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:



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> 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?
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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](http://dedekind.mit.edu/%7Erstan/pubs/pubfiles/66.pdf) $\mathcal{O}(\prec)$.



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


