The Stacks project

Lemma 92.21.1. In the situation above we have

  1. There is a canonical element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}(L_{X/S}, \mathcal{G})$ whose vanishing is a sufficient and necessary condition for the existence of a solution to (

  2. If there exists a solution, then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(L_{X/S}, \mathcal{G})$.

  3. Given a solution $X'$, the set of automorphisms of $X'$ fitting into ( is canonically isomorphic to $\mathop{\mathrm{Ext}}\nolimits ^0_{\mathcal{O}_ X}(L_{X/S}, \mathcal{G})$.

Proof. Via the identifications $\mathop{N\! L}\nolimits _{X/S} = \tau _{\geq -1}L_{X/S}$ (Lemma 92.20.4) and $H^0(L_{X/S}) = \Omega _{X/S}$ (Lemma 92.20.2) we have seen parts (2) and (3) in Deformation Theory, Lemmas 91.7.1 and 91.7.3.

Proof of (1). Roughly speaking, this follows from the discussion in Deformation Theory, Remark 91.7.9 by replacing the naive cotangent complex by the full cotangent complex. Here is a more detailed explanation. By Deformation Theory, Lemma 91.7.8 there exists an element

\[ \xi ' \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{S/S'}, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*L_{S/S'}, \mathcal{G}) \]

such that a solution exists if and only if this element is in the image of the map

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(NL_{X/S'}, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(L_{X/S'}, \mathcal{G}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*L_{S/S'}, \mathcal{G}) \]

The distinguished triangle of Lemma 92.20.3 for $X \to S \to S'$ gives rise to a long exact sequence

\[ \ldots \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(L_{X/S'}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*L_{S/S'}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}(L_{X/S}, \mathcal{G}) \to \ldots \]

Hence taking $\xi $ the image of $\xi '$ works. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08UZ. Beware of the difference between the letter 'O' and the digit '0'.