Ideal of the boundary of $G/U \subset \overline{G/U}$

Question

Let $G$ be a semi simple algebraic group, $B \subset G$ is a Borel subgroup and $U \subset B$ is the unipotent radical of $B$. We can consider the variety $G/U$. Let us also denote $\overline{G/U}:=\operatorname{Spec}(\mathbb{C}[G/U])$. It is known that the natural morphism $G/U \rightarrow \overline{G/U}$ is an open embedding. Let $\partial{G/U}$ be the boundary of $G/U$ inside $\overline{G/U}$. Note now that $\mathbb{C}[G/U]=\bigoplus_{\mu} V(\mu)$, where the sum runs through dominant characters $\mu$ of $G$ (we fix some maximal torus $T \subset B$, here $V(\mu)$ is the irreducible representation of $G$ with highest weight $\mu$).

Claim: the ideal of $\partial{G/U} \subset \overline{G/U}$ is generated by $V(\mu)$ with $\mu$ being regular (strictly dominant). How to prove this claim? Maybe there are any references?

Answer 1

Here is one way to see it, via classifying $G$-invariant radical ideals. (This has the bonus that it implicitly describes the boundary.)

Lemma: $G$-invariant ideals $I$ of $\mathbb{C}[G/U]$ are in bijection with sets of weights $S$ so that for $\lambda\in S$ and $\mu > \lambda$, $\mu\in S$. Such an ideal is radical iff for all $\lambda\notin S,$ we have $n\lambda\notin S$ for all positive integers $n$.

To see this, note that $G$-invariance tells you that $I$ must split as a sum $$\displaystyle\bigoplus_{\lambda\in S}V(\lambda)$$ for some set $S$. Now if $\lambda\in S,$ the multiplication map $V(\mu-\lambda)\otimes V(\lambda)\rightarrow V(\mu)$ is surjective and hence $\mu > \lambda$ must also be in $S$.

The statement about radical ideals follows similarly.

From this statement, you can see that the minimal nonzero $G$-invariant radical ideal (which necessarily cuts out the boundary) corresponds to taking $S$ the set of all regular weights.