# Are two pairs $(M\times M, M\times \{a\})$ and $(M\times M, D_{M})$ homeomorphic?

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.

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.
• Regarding "any", would that argument work if $M$'s tangent bundle is non-stably nontrivial, as in $M=S^2$? – Michael Oct 13 '16 at 20:35
• 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. – Danny Ruberman Oct 13 '16 at 22:28
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 $PD(M\times\{a\})^2=0$.