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Say we have a complex variety given by:

$$x^5+y^4+3=0$$

with $(x,y)\in \mathbb{C}^2$. Then apparently this is not a Kähler surface. (Or is it?) But if we do the projective completion:

$$x^5+y^4z+3z^5=0$$

With $(x,y,z)\in \mathbb{PC}^2$ then this is a Kähler surface as all complex projective varieties in complex-projective space are Kähler as "Every smooth complex projective variety" is Kähler.

But the second case is just the first case with extra points `at infinity'.

So how can you have a complex structure on the second case but not on the first? The second is just the first projected onto the hyper-plane $z=1$ with some points projected to points at infinity

The first case is non-compact but this does not seem to be a condition of Kähler surfaces. Why is the first case not called Kähler also?

zooby
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  • What makes you think that a smooth affine variety is not Kahler (when equipped with the restriction of the flat metric)? – Moishe Kohan Apr 16 '23 at 19:04
  • I read it in this comment: https://mathoverflow.net/questions/108307/are-complex-varieties-kahler-algebraic-non-projective-complex-manifolds I read somewhere which said most complex varieties are not Kahler only projective ones. But I got confused because I can turn any variety into a projective one by projective completion – zooby Apr 16 '23 at 19:06
  • You misunderstood what you have read, they only discussed compact complex manifolds. Read a textbook on Kahler geometry instead. – Moishe Kohan Apr 16 '23 at 19:45
  • @MoisheKohan OK. Sorry for asking math questions in a math forum – zooby Apr 16 '23 at 19:50
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    There is nothing to be sorry about, however, MO and MSE are not substitutes for a systematic learning. – Moishe Kohan Apr 16 '23 at 19:53

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