I'm reading an article by Ricardo Mañé, "The Hausdorff dimension of horseshoes of diffeomorphisms of surfaces" (https://doi.org/10.1007/BF02585431). I'm having a technical problem. Sorry for my ignorance, but I would like a reference which explains how to equip the Grassmann manifolds with a metric.
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13$\begingroup$ Given two $k$-dimensional vector subspaces $W_1$, $W_2$ of a common vector space $V$, let $d(W_1,W_2)$ be the Hausdorff distance between $W_1 \cap S$ and $W_2 \cap S$ where $S$ is the unit sphere in your vector space -- assuming your vector space has an inner product. See: en.wikipedia.org/wiki/Hausdorff_distance $\endgroup$– Ryan BudneyCommented Dec 3, 2010 at 0:51
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14$\begingroup$ Another way is to view the Grassmannian as a homogeneous space, i.e. as a quotient of a Lie group by a compact subgroup. Homogeneous spaces inherit metrics as quotient spaces, since compact Lie groups have invariant metrics. $\endgroup$– Ryan BudneyCommented Dec 3, 2010 at 0:53
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3$\begingroup$ @RyanBudney The Riemannian metric from your second suggestion is explicitly described in this 1967 paper pnas.org/content/57/3/589.full.pdf+html which also describes geodesics and other interesting stuff. $\endgroup$– Jairo BochiCommented Mar 18, 2015 at 21:04
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$\begingroup$ what about the angle between $W_1$ and $W_2$? It seems to be a distance taking values in $[0, \pi/2]$ $\endgroup$– Daniele ZuddasCommented Mar 19, 2015 at 1:02
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2$\begingroup$ "Natural" metrics on the Grassmannian should be invariant under isometries of the undelying Euclidian space. Actually there are many such metrics, and this nice paper gives a general method for constructing them: Qiu, Zhang, Li, "Unitarily invariant metrics on the grassmann space". SIAM J. Matrix. Anal. Appl. vol 27, no 2, 507-531. $\endgroup$– Jairo BochiCommented Jul 14, 2015 at 20:51
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3 Answers
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I found it surprisingly difficult to find a reference for this when I was studying Mane's papers on multiplicative ergodic theorems. My hypothesis was that people working with the Grassmannian in other areas are happy with the fact that the Grassmannian is metrisable for abstract topological reasons, and don't actually care very much about a precise metric, but I might be wrong about this... in my answer I'm going to assume that we're considering a finite-dimensional space equipped with an inner product structure.
If you are interested in precise metrics on the Grassmannian, the most popular definition of which I am aware is this one: $$d(V,W):=\max\left\{\sup_{w \in W, \|w\|=1}\inf \{\|v-w\| \colon v \in V \},\sup_{v \in V, \|v\|=1}\inf \{\|v-w\| \colon w \in W\}\right\}$$ This is I think not quite the same as the one suggested by Ryan Budney, but produces the same topology. This one seems to be the most popular definition for people working in multiplicative ergodic theory (it is in Barreira and Pesin's book, for example).
There are some equivalent ways of describing this metric which seem to be less well-known. If we know a priori that $V$ and $W$ have the same dimension, then the maximum in the expression above is always attained by both expressions simultaneously! Hence if we fix a dimension $r$, then the expression $$d(V,W):=\sup_{v \in V,\|v\|=1}\inf\left\{\|v-w\|\colon w \in W\right\}$$ is actually a metric for the component of the Grassmannian which consists of all $r$-dimensional subspaces. This does not seem to be very well-known; I actually discovered this by reading Kato's book on perturbation theory, which isn't exactly the first place I'd go to to find out about Grassmannian manifolds...
Another way to put a metric on the Grassmannian is as follows. We can identify a subspace $U$ with the unique linear operator of orthogonal projection onto that subspace, and take the metric given by setting the distance between two subspaces to be the operator norm distance between the orthogonal projection operators corresponding to those subspaces. I personally like this approach a great deal, because I think it makes it very obvious that the Grassmannian is compact (well, obvious if you're a functional analyst!). This metric is also, rather pleasantly I think, exactly identical to the first metric I defined above. You can find a proof that the two things are the same in the book on Hilbert spaces by Akhiezer and Glazman.
There's a short discussion on this topic in my paper "A rapidly-converging lower bound for the joint spectral radius via multiplicative ergodic theory", which is basically the result of a gentle argument between myself and the referee over how the metric on the Grassmannian should be defined!
answered Dec 6, 2010 at 13:33
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1$\begingroup$ How are any of the sets above, which are being sup'ed over, bounded? In each expression one vector can have arbitrary magnitude, no? I don't understand. $\endgroup$ Commented Dec 6, 2010 at 18:17
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1$\begingroup$ I thought of the case inf was meant, but then k-planes with non-trivial intersections are distance 0 from each other. Shouldn't it be, maybe, inf over w, sup over v, ||v||=1, similarly to how the Hausdorff metric is defined? $\endgroup$ Commented Dec 6, 2010 at 19:00
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3$\begingroup$ Do I miss something if I say that when $V\cap W$ is nonzero, your $d(V,W)$ vanishes ? $\endgroup$– BS.Commented Dec 6, 2010 at 19:03
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1$\begingroup$ @ABlumenthal: in a Hilbert space this follows from Theorem I.6.34, although some translation is required to get the above statement. (My original post could be rather clearer in this respect.) One must use the fact that $d(V,W)$ equals the norm distance between the orthogonal projections onto $V$ and $W$ respectively (for which result one must refer to Akhiezer and Glazman) and then unpack a couple of definitions, to arrive at this statement. See Lemma 3.2 in bit.ly/18CoDdZ . In a Banach space I have no idea if the result is true or not. $\endgroup$ Commented Dec 4, 2013 at 14:17
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1$\begingroup$ @IanMorris Thank you. I actually suspect it isn't true in Banach space. I can show, though, that these two gaps aren't off by more than a constant factor in $\dim V = \dim W$ (one may use John's theorem to construct a 'good' basis and then fish around for the right inequality). $\endgroup$ Commented Dec 4, 2013 at 14:52
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The $k$-dimensional Grassmannian over a vector space $V$ naturally embeds into the projectivization of the $k$-th exterior power of $V$. If you have a Euclidean structure on $V$, then it extends to its exterior products as well. Then a metric on the projectivization is provided, say, by the angle between the corresponding directions, or by sine of that angle.
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My "answer" is just a (still another) reformulation of Ian's.
Given two subspaces $V$, $W \subset \mathbb{R}^n$ with the same dimension, define their distance as: $$ d(V,W):= \inf_J \| J - i_V\| $$ where $i_V : V \to \mathbb{R}^n$ denotes the inclusion, $\|\mathord{\cdot}\|$ is the Euclidian operator norm, and $J$ runs over the linear maps $V \to \mathbb{R}^n$ whose such that $J(V) \subset W$. It's not too difficult to prove that the the infimum is attained at map $J=P_W|V$ (i.e. the orthogonal projection onto $W$ restricted to $V$). Trivially, $$ \|P_W|V - i_V\| = \sup_{v\in V, \|v\|=1} \inf_{w\in W} \|w-v\|, $$ so my formula is equivalent to Ian's.
As an easy consequence, we have the following interesting fact:
Fact: Let $T$ be a linear automorphism of $\mathbb{R}^n$. Then the induced trasformation of the grassmannian $G(k,n)$ is (bi-)Lipschitz with constant $\|T\|\,\|T^{-1}\|$.
Proof: $$ d(TV,TW) = \inf_{K(TV)\subset TW} \|K-i_{TV}\|=\inf_{J(V)\subset W}\big\|T(J-i_V)T^{-1}|_{TV}\big\| \le \|T\| \, \|T^{-1}\| \, d(V,W). $$ (The proofs of the "fact" using other characterizations of $d$ become messier, I think.)
