Compact complex affine Kähler manifold is a torus












5












$begingroup$


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.






dg.differential-geometry complex-geometry kahler-manifolds complex-manifolds







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$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago
















5












$begingroup$


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.






dg.differential-geometry complex-geometry kahler-manifolds complex-manifolds







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago














5












5








5





$begingroup$


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.






dg.differential-geometry complex-geometry kahler-manifolds complex-manifolds







share|cite|improve this question











$endgroup$




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.






dg.differential-geometry complex-geometry kahler-manifolds complex-manifolds




dg.differential-geometry complex-geometry kahler-manifolds complex-manifolds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Michael Albanese

7,59655192




7,59655192










asked 8 hours ago









J.E.M.SJ.E.M.S

21114




21114








  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago














  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago








1




1




$begingroup$
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.
$endgroup$
– Ben McKay
3 hours ago




$begingroup$
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.
$endgroup$
– Ben McKay
3 hours ago










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Michael Albanese
    19 mins ago



















1












$begingroup$

Consider $S^2$ it is a projective manifold ( a manifold endowed with an atlas $(U_i), f_i:U_irightarrow Pmathbb{R}^n$ such that $f_icirc f_j^{-1}$ 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 $Mtimes S^1$. $S^2$ is endowed with a real projective structure, thus $S^2times S^1$ is a real affine manifold, $S^2times S^1times S^1$ 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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – J.E.M.S
    4 hours ago










  • $begingroup$
    The result says that if $M$ is projective, $Mtimes S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16.
    $endgroup$
    – Tsemo Aristide
    4 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago










  • $begingroup$
    What I am saying is the following: the quotient of $mathbb{R}^3-{0}$ by the homothetic map defined by $h(x)=2x$ is $S^2times S^1$. The quotient of $mathbb{R}^+=mathbb{R}-{0}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2times S^1times S^1$, $(mathbb{R}^3-{0})times mathbb{R}^+$ is contained in $mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2times S^1times S^1$.
    $endgroup$
    – Tsemo Aristide
    5 mins ago










  • $begingroup$
    The argument of abx shows that there does not exists a symplectic structure such that $(S^2times S^1times S^1,omega, J)$ 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.
    $endgroup$
    – Tsemo Aristide
    5 mins ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Michael Albanese
    19 mins ago
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Michael Albanese
    19 mins ago














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









abxabx

23.6k34885




23.6k34885












  • $begingroup$
    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?
    $endgroup$
    – Michael Albanese
    19 mins ago


















  • $begingroup$
    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?
    $endgroup$
    – Michael Albanese
    19 mins ago
















$begingroup$
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?
$endgroup$
– Michael Albanese
19 mins ago




$begingroup$
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?
$endgroup$
– Michael Albanese
19 mins ago











1












$begingroup$

Consider $S^2$ it is a projective manifold ( a manifold endowed with an atlas $(U_i), f_i:U_irightarrow Pmathbb{R}^n$ such that $f_icirc f_j^{-1}$ 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 $Mtimes S^1$. $S^2$ is endowed with a real projective structure, thus $S^2times S^1$ is a real affine manifold, $S^2times S^1times S^1$ 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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – J.E.M.S
    4 hours ago










  • $begingroup$
    The result says that if $M$ is projective, $Mtimes S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16.
    $endgroup$
    – Tsemo Aristide
    4 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago










  • $begingroup$
    What I am saying is the following: the quotient of $mathbb{R}^3-{0}$ by the homothetic map defined by $h(x)=2x$ is $S^2times S^1$. The quotient of $mathbb{R}^+=mathbb{R}-{0}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2times S^1times S^1$, $(mathbb{R}^3-{0})times mathbb{R}^+$ is contained in $mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2times S^1times S^1$.
    $endgroup$
    – Tsemo Aristide
    5 mins ago










  • $begingroup$
    The argument of abx shows that there does not exists a symplectic structure such that $(S^2times S^1times S^1,omega, J)$ 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.
    $endgroup$
    – Tsemo Aristide
    5 mins ago


















1












$begingroup$

Consider $S^2$ it is a projective manifold ( a manifold endowed with an atlas $(U_i), f_i:U_irightarrow Pmathbb{R}^n$ such that $f_icirc f_j^{-1}$ 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 $Mtimes S^1$. $S^2$ is endowed with a real projective structure, thus $S^2times S^1$ is a real affine manifold, $S^2times S^1times S^1$ 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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – J.E.M.S
    4 hours ago










  • $begingroup$
    The result says that if $M$ is projective, $Mtimes S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16.
    $endgroup$
    – Tsemo Aristide
    4 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago










  • $begingroup$
    What I am saying is the following: the quotient of $mathbb{R}^3-{0}$ by the homothetic map defined by $h(x)=2x$ is $S^2times S^1$. The quotient of $mathbb{R}^+=mathbb{R}-{0}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2times S^1times S^1$, $(mathbb{R}^3-{0})times mathbb{R}^+$ is contained in $mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2times S^1times S^1$.
    $endgroup$
    – Tsemo Aristide
    5 mins ago










  • $begingroup$
    The argument of abx shows that there does not exists a symplectic structure such that $(S^2times S^1times S^1,omega, J)$ 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.
    $endgroup$
    – Tsemo Aristide
    5 mins ago
















1












1








1





$begingroup$

Consider $S^2$ it is a projective manifold ( a manifold endowed with an atlas $(U_i), f_i:U_irightarrow Pmathbb{R}^n$ such that $f_icirc f_j^{-1}$ 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 $Mtimes S^1$. $S^2$ is endowed with a real projective structure, thus $S^2times S^1$ is a real affine manifold, $S^2times S^1times S^1$ 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






share|cite|improve this answer











$endgroup$



Consider $S^2$ it is a projective manifold ( a manifold endowed with an atlas $(U_i), f_i:U_irightarrow Pmathbb{R}^n$ such that $f_icirc f_j^{-1}$ 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 $Mtimes S^1$. $S^2$ is endowed with a real projective structure, thus $S^2times S^1$ is a real affine manifold, $S^2times S^1times S^1$ 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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 6 hours ago









Tsemo AristideTsemo Aristide

2,7231616




2,7231616












  • $begingroup$
    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.
    $endgroup$
    – J.E.M.S
    4 hours ago










  • $begingroup$
    The result says that if $M$ is projective, $Mtimes S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16.
    $endgroup$
    – Tsemo Aristide
    4 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago










  • $begingroup$
    What I am saying is the following: the quotient of $mathbb{R}^3-{0}$ by the homothetic map defined by $h(x)=2x$ is $S^2times S^1$. The quotient of $mathbb{R}^+=mathbb{R}-{0}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2times S^1times S^1$, $(mathbb{R}^3-{0})times mathbb{R}^+$ is contained in $mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2times S^1times S^1$.
    $endgroup$
    – Tsemo Aristide
    5 mins ago










  • $begingroup$
    The argument of abx shows that there does not exists a symplectic structure such that $(S^2times S^1times S^1,omega, J)$ 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.
    $endgroup$
    – Tsemo Aristide
    5 mins ago




















  • $begingroup$
    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.
    $endgroup$
    – J.E.M.S
    4 hours ago










  • $begingroup$
    The result says that if $M$ is projective, $Mtimes S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16.
    $endgroup$
    – Tsemo Aristide
    4 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Ben McKay
    3 hours ago










  • $begingroup$
    What I am saying is the following: the quotient of $mathbb{R}^3-{0}$ by the homothetic map defined by $h(x)=2x$ is $S^2times S^1$. The quotient of $mathbb{R}^+=mathbb{R}-{0}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2times S^1times S^1$, $(mathbb{R}^3-{0})times mathbb{R}^+$ is contained in $mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2times S^1times S^1$.
    $endgroup$
    – Tsemo Aristide
    5 mins ago










  • $begingroup$
    The argument of abx shows that there does not exists a symplectic structure such that $(S^2times S^1times S^1,omega, J)$ 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.
    $endgroup$
    – Tsemo Aristide
    5 mins ago


















$begingroup$
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.
$endgroup$
– J.E.M.S
4 hours ago




$begingroup$
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.
$endgroup$
– J.E.M.S
4 hours ago












$begingroup$
The result says that if $M$ is projective, $Mtimes S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16.
$endgroup$
– Tsemo Aristide
4 hours ago




$begingroup$
The result says that if $M$ is projective, $Mtimes S^1$ is affine, it is a radiant affine manifold, look the corollary p. 16.
$endgroup$
– Tsemo Aristide
4 hours ago




1




1




$begingroup$
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.
$endgroup$
– Ben McKay
3 hours ago




$begingroup$
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.
$endgroup$
– Ben McKay
3 hours ago












$begingroup$
What I am saying is the following: the quotient of $mathbb{R}^3-{0}$ by the homothetic map defined by $h(x)=2x$ is $S^2times S^1$. The quotient of $mathbb{R}^+=mathbb{R}-{0}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2times S^1times S^1$, $(mathbb{R}^3-{0})times mathbb{R}^+$ is contained in $mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2times S^1times S^1$.
$endgroup$
– Tsemo Aristide
5 mins ago




$begingroup$
What I am saying is the following: the quotient of $mathbb{R}^3-{0}$ by the homothetic map defined by $h(x)=2x$ is $S^2times S^1$. The quotient of $mathbb{R}^+=mathbb{R}-{0}$ by $g(x)=2x$ is $S^1$. This defines an affine structure on $S^2times S^1times S^1$, $(mathbb{R}^3-{0})times mathbb{R}^+$ is contained in $mathbb{C}^2$ and $h(x)=2x$ preserves the complex structure. This implies the existence of a complex structure $J$ on $S^2times S^1times S^1$.
$endgroup$
– Tsemo Aristide
5 mins ago












$begingroup$
The argument of abx shows that there does not exists a symplectic structure such that $(S^2times S^1times S^1,omega, J)$ 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.
$endgroup$
– Tsemo Aristide
5 mins ago






$begingroup$
The argument of abx shows that there does not exists a symplectic structure such that $(S^2times S^1times S^1,omega, J)$ 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.
$endgroup$
– Tsemo Aristide
5 mins ago




















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