Suppose $X$ is a topological manifold and $Y \subset X$ is a locally flat submanifold. We know that $Y$ doesn't necessarily have a tubular neighborhood. My definition of a tubular neighborhood of $Y$ is a neighborhood $U$ with a map $\pi: U \to Y$ which is identity on $Y$ and $\pi: U \to Y$ is a disk bundle over $Y$.

The locally flatness implies, however, every point $p\in Y$ has a neighborhood in $Y$ which has a tubular neighborhood. So "having a tubular neighborhood" is a property which cannot be simply glued from local to global.

However, I wonder if the following is true. Suppose $Y$ has an open cover $\{Y_i\}$ and each $Y_i$ has a tubular neighborhood $\pi_i: U_i \to Y_i$. Suppose we require that $\pi_i$ agree on overlaps, i.e.,

$\pi_i(U_i \cap U_j) = \pi_j(U_i \cap U_j), \forall i,j$

Can this condition imply the existence of a tubular neighborhood of $Y$? Or, what should be the correct and natural condition to gurantee the gluing from local to global?


Assuming you mean $\pi$ is a closed disk bundle, I think a counterexample is given in the paper that first (as far as I know) constructed open disk bundles without closed disk sub-bundles. If you see Browder here (http://www.jstor.org/stable/1970428?seq=1#page_scan_tab_contents), theorem 3 gives manifolds embedded in euclidean space with normal bundle an open disk bundle (hence locally flat) without a closed disk sub-bundle.

William Browder, Open and Closed Disk Bundles Annals of Mathematics (Mar. 1966), pp.218-230

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  • $\begingroup$ I think this is strange. If the open disk bundle is the unit disk bundle of a vector bundle (with a metric), then just take the radius 1/2 disk bundle gives a subbundle. $\endgroup$ – Guangbo Xu Aug 25 '15 at 13:09
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    $\begingroup$ @GuangboXu That's right. An open disk bundle without a closed disk sub-bundle cannot admit the structure of a vector bundle. $\endgroup$ – PVAL Aug 25 '15 at 15:22
  • $\begingroup$ What if I only consider disk bundles inside vector bundles? $\endgroup$ – Guangbo Xu Aug 28 '15 at 1:27

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