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What is an example of a compact manifold $M$ without boundary which does not satisfy the following property:

For every $a\in M$, two pairs $(M\times M, M\times \{a\})$ and $(M\times M, D_{M})$ are homeomorphic pairs, where $D_{M}=\{(a,a)\mid a\in M\}$

Do all spheres satisfy the above property?

Note that every Lie group satisfies the above property, in the stronger geometric version. That is the homeomorphism can be chosen as an isometry of the product metric.

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Any manifold whose tangent bundle is not topologically trivial gives an example. The normal bundle of $M \times \{a\}$ in $M \times M$ is trivial, but the normal bundle of $D_M$ is isomorphic to the tangent bundle of $M$. For this latter fact, see Milnor-Stasheff, Lemma 11.5.

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    $\begingroup$ Regarding "any", would that argument work if $M$'s tangent bundle is non-stably nontrivial, as in $M=S^2$? $\endgroup$
    – Michael
    Commented Oct 13, 2016 at 20:35
  • $\begingroup$ Yes; the argument as given applies to the actual bundle, rather than the stable tangent bundle. $\endgroup$
    – Danny Ruberman
    Commented Oct 13, 2016 at 21:04
  • $\begingroup$ Could you please clarify, how to get identification of normal bundles, if your map is just homeomorphism and not smooth. $\endgroup$
    – Yury Ustinovskiy
    Commented Oct 13, 2016 at 21:06
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    $\begingroup$ This is a bit complicated, as you need a topologically invariant notion of tangent bundle for manifolds in the absence of a smooth structure. This is the notion is the tangent microbundle, due to Milnor. This is equivalent, by Kister's theorem (Ann. of Math. (2) 80 1964 190–199) to a topological $R^n$ bundle, and a homeomorphism induces an isomorphism of those bundles. Each embedding defines a microbundle, the first is trivial, and the second the tangent microbundle. Certainly, for a concrete example, your explanation is best; my point is that this is a general phenomenon. $\endgroup$
    – Danny Ruberman
    Commented Oct 13, 2016 at 22:28
  • $\begingroup$ @DannyRuberman Thank you very much for your interesting answer.I am sorry if this question trivial:Is it obvious, in general, that the structure of normal bundle of a submanifold is independent of the riemannian metrics defined on the bigger manifold? $\endgroup$
    – Ali Taghavi
    Commented Oct 14, 2016 at 21:12
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If manifold $M$ satisfies this property, then there is an automorphism of $H^*(M\times M,\mathbb Z)$ which takes the (Poincare dual of) class of diagonal to the class of $M\times\{a\}$.

This is not the case, e.g., for even-dimensional sphere, since $PD(D_M)^2=2$, while $PD(M\times\{a\})^2=0$.

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