4
$\begingroup$

Let $H=(H, m, u, \Delta, \epsilon, S)$ be a Hopf algebra, see for example the lecture notes, where $m$ is the multiplication, $u$ is the unit, $\Delta$ is the comultiplication, $\epsilon$ is the counit, $S$ is the antipode. Let $H^{cop} = (H, m, u, \Delta^{op}, \epsilon, S)$, where $m,u,\epsilon,S$ are the same maps as the maps in $H$, and $\Delta^{op}=\tau \circ \Delta$, $\tau$ is the flip map. Is $H^{cop}$ a Hopf algebra?

I am asking this question because in the book: Hopf algebras and their actions on rings, pages 213, 214, it is said that $t_{M,N}$ is a braiding if $H^{cop}$ is a Hopf algebra. Why the condition is not "$H$ is a Hopf algebra" but "$H^{cop}$ is a Hopf algebra"? Are "$H$ is a Hopf algebra " and "$H^{cop}$ is a Hopf algebra" equivalent? Thank you very much.

enter image description here

enter image description here

$\endgroup$
3
$\begingroup$

For a Hopf algebra $H$, $H^{cop}$ may or may not be a Hopf algebra. If it is, then the maps $t_{M,N}$ above define a braiding.

$\endgroup$
  • $\begingroup$ thank you very much. Where do we use the condition $H^{cop}$ is a Hopf algebra? In the book, it is only assumed that $H$ is a bialgebra. According to the lecture notes, $H$ is a bialgebra implies that $H^{cop}$ is a bialgebra. $\endgroup$ – LJR Mar 2 '17 at 16:20
  • $\begingroup$ But isn't $H^{cop}$ always a Hopf algebra? For example $(S\otimes id)\circ\Delta^{op}=(id\otimes S)\circ\Delta$, right? Where does the definition of Hopf algebra fail for $\Delta^{op}$? $\endgroup$ – freakish Mar 2 '17 at 16:20
  • $\begingroup$ @freakish Why equal? the first is $S(c_2)\otimes c_1$ and the second is $c_1\otimes S(c_2)$ $\endgroup$ – David Hill Mar 2 '17 at 16:31
  • $\begingroup$ @DavidHill, but in the wiki page, the condition "$H^{cop}$ is a Hopf algebra" is not required? $\endgroup$ – LJR Mar 2 '17 at 16:58
  • $\begingroup$ I'm a bit suspicious. Note that the compatibility condition in the wiki page is different than (10.6.11) above. It may be that the condition on the wiki page automatically implies that $H^{cop}$ is a Hopf algebra. $\endgroup$ – David Hill Mar 2 '17 at 19:21
3
$\begingroup$

The answer is generally no.
The conditions: "$H$ is a Hopf algebra" and "$H^{cop}$ is a Hopf algebra" are generally not equivalent. But they are, if we confine ourselves at the bialgebra level. To be more detailed:

If $H=(H, m, u, \Delta, \epsilon)$ is a bialgebra, $H^{cop} = (H, m, u, \Delta^{op}, \epsilon)$ is also a bialgebra called co-opposite bialgebra. (and $H^{op} = (H, m^{op}, u, \Delta, \epsilon)$ is called the opposite bialgebra).

However, if $H=(H, m, u, \Delta, \epsilon, S)$ is a hopf algebra, $(H, m, u, \Delta^{op}, \epsilon, S)$ is not necessarily a hopf algebra. (Neither is $(H, m^{op}, u, \Delta, \epsilon, S)$).

A sufficient condition is the antipode to be invertible: If $H=(H, m, u, \Delta, \epsilon, S)$ is a hopf algebra with invertible (i.e.: bijective) antipode then $H^{cop} = (H, m, u, \Delta^{op}, \epsilon, S^{-1})$ is also a hopf algebra. (and $H^{op} = (H, m^{op}, u, \Delta, \epsilon, S^{-1})$ as well). These are now called the co-opposite and the opposite Hopf algebras respectively.
Now, if $S^2=Id$ thus $S=S^{-1}$, then the co-opposite $H^{cop} = (H, m, u, \Delta^{op}, \epsilon, S)$ and the opposi-te $H^{op} = (H, m^{op}, u, \Delta, \epsilon, S)$ are also hopf algebras. For example, this is the situation for either commutative or cocommutative hopf algebras.
(see also: https://math.stackexchange.com/a/1691054/195021 for a related discussion).

$\endgroup$

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.