Let A be a commutative (or graded commutative) algebra over a field k. In some sources, such as Mcleary's book on spectral sequences, Corollary 7.12, pg. 248, it is claimed that TorA(k,k) is a bicommutative Hopf algebra. Is this a typo? Obviously, Tor has a commutative algebra structure, but is it true that the coaddition is cocommutative?
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2What is the comultiplication? To get the multiplication we apply TorA(−,k) to k⊗k→k and then precompose with the map TorA(k,k)⊗TorA(k,k)→TorA(k⊗k,k). But to get the comultiplication it seems like we can't play the same game with k→k⊗k since the canonical map TorA(k,k)⊗TorA(k,k)→TorA(k⊗k,k) is in the wrong direction.– Connor MalinCommented Apr 13, 2020 at 17:21
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2Proposition 7.10 gives the comultiplication: [a1,…,an]→Σ[a1,…,aj]⊗[aj+1,…,an].– Connor MalinCommented Apr 13, 2020 at 17:56
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1 Answer
This is not true. Consider the algebra A=T(V)/V⊗2, it is a commutative algebra whose augmentation ideal has zero multiplication. We have TorA(k,k)≅T(V[1]) with the shuffle product and deconcatenation coproduct, so the coproduct is very much noncommutative (it is the coproduct of the cofree conilpotent coalgebra co-generated by V[1]).
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To me, this coproduct looks like the dual of something like the cup product which makes me think it should be cocommutative. Is this completely wrongheaded? Commented Apr 13, 2020 at 19:19
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@ConnorMalin I can't guess your thinking process but one possibility is that you are confusing the concatenation product with the shuffle product? Commented Apr 13, 2020 at 19:30