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? – Mariano Suárez-Alvarez Apr 19 '11 at 17:15

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. – LJR Apr 19 '11 at 17:23