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?
Best Answer
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.