Topology on the set of linear subspaces

Hello,

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 $x_i \rightarrow x$.

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.

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.

Can you give me hints where to find theory into this direction?

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The projection operators from X onto E_i converge to the identity operator from X to X in, say, the strong operator topology. – Qiaochu Yuan Jan 9 '11 at 17:41

Dear Martin, your intuition is excellent.

Here 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.

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This link is now dead and I am not able to identify the paper concerned: do you remember its name and author? – Ian Morris Nov 12 '13 at 14:38
Dear Ian, I'm afraid not. This will be a lesson: I'll give title and author in the future when I link to a paper. Apologies for this frustrating disappearance of my reference. – Georges Elencwajg Nov 13 '13 at 20:48

Some of the answers to this question 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.

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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 $1$.

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