Let $X_0$ be some algebro-geometric object defined over a field, and suppose its deformation functor is prorepresentable, so there is a universal deformation ring $R$. Then $Aut(X_0)$ acts naturally on the deformation functor, and hence on $R$. One may consider the group cohomology $$H^i(Aut(X_0),R)$$ where $R$ is considered as its underlying additive group.

Do these cohomology groups have a "geometric interpretation"?

  • $\begingroup$ Perhaps i'm being silly but how does $Aut(X_0)$ act on the deformation functor? $\endgroup$ Jan 18, 2018 at 10:55
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    $\begingroup$ act naturally, read well the question $\endgroup$ Jan 18, 2018 at 10:58
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    $\begingroup$ @SaalHardali Usually the deformation functor, evaluated on an artin local ring $A$, outputs a pair $(X,i)$ where $X$ is an object over $A$, and $i$ is an isomorphism $i : X\times_A A/m\cong X_0$. Thus, automorphisms of $X_0$ act naturally on the deformation functor by its natural action on the "$i$" part of such pairs. $\endgroup$
    – Will Chen
    Jan 18, 2018 at 16:06
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    $\begingroup$ I see, well in that case i'm pretty sure the lower cohomologies have interpretations in terms of lifting of automorphisms (and obstructions thereof). I have never found a treatment of this part of deformation theory so i'm looking forward to an answer too. $\endgroup$ Jan 18, 2018 at 16:11
  • $\begingroup$ I don't know if it helps but the derived quotient $spec(R)// Aut(X_0)$, at least for good enough automorphisms group, classify the collection of families with special fiber $X_0$ (but without specified iso.) So these cohomologies can be seen as derived functions on this moduli stack. Usually functions have no immediate geometric interpretation, though (they are exactly the algebra part of the picture). $\endgroup$
    – S. carmeli
    Mar 10, 2018 at 23:03


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