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I am new to the theory of Hopf algebra. So I am sorry if the following question has really trivial answer.

Suppose that we have a quasi triangular Hopf algebra with the universal R-matrix $R$. It satisfies the following equation by definition: $$(\Delta \otimes \mathrm{id})(R)=R_{13}R_{23}$$ where $\Delta$ is the comultiplication and $ R_{13}=\phi_{13}(R)$ with $\phi_{13}(a\otimes b)=a\otimes 1 \otimes b$ and $ R_{23}=\phi_{23}(R)$ with $\phi_{23}(a\otimes b)=1\otimes a \otimes b$.

Let us write $R=\sum e_i\otimes f_i$. Let $R'=\sum f_i \otimes e_i$.

My question is whether we can replace $R$ with $R'$ in the above equation. Namely: Is $$(\Delta \otimes \mathrm{id})(R')=R'_{13}R'_{23}$$ hold true?

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  • $\begingroup$ By definition (in Etingof-Schiffman's lectures on quantum groups) $R$ should also satisfy : $$(Id \otimes \Delta)R=R_{13}R_{12}\qquad (*).$$ Changing the order of the tensors in (*) you get the identity with $R'$. Are you using another definition? or ? $\endgroup$ – user225222 Apr 25 '15 at 1:30
  • $\begingroup$ @user225222 I think this is what I am looking for. Could you write down more detail in answer? I especially want to know how to change the order of the tensors. $\endgroup$ – Snow Apr 25 '15 at 5:21
  • $\begingroup$ I've edited the answers. Hope it's helpfull now. The equation you gave is true if you replace $\Delta$ with $\Delta^{op}$ athoerwise not. There is an archaic counterexample (in the answer). Sorry for the mistakes done before. $\endgroup$ – user225222 Apr 27 '15 at 9:14
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In general the identity : $$(\Delta \otimes id)R'=R'_{13}R'_{23}\qquad (E)$$ does not hold (See Counterexample below). But it is true that $R'$ satisfy the conditions of "quasi-triangularity" with respect to $\Delta^{op}$ and therefore, we have : $$(\Delta^{op} \otimes id)R'=R'_{13}R'_{23}\qquad (E)$$

$\textbf{Counterexample:}$

Le $H$ be the Sweedler's algebra like in http://en.wikipedia.org/wiki/Sweedler%27s_Hopf_algebra.

Sweedler's hopf algebra is quasi-triangular with respect to: $$R=\frac 1 2 (1\otimes 1 +1\otimes g + g\otimes 1-g\otimes g) +\frac \lambda 2 (x\otimes x+x\otimes gx+gx\otimes gx -gx\otimes x), $$ this is given in Kassel's quantum groups.

Let us take $\lambda=1$ ($\lambda\neq 0$ works). We have $R'=R+R''$ with $R''=gx\otimes x-x\otimes gx$.

Assuming that $R'$ satisfies $(E)$ we get : $$ (\Delta \otimes id)R''= R_{13} R''_{23}+R''_{13}R'_{23}\qquad (*).$$

Denote $V_a=a\otimes H\otimes H$ and $W=V_x\oplus V_g \oplus V_{xg}$. Direct computation shows that : $$(\Delta \otimes id)R''\in -(1\otimes x\otimes gx)+W.$$

On the other hand, since $R''_{13} \in W$ and $ R'_{23} \in V_1 $, using that $W\cdot V_1 \subset W$ gives that $R''_{13}R'_{23} \in W$. So if (*) holds we should have : $$ R_{13} R''_{23} \in -(1\otimes x\otimes gx)+W \qquad(**).$$

Let us compute. We see that $R_{13} \in 1\otimes 1\otimes \frac{1+g}{2} +W $ and $R''_{23}\in V_1$. Therefore : $$ R_{13} R''_{23}\in (1\otimes 1\otimes \frac{1+g}{2})R''_{23} +W.$$

Finaly, we check that : $$ (1\otimes 1\otimes \frac{1+g}{2})R''_{23} \notin -(1\otimes x\otimes gx)+W .$$ Wich leads to a contradiction.

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