Given a cotangent bundle of a projective variety $Y=T^*X,$ do we know that its affinization map, $$Y \rightarrow \mathrm{Spec}(H^0(Y,\mathscr{O}_Y))$$ is proper or projective?
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2$\begingroup$ No, that is not true. It can happen that the only global sections of the structure sheaf are constants. Then the morphism is constant, hence non-proper, since $Y$ is not proper. $\endgroup$– Jason StarrCommented Apr 9, 2023 at 0:29
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$\begingroup$ Hmm, I see. An example of this would be helpful? $\endgroup$– FilipCommented Apr 9, 2023 at 0:35
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2$\begingroup$ All positive tensor powers of the cotangent bundle on the projective line have vanishing global sections. All positive tensor powers of the tangent bundle on hyperbolic curves have vanishing global sections. $\endgroup$– Jason StarrCommented Apr 9, 2023 at 0:53
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1$\begingroup$ There are two opposite conventions on the meaning of that: the one in EGA and the other one. That is precisely why I gave one example for each convention. $\endgroup$– Jason StarrCommented Apr 9, 2023 at 14:50
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2$\begingroup$ @Filip In algebraic geometry, the total space of a vector bundle E is defined as the relative Spec of $Sym(E)$ or $Sym(E^{\vee})$ depending on what universal property you want it to satisfy. Let's just take the first convention for simplicity. In the case you are looking at, $E=L$ is a line bundle and $Sym(L)$ will be a direct sum of $L^{\otimes r}$. So, if all of the global sections of $L^{\otimes r}, r>0$ vanish, you will just have constant functions on Tot E. $\endgroup$– onefishtwofishCommented Apr 13, 2023 at 13:27
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