Timeline for Deformation of finite coverings between smooth projective varieties
Current License: CC BY-SA 3.0
10 events
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| Sep 3, 2015 at 10:46 | comment | added | Jason Starr | . . . Thus, when $Z$ is $\text{Spec}(\mathbb{C}[\epsilon]/\langle \epsilon^2 \rangle)$, for a given first-deformation $e(Y)\in H^1(Y,T_Y)$, for a first-order deformation $e(X)$ in $H^1(X,T_X)$, there exists a corresponding first-order deformation of $f$ with respect to $e(Y)$ and $e(X)$ if and only if the image of $e(X)$ in $H^1(X,f^*T_Y)$ equals the pullback of $e(Y)$. | |
| Sep 3, 2015 at 10:42 | comment | added | Jason Starr | I am a little rusty, but it seems that Prop. 2.1.4.5 on p. 189 says, for a given deformation of $Y$ over an (Artinian?) base $Z$ whose associated first order deformation is $e(Y)$ in $H^1(Y,T_Y)$, for a given closed subscheme $\overline{Z}\hookrightarrow Z$ whose ideal sheaf is a square-zero ideal $I$, for a given extension of $(X,f:X\to Y)$ over $\overline{Z}$, the element in $H^1(X,N_f)\otimes I = RHom^2_{\mathcal{O}_X}(L_f,\mathcal{O}_X[0])\otimes I$ measuring existence of a lift is the image of the pullback of $e(Y)$ in $H^1(X,f^*T_Y)\otimes I$ . . . | |
| Sep 3, 2015 at 10:14 | comment | added | Jason Starr | It should be in Illusie -- I will check. I believe the point is that, for a given flat family of Y's over a base, the obstruction to deforming $(X,f:X\to Y)$ to match that $Y$ lives in the same group. So probably that has something to do with the map from $H^1(Y,T_Y)$ to the obstruction group. | |
| Sep 3, 2015 at 10:12 | comment | added | Francesco Polizzi | I've made an edit of the question about this specific point. | |
| Sep 3, 2015 at 10:12 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 3, 2015 at 10:10 | comment | added | Francesco Polizzi | @JasonStarr: This is interesting, thank you for the references. Do you know anything about $\eta$? For instance, can one characterize the first order deformations of $X$ mapping onto the first summand $H^1(Y, \, T_Y)$? | |
| Sep 2, 2015 at 19:12 | comment | added | Jason Starr | I guess the reference is Prop. III.2.1.2.3, p. 186, Luc Illusie, "Complexe cotangent et deformations, I", Lect. Notes in Math. 239, Springer-Verlag, Berlin - New York, 1971. | |
| Sep 2, 2015 at 18:28 | comment | added | Jason Starr | A good reference is Section 6 of K. Behrend, B. Fantechi, "The Intrinsic Normal Cone", Invent. Math. 128 (1997), 45-88. The adjoint of $df$ gives a perfect complex concentrated in amplitude $[-1,0]$ that is quasi-isomorphic to the cotangent complex $L_f$ of $f$. The hyperext group $RHom^1_{\mathcal{O}_X}(L_f,\mathcal{O}_X[0])$ is the space of first-order deformations of the pair $(X,f:X\to Y)$ with $Y$ held fixed. In your setting, this is the same as $H^0(X,N_f)$. Thus the image of an element under $\psi$ equals $0$ if and only if there is a corresponding first-order deformation of $f$. | |
| Sep 2, 2015 at 17:16 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 2, 2015 at 17:02 | history | asked | Francesco Polizzi | CC BY-SA 3.0 |