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Let $A$ be an $R$-algebra.

Suppose $A$ has a $R$-coalgebra structure compatible with the algebra structure. (I.e. there is a comultiplication map $\Delta$ and counit map $\epsilon$ compatible with the multiplication map and unit map of $A$.)

Then I am wondering whether $A$ can have another $R$-coalgebra structure other than $(\Delta, \epsilon)$?

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Yes, $A$ can have multiple compatible $R$-coalgebra structures. Let $G$ be a finite set and let $A=R(G)$ be the commutative algebra of functions on $G$ with pointwise multiplication. Then any group structure on $G$ gives an Hopf structure on $A$ and non-isomorphic group structures on $G$ leads to non-isomorphic Hopf structures on $A$.

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  • $\begingroup$ Is it true for all $R$ algebra not only of the form $R(G)$? I am especially considering the universal enveloping algebra. Is it true for universal enveloping algebra? $\endgroup$ – Monty Apr 16 at 16:12
  • $\begingroup$ Yes, e.g. take the trivial Lie algebra, then you just get a polynomial algebra and those have many Hopf structures. If you start with a simple Lie algebra $\mathfrak g$ and let $R=\mathbb{C}[\epsilon]/(\epsilon^2)$ then the standard Lie bialgebra structure on $\mathfrak g$ gives you another example. I'm sure there are plenty more. $\endgroup$ – Adrien Apr 16 at 16:24
  • $\begingroup$ Sorry, I am not sure whether I understood well your comment. You say that there might be some Lie algebra $\mathfrak{g}$ such that $U(\mathfrak{g})$ may have many Hopf algebra structure? Right? I guessed that $U(\mathfrak{g})$ has only one Hopf algebra structure! $\endgroup$ – Monty Apr 16 at 16:33
  • $\begingroup$ yes that's what I'm saying, and there are examples already when $\mathfrak g$ is abelian, i.e. have the trivial Lie structure. $\endgroup$ – Adrien Apr 16 at 16:42
  • $\begingroup$ If you want specifics : let $\mathfrak g=R$ be the one dimensional Lie algebra with zero bracket. Then $U(\mathfrak g)$ is just $R[x]$ as an algebra, and this has a Hopf structure given by $\Delta(x)=x \otimes x$ which is not the one coming from being an enveloping algebra. $\endgroup$ – Adrien Apr 16 at 16:44

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