The Stacks project

Lemma 89.12.6. Let $\mathcal{F}$ be a predeformation category over $\mathcal{C}_\Lambda $. If $\overline{\mathcal{F}}$ has (S2) then the maps $\gamma _ V$ are $k$-linear and we have $a_ V(D, x) = x + \gamma _ V(D)$.

Proof. In the proof of Lemma 89.12.2 we have seen that the functor $V \mapsto \overline{\mathcal{F}}(k[V])$ transforms $0$ to a singleton and products to products. The same is true of the functor $V \mapsto \text{Der}_\Lambda (k, V)$. Hence $\gamma _ V$ is linear by Lemma 89.11.5. Let $D : k \to V$ be a $\Lambda $-derivation. Set $D_1 : k \to V^{\oplus 2}$ equal to $a \mapsto (D(a), 0)$. Then

\[ \xymatrix{ k[V \times V] \ar[r]_{+} \ar[d]^{f_{1, D_1}} & k[V] \ar[d]^{f_{1, D}} \\ k[V \times V] \ar[r]^{+} & k[V] } \]

commutes. Unwinding the definitions and using that $\overline{F}(V \times V) = \overline{F}(V) \times \overline{F}(V)$ this means that $a_ D(x_1) + x_2 = a_ D(x_1 + x_2)$ for all $x_1, x_2 \in \overline{F}(V)$. Thus it suffices to show that $a_ V(D, 0) = 0 + \gamma _ V(D)$ where $0 \in \overline{F}(V)$ is the zero vector. By definition this is the element $f_{0, *}(x_0)$. Since $f_ D = f_{1, D} \circ f_0$ the desired result follows. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 89.12: Tangent spaces of predeformation categories

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Your email address will not be published.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 06SU. Beware of the difference between the letter 'O' and the digit '0'.