Proposition 5.1 the dimension of the kernel of (28) is b1(X). From Proposition 2.2 (ii) with S = 0 we see that the image of (28) is the sum of ∗ p,κ + p,κ p,κ d L1 and d L1 . The codimensions of these spaces in L are b0(X) and + b2 (X), respectively.
5.2 The case of a single moduli space
Consider the situation of Subsection 1.3. Initially we do not assume Condi- tion (A).
Proposition 5.3 Fix 1 0 be a small constant and ∞ κ: X R a smooth function such that κ(t,y) = σt for all (t,y) R Y . → − ∈ + × Let Ao be a spin connection over X which is translationary invariant over the ends of X . For n = 1, 2,... let Sn = (Ao + an, Φn) be a smooth configuration over X which satisfies the monopole equations (20) with µ = µn , ~p = ~pn .
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Suppose a Lq,κ for every n, and sup Φ < . Then there exist n ∈ 1 n k nk∞ ∞ smooth u : X U(1) such that if k is any non-negative integer with n →
sup ( µn Ck + pn,j Ck ) < (29) j,n k k k k ∞
p′ then the sequence un(Sn) is bounded in Lk+1 over compact subsets of X for every p′ 1. ≥
Before giving the proof we record the following two elementary lemmas:
S T Lemma 5.1 Let E,F,G be Banach spaces, and E F and E G bounded → → linear maps. Set
S + T : E F G, x (Sx,Tx). → ⊕ 7→ Suppose S has finite-dimensional kernel and closed range, and that S + T is injective. Then S + T has closed range, hence there is a constant C > 0 such that x C( Sx + Tx ) k k≤ k k k k for all x E . ∈
Proof Exercise.
Lemma 5.2 Let X be a smooth, connected manifold and x X . Let 0 ∈ Map (X, U(1)) denote the set of smooth maps u: X U(1) such that u(x )= 0 → 0 1, and let V denote the set of all closed 1–forms φ on X such that [φ] ∈ H1(X; Z). Then 1 Map (X, U(1)) V, u u−1du 0 → 7→ 2πi is an isomorphism of Abelian groups.
Proof If φ V define ∈ x u(x) = exp(2πi φ), Zx0 where x φ denotes the integral of φ along any path from x to x. Then x0 0 1 u−1du = φ. The details are left to the reader. 2πi R
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Proof of Proposition 5.3 We may assume X is connected and that (29) holds at least for k = 0. Choose closed 3–forms ω1,...,ωb1(X) which are 3 supported in the interior of X:0 and represent a basis for Hc (X). For any a Ω1(X) define the coordinates of Ja Rb1(X) by ∈ ∈ (Ja) = a ω . k ∧ k ZX Then J induces an isomorphism H1(X) Rb1(X) , by Poincar´eduality. By → Lemma 5.2 we can find smooth v : X U(1) such that J(a v−1dv ) is n → n − n n bounded as n . We can arrange that vn(t,y) is independent of t 0 for → ∞ q,κ ≥ every y Y . Then there are ξ L (X; iR) such that b = a v−1dv dξ ∈ n ∈ 2 n n − n n − n satisfies d∗b = 0; R b = 0, j = 1, . . . , r 1 n j n − where R is as in (27). If r 1 this follows from Proposition 2.4, while if j ≥ r = 0 (ie if X is closed) it follows from Proposition 2.3. (By Stokes’ theorem we have Rrbn = 0 as well, but we don’t need this.) Note that ξn must be ∗ smooth, by elliptic regularity for the Laplacian d d. Set un = exp(ξn)vn . Then un(Ao + an) = Ao + bn . By Proposition 5.1 and Lemma 5.1 there is a C1 > 0 such that r−1 ∗ + b q,κ C1 (d + d )b Lq,κ + Rjb + Jb k kL1 ≤ k k | | k k Xj=1 for all b Lq,κ . From inequality (16) and the curvature part of the Seiberg– ∈ 1 Witten equations we find that sup d+b < , hence n k nk∞ ∞ + bn q,κ C1( d bn Lq,κ + Jbn ) C2 k kL1 ≤ k k k k ≤ for some constant C2 . We can now complete the proof by bootstrapping over compact subsets of X , using alternately the Dirac and curvature parts of the Seiberg–Witten equation.
Combining Proposition 5.3 (with k 1) and Proposition 3.1 we obtain, for ≥ fixed closed 2–forms ηj on Yj :
Corollary 5.1 If (A) holds then for every constant C < there exists a 0 ∞ constant C < with the following significance. Suppose µ 1 , p 1 C 1 ∞ k kC k j kC ≤ 0 for each j . Then for any ~α = (α ,...,α ) with α , and any [S] 1 r j ∈ RYj ∈ M(X; ~α; µ;~p) and t ,...,t [0,C ] one has 1 r ∈ 0 r e λ ϑ(S ) C . j |{tj }×Yj ≤ 1 j=1 X
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Note that if λ ϑ(α ) C then this gives j j j ≥− 0 P r λ ϑ(S ) ϑ(α ) C + C . (30) j |{tj }×Yj − j ≤ 0 1 Xj=1
5.3 Condition (C)
Consider the situation in Subsection 1.4 and suppose γ is simply-connected and equipped with an orientation o. Throughout this subsection (and the next) (co)homology groups will have real coefficients, unless otherwise indicated. We associate to (γ, o) a “height function”, namely the unique integer valued function h on the set of nodes of γ whose minimum value is 0 and which satisfies h(e′)= h(e) + 1 whenever there is an oriented edge from e to e′ . Let Zk and Z[k denote the union of all subspaces Z X# where e has height e ⊂ k and k, respectively. Set ≥ − k − ∂ Z = ∂ Ze. h([e)=k For each node e of γ choose a subspace G H (Z ) such that e ⊂ 1 e H (Z )= G im H (∂−Z ) H (Z ) . 1 e e ⊕ 1 e → 1 e Then the natural map F G∗ is an isomorphism, where G∗ is the dual of e → e e the vector space Ge .
Lemma 5.3 The natural map H1(X#) G∗ is injective. Therefore, this →⊕e e map is an isomorphism if and only if Σ(X, γ, o) = 0.
Proof Let N be the maximum value of h and suppose z H1(X#) lies ∈ in the kernel of the map in the lemma. It is easy to show, by induction on k = 0,...,N , that H1(X#) H1(Zk) maps z to zero for each k. We now → invoke the Mayer–Vietoris sequence for the pair of subspaces (Zk−1,Z[k) of X# : a b H0(Zk−1) H0(Z[k) H0(∂−Zk) H1(Z[k−1) H1(Zk−1) H1(Z[k). ⊕ → → → ⊕ Using the fact that γ is simply-connected it is not hard to see that a is surjec- tive, hence b is injective. Arguing by induction on k = N,N 1,..., 0 we then − find that H1(X#) H1(Z[k) maps z to zero for each k. →
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We will now formulate a condition on (X, γ) which is stronger than (C) and perhaps simpler to verify. A connected, oriented graph is called a tree if it has a unique node (the root node) with no incoming edge, and any other node has a unique incoming edge.
Proposition 5.4 Suppose there is an orientation o of γ such that (γ, o) is a tree and H1(Z[k) H1(Zk) → is surjective for all k. Then Condition (C) holds.
Proof It suffices to verify that Σ(X, γ, o)=0. Set F [k = ker(H1(Z[k) H1(∂−Zk)), → F k = ker(H1(Zk) H1(∂−Zk)). → The Mayer–Vietoris sequence for (Zk,Z[k+1) yields an exact sequence 0 F [k F k H1(Z[k+1) H1(∂−Z[k+1). → → ⊕ → If H1(Z[k) H1(Zk) is surjective then so is F [k F k , hence ker(F [k → → → F k) F [k+1 is an isomorphism, in which case → dim F [k = dim F [k+1 + dim F k. Therefore, Σ(X, γ, o) = dim H1(X#) dim F k = dim H1(X#) dim F [0 = 0. − − Xk
5.4 Neck-stretching II
Consider again the situation in Subsection 1.4. The following set-up will be used in the next two lemmas. We assume that γ is simply-connected and that o is an orientation of γ with Σ(X, γ, o)=0. Let 1
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3 a basis for the image of Ge in Hc (Xe) under the Poincar´eduality isomorphism. For any a Ω1(Z′ ) define J a Rge by ∈ e e ∈ (J a) = a q . e k ∧ ek ZXe For each e let R Y , m = 1,...,h be the outgoing ends of X . For any + × em e e a Ω1(Z′ ) define R a Rhe by ∈ e e ∈ (R a) = a. e m ∗ Z{0}×Yem Set ne = ge + he and L a = (J a, R a) Rne . e e e ∈ For any a Ω1(X(T )) let La V = Rne be the element with components ∈ ∈ ⊕e Lea. For any tubular end R P of X let t: R P R be the projection. + × + × → + Choose a small σ > 0 and for each e a smooth function κ : X R such that e e → σt on incoming ends, κe = ( σt on outgoing and neutral ends. − Let X{T } X be as in Subsection 1.4 and let κ = κ : X(T ) R be a smooth ⊂ T → function such that κ κ is constant on X X{T } for each e. (Such a T − e e ∩ function exists because γ is simply-connected.) This determines κT up to an additive constant. Fix a point x X and define a norm on V by e ∈ e k · kT v = exp(κ (x )) v , k kT T e k ek e X where is the Euclidean norm on Rne and v the components of v. k · k { e} Let denote the operator d∗ + d+ on X(T ) . D −
Lemma 5.4 There is a constant C such that for every r–tuple T with minj Tj p,κ (T ) sufficiently large and every L1 1–form a on X we have
a p,κ C( a Lp,κ + La T ). k kL1 ≤ kD k k k
Note that adding a constant to κ rescales all norms in the above inequality by the same factor.
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Proof Let τ be a function on X which is equal to 2T on the ends R ( Y ) j + × ± j for each j . Choose smooth functions f ,f : R R such that (f (t))2 +(f (1 1 2 → 1 2 − t))2 = 1 for all t, and f (t) = 1 for t 1 , k = 1, 2. For each e define k ≤ 3 β : X R by e e → f1(t/τ) on outgoing ends, βe = f2(t/τ) on incoming ends, 1 elsewhere.
Let β¯ denote the smooth function on X(T ) which agrees with β on X X{T } e e e ∩ and is zero elsewhere.
In the following, C,C1,C2,... will be constants that are independent of T . Set Tˇ = min T . Assume Tˇ 1. j j ≥ Note that β C Tˇ−1 everywhere, and similarly for β¯ . Therefore |∇ e|≤ 1 e βea p,κe C2 a p,κe k kL1 ≤ k kL1 for 1–forms a on Xe . Let denote the operator d∗ + d+ on X . By Proposition 5.1 the Fredholm De − e operator L : Lp,κe Lp,κj Rne De ⊕ e 1 → ⊕ is injective, hence it has a bounded left inverse Pe , P ( L )=Id. e De ⊕ e If a is a 1–form on X(T ) and v V set ∈ β¯e(a, v) = (β¯ea, ve).
Here we regard β¯ea as a 1–form on Xe . Define P = β P β¯ : Lp,κ V Lp,κ. e e e ⊕ → 1 e X If we use the norm on V then P C . Now k · kT k k≤ 3 P ( L)a = β P (β¯ a, L a) D⊕ e e eD e e X = β P ( β¯ a + [β¯ , ]a, L β¯ a) e e De e e D e e e X = (β β¯ a + β P ([β¯ , ]a, 0)) e e e e e D e X = a + Ea,
eometry & opology, Volume 9 (2005) G T 50 Kim A Frøyshov where −1 Ea p,κ C4Tˇ a Lp,κ . k kL1 ≤ k k Therefore, P ( L) I C Tˇ−1, k D⊕ − k≤ 4 so if Tˇ >C then z = P ( L) will be invertible, with 4 D⊕ z−1 (1 z I )−1. k k≤ − k − k In that case we can define a left inverse of L by D⊕ Q = (P ( L))−1P. D⊕ If Tˇ 2C say, then Q 2C , whence for any a Lp,κ we have ≥ 4 k k≤ 3 ∈ 1 a p,κ = Q( a, La) p,κ C( a Lp,κ + La T ). k kL1 k D kL1 ≤ kD k k k Lemma 5.5 Let e be a node of γ and for n = 1, 2,... let T (n) be an r–tuple p,κn (T (n)) and an an L1 1–form on X , where κn = κT (n) . Suppose (i) Σ(X, γ, o) = 0, (ii) min T (n) as n , j j →∞ →∞ (iii) There is a constant C′ < such that ∞ ′ κ (x ′ ) κ (x )+ C n e ≤ n e for all nodes e′ of γ and all n, (iv) sup d+a < . n k nk∞ ∞ (T (n)) Then there are smooth un : X U(1) such that the sequence bn = −1 p → p,κn an un dun is bounded in L1 over compact subsets of Xe , and bn L1 and ∗ − ∈ d bn = 0 for every n.
Note that (iii) implies that e must be a source of (γ, o).
Proof Without loss of generality we may assume that κn(xe) = 1 for all n, in which case sup 1 Lp,κn < . n k k ∞ By Lemmas 5.2 and 5.3 we can find smooth v : X(T (n)) U(1) such that n → ′ −1 sup Je (an vn dvn) < n k − k ∞ for every node e′ , where is the Euclidean norm. (Compare the proof of k · k Proposition 5.3.) Moreover, we can take vn translationary invariant over each
eometry & opology, Volume 9 (2005) G T Monopoles over 4–manifolds containing long necks, I 51 end of X(T (n)) . Proposition 2.4 then provides smooth ξ Lp,κn (X; iR) such n ∈ 2 that b = a v−1dv dξ Lp,κn n n − n n − n ∈ 1 satisfies ∗ d bn = 0, bn = 0 {0}×Y ′ ∗ Z j for j = 1, . . . , r′ 1. Stokes’ theorem shows that the integral vanishes for j = r′ − as well, and since γ is simply-connected we obtain, for j = 1, . . . , r,
b =0 for t T (n). ∗ n | |≤ j Z{t}×Yj ′ In particular, Re′ bn = 0 for all nodes e of γ . Set u = v exp(ξ ), so that b = a u−1du . By Lemma 5.4 we have n n n n n − n n
+ b p,κn C d b p,κn + exp(κ (x ′ )) J ′ (b ) n L1 n L n e e n k k ≤ k k ′ k k! Xe
+ ′ ′ −1 C d an Lp,κn + exp(1 + C ) Je (an vn dvn) , ≤ k k ′ k − k! Xe which is bounded as n . →∞ Proposition 5.5 Suppose γ is simply-connected and that Condition (C) holds for (X, γ). For n = 1, 2,... let [S ] M(X(T (n)); ~α′ ; µ ;~p ;~p′ ), where n ∈ n n n n min T (n) . Then there exist smooth maps w : X U(1) such that j j → ∞ n → if k is any positive integer with ′ sup ( µn Ck + pj,n Ck + pj′,n Ck ) < j,j′,n k k k k k k ∞
p′ then the sequence wn(Sn) is bounded in Lk+1 over compact subsets of X for every p′ 1. ≥ Proof Consider the set-up in the beginning of this subsection where now p> 4 is the exponent used in defining configuration spaces, and o is an orientation of γ for which (C) is fulfilled. By passing to a subsequence we can arrange that κ (x ) κ (x ′ ) converges to a point ℓ(e, e′) [ , ] for each pair of n e − n e ∈ −∞ ∞ nodes e, e′ of γ . Define an equivalence relation on the set of nodes of γ ∼ N by declaring that e e′ if and only if ℓ(e, e′) is finite. Then we have a linear ∼ ordering on / such that [e] [e′] if and only if ℓ(e, e′) > . Here [e] N ∼ ≥ −∞ denotes the equivalence class of e.
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Choose e such that [e] is the maximum with respect to this linear ordering. Let Sn = (Ao + an, Φn). Then all the hypotheses of Lemma 5.5 are satisfied. If un is as in that lemma then, as in the proof of Proposition 5.3, un(Sn) = p′ (Ao + bn, unΦn) will be bounded in Lk+1 over compact subsets of Xe for every p′ 1. ≥ For any r–tuple T let W (T ) be the result of gluing ends of X X according to \ e the graph γ e and (the relevant part of) the vector T . To simplify notation let \ us assume that the outgoing ends of X are R ( Y ), j = 1, . . . , r . Then e + × − j 1 R Y is an end of W T (n) for j = 1, . . . , r . Let b′ be the 1–form on W T (n) + × j 1 n which away from the ends R Y , 1 j r agrees with b , and on each of + × j ≤ ≤ 1 n these ends is defined by cutting off bn :
′ f1(t 2Tj (n) + 1) bn(t,y), 0 t 2Tj(n), bn(t,y)= − · ≤ ≤ (0, t 2Tj(n). ≥ Here f is as in the proof of Lemma 5.4. Then sup d+b′ < . After 1 n k nk∞ ∞ choosing an orientation of γ e for which (C) holds we can apply Lemma 5.5 to \ each component of γ e, with b′ in place of a . Repeating this process proves \ n n the proposition.
Corollary 5.2 If (B2) holds then for every constant C < there ex- 0 ∞ ists a constant C1 < such that for any element [S] of a moduli space (T ) ′ ′ ∞ ′ 1 M(X ; ~α ; µ;~p;~p ) where minj Tj >C1 and µ, pj, pj all have C –norm r r′ ′ λj ϑ(S {−T }×Y ) ϑ(S {T }×Y ) + λ ϑ(S {0}×Y ′ ) The next proposition, which is essentially a corollary of Proposition 5.5, exploits the fact that Condition (C) is preserved under certain natural extensions of (X, γ).
Proposition 5.6 Suppose γ is simply-connected and (C) holds for (X, γ). Then for every constant C < there is a constant C < such that if 0 ∞ 1 ∞ S = (A, Φ) represents an element of a moduli space M(X(T ); ~α′; µ;~p;~p′) where ′ ∞ minj Tj >C1 and µ, pj, pj all have L norm ϑ 2 eometry & opology, Volume 9 (2005) G T Monopoles over 4–manifolds containing long necks, I 53
Proof Given an edge v of γ corresponding to a pair of ends R ( Y ) of + × ± j X , we can form a new pair (X , γ ) where X = X (R Y ), and γ (j) (j) (j) × j (j) is obtained from γ by splitting v into two edges with a common end-point ` representing the component R Y of X . × j (j) ′ Similarly, if e is a node of γ and R+ Yj an end of Xe then we can form a (j) (j) (j) × ′ (j) new pair (X , γ ) where X = X (R Yj ), and γ is obtained from γ ′ × ′ (j) by adding one node ej representing the component R+ Yj of X and one ′ ` × edge joining e and ej . One easily shows, by induction on the number of nodes of γ , that if (C) holds for (j) (j) (X, γ) then (C) also holds for each of the new pairs (X(j), γ(j)) and (X , γ ). Given this observation, the proposition is a simple consequence of Proposi- tion 5.5.
6 Exponential decay
In this section we will prove exponential decay results for genuine monopoles over half-cylinders R Y and long bands [ T, T ] Y . The overall scheme of + × − × proof will be the same as that for instantons in [10], and Subsections 6.1 and 6.3 follow [10] quite closely. On the other hand, the proof of the main result of Subsection 6.2, Proposition 6.1, is special to monopoles (and is new, as far as we know). Throughout this section Y will be a closed, connected Riemannian spinc 3– manifold, and η Ω2(Y ) closed. We will study exponential decay towards ∈ a non-degenerate critical point α of ϑ = ϑη . We make no non-degeneracy assumptions on any other (gauge equivalence classes of) critical points, and we do not assume that (O1) holds, except implicitly in Proposition 6.3. All monopoles will be genuine (ie p = 0). Previously, Nicolaescu [29] has proved a slightly weaker exponential decay result in the case η = 0, using different techniques. Other results (with less complete proofs) were obtained in [28] and [26].
6.1 A differential inequality
We begin by presenting an argument from [10] in a more abstract setting, so that it applies equally well to the Chern–Simons and the Chern–Simons–Dirac functionals.
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Let E be a real Banach space with norm and E′ a real Hilbert space k · k with inner product , . Let E E′ be an injective, bounded operator with h· ·i → dense image. We will identify E as a vector space with its image in E′ . Set x = x,x 1/2 for x E′ . | | h i ∈ Let U E be an open set containing 0 and ⊂ f : U R, g : U E′ → → smooth maps satisfying f(0) = 0, g(0) = 0, and Df(x)y = g(x),y h i for all x U , y E . Here Df(x): E R is the derivative of f at x. Suppose ∈ ∈ → H = Dg(0): E E′ is an isomorphism (of topological vector spaces). Note → that H can be thought of as a symmetric operator in E′ . Suppose E contains a countable set e of eigenvectors for H which forms an orthonormal basis { j} for E′ . Suppose σ, λ are real numbers satisfying 0 λ<σ and such that H ≤ has no positive eigenvalue less than σ.
Lemma 6.1 In the above situation there is a constant C > 0 such that for every x U with x C−1 one has ∈ k k≤ 2σf(x) g(x) 2 + C g(x) 3, ≤ | | | | 2λf(x) g(x) 2. ≤ | | Proof It clearly suffices to establish the first inequality for some C . By Tay- lor’s formula [8, 8.14.3] there is a C > 0 such that for all x U with x C−1 1 ∈ k k≤ 1 one has 1 f(x) Hx,x C x 3, | − 2h i| ≤ 1k k g(x) Hx C x 2. | − |≤ 1k k Let He = λ e , and set x = x, e e . Then j j j j h j i j σ Hx,x = σ λ x 2 λ2 x 2 = Hx 2. h i j| j| ≤ j | j| | | Xj Xj By assumption, there is a C2 > 0 such that x C Hx k k≤ 2| | for all x E . Putting the above inequalities together we get, for r = x −1 ∈ k k ≤ C1 , Hx g(x) + C r2 | | ≤ | | 1 g(x) + rC C Hx . ≤ | | 1 2| |
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−1 If r< (C1C2) this gives Hx (1 rC C )−1 g(x) , | |≤ − 1 2 | | hence 2σf(x) σ Hx,x + 2σC r3 ≤ h i 1 Hx 2 + 2σrC C2 Hx 2 ≤ | | 1 2 | | 1 + 2σrC C2 1 2 g(x) 2 ≤ (1 rC C )2 | | − 1 2 (1 + C r) g(x) 2 ≤ 3 | | g(x) 2 + C g(x) 3 ≤ | | 4| | for some constants C3,C4 .
We will now apply this to the Chern–Simons–Dirac functional. Let α = (B , Ψ ) be a non-degenerate critical point of ϑ. Set K = ker( ∗) Γ(iΛ1 S) 0 0 α Iα ⊂ ⊕ and H = H : K K . Note that any eigenvalue of H is also an eigen- α α|Kα α → α α value of the self-adjoint elliptic operator e 0 ∗ e Iα H Iα α over Y acting on sections of iΛ0 iΛ1 S. Let λ± be positive real numbers ⊕ ⊕ such that H has no eigenvalue in the interval [ λ−, λ+]. α − In the following lemma, Sobolev norms of sections of the spinor bundle SY over e Y will be taken with respect to B0 and some fixed connection in the tangent bundle TY . This means that the same constant ǫ will work if α is replaced with some monopole gauge equivalent to α.
Lemma 6.2 In the above situation there exists an ǫ> 0 such that if S is any smooth monopole over the band ( 1, 1) Y satisfying S0 α 2 ǫ then − × k − kL1(Y ) ≤ 2λ±(ϑ(S ) ϑ(α)) ∂ ϑ(S ). ± 0 − ≤− t|0 t Proof Choose a smooth u: ( 1, 1) Y U(1) such that u(S) is in temporal − × → gauge. Then ∂ ϑ(S )= ∂ ϑ(u (S )) = ϑ(u (S )) 2 = ϑ(S ) 2. t t t t t −k∇ t t k2 −k∇ t k2 If ǫ> 0 is sufficiently small then by the local slice theorem we can find a smooth v : Y U(1) which is L2 close to 1 and such that ∗(v(S ) α)=0. We now → 2 Iα 0 − apply Lemma 6.1 with E the kernel of ∗ in L2 , E′ the kernel of ∗ in L2 , Iα 1 Iα and f(x) = (ϑ(α + x) ϑ(α)). The assumption that α be non-degenerate ± − means that H = H : E E′ is an isomorphism, so the lemma follows. α → e eometry & opology, Volume 9 (2005) G T 56 Kim A Frøyshov
6.2 Estimates over [0, T ] Y × Let α be a non-degenerate critical point of ϑ and α = (B, Ψ) the monopole over R Y that α defines. Throughout this subsection, the same convention × for Sobolev norms of sections of SY will apply as in Lemma 6.2. For Sobolev norms of sections of the spinor bundles over (open subsets of) R Y we will × use the connection B. Throughout this subsection S = (A, Φ) will be a monopole over a band B = [0, T ] Y where T 1. Set s = (a, φ)= S α and × ≥ − δ = s 2 B , k kL2( ) 2 2 (31) ν = ϑ 2 B = ϑ(S ) ϑ(S ). k∇ SkL ( ) 0 − T The main result of this subsection is Proposition 6.1, which asserts in particular that if δ is sufficiently small then S is gauge equivalent to a configuration S which is in Coulomb gauge with respect to α and satisfies S α 2 B k − kL1( ) ≤ const ν . e · e We will assume δ 1. Let a′ denote the contraction of a with the vector field ∂ ≤ ∂1 = ∂t . Quantities referred to as constants or denoted “const” may depend on Y,η, [α], T but not on S . Note that ν const ( s + s 2 ) const, (32) ≤ · k k1,2 k k1,2 ≤ the last inequality because δ 1. ≤ For real numbers t set i : Y R Y, y (t,y). t → × 7→ If ω is any differential form over B set ω = i∗ω, 0 t T . Similar notation t t ≤ ≤ will be used for connections and spinors over B.
Lemma 6.3 There is a constant C0 > 0 such that ∂ φ C (ν + a′ ). k 1 k2 ≤ 0 k k3 Proof We have ∂ φ = ∂ Φ= AΦ a′Φ, 1 1 ∇1 − where A is the covariant derivative with respect to A in the direction of the ∇1 vector field ∂ = ∂ . Now AΦ depends only on the gauge equivalence class of 1 ∂t |∇1 | S = (A, Φ), and if A is in temporal gauge (ie if a′ = 0) then ( AΦ) = ∂ Φ . ∇1 t At t The lemma now follows because δ 1. ≤
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Lemma 6.4 There is a constant C1 > 0 such that if δ is sufficiently small then the following hold: (i) φ C ( ∗s + a′ + ν), k k1,2 ≤ 1 kIα k2 k k1,2 (ii) There is a smooth fˇ: B iR such that sˇ = (ˇa, φˇ) = exp(fˇ)(S) α → − satisfies sˇ C ϑ , 0 t T, k tk1,2 ≤ 1k∇ St k2 ≤ ≤ (iii) da C ν , k k2 ≤ 1 where in (i) and (iii) all norms are taken over B.
Proof The proof will use an elliptic inequality over Y , the local slice theorem for Y , and the gradient flow description of the Seiberg–Witten equations over R Y . × (i) Since α is non-degenerate we have z const ( ∗ + H )z k k1,2 ≤ · k Iα α k2 for L2 sections z of (iΛ S) . Recall that 1 ⊕ Y ϑ = H z + z z ∇ α+z α ⊗ where z z represents a pointwise quadratic function of z. Furthermore, z ⊗ k ⊗ z const z 2 . If z is sufficiently small then we can rearrange to get k2 ≤ · k k1,2 k k1,2 z const ( ∗z + ϑ ). (33) k k1,2 ≤ · kIα k2 k∇ α+zk2 By the Sobolev embedding theorem we have
st 2 const s 2 B , t [0, T ], k kL1(Y ) ≤ · k kL2( ) ∈ for some constant independent of t, so we can apply inequality (33) with z = st when δ is sufficiently small. Because ( ∗s ∂ a′) = ∗s Iα − 1 t Iα t we then obtain T 2 ∗ 2 ′ 2 2 st 2 dt const ( αs 2 B + ∂1a 2 B + ν ). k kL1(Y ) ≤ · kI kL ( ) k kL ( ) Z0 This together with Lemma 6.3 establishes (i). (ii) Choose a base-point y Y . By the local slice theorem there is a constant 0 ∈ C such that if δ is sufficiently small then for each t [0, T ] there is a unique ∈ smooth fˇ : Y iR such that t →
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fˇ Cδ, • k tk2,2 ≤ fˇ(y )=0 if α is reducible, • t 0 sˇ = exp(fˇ)(S ) α satisfies ∗sˇ = 0. • t t t − Iα t It is not hard to see that the function fˇ: B iR given by fˇ(t,y) = fˇ(y) is → t smooth. Moreover, sˇ const s . Part (ii) then follows by taking k tk1,2 ≤ · k tk1,2 z =s ˇt in (33). (iii) Choose a smooth u: B U(1) such that u(S) is in temporal gauge, and → set (a, φ)= u(S) α. Then − da = da = dt ∂ a + d a = dt ϑ + d aˇ , ∧ 1 Y − ∧∇1 St Y t where ϑ is the first component of ϑ. This yields the desired estimate on ∇1 ∇ da.
Lemma 6.5 Let v , . . . , v be a family of closed 2–forms on Y which { 1 b1(Y )} represents a basis for H2(Y ; R). Then there is a constant C such that
∗ ∗ b L2(B) C (d + d)b L2(B) + ( b) ∂B L2 (∂B) + dt π vj b k k 1 ≤ k k k ∗ | k 1/2 B ∧ ∧ j Z X (34) for all L2 1–forms b on B, where π : B Y is the projection. 1 → Proof Let K denote the kernel of the operator 1 0 2 0 ∗ ΩB ΩB ΩB Ω B, b (d b, db, b B). → ⊕ ⊕ ∂ 7→ ∗ |∂ Then we have an isomorphism ρ: K ≈ H1(Y ; R), b [b ]. → 7→ 0 For on the one hand, an application of Stokes’ theorem shows that ρ is injective. On the other hand, any c H1(Y ; R) can be represented by an harmonic 1– ∈ form ω, and π∗ω lies in K , hence ρ is surjective. It follows that every element of K is of the form π∗(ω). Now apply Proposi- tion 4.1 and Lemma 5.1.
Lemma 6.6 There is a smooth map fˆ: B iR, unique up to an additive → constant, such that aˆ = a dfˆ satisfies − ∗ d aˆ = 0, ( aˆ) B = 0. ∗ |∂ Given any such fˆ, if we set sˆ = (ˆa, φˆ) = exp(fˆ)(S) α then − aˆ 2 B C2ν, sˆ 2 B C2δ k kL1( ) ≤ k kL2( ) ≤ for some constant C2 > 0.
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Proof The first sentence of the lemma is just the solution to the Neumann problem. If we fix x B then there is a unique fˆ as in the lemma such that 0 ∈ fˆ(x ) = 0, and we have fˆ const a . Writing 0 k k3,2 ≤ · k k2,2 φˆ = exp(fˆ)Φ Ψ = (exp(fˆ) 1)Φ + φ − − and recalling that, for functions on B, multiplication is a continuous map L2 3 × L2 L2 for 0 k 3, we get k → k ≤ ≤ φˆ C exp(fˆ) 1 Φ + φ k k2,2 ≤ k − k3,2k k2,2 k k2,2 C′ fˆ exp(C′′ fˆ )+ φ ≤ k k3,2 k k3,2 k k2,2 C′′′ s ≤ k k2,2 for some constants C,...,C′′′ , since we assume δ 1. There is clearly a similar 2 2 ≤ L2 bound ona ˆ, so this establishes the L2 bound ons ˆ. We now turn to the L2 bound ona ˆ. Leta ˇ be as in Lemma 6.4. Sincea ˆ aˇ is 1 − exact we have
v aˆ = v aˇ const v aˇ ∧ t ∧ t ≤ · k k2k tk2 ZY ZY for any closed v Ω2 . Now take b =a ˆ in Lemma 6.5 and use Lemma 6.4, ∈ Y remembering that daˆ = da.
Definition 6.1 For any smooth h: Y iR define h, P (h): B iR by → → h(t,y)= h(y) and P (h)=∆h + i iΨ, exp(h)Φ , h i where ∆ = d∗d is the Laplacian over R Y . Let P (h) be the restriction of × t P (h) to t Y . { }×
Note that ∗(exp(h)(S) α)= d∗a + P (h). Iα − −
Lemma 6.7 If α is irreducible then the following hold:
(i) There is a C3 > 0 such that if δ is sufficiently small then there exists a unique smooth h: Y iR satisfying h C δ and P (h) = 0. → k k3,2 ≤ 3 0 (ii) If h: Y iR is any smooth function satisfying P (h) = 0 then → 0 ′ P (h) 2 B const (ν + a 3 B ). k kL ( ) ≤ · k kL ( )
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Proof (i) We will apply Proposition B.1 (ie the inverse function theorem) to the smooth map P : L2 L2, h ∆ h + i iΨ , exp(h)Φ . 0 3 → 1 7→ Y h 0 0i The first two derivatives of this map are DP (h)k =∆ k + k Ψ , exp(h)Φ , 0 Y h 0 0i D2P (h)(k,ℓ)= ikℓ iΨ , exp(h)Φ . 0 h 0 0i The assumption δ 1 gives ≤ D2P (h) const (1 + h ). k 0 k≤ · k∇ k3 Set L = DP0(0). Then (L ∆ Ψ 2)k = k Ψ , φ , − Y − | 0| h 0 0i hence L ∆ Ψ 2 const δ. k − Y − | 0| k≤ · Thus if δ is sufficiently small then L is invertible and L−1 (∆ + Ψ 2)−1 + 1. k k ≤ k Y | 0| k Furthermore, we have P (0) = i iΨ , φ , so 0 h 0 0i P (0) const δ. k 0 k1,2 ≤ · By Proposition B.1 (i) there exists a constant C > 0 such that if δ is sufficiently small then there is a unique h L2 such that h C and P (h) = 0 (which ∈ 3 k k3,2 ≤ 0 implies that h is smooth). Proposition B.1 (ii) then yields h const P (0) const δ. k k3,2 ≤ · k 0 k1,2 ≤ · (ii) Setting Q = P (h) we have, for 0 t T , ≤ ≤ t 2 2 2 Q(t,y) dy = ∂1Q(s,y) ds dy const ∂1Q . | | ≤ · B | | ZY ZY Z0 Z
Now, ∂1Q = i iΨ, exp(h)∂1Φ , hence h i ∂ Q const ∂ Φ const (ν + a′ ) k 1 k2 ≤ · k 1 k2 ≤ · k k3 by Lemma 6.3.
Proposition 6.1 There is a constant C4 such that if δ is sufficiently small then there exists a smooth f : B iR such that s = (a, φ) = exp(f)(S) α → − satisfies ∗ e e e s = 0, ( a) ∂B = 0, s 2 B C4ν,e s e 2 B C4δ, Iα ∗ | k kL1( ) ≤ k kL2( ) ≤ where δ, ν are as in (31). e e e e
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This is analogous to Uhlenbeck’s theorem [32, Theorem 1.3] (with p=2), except that we assume a bound on δ rather than on ν .
Proof To simplify notation we will write = in this proof. I Iα Case 1: α reducible In that case the operator ∗ is given by ∗(b, ψ) = I I d∗b. Let f be the fˆ provided by Lemma 6.6. Then apply Lemma 6.4 (ii), − taking the S of that lemma to be the present exp(f)(S). e Case 2: α irreducible Let f,ˆ Sˆ etc be as in Lemma 6.6. Choose h: Y iR e → such that the conclusions of Lemma 6.7 (i) holds with the S of that lemma taken to be the present Sˆ. Set S´ = (A,´ Φ)´ = exp(h)(Sˆ) ands ´ = (´a, φ´)= S´ α. − By Lemmas 6.7 and 6.4 (ii) we have ∗s´ const ν, s´ const δ, φ´ const ν. kI k2 ≤ · k k2,2 ≤ · k k1,2 ≤ · Since d∗a´ = ∗s´ i iΨ, φ´ we also get − I − h i d∗a´ const ν. k k2 ≤ · Applying Lemma 6.5 as in the proof of Lemma 6.6 we see that a´ const ν. k k1,2 ≤ · It now only remains to make a small perturbation to S´ so as to fulfil the Coulomb gauge condition. To this end we invoke the local slice theorem for B. This says that there is a C > 0 such that if δ is sufficiently small then there exists a unique smooth f : B iR such that setting s = (a, φ) = exp(f)(S´) α → − one has ∗ f Cδ, s = 0, a B = 0. e k k3,2 ≤ I ∗ |∂ e e We will now estimate first f , then s in terms of ν . First note that k k2,2 e k k1,2 e a´ B = aˆ B = 0, and ∗ |∂ ∗ |∂ a =a ´ df, φ =e exp(f)Φ´ Ψ − − by definition. Write the imaginary part of exp(f) as f + f 3u. Then f satisfies e the equations (∂ f) Be=0 and t |∂ ∗ 0= d a + i iΨ, φ R − h i ∗ =∆f d a´ + i iΨ, exp(f)(φ´ + Ψ) R − he i e 2 ∗ 3 =∆f + Ψ f d a´ + i iΨ, exp(f)φ´ + f uΨ R. | | − h i By the Sobolev embedding theorem we have f const f const δ, k k∞ ≤ · k k3,2 ≤ ·
eometry & opology, Volume 9 (2005) G T 62 Kim A Frøyshov and we assume δ 1, so u const. Therefore, ≤ k k∞ ≤ f const ∆f + Ψ 2f k k2,2 ≤ · k | | k2 ν + const f 3 ≤ · k k2 ν + const f 3 , ≤ · k k2,2 cf Subsection 2.5 for the first inequality. If δ is sufficiently small then we can rearrange to get f const ν . Consequently, a const ν . To k k2,2 ≤ · k k1,2 ≤ · estimate φ we write k k1,2 φ = gΨ + exp(f)φ,´ e e where g = exp(f) 1. Then dg = df and g const f . Now − |e | | | | |≤ · | | φ const f + φ´ const ν, k k2 ≤ · k k2 k k2 ≤ · φ const ( g + df φ´ + φ´ ) k∇ek2 ≤ · k k1,2 k ⊗ k2 k∇ k2 const (ν + df 4 φ´ 4) e ≤ · k k k k const (ν + f φ´ ) ≤ · k k2,2k k1,2 const (ν + ν2) ≤ · const ν ≤ · by (32). Therefore, φ const ν . Thus, the proposition holds with k k1,2 ≤ · ˆ e f = f + h + f.
Proposition 6.2 Let k be a positivee integer and V ⋐ int(B) an open subset. Then there are constants ǫk,Ck,V , where ǫk is independent of V , such that if ∗ s = 0, s 2 B ǫk Iα k kL1( ) ≤ then
s L2 (V ) Ck,V s L2(B). k k k ≤ k k 1
Proof The argument in [11, pages 62–63] carries over, if one replaces the op- erator d∗ +d+ with ∗ +DΘ , where DΘ is the linearization of the monopole Iα α α map at α. Note that ∗ + DΘ is injective over S1 Y because α is non- Iα α × degenerate, so if γ : B R is a smooth function supported in int(B) then → γs C′ ( ∗ + DΘ )(γs) k kk,2 ≤ kk Iα α kk−1,2 ′ for some constant Ck .
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6.3 Decay of monopoles
The two theorems in this subsection are analogues of Propositions 4.3 and 4.4 in [10], respectively.
2 Let β be a non-degenerate monopole over Y , and U Y an L –closed subset ⊂B ± which contains no monopoles except perhaps [β]. Choose λ > 0 such that Hβ has no eigenvalue in the interval [ λ−, λ+], and set λ = min(λ−, λ+). Define − e B = [t 1,t + 1] Y. t − ×
Theorem 6.1 For any C > 0 there are constants ǫ,C0,C1,... such that the following holds. Let S = (A, Φ) be any monopole in temporal gauge over ( 2, ) Y such that [S ] U for some t 0. Set − ∞ × t ∈ ≥ ν¯ = ϑ 2 , ν(t)= ϑ 2 B . k∇ SkL ((−2,∞)×Y ) k∇ SkL ( t) If Φ C and ν¯ ǫ then there is a smooth monopole α over Y , gauge k k∞ ≤ ≤ equivalent to β, such that if B is the connection part of α then for every t 1 ≥ and non-negative integer k one has k −λ+t sup B(S α) (t,y) Ck ν(0)e . (35) y∈Y |∇ − | ≤ p
Proof It follows from the local slice theorem that is a (topological) BY →BY principal H1(Y ; Z)–bundle. Choose a small open neighbourhood V of [β] ∈BY which is the image of a convex set in . We definee a continuous function CY f¯: V R by → f¯(x)= ϑ(σ(x)) ϑ(σ([β])) − where σ : V is any continuous cross-section. It is clear that f¯ is inde- → BY pendent of σ. e Given C > 0, let S = (A, Φ) be any monopole over ( 2, ) Y such that − ∞ × Φ C and [S ] U for some t 0. If δ > 0, and k is any non- k k∞ ≤ t ∈ ≥ negative integer, then providedν ¯ is sufficiently small, our local compactness results (Lemmas 4.2 and 4.1) imply that for every t 0 we can find a smooth ≥ u: B U(1) such that 0 → u(S B ) β k B < δ. k | t − kC ( 0) In particular, ifν ¯ is sufficiently small then f : R R, t f¯([S ]) + → 7→ t
eometry & opology, Volume 9 (2005) G T 64 Kim A Frøyshov is a well-defined smooth function. Since f(t) ϑ(S ) is locally constant, and − t f(t) 0 as t , we have → →∞ f(t)= ϑ(S ) L, t − where L = limt→∞ ϑ(St). Ifν ¯ is sufficiently small then Lemma 6.2 gives 2λ+f f ′ , hence ≤− + 0 f(t) e−2λ tf(0), t 0. ≤ ≤ ≥ This yields + ν(t)2 = f(t 1) f(t + 1) const e−2λ tf(0), t 1. − − ≤ · ≥ Ifν ¯ is sufficiently small then by Propositions 6.1 and 6.2 we have k ′ f(t) const ν(t), sup A( ϑS) (t,y) Ckν(t) (36) ≤ · y∈Y |∇ ∇ | ≤ for every t 0 and non-negative integer k, where C′ is some constant. Here ≥ k we are using the simple fact that if E, E′ are Banach spaces, W E an open ⊂ neighbourhood of 0, and h: W E′ a differentiable map with h(0) = 0 then → h(x) ( Dh(0) + 1) x in some neighbourhood of 0. For instance, to k k ≤ k k k k deduce the second inequality in (36) we can apply this to the map 2 2 k h: L L , s = (a, φ) ′ ( ϑ ) k+j+1 → j 7→ ∇B +a ∇ β+s where j 3, say, and B′ is the connection part of β . ≥ Putting the inequalities above together we get k ′′ −λ+t sup A( ϑS) (t,y) Ck ν(0)e , t 1 y∈Y |∇ ∇ | ≤ ≥ ′′ p for some constants Ck . If S is in temporal gauge we deduce, by taking k = 0, that St converges uniformly to some continuous configuration α. One can now prove by induction on k that α is of class Ck and that (35) holds.
Theorem 6.2 For any C > 0 there are constants ǫ,C0,C1,... such that the following holds for every T > 1. Let S = (A, Φ) be any smooth monopole in temporal gauge over the band [ T 2, T + 2] Y , and suppose [S ] U for − − × t ∈ some t [ T, T ]. Set ∈ − ν¯ = ϑ 2 , ν(t)= ϑ 2 B . k∇ SkL ((−T −2,T +2)×Y ) k∇ SkL ( t) If Φ C and ν¯ ǫ then there is a smooth monopole α over Y , gauge k k∞ ≤ ≤ equivalent to β, such that if B is the connection part of α then for t T 1 | |≤ − and every non-negative integer k one has k 1/2 −λ(T −|t|) sup B(S α) (t,y) Ck(ν( T )+ ν(T )) e . y∈Y |∇ − | ≤ −
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Proof Given C > 0, let S = (A, Φ) be any monopole over [ T 2, T + 2] Y − − × such that Φ C and [S ] U for some t [ T, T ]. Ifν ¯ is sufficiently k k∞ ≤ t ∈ ∈ − small then we can define the function f(t) for t T as in the proof of | | ≤ Theorem 6.1, and (36) will hold with f(t) replaced by f(t) , for t T . | | | | ≤ Again, f(t)= ϑ(S ) L for some constant L. Lemma 6.2 now gives t − e−2λ−(T −t)f(T ) f(t) e−2λ+(T +t)f( T ), t T, ≤ ≤ − | |≤ which implies f(t) ( f( T ) + f(T ) )e−2λ(T −|t|), t T, | |≤ | − | | | | |≤ ν(t)2 const (ν( T )+ ν(T ))e−2λ(T −|t|), t T 1. ≤ · − | |≤ − By Propositions 6.1 and 6.2 there is a critical point α gauge equivalent to β such that k ′′′ (S α) ∞ C ν(0) k∇B 0 − kL (Y ) ≤ k ′′′ for some constants Ck . It is now easy to complete the proof by induction on k.
6.4 Global convergence
The main result of this subsection is Proposition 6.3, which relates local and global convergence of monopoles over a half-cylinder. First some lemmas.
Lemma 6.8 Let Z be a compact Riemannian n–manifold (perhaps with boundary), m a non-negative integer, and q n/2. Then there is a real ≥ polynomial Pm,q(x) of degree m + 1 satisfying Pm,q(0) = 0, such that for any smooth u: Z U(1) one has → du P ( u−1du ). k km,q ≤ m,q k km,q Proof Argue by induction on m and use the Sobolev embedding Lr (Z) k ⊂ L2r (Z) for k 1, r n/2. k−1 ≥ ≥ Lemma 6.9 Let Z be a compact, connected Riemannian n–manifold (perhaps with boundary), z Z , m a positive integer, and q 1. Then there is a C > 0 ∈ ≥ such that for any smooth f : Z C one has → (i) f fav C df , k − km,q ≤ k km−1,q (ii) f C( df + f(z) ), k km,q ≤ k km−1,q | | av −1 where f = Vol(Z) Z f is the average of f . R eometry & opology, Volume 9 (2005) G T 66 Kim A Frøyshov
Proof Exercise.
Lemma 6.10 Let Z be a compact Riemannian n–manifold (perhaps with boundary), m a positive integer, and q a real number such that mq >n. Let Φ be a smooth section of some Hermitian vector bundle E Z , Φ 0. Then → 6≡ there exists a C > 0 with the following significance. Let φ1 be a smooth section of E satisfying φ C−1 and w : Z C a smooth map. Define another k 1kq ≤ → section φ2 by w(Φ + φ1)=Φ+ φ2. Then w 1 C( dw + φ φ ). k − km,q ≤ k km−1,q k 2 − 1kq
Proof The equation (w 1)Φ = φ φ (w 1)φ − 2 − 1 − − 1 gives w 1 const ( dw + (w 1)Φ ) k − km,q ≤ · k km−1,q k − kq const ( dw + φ φ + w 1 φ ). ≤ · k km−1,q k 2 − 1kq k − km,qk 1kq Here the first inequality is analogous to Lemma 6.9 (ii). If φ is sufficiently k 1kq small then we can rearrange to get the desired estimate on w 1 . k − km,q
Now let α be a non-degenerate critical point of ϑ. Note that if S = (A, Φ) is any finite energy monopole in temporal gauge over R Y such that Φ < +× k k∞ ∞ and lim inf S α r = 0 t→∞ | − | Z[t,t+1]×Y for some r> 1 then by the results of Section 4 we have [S ] [α] in , hence t → BY S α decays exponentially by Theorem 6.1. In this situation we will simply − say that S is asymptotic to α. (Here we used the fact that for any p> 2 and 1 < r 2p, say, the Lr metric p ≤ on the L configuration space ([0, 1] Y ) is well-defined.) 1 B ×
Lemma 6.11 If S = (A, Φ) is any smooth monopole over R+ Y such that p × Φ < and S α L for some p> 2 then there exists a null-homotopic k k∞ ∞ − ∈ 1 smooth u: R Y U(1) such that u(S) is in temporal gauge and asymptotic + × → to α.
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Proof By Theorem 6.1 there exists a smooth u: R Y U(1) such that + × → u(S) is in temporal gauge and asymptotic to α. Lemma 6.8, Lemma 6.9 (i), and the assumption S α Lp then gives − ∈ 1 u u ∞ 0 as t , k − avkL ([t,t+1]×Y ) → →∞ hence u is null-homotopic.
It follows that all elements of the moduli spaces defined in Subsection 3.4 have smooth representatives that are in temporal gauge over the ends.
Proposition 6.3 Let δ > 0 and suppose ϑ: R has no critical value BY → in the half-open interval (ϑ(α),ϑ(α) + δ] (this implies Condition (O1)). For n = 1, 2,... let S = (A , Φ ) be a smooth monopolee over R Y such that n n n + × p Sn α L1, sup Φn ∞ < , ϑ(Sn(0)) ϑ(α)+ δ, − ∈ n k k ∞ ≤ for some p > 2. Let v : R Y U(1) be a smooth map such that the n + × → sequence v (S ) converges in C∞ over compact subsets of R Y to a con- n n + × figuration S in temporal gauge. Then the following hold: (i) S is asymptotic to a critical point α′ gauge equivalent to α, ′ (ii) If α = α then vn is null-homotopic for all sufficiently large n, and there exist smooth u : R Y U(1) with the following significance: For n + × → every t 0 one has u = 1 on [0,t] Y for all sufficiently large n. ≥ n × Moreover, for any σ < λ+ , q 1 and non-negative integer m one has ≥ u v (S ) S q,σ 0 as n . k n n n − kLm → →∞ Here λ+ is as in Subsection 6.1.
Proof It clearly suffices to prove the proposition when q 2 and mq > 4, ≥ which we assume from now on. By Lemma 4.5 we have
2 ϑSn = ϑ(Sn(0)) ϑ(α) δ (37) R |∇ | − ≤ Z +×Y 2 for each n, hence ϑS δ. Part (i) of the proposition is now a R+×Y |∇ | ≤ consequence of Theorem 6.1 and the following R Claim 6.1 [S(t)] converges in to [α] as t . BY →∞
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Proof of claim For r> 0 let Br Y denote the open r–ball around [α] in 2 ⊂B the L metric, and let Br be the corresponding closed ball. Choose r> 0 such that B2r contains no monopole other than [α]. Assuming the claim does not hold then by Lemma 4.1 one can find a sequence t′ such that t′ as j j j →∞ →∞ and [S(t′ )] B for each j . Because of the convergence of v (S ) it follows j 6∈ 2r n n by a continuity argument that there are sequences n ,t with t ,n as j j j j → ∞ j , such that →∞ [S (t )] B B nj j ∈ 2r \ r for all j . For s R let : R Y R Y be translation by s: ∈ Ts × → × (t,y) = (t + s,y). Ts Again by Lemma 4.1 there are smooth ω : R Y U(1) such that a sub- j + × → sequence of ( )∗(ω (S )) converges in C∞ over compact subsets of R Y Ttj j nj × to some finite energy monopole S′ whose spinor field is pointwise bounded. Moreover, it is clear that ϑ ωj(0) ϑ R must be bounded as j , so ◦ − ∈ −1 → ∞ by passing to a subsequence and replacing ωj by ωjωj0 for some fixed j0 we may arrange that ϑ ω (0) = ϑ for all n. Then ℓ = lim ϑ(S′(t)) must be ◦ j t→−∞ a critical value of ϑ. Since [S′(0)] B B , ∈ 2r \ r S′(0) is not a critical point, whence ∂ ϑ(S′(t)) < 0. Therefore, t|0 ϑ(α)+ δ ℓ>ϑ(S′(0)) >ϑ(α), ≥ contradicting our assumptions. This proves the claim.
We will now prove Part (ii). For τ 0 let ≥ B− = [0,τ] Y, B+ = [τ, ) Y, = [τ,τ + 1] Y. τ × τ ∞ × Oτ × By Lemma 6.11 there is, for every n, a null-homotopic, smooth vn : R+ Y ′′ × → U(1) such that Sn = vn(Sn) is in temporal gauge and asymptotic to α. Note that e e lim lim sup ϑ(Sn(t)) = ϑ(α). t→∞ n→∞ For otherwise we could find an ǫ > 0 and for every natural number j a pair t ,n j such that j j ≥ ϑ(S (t )) ϑ(α)+ ǫ, nj j ≥ and we could then argue as in the proof of Claim 6.1 to produce a critical value of ϑ in the interval (α, α + δ]. Since ϑ = ϑ ′′ it follows from (37) and |∇ Sn | |∇ Sn | Theorem 6.1 that there exists a t 0 such that if τ t then 1 ≥ ≥ 1 ′′ (σ−λ+)τ lim sup Sn α Lq,σ (B+) const e n→∞ k − k m τ ≤ ·
eometry & opology, Volume 9 (2005) G T Monopoles over 4–manifolds containing long necks, I 69 where the constant is independent of τ . Then we also have ′′ (σ−λ+)τ lim sup Sn S Lq,σ(B+) const e . n→∞ k − k m τ ≤ · ′ −1 Set Sn = vn(Sn) and wn = vnvn . Then we get + ′′ ′ q −λ τ lim sup Sn Sn Lm(Oτ ) const e , n→∞ k e − k ≤ · which gives + q −λ τ lim sup dwn Lm(Oτ ) const e n→∞ k k ≤ · by Lemma 6.8. In particular, wn is null-homotopic for all sufficiently large n. Fix y Y and set x = (τ,y ). Choose a sequence τ such that τ as 0 ∈ τ 0 n n → ∞ n and →∞ ′ − Sn S Lq,σ(B ) 0, k − k m τn+1 → (38) ′′ Sn S q,σ + 0 k − kLm (Bτn ) → as n . If α is reducible then by multiplying each vn by a constant and → ∞ ′′ redefining wn,Sn accordingly we may arrange that wn(xτn ) = 1 for all n. (If α is irreducible we keep vn as before.) Then (38)e still holds. Applying Lemma 6.8 together with Lemma 6.9 (ii) (if α is reducible) or Lemma 6.10 (if α is irreducible) we see that e
στn e wn 1 Lq (O ) 0 k − k m+1 τn → as n . →∞ Let β : R R be a smooth function such that β(t)=0 for t 1 and β(t) = 1 → ≤ 3 for t 2 . Set β (t) = β(t τ). Given any function w : C ( , 0] ≥ 3 τ − Oτ → \ −∞ define = exp(β log w) Uw,τ τ where log(exp(z)) = z for complex numbers z with Im(z) < π. Let m′ be | | any integer such that m′q > 4. If w 1 ′ is sufficiently small then k − km ,q 1 ′ const w 1 ′ , kUw,τ − km ,q ≤ · k − km ,q −1 (39) w dw ′ const w 1 ′ . k km −1,q ≤ · k − km ,q To see this recall that for functions on R4 , multiplication defines a continuous q q q ′ map Lm′ Lk Lk for 0 k m . Therefore, if V is the set of all functions q × → ≤ ≤ in Lm′ ( τ , C) that map into some fixed small open ball about 1 C then O ∞ q ∈ w defines a C map V L ′ . This yields the first inequality in (39), 7→ Uw,τ → m and the proof of the second inequality is similar.
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Combining (38) and (39) we conclude that Part (ii) of the proposition holds with − 1 in Bτn , un = in , Uwn,τn Oτn + wn in Bτn+1. This completes the proof of Proposition 6.3.
7 Global compactness
In this section we will prove Theorems 1.3 and 1.4. Given the results of Sec- tions 4 and 5, what remains to be understood is convergence over ends and necks. We will use the following terminology: c-convergence = C∞ convergence over compact subsets.
7.1 Chain-convergence
We first define the notion of chain-convergence. For simplicity we only consider two model cases: first the case of one end and no necks, then the case of one neck and no ends. It should be clear how to extend the notion to the case of multiple ends and/or necks.
Definition 7.1 Let X be a spinc Riemannian 4–manifold with one tubular end R Y , where Y is connected. Let α , α ,... and β ,...,β be elements of +× 1 2 0 k , where k 0 and ϑ(β ) >ϑ(β ) for j = 1,...,k. Let ω M(X; β ) and RY ≥ j−1 j ∈ 0 ~v = (v , . . . , v ), where v Mˇ (β , β ). We say a sequence [S ] M(X; α ) 1 k j ∈ j−1 j n ∈ n chain-convergese to (ω,~v) if there exist, for each n, a smooth map u : X U(1), • n → for j = 1,...,k a smooth map u : R Y U(1), • n,j × → a sequence 0 = t (i) un(Sn) c-converges over X to a representative of ω (in the sense of Sub- section 2.4), (ii) t t as n , n,j − n,j−1 →∞ →∞ (iii) u ( ∗ S ) c-converges over R Y to a representative of v , n,j Ttn,j n × j
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(iv) lim sup [ϑ(S (t + τ)) ϑ(S (t τ))] 0 as τ , n→∞ n n,j−1 − n n,j − → →∞ (v) lim sup [ϑ(S (t + τ)) ϑ(α )] 0 as τ , n→∞ n n,k − n → →∞ where (ii), (iii) and (iv) should hold for j = 1,...,k.
Conditions (iv) and (v) mean, in familiar language, that “no energy is lost in the limit”. As before, denotes translation by s, ie (t,y) = (t + s,y). Ts Ts We now turn to the case of one neck and no ends.
Definition 7.2 In the situation of Subsection 1.4, suppose r =1 and r′ = 0. Let β ,...,β , where k 0 and ϑ(β ) >ϑ(β ), j = 1,...,k. Let ω 0 k ∈ RY ≥ j−1 j ∈ M(X; β , β ) and ~v = (v , . . . , v ), where v Mˇ (β , β ). Let T (n) 0 k 1 k j ∈ j−1 j → ∞ as n . We saye a sequence [S ] M(X(T (n))) chain-converges to (ω,~v) if → ∞ n ∈ there exist, for every n, a smooth map u : X(T (n)) U(1), • n → for j = 1,...,k a smooth map u : R Y U(1), • n,j × → a sequence T (n)= t Definition 7.3 A set of perturbation parameters ~p,~p′ is admissible for a vector ~α′ of critical points if for some t 1 the following holds. Let be the 0 ≥ M disjoint union of all moduli spaces M(X(T ); ~α′;~p;~p′) with min T t . Then j j ≥ 0 we require, for all j, k, that (i) If S is any configuration over [ 1, 1] Y which is a C∞ limit of config- − × j urations of the form S with S and t T 1, then S |[t−1,t+1]×Yj ∈ M | |≤ j − is regular.e e
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(ii) If S is any configuration over [ 1, 1] Y ′ which is a C∞ limit of con- − × k figurations of the form S [t−1,t+1]×Y ′ with S and t 1, then S is | k ∈ M ≥ regular.e e In particular, the zero perturbation parameters are always admissible. The next two propositions describe some properties of chain-convergence.
Proposition 7.1 In the notation of Theorem 1.3, suppose ωn chain-converges to (ω,~v1, . . . ,~vr), where ~αn = β~ for all n, each ~vj is empty, and ~p is admissible for β~. Then ω ω in M(X; β~) with its usual topology. n →
Proof This follows from Proposition 6.3.
In other words, if a sequence ωn in a moduli space M chain-converges to an element ω M , then ω ω in M provided the perturbations are admissible. ∈ n →
Proposition 7.2 In the notation of Theorem 1.4, suppose that the sequence (T (n)) ′ ′ ′ ωn M(X ; ~αn) chain-converges to V = (ω,~v1, . . . ,~vr,~v1, . . . ,~vr′ ), where ∈ ′ minj Tj(n) . Suppose also that the perturbation parameters ~p,~p are → ∞ ′ admissible for each ~αn . Then the following hold: (T (n)) (i) For sufficiently large n there is a smooth map un : X U(1) such ′ ′ ′ → that vn,j = un {0}×Y ′ satisfies vn,j(α )= γ , j = 1, . . . , r . | j n,j j (ii) The chain limit is unique up to gauge equivalence, ie if V, V′ are two chain limits of ω then there exists a smooth u: X# U(1) which is n → translationary invariant over the ends of X# , and such that u(V)= V′ .
In (i), recall that moduli spaces are labelled by critical points modulo null- homotopic gauge transformations. Note that we can arrange that the maps un are translationary invariant over the ends. This allows us to identify the moduli spaces M(X; ~α′ ) and M(X;~γ′), so that we obtain a sequence u (ω ), n 0 n n n ≫ in a fixed moduli space. In (ii) we define u(V) as follows. Let w : R Y U(1) and w′ : R Y ′ U(1) j × j → j × j → be the translationary invariant maps which agree with u on 0 Y and { } × j R Y ′ , respectively. Let w : X U(1) be the map which is translationary + × j → invariant over each end and agrees with u on X:1 . Then u(V) is the result of ′ applying the appropriate maps w, wj, wj to the various components of u(V).
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Proof (i) For simplicity we only discuss the case of one end and no necks, ie the situation of Definition 7.1. The proof in the general case is similar. Using Condition (v) of Definition 7.1 and a simple compactness argument it is easy to see that αn is gauge equivalent to βk for all sufficiently large n. Moreover, Conditions (iv) and (v) of Definition 7.1 ensure that there exist ′ ′ τ,n > 0 such that if n>n then ωn restricts to a genuine monopole on (t + τ, ) Y and on (t + τ,t τ) Y for j = 1,...,k. It then n,k ∞ × n,j−1 n,j − × follows from Proposition 6.3 that v = u satisfies v (α )= β for n n n,k|{0}×Y n n k ≫ 0. (Recall again that α , β are critical points modulo null-homotopic n k ∈ RY gauge transformations, so vn(αn) depends only on the homotopy class of vn .) Similarly, it follows from Theoremse 6.1, 6.2 that u is homotopic to n,j−1|{0}×Y u for j = 1, . . . , r and n 0, where u = u . Therefore, v extends n,j|{0}×Y ≫ n,0 n n over X:0 . (ii) This is a simple exercise.
7.2 Proof of Theorem 1.3
1 By Propositions 3.1,5.6, and 4.3, if each pj has sufficiently small C norm then ~p will be admissible for all ~α. Choose ~p so that this is the case. Set
C0 = inf λjϑ(αn,j) < . − n ∞ Xj
Let Sn be a smooth representative for ωn . The energy assumption on the asymptotic limits of Sn is unaffected if we replace Sn by un(Sn) for some smooth u : X U(1) which is translationary invariant on (t , ) Y for some n → n ∞ × tn > 0. After passing to a subsequence we can therefore, by Proposition 5.3, ′ assume that Sn c-converges over X to some monopole S which is in temporal gauge over the ends. Because ~p is admissible we have that ∂ ϑ(S ) 0 t n|{t}×Yj ≤ for all j, n and t 0. From the energy bound (30) we then see that S′ must ≥ have finite energy. Let γ denote the asymptotic limit of S′ over the end R Y j +× j as guaranteed by Proposition 4.4. Then
lim sup ϑ(αn,j) ϑ(γj) n ≤ for each j . Hence there is a constant C < such that for h = 1, . . . , r and 2 ∞ all n one has C + λ ϑ(α ) λ ϑ(α ) C . 2 h n,h ≥ j n,j ≥− 0 Xj
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Consequently, sup ϑ(αn,j) < . n,j | | ∞ For the remainder of this proof we fix j and focus on one end R Y . For × j simplicity we drop j from notation and write Y, αn instead of Yj, αn,j etc.
After passing to a subsequence we may arrange that ϑ(αn) has the same value L for all n (here we use Condition (O1)). If ϑ(γ)= L then we set k = 0 and the proof is complete. Now suppose ϑ(γ) > L. Then there is an n′ such that ∂ ϑ(S (t)) < 0 for all n n′ , t 0. Set t n ≥ ≥ 1 δ = min x y : x,y are distinct critical values of ϑ: R . 2 {| − | BY → } The minimum exists by (O1). For sufficiently large n we define t 0 e n,1 ≫ implicitly by ϑ(S (t )) = ϑ(γ) δ. n n,1 − It is clear that t as n . Moreover, Definition 7.1 (iv) must hold for n,1 →∞ →∞ j = 1. For otherwise we can find ǫ> 0 and sequences τ , n with τ ,n ℓ ℓ ℓ ℓ →∞ as ℓ , such that →∞ ϑ(S (τ )) ϑ(S (t τ )) >ǫ (40) nℓ ℓ − nℓ nℓ,1 − ℓ for every ℓ. As in the proof of Claim 6.1 there are smooth uℓ : R Y U(1) ∗ × → satisfying ϑ uℓ(0) = ϑ such that a subsequence of uℓ( t 1 Snℓ ) c-converges ◦ T nℓ, over R Y to a finite energy monopole S in temporal gauge.e The asymptotic × limit γ of S ate must satisfy e −∞ ǫ ϑ(γ) ϑe(γ) < δ, e e ≤ − where the first inequality follows from (40). This contradicts the choice of δ. Therefore, Definition 7.1 (iv) holds for j =e 1 as claimed.
After passing to a subsequence we can find un,1 : R Y U(1) such that ∗ R × → ′ un,1( tn,1 Sn) c-converges over Y to some finite energy monopole S1 in T ± × ′ temporal gauge. Let β1 denote the limit of S1 at . A simple compactness − ±∞ argument shows that γ and β1 are gauge equivalent, so we can arrange that γ = β− by modifying the u by a fixed gauge transformation R Y U(1). 1 n,1 × → As in the proof of Proposition 7.2 (i) we see that un,1 must be null-homotopic + + for all sufficiently large n. Hence ϑ(β1 ) L. If ϑ(β1 ) = L then we set ≥ + k = 1 and the proof is finished. If on the other hand ϑ(β1 ) > L then we continue the above process. The process ends when, after passing successively ± + − to subsequences and choosing un,j,tn,j, βj for j = 1,...,k (where βj−1 = βj , and u is null-homotopic for n 0) we have ϑ(β+) = L. This must occur n,j ≫ k after finitely many steps; in fact k (2δ)−1(ϑ(γ) L). ≤ −
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7.3 Proof of Theorem 1.4
For simplicity we first consider the case when there is exactly one neck (ie r = 1), and we write Y = Y1 etc. We will make repeated use of the local compactness results proved earlier.
Let Sn be a smooth representative of ωn . After passing to a subsequence we can find smooth maps u : X(T (n)) ( 0 Y ) U(1) n \ { }× → ′ such that Sn = un(Sn) c-converges over X to some finite energy monopole S which is in temporal gauge over the ends. Introduce the temporary notation S (t)= S e , and similarly for S and u . For 0 τ < T (n) set n n|{t}×Y n n ≤ Θ = ϑ(S ( T (n)+ τ)) ϑ(S (T (n) τ)). τ,n n − e − n − Let u± = u ( T (n)) and n n ± I± = 2π η [u±], n j ∧ n ZY cf Equation (1). Since Θ is bounded as n , it follows that I+ I− 0,n e → ∞ n − n is bounded as n . By Condition (O1) there is a q > 0 such that qI± is → ∞ n integral for all n. Hence we can arrange, by passing to a subsequence, that I+ I− is constant. In particular, n − n I+ I+ = I− I−. n − 1 n − 1 Choose a smooth map w : X U(1) which is translationary invariant over the −1 → ends, and homotopic to u1 over X:0 . After replacing un by wun for every n + − ± we then obtain In = In . Set In = In . We now have Θ = ϑ(S ( T (n))) ϑ(S (T (n))). τ,n n − − n Let β , β′ denote the asymptotic limits of S′ over the ends ι+(R Y ) and 0 e e + × ι−(R Y ), respectively. Set + × ′ L = lim lim Θτ,n = ϑ(β0) ϑ(β ). τ→∞ n→∞ − Since Θ 0 for τ 0 we have L 0. τ,n ≥ ≥ ≥ Suppose L = 0. Then a simple compactness argument shows that there is a smooth v : Y U(1) such that v(β ) = β′ . Moreover, there is an n such → 0 0 that v u− u+ for n n , where means ‘homotopic’. Therefore, we can n ∼ n ≥ 0 ∼ find a smooth z : X U(1) which is translationary invariant over the ends −1→ and homotopic to un0 over X:0 , such that after replacing un by zun for every
eometry & opology, Volume 9 (2005) G T 76 Kim A Frøyshov n we have that β = β′ and u+ u− . In that case we can in fact assume 0 n ∼ n that u is a smooth map X(T (n)) U(1). The remainder of the proof when n → L = 0 (dealing with convergence over the ends) is now a repetition of the proof of Theorem 1.3.
We now turn to the case L > 0. For large n we must then have ∂tSn(t) < 0 for t T (n). Let δ be as in the proof of Theorem 1.3. We define t | | ≤ n,1 ∈ ( T (n), T (n)) implicitly for large n by − ϑ(β0)= ϑ(Sn(tn,1)) + In + δ. Then t T (n) as n . As in the proof of Theorem 1.3 one sees | n,1 ± |→∞ → ∞ that lim sup [ϑ(Sn( T (n)+ τ)) ϑ(Sn(tn,1 τ))] 0 n→∞ − − − → as τ , and after passing to a subsequence we can find smooth u : R → ∞ n,1 × Y U(1) such that u ( ∗ S ) c-converges over R Y to a finite energy → n,1 Ttn,1 n × monopole S′ in temporal gauge whose asymptotic limit at is β . Let β 1 −∞ 0 1 denote the asymptotic limit of S′ at . We now repeat the above process. 1 ∞ The process ends when, after passing successively to subsequences and choosing un,j,tn,j, βj for j = 1,...,k one has that
lim sup [ϑ(Sn(tn,k + τ)) ϑ(Sn(T (n) τ))] 0 n→∞ − − → ′ as τ . As in the case L = 0 one sees that βk, β must be gauge equivalent, →∞ ′ and after modifying un, un,j one can arrange that βk = β . This establishes chain-convergence over the neck. As in the case L = 0 we can in fact assume that u is a smooth map X(T (n)) U(1), and the rest of the proof when L> 0 n → is again a repetition of the proof of Theorem 1.3. In the case of multiple necks one applies the above argument successively to each neck. In this case, too, after passing to a subsequence one ends up with smooth maps u : X(T (n)) U(1) such that u (S ) c-converges over X . One n → n n can then deal with convergence over the ends as before.
8 Transversality
We will address two kinds of transversality problems: non-degeneracy of critical points of the Chern–Simons–Dirac functional, and regularity of moduli spaces over 4–manifolds. In this section we do not assume Condition (O1). Recall that a subset of a topological space Z is called residual if it contains a countable intersection of dense open subsets of Z .
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8.1 Non-degeneracy of critical points
Lemma 8.1 Let Y be a closed, connected, Riemannian spinc 3–manifold and η any closed (smooth) 2–form on Y . Let G∗ be the set of all ν Ω1(Y ) ∈ such that all irreducible critical points of ϑη+dν are non-degenerate. Then G∗ Ω1(Y ) is residual, hence dense (with respect to the C∞ topology). ⊂ Proof The proof is a slight modification of the argument in [15]. For 2 k k ≤ ≤ and δ > 0 let Wk,δ be the space of all 1–forms ν on Y of class C which ∞ k satisfy dν 1 < δ. Let W have the C topology. For 1 k< we define k kC k,δ ≤ ∞ a –equivariant smooth map G Υ : ∗ L2(Y ; iR) W L2(Y ; iΛ1 S), k C × 1 × k,δ → ⊕ (B, Ψ,ξ,ν) ξ + ϑ (B, Ψ), 7→ IΨ ∇ η+dν where acts trivially on forms, and by multiplication on spinors. Now if G Υk(B, Ψ,ξ,ν)=0 then
ξ 2 = ϑ (B, Ψ), ξ = 0 kIΨ k2 − h∇ η+dν IΨ i ZY by (12), which implies ξ = 0 since Ψ = 0. The derivative of Υ at a point 6 k x = (B, Ψ, 0,ν) is DΥ (x)(b, ψ, f, v)= H (b, ψ)+ f + (i dv, 0). (41) k (B,Ψ) IΨ ∗
Let (B, Ψ) be any irreducible critical point of ϑη . We will show that P = 2 DΥk(B, Ψ, 0, 0) is surjective. Note that altering (B, Ψ) by an L2 gauge trans- formation u has the effect of replacing P by uP u−1 . We may therefore assume that (B, Ψ) is smooth. Since P = + H has surjective symbol, the 1 IΨ (B,Ψ) image of the induced operator L2 L2 is closed and has finite codimension. 1 → The same must then hold for im(P ). Suppose (b, ψ) L2 is orthogonal to ∗ ∈ im(P ), ie db =0 and P1 (b, ψ) = 0. The second equation implies that b and ψ are smooth, by elliptic regularity. Writing out the equations we find as in [15] that on the complement of Ψ−1(0) we have b = idr for some smooth function − r : Y Ψ−1(0) R. We now invoke a result of B¨ar [4] which says that, because \ → B is smooth and Ψ 0, the equation ∂ Ψ = 0 implies that the zero-set of Ψ 6≡ B is contained in a countable union of smooth 1–dimensional submanifolds of Y . In particular, any smooth loop in Y can be deformed slightly so that it misses −1 Ψ (0). Hence b is exact. From B¨ar’s theorem (or unique continuation for ∂B , which holds when B is of class C1 , see [19]) we also deduce that the comple- ment of Ψ−1(0) is dense and connected. Therefore, f has a smooth extension to all of Y , and as in [15] this gives (b, ψ) = 0. Hence P is surjective.
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Consider now the vector bundle ∗ 2 1 2 ∗ 2 E = ( L (Y ; iΛ S)) L1(Y ; iR) L1(Y ; iR). C ×G ⊕ × →B × For 1 k < the map Υ defines a smooth section σ of the bundle ≤ ∞ k k,δ E W ∗ L2(Y ; iR) W . × k,δ →B × 1 × k,δ By the local slice theorem, a zero of Υk is a regular point of Υk if and only if the corresponding zero of σk,δ is regular. Since surjectivity is an open property for bounded operators between Banach spaces, a simple compactness argument shows that the zero-set of σ2,δ is regular when δ > 0 is sufficiently small. Fix such a δ. Observe that the question of whether the operator (41) is surjective for a given x is independent of k. Therefore, the zero-set Mk,δ of σk,δ is regular for 2 k < . In the remainder of the proof assume k 2. ≤ ∞ ≥ For any ρ> 0 let be the set of elements [B, Ψ] satisfying Bρ ∈B Ψ ρ. | |≥ ZY Define M M similarly. For any given ν , the formula for Υ defines a k,δ,ρ ⊂ k,δ k Fredholm section of E which we denote by σν . Let Gk,δ,ρ be the set of those ν W such that σ has only regular zeros in 0 . For k< let ∈ k,δ ν Bρ × { } ∞ π : M W k,δ → k,δ be the projection, and Σ M the closed subset consisting of all singular ⊂ k,δ points of π. A compactness argument shows that π restricts to a closed map on Mk,δ,ρ , hence G = W π(M Σ) k,δ,ρ k,δ \ k,δ,ρ ∩ is open in Wk,δ . On the other hand, applying the Sard–Smale theorem as in [11, Section 4.3] we see that Gk,δ,ρ is residual (hence dense) in Wk,δ . Because W∞,δ is dense in Wk,δ , we deduce that G∞,δ,ρ is open and dense in W∞,δ . But then G 1 ∞,δ, n N n\∈ is residual in W , and this is the set of all ν W such that σ has only ∞,δ ∈ ∞,δ ν regular zeros.
An irreducible critical point of ϑη+dν is non-degenerate if and only if the cor- responding zero of σν is regular. Thus we have proved that among all smooth 1–forms ν with dν C1 < δ, those ν for which all irreducible critical points of k k ∞ ϑη+dν are non-degenerate make up a residual subset in the C topology. The
eometry & opology, Volume 9 (2005) G T Monopoles over 4–manifolds containing long necks, I 79 same must hold if η is replaced with η + dν for any ν Ω1(Y ), so we conclude ∈ that G∗ is locally residual in Ω1(Y ), ie any point in G∗ has a neighbourhood V such that G∗ V is residual in V . Hence G∗ is residual in Ω1(Y ). (This ∩ last implication holds if Ω1(Y ) is replaced with any second countable, regular space.)
Proposition 8.1 Let Y be a closed, connected, Riemannian spinc 3–manifold and η any closed 2–form on Y such that either b1(Y ) = 0 or η = 0. Let G be 1 6 the set of all ν Ω (Y ) such that all critical points of ϑη+dν are non-degenerate. ∈ 1 ∞ Then G is open and dense in Ω (Y ) with respect to the C topology.e
Proof A compactness argument shows that G is open. If b1(Y ) > 0 then ϑη+dν has no reducible critical points and the result follows from Lemma 8.1.
Now suppose b1(Y ) = 0. Then we may assume η = 0. For 0 k k k ≤ ≤ ∞ let Wk be the space of 1–forms on Y of class C , with the C topology. If ν W then ϑ has up to gauge equivalence a unique reducible critical point, ∈ 1 dν represented by (B iν, 0) for any smooth spin connection B over Y with Bˇ − flat. This critical point is non-degenerate precisely when 2 ker(∂B−iν )=0 in L1 . (42) ′ Let Gk be the set of all ν Wk such that (42) holds. This is clearly an open ∈ ′ subset of Wk . The last part of the proof of [15, Proposition 3] shows that Gk ′ is residual (hence dense) in Wk for 2 k < . Hence G∞ is open and dense 1 ≤ ∞ in Ω (Y )= W∞ . Now apply Lemma 8.1.
Marcolli [25] proved a weaker result in the case b1(Y ) > 0, allowing η to vary freely among the closed 2–forms.
8.2 Regularity of moduli spaces
The following lemma will provide us with suitable Banach spaces of perturba- tion forms.
Lemma 8.2 Let X be a smooth n–manifold, K X a compact, codimen- ⊂ sion 0 submanifold, and E X a vector bundle. Then there exists a Banach → space W consisting of smooth sections of E supported in K , such that the following hold: (i) The natural map W Γ(E ) is continuous with respect to the C∞ → |K topology on Γ(E ). |K
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(ii) For every point x int(K) and every v E there exists a section ∈ ∈ x s Γ(E) with s(x) = v and a smooth embedding g : Rn X with ∈ → g(0) = x such that for arbitrarily small ǫ > 0 there are elements of W of the form fs where f : X [0, 1] is a smooth function which vanishes → outside g(Rn) and satisfies 0, z 2ǫ, f(g(z)) = | |≥ (1, z ǫ. | |≤ Proof Fix connections in E and T X , and a Euclidean metric on E . For any sequence a = (a , a ,... ) of positive real numbers and any s Γ(E) set 0 1 ∈ k s a = sup ak s ∞ k k 0≤k<∞ k∇ k and W = s Γ(E) : supp(s) K, s < . a { ∈ ⊂ k ka ∞} Then W = W , equipped with the norm , clearly satisfies (i) for any a k · ka a. We claim that one can choose a such that (ii) also holds. To see this, first observe that there is a finite dimensional subspace V Γ(E) such that ⊂ V E , s s(x) → x 7→ is surjective for every x K . Fix a smooth function b: R [0, 1] satisfying ∈ → 1, t 1, b(t)= ≤ 0, t 4. ( ≥ We use functions f that in local coordinates have the form f (z)= b(r z 2), r | | k where r 0. Note that for each k there is a bound f k const r where ≫ k rkC ≤ · the constant is independent of r 1. It is now easy to see that a suitable ≥ sequence a can be found.
In the next two propositions, X, ~α, µ will be as in Subsection 1.3. Let K X be ⊂ any non-empty compact codimension 0 submanifold. Let W be a Banach space of smooth self-dual 2–forms on X supported in K , as provided by Lemma 8.2. The following proposition will be used in the proof of Theorem 1.2.
Proposition 8.2 In the above situation, let G be set of all ν W such that ∈ all irreducible points of the moduli space M(X; ~α; µ + ν; 0) are regular (here p = 0 for each j ). Then G W is residual, hence dense. j ⊂
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There is another version of this proposition where W is replaced with the Fr´echet space of all smooth self-dual 2–forms on X supported in K , at least if one assumes that (O1) holds for each pair Yj, ηj and that (A) holds for X, ηj, λj . The reason for the extra assumptions is that the proof then seems to require global compactness results (cf the proof of Lemma 8.1). e
Proof We may assume X is connected. Let Θ be as in Subsection 3.4. Then (S,ν) Θ(S,µ + ν, 0) 7→ e defines a smooth map e f : ∗ W Lp,w(X; iΛ+ S−), C × → ⊕ where ∗ = ∗(X; ~α). We will show that 0 is a regular value of f . Suppose C C f(S,ν) = 0 and write S = (A, Φ). We must show that the derivative P = Df(S,ν) is surjective. Because of the gauge equivariance of f we may assume that S is smooth. Let P1 denote the derivative of f( ,ν) at S . Since the p,w · image of P1 in L is closed and has finite codimension, the same holds for the image of P . Let p′ be the exponent conjugate to p and suppose (z, ψ) ′ ∈ Lp ,−w(X; iΛ+ S−) is L2 orthogonal to the image of P , ie ⊕ P (a, φ, ν′), (z, ψ) = 0 h i ZX p,w ′ ′ ∗ for all (a, φ) L1 and ν W . Taking ν = 0 we see that P1 (z, ψ) = 0. ∗ ∈ ∈ Since P1 has injective symbol, z, ψ must be smooth. On the other hand, taking a, φ = 0 and varying ν′ we find that z = 0 by choice of W . By assumption, |K Φ is not identically zero. Since DAΦ = 0, the unique continuation theorem 2 in [19] applied to DA says that Φ cannot vanish in any non-empty open set. Hence Φ must be non-zero at some point x in the interior of K . Varying a ∗ alone near x one sees that ψ vanishes in some neighbourhood of x. But P1P1 has the same symbol as D2 d+(d+)∗ , so another application of the same A ⊕ unique continuation theorem shows that (z, ψ) = 0. Hence P is surjective. Consider now the vector bundle E = ∗ Lp,w(X; iΛ+ S−) C ×G ⊕ over ∗ . The map f defines a smooth section σ of the bundle B E W ∗ W. × →B × Because of the local slice theorem and the gauge equivariance of f , the fact that 0 is a regular value of f means precisely that σ is transverse to the zero-section. Since σ( ,ν) is a Fredholm section of E for any ν , the proposition follows by · another application of the Sard–Smale theorem.
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We will now establish transversality results for moduli spaces of the form M(X, ~α) or M(α, β) involving perturbations of the kind discussed in Subsec- tion 3.3. For the time being we limit ourselves to the case where the 3–manifolds Y,Yj are all rational homology spheres. We will use functions hS that are a small modification of those in [15]. To define these, let Y be a closed Rieman- c nian spin 3–manifold satisfying b1(Y ) = 0, and ϑ the Chern–Simons–Dirac functional on Y defined by some closed 2–form η. Choose a smooth, non- negative function χ: R R which is supported in the interval ( 1 , 1 ) and → − 4 4 satisfies χ =1. If S is any L2 configuration over a band (a 1 , b + 1 ) where 1 − 4 4 a b define the smooth function ϑ : [a, b] R by ≤ R S →
ϑS(T )= e χ(T t)ϑ(St) dt, R − Z where we interpret the righte hand side as an integral over R Y . A simple × exercise, using the Sobolev embedding theorem, shows that if S S weakly n → in L2 over (a 1 , b + 1 ) Y then ϑ ϑ in C∞ over [a, b]. 1 − 4 4 × Sn → S Choose a smooth function c: R R with the following properties: → e e c′ > 0, • c and all its derivatives are bounded, • c(t)= t for all critical values t of ϑ, • where c′ is the derivative of c. The last condition is added only for convenience.
For any L2 configuration S over (a 1 , b + 1 ) Y with a b define 1 − 2 2 × ≤
hS(t)= χ(t1)c(ϑS (t t1))dt1. R − Z It is easy to verify that hS satisfies the propertiese (P1)–(P3). It remains to choose Ξ and P. Choose one compact subinterval (with non- empty interior) of each bounded connected component of R crit(ϑ), where \ crit(ϑ) is the set of critical values of ϑ. Let Ξ be the union of these compact subintervals. Let P = P be a Banach space of 2–forms on R Y supported Y × in Ξ Y as provided by Lemma 8.2. × We now return to the situation described in the paragraph preceding Proposi- tion 8.2. Let W ′ W be the open subset consisting of those elements ν that ⊂ satisfy ν 1 < 1. Let Π denote the set of all ~p = (p ,..., p ) where p P k kC δ 1 r j ∈ Yj and p 1 < δ for each j . k j kC
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Proposition 8.3 Suppose each Y is a rational homology sphere and K X . j ⊂ :0 Then there exists a δ > 0 such that the following holds. Let G be the set of all (ν,~p) W ′ Π such that every irreducible point of the moduli space ∈ × δ M(X; ~α; µ + ν;~p) is regular. Then G W ′ Π is residual, hence dense. ⊂ × δ
It seems necessary here to let ~p vary as well, because if any of the pj is non-zero then the linearization of the monopole map is no longer a differential operator, and it is not clear whether one can appeal to unique continuation as in the proof of Proposition 8.2.
Proof To simplify notation assume r = 1 and set Y = Y1 , α = α1 etc. (The proof in the general case is similar.) Note that (A) is trivially satisfied, since each Yj is a rational homology sphere. Therefore, by Propositions 4.3 and 5.6, if δ > 0 is sufficiently small then for any (ν, p) W ′ Π and ∈ × δ [S] M(X; α; µ + ν; p) one has that either ∈ (i) [S ]= α for t 0, or t ≥ (ii) ∂ ϑ(S ) < 0 for t 0. t t ≥ As in the proof of Proposition 8.2 it suffices to prove that 0 is a regular value of the smooth map f : ∗ W ′ Π Lp,w, C × × δ → (S,ν, p) Θ(S,µ + ν, p). e 7→ The smoothness of the perturbation term g(S, p) = qh follows from the e S,p smoothness of the map (19), for by (P1) there exist a t0 and a neighbourhood ′ U of S such that hS′ (t) Ξ for all t>t0 and S U . ⊂C 6∈ ′ ∈ Now suppose f(S,ν, p)=0 and (z, ψ) Lp ,−w is orthogonal to the image of ∈ Df(S,ν, p). We will show that z is orthogonal to the image of T = Dg(S, p), or equivalently, thate (z, ψ) is orthogonal to the image of Df(S,ν), where f = f g − ase before. The latter implies (z, ψ) = 0 by the proof of Proposition 8.2. e Let h : [ 1 , ) R be defined in terms of the restriction of S to R Y . S 2 ∞ → + × If h (J) R Ξ for some compact interval J then by (P1) one has that S ⊂ \ h ′ (J) R Ξ for all S′ in some neighbourhood of S in . Therefore, all S ⊂ \ C elements of im(T ) vanish on h−1(R Ξ) Y . S \ × We now digress to recall that if u is any locally integrable function on Rn then the complement of the Lebesgue set of u has measure zero, and if v is any continuous function on Rn then any Lebesgue point of u is also a Lebesgue point of uv. The notion of Lebesgue set also makes sense for sections τ of a
eometry & opology, Volume 9 (2005) G T 84 Kim A Frøyshov vector bundle of finite rank over a finite dimensional smooth manifold M . In that case a point x M is called a Lebesgue point of τ if it is a Lebesgue point ∈ in the usual sense for some (hence any) choice of local coordinates and local trivialization of the bundle around x. Returning to our main discussion, there are now two cases: If (i) above holds 1 then hS(t) = ϑ(α) Ξ for t 2 , whence T = 0 and we are done (recall 6∈ −1 ≥ the overall assumption q (0) = X 3 made in Subsection 3.4). Otherwise (ii) : 2 1 must hold. In that case we have ∂tc(ϑ(t)) < 0 for t 4 and ∂thS(t) < 0 1 ′ ≥ for t . Since z is orthogonal to qh p′ for all p P we conclude that ≥ 2 S, ∈ Y z(t,y) = 0 for every Lebesgue point (t,ye) of z with t> 3 and h (t) int(Ξ). 2 S ∈ Since h−1(∂Ξ) ( 3 , ) is a finite set, z must vanish almost everywhere in S ∩ 2 ∞ [h−1(Ξ) ( 3 , )] Y . Combining this with our earlier result we deduce that S ∩ 2 ∞ × z is orthogonal to im(T ).
In the next proposition (which is similar to [15, Proposition 5]) let Πδ be as above with r = 1, and set Y = Y1 .
Proposition 8.4 In the situation of Subsection 1.2, suppose Y is a rational homology sphere and α, β = . Then there exists a δ > 0 such that ∈ RY RY the following holds. Let G be the set of all p Π such that every point in ∈ δ M(α, β; p) is regular. Then G Π eis residual, hence dense. ⊂ δ
Proof If α = β then an application of Proposition 4.3 shows that if p 1 is k kC sufficiently small then M(α, β; p) consists of a single point represented by α, which is regular because α is non-degenerate.
If α = β and p 1 is sufficiently small then for any [S] M(α, β; p) one has 6 k kC ∈ ∂tϑ(St) < 0 for all t. Moreover, the moduli space contains no reducibles, since α, β cannot both be reducible. The proof now runs along the same lines as that of Proposition 8.3. Note that the choice of Ξ is now essential: it ensures that im(hS) = (ϑ(α),ϑ(β)) contains interior points of Ξ.
9 Proof of Theorems 1.1 and 1.2
In these proofs we will only use genuine monopoles. Proof of Theorem 1.1 We may assume Y is connected. Let η be a closed non-exact 2–form on Y which is the restriction of a closed form on Z . Let Y have a metric of positive scalar curvature. If s = 0 is a small real number 6
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then ϑsη will have no irreducible critical points, by the a priori estimate on the spinor fields and the positive scalar curvature assumption. If in addition s[η] πc ( ) = 0 then ϑ will have no reducible critical points either. − 1 LY 6 sη Choose a spinc Riemannian 4–manifold X as in Subsection 1.4, with r = 1, r′ = 0, such that there exists a diffeomorphism X# Z which maps 0 Y → { }× 1 isometrically onto Y . Let η1 be the pull-back of sη. Then (B1) is satisfied (but perhaps not (B2)), so it follows from Theorem 1.4 that M(X(T )) is empty for T 0. ≫
We will now define an invariant h for closed spinc 3–manifolds Y that satisfy b1(Y ) = 0 and admit metrics with positive scalar curvature. Let g be such a metric on Y . Recall that for the unperturbed Chern–Simons–Dirac functional ϑ the space of critical points modulo gauge equivalence consists of a single RY point θ, which is reducible. Let (B, 0) be a representative for θ. Let Y1,...,Yr be the connected components of Y and choose a spinc Riemannian 4–manifold X with tubular ends R Y , j = 1, . . . , r (in the sense of Subsection 1.3) and + × j a smooth spin connection A over X such that the restriction of Aˇ to R Y is + × equal to the pull-back of Bˇ . (The notation here is explained in Subsection 3.1.) Define
1 2 h(Y, g) = indC(D ) (c ( ) σ(X)) A − 8 1 LX − 1 = (dim M(X; θ) d(X)+ b (X)), 2 − 0 2 2 where DA : L1 L , ‘dim’ is the expected dimension, and d(X) is the quantity → 1 defined in Subsection 1.1. Since indC(D ) = (c ( )2 σ(X)) when X A 8 1 LX − is closed, it follows easily from the addition formula for the index (see [10, Proposition 3.9]) that h(Y, g) is independent of X and that h( Y, g)= h(Y, g). − − Clearly,
h(Y, g)= h(Yj, gj), Xj where gj is the restriction of g to Yj . To show that h(Y, g) is independent of g we may therefore assume Y is connected. Suppose g′ is another positive scalar curvature metric on Y and consider the spinc Riemannian manifold X = R Y where the metric agrees with 1 g on ( , 1] Y and with × × −∞ − × 1 g′ on [1, ) Y . Theorem 1.3 and Proposition 7.2 (ii) say that M(X; θ,θ) is × ∞ × compact. This moduli space contains one reducible point, and arguing as in the
eometry & opology, Volume 9 (2005) G T 86 Kim A Frøyshov proof of [15, Theorem 6] one sees that the moduli space must have non-positive (odd) dimension. Thus, 1 h(Y, g′)+ h( Y, g)= (dim M(X; θ,θ) + 1) 0. − 2 ≤ This shows h(Y )= h(Y, g) is independent of g.
Proof of Theorem 1.2 Let each Yj have a positive scalar curvature metric. Choose a spinc Riemannian 4–manifold X as in Subsection 1.4, with r′ = 0 and with the same r, such that there exists a diffeomorphism f : X# Z → which maps 0 Y isometrically onto Y . Then (B1) is satisfied (but perhaps { }× j j not (B2)). Let X0 be the component of X such that W = f((X0):1). For each j set η = 0 and let α be the unique (reducible) critical point. Choose j j ∈ RYj a reference connection A as in Subsection 3.4 and set A = A . Since o 0 o|X0 each αj has representatives of the form (B, 0) where Bˇ is flat it follows that Fˆ(Ao) is compactly supported. In the following, µ will denote the (compactly supported) perturbation 2–form on X and µ0 its restriction to X0 . Let + be the space of self-dual closed L2 2–forms on X . Then dim + = H 0 H b+(X ) > 0, so + contains a non-zero element z. By unique continuation 2 0 H for harmonic forms we can find a smooth 2–form µ0 on X0 , supported in any ˆ+ + 2 ˆ+ given small ball, such that F (A0)+ iµ0 is not L orthogonal to z. (Here F is the self-dual part of Fˆ .) Then Fˆ+(A )+ iµ+ im(d+ : Lp,w Lp,w), 0 0 6∈ 1 → where w is the weight function used in the definition of the configuration space. Hence M(X0; ~α) contains no reducible monopoles. After perturbing µ0 in a small ball we can arrange that M(X0; ~α) is transversally cut out as well, by Proposition 8.2. To prove (i), recall that dim M(X ; ~α)= d(W ) 1 + 2 h(Y ), (43) 0 − j Xj so the inequality in (i) simply says that
dim M(X0; ~α) < 0, hence M(X0; ~α) is empty. Since there are no other moduli spaces over X0 , it follows from Theorem 1.4 that M(X(T )) is empty when min T 0. j j ≫ We will now prove (ii). If M(X(T )) has odd or negative dimension then there is nothing to prove, so suppose this dimension is 2m 0. Since M(X ; ~α) ≥ 0 contains no reducibles we deduce from Theorem 1.4 that M(X(T )) is also free
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of reducibles when minj Tj is sufficiently large. Let B X0 be a compact ∗ p⊂ 4–ball and (B) the Banach manifold of irreducible L1 configurations over Bp B modulo L2 gauge transformations. Here p > 4 should be an even integer to ensure the existence of smooth partitions of unity. Let L ∗(B) be the → B natural complex line bundle associated to some base-point in B, and s a generic section of the m–fold direct sum mL. For min T 0 let j j ≫ S(T ) M(X(T )), S M(X ; ~α) ⊂ 0 ⊂ 0 be the subsets consisting of those elements ω that satisfy s(ω B) =0. By | (T ) assumption, S0 is a submanifold of codimension 2m. For any T for which S is transversely cut out the Seiberg–Witten invariant of Z is equal to the number of points in S(T ) counted with sign. Now, the inequality in (ii) is equivalent to
d(W ) + 2 h(Yj) < d(Z) = 2m + 1, Xj which by (43) gives dim S = dim M(X ; ~α) 2m< 0. 0 0 − Therefore, S is empty. By Theorem 1.4, S(T ) is empty too when min T 0, 0 j j ≫ hence SW(Z)=0.
A Patching together local gauge transformations
In the proof of Lemma 4.1 we encounter sequences Sn of configurations such that for any point x in the base-manifold there is a sequence vn of gauge trans- formations defined in a neighbourhood of x such that vn(Sn) converges (in some Sobolev norm) in a (perhaps smaller) neighbourhood of x. The problem then is to find a sequence un of global gauge transformations such that un(Sn) converges globally. If vn, wn are two such sequences of local gauge transfor- −1 mations then vnwn will be bounded in the appropriate Sobolev norm, so the problem reduces to the lemma below. This issue was discussed by Uhlenbeck in [32, Section 3]. Our approach has the advantage that it does not involve any “limiting bundles”.
Lemma A.1 Let X be a Riemannian manifold and P X a principal G– → bundle, where G is a compact subgroup of some matrix algebra Mr(R). Let Mr(R) be equipped with an AdG –invariant inner product, and fix a connection in the Euclidean vector bundle E = P Ad M (R) (which we use to define × G r
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Sobolev norms of automorphisms of E ). Let U ∞ , V ∞ be open covers { i}i=1 { i}i=1 of X such that Ui ⋐ Vi for each i. We also assume that each Vi is the interior of a compact codimension 0 submanifold of X , and that ∂Vi and ∂Vj intersect transversally for all i = j . For each i and n = 1, 2,... let v be a continuous 6 i,n automorphism of P . Suppose v v−1 converges uniformly over V V for |Vi i,n j,n i ∩ j each i, j (as maps into E ). Then there exist a sequence of positive integers n n , • 1 ≤ 2 ≤··· for each positive integer k an open subset W X with • k ⊂ k k U ⋐ W V , i k ⊂ i i[=1 i[=1 for each k and n n a continuous automorphism w of P , • ≥ k k,n |Wk such that (i) If 1 j k and n n then w = w on j U , ≤ ≤ ≥ k j,n k,n i=1 i (ii) For each i, k the sequence w v−1 converges uniformly over W V , k,n i,n S k ∩ i (iii) If 1 p< , and m> n is an integer such that v Lp for all i, n ≤ ∞ p i,n ∈ m,loc then w Lp for all k and n n . If in addition k,n ∈ m,loc ≥ k
−1 p sup vi,nvj,n Lm(Vi∩Vj ) < for all i, j n k k ∞ then −1 p sup wk,nvi,n Lm(Wk∩Vi) < for all k,i. n≥nk k k ∞
The transversality condition ensures that the Sobolev embedding theorem holds for V V (see [1]). Note that this condition can always be achieved by shrinking i∩ j the Vi ’s a little.
Proof Let N ′ LG be a small Ad invariant open neighbourhood of 0. ⊂ G Then exp: LG G maps N ′ diffeomorphically onto an open neighbourhood → N of 1. Let f : N N ′ denote the inverse map. Let Aut(P ) the bundle of → fibre automorphisms of P and gP the corresponding bundle of Lie algebras. Set N = P N g and let exp−1 : N Aut(P ) be the map defined by ×AdG ⊂ P → f .
Set w1,n = v1,n and W1 = V1 . Now suppose wk,n , Wk have been chosen for 1 k<ℓ, where ℓ 2, such that (i)–(iii) hold for these values of k. Set ≤ ≥ z = w (v )−1 on W V . According to the induction hypothesis the n ℓ−1,n ℓ,n ℓ−1 ∩ ℓ
eometry & opology, Volume 9 (2005) G T Monopoles over 4–manifolds containing long necks, I 89 sequence z converges uniformly over W V , hence there exists an integer n ℓ−1 ∩ ℓ n n such that y = (z )−1z takes values in N for n n . ℓ ≥ ℓ−1 n nℓ n ≥ ℓ Choose an open subset X which is the interior of a compact codimension 0 W ⊂ submanifold of X , and which satisfies ℓ−1 U ⋐ ⋐ W . i W ℓ−1 i[=1 We also require that ∂ intersect ∂V ∂V transversally for all i, j . (For W i ∩ j instance, one can take = α−1([0,ǫ]) for suitable ǫ, where α: X [0, 1] is W → any smooth function with α =0 on ℓ−1U and α =1 on W .) Choose also ∪i=1 i ℓ−1 a smooth, compactly supported function φ: W R with φ = 1. Set ℓ−1 → |W W = V and for n n define an automorphism w of P by ℓ W∪ ℓ ≥ ℓ ℓ,n |Wℓ wℓ−1,n on , −1 W wℓ,n = z exp(φ exp y )v on W V , nℓ n ℓ,n ℓ−1 ∩ ℓ z v on V supp(φ). nℓ ℓ,n ℓ \ Then (i)–(iii) hold for k = ℓ as well. To see that (iii) holds, note that our transversality assumptions guarantee that the Sobolev embedding theo- p rem holds for W V and for all V V . Since mp>n, Lm is therefore a ℓ−1 ∩ ℓ i ∩ j Banach algebra for these spaces (see [1]). Recalling the proof of this fact, and p the behaviour of Lm under composition with smooth maps on the left (see [27, p 184]), one obtains (iii).
B A quantitative inverse function theorem
In this section E, E′ will be Banach spaces. We denote by (E, E′) the B Banach space of bounded operators from E to E′ . If T (E, E′) then ∈ B T = sup Tx . If U E is open and f : U E′ smooth then k k kxk≤1 k k ⊂ → Df(x) (E, E′) is the derivative of f at x U . The second derivative ∈ B ∈ D(Df)(x) (E, (E, E′)) is usually written D2f(x) and can be identified ∈ B B with the symmetric bilinear map E E E′ given by × → ∂2 D2f(x)(y, z)= f(x + sy + tz). ∂s∂t (0,0)
The norm of the second derivative is D2f(x) = sup D2f(x)(y, z) . k k kyk,kzk≤1 k k
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For r> 0 let B = x E : x < r . r { ∈ k k } Lemma B.1 Let ǫ,M > 0 be positive real numbers such that ǫM < 1, and suppose f : B E is a smooth map satisfying ǫ → f(0) = 0; Df(0) = I; D2f(x) M for x B . k k≤ ∈ ǫ −1 ≈ Then f restricts to a diffeomorphism f B ǫ B ǫ . 2 → 2 The conclusion of the lemma holds even when ǫM = 1, see Proposition B.1 below.
Proof The estimate on D2f gives Df(x) I = Df(x) Df(0) M x . (44) k − k k − k≤ k k Therefore the map 1 h(x)= f(x) x = (Df(tx) I)x dt − − Z0 satisfies M h(x) x 2, k k≤ 2 k k h(x ) h(x ) ǫM x x k 2 − 1 k≤ k 2 − 1k for all x,x1,x2 Bǫ . Hence for every y B ǫ the assignment x y h(x) ∈ ∈ 2 7→ − defines a map Bǫ Bǫ which has a unique fix-point. In other words, f maps −1 → f B ǫ bijectively onto B ǫ . Moreover, Df(x) is an isomorphism for every 2 2 x B , by (44). Applying the contraction mapping argument above to f ∈ ǫ around an arbitrary point in Bǫ then shows that f is an open map. It is then a −1 simple exercise to prove that the inverse g : B ǫ f B ǫ is differentiable and 2 → 2 Dg(y) = (Df(g(y)))−1 (see [8, 8.2.3]). Repeated application of the chain rule then shows that g is smooth.
For r> 0 let B E be as above, and define B′ E′ similarly. r ⊂ r ⊂ Proposition B.1 Let ǫ, M be positive real numbers and f : B E′ a ǫ → smooth map such that f(0) = 0, L = Df(0) is invertible, and D2f(x) M for all x B . k k≤ ∈ ǫ Set κ = L−1 −1 ǫM and ǫ′ = ǫ L−1 −1 . Then the following hold: k k − k k
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(i) If κ 0 then f is a diffeomorphism onto an open subset of E′ containing ′ ≥ Bǫ′/2 . ′ (ii) If κ> 0 and g : Bǫ′/2 Bǫ is the smooth map satisfying f g = I then → ′ ◦ for all x B and y B ′ one has ∈ ǫ ∈ ǫ /2 Df(x)−1 , Dg(y) < κ−1, D2g(y) < Mκ−3. k k k k k k The reader may wish to look at some simple example (such as a quadratic polynomial) to understand the various ways in which these results are optimal.
Proof (i) For every x B we have ∈ ǫ Df(x)L−1 I Df(x) L L−1 < ǫM L−1 1, k − k ≤ k − k · k k k k≤ hence Df(x) is invertible. Thus f is a local diffeomorphism by Lemma B.1. Set h(x)= f(x) Lx. If x ,x B and x = x then − 1 2 ∈ ǫ 1 6 2 f(x ) f(x ) L(x x ) h(x ) h(x ) > κ x x , k 2 − 1 k ≥ k 2 − 1 k − k 2 − 1 k k 2 − 1k hence f is injective. By choice of ǫ′ the map −1 ′ ′ f = f L : B ′ E ◦ ǫ /2 → ′ is well defined, and for every y B ′ one has e ∈ ǫ /2 D2f(y) M L−1 2. k k≤ k k Because ǫ′M L−e1 2 = ǫM L−1 1, k k k k≤ ′ ′ Lemma B.1 says that the image of f contains every ball Bδ/2 with 0 <δ<ǫ , ′ hence also Bǫ′/2 . e (ii) Set c = I Df(x)L−1 . Then − ∞ Df(x)−1 = L−1 cn, n=0 X hence L−1 L−1 Df(x)−1 k k < k k = κ−1. k k≤ 1 c 1 ǫM L−1 − k k − k k This also gives the desired bound on Dg(y)= Df(g(y))−1 . To estimate D2g, let Iso(E, E′) (E, E′) be the open subset of invertible ⊂ B operators, and let ι: Iso(E, E′) (E′, E) be the inversion map: ι(a)= a−1 . →B Then ι is smooth, and its derivative is given by Dι(a)b = a−1ba−1, −
eometry & opology, Volume 9 (2005) G T 92 Kim A Frøyshov see [8]. The chain rule says that Dg = ι Df g, ◦ ◦ D(Dg)(y)= Dι(Df(g(y))) D(Df)(g(y)) Dg(y). ◦ ◦ This gives D(Dg)(y) Df(g(y))−1 2 D(Df)(g(y)) Dg(y) < κ−2 M κ−1. k k ≤ k k ·k k·k k · ·
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