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The question is already in the title. Less succinctly, let's call a map $f:X \to Y$ of schemes $L$-trivial if its cotangent complex is quasi-isomorphic to $0$. Such maps have striking deformation-theoretic consequences; for example, any deformation of $Y$ can be followed uniquely by a deformation of $X$.

My primary (and probably naive) question is:

Is there a classification of $L$-trivial maps?

I am sure this question has been asked before, but I did not find any literature that deals with it. The three examples of $L$-trivial maps I am familiar with are:

  • Etale morphisms (and these are the only examples under finiteness constraints).
  • Any map between perfect $\mathbb{F}_p$-schemes.
  • The inclusion of the closed point in the spectrum of a valuation ring with divisible value group, or similar "divisible" constructions. For example, $\mathrm{Spec}(\mathbb{C}) \hookrightarrow \mathrm{Spec}(\mathbb{C}[ t^{\mathbb{Q}_{\geq 0}}])$ is $L$-trivial.

[ Edit: I learnt the last one in conversation after positing the first version of this question. ]

More examples can be obtained by taking filtered colimits of the above examples, but those are only slightly different. Hence, a second question is: are there other fundamentally different examples of $L$-trivial maps?

Perhaps a classification is unreasonable to expect, so I am also happy to learn more about $L$-trivial maps in other geometric categories, like algebraic stacks, or derived/spectral schemes/stacks, or (complex/rigid) analytic spaces, etc.. In particular, I am especially curious to know if $L$-trivial maps can be better understood using derived algebraic geometry.

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Isn't this notion the same as "formally etale"? In particular, if $f$ is of finite presentation, isn't this the same as just being etale? –  Piotr Achinger Jan 13 '13 at 18:11
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Yes for the second question about finitely presented maps (as I indicated parenthetically in the question), but I am interested in the general case. I do not think $L$-trivial maps are the same as formally etale maps; the latter only corresponds to the vanishing of the first couple of cohomology sheaves of the cotangent complex, and I see no reason why this implies vanishing of the full complex without additional strong finiteness assumptions (like Quillen's conjecture proven by Avramov), but maybe I am missing something simpler? –  anon Jan 13 '13 at 18:36

1 Answer 1

Edit: Here is a possible characterization. As mentioned in the comments above, the vanishing of the 1-truncated cotangent complex $\tau_{\leq 1}L_{B/A}$ of a map of rings $f \colon A \to B$ is equivalent to a lifting property with respect to square-zero extensions $T' \to T$. This follows from the fact that the space $Map(L_{T/A},M[1])$ is equivalent to the groupoid of square-zero extensions of $T$ over $A$ with kernel $M$. Here $M$ is a $T$-module.

Rephrasing this, $\tau_{\leq 1} L_{B/A}$ vanishes if and only if $f$ has a lifting property with respect to morphisms of 0-truncated simplicial algebras such that the kernel is concentrated in degree 0 and squares to 0. Let's call these 0-concentrated.

Then we can go on to look at morphisms of 1-truncated simplicial algebras with kernel $K$ concentrated in degree 1. The squaring-to-zero property is vacuous here, because a product of two elements in $\pi_1(K)$ will be in $\pi_2(K)$, which is zero by assumption. Let's call these 1-concentrated Then we find that $\tau_{\leq 2} L_{B/A}$ vanishes if and only if $f$ has a lifting property with respect to all 0- and 1-concentrated maps. This again holds because the cotangent complex classifies 0- and 1-concentrated maps.

I think now it's clear how to go on: $\tau_{\leq n+1}L_{B/A}$ vanishes if and only if $f$ has a lifting property with respect to all $m$-concentrated maps with $m \leq n$. And the full cotangent complex vanishes if and only if $f$ has the lifting property with respect to $n$-concentrated maps for all $n$.

These directions one looks at if one starts to check with respect to $n$-concentrated maps for $n \geq 1$ are sometimes called the derived directions. So Avramov's theorem might be rephrased as saying that under strong finiteness assumptions, unobstructedness in the classical directions implies unobstructedness in all derived directions.


This is an anwer to your last paragraph about L-trivial maps in other geometric categories. If you are only interested in schemes it doesn't tell you anything interesting.

One thing that the cotangent complex is good at is measuring connectivity of a morphism of simplicial rings. This also holds without any finiteness assumptions on the ring.

Recall that a morphism $f \colon A \to B$ of simplicial rings is called n-connective if it induces isomorphisms $\pi_i (A) \to \pi_i (B)$ in degrees $< n$ and a surjection $\pi_n (A) \to \pi_n (B)$ in degree $n$. There then is a result that states that if $f$ is $n$-connective, then the homology of the relative contangent complex $L_{B/A}$ vanishes in degrees $\leq n$. (I hope I got all the indices right.) So in particular, any equivalence of simplicial rings is L-trivial.

One way of intepreting your question is to ask when the converse holds. What do I know if a morphism of simiplicial rings is L-trivial? There is a partial converse to the statement above. Namely, if a morphism $f \colon A \to B$ induces an isomorphism $\pi_0(A) \to \pi_0(B)$ and is L-trivial, then it is an equivalence! I find this pretty suprprising, as L is only a linear piece of data, but still manages to detect equivalences.

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Ah, thanks. This observation strongly supports the (well-advertised) point of view that a simplicial commutative ring is just a ring with additional "nilpotent" data. It also raises the following question: if $A \to B$ is an $L$-trivial map of ordinary rings that is additionally an isomorphism modulo nilpotents, then is $A \simeq B$? –  anon Jan 14 '13 at 11:19
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I think this true, at least with finiteness ass. By the way, I'd be surprised if your question has a complete answer in terms of only ordinary rings. As evidence, one can define something like L-finite maps having perfect rel.cot.complex. Then this notion is completely unrelated to being of finite presentation. On the contrary, if you look at maps that are homotopically of finite presentation, then you can again detect these on the cotangent complex (if you add $\pi_0(B)$ finitely presented over $\pi_0(A)$). In short, I think it's hard to say something about the cot.complex using ord. rings. –  Timo Schürg Jan 14 '13 at 14:02

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