Let $X$ be a fixed closed manifold,$S(X)$ the structure set and $Aut(X)$ the group of self homotopy equivalence of $X$.

surgery theory tells us that $\mathcal{M}(X):=S(X)/Aut(X)$ is in bijection with the set of $h$-cobordism classes of manifolds homotopy equivalent to $X$.

If $X$ is simply connected and dim$X\geq 5$,then by the $h$-cobordism theorem,$S^{Top}(X)/Aut(X)$ is in bijection with the homeomorphism classes of manifolds homotopy equivalent to $X$.

We call manifold $M$ a ** fake** $X$ if $M$ is homotopy equivalent but not homeomorphic to $X$.

$S^{Top}(S^{2k+1}\times S^{2k+1})=0$,hence there is no fake $S^{2k+1}\times S^{2k+1}$.

For $S^{4k}\times S^{4k}$,we have $S^{Top}(S^{4k}\times S^{4k})\cong\mathbb{Z}\oplus\mathbb{Z}$ and $Aut(S^{4k}\times S^{4k})$ is finite group. This means $\mathcal{M}(S^{4k}\times S^{4k})$ is infinite.

How much do we know about fake $S^{4k}\times S^{4k}$? what is the general procedure of constructing fake $S^{4k}\times S^{4k}$?

For $S^{4k+2}\times S^{4k+2}$, $S^{Top}(S^{4k+2}\times S^{4k+2})\cong\mathbb{Z}_2\oplus\mathbb{Z}_2$ and $Aut(S^{4k+2}\times S^{4k+2})$ is still finite.since i do not know the action of $Aut(S^{4k+2}\times S^{4k+2})$ on $S^{Top}(S^{4k+2}\times S^{4k+2})$,i have no idea if $\mathcal{M}(S^{4k+2}\times S^{4k+2})$ is trivial or not,so

Is there a fake $S^{4k+2}\times S^{4k+2}$?