TITLE: Intuition about the cotangent complex?
QUESTION [63 upvotes]: 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?

REPLY [36 votes]: 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 $Ext^0(L_{X/S}, I)$.