Chapter II: Self-Gluing formula of the Seiberg-Witten Invariants
3.4 Proof of Theorem 3.1
and then scaleω properly so that[ω] ∈ H2(XY,Z). We show thatKω satisfies the pull-back condition. Based on adjunction inequality, for any closed curveαwhich is homologically nontrivial inC,T2= S1×α ∈S1×C, hKω,S1×αi =0.Moreover, from the assumption
hKω,[Σ]i =2g(Σ) −2. Consequently,Kω satisfies the pull-back condition.
LetΩbe the pull back 2-form ofωon ˜XY.LetKΩ∈ H2(X˜Y,Z)be the canonical class of the symplectic structureΩ on ˜XY. KΩ = p∗(Kω). Likewise, KΩ ∈ H2(S1×Σ)˜ fulfills the pull-back condition on every component. In conclusion, we can use Theorem 3.1.2 for this canonical structure KΩ. i∗
1,i∗
2 denote the natural maps H2(X˜Y,Z) → H2(M0
1) and H2(X˜Y,Z) → H2(M0
2) induced by inclusion maps M0
1,M0
2 ,→ X˜Y. Henceforth, KM0
1,KM0
2 denotei∗
1(KΩ),i∗
2(KΩ)respectively. We pick k ∈ H2(X˜Y,Z) which also satisfies the condition that k|S1×Σ˜ = ρ∗(k0). Then, the restriction(k−KΩ)onS1×Σ˜ becomes zero. Moreover, letkM0
1 =i∗
1(k),kM0
2 =i∗
2(k). Then, kM0
1
−KM0
1 andkM0
2
−KM0
2 vanish on the boundary.
We focus on the element (k − KΩ) ^ [Ω]in H4(X˜Y,Z) Z. We can decompose the integer corresponding to (k −KΩ) ^ [Ω] in the following way. First, [F] = PD[(k − KΩ)] ∈ H2(X˜Y), where a surface F ∈ C2(X˜Y) has no intersection with S1×Σ.˜ Next, F can be decomposed into F1+F2 for F1 ∈ C2(M0
1),F2 ∈ C2(M0
2).
Then,
(k −KΩ)^[Ω]= PD[F1]^i∗
1[Ω]+PD[F2]^i∗
2[Ω]. (3.4.1) The equation 3.4.1 verifies the relationship betweenSpinc-structuresk ∈H2(X˜Y,Z) and its restrictions toM0
1,M0
2.
We add the equation in Theorem 2.5.1,(−1)?Í
s∈K(k)SWX(s)=Í
SWM1(s1)SWM2(s2) fork ∈H2(X˜Y)satisfying that
1. k|S1×Σ˜ = ρ∗(k0) where ρ : S1 × Σ˜ → Σ˜ and k0 ∈ H2(Σ)˜ satisfies that hk0,[Σ]i˜ = 2g(Σ) −˜ 2.
2. k ^[Ω]= KΩ^[Ω] ∈ Z 3. k2= KΩ2.
It is known that only KΩ satisfies all of the above properties (1)-(3) and SW , 0.
Therefore, the left hand side is exactly equal to SW(KΩ) which is +1 or −1 from
Theorem 1.9.1. The right hand side becomes
Õ
(zˆ1,zˆ2)∈H2(M1)×H2(M2)
SWM1(zˆ1)SWM2(zˆ2) for(zˆ1,zˆ2) ∈H2(M1) ×H2(M2)satisfying
1. ˆz12+zˆ22= k2− (4g−4)l 2. There exist[F1],[F2] ∈ H2(M0
1),H2(M0
2)such that
• F1∩∂M0
1= F2∩∂M0
2 =φ
• (zˆ1|M0
1
− KM0
1
) = j1(PD[F1]) and (zˆ2|M0
2
− KM0
2
) = j2(PD[F2]) where j1: H2(M0
1, ∂M0
1) →H2(M0
1), j2 : H2(M0
2, ∂M0
2) → H2(M0
2)are natural maps.
3. [F1],[F2]determined in (2) satisfy thatPD[F1]^i∗
1[Ω]+PD[F2]^i∗
2[Ω]= 0.
Now we see Õ
(zˆ1,zˆ2)∈H2(M1)×H2(M2)
SWM1(zˆ1)SWM2(zˆ2)=Õ
zˆ2
SWM2(zˆ2)(Õ
zˆ1
SWM1(zˆ1)). (3.4.2)
We want to prove that the inner sum Õ
SWM1(zˆ1)in the right hand side becomes zero. Remark that ifSWM1(zˆ1),0, then ˆz12 =0. Therefore, the first condition does not need to be considered in the inner sum of Equation 3.4.2. Let Zm be a set of zˆ1 ∈ H2(S1×Y˜)such that the corresponding F1satisfiesPD[F1] ^i∗
1[Ω] = m. In other words, we show that
Õ
zˆ1∈Zm
SWM1(zˆ1)=0 for allm.
We have the formula for the Seiberg Witten invariants. For h ∈ H2(S1×Y˜), [h] denotes the quotient element in H(S1×Y˜) H2(S1×Y˜)/T or s. If two elements x,y ∈ H2(S1×Y˜)satisfy that[x] =[y], thenx ∈ Zm is equivaltent to y ∈ Zm since the result of cup products does not depend on the torsion part. Let
∆Y˜ = Õ
h∈H(Y)˜
gh·h
for gh ∈ Z. We have the natural projection p : S1×Y˜ → Y˜ and the induced map p∗ :H(Y˜) → H(S1×Y˜) H(Y˜) ⊕ (Z⊗H1(Y˜)/T or s). The image ofp∗ is included in the first summand.
First, suppose thatb1(Y˜)> 1. Then from Theorem 3.3.4, SWS1×Y˜ =ξΦ2(∆Y˜).
From Kunneth formula, H2(S1 ×Y˜) = H2(Y˜) ⊕ (H1(S1) ⊗ H1(Y˜)). Since ξ ∈
±p∗(H(Y˜)),ξ supports on the first summand ofH(S1×Y˜). Conclusively, SWM
1 = ξΦ2(∆Y˜)
=⇒ Õ
h∈H(M1)
SWM1(h)[h]= Õ
h∈H(S1×Y˜)
gh· (2h+ξ).
Therefore, we can summarize the equality:
1.
Õ
h∈H2(S1×Y),˜ [h]=2l+ξ∈H(S1×Y)˜
SW(h)=gl forl ∈H(Y˜) ⊂ H(S1×Y˜).
2. Otherwise, Õ
[h]=k
SW(h)= 0 wherek is not represented by 2l+ξ.
We defineφ : H2(S1×Y˜) → H(Y)to be the composition of trivial quotient maps H2(S1×Y˜) → H(S1×Y˜) →H(Y˜)andπ∗ :H(Y˜) → H(Y). Therefore,
Õ
h∈π−1∗ (x)
gh =0 (3.4.3)
for each x ∈H(Y). This is equivalent with Õ
h∈H2(S1×Y)˜ φ(h)=x
SWS1×Y˜(h)=0 (3.4.4)
for each x ∈H(Y).
Lemma 3.4.1. Ifz,ˆ wˆ ∈ H2(Y˜) ⊂ H2(S1×Y˜)satisfy thatφ(z)ˆ = φ(w)ˆ , then zˆ∈ Zk ⇐⇒ wˆ ∈ Zk.
Proof. LetFz,Fwbe the 2-chain corrsponding to ˆz,wˆ respectively. We observe that the covering prestricted to M0
1is isomorphic to the covering p1: S1×Y˜ → S1×Y restricted to M0
1. Let p0 : M0
1 → p(M0
1) = Z. Recall that Z S1 ×Y \ D2× Σ.
The diagram 3.2 commutes. The horizontal maps are covering maps p,p0and the vertical maps are natural inclusion maps.
Recall that Ω = p∗[ω]. Therefore,i∗
1[Ω] = p0∗[ω|Z]. Moreover, we also have the following commuting diagram 3.3.
X˜Y XY
M0
1 Z
p
p
Figure 3.2: Covering diagram M˜1 S1×Y
M0
1 Z
p
p
Figure 3.3: Covering diagram
Hence,
PD[Fz]^i∗1[Ω]= h[Fz],i∗1[Ω]i
= h[Fz],p0∗[ω|Z]i
= h[p0(Fz)],[ω|Z]i
= PD[p0(Fz)]^[ω|Z]
= hp0(Fz),[ω|Z]i.
The second commutative diagram indicates that p0(Fz) = p0(Fw). Therefore, the statement is true.
Finally, based on Lemma 3.4.1, the inner sum of Equation 3.4.2 can be decomposed into the sum of Equation 3.4.4 for some x. Therefore, the inner sum of Equation 3.4.2 becomes zero.
Second, suppose that b1(Y˜) = 1. Since b1(Y) = 1, π∗ : H(Y˜) → H(Y)is a homo- morphism from Z → Z. Since this homomorphism is not trivial, π∗ is injective.
Therefore, π∗(∆Y˜) = 0 implies that ∆Y˜ = 0. Therefore, Õ
[h]=l
SW(h) = 0 for all l ∈ H(S1×Y˜).Therefore, the inner sum of the right hand side is also zero.
Therefore, this is a contradiction because Equation 3.3.2 is not true. This implies that XY does not have a symplectic structure and hence concludes the proof of Theorem 3.1.2.
Remark 3.4.2. We add remarks in a more general setting. First, in case ofb1(Y) >1, according to Remark 3.3.6, we need to show the corresponding equation to Equation 3.4.3: for eachm ∈Z,
Õ
ψ^π∗(h)=m
gh =0
for a non-fiberedψ ∈H1(Y).Moreover, to extend the results to the case ofb1(Y) >1, Lemma 3.4.1 is transformed into: for twoz,ˆ wˆ satisfying that
ψ ^ π(zˆ)= ψ ^ π(wˆ),
the statementzˆ∈ Zk ⇐⇒ wˆ ∈ Zk is true. However, we cannot use the same proof in Lemma 3.4.1 since the2-formω onXY does not have the unique corresponding 2-form onS1×Y.
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