Kahler manifolds and algebraic varieties

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Let $X$ be a smooth complete algebraic variety over $mathbbC$. Can it happen that the underlying complex manifold is not Kahler? If yes, are there explicit examples? If not - how to prove this?






ag.algebraic-geometry complex-geometry







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    Yes, that can and does happen. Please see "Hironaka's example" in the appendices of Hartshorne's "Algebraic geometry".
    – Jason Starr
    1 hour ago






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    Relevant: mathoverflow.net/questions/108307/…
    – M.G.
    1 hour ago














up vote
2
down vote

favorite












Let $X$ be a smooth complete algebraic variety over $mathbbC$. Can it happen that the underlying complex manifold is not Kahler? If yes, are there explicit examples? If not - how to prove this?






ag.algebraic-geometry complex-geometry







share|cite|improve this question

















  • 2




    Yes, that can and does happen. Please see "Hironaka's example" in the appendices of Hartshorne's "Algebraic geometry".
    – Jason Starr
    1 hour ago






  • 2




    Relevant: mathoverflow.net/questions/108307/…
    – M.G.
    1 hour ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $X$ be a smooth complete algebraic variety over $mathbbC$. Can it happen that the underlying complex manifold is not Kahler? If yes, are there explicit examples? If not - how to prove this?






ag.algebraic-geometry complex-geometry







share|cite|improve this question













Let $X$ be a smooth complete algebraic variety over $mathbbC$. Can it happen that the underlying complex manifold is not Kahler? If yes, are there explicit examples? If not - how to prove this?






ag.algebraic-geometry complex-geometry




ag.algebraic-geometry complex-geometry






share|cite|improve this question













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asked 2 hours ago









Alexander Braverman

3,8941239




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  • 2




    Yes, that can and does happen. Please see "Hironaka's example" in the appendices of Hartshorne's "Algebraic geometry".
    – Jason Starr
    1 hour ago






  • 2




    Relevant: mathoverflow.net/questions/108307/…
    – M.G.
    1 hour ago












  • 2




    Yes, that can and does happen. Please see "Hironaka's example" in the appendices of Hartshorne's "Algebraic geometry".
    – Jason Starr
    1 hour ago






  • 2




    Relevant: mathoverflow.net/questions/108307/…
    – M.G.
    1 hour ago







2




2




Yes, that can and does happen. Please see "Hironaka's example" in the appendices of Hartshorne's "Algebraic geometry".
– Jason Starr
1 hour ago




Yes, that can and does happen. Please see "Hironaka's example" in the appendices of Hartshorne's "Algebraic geometry".
– Jason Starr
1 hour ago




2




2




Relevant: mathoverflow.net/questions/108307/…
– M.G.
1 hour ago




Relevant: mathoverflow.net/questions/108307/…
– M.G.
1 hour ago










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Nonprojective compact algebraic manifolds are never Kähler. Any compact algebraic manifold is Moishezon, and Moishezon's theorem says that a Moishezon manifold is Kähler if and only if it is a projective variety.






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    Nonprojective compact algebraic manifolds are never Kähler. Any compact algebraic manifold is Moishezon, and Moishezon's theorem says that a Moishezon manifold is Kähler if and only if it is a projective variety.






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      Nonprojective compact algebraic manifolds are never Kähler. Any compact algebraic manifold is Moishezon, and Moishezon's theorem says that a Moishezon manifold is Kähler if and only if it is a projective variety.






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        Nonprojective compact algebraic manifolds are never Kähler. Any compact algebraic manifold is Moishezon, and Moishezon's theorem says that a Moishezon manifold is Kähler if and only if it is a projective variety.






        share|cite|improve this answer












        Nonprojective compact algebraic manifolds are never Kähler. Any compact algebraic manifold is Moishezon, and Moishezon's theorem says that a Moishezon manifold is Kähler if and only if it is a projective variety.







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        answered 1 hour ago









        Dan Petersen

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