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.
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1$\begingroup$ The Grassmanian is a homogeneous space for the orthogonal group (unitary group in the complex case) and hence inherits a natural metric. $\endgroup$– Paul SiegelCommented Aug 14, 2014 at 23:28
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1$\begingroup$ If you want an explicit formula, see mathoverflow.net/questions/141483/… $\endgroup$– David E SpeyerCommented Aug 15, 2014 at 1:46
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$\begingroup$ See montefiore.ulg.ac.be/systems/Publi/Grass_geom.pdf, for the Grassmann manifold of $p$-planes in $\mathbf{R}^n$. $\endgroup$– user62675Commented Aug 15, 2014 at 2:33
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9$\begingroup$ The space $\mathrm{Gr}(p,V)$ of $p$-dimensional subspaces in a vector space $V$ does not have a nontrivial canonical metric. There is no Riemannian metric on this space that is invariant under the natural action of $\mathrm{Aut}(V)=\mathrm{GL}(V)$. Upon fixing an additional structure on $V$, namely, a positive definite inner product $q$, there is a Riemannian metric on this space that is invariant under the natural action of $\mathrm{Aut}(V,q)=\mathrm{O}(q)$, unique up to a constant multiple. This multiple can be uniquely determined by specifying the total volume or diameter, for example. $\endgroup$– Robert BryantCommented Aug 15, 2014 at 11:18
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1$\begingroup$ @PaulSiegel , Aaron asked canonical metric not natural metric, ;) . See my answer $\endgroup$– user21574Commented Jan 18, 2016 at 16:16
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3 Answers
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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.
answered Aug 15, 2014 at 2:54
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$\begingroup$ he asking canonical metric, $\endgroup$– user21574Commented Jan 18, 2016 at 16:26
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$\begingroup$ @HassanJolany This metric is actually symmetric and unique up to a constant multiple, as stated above. Any normalisation you wish to choose makes it canonical. $\endgroup$ Commented Jan 26, 2016 at 12:40
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$\begingroup$ what is the your definition about canonical metric,? $\endgroup$– user21574Commented Jan 26, 2016 at 15:25
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1$\begingroup$ To repeat what Sebastian already said, the question refers to Grassmannians, which most of us would interpret as real Grassmannians, i.e., $G(n,k) = $ the space of $k$-linear subspaces in an $n$-dimensional real vector space. In general there is no complex structure on such a space, for example if the Grassmannian is odd-dimensional. However, it is a homogeneous space, and the standard meaning of a "canonical metric" on a homogeneous space is one that is invariant under the group action. $\endgroup$ Commented Jan 26, 2016 at 16:33
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1$\begingroup$ I didn't know the book of Besse, I just checked it. It seems it had long history and I didn't know anything about it. $\endgroup$– user21574Commented Jan 26, 2016 at 17:47
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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.
answered Aug 30, 2014 at 7:57
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$\begingroup$ So is there any exception to what Yurii has shown? Should any invariant metric on the grassmannian look like a symmetric norm function of Jordan angles? $\endgroup$– Troy WooCommented May 14, 2015 at 21:53
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$\begingroup$ Since it is an irreducible symmetric space of compact type, there is not a lot of choice: any invariant metric is a constant multiple of this one. $\endgroup$ Commented May 15, 2015 at 4:39
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$\begingroup$ Great!...do you mind providing a reference? i.e. uniqueness of invariant Finsler metric...I'm kinda slow...I am an engineer... $\endgroup$– Troy WooCommented May 15, 2015 at 8:38
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1$\begingroup$ Misha Verbitsky's answer gives you a proof in 3 lines. Helgason: "Differential Geometry, Symmetric Spaces, and Lie groups" is a standard reference, but by far too much for this simple question. As a Finsler metric it is not unique: any power of the Riemannian norm is an invariant Finsler metric. $\endgroup$ Commented May 15, 2015 at 8:54
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$\begingroup$ Oh, by the way, is it possible to characterize all invariant Finsler metrics in some way? $\endgroup$– Troy WooCommented May 15, 2015 at 10:08
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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.
answered Aug 30, 2014 at 3:50