Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $U$ be the complement of a closed star-shaped subset in a separable infinite-dimensional Frechet space. Since every separable Frechet space is homeomorphic to $l_2$, one knows that $U$ is a Hilbert manifold.

Question: what else can be said about the topology of $U$?

share|improve this question
add comment

1 Answer

I guess you want to impose some additional conditions on $U$. If $U$ is unbounded, then the topology of $l_2\setminus U$ can be as complicated as you wish. For example if you take U to be the product of a complex hyperplane arrangement $A\subset \mathbb C^n$ with $l_2$, i.e $U= A\times l_2\subset \mathbb C^n\times l_2$ you get a manifold with a complicated fundamental group. You can also throw away from $l_2$ a countable set of planes of finite (but growing codimention), this will give you holomogy in all dimensions.

Of course all this is true even you ask your question for complements to closed star-shaped sets in finite dimensional spaces. So, it would be nice to know what kind of finite-dimensional phenomena you want to generalise. Because obviously, if you assume that $U$ is bounded and contains a neighbourhood of $0$, then the complement $l_2\setminus U$ deformation retracts to $S^{\infty}$, so it is contractible.

share|improve this answer
    
In a finite dimensional space one has Alexander duality, relating cohomology of the closed subset and homology of the complement; it does not seem to generalise in any way to infinite dimension. I just hope that the assumption "closed star-shaped subset" is quite restrictive, and not every open subset of $l_2$ occurs as the complement of a closed star-shaped subset. –  Igor Belegradek May 6 '11 at 1:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.