17.31 The naive cotangent complex
This section is the analogue of Algebra, Section 10.134 for morphisms of ringed spaces. We urge the reader to read that section first.
Let $X$ be a topological space. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings. In this section, for any sheaf of sets $\mathcal{E}$ on $X$ we denote $\mathcal{A}[\mathcal{E}]$ the sheafification of the presheaf $U \mapsto \mathcal{A}(U)[\mathcal{E}(U)]$. Here $\mathcal{A}(U)[\mathcal{E}(U)]$ denotes the polynomial algebra over $\mathcal{A}(U)$ whose variables correspond to the elements of $\mathcal{E}(U)$. We denote $[e] \in \mathcal{A}(U)[\mathcal{E}(U)]$ the variable corresponding to $e \in \mathcal{E}(U)$. There is a canonical surjection of $\mathcal{A}$-algebras
17.31.0.1
\begin{equation} \label{modules-equation-canonical-presentation} \mathcal{A}[\mathcal{B}] \longrightarrow \mathcal{B},\quad [b] \longmapsto b \end{equation}
whose kernel we denote $\mathcal{I} \subset \mathcal{A}[\mathcal{B}]$. It is a simple observation that $\mathcal{I}$ is generated by the local sections $[b][b'] - [bb']$ and $[a] - a$. According to Lemma 17.28.9 there is a canonical map
17.31.0.2
\begin{equation} \label{modules-equation-naive-cotangent-complex} \mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega _{\mathcal{A}[\mathcal{B}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{B}]} \mathcal{B} \end{equation}
whose cokernel is canonically isomorphic to $\Omega _{\mathcal{B}/\mathcal{A}}$.
Definition 17.31.1. Let $X$ be a topological space. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings. The naive cotangent complex $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}$ is the chain complex (17.31.0.2)
\[ \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} = \left(\mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega _{\mathcal{A}[\mathcal{B}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{B}]} \mathcal{B}\right) \]
with $\mathcal{I}/\mathcal{I}^2$ placed in degree $-1$ and $\Omega _{\mathcal{A}[\mathcal{B}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{B}]} \mathcal{B}$ placed in degree $0$.
This construction satisfies a functoriality similar to that discussed in Lemma 17.28.8 for modules of differentials. Namely, given a commutative diagram
17.31.1.1
\begin{equation} \label{modules-equation-commutative-square-sheaves} \vcenter { \xymatrix{ \mathcal{B} \ar[r] & \mathcal{B}' \\ \mathcal{A} \ar[u] \ar[r] & \mathcal{A}' \ar[u] } } \end{equation}
of sheaves of rings on $X$ there is a canonical $\mathcal{B}$-linear map of complexes
\[ \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \longrightarrow \mathop{N\! L}\nolimits _{\mathcal{B}'/\mathcal{A}'} \]
Namely, the maps in the commutative diagram give rise to a canonical map $\mathcal{A}[\mathcal{B}] \to \mathcal{A}'[\mathcal{B}']$ which maps $\mathcal{I}$ into $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{A}'[\mathcal{B}'] \to \mathcal{B}')$. Thus a map $\mathcal{I}/\mathcal{I}^2 \to \mathcal{I}'/(\mathcal{I}')^2$ and a map between modules of differentials, which together give the desired map between the naive cotangent complexes. The map is compatible with compositions in the following sense: given a commutative diagram
\[ \xymatrix{ \mathcal{B} \ar[r] & \mathcal{B}' \ar[r] & \mathcal{B}'' \\ \mathcal{A} \ar[u] \ar[r] & \mathcal{A}' \ar[u] \ar[r] & \mathcal{A}'' \ar[u] } \]
of sheaves of rings then the composition
\[ \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \longrightarrow \mathop{N\! L}\nolimits _{\mathcal{B}'/\mathcal{A}'} \longrightarrow \mathop{N\! L}\nolimits _{\mathcal{B}''/\mathcal{A}''} \]
is the map for the outer rectangle.
We can choose a different presentation of $\mathcal{B}$ as a quotient of a polynomial algebra over $\mathcal{A}$ and still obtain the same object of $D(\mathcal{B})$. To explain this, suppose that $\mathcal{E}$ is a sheaves of sets on $X$ and $\alpha : \mathcal{E} \to \mathcal{B}$ a map of sheaves of sets. Then we obtain an $\mathcal{A}$-algebra homomorphism $\mathcal{A}[\mathcal{E}] \to \mathcal{B}$. If this map is surjective, i.e., if $\alpha (\mathcal{E})$ generates $\mathcal{B}$ as an $\mathcal{A}$-algebra, then we set
\[ \mathop{N\! L}\nolimits (\alpha ) = \left( \mathcal{J}/\mathcal{J}^2 \longrightarrow \Omega _{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}]} \mathcal{B}\right) \]
where $\mathcal{J} \subset \mathcal{A}[\mathcal{E}]$ is the kernel of the surjection $\mathcal{A}[\mathcal{E}] \to \mathcal{B}$. Here is the result.
Lemma 17.31.2. In the situation above there is a canonical isomorphism $\mathop{N\! L}\nolimits (\alpha ) = \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}$ in $D(\mathcal{B})$.
Proof.
Observe that $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} = \mathop{N\! L}\nolimits (\text{id}_\mathcal {B})$. Thus it suffices to show that given two maps $\alpha _ i : \mathcal{E}_ i \to \mathcal{B}$ as above, there is a canonical quasi-isomorphism $\mathop{N\! L}\nolimits (\alpha _1) = \mathop{N\! L}\nolimits (\alpha _2)$ in $D(\mathcal{B})$. To see this set $\mathcal{E} = \mathcal{E}_1 \amalg \mathcal{E}_2$ and $\alpha = \alpha _1 \amalg \alpha _2 : \mathcal{E} \to \mathcal{B}$. Set $\mathcal{J}_ i = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}_ i] \to \mathcal{B})$ and $\mathcal{J} = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}] \to \mathcal{B})$. We obtain maps $\mathcal{A}[\mathcal{E}_ i] \to \mathcal{A}[\mathcal{E}]$ which send $\mathcal{J}_ i$ into $\mathcal{J}$. Thus we obtain canonical maps of complexes
\[ \mathop{N\! L}\nolimits (\alpha _ i) \longrightarrow \mathop{N\! L}\nolimits (\alpha ) \]
and it suffices to show these maps are quasi-isomorphism. To see this it suffices to check on stalks (Lemma 17.3.1). If $x \in X$ then the stalk of $\mathop{N\! L}\nolimits (\alpha )$ is the complex $\mathop{N\! L}\nolimits (\alpha _ x)$ of Algebra, Section 10.134 associated to the presentation $\mathcal{A}_ x[\mathcal{E}_ x] \to \mathcal{B}_ x$ coming from the map $\alpha _ x : \mathcal{E}_ x \to \mathcal{B}_ x$. (Some details omitted; use Lemma 17.28.7 to see compatibility of forming differentials and taking stalks.) We conclude the result holds by Algebra, Lemma 10.134.2.
$\square$
Lemma 17.31.3. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings on $Y$. Then $f^{-1}\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} = \mathop{N\! L}\nolimits _{f^{-1}\mathcal{B}/f^{-1}\mathcal{A}}$.
Proof.
Omitted. Hint: Use Lemma 17.28.6.
$\square$
Lemma 17.31.4. Let $X$ be a topological space. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings on $X$. Let $x \in X$. Then we have $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}, x} = \mathop{N\! L}\nolimits _{\mathcal{B}_ x/\mathcal{A}_ x}$.
Proof.
This is a special case of Lemma 17.31.3 for the inclusion map $\{ x\} \to X$.
$\square$
Lemma 17.31.5. Let $X$ be a topological space. Let $\mathcal{A} \to \mathcal{B} \to \mathcal{C}$ be maps of sheaves of rings. Let $C$ be the cone (Derived Categories, Definition 13.9.1) of the map of complexes $\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}$. There is a canonical map
\[ c : \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C} \longrightarrow C[-1] \]
of complexes of $\mathcal{C}$-modules which produces a canonical six term exact sequence
\[ \xymatrix{ H^0(\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C}) \ar[r] & H^0(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}}) \ar[r] & H^0(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}) \ar[r] & 0 \\ H^{-1}(\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C}) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}}) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}) \ar[llu] } \]
of cohomology sheaves.
Proof.
To give the map $c$ we have to give a map $c_1 : \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}}$ and an explicit homotopy between the composition
\[ \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}} \]
and the zero map, see Derived Categories, Lemma 13.9.3. For $c_1$ we use the functoriality described above for the obvious diagram. For the homotopy we use the map
\[ \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}^0 \otimes _\mathcal {B} \mathcal{C} \longrightarrow \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}^{-1},\quad \text{d}[b] \otimes 1 \longmapsto [\varphi (b)] - b[1] \]
where $\varphi : \mathcal{B} \to \mathcal{C}$ is the given map. Please compare with Algebra, Remark 10.134.5. To see the consequence for cohomology sheaves, it suffices to show that $H^0(c)$ is an isomorphism and $H^{-1}(c)$ surjective. To see this we can look at stalks, see Lemma 17.31.4, and then we can use the corresponding result in commutative algebra, see Algebra, Lemma 10.134.4. Some details omitted.
$\square$
The cotangent complex of a morphism of ringed spaces is defined in terms of the cotangent complex we defined above.
Definition 17.31.6. The naive cotangent complex $\mathop{N\! L}\nolimits _ f = \mathop{N\! L}\nolimits _{X/Y}$ of a morphism of ringed spaces $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ is $\mathop{N\! L}\nolimits _{\mathcal{O}_ X/f^{-1}\mathcal{O}_ Y}$.
Given a commutative diagram
\[ \xymatrix{ X' \ar[r]_ g \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ h & Y } \]
of ringed spaces, there is a canonical map $c : g^*\mathop{N\! L}\nolimits _{X/Y} \to \mathop{N\! L}\nolimits _{X'/Y'}$. Namely, it is the map
\[ g^*\mathop{N\! L}\nolimits _{X/Y} = \mathcal{O}_{X'} \otimes _{g^{-1}\mathcal{O}_ X} \mathop{N\! L}\nolimits _{g^{-1}\mathcal{O}_ X/g^{-1}f^{-1}\mathcal{O}_ Y} \longrightarrow \mathop{N\! L}\nolimits _{\mathcal{O}_{X'}/(f')^{-1}\mathcal{O}_{Y'}} = \mathop{N\! L}\nolimits _{X'/Y'} \]
where the arrow comes from the commutative diagram of sheaves of rings
\[ \xymatrix{ g^{-1}\mathcal{O}_ X \ar[r]_{g^\sharp } & \mathcal{O}_{X'} \\ g^{-1}f^{-1}\mathcal{O}_ Y \ar[r]^{g^{-1}h^\sharp } \ar[u]^{g^{-1}f^\sharp } & (f')^{-1}\mathcal{O}_{Y'} \ar[u]_{(f')^\sharp } } \]
as in (17.31.1.1) above. Given a second such diagram
\[ \xymatrix{ X'' \ar[r]_{g'} \ar[d] & X' \ar[d] \\ Y'' \ar[r] & Y' } \]
the composition of $(g')^*c$ and the map $c' : (g')^*\mathop{N\! L}\nolimits _{X'/Y'} \to \mathop{N\! L}\nolimits _{X''/Y''}$ is the map $(g \circ g')^*\mathop{N\! L}\nolimits _{X''/Y''} \to \mathop{N\! L}\nolimits _{X/Y}$.
Lemma 17.31.7. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of ringed spaces. Let $C$ be the cone of the map $\mathop{N\! L}\nolimits _{X/Z} \to \mathop{N\! L}\nolimits _{X/Y}$ of complexes of $\mathcal{O}_ X$-modules. There is a canonical map
\[ f^*\mathop{N\! L}\nolimits _{Y/Z} \to C[-1] \]
which produces a canonical six term exact sequence
\[ \xymatrix{ H^0(f^*\mathop{N\! L}\nolimits _{Y/Z}) \ar[r] & H^0(\mathop{N\! L}\nolimits _{X/Z}) \ar[r] & H^0(\mathop{N\! L}\nolimits _{X/Y}) \ar[r] & 0 \\ H^{-1}(f^*\mathop{N\! L}\nolimits _{Y/Z}) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{X/Z}) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) \ar[llu] } \]
of cohomology sheaves.
Proof.
Consider the maps of sheaves rings
\[ (g \circ f)^{-1}\mathcal{O}_ Z \to f^{-1}\mathcal{O}_ Y \to \mathcal{O}_ X \]
and apply Lemma 17.31.5.
$\square$
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