# If $H$ is a Hopf algebra, do we have $H^{cop}$ is a Hopf algebra?

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.  ## 2 Answers

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.

• 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. – LJR Mar 2 '17 at 16:20
• 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 $\Delta^{op}$? – freakish Mar 2 '17 at 16:20
• @freakish Why equal? the first is $S(c_2)\otimes c_1$ and the second is $c_1\otimes S(c_2)$ – David Hill Mar 2 '17 at 16:31
• @DavidHill, but in the wiki page, the condition "$H^{cop}$ is a Hopf algebra" is not required? – LJR Mar 2 '17 at 16:58
• 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. – David Hill Mar 2 '17 at 19:21

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).