# How to write the map $ℂ[G/U]↪ℂ[B]$ explicitly?

Let $G$ be a reductive algebraic group and $B$ a Borel subgroup of $G$. Let $T$ be a maximal torus of $G$ contained in $B$. The $B=UT=TU$ for some unipotent subgroup $U$ of $G$. We have Bruhat decomposition $G = \cup_{w \in W} BwB = \cup_{w \in W} BwU$. Let $U$ act on $G$ by right multiplication. Then we have an embedding $B \hookrightarrow G/U$. Therefore there is an embedding $ℂ[G/U]↪ℂ[B]$.

My question is: how to write the map $ℂ[G/U]↪ℂ[B]$ explicitly? For example, let $G=GL_2$, $B$ the subgroup of all upper triangular matrices, and $U$ the subgroup consisting of all upper triangular matrices. Then we have $$ℂ[G/U]=ℂ[g_{11},g_{21},g_{11}g_{22}−g_{12}g_{21}]$$ and $$ℂ[B]=ℂ[b_{11},b_{12},b_{22}].$$ What are the images of $g_{11},g_{21},g_{11}g_{22}−g_{12}g_{21}$? Thank you very much.

First of all, for $GL(2)$ you have to add $(g_{11}g_{22}-g_{12}g_{21})^{-1}$ to your generators. Same for $B$: also $b_{11}^{-1}$ and $b_{22}^{-1}$ are generators.
The map $B\to G/U$ is $b\mapsto bw_0U$ where $w_0$ is the longest Weyl group element. The map on rings is just restriction. Thus $f(gU)$ is mapped to $f(bw_0)$. For $GL(2)$ this reads $$g_{11}\mapsto b_{12},g_{21}\mapsto b_{22},g_{11}g_{22}-g_{12}g_{21}\mapsto -b_{11}b_{22}$$ where I chose $w_0=\begin{pmatrix}0&1\\1&0\end{pmatrix}$.