Are these 2 definitions of $K$ and $H$ on $U(sl_q(2))$ coherent?

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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.
$$H=fracK-K^-1q-q^-1$$
$$K=q^H$$
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




abstract-algebra complex-analysis representation-theory hopf-algebras quantum-groups




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    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.
    $$H=fracK-K^-1q-q^-1$$
    $$K=q^H$$
    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




    abstract-algebra complex-analysis representation-theory hopf-algebras quantum-groups




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      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.
      $$H=fracK-K^-1q-q^-1$$
      $$K=q^H$$
      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




      abstract-algebra complex-analysis representation-theory hopf-algebras quantum-groups




      share|cite|improve this question











      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.
      $$H=fracK-K^-1q-q^-1$$
      $$K=q^H$$
      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





      abstract-algebra complex-analysis representation-theory hopf-algebras quantum-groups





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      asked Jul 16 at 16:41









      Dac0

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          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 $H = fracq^h - q^-hq-q^-1$ acts on it by the "q-integer" $[n]_q = fracq^n - q^-nq-q^-1 = q^n + q^n-2 + dots + q^-n$. Then it's clear that if we specialize $q$ to $1$ then $H$ just acts by the same thing as $h$.






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            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 $H = fracq^h - q^-hq-q^-1$ acts on it by the "q-integer" $[n]_q = fracq^n - q^-nq-q^-1 = q^n + q^n-2 + dots + q^-n$. Then it's clear that if we specialize $q$ to $1$ then $H$ just acts by the same thing as $h$.






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              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 $H = fracq^h - q^-hq-q^-1$ acts on it by the "q-integer" $[n]_q = fracq^n - q^-nq-q^-1 = q^n + q^n-2 + dots + q^-n$. Then it's clear that if we specialize $q$ to $1$ then $H$ just acts by the same thing as $h$.






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                up vote
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                up vote
                3
                down vote









                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 $H = fracq^h - q^-hq-q^-1$ acts on it by the "q-integer" $[n]_q = fracq^n - q^-nq-q^-1 = q^n + q^n-2 + dots + q^-n$. Then it's clear that if we specialize $q$ to $1$ then $H$ just acts by the same thing as $h$.






                share|cite|improve this answer













                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 $H = fracq^h - q^-hq-q^-1$ acts on it by the "q-integer" $[n]_q = fracq^n - q^-nq-q^-1 = q^n + q^n-2 + dots + q^-n$. Then it's clear that if we specialize $q$ to $1$ then $H$ just acts by the same thing as $h$.







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                answered Jul 16 at 18:08









                Nate

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