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We say that a projective variety $X$ is of general type if the Kodaira dimension is equal to the dimension of $X$., i.e. $\text{kod}(X)=\dim X$.

When $K_X$ is positive then by the result of S.T.Yau we have a Kaehler-Einstein metric on $X$, i.e., $Ric(\omega)=-\omega$,

For varieties of general type, we have the finite generation of the canonical ring by recent breakthrough prize winners and hence they solved that we still have Kaehler-Einstein metric on varieties of general type i.e., the extension of S.T.Yau result.

OK, Which type of questions can be solved by using finite generation of the canonical ring in the geometric analysis? (a summary of important results could be enough!)

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  • $\begingroup$ Let me depress you! when $K_X$ is positive then by celebrated result of Yau-Aubin, we have a Kahler-Einstein metric, but for varieties of general type we still have a KE etric on $X\setminus Z$ but the sad part is that you may not be able to extend such canonical line bundle (in metric sense as current) to whole of $X$, this corresponds to $L^2$-Extension theory "à la Siu" , the second sad part of story is that such canonical line bundle smoothly can be extended to whole of $X$ ?, last disaster part is that this new current again sit in KE metric !? $\endgroup$
    – user21574
    Commented Dec 12, 2017 at 14:23
  • $\begingroup$ OK OK ,So for varieties of general type we just can say that we can have some sort of twisted Kahler-Einstein metric $Ric(\omega)=-\omega+[PD(E)]$. OK, OK but when $K_X$ is positive and $X$ is smooth then we have KE metric. $\endgroup$
    – user21574
    Commented Dec 12, 2017 at 14:23
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    $\begingroup$ Two points: 1. the finite generation result you refer to was proved by four people, of whom two won the Breakthrough prize. 2. They proved that varieties of general type have minimal models. (I don't know what "minimal general type" is.) But that implies finite generation of the canonical ring for all varieties. $\endgroup$
    – Tippi
    Commented Dec 12, 2017 at 16:53
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    $\begingroup$ @HassanJolany , In which case when $X$ is of general type, then $K_X$ is positive? $\endgroup$
    – Dima
    Commented Dec 13, 2017 at 10:04
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    $\begingroup$ @Dima You can impose some condition, for instance, Let $X$ be a smooth projective variety of general type $ kod(X)=dimX$, which contains no rational curves. Then, $K_X$ is ample hence it is positive and your variety admits Kahler-Einstein metric of negative Ricci curvature on $X$. See p.219, Exercise 8, of the book Olivier Debarre. Higher-dimensional algebraic geometry. Universitext. Springer-Verlag, New York, 2001 $\endgroup$
    – user21574
    Commented Dec 13, 2017 at 10:38

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