We will adopt the general description of a hyperkähler metric in terms of a triple of three symplectic forms due to Donaldson [Don06], a description which was also used, for example, in [FLS17, Fos16]. Namely, we will not directly construct a Ricci-flat metric on $\mathcal{M}$. Instead we will glue together triples of symplectic forms, which we will then perturb to obtain a hyperkähler triple, which will then yield in particular a Ricci-flat Kähler metric.

Let $M^{4}$ be an oriented $4$-manifold with a volume form $\dvol_{0}$. A triple of $2$-forms $\bm{\omega}=(\omega_{1},\omega_{2},\omega_{3})$ is called a definite triple if the matrix $Q=(Q_{ij})$ defined by

(1.12)

$$\frac{1}{2}\omega_{i}\wedge\omega_{j}=Q_{ij}\dvol_{0}$$

is positive. Given a definite triple $\bm{\omega}$, the associated volume form is defined as

(1.13)

$$\dvol_{\bm{\omega}}=(\det(Q))^{\frac{1}{3}}\dvol_{0},$$

which is independent of the choice of volume form $\dvol_{0}$. We denote by $Q_{\bm{\omega}}\equiv(\det(Q))^{-\frac{1}{3}}Q$ the renormalized matrix with unit determinant. Furthermore, a definite triple $\bm{\omega}=(\omega_{1},\omega_{2},\omega_{3})$ is called a hyperkähler triple if $d\omega_{1}=d\omega_{2}=d\omega_{3}=0$ and the renormalized coefficient matrix satisfies

(1.14)

$\displaystyle Q_{\bm{\omega}}=\Id.$

Note that equation (1.14) is equivalent to

(1.15)

$$\frac{1}{2}\omega_{i}\wedge\omega_{j}=\frac{1}{6}\delta_{ij}(\omega_{1}^{2}+\omega_{2}^{2}+\omega_{3}^{2}),$$

for every $1\leq i\leq j\leq 3$.

Each definite triple $\bm{\omega}$ defines a Riemannian metric $g_{\bm{\omega}}$ such that each $\omega_{j}$ is self-dual with respect to $g_{\bm{\omega}}$. If moreover $\bm{\omega}=(\omega_{1},\omega_{2},\omega_{3})$ is a hyperkähler triple, then $g_{\bm{\omega}}$ is a hyperkähler metric. Furthermore, $\omega_{2}+i\omega_{3}$ is a holomorphic $2$-form with respect to the complex structure determined by $\omega_{1}$. See Section 2 for more concrete examples of such triples.

Since in our construction each piece of the manifold $\mathcal{M}$ is hyperkähler, hence carries a hyperkähler triple, we will first show that there exists a glued triple of closed $2$-forms $\bm{\omega}$ on $\mathcal{M}$ which is very close to being a hyperkähler triple, i.e.,

(1.16)

$$\|Q_{\bm{\omega}}-\Id\|_{C^{0}(\mathcal{M})}<\epsilon$$

for some sufficiently small constant $\epsilon>0$. A precise statement will be proved in Section 6. Starting from such a glued approximately hyperkähler triple $\bm{\omega}$, the goal of the perturbation procedure is to find a triple of closed $2$-forms $\bm{\theta}=(\theta_{1},\theta_{2},\theta_{3})$ such that $\underline{\bm{\omega}}\equiv\bm{\omega}+\bm{\theta}$ is an actual hyperkähler triple on $\mathcal{M}$. Precisely, we will solve the system

(1.17)

$$\frac{1}{2}(\omega_{i}+\theta_{i})\wedge(\omega_{j}+\theta_{j})=\delta_{ij}\dvol_{\bm{\omega}+\bm{\theta}},$$

which is equivalent to

(1.18)

$$\frac{1}{2}(\omega_{i}+\theta_{i})\wedge(\omega_{j}+\theta_{j})=\frac{1}{6}\delta_{ij}\sum\limits_{j=1}^{3}(\omega_{j}+\theta_{j})^{2}.$$

We expand equation (1.18):

(1.19)

$$\frac{1}{2}(\omega_{i}\wedge\omega_{j}+\omega_{i}\wedge\theta_{j}+\omega_{j}\wedge\theta_{i}+\theta_{i}\wedge\theta_{j})=\frac{1}{6}\delta_{ij}\sum\limits_{j=1}^{3}\Big(\omega_{j}^{2}+\theta_{j}^{2}+2\omega_{j}\wedge\theta_{j}\Big).$$

Now split $\bm{\theta}$ into its self-dual and anti-self-dual parts with respect to $g_{\bm{\omega}}$, $\bm{\theta}=\bm{\theta}^{+}+\bm{\theta}^{-}$. We define a matrix $A=(A_{ij})$ by $\theta^{+}_{i}=\sum\limits_{j=1}^{3}A_{ij}\omega_{j}$ and also define the matrix $S_{\bm{\theta}^{-}}=(S_{ij})$ by

(1.20)

$$\frac{1}{2}\theta_{i}^{-}\wedge\theta_{j}^{-}=S_{ij}\dvol_{\bm{\omega}},\ 1\leq i\leq j\leq 3.$$

Then in terms of the volume form $\dvol_{\bm{\omega}}$ given by the approximately hyperkähler triple $\bm{\omega}$, the expansion (1.19) can be rewritten as the matrix equation

(1.21)

$\displaystyle\begin{split}&(Q_{\bm{\omega}}+Q_{\bm{\omega}}A^{T}+AQ_{\bm{\omega}}+AQ_{\bm{\omega}}A^{T})+S_{\bm{\theta}^{-}}\\ &=\frac{1}{3}\Id\cdot\Big(\Tr(Q_{\bm{\omega}})+\Tr(AQ_{\bm{\omega}}A^{T})+\Tr(S_{\bm{\theta}^{-}})+\Tr(AQ_{\bm{\omega}})+\Tr(Q_{\bm{\omega}}A^{T})\Big).\end{split}$

For any $3\times 3$ real matrix $B$ denote by

(1.22)

$$\TF(B)=B-\frac{1}{3}\Tr(B)\Id$$

the trace-free part of $B$. Then we get

(1.23)

$$\TF(Q_{\bm{\omega}}A^{T}+Q_{\bm{\omega}}A+AQ_{\bm{\omega}}A^{T})=\TF(-Q_{\bm{\omega}}-S_{\bm{\theta}^{-}}).$$

For simplicity, in our context, we always identify a $3\times 3$-matrix with a triple of self-dual $2$-forms. Then observe that a solution of following gauge-fixed system

(1.24)

$\displaystyle d^{+}\bm{\eta}+\bm{\xi}=\mathfrak{F}_{0}\Big(\TF(-Q_{\bm{\omega}}-S_{d^{-}\bm{\eta}})\Big),\ d^{*}\bm{\eta}=0,$

is also a solution of (1.23). Here $\mathfrak{F}_{0}$ denotes the local inverse near zero to the local diffeomorphism $\mathfrak{G}_{0}:\mathscr{S}_{0}(\mathbb{R}^{3})\to\mathscr{S}_{0}(\mathbb{R}^{3})$ on the space of trace-free symmetric $3\times 3$-matrices defined by

(1.25)

$\displaystyle\mathfrak{G}_{0}(A)=\TF(Q_{\bm{\omega}}A^{T}+AQ_{\bm{\omega}}+AQ_{\bm{\omega}}A^{T}).$

Moreover, $\bm{\eta}=(\eta_{1},\eta_{2},\eta_{3})$ with $\eta_{j}\in\Omega^{1}(\mathcal{M})$, $\bm{\xi}=(\xi_{1},\xi_{2},\xi_{3})$ with $\xi_{i}\in\mathcal{H}^{+}(\mathcal{M})$ (the space of self-dual harmonic $2$-forms with respect to $g_{\bm{\omega}}$), and $d^{\pm}\bm{\eta}$ is the self-dual or anti-self-dual part of $d\bm{\eta}=\bm{\theta}-\bm{\xi}$ with respect to $g_{\bm{\omega}}$, respectively.

The linearization of the elliptic system (1.24) at $\bm{\eta}=0$ is

(1.26)

$$\mathscr{L}=(\mathscr{D}\oplus\Id)\otimes\mathbb{R}^{3}:(\Omega^{1}(\mathcal{M})\oplus\mathcal{H}^{+}(\mathcal{M}))\otimes\mathbb{R}^{3}\longrightarrow(\Omega^{0}(\mathcal{M})\oplus\Omega^{2}_{+}(\mathcal{M}))\otimes\mathbb{R}^{3},$$

where

(1.27)

$$\mathscr{D}\equiv d^{*}+d^{+}:\Omega^{1}(\mathcal{M})\longrightarrow(\Omega^{0}(\mathcal{M})\oplus\Omega^{2}_{+}(\mathcal{M}))$$

is a Dirac-type operator.

In order to solve the elliptic system (1.24) we will use the implicit function theorem (see Lemma 9.3). This requires finding a bounded right inverse to the linearized operator $\mathscr{L}$.

The first step is to define certain weighted Banach norms whose setup requires a careful understanding of the collapsing geometries in our construction. The construction of a suitable weight function will be more complicated than in almost all known gluing constructions because our neck region is topologically nontrivial. Geometrically, a crucial step is to assign to each $p\in\mathcal{M}$ an appropriate scale $r_{p}$ such that the geodesic ball $B_{r_{p}}(p)$ captures enough geometric information and curvature remains uniformly bounded for most points in $B_{r_{p}}(p)$. This is equivalent to finding a canonical rescaling factor for each $p\in\mathcal{M}$ which reflects the collapsing behavior of the glued metric near $p$. This leads to a decomposition of $\mathcal{M}$ (see Section 7.1 and 7.3) into 9 different regions, labeled $\I$, $\II$, $\III$, $\IV_{\pm}$, $\V_{\pm}$, $\VI_{\pm}$, with each of these regions exhibiting different regularity and convergence behaviors.

Second, in order to prove uniform boundedness of the right inverse by contradiction (Proposition 9.2), we then need to analyze the kernel of the linearized operator in suitable weighted spaces on the building blocks of the gluing construction. Here a crucial ingredient is a new Liouville theorem for half-harmonic 1-forms, i.e., $1$-forms $\phi$ satisfying

(1.28)

$$d^{*}\phi=0,\ \ d^{+}\phi=0,$$

on a Tian-Yau space, see Theorem 5.1.