The Stacks project

Lemma 88.9.4. Let $\mathcal{F}$ be a predeformation category. Let $\xi $ be a versal formal object of $\mathcal{F}$ lying over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. The following are equivalent

  1. $\mathcal{F}$ is unobstructed, and

  2. $\Lambda \to R$ is formally smooth in the $\mathfrak m_ R$-adic topology.

In the classical case these are also equivalent to

  1. $R \cong \Lambda [[x_1, \ldots , x_ n]]$ for some $n$.

Proof. If (1) holds, i.e., if $\mathcal{F}$ is unobstructed, then the composition

\[ \underline{R}|_{\mathcal{C}_\Lambda } \xrightarrow {\underline{\xi }} \mathcal{F} \to \mathcal{C}_\Lambda \]

is smooth, see Lemma 88.8.7. Hence we see that (2) holds by Lemma 88.9.3. Conversely, if (2) holds, then the composition is smooth and moreover the first arrow is essentially surjective by Lemma 88.8.11. Hence we find that the second arrow is smooth by Lemma 88.8.7 which means that $\mathcal{F}$ is unobstructed by definition. The equivalence with (3) in the classical case follows from Lemma 88.9.3. $\square$


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Your email address will not be published.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 0DZK. Beware of the difference between the letter 'O' and the digit '0'.