We write the manifold $\mathcal{M}$ as the union of open sets $U\cup V$, where

(6.59)

$\displaystyle U=\mathcal{N}_{m_{0}}^{4}(-T_{-},T_{+}),\ V=X_{b_{-}}(T_{-}+1)\sqcup X_{b_{+}}(T_{-}+1),$

where $X_{b_{\pm}}=Q_{b_{\pm}}\setminus\mathbb{T}^{2}$, $m_{0}=b_{-}+b_{+}$, with $Q_{b_{\pm}}$ a del Pezzo surface of degree $b_{\pm}$. Clearly, $U\cap V$ deformation retracts onto $\Nil_{b_{-}}^{3}\sqcup\Nil_{b_{+}}^{3}$.

Next, we claim that the de Rham cohomology $H^{1}(X_{b_{\pm}})=0$. To see this, we use the long exact sequence of a pair in de Rham cohomology

(6.60)

$\displaystyle\cdots\rightarrow H^{k}_{c}(Q_{b_{\pm}}\setminus\mathbb{T}^{2})\rightarrow H^{k}(Q_{b_{\pm}})\rightarrow H^{k}(\mathbb{T}^{2})\xrightarrow{\phi}H^{k+1}_{c}(Q_{b_{\pm}}\setminus\mathbb{T}^{2})\rightarrow\cdots,$

see [Spi79, Chapter 11]. Since $H^{3}(Q_{b_{\pm}})=0$, (6.60) yields an exact sequence

(6.61)

$\displaystyle\dots\rightarrow H^{2}(Q_{b_{\pm}})\xrightarrow{i^{*}}H^{2}(\mathbb{T}^{2})\rightarrow H^{3}_{c}(Q_{b_{\pm}}\setminus\mathbb{T}^{2})\rightarrow 0.$

Here the mapping $i^{*}:H^{2}(Q_{b_{\pm}})\rightarrow H^{2}(\mathbb{T}^{2})$ is just the pullback under inclusion, which is dual to the mapping on homology $i_{*}:H_{2}(\mathbb{T}^{2};\mathbb{R})\rightarrow H_{2}(Q_{b_{\pm}};\mathbb{R})$. Since $\mathbb{T}^{2}$ is a complex submanifold of a Kähler manifold, this latter mapping is injective, so the mapping $i^{*}$ is surjective, and by Poincaré duality we conclude that

(6.62)

$\displaystyle H^{1}(Q_{b_{\pm}}\setminus\mathbb{T}^{2})\cong H^{3}_{c}(Q_{b_{\pm}}\setminus\mathbb{T}^{2})=0.$

Since we just showed that $H^{1}(X_{b_{\pm}})=0$, the Mayer-Vietoris sequence in cohomology for the pair $\{U,V\}$ yields an exact sequence

(6.63)

$\displaystyle 0\rightarrow H^{1}(\mathcal{M})\rightarrow H^{1}(\mathcal{N}_{m_{0}})\xrightarrow{i^{*}}H^{1}(\Nil_{b_{-}}^{3}\sqcup\Nil_{b_{+}}^{3})\cong H^{1}(\Nil_{b_{-}}^{3})\oplus H^{1}(\Nil_{b_{+}}^{3}).$

The mapping $i^{*}$ is the pullback under inclusion of the two nilmanifold fibers of the neck at each end. We claim that this mapping is injective. To see this, let $\mathcal{P}_{m_{0}}\equiv\{p_{1},\ldots,p_{m_{0}}\}$ denote the monopole points in $B=\mathbb{T}^{2}\times(-T_{-},T_{+})$, where $m_{0}=b_{-}+b_{+}$. Then there are $\tilde{p}_{j}\in\mathcal{N}_{m_{0}}^{4}$ such that $\mathcal{N}_{m_{0}}^{4}\setminus\widetilde{\mathcal{P}}_{m_{0}}$ is a circle bundle over $B\setminus\mathcal{P}_{m_{0}}$,

(6.64)

$\displaystyle S^{1}\longrightarrow\mathcal{N}_{m_{0}}^{4}\setminus\widetilde{\mathcal{P}}_{m_{0}}\xrightarrow{\ \pi\ }B\setminus\mathcal{P}_{m_{0}}.$

The Gysin sequence of (6.64) begins with

(6.65)

$\displaystyle 0\rightarrow H^{1}(B\setminus\mathcal{P}_{m_{0}})\xrightarrow{\pi^{*}}H^{1}(\mathcal{N}_{m_{0}}^{4}\setminus\widetilde{\mathcal{P}}_{m_{0}})\rightarrow\cdots$

It is easy to see inclusion induces an isomorphism $H^{1}(B\setminus\mathcal{P}_{m_{0}})\cong H^{1}(B)\cong\mathbb{R}\oplus\mathbb{R}$, and similarly, $H^{1}(\mathcal{N}_{m_{0}}^{4}\setminus\widetilde{\mathcal{P}}_{m_{0}})\cong H^{1}(\mathcal{N}_{m_{0}}^{4})$. Then (6.65) becomes

(6.66)

$\displaystyle 0\rightarrow\mbox{span}\{dx,dy\}\xrightarrow{\pi^{*}}H^{1}(\mathcal{N}_{m_{0}}^{4})\rightarrow\cdots$

Together with Proposition 2.3, and the exact sequence (6.63), we conclude that $i^{*}\pi^{*}dx$ and $i^{*}\pi^{*}dy$ are both nontrivial and are linearly independent in $H^{1}(\Nil^{3}_{b_{-}}\sqcup\Nil^{3}_{b_{+}})$, so $i^{*}$ is injective as claimed. Then (6.63) implies that $b_{1}(\mathcal{M})=0$. Since $\mathcal{M}$ is a compact orientable $4$-manifold, Poincaré duality also implies that $b_{3}(\mathcal{M})=0$.

Next, it follows from the fibration (6.64) that $\chi(\mathcal{N}_{m_{0}}^{4}\setminus\widetilde{\mathcal{P}}_{m_{0}})=0$, and therefore

(6.67)

$\displaystyle\chi(\mathcal{N}_{m_{0}}^{4})=\#{\mbox{ of monopole points }}=m_{0}=b_{-}+b_{+}.$

For a Tian-Yau space, it follows that

(6.68)

$\displaystyle\chi(X_{b}^{4})=\chi(Q_{b}\setminus\mathbb{T}^{2})=\chi(Q_{b})-\chi(\mathbb{T}^{2})=\chi(Q_{b}),$

where $Q_{b}$ is a degree $b$ del Pezzo surface, so

(6.69)

$\displaystyle\chi(X_{b}^{4})=\chi(Q_{b}\setminus\mathbb{T}^{2})=12-b$

Note also that $\chi(\Nil_{b_{-}}^{3})=\chi(\Nil_{b_{+}}^{3})=0$ since it is an orientable 3-manifold. Then we have

(6.70)

$\displaystyle\chi(\mathcal{M})=\chi(X_{b_{-}}^{4})+\chi(\mathcal{N})+\chi(X_{b_{+}}^{4})=(12-b_{-})+(b_{+}+b_{-})+(12-b_{+})=24.$

Since we have shown above that $b_{1}(\mathcal{M})=b_{3}(\mathcal{M})=0$, this proves that $b_{2}(\mathcal{M})=22$.

Next, as we constructed in (6.55) the approximate definite triple $\bm{\omega}^{\mathcal{M}}\equiv(\omega_{1},\omega_{2},\omega_{3})$, which are everywhere non-zero self-dual 2-forms forming a basis of $\Lambda^{2}_{+}$ at every point. This implies the bundle $\Lambda^{2}_{+}(\mathcal{M})$ is a trivial rank $3$ bundle. Also, $\omega_{1}$ being non-zero everywhere means that there is an almost complex structure ($\omega_{1}/|\omega_{1}|$ is a unit norm self-dual 2-form, which is equivalent to an orthogonal almost complex structure). By Corollary 6.5, for $\beta\gg 1$, the rank 2 subbundle $V\subset\Lambda^{2}_{0}$, given by the orthogonal complement of $\omega_{1}/|\omega_{1}|$ is trivial. Then $0=c_{1}(V\otimes\mathbb{C})=c_{1}(T\mathcal{M},J)^{2}$, and the Hirzebruch signature theorem implies that

(6.71)

$\displaystyle 2\chi(\mathcal{M})+3\tau(\mathcal{M})=\int_{\mathcal{M}}c_{1}^{2}=0,$

from which it follows that $\tau(\mathcal{M})=-16$. Therefore, $b_{2}^{+}(\mathcal{M})=3$ and $b_{2}^{-}(\mathcal{M})=19$.

∎