Does anyone have an answer to the question "What does the cotangent complex measure?"

Algebraic intuitions (like "homology measures how far a sequence is from being exact") are as welcome as geometric ones (like "homology detects holes"), as are intuitions which do not exactly answer the above question.

In particular: Do the degrees have a meaning? E.g. if an ideal $I$ in a ring $A$ is generated by a regular sequence, the cotangent complex of the quotient map $A\twoheadrightarrow A/I$ is $(I/I^2)[-1]$. Why does it live in degree 1?


One thing the cotangent complex measures is what kind of deformations a scheme has. The precise statements are in Remark 5.30 and Theorem 5.31 in Illusie's article in "FGA explained". Here's the short simplified version in the absolute case:

If you have a scheme $X$ over $k$, a first order deformation is a space $\mathcal{X}$ over $k[\epsilon]/(\epsilon)^2$ whose fiber over the only point of $k[\epsilon]/(\epsilon)^2$ is $X$ again. You can imagine $k[\epsilon]/(\epsilon)^2$ as a point with an infinitesimal arrow attached to it and $\mathcal{X}$ as an infinitesimal thickening of $X$. The cotangent complex gives you precise information on how many such thickenings there are: The set of such thickenings is isomorphic $\mathop{Ext}^1(L_X, \epsilon^2)$.

Now let's assume that we have chosen one such infinitesimal thickening $\mathcal{X}$ over $k[\epsilon]/(\epsilon)^2$. It is not always true that you can go on and make this thickening into a thickening to the next order. Whether or not you can do this is measured precisely by the cotangent complex: There is a map that takes as input your chosen thickening $\mathcal{X}$ and spits out an element in $\mathop{Ext}^2(L_X, \epsilon^3)$. If the element in the Ext group is zero you can go on to the next level. If it is not zero, it's game over and your stuck.

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    $\begingroup$ Perhaps it would be useful to note that "L_X" denotes the cotangent complex. $\endgroup$ Oct 27 '09 at 11:19
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    $\begingroup$ A question: wha does it mean Ext^1(L_X,epsilon^2)? Isn't epsilon^2=0? How can you see "epsilon^2" as a sheaf of O_X-modules over X? $\endgroup$
    – Qfwfq
    Mar 23 '10 at 12:54
  • $\begingroup$ Shouldn't that be the trivial sheaf instead? $\endgroup$
    – Will Sawin
    Aug 2 '13 at 14:04
  • $\begingroup$ It should be $\text{Ext}^1(L_X,\epsilon)$, this is theorem 1.7 of Illusie's paper "Complexe Cotangent". More precisely, $\epsilon$ is the O module pullback by $X\to \text{Spec}k$ of the $k$ module $\epsilon k[\epsilon]/\epsilon^2$. $\endgroup$
    – Meow
    Sep 28 '20 at 10:12

First a correction: the cotangent complex of a local complete intersection embedding is concentrated in degree -1, not in degree 1.

In general, the cotangent complex of an algebraic space can be supported in arbitrary non-positive degrees. The cotangent complex of an Artin stack can be nonzero in degree 1. The degrees in which the cotangent complex is concentrated imply various things about a morphism of schemes:

it is perfect in degree 0 if and only if the map is smooth;

it is perfect in $[-1,0]$ if and only if the map is lci;

$H^1 = 0$ if and only if it is a DM stack;

$H^0 = H^1 = 0$ if and only if it is an etale local immersion.

Other people have already said some things about the relationship to deformation theory. The cotangent complex actually has two immediate relationships to deformation theory: one to the deformations of morphisms and one to the deformation of spaces.

In what's written below, $L_X$ is the absolute cotangent complex and $L_{X/S}$ is the relative cotangent complex.

If $f : S \to X$ is a map of schemes and $S'$ is a square-zero extension of $S$ with ideal $J$, there is an obstruction to extending $f$ to $S'$ in the group $Ext^1(f^\ast L_X, J)$. If this obstruction vanishes, such extensions have a canonical structure of a torsor under $Ext^0(f^\ast L_X, J)$.

If $p : X\to S$ is a morphism, $S'$ is a square-zero extension with ideal $J$, and $p^\ast J \rightarrow I$ is a homomorphism of quasi-coherent sheaves on $X$, then the problem of finding a square-zero extension $X'$ with ideal $I$ and a map $X' \to S'$ extending $X \to S$ compatible with the given map on ideals is obstructed by a class in $Ext^2(L_{X/S}, I)$. If this class is zero, isomorphism classes of solutions form a torsor under $Ext^1(L_{X/S}, I)$ and isomorphisms between any two solutions form a torsor under $Ext^0(L_{X/S}, I)$.

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    $\begingroup$ Great post, thank you! As for the degree convention: I actually learnt it the other way, which is more convenient to make the cotangent complex Dold-Kan-equivalent to something simplicial. But the way you stated it it's certainly more standard. $\endgroup$ Nov 10 '09 at 14:06

One reason why the cotangent complex should live in degree $1$ in this case is that you should always think of it as a relative cotangent complex.

A general ring map $R \to S$ gives rise to a map of absolute cotangent spaces $S \otimes_R \Omega_R \to \Omega_S$ on $S$, and the cotangent complex should contain information about some "derived" cokernel of this map, like a mapping cylinder chain complex. If the map is not surjective on cotangent spaces (such as $k \to k[x]$), then you have a cokernel living in degree $0$. If the map is not injective on cotangent spaces (such as $k[x] \to k$), then the mapping cylinder should detect the kernel of the map on cotangent spaces in degree $1$.

The degrees have a meaning in terms of deformation theory, but (say in characteristic zero) the terms not in degree $1$ may measure deformations to a differential graded algebra with terms not in degree $0$.

  • $\begingroup$ Great, this totally settles my degree question! Thanks! $\endgroup$ Oct 26 '09 at 15:25
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    $\begingroup$ I only answered the degree question, not the main question. $\endgroup$ Oct 28 '09 at 0:45

I'm really glad that this question was posted, because I am currently in the process of trying to learn about the cotangent complex myself.

First of all, the wikipedia entry on the cotangent complex is pretty good, so you should take a look at it if you haven't already.

I'll try to make some naive and vague comments about the relation to deformation theory, as Charles has hinted at. The two examples on the bottom of the wikipedia page are already instructive. The first example of a cotangent complex is that of a smooth $S$-scheme $X$. Then the cotangent complex in this case is just the cotangent bundle $\Omega_{X/S}$. By Kodaira-Spencer theory, we know that the tangent bundle $T_{X/S}$ has a lot to do with the deformation theory of $X$. The second example is the cotangent complex of a closed embedding $X\to Y$ of smooth $S$-schemes; it is the conormal bundle of $X$ in $Y$. Similarly to how the tangent bundle $T_{X/S}$ tells us about deformations of $X$, the normal bundle $N_{X/Y}$ tells us about deformations of $X$ inside of $Y$. If you look at e.g. chapter 3 of Deformations of Algebraic Schemes, you will find precise versions of the vague statements I've made above. My understanding is that the cotangent complex is a rather general gadget that contains information about deformations, in particular, it works for situations where your schemes (or stacks) are not necessarily smooth. One such situation of interest is in Gromov-Witten theory, where we consider moduli of stable maps, which can be very singular, even when considered as Deligne-Mumford stacks.


I will admit flat out that I don't know what I'm saying, really, so take it with a grain of salt, but I've been told (numerous times) that the cotangent complex measures the deformation theory of your (variety, scheme, stack, whatever).


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