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 $(\Delta, \epsilon)$?


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 $A$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.