1
$\begingroup$

I'm studing $U(sl_q(2))$ and studying how to recover $U(sl(2))$ from $U(sl_q(2))$ I found these two definition for both $H$ and $K$ as formal generators. $$H=\frac{K-K^{-1}}{q-q^{-1}}$$ $$K=q^{H}$$ I'd like to know if they are coherent to each other and how could I derive the first one from the second one if needed? Thank you in advance

$\endgroup$
3
$\begingroup$

I think it's better to think of it as $K = q^h$ where $h$ is the usual generator of $sl_2$. If $V$ is a finite dimensional representation of $sl_2$ then $h$ acts with integer eigenvalues. If $h$ acts on a vector with eigenvalue $n$ then the operator $H = \frac{q^h - q^{-h}}{q-q^{-1}}$ acts on it by the "q-integer" $[n]_q = \frac{q^n - q^{-n}}{q-q^{-1}} = q^n + q^{n-2} + \dots + q^{-n}$. Then it's clear that if we specialize $q$ to $1$ then $H$ just acts by the same thing as $h$.

$\endgroup$

Your Answer

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

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