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I'm reading $\textit{The Hopf Ring for Complex Cobordism}$ by Ravenel and Wilson where they discuss the notion of group and ring objects over a category. They say that a hopf algebra over a ring in this context is a group object in the category of coalgebras. Here's my problem. I assume that the group operation for a hopf algebra must be the algebra multiplication, since the addition is presupposed from the coalgebra structure. Not all hopf algebras are division algebras. But, shouldn't a group object have inverses?

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That's what the antipode is for, right? –  Qiaochu Yuan Dec 2 '10 at 17:04

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The group operation corresponds to the multiplication map $\mu:A\otimes A\to A$ and the identity should be the natural map $\iota:k\to A$. Both these should be coalgebra maps. The inverse should correspond to a map $S:A\to A$ with $\mu\circ(\rm{id}\otimes S)\circ\Delta=\iota\circ\epsilon =\mu\circ(S\otimes\rm{id})\circ\Delta$, so $S$ is the antipode.

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The trouble is that the tensor product is not the cartesian product in the category of coalgebras (it is the cartesian product in the category of cocommutative coalgebras), so I'm wondering whether the statement by Ravenel and Wilson is actually technically correct. –  Todd Trimble Dec 2 '10 at 18:08
    
Thanks Todd, but doesn't one also have group objects in monoidal categories? –  Robin Chapman Dec 2 '10 at 18:35
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Robin, no, you need a diagonal to even state the axiom for inverses. But you can define group objects in monoidal categories with a diagonal. That's what Ravenel and Wilson mean. Todd, does it also worry you that if you want to define a ring as a monoid in abelian groups, i.e. as a map $A \otimes A \to A$, that $\otimes$ is not the product in abelian groups? –  Tilman Dec 2 '10 at 18:44
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Todd, (I don't have the paper in front of me to check this but) I suspect that Ravenel and Wilson are working with cocommutative coalgebras since they are working with homology groups. Tilman, technically, rings are Tall-Wraith monoids in the category of abelian groups. –  Loop Space Dec 2 '10 at 19:41
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They are working with cocommutative coalgebras. Thanks everybody. –  Joe Johnson Dec 2 '10 at 23:07

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