Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be an algebra, $H$ a Hopf algebra, and $$ \beta_A: A \to A \otimes H, ~~~~~ a \mapsto a^{(1)} \otimes a^{(2)} $$ a right $H$-coaction. This induces a right $H$-coaction on $A \otimes A$ defined by $$ \beta_{A \otimes A}: a \otimes b \mapsto a^{(1)} \otimes b^{(1)} \otimes a^{(2)}b^{(2)}. $$ My question is: Does this restrict to a coaction on the universal calculus over $A$, namely to a $H$-coaction on the kernel of the multiplication map $m:A \otimes A \to A$? I feel this is a very simple question but I can't seem to find an answer.

If the construction does not work, does anyone know of a way to induce a coaction on the universal calculus over $A$ from $\beta_{A}$?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

If $A$ is an $H$-comodule algebra (that is, if the multiplication map $\mu$ is a map of comodules), then the answer is yes (trivially, because the category of $H$-comodules has kernels). If it isn't, then probably not, as then you have no compatibility between the algebra structure on $A$ and the comodule structure.

share|improve this answer
    
I'm sorry, but I don't understand your comment the category of $H$-comodules having kernels. –  Abtan Massini Jan 25 '10 at 22:57
    
In a category with zero-morphisms, the kernel of a map is the equalizer of the map with the zero morphism. –  Harry Gindi Jan 25 '10 at 23:02
    
Moreover, I see that if $A$ is a $H$-comodule algebra, then $m(a \otimes b) = 0$ implies that $(m \otimes $id$) \beta_{A \otimes A} (a\otimes b) = 0$. What I do not see is why this should imply that $a^{(1)} \otimes b^{(1)} \otimes a^{(2)} b^{(2)} \in \Omega_u^1 (A) \otimes H$. –  Abtan Massini Jan 25 '10 at 23:15
    
That is precisely what I meant by my second parenthetical remark: If $f:M\to N$ is a morphism of $H$-comodules, then $\ker f$ is a subcomodule of $M$, and this means, among other things, that if $\rho:M\to M\otimes H$ is the coaction on $M$, then $\rho(\ker f)\subseteq (\ker f)\otimes H$. –  Mariano Suárez-Alvarez Jan 25 '10 at 23:19
    
(That is proved when one shows that the category of comodules is abelian. I imagine this is in Sweedler's book on Hopf Algebras, for example) –  Mariano Suárez-Alvarez Jan 25 '10 at 23:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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