6
$\begingroup$

I would like to know if one can compute all the cohomology sheaves of the cotangent complex of a subvariety of the affine space once a resolution of its ideal sheaf is given?

In my precise situation, I work with $X \subset \mathbb{A}^n$ Gorenstein of codimension 3 and I have a resolution:

$$0 \rightarrow \mathcal{O}_{\mathbb{A}^n} \rightarrow E^* \rightarrow E \rightarrow I_X \rightarrow 0$$

Is it possible to compute the cotangent complex of the closed immersion $ X \hookrightarrow \mathbb{A}^n$ from this resolution?

Many thanks!

EDIT : I really want the cotangent complex of the closed immersion $X \hookrightarrow \mathbb{A}^n$ (and not of the map $X \rightarrow spec(k)$).

$\endgroup$
  • $\begingroup$ I doubt there is any "simple" formula for the cotangent complex that comes directly from the resolution (except in the sense that you can recover your scheme from the resolution, and the cotangent complex is determined by the scheme). For instance, if memory serves, for a hypersurface, the form of the cotangent complex depends on whether or not the hypersurface is reduced. $\endgroup$ – Jason Starr Oct 2 '13 at 18:25
  • 1
    $\begingroup$ @Jason Starr : I might be mistaken, but I read somewhere that for a local complete intersection $X \subset Y$ the cotangent complex is isomorphic to $I/I^2[1] $ (where $I$ is the ideal sheaf of $X$ in $Y$). $\endgroup$ – Libli Oct 2 '13 at 21:42
4
$\begingroup$

Your X is not a local complete intersection in A^n. Then the cotangent complex is going to have cohomology in infinitely many degrees. (Follows from a conjecture by Quillen proved by Avramov IIRC.) Your best bet would be to use Quillen's spectral sequence relating Tor_*(O/I, O/I) to the cohomology of the cotangent complex. You should read the original, but if you are in a hurry, take a look at Example Tag 08RG of the Stacks project. If you only want to compute the first 3 terms, then I suggest looking at the method of Lichtenbaum-Schlessinger, see Section Tag 09AM.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.