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I am interested to know an example of a simply connected smooth projective 3-fold $X$ (over $\mathbb{C}$) satisfying the following two constraints:

  1. $X$ has the same Betti numbers as $\mathbb{C}\mathbb{P}^{3}$ i.e. $b_{1}(X) = b_{3}(X) = 0$ and $b_{2}(X) = 1$ and all of its cohomology groups are torsion-free.

  2. $\mathrm{Kod}(X) \geq 0$.

($\mathrm{Kod}(X)$ denotes the Kodaira dimension).

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    $\begingroup$ There is an argument that such a threefold cannot exist in section 3 of "Uniformization of Fake Projective Four Spaces", by Sai-Kee Yeung. $\endgroup$
    – dhy
    Commented Aug 25, 2017 at 15:14
  • $\begingroup$ Very helpful, Thanks!!, I think I will change the question to ask about positive Kodiara dimension, since this is also interesting for me. $\endgroup$
    – Nick L
    Commented Aug 25, 2017 at 15:19
  • $\begingroup$ Actually the discussion there completely answers this question also! $\endgroup$
    – Nick L
    Commented Aug 25, 2017 at 15:34
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    $\begingroup$ Since $\mathrm{Pic}(X)=\mathbb{Z}$, $\mathrm{Kod}(X)\geq 0$ implies that $K_X$ is either ample or trivial, and the latter case would imply $b_3>1$. $\endgroup$
    – abx
    Commented Aug 25, 2017 at 16:51

2 Answers 2

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Let me just mention that the non-existence of such a threefold is an immediate consequence of Yau's inequality. First, as explained in the above comment, the conditions $b_2=1$ and $\mathrm{Kod}(X)\geq 0$ imply that $K_X$ is ample. Then Yau gives $c_1^3\geq \frac{8}{3}c_1c_2 $, which is equivalent by Riemann-Roch to $K_X^3\leq 64 \chi (K_X)$. But the conditions on the Betti numbers imply $H^i(X,\mathcal{O}_X)=0$ for $i>0$, hence $\chi (K_X)=-\chi (\mathcal{O}_X)=-1$, a contradiction.

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As pointed out by dhy, the question is completely resolved in section 3 of "Uniformization of Fake Projective Four Spaces", by Sai-Kee Yeung. The conclusion is that there is no such $3$-fold.

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  • $\begingroup$ would you 1) complete your answer saying that the answer is that there is no such example 2) tick your own answer as accepted? (otherwise the question is considered by the robot as unsolved) $\endgroup$
    – YCor
    Commented Aug 25, 2017 at 19:22
  • $\begingroup$ When I click the tick, I get a message "You can accept your own answer in two days" $\endgroup$
    – Nick L
    Commented Aug 25, 2017 at 19:24
  • $\begingroup$ Ah OK, I didn't know this feature. $\endgroup$
    – YCor
    Commented Aug 25, 2017 at 19:24

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