Straight things on Grassmanians I am new to Grassmanian, so this question may be too easy. But I didn't find it in any books I know.
When we talk about Grassmanian, it should not be only a manifold or a variety. At least I think we can talk about the straight things on it. For example, let $V$ be a fixed linear space, we have the following two natrual ways to define straight things on the Grassmanian $Gr(m,V)$.
(a) Use the natural corresponding between $Gr(m,V)$ and $Gr(m+k,V)$, we can define a $k$-plane on $Gr(m,V)$ to be any element in $Gr(m+k,V)$.
(b) Use the Plücker embedding $Gr(m,V)\hookrightarrow \mathbb P^N$, we can pullback the hyperplane line bundles $\mathcal O(1)$ on $\mathbb P^N$ to define the hyperplanes on $Gr(m,V)$, and then define $k$-planes to be the intersection of $l$ hyperplanes, where $l=dim(Gr(m,V))-k$.
I want to know are these two defintions consistent? And where can I find references about this?
Thanks in advance!
 A: In fact, for general cases, (a) is not a good definition. More precisely, for $Gr(m,V)$ where $m>1$, two points on $Gr(m,V)$, which correspond to two $m$-dimensional linear subspaces of $V$, do not contain in any "lines" on $Gr(m,V)$, which correspond to $(m+1)$-dimensional linear subspaces of $V$, unless the very special case that the two $m$-dimensional linear subspaces have a codimension one intersection. Also, let $n=dim(V)$, then definition (a) only produces $k$-planes with $k=0,1,\ldots,n-m$, which are much fewer compared to $dim(Gr(m,V))$ which equals $m(n-m)$.
I think (b) looks like a sutible definition. By simple computations, one can show that in definition (b), a (at least some, see below) hyperplane on $Gr(m,V)$ contains all the $m$-dimensional linear subspaces of $V$ which are vertical to a certain $m$-dimensional linear subspace. More precisely, let $x_i$'s be the standard coordinates of $V$, then the Plücker embedding $\mathbb P^N$ has the standard coordinates $x_{i_1}\wedge \ldots\wedge x_{i_m}$'s, and a hyperplane will correspond to 
$$\sum a_Ix_{i_1}\wedge \ldots\wedge x_{i_m}=0.$$
 For the hyperplane $x_1\wedge\ldots\wedge x_m=0$ on $\mathbb P^N$, it can be restricted to give a hyperplane on $Gr(m,V)$ which contains all $m$-dimensional linear subspace of $V$ vertical to $x_1\wedge\ldots\wedge x_m$. For a hyperplanes $\sum a_Ix_{i_1}\wedge \ldots\wedge x_{i_m}=0$ not decomposable, their geometric interpretations are still not clear to me. Even I am not sure will they concide with the hyperplanes we already get, or it will produces (much) more new hyperplanes. Or maybe we should only define hyperplanes on $Gr(m,V)$ to be the decomposible ones? It would be very nice if someone knew this could give a comment.
