In Section 5 we will construct closed definite triples $\bm{\underline{\omega}}$ which are approximately hyperkähler, in the sense that the intersection matrix $Q_{\bm{\underline{\omega}}}$ is close to the identity. We now explain how to formulate the problem of deforming such a triple $\bm{\underline{\omega}}$ into a hyperkähler structure.

Let $\bm{\underline{\omega}}$ be a closed definite triple on a $4$–manifold $M$ and assume that $\|Q_{\bm{\underline{\omega}}}-\text{id}\|_{C^{0}}<\sigma$ for some small $\sigma>0$. We want to deform $\bm{\underline{\omega}}$ into a hyperkähler triple, i.e. we look for a triple of closed $2$–forms $\bm{\underline{\eta}}=(\eta_{1},\eta_{2},\eta_{3})$ on $M$ such that

(2.5)

$$\tfrac{1}{2}\left(\omega_{i}+\eta_{i}\right)\wedge\left(\omega_{j}+\eta_{j}\right)=\delta_{ij}\,\mu_{\bm{\underline{\omega}}}.$$

Decompose $\bm{\underline{\eta}}$ into self-dual and anti-self dual parts $\bm{\underline{\eta}}=\bm{\underline{\eta}}^{+}+\bm{\underline{\eta}}^{-}$ with respect to $g_{\bm{\underline{\omega}}}$. The self-dual part can be written in terms of a $M_{3\times 3}(\mathbb{R})$–valued function $A$ by

$$\eta_{i}^{+}=\sum_{j=1}^{3}{A_{ij}\,\omega_{j}}.$$

Denote by $\bm{\underline{\eta}}^{-}\ast\bm{\underline{\eta}}^{-}$ the symmetric $(3\times 3)$–matrix with entries $(\tfrac{1}{2}\,\eta_{i}^{-}\wedge\eta_{j}^{-})/\mu_{\bm{\underline{\omega}}}$. Then we can rewrite (2.5) as

(2.6)

$$Q_{\bm{\underline{\omega}}}+Q_{\bm{\underline{\omega}}}\,A^{T}+A\,Q_{\bm{\underline{\omega}}}+A\,Q_{\bm{\underline{\omega}}}\,A^{T}+\bm{\underline{\eta}}^{-}\ast\bm{\underline{\eta}}^{-}=\text{id}.$$

Now, consider the map

$$M_{3\times 3}(\mathbb{R})\longrightarrow Sym^{2}(\mathbb{R}^{3});\qquad A\longmapsto Q_{\bm{\underline{\omega}}}\,A^{T}+A\,Q_{\bm{\underline{\omega}}}+A\,Q_{\bm{\underline{\omega}}}\,A^{T}$$

and its differential $A\mapsto Q_{\bm{\underline{\omega}}}\,A^{T}+A\,Q_{\bm{\underline{\omega}}}$. Since $Q_{\bm{\underline{\omega}}}$ is arbitrarily close to the identity, this linear map induces an isomorphism $Sym^{2}(\mathbb{R}^{3})\rightarrow Sym^{2}(\mathbb{R}^{3})$ for $\sigma$ sufficiently small. We can therefore define a smooth function $\mathcal{F}\colon\thinspace Sym^{2}(\mathbb{R}^{3})\rightarrow Sym^{2}(\mathbb{R}^{3})$ such that $Q_{\bm{\underline{\omega}}}\,A^{T}+A\,Q_{\bm{\underline{\omega}}}+A\,Q_{\bm{\underline{\omega}}}\,A^{T}=S$ if and only if $A=\mathcal{F}(S)$.

Hence we reformulate (2.6) as

(2.7)

$$\bm{\underline{\eta}}^{+}=\mathcal{F}\left((\text{id}-Q_{\bm{\underline{\omega}}})-\bm{\underline{\eta}}^{-}\ast\bm{\underline{\eta}}^{-}\right).$$

Now, let $\mathcal{H}^{+}_{\bm{\underline{\omega}}}$ be the space of self-dual harmonic $2$–forms with respect to $g_{\bm{\underline{\omega}}}$. If a solution of (2.6) exists on a compact manifold $M$ then $M$ must be either a $4$–torus or a K3 surface with the standard orientation and therefore $\mathcal{H}^{+}_{\bm{\underline{\omega}}}$ is $3$–dimensional. Since $\omega_{1},\omega_{2},\omega_{3}$ are closed and self-dual (therefore harmonic) and linearly independent (since $\bm{\underline{\omega}}$ is a definite triple) we deduce that $\mathcal{H}^{+}_{\bm{\underline{\omega}}}$ consist of constant linear combinations of $\omega_{1},\omega_{2},\omega_{3}$.

By Hodge theory with respect to $g_{\bm{\underline{\omega}}}$ we can finally rewrite (2.7) as the elliptic equation

(2.8)

$$d^{+}\bm{\underline{a}}+\bm{\underline{\zeta}}=\mathcal{F}\left((\text{id}-Q_{\bm{\underline{\omega}}})-\bm{\underline{\eta}}^{-}\ast\bm{\underline{\eta}}^{-}\right),\qquad d^{\ast}\bm{\underline{a}}=0,$$

for a triple $\bm{\underline{a}}$ of $1$–forms on $M$ and a triple $\bm{\underline{\zeta}}\in\mathcal{H}^{+}_{\bm{\underline{\omega}}}\otimes\mathbb{R}^{3}$. Here $2\,d^{+}a=da+\ast da$ is the self-dual part of $da$.

The linearisation of (2.8) is

(2.9)

$$(D\oplus\text{id})\otimes\mathbb{R}^{3}\colon\thinspace\left(\Omega^{1}(M)\oplus\mathcal{H}^{+}_{\bm{\underline{\omega}}}\right)\otimes\mathbb{R}^{3}\longrightarrow\left(\Omega^{0}(M)\oplus\Omega^{+}(M)\right)\otimes\mathbb{R}^{3},$$

where $D$ is the Dirac-type operator

(2.10)

$$D=d^{\ast}\oplus d^{+}\colon\thinspace\Omega^{1}(M)\longrightarrow\Omega^{0}(M)\oplus\Omega^{+}(M).$$

Note that the operator in (2.9) is always surjective with kernel consisting of harmonic $1$–forms.