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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.

share|improve this question
    
What do you know about tensor products? –  Qiaochu Yuan May 8 '12 at 3:58
    
is it the same as "outer product", which produces a matrix from two vectors? –  yoyostein May 9 '12 at 4:50

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
    
@yoyostein, I think you are already familiar with tensor product since your question were asked quite a long ago –  userNaN Mar 24 '13 at 19:24

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