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Let $G$ be an algebraic group and $N$ its maximal unipotent subgroup consisting of all upper triangular unipotent matrices. Let $\mathfrak{n}$ be the Lie algebra of $N$. It is said that $\mathbb{C}_q[N]$ is isomorphic to $U_q(\mathfrak{n})$? Here $\mathbb{C}_q[N]$ is the quantum coordinate ring of $N$ and $U_q(\mathfrak{n})$ is the quantized universal enveloping algebra of $\mathfrak{n}$.

Let $\mathbb{C}[N]$ be the coordinate ring of $N$ and $U(\mathfrak{n})$ the universal enveloping algebra of $\mathfrak{n}$. Is $\mathbb{C}[N]$ isomorphic to $U(\mathfrak{n})$? Thank you very much.

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Suppose that $N$ is abelian. For instance, one could take $N$ to be the unipotent radical of a maximal parabolic subgroup of $\operatorname{GL}_n(\mathbb{C})$ with $n \geq 2$. In this case, the Lie algebra $\mathfrak{n}$ is abelian as well, and the exponential map $\exp : \mathfrak{n} \to N$ is an isomorphism of commutative algebraic groups ($\mathfrak{n}$ being considered as an additive group). On the other hand, in this case $U(\mathfrak{n})$ is canonically isomorphic with the symmetric algebra $\operatorname{S}(\mathfrak{n})$ of $\mathfrak{n}$, which in turn is canonically isomorphic with the polynomial ring $\mathbb{C}[\mathfrak{n}^*]$, where $\mathfrak{n}^*$ is the dual vector space of $\mathfrak{n}$. Identifying $\mathfrak{n}$ with $\mathfrak{n^*}$ in some (non-canonical) manner, one obtains an isomorphism $U(\mathfrak{n}) \simeq \mathbb{C}[\mathfrak{n}] \simeq \mathbb{C}[N]$.

If $N$ is not abelian (e.g. if $N$ is the unipotent radical of a Borel subgroup of $\operatorname{GL}_n(\mathbb{C})$ for $n\geq3$), then $U(\mathfrak{n})$ is a non-commutative associative algebra, so it cannot be isomorphic to the coordinate ring $\mathbb{C}[N]$. On the other hand, for any algebraic group $G$, one can identify $U(\mathfrak{g})$ with the algebra of distributions on $G$ supported at the identity, or with the algebra of left-inavriant linear differential operators (of any order) on $G$.

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