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Apr 6, 2015 at 16:38 comment added Romie Banerjee @ZhaotingWei, Thanks for the Toen reference!
Apr 6, 2015 at 14:50 comment added Zhaoting Wei Yes I agree. I think that it is a little bit more complicated to express the quasi-representability under the self-duality but I'm not sure about the details.
Apr 6, 2015 at 14:17 comment added Daniel Pomerleano To avoid possible competing definitions/confusion, let's assume $X$ is a sufficiently nice scheme over a field $k$. The self-duality you need here is not a very complicated thing---- you take a perfect complex $P$ to it's linear dual $RHom(P,O_X)$ to obtain a functor $Perf(X) \to Perf(X)^{op}$. Passing to co-limit completions gives the required self-duality. Being able to perform this last step is one of the key points of the perfect definition.
Apr 6, 2015 at 13:59 history edited Zhaoting Wei CC BY-SA 3.0
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Apr 6, 2015 at 12:44 history answered Zhaoting Wei CC BY-SA 3.0