Rozansky-Witten invariants of hyperkahler manifolds and independence of complex structure

Question

Recently I have been learning about Rozansky-Witten invariants, mainly through Hitchin-Sawon's paper "curvature and characteristic numbers of hyperkahler manifolds" and through Justin Sawon's phD thesis. I wanted to ask for explanations about a specific point in these references, namely the independence of choice of complex structure compatible with the hyperkahler structure. What follows is a more detailed version of this question; any help would be much appreciated!

Let $(X,g,I,J,K)$ be a compact hyperkahler manifold of real dimension $4n$. We make a choice and use $I$ to decompose the complexified tangent bundle $TX\otimes_\mathbb{R}\mathbb{C}$ as $T\oplus \overline{T}$, and similarly for complex differential forms on $X$. Fixing notation, we write $\omega_I,\omega_J,\omega_K$ the Kahler forms associated to $I,J,K$ and $\omega = \omega_J + i\omega_K\in \Omega^{2,0}(X)$, which is a holomorphic symplectic form. The curvature tensor can be written as $K\in\Omega^{1,1}(X,\mathrm{End}T)$, or using $\omega$ to identify the holomorphic tangent bundle $T$ with the holomorphic cotangent bundle $T'$ we can write it as $\Phi\in\Omega^{0,1}(X,\mathrm{Sym}^3T')$.

For any trivalent graph $\Gamma$ with $2n$ vertices, we define a $(0,2n)$ differential form $\Gamma(\Phi)$ by contracting the tensor $\Phi^{\otimes 2n}\otimes \tilde{\omega}^{\otimes 3n}$ along the edges of $\Gamma$, where $\tilde{\omega}\in T\wedge T$ is the dual form to $\omega$, and then anti-symmetrizing to obtain a differential form. See the references for details.The Rozansky-Witten invariant associated with $\Gamma$ and $X$ is then defined as $$ b_{\Gamma}(X) = \frac{1}{(8\pi^2)^n n!}\int_X \Gamma(\Phi)\wedge\omega^{n}. $$ My question is the following: why doesn't $b_{\Gamma}(X)$ depend on the choice of $(I,J,K)$? More precisely, we could have chosen $(I',J',K')$ related to $(I,J,K)$ by any rotation in $SO(3)$ and the claim is that the invariant above stays the same.

Here is how I have interpreted the explanation given in Hitchin-Sawon and other papers by Sawon such as "Generalisations of Rozansky-Witten invariants" with Justin Roberts. The hyperkahler structure on $X$ can be expressed as $TX\otimes_\mathbb{R}\mathbb{C} \approx E_{n}\otimes T$ where $E_n$ is a complex vector bundle over $X$ of rank $2n$ endowed with non-degenerate form $\epsilon\in\Lambda^2E_n^*$ and $T$ is a trivialisable complex vector bundle of rank $2$ - some explanations about this would be very helpful, but mostly what I'm interested in is how to use this to prove the following claim.

Write the curvature tensor as $\Omega\in\mathrm{Sym}^4E_n$ using the isomorphism $TX\otimes_\mathbb{R}\mathbb{C} \approx E_{n}\otimes T$. We may use $(\Omega,\epsilon)$ as weights for our graphs instead of $(\Phi,\tilde\omega)$, so proceeding by before we associate to any $\Gamma$ trivialent graph of $2n$ vertices a form $\Gamma(\Omega)\in\Lambda^{2n}E_n$.Denoting by $\tilde{\epsilon}\in\Lambda^2 E_n$ the dual form of $\epsilon$, we have $\Gamma(\Omega)=\Gamma_\epsilon(\Omega)\tilde{\epsilon}^n$ for some function $\Gamma_\epsilon(\Omega)$ and then, denoting by $d\mathrm{vol}$ the volume form of $(X,g)$, $$ b_\Gamma(X) = \frac{1}{(8\pi^2)^n n!}\int_X \Gamma_\epsilon(\Omega) d\mathrm{vol} $$ The claim is that the right-hand side is equal to the left-hand side, which is defined as before by making a choice of $(I,J,K)$. Since the right-hand side does not depend on this choice, we can deduce that $b_\Gamma(X)$ depends only on the hyperkahler structure and not on arbitrary choices of the generators $I,J,K$. How to prove this claim?

Answer 1

There is an alternative approach which was explained to me 2 days ago, I believe it provides a proof of "independence of choice of complex structure compatible with the hyperkahler structure". I will attempt to explain this here.

From the hypercomplex structure, any hyperkahler manifold $X$ admits an action of $SU(2)$ on $TX$, which naturally extends to the complexification $TX\otimes_\mathbb{R}\mathbb{C}$ and to all tensors. Let $T=T(\sigma)$ be a tensor obtained after fixing generators of the quaternionic structure $\sigma=(I,J,K)$. We say $T$ is $SU(2)$-equivariant if $(U\cdot T)(\sigma) = T(U\cdot\sigma)$, where $U\cdot\sigma=(UIU^*,UJU^*,UKU^*)$ and $U\cdot T$ denotes the action of $U$ on the tensor $T$. The statement is proved once we show that $\Phi$, $\omega$ and $\tilde{\omega}$ are $SU(2)$-equivariant in the sense above.

Indeed, assuming this then for any oriented trivalent graph $\Gamma$ with $2n$ vertices, where $\dim_\mathbb{R}(X)=4n$, we have that $\Gamma(\Phi)\wedge\omega^n$ is $SU(2)$-equivariant. Since it is a volume form, and the $SU(2)$ action is trivial on volume forms, we conclude that $\Gamma(\Phi)\wedge\omega^n$ is $SU(2)$-invariant - therefore the Rozansky-Witten invariants does not depend on the choice of $\sigma$. I am left with the exercise of proving that $\Phi$, $\omega$ and $\tilde{\omega}$ are $SU(2)$-equivariant.

For example, $SU(2)$-equivariance of $\omega$ seems to follow easily. Let $\omega_L(u,v) = g(u,Lv)$ for any complex structure $L=aI+bJ+cK$, parameterized by $\mathbb{S}^2$. Then: $$ L = UIU^*\Rightarrow \omega_L(u,v) = \omega_{I}(U^* u , U^*v), $$ which shows $SU(2)$-equivariance since the right-hand side is $(U\cdot\omega_I)(u,v)$. The relation $\omega_L(u,v) = \omega_{I}(U^* u , U^*v)$ is straightforward to check: $$ \omega_L(u,v) = g(u,Lv) = g(u,UIU^*v) = g(U^*u,IU^*v) = \omega_I(U^*u,U^*v). $$ This implies that $\omega=\omega_J + i\omega_K$ is $SU(2)$-equivariant; the equivariance of $\tilde{\omega}$ follows. Similarly, the equivariance of $\Phi$ follows from the invariance of the curvature tensor of a hyperkahler manifold, and this seems to answer my question completely!