The coproducts $\mathbb{C}_q[U] \to \mathbb{C}_q[U] \otimes \mathbb{C}_q[U]$ and $\mathbb{C}[U] \to \mathbb{C}[U] \otimes \mathbb{C}[U]$

Question

A coproduct $\varphi: \mathbb{C}_q[U] \to \mathbb{C}_q[U] \otimes \mathbb{C}_q[U]$ is given by: $x \mapsto 1 \otimes x + x \otimes 1$, where $x$ is a generator of $\mathbb{C}_q[U]$. There is a coproduct $\mathbb{C}[U] \to \mathbb{C}[U] \otimes \mathbb{C}[U]$ which is the pull-back of the multiplication map: $U \times U \to U$. The coproduct $\psi: \mathbb{C}[U] \to \mathbb{C}[U] \otimes \mathbb{C}[U]$ is given by: $x_{ik} \mapsto \sum_{j} x_{ij} \otimes x_{jk}$.

Are there some relation between the map $\varphi$ and the map $\psi$? Thank you very much.

Edit: $U$ is the group of unipotent upper triangular matrices.

Answer 1

The algebra $ \mathbb{C}[U] $ is a polynomial ring on generators $ x_{ij} $ where $ i < j $. The coproduct is given (as you say) by $ \psi(x_{ik})= \sum_j x_{ij} \otimes x_{jk} $ where where interpret $ x_{ij} = 0 $ if $ i > j $ and $ x_{ii} = 1 $. In particular, we see that $$ \psi(x_{i i+1})= 1 \otimes x_{i i+1} + x_{i i+1} \otimes 1. $$ Of course, these $ x_{i i+1} $ do not generate $ \mathbb{C}[U] $, since they are only some of the generators of this polynomial ring.

However, when we move to the quantization $ \mathbb{C}_q[U] $, then $ x_{i i+1} $ are now generators of this non-commutative algebra and as you say $ \phi(x_{i i+1}) = 1 \otimes x_{i i+1} + x_{i i+1} \otimes 1$. So we see that $ \phi $ and $ \psi $ do agree on these elements.