Theorem 89.18.2. Let $F: \mathcal{C}_\Lambda \to \textit{Sets}$ be a functor. Then $F$ is prorepresentable if and only if (a) $F$ is a deformation functor, (b) $\dim _ k TF$ is finite, and (c) $\gamma : \text{Der}_\Lambda (k, k) \to TF$ is injective.
Proof. Assume $F$ is prorepresentable by $R \in \widehat{\mathcal{C}}_\Lambda $. We see $F$ is a deformation functor by Example 89.16.10. We see $\dim _ k TF$ is finite by Example 89.11.11. Finally, $\text{Der}_\Lambda (k, k) \to TF$ is identified with $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (R, k)$ by Example 89.11.14 which is injective because $R \to k$ is surjective.
Conversely, assume (a), (b), and (c) hold. By Lemma 89.16.6 we see that (S1) and (S2) hold. Hence by Theorem 89.15.5 there exists a minimal versal formal object $\xi $ of $F$ such that (89.15.0.2) holds. Say $\xi $ lies over $R$. The map
is bijective on $\text{Der}_\Lambda (k, k)$-orbits. Since the action of $\text{Der}_\Lambda (k, k)$ on the left hand side is free by (c) and Lemma 89.12.6 we see that the map is bijective. Thus we see that $\underline{\xi }$ is an isomorphism by Lemma 89.18.1. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)