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Sep 16, 2019 at 23:42 history edited Dmitry Vaintrob CC BY-SA 4.0
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Sep 16, 2019 at 20:35 comment added Dmitry Vaintrob @crystalline Thanks! That's exactly what I was hoping for -- will read the paper. If you can turn that into an answer, I'll approve it.
Sep 16, 2019 at 17:44 comment added crystalline Well what STV show is that the derived enhancement of $SM$ is a quasi-smooth proper derived DM stack (ie representable by a derived DM stack), using Lurie’s represetabikity theorem using the fact that the 0-truncation is known to be representable by Kontsevich.
Sep 16, 2019 at 3:48 comment added Dmitry Vaintrob @crystalline I meant of course "quasi-smooth". "Complicated" is in the eye of the beholder, but the definitions I knew of seemed so to me. Thank you for the Schurg, Toen, Vezzosi reference -- that seems very close to what I need (even closer, though probably non-existent, is Lurie's assertion in HAG that stable maps are a representable DG moduli problem.
Sep 15, 2019 at 3:48 comment added crystalline I don’t folllow your construction but for the record, smooth derived schemes are automatically 0-truncated. Also, it is not “complicated” to define $\tilde{SM}$, it is just the obvious extension of the moduli problem from ordinary to simplicial commutative rings. See the paper by Schurg, Toën, Vezzosi in Crellle journal.
Sep 11, 2019 at 22:21 comment added Ben Wieland Kontsevich's original construction was very similar. He worked analytically and broke the curve into compact pieces with boundary. I think it's valuable motivation that seems to be largely forgotten. But infinite dimensions are hard! I don't think anyone claimed to make it work. Working algebraically and dropping compactness sounds even more difficult.
Sep 11, 2019 at 19:27 history edited Dmitry Vaintrob CC BY-SA 4.0
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Sep 11, 2019 at 19:10 history asked Dmitry Vaintrob CC BY-SA 4.0