ScalingStacks

Proof. [051J]

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Proof.

As mentioned in the beginning of this section, we identify a tubular neighborhood of PP in QQ with a neighborhood of the zero section in its normal bundle N=N0⊕ℝN=N_{0}\oplus\mathbb{R}. For simplicity we may assume this neighborhood is given by ℬϵ\mathcal{B}_{\epsilon}, the 2-ball bundle over PP consisting of the set of all elements in N0⊕ℝN_{0}\oplus\mathbb{R} with norm smaller than or equal to ϵ\epsilon, and we denote by 𝒮ϵ\mathcal{S}_{\epsilon} the boundary of ℬϵ\mathcal{B}_{\epsilon}.

Fix z0>0z_{0}>0, then the composition of the natural maps

(4.22) D≃D×{z0}↪Q∖P↪Q→DD\simeq D\times\{z_{0}\}\hookrightarrow Q\setminus P\hookrightarrow Q\rightarrow D

is the identity map, which implies that for all kk, the map Hk​(Q∖P,ℤ)→Hk​(Q,ℤ)H_{k}(Q\setminus P;\mathbb{Z})\rightarrow H_{k}(Q;\mathbb{Z}) is surjective and we have a natural splitting

(4.23) H2​(Q∖P,ℤ)=H2​(D,ℤ)⊕KH_{2}(Q\setminus P;\mathbb{Z})=H_{2}(D;\mathbb{Z})\oplus K

for some KK. By assumption for z>0z>0,

(4.24) [∂zω~​(z)]=[∂zψ⁡(z)]=k+​[ωD]=2​π​k+​c1​(L),[\partial_{z}\tilde{\omega}(z)]=[\partial_{z}\psi(z)]=k_{+}[\omega_{D}]=2\pi k_{+}c_{1}(L),

so 12​π​[Υ]|D×{z0}=k+​c1​(L)\frac{1}{2\pi}[\Upsilon]|_{D\times\{z_{0}\}}=k_{+}c_{1}(L) is integral. Hence it suffices to show the integral of 12​π​Υ\frac{1}{2\pi}\Upsilon over any element in KK is also an integer.

By the Mayer-Vietoris sequence applied to Q=(Q∖P)∪ℬϵQ=(Q\setminus P)\cup\mathcal{B}_{\epsilon}, we get

(4.25) 0→H2​(𝒮ϵ,ℤ)→H2​(Q∖P,ℤ)⊕H2​(ℬϵ,ℤ)→H2​(Q,ℤ)≃H2​(D,ℤ)→0.0\rightarrow H_{2}(\mathcal{S}_{\epsilon};\mathbb{Z})\rightarrow H_{2}(Q\setminus P;\mathbb{Z})\oplus H_{2}(\mathcal{B}_{\epsilon};\mathbb{Z})\rightarrow H_{2}(Q;\mathbb{Z})\simeq H_{2}(D;\mathbb{Z})\rightarrow 0.

So we obtain the exact sequence

(4.26) 0→K→H2​(𝒮ϵ,ℤ)→H2​(ℬϵ,ℤ)≃H2​(P,ℤ).0\rightarrow K\rightarrow H_{2}(\mathcal{S}_{\epsilon};\mathbb{Z})\rightarrow H_{2}(\mathcal{B}_{\epsilon};\mathbb{Z})\simeq H_{2}(P;\mathbb{Z}).

On the other hand, by the Gysin sequence applied to the 2-sphere bundle p:𝒮ϵ→Pp:\mathcal{S}_{\epsilon}\rightarrow P we get

(4.27) 0→H2​(P,ℤ)→p∗H2​(𝒮ϵ,ℤ)→∫H0​(P,ℤ)→∧eH3​(P,ℤ)→⋯0\rightarrow H^{2}(P;\mathbb{Z})\xrightarrow{p^{*}}H^{2}(\mathcal{S}_{\epsilon};\mathbb{Z})\xrightarrow{\int}H^{0}(P;\mathbb{Z})\xrightarrow{\wedge e}H^{3}(P;\mathbb{Z})\rightarrow\cdots

where ∫\int denotes integration over the 2-sphere fibers, and ∧e\wedge e denotes the wedge product with Euler class of 𝒮ϵ\mathcal{S}_{\epsilon}. Since the Euler class ee of N0⊕ℝN_{0}\oplus\mathbb{R} vanishes, the above becomes

(4.28) 0→H2​(P,ℤ)→p∗H2​(𝒮ϵ,ℤ)→∫H0​(P,ℤ)≃ℤ→0.0\rightarrow H^{2}(P;\mathbb{Z})\xrightarrow{p^{*}}H^{2}(\mathcal{S}_{\epsilon};\mathbb{Z})\xrightarrow{\int}H^{0}(P;\mathbb{Z})\simeq\mathbb{Z}\rightarrow 0.

(4.26) and (4.28) together imply that modulo torsion, KK is generated by the homology class of a 2-sphere fiber of pp. So we just need to show ∫12​π​[Υ]|𝒮ϵ\int\frac{1}{2\pi}[\Upsilon]|_{\mathcal{S}_{\epsilon}} is an integer.

By the expansion of ψ\psi and hh in Proposition 3.24 and Proposition 3.28, it is easy to check that by restricting to the fiber of NN over pp, we have

(4.29) Υ|N⁡(p)=−−14​r3​(z​d​y​d​y¯+(y​d​y¯−y¯​d​y)​d​z)+O⁡(1).\Upsilon|_{N(p)}=-\frac{\sqrt{-1}}{4r^{3}}(zdyd\bar{y}+(yd\bar{y}-\bar{y}dy)dz)+O(1).

Further restricting to the 22-sphere with radius ϵ\epsilon, we get

(4.30) Υ|𝒮ϵ​(p)=−12​ϵ2​dvolSϵ2+O⁡(1),\Upsilon|_{\mathcal{S}_{\epsilon}(p)}=-\frac{1}{2\epsilon^{2}}\dvol_{S^{2}_{\epsilon}}+O(1),

where dvolSϵ2\dvol_{S^{2}_{\epsilon}} is the area form of the standard ϵ\epsilon-sphere in ℝ3\mathbb{R}^{3}. Taking the integral and let ϵ→0\epsilon\rightarrow 0 gives that

(4.31) ∫𝒮ϵ​(p)Υ=−2​π.\int_{\mathcal{S}_{\epsilon}(p)}\Upsilon=-2\pi.

∎

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