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So the category of affine schemes is dual to the category of commutative rings, Stone spaces are dual to Boolean algebras, localizable measurable spaces are dual to commutative Von Neumann algebras, and I'm sure there are many more examples. In general, a category of algebraic structures is going to be dual to some related category of geometric structures.

My question is, then: is there an analogous story for coalgbraic things? If I take a category of coalgebras for some comonad and flip the arrows around, will I get something interesting? Are there any good examples of this over familiar comonads (say, the costate comonad)?

David Myers
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    "In general, a category of geometric structures is going to be dual to some related category of algebraic structures." - No. For example, the category of schemes (or the category of manifolds) has no reasonable algebraic dual category. It is vice versa: Often, starting with an algebraic category, the dual turns out to be "affine" geometric. (Whatever this means.) – Martin Brandenburg Jun 23 '15 at 13:03
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    (Cocommutative) coalgebras are formal spaces. This is not a duality: the correspondence is covariant rather than contravariant. – Qiaochu Yuan Jun 23 '15 at 16:29
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    @MartinBrandenburg Thanks I'll change the question to reflect that. – David Myers Jun 23 '15 at 18:53

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If $R$ is a commutative ring, then the category of commutative bialgebras (resp. Hopf algebras) over $R$ is dual to the category of affine monoid (resp. group) schemes over $R$. This may be seen as an extension of the duality between commutative algebras over $R$ and affine schemes over $R$. However, I don't know any geometric description of coalgebras over $R$.

Martin Brandenburg
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    If $R$ is a field, cocommutative coalgebras over $R$ are the ind-category of finite-dimensional cocommutative coalgebras over $R$, which is dual to the category of finite-dimensional commutative algebras over $R$; this is itself dual to a category of formal varieties, and hence the original category of coalgebras is a category of "ind-formal" varieties. – Qiaochu Yuan Jun 23 '15 at 20:42
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    Why don't you make this to an answer? This is not a comment to my answer. – Martin Brandenburg Jun 23 '15 at 22:39