The Stacks project

91.26 The cotangent complex of a morphism of algebraic spaces

We define the cotangent complex of a morphism of algebraic spaces using the associated morphism between the small étale sites.

Definition 91.26.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The cotangent complex $L_{X/Y}$ of $X$ over $Y$ is the cotangent complex of the morphism of ringed topoi $f_{small}$ between the small étale sites of $X$ and $Y$ (see Properties of Spaces, Lemma 65.21.3 and Definition 91.22.1).

In particular, the results of Section 91.22 apply to cotangent complexes of morphisms of algebraic spaces. The next lemmas show this definition is compatible with the definition for ring maps and for schemes and that $L_{X/Y}$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$.

Lemma 91.26.2. Let $S$ be a scheme. Consider a commutative diagram

\[ \xymatrix{ U \ar[d]_ p \ar[r]_ g & V \ar[d]^ q \\ X \ar[r]^ f & Y } \]

of algebraic spaces over $S$ with $p$ and $q$ étale. Then there is a canonical identification $L_{X/Y}|_{U_{\acute{e}tale}} = L_{U/V}$ in $D(\mathcal{O}_ U)$.

Proof. Formation of the cotangent complex commutes with pullback (Lemma 91.18.3) and we have $p_{small}^{-1}\mathcal{O}_ X = \mathcal{O}_ U$ and $g_{small}^{-1}\mathcal{O}_{V_{\acute{e}tale}} = p_{small}^{-1}f_{small}^{-1}\mathcal{O}_{Y_{\acute{e}tale}}$ because $q_{small}^{-1}\mathcal{O}_{Y_{\acute{e}tale}} = \mathcal{O}_{V_{\acute{e}tale}}$ (Properties of Spaces, Lemma 65.26.1). Tracing through the definitions we conclude that $L_{X/Y}|_{U_{\acute{e}tale}} = L_{U/V}$. $\square$

Lemma 91.26.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $X$ and $Y$ representable by schemes $X_0$ and $Y_0$. Then there is a canonical identification $L_{X/Y} = \epsilon ^*L_{X_0/Y_0}$ in $D(\mathcal{O}_ X)$ where $\epsilon $ is as in Derived Categories of Spaces, Section 74.4 and $L_{X_0/Y_0}$ is as in Definition 91.24.1.

Proof. Let $f_0 : X_0 \to Y_0$ be the morphism of schemes corresponding to $f$. There is a canonical map $\epsilon ^{-1}f_0^{-1}\mathcal{O}_{Y_0} \to f_{small}^{-1}\mathcal{O}_ Y$ compatible with $\epsilon ^\sharp : \epsilon ^{-1}\mathcal{O}_{X_0} \to \mathcal{O}_ X$ because there is a commutative diagram

\[ \xymatrix{ X_{0, Zar} \ar[d]_{f_0} & X_{\acute{e}tale}\ar[l]^\epsilon \ar[d]^ f \\ Y_{0, Zar} & Y_{\acute{e}tale}\ar[l]_\epsilon } \]

see Derived Categories of Spaces, Remark 74.6.3. Thus we obtain a canonical map

\[ \epsilon ^{-1}L_{X_0/Y_0} = \epsilon ^{-1}L_{\mathcal{O}_{X_0}/f_0^{-1}\mathcal{O}_{Y_0}} = L_{\epsilon ^{-1}\mathcal{O}_{X_0}/\epsilon ^{-1}f_0^{-1}\mathcal{O}_{Y_0}} \longrightarrow L_{\mathcal{O}_ X/f^{-1}_{small}\mathcal{O}_ Y} = L_{X/Y} \]

by the functoriality discussed in Section 91.18 and Lemma 91.18.3. To see that the induced map $\epsilon ^*L_{X_0/Y_0} \to L_{X/Y}$ is an isomorphism we may check on stalks at geometric points (Properties of Spaces, Theorem 65.19.12). We will use Lemma 91.18.9 to compute the stalks. Let $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X_0$ be a geometric point lying over $x \in X_0$, with $\overline{y} = f \circ \overline{x}$ lying over $y \in Y_0$. Then

\[ L_{X/Y, \overline{x}} = L_{\mathcal{O}_{X, \overline{x}}/\mathcal{O}_{Y, \overline{y}}} \]

and

\[ (\epsilon ^*L_{X_0/Y_0})_{\overline{x}} = L_{X_0/Y_0, x} \otimes _{\mathcal{O}_{X_0, x}} \mathcal{O}_{X, \overline{x}} = L_{\mathcal{O}_{X_0, x}/\mathcal{O}_{Y_0, y}} \otimes _{\mathcal{O}_{X_0, x}} \mathcal{O}_{X, \overline{x}} \]

Some details omitted (hint: use that the stalk of a pullback is the stalk at the image point, see Sites, Lemma 7.34.2, as well as the corresponding result for modules, see Modules on Sites, Lemma 18.36.4). Observe that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X_0, x}$ and similarly for $\mathcal{O}_{Y, \overline{y}}$ (Properties of Spaces, Lemma 65.22.1). Thus the result follows from Lemma 91.8.7. $\square$

Lemma 91.26.4. Let $\Lambda $ be a ring. Let $X$ be an algebraic space over $\Lambda $. Then

\[ L_{X/\mathop{\mathrm{Spec}}(\Lambda )} = L_{\mathcal{O}_ X/\underline{\Lambda }} \]

where $\underline{\Lambda }$ is the constant sheaf with value $\Lambda $ on $X_{\acute{e}tale}$.

Proof. Let $p : X \to \mathop{\mathrm{Spec}}(\Lambda )$ be the structure morphism. Let $q : \mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}\to (*, \Lambda )$ be the obvious morphism. By the distinguished triangle of Lemma 91.22.3 it suffices to show that $L_ q = 0$. To see this it suffices to show (Properties of Spaces, Theorem 65.19.12) for a geometric point $\overline{t} : \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(\Lambda )$ that

\[ (L_ q)_{\overline{t}} = L_{\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}, \overline{t}}/\Lambda } \]

(Lemma 91.18.9) is zero. Since $\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}, \overline{t}}$ is a strict henselization of a local ring of $\Lambda $ (Properties of Spaces, Lemma 65.22.1) this follows from Lemma 91.8.4. $\square$


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