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    Infinitesimal automorphisms and first-order deformations

    T-DEFEXT

    $H^0(T_X)$ controls infinitesimal automorphisms, $H^1(T_X)$ first-order deformations, and $H^2(T_X)$ their obstructions.Suggest a slogan

    Let \(k\) be a field, \(D = \Spec k[\varepsilon]/(\varepsilon^2)\), and \(X\) a smooth separated scheme of finite type over \(k\) with tangent sheaf \(T_X = \mathcal{H}om(\Omega_{X/k}, \OO_X)\).

    1. \(\operatorname{Ext}^0(\Omega_{X/k}, \OO_X) = H^0(X, T_X)\), and it is in bijection with the automorphisms of \(X \times D\) over \(D\) that restrict to the identity on \(X\).

    2. \(\operatorname{Ext}^1(\Omega_{X/k}, \OO_X) = H^1(X, T_X)\), and it is in bijection with the isomorphism classes of first-order deformations: flat \(D\)-schemes \(\mathcal{X}\) with \(\mathcal{X} \times_D \Spec k \cong X\).

    3. Let \(A_n = k[t]/(t^{n+1})\) for \(n \geq 1\), so \(A_1 \cong k[\varepsilon]/(\varepsilon^2)\). A flat deformation \(\mathcal{X}_n\) of \(X\) over \(A_n\) determines an obstruction class \(o(\mathcal{X}_n) \in H^2(X, T_X)\), and \(o(\mathcal{X}_n) = 0\) if and only if there is a flat deformation \(\mathcal{X}_{n+1}\) over \(A_{n+1}\) with \(\mathcal{X}_{n+1} \times_{A_{n+1}} \Spec A_n \cong \mathcal{X}_n\).

    For a smooth projective curve \(C\) of genus \(g \geq 2\), \(H^0(T_C) = 0\), so \(C\) has no infinitesimal automorphisms, \(H^2(T_C) = 0\), so deformations are unobstructed, and \(h^1(T_C) = h^0(\omega_C^{\otimes 2}) = 3g - 3\) by Serre duality and Riemann–Roch.

    For $X \to B$ smooth, why is $\operatorname{Ext}^0(\Omega_{X/B}, \OO_X) \cong H^0(T_{X/B})$, and why does it measure infinitesimal automorphisms?
    Why does $\operatorname{Ext}^1(\Omega_{X/B}, \OO_X)$ measure first-order deformations?

    Dependencies

    • D-COTCPLXThe cotangent complex and its dualuses

    Backlinks

    • T-MGSMOOTHThe moduli stacks of smooth and stable curvesuses

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