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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 $\beta_{A}$?

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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.

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  • $\begingroup$ I'm sorry, but I don't understand your comment the category of $H$-comodules having kernels. $\endgroup$ – Abtan Massini Jan 25 '10 at 22:57
  • $\begingroup$ In a category with zero-morphisms, the kernel of a map is the equalizer of the map with the zero morphism. $\endgroup$ – Harry Gindi Jan 25 '10 at 23:02
  • $\begingroup$ 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 $a^{(1)} \otimes b^{(1)} \otimes a^{(2)} b^{(2)} \in \Omega_u^1 (A) \otimes H$. $\endgroup$ – Abtan Massini Jan 25 '10 at 23:15
  • $\begingroup$ 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 $\rho(\ker f)\subseteq (\ker f)\otimes H$. $\endgroup$ – Mariano Suárez-Álvarez Jan 25 '10 at 23:19
  • $\begingroup$ (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) $\endgroup$ – Mariano Suárez-Álvarez Jan 25 '10 at 23:24

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