Lemma 93.11.3. In Example 93.11.1 let $X$ be an algebraic space over $k$. Then
and
Lemma 93.11.3. In Example 93.11.1 let $X$ be an algebraic space over $k$. Then
and
Proof. Recall that $\text{Inf}(\mathcal{D}\! \mathit{ef}_ X)$ is the set of automorphisms of the trivial deformation $X' = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ])$ of $X$ to $k[\epsilon ]$ equal to the identity modulo $\epsilon $. By Deformation Theory, Lemma 91.14.2 this is equal to $\text{Ext}^0_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/k}, \mathcal{O}_ X)$. The equality $\text{Ext}^0_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/k}, \mathcal{O}_ X) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\Omega _{X/k}, \mathcal{O}_ X)$ follows from More on Morphisms of Spaces, Lemma 76.21.4. The equality $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\Omega _{X/k}, \mathcal{O}_ X) = \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$ follows from More on Morphisms of Spaces, Definition 76.7.2 and Modules on Sites, Definition 18.33.3.
Recall that $T_{x_0}\mathcal{D}\! \mathit{ef}_ X$ is the set of isomorphism classes of flat deformations $X'$ of $X$ to $k[\epsilon ]$, more precisely, the set of isomorphism classes of $\mathcal{D}\! \mathit{ef}_ X(k[\epsilon ])$. Thus the second statement of the lemma follows from Deformation Theory, Lemma 91.14.2. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)