# Cotangent complex

In mathematics the cotangent complex is roughly a universal linearization of a morphism of geometric or algebraic objects. Cotangent complexes were originally defined in special cases by a number of authors. Luc Illusie, Daniel Quillen, and M. André independently came up with a definition that works in all cases.

## Motivation

Suppose that X and Y are algebraic varieties and that f : XY is a morphism between them. The cotangent complex of f is a more universal version of the relative Kähler differentials ΩX/Y. The most basic motivation for such an object is the exact sequence of Kähler differentials associated to two morphisms. If Z is another variety, and if g : YZ is another morphism, then there is an exact sequence

$f^*\Omega_{Y/Z} \to \Omega_{X/Z} \to \Omega_{X/Y} \to 0.$

In some sense, therefore, relative Kähler differentials are a right exact functor. (Literally this is not true, however, because the category of algebraic varieties is not an abelian category, and therefore right-exactness is not defined.) In fact, prior to the definition of the cotangent complex, there were several definitions of functors that might extend the sequence further to the left, such as the Lichtenbaum–Schlessinger functors Ti and imperfection modules. Most of these were motivated by deformation theory.

This sequence is exact on the left if the morphism f is smooth. If Ω admitted a first derived functor, then exactness on the left would imply that the connecting homomorphism vanished, and this would certainly be true if the first derived functor of f, whatever it was, vanished. Therefore a reasonable speculation is that the first derived functor of a smooth morphism vanishes. Furthermore, when any of the functors which extended the sequence of Kähler differentials were applied to a smooth morphism, they too vanished, which suggested that the cotangent complex of a smooth morphism might be equivalent to the Kähler differentials.

Another natural exact sequence related to Kähler differentials is the conormal exact sequence. If f is a closed immersion with ideal sheaf I, then there is an exact sequence

$I/I^2 \to f^*\Omega_{Y/Z} \to \Omega_{X/Z} \to 0.$

This is an extension of the exact sequence above: There is a new term on the left, the conormal sheaf of f, and the relative differentials ΩX/Y have vanished because a closed immersion is formally unramified. If f is the inclusion of a smooth subvariety, then this sequence is a short exact sequence.[1] This suggests that the cotangent complex of the inclusion of a smooth variety is equivalent to the conormal sheaf shifted by one term.

## Early work on cotangent complexes

The cotangent complex dates back at least to SGA 6 VIII 2, where Pierre Berthelot gave a definition when f is a smoothable morphism, meaning there is a scheme V and morphisms i : XV and h : VY such that f = hi, i is a closed immersion, and h is a smooth morphism. (For example, all projective morphisms are smoothable, since V can be taken to be a projective bundle over Y.) In this case, he defines the cotangent complex of f as an object in the derived category of coherent sheaves X as follows:

• $L^{X/Y}_0 = i^*\Omega_{V/Y},$
• If J is the ideal of X in V, then $L^{X/Y}_1 = J/J^2 = i^*J$,
• $L^{X/Y}_i = 0$ for all other i,
• The differential $L^{X/Y}_1 \to L^{X/Y}_0$ is the pullback along i of the inclusion of J in the structure sheaf $\mathcal{O}_V$ of V followed by the universal derivation $d : \mathcal{O}_V \to \Omega_{V/Y}$.
• All other differentials are zero.

Berthelot proves that this definition is independent of the choice of V[2] and that for a smoothable complete intersection morphism, this complex is perfect.[3] Furthermore, he proves that if g : YZ is another smoothable complete intersection morphism and if an additional technical condition is satisfied, then there is an exact triangle

$\mathbf{L}f^*L^{Y/Z}_\bullet \to L^{X/Z}_\bullet \to L^{X/Y}_\bullet \to \mathbf{L}f^*L^{Y/Z}_\bullet[1].$

## The definition of the cotangent complex

The correct definition of the cotangent complex begins in the homotopical setting. Quillen and André worked with the simplicial commutative rings, while Illusie worked with simplicial ringed topoi. For simplicity, we will consider only the case of simplicial commutative rings. Suppose that A and B are simplicial rings and that B is an A-algebra. Choose a resolution r : PB of B by simplicial free A-algebras. Applying the Kähler differential functor to P produces a simplicial B-module. The total complex of this simplicial object is the cotangent complex LB/A. The morphism r induces a morphism from the cotangent complex to ΩB/A called the augmentation map. In the homotopy category of simplicial A-algebras (or of simplicial ringed topoi), this construction amounts to taking the left derived functor of the Kähler differential functor.

Given a commutative square as follows:

there is a morphism of cotangent complexes LB/AB DLD/C which respects the augmentation maps. This map is constructed by choosing a free simplicial C-algebra resolution of D, say s : QD. Because P is a free object, the composite hr can be lifted to a morphism PQ. Applying functoriality of Kähler differentials to this morphism gives the required morphism of cotangent complexes. In particular, given homomorphisms ABC, this produces the sequence

$L^{B/A} \otimes_B C \to L^{C/A} \to L^{C/B}.$

There is a connecting homomorphism $L^{C/B} \to (L^{B/A} \otimes_B C)[1]$ which turns this sequence into an exact triangle.

The cotangent complex can also be defined in any combinatorial model category M. Suppose that $f\colon A\rightarrow B$ is a morphism in M. The cotangent complex $L^f$ (or $L^{B/A}$) is an object in the category of spectra in $M_{B//B}$. A pair of composable morphisms $A\xrightarrow{f} B\xrightarrow{g} C$ induces an exact triangle in the homotopy category, $L^{B/A}\otimes_BC\rightarrow L^{C/A}\rightarrow L^{C/B}\rightarrow (L^{B/A}\otimes_BC)[1]$.

## Properties of the cotangent complex

### Flat base change

Suppose that B and C are A-algebras such that TorAq(B, C) = 0 for all q > 0. Then there are quasi-isomorphisms[4]

$L^{B \otimes_A C/C} \cong B \otimes_A L^{C/A},$
$L^{B \otimes_A C/A} \cong (L^{B/A} \otimes_A C) \oplus (B \otimes_A L^{C/A}).$

If C is a flat A-algebra, then the condition that TorAq(B, C) vanishes for q > 0 is automatic. The first formula then proves that the construction of the cotangent complex is local on the base in the flat topology.

### Vanishing properties

Let f : AB. Then:[5][6]

## Examples

• Let X be smooth over S. Then the cotangent complex is ΩX/S. In Berthelot's framework, this is clear by taking V = X. In general, étale locally on S, X is a finite dimensional affine space and the morphism from X to S is projection, so we may reduce to the situation where S = Spec A and X = Spec A[x1, ..., xn]. We can take the resolution of A[x1, ..., xn] to be the identity map, and then it is clear that the cotangent complex is the same as the Kähler differentials.
• Let X and Y be smooth over S, and assume that i : XY is a closed embedding. Using the exact triangle corresponding to the morphisms XYS, we may determine the cotangent complex LX/Y. To do this, note that by the previous example, the cotangent complexes LX/S and LY/S consist of the Kähler differentials ΩX/S and ΩY/S in the zeroth degree, respectively, and are zero in all other degrees. The exact triangle implies that LX/Y is nonzero only in the first degree, and in that degree, it is the kernel of the map i*ΩY/S → ΩX/S. This kernel is the conormal bundle, and the exact sequence is the conormal exact sequence, so in the first degree, LX/Y is the conormal bundle of X in Y.