Let $W = \prod_{k \in \mathbb N} V_k$ be an infinite product of vector spaces, and let $V = \oplus_{k \in \mathbb N} V_k$ be the corresponding sum. Already the case where $V_k$ is 1-dimensional for each $k$ seems interesting.
Let $\mathcal B \subset GL(W)$ be the "lower-triangular" group, i.e. $B \in \mathcal B$ iff $B$ restricts to an automorphism of $W_n = \prod_{k \geq n} V_k$ for each $n$ (or equivalently, $B$ induces an automorphism of $W/W_n$ for each $n$).
Let $\mathcal S$ be the set of subspaces $S \subseteq W$ which are "full" in the sense that $S / (S \cap W_n) \cong W/W_n$ for each $n$ (via the canonical map). $\mathcal B$ acts naturally on $\mathcal S$.
If we were dealing with a finite sum / product, then $V = W$ and the only full subspace would be $W$ itself. As it is, a subspace $S \subseteq W$ is full if and only if some $\mathcal B$-conjugate of $S$ contains $V$ (where $V$ is regarded as a subspace of $W$ in the obvious way).
Question: How might one classify full subspaces of $W$ up to $\mathcal B$-conjugacy? That is,
When are two full subspaces $S,S' \subseteq W$ conjugate by $\mathcal B$?
Does the orbit set $\mathcal S / \mathcal B$ have some nicer description?
Note the following cardinality considerations: if the field is finite and each $V_k$ is 1-dimensional, then $|\mathcal B| = 2^{\aleph_0}$ whereas $|\mathcal S| = 2^{2^{\aleph_0}}$, so that $|\mathcal S / \mathcal B| = 2^{2^{\aleph_0}}$.
Motivation 1: This might be seen as a first attempt to understand the difference between linear algebra in finite and infinite dimensions: if the product were finite, then the only invariant of a subspace $S \subseteq W$ up to $\mathcal B$-conjugacy would be the sequence of dimensions $\dim(S \cap W_n)_n$, but the "fullness" condition trivializes these invariants.
Motivation 2: When the base field is $\mathbb F_p$ for a prime $p$, the data of such a product of vector spaces $W = \prod_k V_k$ and a full subspace $S \subseteq W$ (up to $\mathcal B$-conjugacy) classifies (up to isomorphism) a $p$-complete group of Ulm rank $\omega+1$ with no torsion-free summand.
