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Suppose I have a manifold $\mathcal{M}$ and a closed submanifold $\mathcal{N} \subset \mathcal{M}$ of codimension 1. If I remove the closed submanifold $\mathcal{N}$ from $\mathcal{M}$ will I be left with a manifold?

I am not sure if it is true but it looks very plausible. However, I am pretty sure that if would only hold for codimension 1. For example in the manifold $\mathbb{R}^2$ I can take the submanifold complement to a figure "8".

Furthermore, would the statement also be true for smooth manifolds or symplectic manifolds?

Any help is welcome.

Thanks in advance.

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    @user29751 In the question you talked about removing things that were manifolds. That's all I meant.2012-06-03

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As pointed out in comments, removing any closed set from a manifold (topological, smooth, complex, symplectic...) leaves a manifold of the same kind, because the charts restrict to open subsets. One should note that the remaining manifold may be disconnected.

Removing a set which is not closed, generally, does not preserve the manifold structure, even if the removed set is itself a manifold. For example, remove an open line segment from a plane (example given by Chris Eagle).