Vrftsjtryk

What are some examples of (compact, say) manifolds $X$ and $Y$ that are stably equivalent, i.e. $Sigma^{infty}X_+simeqSigma^{infty}Y_+$, but are not homotopy equivalent?

The easiest examples are given by complements of open neighborhoods of distinct knots in $S^n$.

More generally, Spanier-Whitehead duality says that if $X$ is a compact simplicial complex and if $f,g:X rightarrow S^n$ are embeddings, then $S^n - Im(f)$ is stabily equivalent to $S^n - Im(g)$.

See my notes here for an elementary discussion.

Maybe it is worth adding some simply-connected examples.

Every simply connected closed 4-manifold may be described as $X = D^4 cup_f (S^2 vee cdots vee S^2)$, where $f$ is a map $S^3 to S^2 vee cdots vee S^2$; $pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;Bbb Z) otimes H^2(X;Bbb Z) to H^4(X;Bbb Z) = Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $Bbb Z$.

Suspending takes this matrix mod 2, and so the homotopy type of the suspension is dictated by the isomorphism class of $A$ modulo $2$ as a symmetric bilinear form. These have a classification (see eg 3.7 here). In particular, you find that every even form of rank $2n$ becomes equivalent to the intersection form of $#^n (S^2 times S^2).$ So you find that every simply connected spin 4-manifold is stably equivalent to $#^{b_2/2} (S^2 times S^2)$.

The odd case is even easier, as if the form was indefinite in the first place, it is diagonalizable over the integers, and so your manifold is stably homotopy equivalent to $#^{b_2} Bbb{CP}^2$.

In particular, every simply connected closed smooth 4-manifold is stably equivalent to either $#^{b_2/2} (S^2 times S^2)$ or $#^{b_2} Bbb{CP}^2$, determined by whether or not its intersection form is even. These are not stably equivalent (the former has trivial Steenrod squares, the second has a nontrivial square).

The definite case is probably harder, but Donaldson says those aren't smoothable, so I will not think about it.