When does the cotangent complex vanish? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:18:34Z http://mathoverflow.net/feeds/question/118834 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118834/when-does-the-cotangent-complex-vanish When does the cotangent complex vanish? X-curious 2013-01-13T17:40:03Z 2013-01-16T17:05:58Z <p>The question is already in the title. Less succinctly, let's call a map $f:X \to Y$ of schemes <em>$L$-trivial</em> 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 $X$. </p> <p>My primary (and probably naive) question is:</p> <p> Is there a classification of $L$-trivial maps? </p> <p>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:</p> <ul> <li>Etale morphisms (and these are the <em>only</em> examples under finiteness constraints).</li> <li>Any map between perfect $\mathbb{F}_p$-schemes.</li> <li>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.</li> </ul> <p>[ <strong>Edit</strong>: I learnt the last one in conversation after positing the first version of this question. ]</p> <p>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? </p> <p>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. </p> http://mathoverflow.net/questions/118834/when-does-the-cotangent-complex-vanish/118866#118866 Answer by Timo Schürg for When does the cotangent complex vanish? Timo Schürg 2013-01-14T10:51:18Z 2013-01-16T17:05:58Z <p><strong>Edit:</strong> 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.</p> <p>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.</p> <p>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.</p> <p>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 $n$.</p> <p>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.</p> <hr> <p>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.</p> <p>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.</p> <p>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 $&lt; n$ 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.</p> <p>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.</p>