Let $X$ be a fixed closed manifold,define$S(X)$ the structure set and $Aut(X)$ the group of self homotopy equivalence of $X$.
$\mathcal{M}(X):=\{M\mid M$ surgery theory tells us that $\mathcal{M}(X):=S(X)/Aut(X)$ is closed manifold whichin 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\}/\sim_{\text{homeo}}$$X$.
AWe call manifold $M$ is called a fakefake $X$ if $M\in\mathcal{M}(X)-\{X\}$,i.e.a fake $X$ is a manifold which$M$ is homotopy equivalent but not homeomorphic to $M$.
We know $\mathcal{M}(X)\cong S(X)/Aut(X)$ where $S(X)$ is the structure set and $Aut(X)$ is the group of self homotopy equivalence of $X$. $Aut(X)$ acts on $S(X)$ by composition.
$S^{Top}(S^{2k+1}\times S^{2k+1})=0$,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 $S^{4k+2}\times S^{4k+2}$?