Cobraided and coquasitriangular Hopf algebras

Question

In a Hopf algebra, are the notions "cobraided" and "coquasitriangular" the same? Kassel (Hopf algebras) uses "cobraided" and Montgomery (HAs and their actions on rings) uses the other -- both on page 184. See the note at the top of Kassel page 174. Their definitions look different but the contexts seem similar. In Montgomery equation (10.2.4) the coproduction of "l" is twisted: l_2 precedes l_1, unlike anything in Kassel.

Answer 1

The definitions of:

• cobraided (according to the terminology of Kassel's book; see (5.1)-(5.2)-(5.3) p. 184-185 or equivalently (5.4)-(5.5)-(5.6)-(5.7), p.185) and
• coquasitriangular (according to Montgomery's book; (10.2.2)-(10.2.3)-(10.2.4), p. 184-185)

are the same:

Rel (5.1)$\Leftrightarrow$(5.4) in Kassel's book is equivalent to the demand that the bilinear form (denoted $\langle .. |..\rangle$ in Montg and $r$ in Kassel) is invertible with respect to the convolution product $\star$. Montgomery's relations are written for the inverse $\bar{r}$ of $r$ (wrt to the convolution product).
For example, if you take rel. (5.2) from Kassel you have $ \mu^{op}=r\star\mu\star\bar{r} $ which can be written either as $$ \mu^{op}\star r=r\star\mu $$ or as $$ \bar{r}\star\mu^{op}=\mu\star\bar{r} $$ Applying the first of these to an arbitrary element $h\otimes k$ gives: $$ \sum k_1h_1r(h_2\otimes k_2)=\sum r(h_1\otimes k_1)h_2k_2 $$ while applying the second one gives: $$ \sum \bar{r}(h_1\otimes k_1)k_2h_2=\sum h_1k_1 \bar{r}(h_2\otimes k_2) $$ The last relation is identical to rel (10.2.2) from Montgomery's definition 10.2.1.

(In your last sentence you say:

In Montgomery equation (10.2.4) the coproduction of "l" is twisted: l_2 precedes l_1, unlike anything in Kassel.

I suggest you look carefully at relations (5.6)-(5.7) from Kassel where the same thing is happening. Rel (5.7) is the "twisted" one now. If you try to write Kassel's (5.6)-(5.7) for the inverse form $\bar{r}$ you will get Montgomery's rel (10.2.4)-(10.2.3) respectively).