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May 3, 2022 at 16:24 vote accept Will Chen
Apr 14, 2022 at 18:03 history edited Will Chen CC BY-SA 4.0
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Apr 14, 2022 at 17:53 history edited Will Chen CC BY-SA 4.0
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Apr 10, 2022 at 20:13 history edited Will Chen CC BY-SA 4.0
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Apr 10, 2022 at 20:07 comment added Will Chen @PiotrAchinger Sure, I'd be happy for an answer in the case of smooth $X$.
Apr 9, 2022 at 22:12 answer added Donu Arapura timeline score: 5
Apr 9, 2022 at 15:09 comment added Will Chen @JasonStarr Remy is right - $\chi(X_{y_0},\mathcal{O}_{X_{y_0}})$ would be fine, but $\chi(X, \mathcal{O}_X(X_{y_0}))$ wouldn't be, unless it can somehow be expressed in terms of data which is local on $Y$.
Apr 9, 2022 at 14:12 comment added R. van Dobben de Bruyn @JasonStarr I imagine $\chi(X_{y_0},\mathcal O_{X_{y_0}})$ would be allowed as it is 'fibral', but not $\chi(X,\mathcal O_X(X_{y_0}))$ as it is 'global' on $X$.
Apr 9, 2022 at 7:42 comment added Piotr Achinger Are you willing to assume that $X$ (not $f$) is smooth?
Apr 9, 2022 at 0:31 comment added Jason Starr I am trying to understand how your "nonexample" is not an example. What kind of formula do you envision when $X$ equals $Y\times Z$ if you are not allowing $\chi(Z,\mathcal{O}_Z)$ as part of your formula?
Apr 8, 2022 at 20:30 history edited Will Chen CC BY-SA 4.0
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Apr 8, 2022 at 19:48 history edited Will Chen CC BY-SA 4.0
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Apr 8, 2022 at 19:05 history asked Will Chen CC BY-SA 4.0