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I am a beginner learning Quantum Groups, I have a question of how to show that that twist map $\tau_{M,M}:M\bigotimes M \rightarrow M\bigotimes M$ is a solution to the QYBE.

I tried to prove it by definition: When $R=\tau$,

$R_{(1,2)}$R_{(1,3)}$R_{(2,3)}$=

$(\tau \bigotimes 1_M)(1_M\bigotimes \tau)(\tau \bigotimes 1_M)(1_M \bigotimes \tau)(1_M \bigotimes \tau)$=

$(1_M \bigotimes \tau)(1_M\bigotimes \tau)(\tau \bigotimes 1_M)(1_M \bigotimes \tau)(\tau \bigotimes 1_M)$=

$R_{(2,3)}$R_{(1,3)}$R_{(1,2)}$

I have some questions: 1) Is the above method correct?

2) What does $\bigotimes$ mean? I only know vaguely that it is "tensor product".

Sincere thanks for help.

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    is it the same as "outer product", which produces a matri$x$ from two vectors?2012-05-09

1 Answers 1

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Let $m_1\otimes m_2\otimes m_3\in M\otimes M\otimes M$, then $ \begin{align} R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1\otimes m_2\otimes m_3)&= (\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(1_M\otimes\tau)(m_1\otimes m_2\otimes m_3)\\ &=(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(m_1\otimes m_3\otimes m_2)\\ &=(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(m_1\otimes m_2\otimes m_3)\\ &=(\tau\otimes 1_M)(1_M\otimes\tau)(m_2\otimes m_1\otimes m_3)\\ &=(\tau\otimes 1_M)(m_2\otimes m_3\otimes m_1)\\ &=m_3\otimes m_2\otimes m_1\\ \end{align} $ $ \begin{align} R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1\otimes m_2\otimes m_3)&= (1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(m_1\otimes m_2\otimes m_3)\\ &=(1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(m_2\otimes m_1\otimes m_3)\\ &=(1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(m_2\otimes m_3\otimes m_1)\\ &=(1_M \otimes \tau)(1_M\otimes\tau)(m_3\otimes m_2\otimes m_1)\\ &=(1_M \otimes \tau)(m_3\otimes m_1\otimes m_2)\\ &=m_3\otimes m_2\otimes m_1\\ \end{align} $ so we conclude $ R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1\otimes m_2\otimes m_3)=R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1\otimes m_2\otimes m_3)\tag{1} $ Now take arbitrary $u\in M\otimes M\otimes M$, then we have representation $ u=\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)} $ Hence using $(1)$ we get $ \begin{align} R_{(1,2)}R_{(1,3)}R_{(2,3)}(u) &=R_{(1,2)}R_{(1,3)}R_{(2,3)}\left(\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)}\right)\\ &=\sum\limits_{i=1}^n R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)})\\ &=\sum\limits_{i=1}^n R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)})\\ &=R_{(2,3)}R_{(1,3)}R_{(1,2)}\left(\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)}\right)\\ &=R_{(2,3)}R_{(1,3)}R_{(1,2)}(u)\\ \end{align} $ Since $u\in M\otimes M\otimes M$ is arbitrary we conclude $ R_{(1,2)}R_{(1,3)}R_{(2,3)}=R_{(2,3)}R_{(1,3)}R_{(1,2)} $

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    @yoyostein, I think you are already familiar with tensor product since your question were asked quite a long ago2013-03-24