Let $X$ be a complex smooth projective variety with trivial topological Euler characteristic $\chi_{\text{top}}(X)=0$. We assume that $D$ is a smooth irreducible divisor in the linear system $|K_X|$ of the canonical divisor $K_X$ of $X$. Is $\chi_{\text{top}}(D)=0$?
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3$\begingroup$ What happens if you consider the blow-up at 4 points of the product $C \times C \times C$, where $C$ is a smooth curve of genus $2$? $\endgroup$– Francesco PolizziCommented Nov 16, 2022 at 17:47
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1$\begingroup$ @FrancescoPolizzi Thank you for your example! $\endgroup$– user283487Commented Nov 16, 2022 at 18:25
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1$\begingroup$ @FrancescoPolizzi. Why is there a smooth irreducible divisor in the canonical linear system of that blowing up? $\endgroup$– Jason StarrCommented Nov 16, 2022 at 21:33
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1$\begingroup$ @JasonStarr: Actually, I do not know whether it exists. It is part of my "What happens..." $\endgroup$– Francesco PolizziCommented Nov 16, 2022 at 22:16
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1$\begingroup$ @Francesco Polizzi: it doesn't exist. The canonical system of the blown-up variety has a fixed part, namely the 4 exceptional divisors. $\endgroup$– abxCommented Nov 17, 2022 at 4:26
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This is not true. Consider, for instance a Calabi--Yau threefold $Y$ with $h^{2,1}(Y) = h^{1,1}(Y) + 1$ (an example of such can be found in https://arxiv.org/abs/1602.06303, see page 29) and let $X$ be the blowup of $Y$ in a point. Then $\chi_{\mathrm{top}}(X) = 0$, but the canonical class of $X$ is equal to the exceptional divisor of the blowup, hence the (unique) canonical divisor is isomorphic to $\mathbb{P}^2$.
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1$\begingroup$ Thank you for your conterexample. $\endgroup$ Commented Nov 17, 2022 at 9:05