4

These two links: What is the cotangent complex good for? and Intuition about the cotangent complex? are quite helpful in giving intution for the cotangent complex in terms of deformations but I don't believe any of the answers touch on the following use case.

Fix a ring $k$. For simplicitly, you're welcome to assume $k$ is nonanimated/discrete, if necessary.

Suppose I have an animated $k$-algebra $R$. If I know information about the homotopy of the relative cotangent complex $L_{R/k}$, what, if anything, can I say about the homotopy of $R$ itself?

My understanding of the cotangent complex in that it describes deformation theory, along with the idea that animation describes infinitesimal nilpotents (a la the introduction to Lurie's thesis), suggests to me that there must be some information about $R$ (or $\pi_{i}R$, or $k\to R$) that is carried by $\pi_{i}L_{R/k}$, but I haven't found this written down anywhere. Most of the time it seems notes plug ordinary rings into the cotangent complex, not animated ones. (I don't blame them :))

For instance, here is a specific example to calibrate: suppose I know that $L_{R/k}$ is discrete; i.e., $\pi_{i}L_{R/k}=0$ for $i>0$. What can I say about $R$?

Eric
  • 301
  • If you complete at a point, there are spectral sequences going both ways. See for instance section 5.2 of http://arxiv.org/abs/0705.0344 for one direction, where $H^i(F)^{\vee}$ is just $\pi_i(\mathbb{L}^R\otimes^{\mathbf{L}}_Rk)$. – Jon Pridham Mar 08 '22 at 20:56
  • I don't know how to understand $L_{R/k}$ being concentrated in degree 0. However, if you assume that the $R$-module $L_{R/k}$ is flat, the $k$-algebra $R$ is flat, and that $R$ is locally almost of finite presentation, then the $k$-algebra $R$ is smooth (ref: Lurie's DAG). – Z. M Mar 11 '22 at 11:07

0 Answers0