TITLE: Is there a good way of visualizing Grassmann manifolds?
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: 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 $\mathbb{R}P^6$.)