Yes! It is indeed, to get the quantum group $\mathcal{U}_q(\mathfrak{g})$, you have to...

• start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{Alg}$ representing the Cartan algebra $\mathcal{U}^0(\mathfrak{g})$, where $n$ is the rank of $\mathfrak{g}$

• consider the vectorspace $M=\langle E_1\ldots E_n\rangle_{Vect}$ for the simple roots $\alpha_1\ldots \alpha_n$ with

• an action of $\mathbb{k}[\Gamma]$, namely $K_i\otimes E_j\mapsto q^{(\alpha_i,\alpha_j)}E_j=:q_{ij}E_j$

• a graduation/coaction to $\mathbb{k}[\Gamma]$, namely $E_i\mapsto K_i\otimes E_i$

• therefore a braiding $E_i\otimes E_j\mapsto q_{ij}E_j\otimes E_i$

• form the tensor algebra $\mathcal{T}M$ modulo the Serre relations for the Cartan matrix a_{ij} $$ad_{E_i}^{1-a_{ij}}E_j=0$$ with the braided adjoint action resp. commutator $$ad_{E_i}(E_j)=[E_i,E_j]:=EiE_j-q_{ij}E_jE_i$$ This is the Borel part $\mathcal{U}_q(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^+)$, in this stage only a so-called braided Hopf algebra over $\mathbb{k}[\Gamma]$ with $\Delta(E_i)=1\otimes E_i + E_i\otimes 1$

• glue Borel part and Cartan algebra to a Radford biproduct $$\mathcal{U}^{\geq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})\ltimes\mathcal{U}^+(\mathfrak{g})$$ Especially the action and coaction of $\mathbb{k}[\Gamma]$ on $M$ automatically yield the relations $$K_iE_j=q_{ij}E_jK_j$$ $$\Delta(E_i)=K_i\otimes E_i\otimes 1$$
• Form the generalized Drinfel'd double, which means...
• Do the dual construction with $M^*=\langle F_1,\ldots F_n\rangle$ with action $K_i\otimes F_j\mapsto q^{-(\alpha_i,\alpha_j)}F_j$ to yield another Borel part $\mathcal{U}^-(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^-)$ and another Radford biproduct $$\mathcal{U}^{\leq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})'\ltimes\mathcal{U}^-(\mathfrak{g})$$
• Quotient out an identification of both Cartan algebras $$U_q(\mathfrak{g})=U_q^{\geq}(\mathfrak{g})\otimes U_q^{\leq}(\mathfrak{g})/(U^0_q(\mathfrak{g})= U^0_q(\mathfrak{g})')$$ (the last step is called linking can nowadays be done via a Doi twist)

LITERATURE: e.g. Heckenberger: Nichols Algebras (Lecture Notes), 2008 (http://www.mi.uni-koeln.de/~iheckenb/na.pdf) page 35ff.

PS: This works equivalently for the truncated quantum groups. Here, the role of the Borel part $\mathcal{T}M/Serre$ is taken by a quotient Nichols algebra $\mathfrak{B}(M)$. There also appear some exotic example associated to Dynkin diagrams, that are impossible for semisimple Lie algebras (e.g. a certain triangle). See also the Wikipedia page "Nichols algebras" for links to more papers.

Yes! It is indeed, to get the quantum group $\mathcal{U}_q(\mathfrak{g})$, you have to...

• start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{Alg}$ representing the Cartan algebra $\mathcal{U}^0(\mathfrak{g})$, where $n$ is the rank of $\mathfrak{g}$

• consider the vectorspace $M=\langle E_1\ldots E_n\rangle_{Vect}$ for the simple roots $\alpha_1\ldots \alpha_n$ with

• an action of $\mathbb{k}[\Gamma]$, namely $K_i\otimes E_j\mapsto q^{(\alpha_i,\alpha_j)}E_j=:q_{ij}E_j$

• a graduation/coaction to $\mathbb{k}[\Gamma]$, namely $E_i\mapsto K_i\otimes E_i$

• therefore a braiding $E_i\otimes E_j\mapsto q_{ij}E_j\otimes E_i$

• form the tensor algebra $\mathcal{T}M$ modulo the Serre relations for the Cartan matrix a_{ij} $$ad_{E_i}^{1-a_{ij}}E_j=0$$ with the braided adjoint action resp. commutator $$ad_{E_i}(E_j)=[E_i,E_j]:=EiE_j-q_{ij}E_jE_i$$ This is the Borel part $\mathcal{U}_q(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^+)$, in this stage only a so-called braided Hopf algebra over $\mathbb{k}[\Gamma]$ with $\Delta(E_i)=1\otimes E_i + E_i\otimes 1$

• glue Borel part and Cartan algebra to a Radford biproduct $$\mathcal{U}^{\geq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})\ltimes\mathcal{U}^+(\mathfrak{g})$$ Especially the action and coaction of $\mathbb{k}[\Gamma]$ on $M$ automatically yield the relations $$K_iE_j=q_{ij}E_jK_j$$ $$\Delta(E_i)=K_i\otimes E_i\otimes 1$$
• Form the generalized Drinfel'd double, which means...
• Do the dual construction with $M^*=\langle F_1,\ldots F_n\rangle$ with action $K_i\otimes F_j\mapsto q^{-(\alpha_i,\alpha_j)}F_j$ to yield another Borel part $\mathcal{U}^-(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^-)$ and another Radford biproduct $$\mathcal{U}^{\leq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})'\ltimes\mathcal{U}^-(\mathfrak{g})$$
• Quotient out an identification of both Cartan algebras $$U_q(\mathfrak{g})=(U_q^{\geq}(\mathfrak{g})\otimes U_q^{\leq}(\mathfrak{g}))^{\sigma}/(U^0_q(\mathfrak{g})= U^0_q(\mathfrak{g})')$$ the last step is called linking can nowadays (by A. Masuoka) be done via a Doi twist with a 2-cocycle, which causes the additional relations $$E_i F_i-F_i E_i=\frac{K_i-K_i^{-1}}{q_{ii}-q_{ii}^{-1}}$$

LITERATURE: e.g. Heckenberger: Nichols Algebras (Lecture Notes), 2008 (http://www.mi.uni-koeln.de/~iheckenb/na.pdf) page 35ff.

PS: This works equivalently for the truncated quantum groups. Here, the role of the Borel part $\mathcal{T}M/Serre$ is taken by a quotient Nichols algebra $\mathfrak{B}(M)$. There also appear some exotic example associated to Dynkin diagrams, that are impossible for semisimple Lie algebras (e.g. a certain triangle). See also the Wikipedia page "Nichols algebras" for links to more papers.

Yes! It is indeed, to get the quantum group $\mathcal{U}_q(\mathfrak{g})$, you have to...

• start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{Alg}$ representing the Cartan algebra $\mathcal{U}^0(\mathfrak{g})$, where $n$ is the rank of $\mathfrak{g}$

• consider the vectorspace $M=\langle E_1\ldots E_n\rangle_{Vect}$ for the simple roots $\alpha_1\ldots \alpha_n$ with

• an action of $\mathbb{k}[\Gamma]$, namely $K_i\otimes E_j\mapsto q^{(\alpha_i,\alpha_j)}E_j=:q_{ij}E_j$

• a graduation/coaction to $\mathbb{k}[\Gamma]$, namely $E_i\mapsto K_i\otimes E_i$

• therefore a braiding $E_i\otimes E_j\mapsto q_{ij}E_j\otimes E_i$

• form the tensor algebra $\mathcal{T}M$ modulo the Serre relations for the Cartan matrix a_{ij} $$ad_{E_i}^{1-a_{ij}}E_j=0$$ with the braided adjoint action resp. commutator $$ad_{E_i}(E_j)=[E_i,E_j]:=EiE_j-q_{ij}E_jE_i$$ This is the Borel part $\mathcal{U}_q(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^+)$, in this stage only a so-called braided Hopf algebra over $\mathbb{k}[\Gamma]$ with $\Delta(E_i)=1\otimes E_i + E_i\otimes 1$

• glue Borel part and Cartan algebra to a Radford biproduct $$\mathcal{U}^{\geq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})\ltimes\mathcal{U}^+(\mathfrak{g})$$ Especially the action and coaction of $\mathbb{k}[\Gamma]$ on $M$ automatically yield the relations $$K_iE_j=q_{ij}E_jK_j$$ $$\Delta(E_i)=K_i\otimes E_i\otimes 1$$
• For the generalized Drinfel'd double, which means...
• Do the dual construction with $M^*=\langle F_1,\ldots F_n\rangle$ with action $K_i\otimes F_j\mapsto q^{-(\alpha_i,\alpha_j)}F_j$ to yield another Borel part $\mathcal{U}^-(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^-)$ and another Radford biproduct $$\mathcal{U}^{\leq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})'\ltimes\mathcal{U}^-(\mathfrak{g})$$
• Quotient out an identification of both Cartan algebras $$U_q(\mathfrak{g})=U_q^{\geq}(\mathfrak{g})\otimes U_q^{\leq}(\mathfrak{g})/(U^0_q(\mathfrak{g})= U^0_q(\mathfrak{g})')$$ (the last step is called linking can nowadays be done via a Doi twist)

LITERATURE: e.g. Heckenberger: Nichols Algebras (Lecture Notes), 2008 (http://www.mi.uni-koeln.de/~iheckenb/na.pdf) page 35ff.

PS: This works equivalently for the truncated quantum groups. Here, the role of the Borel part $\mathcal{T}M/Serre$ is taken by a quotient Nichols algebra $\mathfrak{B}(M)$. There also appear some exotic example associated to Dynkin diagrams, that are impossible for semisimple Lie algebras (e.g. a certain triangle). See also the Wikipedia page "Nichols algebras" for links to more papers.

Yes! It is indeed, to get the quantum group $\mathcal{U}_q(\mathfrak{g})$, you have to...

• start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{Alg}$ representing the Cartan algebra $\mathcal{U}^0(\mathfrak{g})$, where $n$ is the rank of $\mathfrak{g}$

• consider the vectorspace $M=\langle E_1\ldots E_n\rangle_{Vect}$ for the simple roots $\alpha_1\ldots \alpha_n$ with

• an action of $\mathbb{k}[\Gamma]$, namely $K_i\otimes E_j\mapsto q^{(\alpha_i,\alpha_j)}E_j=:q_{ij}E_j$

• a graduation/coaction to $\mathbb{k}[\Gamma]$, namely $E_i\mapsto K_i\otimes E_i$

• therefore a braiding $E_i\otimes E_j\mapsto q_{ij}E_j\otimes E_i$

• form the tensor algebra $\mathcal{T}M$ modulo the Serre relations for the Cartan matrix a_{ij} $$ad_{E_i}^{1-a_{ij}}E_j=0$$ with the braided adjoint action resp. commutator $$ad_{E_i}(E_j)=[E_i,E_j]:=EiE_j-q_{ij}E_jE_i$$ This is the Borel part $\mathcal{U}_q(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^+)$, in this stage only a so-called braided Hopf algebra over $\mathbb{k}[\Gamma]$ with $\Delta(E_i)=1\otimes E_i + E_i\otimes 1$

• glue Borel part and Cartan algebra to a Radford biproduct $$\mathcal{U}^{\geq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})\ltimes\mathcal{U}^+(\mathfrak{g})$$ Especially the action and coaction of $\mathbb{k}[\Gamma]$ on $M$ automatically yield the relations $$K_iE_j=q_{ij}E_jK_j$$ $$\Delta(E_i)=K_i\otimes E_i\otimes 1$$
• Form the generalized Drinfel'd double, which means...
• Do the dual construction with $M^*=\langle F_1,\ldots F_n\rangle$ with action $K_i\otimes F_j\mapsto q^{-(\alpha_i,\alpha_j)}F_j$ to yield another Borel part $\mathcal{U}^-(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^-)$ and another Radford biproduct $$\mathcal{U}^{\leq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})'\ltimes\mathcal{U}^-(\mathfrak{g})$$
• Quotient out an identification of both Cartan algebras $$U_q(\mathfrak{g})=U_q^{\geq}(\mathfrak{g})\otimes U_q^{\leq}(\mathfrak{g})/(U^0_q(\mathfrak{g})= U^0_q(\mathfrak{g})')$$ (the last step is called linking can nowadays be done via a Doi twist)

LITERATURE: e.g. Heckenberger: Nichols Algebras (Lecture Notes), 2008 (http://www.mi.uni-koeln.de/~iheckenb/na.pdf) page 35ff.

PS: This works equivalently for the truncated quantum groups. Here, the role of the Borel part $\mathcal{T}M/Serre$ is taken by a quotient Nichols algebra $\mathfrak{B}(M)$. There also appear some exotic example associated to Dynkin diagrams, that are impossible for semisimple Lie algebras (e.g. a certain triangle). See also the Wikipedia page "Nichols algebras" for links to more papers.