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89.12 Tangent spaces of predeformation categories

We will define tangent spaces of predeformation functors using the general Definition 89.11.9. We have spelled this out in Example 89.11.10. It applies to predeformation categories by looking at the associated functor of isomorphism classes.

Definition 89.12.1. Let $\mathcal{F}$ be a predeformation category. The tangent space $T \mathcal{F}$ of $\mathcal{F}$ is the set $\overline{\mathcal{F}}(k[\epsilon ])$ of isomorphism classes of objects in the fiber category $\mathcal F(k[\epsilon ])$.

Thus $T \mathcal{F}$ is nothing but the tangent space of the associated functor $\overline{\mathcal{F}}: \mathcal{C}_\Lambda \to \textit{Sets}$. It has a natural vector space structure when $\mathcal{F}$ satisfies (S2), or, in fact, as long as $\overline{\mathcal{F}}$ does.

Lemma 89.12.2. Let $\mathcal{F}$ be a predeformation category such that $\overline{\mathcal{F}}$ satisfies (S2)1. Then $T \mathcal{F}$ has a natural $k$-vector space structure. For any finite dimensional vector space $V$ we have $\overline{\mathcal{F}}(k[V]) = T\mathcal{F} \otimes _ k V$ functorially in $V$.

Proof. Let us write $F = \overline{\mathcal{F}} : \mathcal{C}_\Lambda \to \textit{Sets}$. This is a predeformation functor and $F$ satisfies (S2). By Lemma 89.10.4 (and the translation of Remark 89.10.3) we see that

\[ F(A \times _ k k[V]) \longrightarrow F(A) \times F(k[V]) \]

is a bijection for every finite dimensional vector space $V$ and every $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_\Lambda )$. In particular, if $A = k[W]$ then we see that $F(k[W] \times _ k k[V]) = F(k[W]) \times F(k[V])$. In other words, the hypotheses of Lemma 89.11.8 hold and we see that $TF = T \mathcal{F}$ has a natural $k$-vector space structure. The final assertion follows from Lemma 89.11.15. $\square$

A morphism of predeformation categories induces a map on tangent spaces.

Definition 89.12.3. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism predeformation categories. The differential $d \varphi : T \mathcal{F} \to T \mathcal{G}$ of $\varphi $ is the map obtained by evaluating the morphism of functors $\overline{\varphi }: \overline{\mathcal{F}} \to \overline{\mathcal{G}}$ at $A = k[\epsilon ]$.

Lemma 89.12.4. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of predeformation categories. Assume $\overline{\mathcal{F}}$ and $\overline{\mathcal{G}}$ both satisfy (S2). Then $d \varphi : T \mathcal{F} \to T \mathcal{G}$ is $k$-linear.

Proof. In the proof of Lemma 89.12.2 we have seen that $\overline{\mathcal{F}}$ and $\overline{\mathcal{G}}$ satisfy the hypotheses of Lemma 89.11.8. Hence the lemma follows from Lemma 89.11.13. $\square$

Remark 89.12.5. We can globalize the notions of tangent space and differential to arbitrary categories cofibered in groupoids as follows. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $, and let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. As in Remark 89.6.4, we get a predeformation category $\mathcal{F}_ x$. We define

\[ T_ x\mathcal{F} = T\mathcal{F}_ x \]

to be the tangent space of $\mathcal{F}$ at $x$. If $\varphi : \mathcal{F} \to \mathcal{G}$ is a morphism of categories cofibered in groupoids over $\mathcal{C}_\Lambda $ and $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$, then there is an induced morphism $\varphi _ x: \mathcal{F}_ x \to \mathcal{G}_{\varphi (x)}$. We define the differential $d_ x \varphi : T_ x \mathcal{F} \to T_{\varphi (x)} \mathcal{G}$ of $\varphi $ at $x$ to be the map $d \varphi _ x: T \mathcal{F}_ x \to T \mathcal{G}_{\varphi (x)}$. If both $\mathcal{F}$ and $\mathcal{G}$ satisfy (S2) then all of these tangent spaces have a natural $k$-vector space structure and all the differentials $d_ x \varphi : T_ x \mathcal{F} \to T_{\varphi (x)} \mathcal{G}$ are $k$-linear (use Lemmas 89.10.6 and 89.12.4).

The following observations are uninteresting in the classical case or when $k/k'$ is a separable field extension, because then $\text{Der}_\Lambda (k, k)$ and $\text{Der}_\Lambda (k, V)$ are zero. There is a canonical identification

\[ \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}_\Lambda }(k, k[\epsilon ]) = \text{Der}_\Lambda (k, k). \]

Namely, for $D \in \text{Der}_\Lambda (k, k)$ let $f_ D : k \to k[\epsilon ]$ be the map $a \mapsto a + D(a)\epsilon $. More generally, given a finite dimensional vector space $V$ over $k$ we have

\[ \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}_\Lambda }(k, k[V]) = \text{Der}_\Lambda (k, V) \]

and we will use the same notation $f_ D$ for the map associated to the derivation $D$. We also have

\[ \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}_\Lambda }(k[W], k[V]) = \mathop{\mathrm{Hom}}\nolimits _ k(V, W) \oplus \text{Der}_\Lambda (k, V) \]

where $(\varphi , D)$ corresponds to the map $f_{\varphi , D} : a + w \mapsto a + \varphi (w) + D(a)$. We will sometimes write $f_{1, D} : a + v \to a + v + D(a)$ for the automorphism of $k[V]$ determined by the derivation $D : k \to V$. Note that $f_{1, D} \circ f_{1, D'} = f_{1, D + D'}$.

Let $\mathcal{F}$ be a predeformation category over $\mathcal{C}_\Lambda $. Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. By the above there is a canonical map

\[ \gamma _ V : \text{Der}_\Lambda (k, V) \longrightarrow \overline{\mathcal{F}}(k[V]) \]

defined by $D \mapsto f_{D, *}(x_0)$. Moreover, there is an action

\[ a_ V : \text{Der}_\Lambda (k, V) \times \overline{\mathcal{F}}(k[V]) \longrightarrow \overline{\mathcal{F}}(k[V]) \]

defined by $(D, x) \mapsto f_{1, D, *}(x)$. These two maps are compatible, i.e., $f_{1, D, *}f_{D', *}x_0 = f_{D + D', *}x_0$ as follows from a computation of the compositions of these maps. Note that the maps $\gamma _ V$ and $a_ V$ are independent of the choice of $x_0$ as there is a unique $x_0$ up to isomorphism.

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. $\square$

A special case of the constructions above are the map

89.12.6.1
\begin{equation} \label{formal-defos-equation-map} \gamma : \text{Der}_\Lambda (k, k) \longrightarrow T\mathcal{F} \end{equation}

and the action

89.12.6.2
\begin{equation} \label{formal-defos-equation-action} a : \text{Der}_\Lambda (k, k) \times T\mathcal{F} \longrightarrow T\mathcal{F} \end{equation}

defined for any predeformation category $\mathcal{F}$. Note that if $\varphi : \mathcal{F} \to \mathcal{G}$ is a morphism of predeformation categories, then we get commutative diagrams

\[ \vcenter { \xymatrix{ \text{Der}_\Lambda (k, k) \ar[r]_-\gamma \ar[rd]_\gamma & T\mathcal{F} \ar[d]_{d\varphi } \\ & T\mathcal{G} } } \quad \text{and}\quad \vcenter { \xymatrix{ \text{Der}_\Lambda (k, k) \times T\mathcal{F} \ar[r]_-a \ar[d]_{1 \times d\varphi } & T\mathcal{F} \ar[d]^{d\varphi } \\ \text{Der}_\Lambda (k, k) \times T\mathcal{G} \ar[r]^-a & T\mathcal{G} } } \]
[1] For example if $\mathcal{F}$ satisfies (S2), see Lemma 89.10.5.

Comments (2)

Comment #1422 by Evan Warner on

small "typo" in the first sentence after remark 69.11.5: although it is true that in the classical case Der(V,k) is trivial presumably we care here more about the fact that Der(k,V) is trivial.

Comment #1435 by on

Fixed this and previous typo you pointed out. Thanks. See here.


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