Timeline for Is this a metric on the Grassmannian Manifold?
Current License: CC BY-SA 3.0
15 events
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| Oct 17, 2019 at 21:40 | comment | added | David E Speyer | The answer is if and only if $A=BQ$ for $Q \in SO(n)$. As I note above, if $A = BQ$ with $Q$ an orthogonal matrix of determinant $-1$, then $d(A,B) = \sqrt{2}$, not $0$. In other words, $d(A,B)=0$ if the columns of $A$ and $B$ are orthonormal bases for the same $n$-plane WITH THE SAME ORIENTATION. So $X/\sim$ is the set of oriented $n$-planes in $\mathbb{R}^m$. | |
| Oct 17, 2019 at 21:39 | comment | added | David E Speyer | @user27493 Whenever you have a set $X$ with a pseudometric $d$, you can define an equivalence relation on $X$ by $x \sim y$ iff $d(x,y)=0$. Then $d$ is a metric on the set of equivalence classes of $\sim$. Let $X$ be the set of $m \times n$ matrices obeying $A^T A = \mathrm{Id}_n$ or, in other words, $n$-tuples of orthonormal vectors in $\mathbb{R}^m$, equipped with the above pseudo-metric. What are the equivalence classes of $\sim$ or, in other words, when is $d(A,B)=0$? (continued) | |
| Oct 17, 2019 at 21:10 | comment | added | user27493 | @DavidESpeyer Can you please give a reference, why this is a correct distance for oriented Grassmannian? Thanks! | |
| Sep 10, 2013 at 15:38 | comment | added | Suvrit | @user35593: but to show this to be a metric (with the abs) value, requires slightly more work than just an invocation of CB.... | |
| Sep 8, 2013 at 13:54 | comment | added | user35593 | @David: to get a distance function for the grassmanian one has to add an absolute value, i.e. $$d(A,B)=\sqrt{1-|\det(A^TB)|}.$$ | |
| Sep 7, 2013 at 19:17 | vote | accept | user35593 | ||
| Sep 7, 2013 at 17:28 | answer | added | Suvrit | timeline score: 9 | |
| Sep 7, 2013 at 15:23 | answer | added | yaoxiao | timeline score: 4 | |
| Sep 7, 2013 at 15:05 | comment | added | David E Speyer | Minor observation: You are building a metric on the oriented Grassmannian, not the Grassmannian. If $S$ is an $n \times n$ orthogonal matrix with determinant $-1$, then $A$ and $AS$ represent the same point of the Grassmannian, but $\det(A^T (AS)) = \det(S) = -1$, so your formula gives $\sqrt{2}$, not $0$. | |
| Sep 7, 2013 at 14:20 | comment | added | yaoxiao | It is obvious that when n=1. | |
| Sep 7, 2013 at 12:59 | history | edited | user35593 | CC BY-SA 3.0 |
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| Sep 7, 2013 at 12:54 | history | edited | user35593 | CC BY-SA 3.0 |
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| Sep 7, 2013 at 12:47 | history | edited | user35593 | CC BY-SA 3.0 |
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| Sep 7, 2013 at 8:20 | review | Close votes | |||
| Sep 7, 2013 at 14:14 | |||||
| Sep 7, 2013 at 8:02 | history | asked | user35593 | CC BY-SA 3.0 |