# 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?

• Possibly related post math.stackexchange.com/questions/1521370/… (which is interesting in itself.) – Henry.L Dec 14 '17 at 20:43
• Connection with the SVD decomposition : have a look to (www4.ncsu.edu/~aszanto/FoCM14/Talk-slides/FoCM/peterson.pdf) – Jean Marie Dec 14 '17 at 21:07
• About lower dimensions: as for $n=1$ it is a projective space, for $n=2$ it is a space of lines in this projective space, though I don’t know if it’s of any visual help. Also, as $\mathrm{Gr}_{\mathbf R}(3, 2)$, you could consider a set of all great circles of a sphere $S^2$. There you can easily spot an isomorphism with $\mathrm{Gr}_{\mathbf R}(3, 3-2)$. – arseniiv Dec 14 '17 at 21:21
• You can visualize the tangent space as $n$ dimensions wiggling around in $m-n$ dimensions. I guess you're looking for the global shape, though, which is a lot less trivial. – Chris Culter Dec 14 '17 at 21:59
• How about Schubert cells to get a schematic picture of a CW structure? I believe the cells are described in Milnor and Stasheff's book on characteristic classes. – Kyle Miller Dec 15 '17 at 1:31

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

• I am currently thinking of using the equivalent representation of Grassmann manifold in Lie groups and visualize them using matroids, which has relatively systematic way of visualization. But what you suggest is certainly sth I would try. Thanks! – Henry.L Dec 15 '17 at 2:47

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.

• Sounds very interesting, let me work it out first before I accept it. Thank you! – Henry.L Dec 15 '17 at 2:25