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Let $X$ be a variety, the triangulated category of singularities $D_{sg}^b(X)$ is obtained by taking the quotient of $D^b(X)$ by the category of perfect complexes. Suppose there is a group $G$ acting on $X$. Is there any relation between $D_{sg}^b(X)$ and $D_{sg}^b(X/G)$ (notice, I really interest in the category of singularities NOT their derived categories)? My wishful thinking is they are derived equivalence, but have no evidence this has to be true.

I am sloppy on the conditions of $X$ and group action (because I don't know what to put), but in my case they are reasonably good: let $E$ be a vector bundle on $S$, and $P = tot(E^\vee)$ be the total space of the vector bundle, let $X$ be a hypersurface which comes from $|\mathcal{O}_P(1)|$. Let $G$ be an action on the base $S$, and also acts on $X$. Then I am looking for the relation between $D_{sg}^b(X)$ and $D_{sg}^b(X/G)$.

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  • $\begingroup$ The right term is "triangulated category of singularities" $\endgroup$
    – Anton Fonarev
    Dec 25, 2014 at 14:20
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    $\begingroup$ I think I'm confused, how can they be equivalent? For example, let X be smooth and X/G be singular (eg you can look at the McKay correspondence for such examples), then the former category of singularities will be zero while the other will have interesting elements. No? Unless by X/G you mean the quotient stack? $\endgroup$
    – bananastack
    Dec 25, 2014 at 16:20
  • $\begingroup$ The current question is in need of some clarification. It's hard for one to interpret "Is there any intelligence could be said for the singular categories of Dbsg(X) and Dbsg(X/G)?" even being sensitive to language issues. Is your question, given a group action on X is there a formal categorical construction that allows us to obtain the equvariant category of coherent sheaves (or category of singularities) from the ordinary category? As a concrete question what does the symbol "X/G" mean for you... quotient stack, GIT quotient(is our variety proj. over aff... what is O(1)??).. $\endgroup$
    – Daniel Pomerleano
    Dec 25, 2014 at 18:23
  • $\begingroup$ @DanielPomerleano: I assume the OP is looking for some relation between D(X) and D(X/G) which he hopes may be known to experts (I should really be writing D(Coh)/Perf). I also bet that by X/G he means some form of GIT quotient. But in this generality and vagueness I don't really know of anything intelligent one could say. $\endgroup$
    – bananastack
    Dec 25, 2014 at 19:19
  • $\begingroup$ @bananastack I have edited my post to make it less vague. $\endgroup$
    – Li Yutong
    Dec 26, 2014 at 0:24

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