Topology on the set of linear subspaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:08:22Z http://mathoverflow.net/feeds/question/51563 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51563/topology-on-the-set-of-linear-subspaces Topology on the set of linear subspaces Martin 2011-01-09T17:37:38Z 2011-01-13T09:21:50Z <p>Hello,</p> <p>let $X$ be a seperable Hilbert space. Let $(e_i)_i$ be a Hilbert basis, and for each index let $E_i = \langle e_1,\dots,e_i \rangle \subset X$ the span of the first $i$ basis vectors. For any $x \in X$, let $x_i$ be the best-approximation of $x$ in $E_i$, and it is clear that $x_i \rightarrow x$.</p> <p>It seems intuitive to say, that the $(E_i)$ approximate $X$ in a certain sense. Nevertheless, I am not aware of a topology on the set of linear subspaces, which would give such a result rigorously.</p> <p>A first attempt might be to identify each linear subspace with its projection-onto, and inspect these projections as a topological (no more linear) space. A next step might be to take into account the order of the basis vectors for each linear subspace (which might be crucial for stability in numerical analysis). I am not known a theory Grassmannian manifolds in infitinte-dimensional vector spaces, nor how to relate non-equidimensional Grassmannian manifolds.</p> <p>Can you give me hints where to find theory into this direction?</p> http://mathoverflow.net/questions/51563/topology-on-the-set-of-linear-subspaces/51573#51573 Answer by Georges Elencwajg for Topology on the set of linear subspaces Georges Elencwajg 2011-01-09T19:14:08Z 2011-01-09T19:14:08Z <p>Dear Martin, your intuition is excellent.</p> <p><a href="http://www.mrlonline.org/jot/2009-061-001/2009-061-001-002.pdf" rel="nofollow">Here</a> is a paper that indeed identifies closed subspaces of Hilbert spaces with the orthogonal projection onto them and thus studies the Grassmannian you are interested in by embedding it in operator spaces. The article sems interesting, well written and fairly elementary( else I wouldn't even understand what it is about since I know so little about Functional Analysis...). There is a bibliography that might be useful too. Good luck.</p> http://mathoverflow.net/questions/51563/topology-on-the-set-of-linear-subspaces/51639#51639 Answer by Ian Morris for Topology on the set of linear subspaces Ian Morris 2011-01-10T11:04:11Z 2011-01-10T11:04:11Z <p>Some of the answers to <a href="http://mathoverflow.net/questions/48118" rel="nofollow">this question</a> might be helpful for your question also. It deals with finite-dimensional Hilbert spaces, but most of my answer to that question applies to the infinite-dimensional case too, with one or two obvious exceptions (e.g. the metric space of 1-dimensional subspaces of an infinite-dimensional Hilbert space is not compact). In particular, the book on Hilbert spaces by Akhiezer and Glazman has a short (5 pages?) section on the Grassmannian of a Hilbert space, and shows that the metric on the Grassmannian given by `aperture' is the same as the metric given by the operator difference between orthogonal projections.</p> http://mathoverflow.net/questions/51563/topology-on-the-set-of-linear-subspaces/51937#51937 Answer by KP Hart for Topology on the set of linear subspaces KP Hart 2011-01-13T09:21:50Z 2011-01-13T09:21:50Z <p>Denote the intersection of a closed linear subspace $A$ with the unit sphere by $S_A$, say. You can define the distance of $A$ and $B$ to be the Hausdorff distance between $S_A$ and $S_B$. This will give you a metric topology on the set of closed linear subspaces but it seems that it does not quite do what you want: the distance between a proper closed subspace and the ambient space is always equal to $1$.</p>