Coordinates on $N_+ \backslash \overline{B_+ w B_+} / N_+$

Question

Let $G = \text{GL}_n(\mathbb{C})$ and let $N_+$ be the subgroup of upper triangular matrices with $1$'s on the diagonal. Let $w$ be a permutation, let $B_+ w B_+$ be the Bruhat cell and let $\overline{B_+ w B_+}$ be its closure in $G$. I want to describe the ring of functions on $\overline{B_+ w B_+}$ which are invariant for $N_+ \times N_+$ acting on the left and right. I believe I know what the answer is, and I would like to know if I am right and get references to previous work.

Additional notation: Let $[n] := \{ 1,2,\ldots, n \}$. Let $T$ be the torus of diagonal matrices in $G$, and let $(t_1, t_2, \dots, t_n)$ be the entries of the diagonal matrix. Recall that the permutation matrix $w$ has $1$'s in position $(w(j), j)$.

Here is what I believe the answer to be. For any $I$, $J \subset [n]$ with $|I| = |J|$, and $g \in \text{GL}_n$, let $\Delta_{I,J}(g)$ be the minor with rows $I$ and columns $J$. Define a partial order $\prec$ on $[n]$ by $j_1 \prec j_2$ iff $j_1 < j_2$ and $w(j_1) > w(j_2)$. So, for $w= w_0$, this is the total order $1 \prec 2 \prec \cdots \prec n$ and, for $w = e$, all the elements of $[n]$ are incomparable.

Let $J$ be any lower order ideal for this partial order. Then one can verify that the minor $\Delta_{w(J), J}(g)$ is invariant for the $N_+ \times N_+$ action on $\overline{B_+ w B_+}$. We'll call this $\Delta(J)$. In addition, we have $\Delta([n]) = \Delta_{[n], [n]}(g)=\det(g)$, so $\Delta([n])$ is a unit and we have to include $\Delta([n])^{-1}$ in our ring of invariants. So, question:

Is the ring of $N_+ \times N_+$ invariant functions genearted by the $\Delta(J)$'s, and by $\Delta([n])^{-1}$? What is a reference for this?

I'll close by noting there is a good, explicit description of the ring generated by the $\Delta(J)$'s. If we restrict $\Delta(J)$ to $wT \subset B_+ w B_+$, we see that $\Delta(J) = \pm \prod_{j \in J} t_j$, so we can think of this as the monomial $\prod_{j \in J} t_j$ on $wT \cong N_+ \backslash B_+ w B_+ / N_+$. Since each orbit of $N_+ \times N_+$ on $B_+ w B_+$ contains a unique point in $wT$, the relations between the $\Delta(J)$'s are exactly the same as the relations between these monomials on $T$. Any product of the $\Delta(J)$'s gives a monomial of the form $\prod t_j^{a_j}$ where $j_1 \prec j_2$ implies $a_{j_1} \geq a_{j_2}$. In other words, the ring generated by the $\Delta(J)$'s is the semigroup ring corresponding to the semigroup of linear extensions of $([n], \prec)$, and is closely related to Stanley's order polyope $\mathcal{O}(\prec)$.

Answer 1

I have proved that the invariant ring is as I expected. As discussed in the question, $N_+ \backslash B_+ w B_+ / N_+ \cong T$, so the ring of $N_+ \times N_+$ invariants on $B_+ w B_+$ is the coordinate ring of $T$, which is the Laurent polynomial ring in the $t_i$.

Note, let $R$ be the ring of $N_+ \times N_+$ invariants on $\overline{B_+ w B_+}$, so $R$ is a subring of $\mathbb{C}[t_1^{\pm}, t_2^{\pm}, \ldots, t_n^{\pm}]$. Moreover, the $T$ action on $\overline{B_+ w B_+}$ normalizes $N_+ \times N_+$, so we get a $T$-action on $R$, so $R$ must be a $T$-invariant subspace of $\mathbb{C}[t_1^{\pm}, t_2^{\pm}, \ldots, t_n^{\pm}]$. So $R$ is a monomial subring of $\mathbb{C}[t_1^{\pm}, t_2^{\pm}, \ldots, t_n^{\pm}]$. We need to show that a monomial $t_1^{a_1} t_2^{a_2} \cdots t_n^{a_n}$ extends to a function on $\overline{B_+ w B_+}$ if and only if $j_1 \prec j_2$ implies $a_{j_1} \geq a_{j_2}$. The question shows that, if $a$ obeys these inequalities, then $t_1^{a_1} t_2^{a_2} \cdots t_n^{a_n}$ does extend to a function on $\overline{B_+ w B_+}$ , so what we need is the reverse implication.

So, suppose that there are $j_1$ and $j_2$ with $j_1 \prec j_2$ and $a_{j_1} < a_{j_2}$. We may assume that $(j_1, j_2)$ is a cover of $\prec$, meaning that there is no $j$ with $j_1 \prec j \prec j_2$. In this case, $w$ covers $w (j_1 j_2)$ in the strong Bruhat order, so $B_+ w (j_1 j_2) B_+$ is a divisor in $\overline{B_+ w B_+}$. We will show that the function $\prod t_j^{a_j}$ does not extend to $B_+ w (j_1 j_2) B_+$.

Put $i_1 = w(j_1)$, $i_2 = w(j_2)$. Consider matrices $X$ whose only nonzero entries are in positions $(w(j), j)$, together with $(i_1, j_2)$ and $(i_2, j_1)$. As long as $X_{i_1 j_1}$ is nonzero, this is in $B_+ w B_+$; when $X_{i_1 j_1}$ becomes $0$, we pass into $B_+ w (j_1 j_2) B_+$. We evaluate the functions $t_j$ on such a matrix: We have $t_j = X_{w(j) j}$ for $j \neq j_1$, $j_2$, we have $t_{j_1} = X_{i_1 j_1}$, and we have $t_{j_2} = \frac{X_{i_2 j_1} X_{i_1 j_2} - X_{i_1 j_1} X_{i_2 j_2}}{X_{i_1 j_1}}$.

We assumed that $a_{j_1} < a_{j_2}$. Plugging the above formulas into the product $\prod t_j^{a_j}$, we see that $X_{i_1 j_1}$ appears with exponent $a_{j_2} - a_{j_1}$ in the denominator. So, as $X_{i_1 j_1} \to 0$, the monomial $\prod t_j^{a_j}$ blows up, and thus doesn't extend to $B_+ w (j_1 j_2) w B_+$.

Still glad to hear references!