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Let $\Bbbk$ be a field, $X$ affine scheme of finite type over $\Bbbk$. Let $\mathcal C_X$ be the category of closed embeddings of $X$ into (say affine) smooth $Y$'s of finite type over $\Bbbk$, morphisms being closed embeddings of $Y$'s (forming commutative triangle). We have a functor $\mathcal C_X^{\mathrm{op}}\to \text{Vect}_\Bbbk$ by $[X\hookrightarrow Y] \mapsto\Omega^1(Y)$ You can also post-compose it with the embedding $\text{Vect}_\Bbbk\to\text{Ch}_\Bbbk$ to the $\infty$-category of chain complexes.

I would like to claim that if you take the direct limit over $\mathcal C^{\mathrm{op}}_X$ of both versions of this functor (the vector-space and the chain-complex ones), what you get is the module of Kähler differentials, resp. the Illusie cotangent complex, of $X$. (There is a natural map in one direction and I claim it's an isomorphism of $\Bbbk$-vector spaces, resp. of objects of the derived category of $\text{Vect}_\Bbbk$.)

Are such statements known? Or is there a reference for something similar?

(I think I have an argument at least for the first version, but I couldn't find any references.)

Edit: (some additional thoghts)

  1. Clearly the first statment follows from the second, since the functor $H_0$ from non-negatively (homologically) graded complexes to vector spaces preserves colimits.

  2. The cotangent complex is defined as a similar colimit for some chosen "cosimplicial resolution" of $X$ by smooth affine varietis. The colimit in question can thus be thought of as "taking all such reslutions at once", but right now I can't formalize this.

  3. It looks like instead of closed embeddings, one could take all maps to affines and (hope to) get the same answer.

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    $\begingroup$ The Kahler differentials are an $\mathcal{O}_X$-module; your direct limit is only a vector space a priori. How are you endowing it with the structure of an $\mathcal{O}_X$-module? $\endgroup$
    – Daniel Litt
    Dec 10, 2013 at 8:33
  • $\begingroup$ well, my (say, first) claim is that the evident map from the direct limit to Kähler differentials is an isomorphism. But I think what you're asking is possible. Let's say we want to act by a function $f\in\mathcal O(X)$ on some element in the colimit coming from some 1-form $\alpha\in\Omega^1(Y)$. The obvious recipe is to lift $f$ to $\tilde f\in \mathcal O(Y)$ and take $\tilde f\alpha$. The you'd of course have to prove that the image of this 1-form in the direct limit is independent of $\tilde f$ (which you can prove by embedding $Y\sqcup_XY$ into a smooth $Z$). $\endgroup$
    – rrrrrttttttt
    Dec 10, 2013 at 19:36
  • $\begingroup$ I wonder how this is related to the infinitesimal site. An embedding into a smooth $Y$ should give you a "covering of the final object" in the infinitesimal topos. $\endgroup$
    – Piotr Achinger
    Dec 11, 2013 at 6:40
  • $\begingroup$ @Piotr: I am not really experienced with sites and topoi, but is there a definition of cotangent complex in terms of infinitesimal topos? $\endgroup$
    – rrrrrttttttt
    Dec 12, 2013 at 0:29
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    $\begingroup$ Something like this does work (modulo Daniel's comment). You should look at Gaitsgory's "Grothendieck topologies and deformation theory II", Rim's expose in SGA7, and Quillen's "On the (co)homology of commutative rings". I've also written a couple of papers about the site. It has something in common with the infinitesimal site, but the arrows are backwards. $\endgroup$
    – Jonathan Wise
    Dec 12, 2013 at 4:39

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This is the approach to the cotangent complex in the Stacks project; there one uses all maps as in your Remark 3 (yes this gives the same answer). See the Stacks project chapter on the cotangent complex. In particular for schemes over a ring in particular see Section Tag 08V7.

Your remark 2 can be made more precise by saying that the homology over your category is computed by taking the usual simplicial resolution of the ring and then applying the functor. The simplest case of this is discussed in Section Tag 08PQ.

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    $\begingroup$ Where can I find the source code for the 'answer_bot'? -- Is it open-source? $\endgroup$
    – Stefan Kohl ♦
    Dec 21, 2013 at 14:08
  • $\begingroup$ @answer_bot: thanks for additional references (to those by Jonathan Wise in the comments to my OP). I think I can accept your answer after I read your refs. I'm actually a bit confused by the fact that for non-affine $X$ one would have to use a mix of homology and cohomology (although I guess it's a bit out of scope of OP) $\endgroup$
    – rrrrrttttttt
    Dec 22, 2013 at 4:26

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