Does $X\times S^1\cong Y\times S^1$ imply that $X\times\mathbb R\cong Y\times\mathbb R$?

Question

This question came up in a recent video series of lectures by Mike Freedman available through Max Planck Institut's website. He proves the "difficult" converse direction, that $X\times \mathbb R\cong Y\times \mathbb R$ implies $X\times S^1\cong Y\times S^1$ using a subtle "push-pull" argument, when $X$ and $Y$ are compact. He goes on to make the remark that the converse holds in the simply connected case by taking universal covers, but it is not obvious in general and there may even be a counterexample. So my question is whether anyone has any ideas about this question. Maybe someone can spot a counterexample in the non-simply connected case?

Edit: Here's a link to the video series. If I recall correctly, the push-pull argument is in the first video.

Answer 1

No, it is not possible to conclude that $X\times\mathbb{R}\cong Y\times\mathbb{R}$ from $X\times S^1\cong Y\times S^1$, even in the case where $X$ and $Y$ are compact differentiable manifolds. In fact, Charlap¹ showed the following in 1965.

There exist compact manifolds $X$, $Y$ of different homotopy types, but with $X\times S^1$ diffeomorphic to $Y\times S^1$.

As $X\times\mathbb{R}$ has the same homotopy type as $X$ for any topological space $X$, this implies that $X\times\mathbb{R}$ and $Y\times\mathbb{R}$ have different homotopy types, so they are certainly not homeomorphic. See also Hilton, Mislin & Roitberg² for examples of non-homotopy equivalent manifolds with diffeomorphic products with higher dimensional spheres.

I just found those references by a bit of searching around, so I am not familiar with all of the details. However, I will now construct an explicit example of compact manifolds with different fundamental groups (hence, different homotopy types), but with diffeomorphic products with the circle. I think this example works along similar lines to the examples which could be constructed with the results of Charlap. In my example, $X$ and $Y$ will be 10-dimensional manifolds with covering space $S^9\times\mathbb{R}$.

Since the products $X\times S^1$ and $Y\times S^1$ are to be diffeomorphic, they must have the same fundamental group. So, the first thing is to find two non-isomorphic groups $G$, $H$ whose direct products with the infinite cyclic group $Z$ (i.e., the integers under addition) are isomorphic. As it is always possible to cancel direct products with $Z$ for abelian groups, it is necessary that $G$ and $H$ are non-abelian. Expressed in terms of generators and relations, one such example is, $$ \begin{align} G &= \langle x,y\mid yx = xy^2, y^{32}=y\rangle,\cr H &= \langle x,y\mid yx = xy^4, y^{32}=y\rangle. \end{align} $$ This is a very slightly simplified version of the example given in Hirshon³. Note that each element of $G$ (resp., $H$) can be uniquely expressed as $x^ry^s$ with $s$ taken modulo 31, and satisfy the multiplication rule $(x^{r_1}y^{s_1})(x^{r_2}y^{s_2})=x^{r_1+r_2}y^{2^{r_2}s_1+s_2}$ (resp., $x^{r_1+r_2}y^{4^{r_2}s_1+s_2}$). These groups are not isomorphic. Indeed, if $\bar x$, $\bar y$ are elements generating $G$ with $\bar y$ of order 31, then we must have $\bar y$ a power of $y$ and $\bar x=x^{\pm1}y^s$, so that $\bar x^{-1}\bar y\bar x = \bar y^2$ or $\bar y^{16}$. In either case, this does not equal $\bar y^4$, so $G\not\cong H$.

On the other hand, the direct product $G\times Z$ can be formed by adding an additional generator $z$ to $G$ which commutes with $x$ and $y$ and satisfies no further relations. Setting $\bar x = x^2z$ and $\bar z = x^5z^2$, then $\bar x, y, \bar z$ also generate $G\times Z$, $\bar z$ commutes with $\bar x$ and $y$, and $\bar x,y$ satisfy the relation $y\bar x=\bar xy^4$. From this it can be seen that $G\times Z\cong H\times Z$.

I'll now define the spaces $X,Y$. The simply connected manifold $\hat X\equiv S^9\times\mathbb{R}$ can be realized as the submanifold of $\mathbb{C}^5\times\mathbb{R}$ consisting of the points $(z,s)$ with $\lVert z\rVert = 1$. Define the diffeomorphisms $R,S$ on $\hat X$ as $$ \begin{align} & R(z_0,z_1,z_2,z_3,z_4,s)=(\omega z_0,\omega^2z_1,\omega^4z_2,\omega^8z_3,\omega^{16}z_4,s),\cr & S(z_0,z_1,z_2,z_3,z_4,s)=(z_4,z_0,z_1,z_2,z_3,s+1) \end{align} $$ where $\omega=e^{2\pi i/31}$. These satisfy the relations $RS=SR^2$ and $R^{32}=R$, so the groups $\Lambda_G\equiv\langle S,R\rangle$ and $\Lambda_H\equiv\langle S^2,R\rangle$ are isomorphic to $G$ and $H$ respectively. Define $X$ and $Y$ to be the quotients $\hat X/\Lambda_G$ and $\hat X/\Lambda_H$. These are compact manifolds with non-isomorphic fundamental groups $\Lambda_G\cong G$ and $\Lambda_H\cong H$.

Finally, I'll show that $X\times S^1$ and $Y\times S^1$ are diffeomorphic. Note that $X\times S^1$ is just the manifold $X\times \mathbb{R}$ quotiented out by the translations $(x,t)\mapsto(x,t+n)$ ($n\in\mathbb{Z}$). This can be written as a quotient $\hat X\times\mathbb{R}/\Lambda^\prime$, where $\Lambda^\prime=\langle S\times I, R\times I, \hat I\times T\rangle$. Here, $\hat I, I$ are the identities on $\hat X$ and $\mathbb{R}$, and $T$ is the translation on $\mathbb{R}$ given by $t\mapsto t+1$. However, writing $\bar S=S^2\times T$ and $\bar T=S^5\times T^2$, then $\bar S, R\times I,\bar T$ generate $\Lambda^\prime$ and we have $$ \begin{align} U^{-1}\bar SU&=S^2\times I,\cr U^{-1}(R\times I)U&=R\times I,\cr U^{-1}\bar TU&=\hat I\times T, \end{align} $$ where $U$ is the diffeomorphism on $S^9\times\mathbb{R}\times\mathbb{R}$ given by $U(z,s,t)=(z,s+5t,s/2+2t)$. So, with $\cong$ denoting diffeomorphism, $$ \begin{align} X\times S^1 &\cong S^9\times\mathbb{R}\times\mathbb{R}/\langle S\times I,R\times I,\hat I\times T\rangle\cr &\cong S^9\times\mathbb{R}\times\mathbb{R}/\langle S^2\times I,R\times I,\hat I\times T\rangle\cr &\cong Y\times S^1. \end{align} $$

Aside: As was noted in the question and shown in the linked lecture series, the implication $X\times\mathbb{R}\cong Y\times\mathbb{R}\Rightarrow X\times S^1\cong Y\times S^1$ holds for compact spaces $X,Y$. However, it does not hold if the spaces are not compact. Although there are rather involved counterexamples, such as $X=\mathbb{R}^3$ and $Y$ being the Whitehead manifold, there are also much simpler counterexamples, and I'll just mention one here that I thought of. Take $X$ to be the sphere minus three points and $Y$ to be a torus minus one point. Then, $X$ is just the same as the open disc minus two points, so $X\times S^1$ embeds in $\mathbb{R}^3$. On the other hand, there exists an embedded closed surface and curve in $Y\times S^1$ with intersection number 1 (if it was embeddable in $\mathbb{R}^3$, the intersection number would have to be zero). So, $X\times S^1\not\cong Y\times S^1$. I'll leave the construction of the closed surface and curve as an interesting exercise, and also the homeomorphism showing that $X\times\mathbb{R}\cong Y\times\mathbb{R}$.

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¹Compact Flat Riemannian Manifolds: I, Leonard S. Charlap, Annals of Mathematics Second Series, Vol. 81, No. 1 (Jan., 1965), pp. 15-30. (link)

²Sphere Bundles Over Spheres and Non-Cancellation Phenomena, P. Hilton, G. Mislin & J. Roitberg. J. London Math. Soc. (1972) s2-6(1): 15-23 (link)

³On Cancellation in Groups, R. Hirshon, The American Mathematical Monthly. Vol. 76, No. 9 (Nov., 1969), pp. 1037-1039. (link) (Alt. freely available link from R. Hirshon's home page)