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I am reading Lusztig's book Introduction to quantum groups. I have a question on page 3. In the fourth line of section 1.2.2, it is said that 'f \otimes 'f is associative. I don't know why.

I think that ((x_1\otimes x_2)(x'_1\otimes x'_2))(x''_1\otimes x''_2) = v^{|x_2||x'_1|}(x_1x'_1\otimes x_2x'_2) \otimes (x''_1\otimes x''_2) = v^{|x_2||x'_1|+|x_2x'_2||x''_1|}x_1x'_1x''_1\otimes x_2x'_2x''_2. But it seems that (x_1\otimes x_2)((x'_1\otimes x'_2)(x''_1\otimes x''_2)) does not equal this. Why ((x_1\otimes x_2)(x'_1\otimes x'_2))(x''_1\otimes x''_2) = (x_1\otimes x_2)((x'_1\otimes x'_2)(x''_1\otimes x''_2))?

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    What exactly are you asking?2011-04-19

1 Answers 1

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Your exponent |x_2|\cdot|x_1'|+|x_2x_2'|\cdot|x_1''| is correct. The exponent for the other grouping is |x_2'|\cdot|x_1''|+|x_2|\cdot|x_1'x_1''|. To show that these are equal, you need the two facts that the product is bilinear and that $|xy|=|x|+|y|$ (corresponding to '\mathbf f_\nu'\mathbf f_{\nu'}\subset'\mathbf f_{\nu+\nu'} at the bottom of page 2):

\begin{eqnarray} && |x_2|\cdot|x_1'|+|x_2x_2'|\cdot|x_1''| \\ &=& |x_2|\cdot|x_1'|+(|x_2|+|x_2'|)\cdot|x_1''| \\ &=& |x_2|\cdot|x_1'|+|x_2|\cdot|x_1''|+|x_2'|\cdot|x_1''| \\ &=& |x_2'|\cdot|x_1''|+|x_2|\cdot|x_1'|+|x_2|\cdot|x_1''| \\ &=& |x_2'|\cdot|x_1''|+|x_2|\cdot(|x_1'|+|x_1''|) \\ &=& |x_2'|\cdot|x_1''|+|x_2|\cdot|x_1'x_1''|\;. \\ \end{eqnarray}

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    Nice answer. Thanks.2011-04-19