Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $((x_1\otimes x_2)(x'_1\otimes x'_2))(x''_1\otimes x''_2) = (x_1\otimes x_2)((x'_1\otimes x'_2)(x''_1\otimes x''_2))$?

share|cite|improve this question
What exactly are you asking? – Mariano Suárez-Alvarez Apr 19 '11 at 17:15
up vote 4 down vote accepted

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):

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

share|cite|improve this answer
Nice answer. Thanks. – LJR Apr 19 '11 at 17:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.