# fake $S^{2k}\times S^{2k}$

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 $S^{4k+2}\times S^{4k+2}$?

• Sorry for my ignorance :) I did not understand what do you mean by $S^{Top}(X)$ ? What is $Aut(X)$ ? is it the group $\pi_{0}$ of the space of self equivalences of $X$, thank you. – Max Feb 17 '15 at 15:43
• For $S(X)$,please check en.wikipedia.org/wiki/Surgery_structure_set. For $Aut(X)$, this is the group of self homotopy equivalence of $X$ under composition. – user2015 Feb 17 '15 at 15:48

The following paper may be helpful:

Wall, C. T. C. Classification of $(n−1)$-connected $2n$-manifolds. Ann. of Math. (2) 75 1962 163–189.

Let me quote one result from this paper:

Theorem 5. If $\pi_{n-1}(SO)=0$ ,$n\geq 3$,and $M_1$ and $M_2$ are differential $(n-1)$ connected $2n$-manifolds of the same homotopy type,then for some manifold $T$ homeomorphic (and so combinatorially equivalent) to $S^{2n}$, $M_1$ is diffeomorphic to $M_2\sharp T$. If $n=3,6$, $M_1$ is diffeomorphic to $M_2$.

This tells you that there is no smoothable fake $S^n\times S^n$ for $n\equiv 6\mod8$

You may want to have a look of the following paper by Kreck and Lueck:

Topological Rigidity for Non-aspherical Manifolds

where they showed that a necessary condition for $S^d\times S^d$ ($d>2$) being Borel (which means $Aut(S^d \times S^d)$ acts on $S^{Top}(S^d \times S^d)$ transitively,or equivalently,there is no "fake" $S^d\times S^d$ in your sense) is that $d$ is odd or $2d+2=2^l$ for some $l$.

Now for $S^{4k+2}\times S^{4k+2}$ ($k>0$),we know the necessary condition above is not satisfied,hence,there is some fake $S^{4k+2}\times S^{4k+2}$.

For $S^2\times S^2$,it is indeed Borel,i.e.Every manifold which is homotopy equivalent to $S^2\times S^2$ is actually homeomorphic to it.