Timeline for Coaction on the Universal Calculus
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2010 at 15:29 | comment | added | Mariano Suárez-Álvarez | Abtan, nothing is completely obvious---obviousness is a relative notion; in any case, proving that the kernel is a subcomodule is not very difficult, and quite similar to showing that the kernel of a morphism of modules over an algebra is a submodule. On the other hand, it is quite safe to assume that the paper is talking about the same comodule structure as the one I mentioned. | |
| Jan 25, 2010 at 23:53 | vote | accept | Abtan Massini | ||
| Jan 25, 2010 at 23:47 | comment | added | Abtan Massini | Finally, I'm reading a paper at the moment that talks about a coaction induced on the universal calculus of a comodule algebra, is it safe to assume that the one I outlined above is the one implied? | |
| Jan 25, 2010 at 23:38 | comment | added | Abtan Massini | ... so it's not completely obvious? | |
| Jan 25, 2010 at 23:24 | comment | added | Mariano Suárez-Álvarez | (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) | |
| Jan 25, 2010 at 23:19 | comment | added | Mariano Suárez-Álvarez | 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 $\rho(\ker f)\subseteq (\ker f)\otimes H$. | |
| Jan 25, 2010 at 23:15 | comment | added | Abtan Massini | 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 $a^{(1)} \otimes b^{(1)} \otimes a^{(2)} b^{(2)} \in \Omega_u^1 (A) \otimes H$. | |
| Jan 25, 2010 at 23:02 | comment | added | Harry Gindi | In a category with zero-morphisms, the kernel of a map is the equalizer of the map with the zero morphism. | |
| Jan 25, 2010 at 22:57 | comment | added | Abtan Massini | I'm sorry, but I don't understand your comment the category of $H$-comodules having kernels. | |
| Jan 25, 2010 at 22:35 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |