Skip to main content
4 events
when toggle format what by license comment
Mar 11, 2017 at 21:32 history edited Yonatan Harpaz CC BY-SA 3.0
added 100 characters in body
Mar 11, 2017 at 20:49 comment added Yonatan Harpaz You're right, there is a gap here. I'm not sure how to reduce from $H^i(L)$ finite to $L^i$ finite. Maybe one can try to argue by comparing the lie-commutative Koszul duality with the associative-associative Koszul duality (and then use Corollary 14.1.3.3.of SAG).
Mar 10, 2017 at 14:23 comment added Louis-Clément LEFÈVRE First, thank you for answering and confirming this is true. However, Lurie's theory is not an easy thing ! I would like to understand it with more down-to-earth arguments. In particular in section 13.3 you quote, Koszul duality seems very abstract but it is written that it can be be more explicitly obtained via $L^\infty$ algebras.. ? By the way the theorem you quote is for $L^i$ finite-dimensional. In that case I already have a satisfying proof (use Hinich + duality algebras-coalgebras). But I have $H^i(L)$ finite-dimensional, how to reduce to that ?
Mar 7, 2017 at 22:08 history answered Yonatan Harpaz CC BY-SA 3.0