In a sense this is a followup of my earlier question Are there nonlinear projective spaces?. In any case, if there is anything interesting about this question it is all by virtue of Guram Berishvili.
Fix a natural number $n$. Moreover fix your favorite field and category to work in - algebraic, complex, real, analytic, smooth, topological, whatever.
For a vector space $V$ let Gr$_n(V)$ be the Grassmanian of all $n$-dimensional subspaces of $V$.
Every morphism $f:X\to$ Gr$_n(V)$ gives rise to a vector bundle on $X$. It also gives rise to a family $(V_x)_{x\in X}$ of $n$-dimensional subspaces of $V$.
Let us impose on this family the condition that for any nonzero $v\in V$ there is a unique $x\in X$ with $v\in V_x$.
Call a vector bundle "special" (just to call it something) if it can be obtained from some morphism $f$ (for some $V$) with the property that the corresponding family $(V_x)_{x\in X}$ satisfies the above condition.
The only examples of "special" vector bundles that I know are related to that question I link to.
Let $k\subset K$ be a finite field extension of degree $n$. Then, for $V$ a $K$-vector space, we have embedding
P$(V)\hookrightarrow$ Gr$_n(V_k)$
(view $K$-lines in $V$ as $n$-dimensional $k$-subspaces of the $k$-vector space $V$). The corresponding family is the tautological one, and the $k$-vector bundle on P$(V)$ so obtained is the tautological $K$-line bundle viewed as an $n$-dimensional $k$-bundle.
More generally, $K$ might be a non-commutative (associative) division ring, or a (non-associative) octonion algebra (but then dimension must be restricted).
Are there any other examples?
Does for every $X$ exist a vector bundle on $X$ that is "special" in this sense?
In particular, which manifolds have the property that their tangent bundles are "special"?
For $n=1$, and only imposing uniqueness of $x$ with $v\in V_x$ without its existence, this more or less amounts to very ample line bundles. However for $n>1$ it is essentially more than requiring $f$ to be an embedding into the Grassmanian, not only because of existence, but also because there is additional restriction that all $V_x$ are not only pairwise different but also must have pairwise zero intersections.
