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Aug 18, 2017 at 2:27 vote accept darij grinberg
Aug 18, 2017 at 2:22 answer added darij grinberg timeline score: 1
Jun 10, 2017 at 20:10 comment added John Palmieri I think "graded" means graded by the integers, and then "connected" means graded in the nonnegative integers with degree 0 component equal to $k$.
Jun 10, 2017 at 19:24 comment added darij grinberg @YCor: Nonnegative integers, to be precise.
Jun 10, 2017 at 19:22 comment added YCor For you "graded" means graded in positive integers?
Jun 10, 2017 at 15:33 comment added John Palmieri Precisely, the degrees of the chosen generators are crucial.
Jun 10, 2017 at 15:03 comment added darij grinberg Oh. It's not just "all but one of the generators", but actually "all but one of the highest-degree generators". And now I see why the comultiplication restricts to a map $A' \to A' \otimes A'$ ! I'm going to check in more detail, but it definitely makes more sense now.
Jun 10, 2017 at 14:53 comment added darij grinberg In the proof of Proposition 4.17 in [B], they define $A'$ to be the subalgebra of $A$ (which is my $H$) generated by $x_1, x_2, \ldots, x_m$ (which are all but one of the generators of $A$ lifted from a homogeneous basis of $Q\left(A\right) = A^+ / \left(A^+\right)^2$). Why is $A'$ a coalgebra? Or do they not use this at all? They seem to apply the induction hypothesis to $A'$ instead of $A$, which seems to require it to be a coalgebra.
Jun 10, 2017 at 14:46 comment added John Palmieri Re "I don't think that Hopf algebras can be deconstructed in such a simple way". They are dealing with graded connected Hopf algebras, and that extra hypothesis gives you all sorts of advantages. What part of their proof (in [B], which is all I have access to) are you concerned about?
Jun 10, 2017 at 13:44 history asked darij grinberg CC BY-SA 3.0