Vrftjkry

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?

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 $Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book.

On the other hand, if $n > 2$, then $$pi_{n+1}(vee^k S^n) = (pi_{n+1} S^n)^k = (Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer.

Furthermore, any map in $GL_k(Bbb Z)$ may be realized as an automorphism of $H_n(vee^k S^n)$ by some autoequivalence of $vee^k S^n$, and the map $GL_k(Bbb Z) to GL_k(Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(Bbb Z/2)^k$ are related by some matrix in $GL_k(Bbb Z/2)$, the homotopy type of $D^{n+2} cup_f (vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial.

Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$Sigma X simeq S^5 vee^{b_2} S^3;$$ if $X$ was not spin, then $$Sigma X simeq Sigma Bbb{CP}^2 vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.

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.