Is there a good way of visualizing Grassmann manifolds? I am trying to explain $Gr_{\mathbb{R}}(m,n)$, the Grassmann manifold of the collection of $n$-dimensional linear subspaces of the $\mathbb{R}^m$ 
 space. 
After I used the projective space homeomorphic example $Gr_{\mathbb{R}}(3,1)$ and $Gr_{\mathbb{R}}(2,1)\cong S^1$ to illustrate the "shape" of Grassmann manifolds, I was then asked if there is an uniformly used or visually intuitive way of visualizing $Gr_{\mathbb{R}}(m,n)$? At least for low dimensions?
 A: First and foremost, $Gr_{\mathbb{R}}(m,n)$ is canonically diffeomorphic to $Gr_{\mathbb{R}}(m,m-n)$.  The map assigns to each $n$-plane in $\mathbb{R}^m$ its orthogonal complement.  Thus, $Gr_{\mathbb{R}}(m,m-1)$ is also a projective space.
Beyond this, I only know of a few other "nice" examples.  For example, $Gr_{\mathbb{R}}(4,2)$ is diffeomorphic to the quotient of $S^2\times S^2$ by the diagonal antipodal map:  $(x,y)\mapsto (-x,-y)$.  Very closely related, the Grassmannian of oriented $2$-planes in $\mathbb{R}^4$ is diffeomorphic to $S^2\times S^2$.
Next, the Grassmannian of oriented $2$-planes in $\mathbb{R}^7$ has a nice description coming from the octionions:  $Gr_{\mathbb{R}}(7,2)$ is a bundle over $S^6$ with fiber $\mathbb{C}P^2$.  Here, the projection map from a $2$-plane in $\mathbb{R}^7$ to $S^6$ is defined as follows.  Given an oriented plane $P\subseteq \mathbb{R}^7 \cong \operatorname{Im}\mathbb{O}$, pick a positively oriented orthonormal basis $\{x,y\}$ for $P$.  Then map $P$ to $xy \in S^6\subseteq \operatorname{Im}\mathbb{O}$.  (If one uses the unoriented Grassmanian, it is instead a bundle over $\mathbb{R}P^6$.)
A: Recall that an $n$-dimensional subspace of $\mathbb{R}^m$ can be described using a basis of $n$ vectors.  Putting the vectors into an $m\times n$ matrix, one can do column reduction to put it into reduced column echelon form.  A representative example would be
$$\begin{pmatrix}
1&0&0\\
*&0&0\\
0&1&0\\
*&*&0\\
0&0&1\\
0&0&0
\end{pmatrix}$$
with some numbers replacing the asterisks to represent a point in $Gr_{\mathbb{R}}(6,3)$.  The set of all such replacements gives an open 3-dimensional topological subspace of $Gr_{\mathbb{R}}(6,3)$, called a Schubert cell for the particular pivot positions.
As an example, consider $\mathbb{R}\mathrm{P}^2=Gr_{\mathbb{R}}(3,1)$.  There are three possible sets of pivot positions:
$$
\begin{pmatrix}
0\\0\\1
\end{pmatrix},
\begin{pmatrix}
0\\1\\*
\end{pmatrix},
\begin{pmatrix}
1\\*\\*
\end{pmatrix}
.
$$
This corresponds to a decomposition of the projective plane into a point, a line, and a plane (a $0$-cell, a $1$-cell, and a $2$-cell).  Seeing how they are attached to each other takes some thinking about column echelon form.  Consider $\begin{pmatrix}0\\1\\x\end{pmatrix}$ as $x\to\pm\infty$.  This does not converge as a matrix, but an an equivalent matrix is $\begin{pmatrix}0\\1/x\\1\end{pmatrix}$ for $x\neq 0$, and this converges to the first of the three matrices.  This shows that the ends of the line are attached to the point, giving a circle ("the circle at infinity").  One can make a similar argument to show that the plane is attached to this circle (though the attachment is more complicated: a degree-two map).  The analysis of $Gr_{\mathbb{R}}(3,2)$ is very similar.
More interesting of an example would be $Gr_{\mathbb{R}}(4,2)$.  Going through all the pivot positions, there would be one each of $0$-, $1$-, $3$-, and $4$-cells, and two $2$-cells.  One could work out what is attached to what and how, but visualizing the resulting $4$-dimensional manifold is possibly out of the question.
