T^i functors are isomorphic for analytically isomorphic isolated singular points - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:39:50Z http://mathoverflow.net/feeds/question/104436 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104436/ti-functors-are-isomorphic-for-analytically-isomorphic-isolated-singular-points T^i functors are isomorphic for analytically isomorphic isolated singular points HNuer 2012-08-10T20:46:21Z 2012-09-17T19:22:00Z <p>I've been having trouble proving the following: Let $B$ and $B'$ be local rings, essentially of finite type over $k$, having isolated singularities at the closed points. Suppose that they are analytically isomorphic, i.e. have isomorphic completions, then the deformation theory modules $T^i_{B/k}$ and $T^i_{B'/k}$ are isomorphic for $i=1,2$ as modules over the completions of $B$ and $B'$.</p> <p>A good answer would only use the definitions and basic properties of these functors, and/or would give a reference where this is worked out in some amount of detail, i.e. not where it is just stated as a fact.</p> <p>Thanks!</p> http://mathoverflow.net/questions/104436/ti-functors-are-isomorphic-for-analytically-isomorphic-isolated-singular-points/104605#104605 Answer by Damian Rössler for T^i functors are isomorphic for analytically isomorphic isolated singular points Damian Rössler 2012-08-13T09:43:36Z 2012-08-13T09:43:36Z <p>Here is a stab at an answer, but it is incomplete. </p> <p>Let $S:={\rm Spec}\, B$ and let $\widehat{S}:={\rm Spec}\,\widehat{B}$ be the completion of $B$ along its maximal ideal $m$. Let $\phi:\widehat{S}\to S$ be the natural morphism (which is faithfully flat). The composition of morphisms $\widehat{S}\to S\to{\rm Spec}\,k$ gives rise to a triangle of cotangent complexes and hence to an exact sequence $$\dots\to T^1_{\phi}\to T^1_{\widehat{S}/k}\to \phi^*T^1_{S/k}\to T^2_{\phi}\to T^2_{\widehat{S}/k}\to \phi^*T^2_{S/k}\to T^3_\phi\to\dots {\rm (*)}$$ so what you need to show is that $T^i_\phi$ vanishes for $i=1,2,3$. You would then get isomorphisms $T^i_{\widehat{S}/k}\to \phi^*T^i_{S/k}$ and since you can repeat this for $B'$ instead of $B$, you would get the required isomorphism. </p> <p>Let $L_\phi$ be the cotangent complex of $\widehat{S}/S$. This complex $L_\phi$ is concentrated in degree $0$. A quick way to see this is to notice that </p> <ul> <li><p>the formation of the cotangent complex is compatible with direct limits of rings (see Quillen, "Cohomology of commutative rings", eq. (4.11))</p></li> <li><p>the ring $B$ is excellent because it is essentially of finite type over a field (Grothendieck) and thus the fibres of $\phi$ are geometrically regular;</p></li> <li><p>a (deep...) result of Popescu (see for instance Th. 1.3 in "Approximations of versal deformations", by B. Conrad and AJ de Jong) then implies that $\widehat{B}$ is a direct limit of smooth $B$-algebras, and for the latter the cotangent complex is clearly concentrated at $0$. </p></li> </ul> <p>By considering the sequence analogous to (*) for the modules $T_i$ instead of $T^i$, this proves the analog of your assertion for the $T_i$. Furthermore, we see that $T^i_\phi={\rm Ext}^i(L_\phi,\widehat{B})={\rm Ext}^i(\Omega_\phi,\widehat{B})$.</p> <p>So the issue is to show that ${\rm Ext}^i(\Omega_\phi,\widehat{B})=0$ for $i>0$. Maybe your hypothesis on the fact that the singularity of $S$ is isolated at the closed point plays a role here.</p>