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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asked Jul 16 at 16:41

Dac0

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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

asked Jul 16 at 16:41

Dac0

5,7041734

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

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

asked Jul 16 at 16:41

Dac0

5,7041734

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

asked Jul 16 at 16:41

Dac0

5,7041734

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 16:41

Dac0

5,7041734

asked Jul 16 at 16:41

Dac0

5,7041734

asked Jul 16 at 16:41

Dac0

5,7041734

5,7041734

add a comment | 

add a comment | 

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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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answered Jul 16 at 18:08

Nate

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1 Answer
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1 Answer
1

active

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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

answered Jul 16 at 18:08

Nate

6,57011020

add a comment | 

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

answered Jul 16 at 18:08

Nate

6,57011020

add a comment | 

up vote
3
down vote

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

answered Jul 16 at 18:08

Nate

6,57011020

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

answered Jul 16 at 18:08

Nate

6,57011020

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 18:08

Nate

6,57011020

answered Jul 16 at 18:08

Nate

6,57011020

answered Jul 16 at 18:08

Nate

6,57011020

6,57011020

add a comment | 

add a comment | 

 

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