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?
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 $\mathbb{R}P^6$.)
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.
