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The thing is that in the infinite dimensional case there is no such thing as "the" double: for any Hopf algebra $K$, given a non-degenerate Hopf pairing on $H\times K$ you can define a double $D(H,K)$.

A natural choice in that case is to take $K$ to be the full dual of $H$: this is a so-called Quantum Formal Series Hopf Algebra (QFSHA), a quantization of the formal Poisson group associated with $\mathfrak g$. This gives I think the double of Klimyk-Schmiidgen.

Now by the so-called quantum duality principle (https://arxiv.org/abs/math/9909071) there is a functor $K\mapsto K^\vee$ from QFSHA's to QUEA's given by, roughly speaking, dividing the generators of $K$ by $\hbar$. $K^\vee$ is then a quantization of the Lie bialgebra $\mathfrak g^*$. The pairing should extend to $K^{\vee}$ and the double in that case should be a QUEA quantizing the double of $\mathfrak g$. This I think gives the algebra with the same relations as the one you gave but without those extra $\hbar$.

The thing is that in the infinite dimensional case there is no such thing as "the" double: for any Hopf algebra $K$, given a non-degenerate Hopf pairing on $H\times K$ you can define a double $D(H,K)$.

A natural choice in that case is to take $K$ to be the full dual of $H$: this is a so-called Quantum Formal Series Hopf Algebra (QFSHA), a quantization of the formal Poisson group associated with $\mathfrak g$. This gives I think the double of Klimyk-Schmiidgen.

Now by the so-called quantum duality principle (https://arxiv.org/abs/math/9909071) there is a functor $K\mapsto K^\vee$ from QFSHA's to QUEA's given by, roughly speaking, dividing the generators of $K$ by $\hbar$. $K^\vee$ is then a quantization of the Lie bialgebra $\mathfrak g^*$. The pairing should extend to $K^{\vee}$ and the double in that case should be a QUEA quantizing the double of $\mathfrak g$. This I think gives the algebra with the same relations as the one you gave but without those extra $\hbar$. This is spelled out in section 4.4 of https://arxiv.org/abs/q-alg/9506005.

The thing is that in the infinite dimensional case there is no such thing as "the" double: for any Hopf algebra $K$, given a non-degenerate Hopf pairing on $H\times K$ you can define a double $D(H,K)$.

A natural choice in that case is to take $K$ to be the full dual of $H$: this is a so-called Quantum Formal Series Hopf Algebra (QFSHA), a quantization of the formal group associated with $\mathfrak g$. This gives I think the double of Klimyk-Schmiidgen.

Now by the so-called quantum duality principle (https://arxiv.org/abs/math/9909071) there is a functor $K\mapsto K^\vee$ from QFSHA's to QUEA's given by, roughly speaking, dividing the generators of $K$ by $\hbar$. $K^\vee$ is then a quantization of the Lie bialgebra $\mathfrak g^*$. The pairing should extend to $K^{\vee}$ and the double in that case should be a QUEA quantizing the double of $\mathfrak g$. This I think gives the algebra with the same relations as the one you gave but without thos extra $\hbar$.

The thing is that in the infinite dimensional case there is no such thing as "the" double: for any Hopf algebra $K$, given a non-degenerate Hopf pairing on $H\times K$ you can define a double $D(H,K)$.

A natural choice in that case is to take $K$ to be the full dual of $H$: this is a so-called Quantum Formal Series Hopf Algebra (QFSHA), a quantization of the formal Poisson group associated with $\mathfrak g$. This gives I think the double of Klimyk-Schmiidgen.

Now by the so-called quantum duality principle (https://arxiv.org/abs/math/9909071) there is a functor $K\mapsto K^\vee$ from QFSHA's to QUEA's given by, roughly speaking, dividing the generators of $K$ by $\hbar$. $K^\vee$ is then a quantization of the Lie bialgebra $\mathfrak g^*$. The pairing should extend to $K^{\vee}$ and the double in that case should be a QUEA quantizing the double of $\mathfrak g$. This I think gives the algebra with the same relations as the one you gave but without those extra $\hbar$.

The thing is that in the infinite dimensional case there is no such thing as "the" double: for any Hopf algebra $K$, given a non-degenerate Hopf pairing on $H\times K$ you can define a double $D(H,K)$.

A natural choice in that case is to take $K$ to be the full dual of $H$: this is a so-called Quantum Formal Series Hopf Algebra (QFSHA), a quantization of the formal group associated with $\mathfrak g$. This gives I think the double of Klimyk-Schmiidgen.

Now by the so-called quantum duality principle (https://arxiv.org/abs/math/9909071) there is a functor $K\mapsto K^\vee$ from QFSHA's to QUEA's given by, roughly speaking, dividing the generators of $K$ by $\hbar$. $K^\vee$ is then a quantization of the Lie bialgebra $\mathfrak g^*$. The pairing should extend to $K^{\vee}$ and the double in that case should be a QUEA quantizing the double of $\mathfrak g$.

The thing is that in the infinite dimensional case there is no such thing as "the" double: for any Hopf algebra $K$, given a non-degenerate Hopf pairing on $H\times K$ you can define a double $D(H,K)$.

A natural choice in that case is to take $K$ to be the full dual of $H$: this is a so-called Quantum Formal Series Hopf Algebra (QFSHA), a quantization of the formal group associated with $\mathfrak g$. This gives I think the double of Klimyk-Schmiidgen.

Now by the so-called quantum duality principle (https://arxiv.org/abs/math/9909071) there is a functor $K\mapsto K^\vee$ from QFSHA's to QUEA's given by, roughly speaking, dividing the generators of $K$ by $\hbar$. $K^\vee$ is then a quantization of the Lie bialgebra $\mathfrak g^*$. The pairing should extend to $K^{\vee}$ and the double in that case should be a QUEA quantizing the double of $\mathfrak g$. This I think gives the algebra with the same relations as the one you gave but without thos extra $\hbar$.