Orbits of $GL(n, \mathcal{O})$ on pairs of linear subspaces over non-Archimedean local fields

The $\text{GL}(n,\mathcal{O})$-orbits refine the $\text{GL}(n,F)$-orbits, i.e., the Bruhat cells. Up to replacing $i$ by $n-i$, assume that $2i\leq n.$ For every integer $m$ with $0\leq m\leq i,$ the Bruhat cell $U_m$ in $\text{Gr}_{i,n}\times \text{Gr}_{i,n}$ is the set of pairs $([V],[W])$ of $F$-vector subspaces of $F^{\oplus n},$ $V\cong F^{\oplus i}$ and $W\cong F^{\oplus i},$ such that $V\cap W$ is an $F$-vector subspace of codimension $m$ in both $V$ and in $W$. The Bruhat cell $U_m$ is an affine space over $F$ of dimenion $i(n-i)+m(n-m-i).$

For every finite, free $\mathcal{O}$-module, say $\mathcal{O}^{\oplus n}$, with its associated $F$-vector space, $\mathcal{O}^{\oplus n} \otimes_{\mathcal{O}} F = F^{\oplus n},$ for every $F$-vector subspace $V\subset F^{\oplus n},$ denote by $V_\mathcal{O}$ the intersection $V\cap \mathcal{O}^{\oplus n}.$

Lemma 1. The $\mathcal{O}$-submodule $V_{\mathcal{O}}$ of $\mathcal{O}^{\oplus n}$ is a finite, free $\mathcal{O}$-module that is a direct summand of $\mathcal{O}^{\oplus n}.$

Proof. Since $\mathcal{O}$ is a Noetherian ring, the $\mathcal{O}$-submodule $V_{\mathcal{O}}$ of $\mathcal{O}^{\oplus n}$ is a finitely generated $\mathcal{O}$-module. Since $V_\mathcal{O}$ is an $\mathcal{O}$-submodule of $\mathcal{O}^{\oplus n},$ it is also torsion-free. Since $\mathcal{O}$ is a DVR, also $V_{\mathcal{O}}$ is a finite, free $\mathcal{O}$-module. Finally, since $V_{\mathcal{O}}$ is a saturated $\mathcal{O}$-submodule of $\mathcal{O}^{\oplus n}$, also the cokernel $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}$ is also a finitely generated and torsion-free $\mathcal{O}$-module. Thus, also $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}$ is a finite, free $\mathcal{O}$-module. Altogether, this implies that $V_{\mathcal{O}}$ is a direct summand of $\mathcal{O}^{\oplus n}.$ QED

Denote by $\pi$ a uniformizing element of $\mathcal{O}.$

Lemma 2. For every pair of $F$-linear subspaces of $F^{\oplus n},$ $V\cong F^{\oplus \ell}$ and $W\cong F^{\oplus m},$ giving a direct sum decomposition, there exists a free basis $(b_1,\dots,b_m)$ of $W_{\mathcal{O}},$ and an ordered sequence of integers $0\leq e_1\leq \dots \leq e_m$ such that for every $i=1,\dots,m,$ the image of $b_i$ in $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}$ equals $\pi^{e_i}\cdot c_i$ for some $c_i,$ and $(c_1,\dots,c_m)$ is a free basis for $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}.$ The ordered sequence of integers is independent of the choice of basis.

Proof. This is part of the theory of finitely generated modules over a Principal Ideal Domain. The sequence of powers $(\pi^{e_1},\dots,\pi^{e_m})$ are the elementary divisors. QED

Denote $i-m$ by $r$. By Lemma 1, the quotient $\mathcal{O}$-module $(V+W)_{\mathcal{O}}/(V\cap W)_{\mathcal{O}}$ is a finite, free $\mathcal{O}$-module with associated $F$-vector space $(V+W)/(V\cap W).$ The $F$-vector subspaces, $V/(V\cap W)\cong F^{\oplus m}$ and $W/(V\cap W)\cong F^{\oplus m}$ give a direct sum decomposition of $(V+W)/(V\cap W).$ For every sequence $\underline{e}=(e_1,\dots,e_m)$ of nonnegative integers, denote by $U_{\underline{e},m}\subset U_m$ the set of pairs such that the elementary divisors of the $\mathcal{O}$-module homomorphism, $$(W/(V\cap W))_{\mathcal{O}} \to (V+W)_{\mathcal{O}}/V_{\mathcal{O}},$$ equal $(\pi^{e_1},\dots,\pi^{e_m}).$

Proposition. The group $\text{GL}(n,\mathcal{O})$ acts transitively on $U_{\underline{e},m}.$

Proof. By Lemma 1, $(V\cap W)_{\mathcal{O}}$ is a finite, free direct summand in $V_{\mathcal{O}},$ in $W_{\mathcal{O}},$ and in $\mathcal{O}^{\oplus n}.$ Similarly, $(V+W)_{\mathcal{O}}$ is a finite, free direct summand in $\mathcal{O}^{\oplus n}.$ Thus, there exists a free basis $(d_1,\dots,d_r)$ for $(V\cap W)_{\mathcal{O}},$ and there exists an extension $(d_1,\dots,d_r,a_1,\dots,a_m)$ to a free basis of $V_{\mathcal{O}}.$ By Lemma 2, there exists an extension $(d_1,\dots,d_r,a_1,\dots,a_m,c_1,\dots,c_m)$ to a free basis of $(V+W)_{\mathcal{O}}$ and there exists a free basis $(d_1,\dots,d_r,b_1,\dots,b_m)$ of $W_{\mathcal{O}}$ such that for the images in $(V+W)_{\mathcal{O}}/V_{\mathcal{O}}$, each $\overline{b}_i$ equals $\pi^{e_i}\overline{c}_i.$ Finally, there exists an extension $(d_1,\dots,d_r,a_1,\dots,a_m,c_1,\dots,c_m,g_1,\dots,g_{n-(2m+r)})$ to a free basis for $\mathcal{O}^{\oplus n}.$

For any other pair $(V',W')$ with associated bases $(d'_i),$ etc., there exists a unique $\mathcal{O}$-module isomorphism $V_{\mathcal{O}}\to V'_{\mathcal{O}}$ sending the free basis $(d_1,\dots,d_r,a_1,\dots,a_m)$ to $(d'_1,\dots,d'_r,a'_1,\dots,a'_m).$ There exists a unique $\mathcal{O}$-module isomorphism $(V+W)_{\mathcal{O}}/V_{\mathcal{O}}\to (V'+W')_{\mathcal{O}}/W'_{\mathcal{O}}$ sending $(\overline{c}_1,\dots,\overline{c}_m)$ to $(\overline{c}'_1,\dots,\overline{c}'_m).$ There is a unique lift of these moduli isomorphisms to an $F$-vector space isomorphism $V+W\to V'+W'$ that maps $(b_1,\dots,b_m)$ to $(b'_1,\dots,b'_m).$ Because the elementary divisors are equal, this $F$-vector space isomorphism maps the $\mathcal{O}$-module saturation $(V+W)_{\mathcal{O}}$ of $V_{\mathcal{O}}+W_{\mathcal{O}}$ to the $\mathcal{O}$-module saturation $(V'+W')_{\mathcal{O}}$ of $V'_{\mathcal{O}}+W'_{\mathcal{O}}.$ Finally, there is a unique extension to a $\mathcal{O}$-module isomorphism of $\mathcal{O}^{\oplus n}$ to $\mathcal{O}^{\oplus n}$ sending $(d_1,\dots,d_r,a_1,\dots,a_m,b_1,\dots,b_m,g_1,\dots,g_{n-(r+2m)})$ to $(d'_1,\dots,d'_r,a'_1,\dots,a'_m,b'_1,\dots,b'_m,g'_1,\dots,g'_{n-(r+2m)}).$ QED

Edit. As Uri Bader points out, the data of an integer $m$ satisfying $0\leq m \leq i$ together with the sequence of elementary divisors is equivalent to the data of the $\mathcal{O}$-module isomorphism class of the finitely generated $\mathcal{O}$-module, $$\mathcal{O}^{\oplus n}/(V_{\mathcal{O}}+W_{\mathcal{O}}).$$ So, as in Uri Bader's answer, a more intrinsic index for the $\text{GL}(n,\mathcal{O})$-orbits of $\text{Gr}_{i,n}\times \text{Gr}_{i,n}$ is an isomorphism class of an $\mathcal{O}$-module that is generated by $\leq n-i$ generators and whose free rank is at least $n-2i$.

I assume $2i\leq n$ (otherwise use duality and replace $\text{Gr}_{i,n}$ with $\text{Gr}_{n-i,n}$). I claim that the set of orbits is parametrized by $i$-tuples $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_i)$, where $\lambda_j\in \{0,1,2,\ldots,\infty\}$ and $\lambda_j\leq \lambda_{j+1}$ (for $2i\geq n$ it will be $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_{n-i})$).

For a commutative ring $R$ I use here the term $\text{Gr}_{i,n}(R)$ to denote the space of direct summands of $R^n$ which are isomorphic to $R^i$. Note that there is a natural (and $\text{GL}_n(\mathcal{O})$-equivariant) identification of $\text{Gr}_{i,n}(F)$ with $\text{Gr}_{i,n}(\mathcal{O})$, thus we may forget about $F$ and focus on $\text{Gr}_{i,n}(\mathcal{O})$.

Let $p<\mathcal{O}$ be the maximal ideal. Note that two pairs of modules in $\text{Gr}_{i,n}(\mathcal{O})$ are in the same $\text{GL}_n(\mathcal{O})$-orbit iff their reductions mod $p^k$ are in the same $\text{GL}_n(\mathcal{O}/p^k)$-orbit for every natural $k$.

Fix $k$ and consider $\mathcal{O}/p^k$-modules. You can check that two pairs $(x,y),(x',y') \in \text{Gr}_{i,n}(\mathcal{O}/p^k) \times \text{Gr}_{i,n}(\mathcal{O}/p^k)$ are in the same $\text{GL}_n(\mathcal{O}/p^k)$-orbit iff $x\cap y \simeq x' \cap y'$. Up to an isomorphism, the intersection module $x\cap y$ could be an arbitrary $\mathcal{O}/p^k$-module of rank bounded by $i$ (here we use $2i\leq n$). The isomorphism types of such modules are parametrized by $i$-tuples $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_i)$, where $\lambda_j\in \{0,1,2,\ldots, k\}$ and $\lambda_j\leq \lambda_{j+1}$, where a module of type $\lambda$ is isomorphic to $\oplus_j \mathcal{O}/p^{\lambda_j}$.

A discussion and justifications are given in claim 2.2 here.

Now, for $(x,y)\in \text{Gr}_{i,n}(\mathcal{O})\times \text{Gr}_{i,n}(\mathcal{O})$, consider the limit of the tuples $\lambda$ associated with their reduction mod $p^k$ where $k$ tends to $\infty$.