# Compact complex affine Kähler manifold is a torus

Before giving a motivation let me ask the precise question firstly.

By a complex affine manifold I mean a complex manifold $$M$$ with the property that there exists an holomorphic atlas for which transition functions are restrictions of functions belonging to $$Aff(\mathbb{C}^{\dim_\mathbb{C}M})$$, the group of complex affine motions of $$\mathbb{C}^{\dim_\mathbb{C}M}$$.

Question: Suppose $$M$$ is a compact complex affine manifold admitting Kähler metric. Does it imply that $$M$$ has a complex torus as a finite covering? What restriction does it imply on a Kähler metric? Does it have to be a flat metric induced from the torus?

The reason for such a question is Remark 2 at the end of Ma. Kato's paper Compact Differentiable 4-Folds with Quaternionic Structures. Apparently Calabi-Yau theorem seems to be of use here. Since I do not understand the explanation given there, the clarification of an argument in that case would be appreciated as well.

• The complete classification of compact complex surfaces admitting holomorphic affine connections was worked out by Kobayashi and Ochiai; see Kobayashi and Horst, Topics in Complex Differential Geometry, in the book Kobayashi and Wu, Complex Differential Geometry, for an overview. Also see Bruno Klingler, Structures affine et projectives sur les surfaces complexes, Annales de l'Institut Fourier, for more information on the possible complex affine structures. – Ben McKay Feb 17 at 21:23

If a compact Kähler manifold $$M$$ admits a holomorphic affine connection, its Atiyah class and therefore all its Chern classes are zero. By Yau's solution of the Calabi conjecture, this implies that a finite covering of $$M$$ is a complex torus.

• Sorry if I am missing something simple, but could you explain how Yau's solution of the Calabi conjecture implies that $M$ is finitely covered by a complex torus? – Michael Albanese Feb 18 at 0:52
• I wouldn't say this is simple, but I think well-known in differential geometry. See for instance this Bourbaki seminar on Yau's theorem by J.-P. Bourguignon, §2, Cor. 2. – abx Feb 18 at 6:02

Consider $$S^2$$ it is a projective manifold ( a manifold endowed with an atlas $$(U_i), f_i:U_i\rightarrow P\mathbb{R}^n$$ such that $$f_i\circ f_j^{-1}$$ is a restriction of an element of $$PGl(n,\mathbb{R})$$. Benzecri has shown in its thesis that if $$M$$ is a projective real manifold, so is $$M\times S^1$$. $$S^2$$ is endowed with a real projective structure, thus $$S^2\times S^1$$ is a real affine manifold, $$S^2\times S^1\times S^1$$ is also a real affine manifold. The holonomy of this manifold can be defined by homothetic maps so preserves a complex structure on $$\mathbb{R}^4$$. This manifold is endowed with a Kahler structure, the complex structure here is not necessarily associated with a symplectic structure such that $$(M,J,\omega)$$ is Kahler.

http://www.numdam.org/article/BSMF_1960__88__229_0.pdf

• Actually I have a problem with understanding this example. Can You elaborate on how does one get $S^2 \times S^1$ is affine? From the result You have cited it follows it's projective. – J.E.M.S Feb 17 at 20:26
• The result says that if $M$ is projective, $M\times S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16. – Tsemo Aristide Feb 17 at 20:39
• This doesn't actually give an example, because there is no Kaehler metric. As abx says, the compact Kaehler examples admit finite covering by complex tori. – Ben McKay Feb 17 at 21:19
• What I am saying is the following: the quotient of $\mathbb{R}^3−\{0\}$ by the homothetic map defined by$h(x)=2x$ is $S^2×S^1$. The quotient of $R^+=\{x\in \mathbb{R}, x>0\}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2×S^1×S^1$, $(\mathbb{R}^3−{0})×\mathbb{R}^+$ is contained in $\mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2×S^1×S^1.$ – Tsemo Aristide Feb 18 at 1:21
• The argument of abx shows that there does not exists a symplectic structure $\omega$ such that $(S^2\times S^1×S^1,ω,J)$ is a khaler structure. I dont know if there exists an easy way to show that. It may be interesting to construct examples of complex structures on khaler manifolds which are not associated to a khaler structure. – Tsemo Aristide Feb 18 at 1:24