Lemma 90.12.6. 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 90.12.2 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 90.11.5. 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. $\square$
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Your email address will not be published.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)
There are also: