Let $G_n(V)$ the Grassmannian of $n$-dimensional subspaces of a finite-dimensional $V$, and $l<n$. I've noticed that it is easy to associate a (possibly singular) submanifold $\tilde{M}\subseteq G_n(V)$ to a manifold $M\subseteq G_l(V)$ by
$$
\tilde{M}:=\{W\in G_n(V)\mid W \textrm{ contains some L}\in M\}.
$$
**Preliminary question 1:** does the procedure $M\mapsto \tilde{M}$ (or something similar to it) have a name?

Let now $D\leq V$ be a subspace, and call $M_D$ the Grassmannian $G_l(D)$, understood as a submanifold of $G_l(V)$: then, if I'm not mistaken, $\widetilde{M_D}=\{W\in G_n(V)\mid \textrm{dim} W\cap D\geq l\}$. Hence, any Schubert cell can be written as the intersection of certain $\widetilde{M_D}$'s: this observation convinced me that the submanifolds of the form $\widetilde{M}$ must be known to the experts in Algrebraic Geometry, whose counsel I'm seeking.

**Preliminary question 2:** is the submanifold $\widetilde{M_D}$ (and/or the equation defining it) a well-known one?

As my usage of the term "manifold" clearly shows, I'm no expert in Algebraic Geometry: I came across these "objects" while studying Monge-Ampére equations from a geometric perspective.

**Actual Question:** Is there any criterion to establish whether a submanifold $E\subseteq G_n(V)$, given either implicitly or parametrically, is of the form $\widetilde{M}$?

An ideal answer would read like this: "compute a prescribed set of invariants of $E$: if they take the right values, then $E=\widetilde{M}$, for some $M$, whose ideal (or parametrization) can be obtained by that of $E$ by algebraic manipulations" - but I really do not expect this much, though I'm sure that some obvious necessary conditions for $E$ to be of the form $\widetilde{M}$ can be easily found, possibly under (even severe) restrictions (i.e. $M$ to be 0-dimensional). If this problem has been considered before, please point a reference for me! Thanks a lot in advance!

**Reformulating the problem in view of the feedbacks received so far:** it seems that there is a way to check whether $E$ is the "double fibration transform" of some $M$, namely

- Lift $E$ to $F_{l,n}(V)$;
- Construct the largest full sub-bundle $\overline{E}$ of $\lambda$ contained into $\nu^{-1}(E)$; [by
*full*I mean that the fibers are the same, but the base may be smaller] - Project $\overline{E}$ back to $G_n(V)$: if the result coincides with $E$, then the answer is YES, and $M$ is the base of $\overline{E}$.

Even if what I wrote is correct, I'm not yet satisfied, since I'd like to see it algebraically. I know how to formulate algebraically steps 1. and 3., but, concerning 2., I can only see an "infinitesimal" way to carry it out. More precisely, I can enlarge the ideal of $\nu^{-1}(E)$ by adding the derivatives of its elements w.r.t. the $\lambda$-vertical differential operators: with this larger ideal, I get the submanifold of $\nu^{-1}(E)$ made of the points of tangency to the $\lambda$-fibers (which correspond to the $\alpha$-distribution mentioned by @alvarezpaiva), which is the "infinitesimal counterpart" of the $\overline{E}$ I needed in step 2..

Perhaps, as @Francois Ziegler pointed out, it is more interesting to work with "infinitesimal objects" rather then mere submanifolds... Yet I'm sure that a simple example of characterization of the double fibration transform of a submanifold can be found, so I keep waiting...