Canonical Metric on Grassmann Manifold I was curious and quite clueless as to how we can equip the Grassmann Manifold with a canonical metric - I have yet to find anything upon this subject. 
 A: In fact, $Gr(n,m)$ with its canonical metric induced from an Euclidean structure is one of the few spaces where you can write down solutions of the geodesic equations explicitly by a formula and write down a formula for the geodesic distance. See the following paper for this


*

*MR1856419 Neretin, Yurii A. On Jordan angles and the triangle inequality in Grassmann manifolds. Geom. Dedicata 86 (2001), no. 1-3, 81–92. 

A: A nice geometric way of endowing a Grassmann manifold with a metric (understood here as a distance, and not directly as a Riemannian metric) is to use the Hausdorff distance for subsets of the round sphere.
Consider $V$ a real vector space of dimension $n$ endowed with an inner product, and let $Gr_k(V)$ be the Grassmannian of $k$-planes on $V$. Let $x,y\in Gr_k(V)$, and denote by $S_x$ and $S_y$ the subspheres of the unit sphere $S_V=\{v\in V:\|v\|=1\}$ defined by intersecting it with the subspaces $x$ and $y$ respectively. Then $S_x,S_y\subset S_V$ are closed subsets and the distance between $x,y\in Gr_k(V)$ can be defined as the Hausdorff distance between these sets:
$$dist(x,y)=dist_H(S_x,S_y).$$
Note that this distance does not coincide with the symmetric space distance on $Gr_k(V)=SO(n)/SO(k)SO(n-k)$. As explained in the paper of Neretin in Michor's answer, the symmetric space distance is computed in terms of the principal angles between two subspaces (namely, it is the square root of the sum of the squares of these angles); while the above distance defined in terms of the Hausdorff distance of closed sets in the unit sphere measures exactly the largest principal angle between subspaces, ignoring all the other smaller principal angles.
A: Since Grassmannian $Gr(n,m)=SO(n+m)/SO(n)\times SO(m)$ is a homogeneous manifold,
you can take any Riemannian metric, and average with $SO(n+m)$-action. Then you show that an $SO(n+m)$-invariant metric is unique up to a constant. This is easy, because the tangent space $T_VGr(n,m)$ (tangent space to a plane $V\subset W$)
is $Hom(V,V^\bot)$, and your metric must be $SO(V)\times SO(V^\bot)$-invariant.
Such a metric is unique (up to a constant multiplier), which follows, e.g., from Schur's lemma.
