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I am reading Dominic Joyce's book 'Compact Manifolds with Special Holonomy' and I am struggling to understand a remark he makes at the end of chapter 8. The assertion is (I think) the following:

Suppose $G$ is a finite subgroup of $U(n)$ acting freely on $\mathbb{C}^n\setminus \{0\}$. Let $(X,J,g)$ be a Kähler manifold such that there exists a compact set $K\subset X$ together with a diffeomorphism $\phi:X\setminus K\to \mathbb{C}^n/G\setminus B_R(0)$ satisfying (for $r>R$, and all $k\in \mathbb{N}_{0}$)

1) $|\nabla^k(\phi_{\ast}g-g_0)|_{C^0(\mathbb{C}^n/G\setminus B_r(0))}=O(r^{-2n-k})$

2) $|\nabla^k(\phi_{\ast}J-J_0)|_{C^0(\mathbb{C}^n/G\setminus B_r(0))}=O(r^{-2n-k})$

where $g_0$ is the Euclidean metric on $\mathbb{C}^n/G$ and $J_0$ is the complex structure on $\mathbb{C}^n/G$.

Then:

3) If $n\geq 2$: $(X,J)$ is birational to a deformation of $\mathbb{C}^n/G$.

4) If $n\geq 3$: $(X,J)$ is birational to $\mathbb{C}^n/G$.

I am willing to believe that 4) follows from 3) (At least there is a reference in the book). I would be interested in the proof of 3). Does this follow from a more general theorem in the literature?

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  • $\begingroup$ There are two parts in Joyce's sketch of proof of 3): first prove that there are enough polynomial growth holomorphic functions on $X$ which can be used to embed $X$ into $\mathbb{C}^N$ as an affine algebraic variety. Second: given this, show that $X$ is birational to a deformation of $\mathbb{C}^n/G$. The second part should boil down to more standard complex geometric-arguments, but I don't know exactly how off the top of my head. $\endgroup$
    – YangMills
    Nov 26, 2012 at 19:49
  • $\begingroup$ Yes, I think I can follow the argument for the first part of the proof. What I'm looking for is the details or a reference for the second part of the proof. First I thought I could extract the proof from Kronheimer's paper (for n=2 at least) but I think it's actually much easier (i.e. stanard) than that. I was wondering where in the literature I could find a result like this. $\endgroup$
    – oydeis
    Nov 27, 2012 at 9:57
  • $\begingroup$ In fact, this apparently was not proved by Joyce, but rather was shown more recently here arxiv.org/pdf/1610.05239.pdf $\endgroup$
    – YangMills
    Jun 9, 2019 at 0:19

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