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In Hartshorne's book 《Algebraic geomery》 p.443, the author gives an explanation of Hironaka's example on non-Kähler deformation of compact Kähler manifolds, his construction can be summarised as follows:

Let $X$ be a complex projective manifold of dimension 3, take 2 non-singular curves $c, d\subset X$ which meet transversally at 2 points $P,Q$, and nowhere else. On $X-Q$, first blow up $c$, then the strict transform of $d$. On $X-P$, first blow up $d$, then the strict transform of $c$. Glue these two manifolds together, then we get Hironaka's example$-$a Moishezon manifold $\tilde X$ which is non-Kähler.

Regarding his construction, I have 2 questions:

  1. What if we change the projective manifold $X$ to an arbitrary compact complex manifold with the blow-up process remains the same, can we still get a non-Kähler $\tilde X$ as in the projective case, or some restrictions should be added on $X$?

  2. Recall that, in Hartshorne's construction, $l_0$ is the inverse image of $P$ when blow up in $c$ in $X-Q$, and $m_0'$ the inverse image of $Q$ when blow up $d$ in $X-P$, and by some algebraic equivalences, he gets the conclusion that $l_0+m_0'\sim 0$ which is impossible on a projective variety. Here comes my question: is $l_0+m_0'\sim 0$ equivalent to $l_0\sim 0$ and $m_0'\sim 0$? And either $l_0\sim 0$ or $m_0'\sim 0$ suffices to imply that $\tilde X$ is non-Kähler?

Any comment is welcome.

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    $\begingroup$ You are mixing the two examples given by Hartshorne. In order to get a Moishezon manifold, he just blows up one curve with a double point, and obtains indeed $l_0\sim 0$. $\endgroup$
    – abx
    Nov 19, 2021 at 9:27
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    $\begingroup$ OK, you are right. Then the answer to 2. is no: if $l_0+m'_0$ implies $l_0\sim m'_0 \sim0$, you get also $l'_0\sim m_0\sim 0$, hence $l\sim l_0+m_0\sim 0$, which is false. $\endgroup$
    – abx
    Nov 19, 2021 at 13:22
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    $\begingroup$ @abx, I think there is a misprint in Hartshorne's book p.444, actually, it should be $l\sim l_0'$ not $l\sim l_0$, so $l_0\sim m_0'\sim 0$ only implies $l\sim m\sim m_0\sim l_0'$ which are all not $\sim 0$, thus there is no conclusion that $l\sim 0$, is that right? $\endgroup$
    – Tom
    Nov 19, 2021 at 13:43
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    $\begingroup$ Right again, there is indeed a misprint in Hartshorne. However $l\sim m$ is impossible: if $e$ is the exceptional divisor above $l$, we have $e\cdot l=-1$ but $e\cdot m=0$. $\endgroup$
    – abx
    Nov 19, 2021 at 14:41
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    $\begingroup$ No, algebraic equivalence is more complicated than what you think. In our case you can just think of $l,m,$ etc. as cohomology classes. $\endgroup$
    – abx
    Nov 20, 2021 at 7:01

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