3
$\begingroup$

I've seen two defining relation for $U_q(\mathfrak{sl}_2)$ by the Serre relations $$[H,E]=E,\quad[H,F]=-F, \quad [E,F]=\frac{q^H-q^{-H}}{q-q^{-1}}, $$ or by taking $K=q^H$ $$KK^{-1}=K^{-1}K=1,\quad [E,F]=\frac{K-K^{-1}}{q-q^{-1}},\\ KE=q^2EK,\quad q^2KF=FK$$ I have two problems:

  1. In the second set of defining relations, $KK^{-1}=K^{-1}K=1$ seems trivial, why do we need them?
  2. When I try to get the first set of relation from the second one, by taking $K=q^H$, I anly get the right equation in first order in $q$. Is that right?
$\endgroup$
3
  • 1
    $\begingroup$ Do you have a reference for the first presentation? I can't find it in any of the standard texts that I have beside me (Kassel, Jantzen, Klimyk and Schmudgen, Brown and Goodearl), but maybe I've just missed it. $\endgroup$
    – Casteels
    Mar 31, 2015 at 10:10
  • 1
    $\begingroup$ I cannot answer the second question, but for the first: $K^{-1}$ is just a name. You could call it $K'$; then $U_q(\mathfrak{sl}_2)$ is the algebra generated by variables $E,F,K,K'$ and subject to relations $KK' = 1 = K'K, [E,F] = \dots$ etc. It then follows that $K'$ is the multiplicative inverse of $K$, so you may as well call it $K^{-1}$ from the start. $\endgroup$ Apr 11, 2015 at 11:11
  • $\begingroup$ 2. No, you made a mistake. Show your work. $\endgroup$ Jun 16, 2018 at 15:20

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.