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Does there exist a topological space $X$ such that $X \ncong Y\times Y$ for every topological space $Y$ but $$X\times X \times X \cong Z\times Z$$ for some topological space $Z$ ?

Here $\cong$ means homeomorphic.

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  • $\begingroup$ There can be no example in which $X$ is finite. $\endgroup$ – Paul Frost Dec 25 '18 at 12:01
  • $\begingroup$ @PaulFrost: How do you prove this if $X$ is a finite non-Hausdorff space? The Hausdorff case is indeed trivial. $\endgroup$ – Moishe Kohan Dec 25 '18 at 14:37
  • $\begingroup$ @MoisheCohen Yes, I was too fast. It is true if $X$ has the trivial or the discrete topology, but in the general case I do not have a proof. So perhaps it is worth looking closer to the finite case. $\endgroup$ – Paul Frost Dec 25 '18 at 15:03
  • $\begingroup$ Is the corresponding problem solvable for groups? $\endgroup$ – ThorbenK Dec 25 '18 at 21:01
  • $\begingroup$ @MoisheCohen For finite algebraic structures (for example, topological spaces), "root extraction" is unique, and $X$ will be isomorphic to "$Z^{\frac 2 3}$" $\endgroup$ – xsnl Dec 27 '18 at 7:13
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Yes.

Here are two pieces of input data.

1) Let $M$ be any noncompact space. Beyond the usual invariants like homology and homotopy groups, there is a further invariant of the homeomorphism type (or proper homotopy type) of $M$, called the fundamental group at infinity: if you choose a proper (inverse image of compact sets is compact) map $\gamma: [0,\infty) \to M$, and let $K_n$ be an increasing compact exhaustion of $M$ (that is, $K_n \subset K_{n+1}$ and $\bigcup K_n = M$) so that $\gamma(t) \not \in K_n$ for $t \in [n, n+1]$, then one may write the inverse limit $$\pi_1^\infty(M,\gamma) := \lim \pi_1(M - K_n, \gamma(n));$$ strictly speaking we have restriction maps $$\pi_1(M - K_n, \gamma(n)) \to \pi_1(M - K_{n-1}, \gamma(n)),$$ but we may use the path $\gamma$ from $\gamma(n)$ to $\gamma(n-1)$ to get a natural isomorphism $\pi_1(M - K_{n-1}, \gamma(n)) \to \pi_1(M - K_{n-1}, \gamma(n-1))$ so that we may take the inverse limit over a sequence of maps as above. This is essentially independent of the choice of sequence $K_n$. It only depends on the ray $\gamma$ up to a proper homotopy.

(Similarly, there is a notion of the set of ends of a space - this is the inverse limit over $\pi_0(M - K_n)$. This is the set we choose $\gamma$ from, in the sense that we choose a connected component for the usual fundamental group.)

2) If $M$ is a smooth, connected, noncompact manifold of dimension $n \geq 5$, a theorem of Stallings (the piecewise linear structure of Euclidean space, here) says that if $M$ is both contractible and $\pi_1^\infty(M,\gamma) = 0$ for the unique end $\gamma$ of $M$, then $M \cong \Bbb R^n$.

Our strategy, therefore, is to find a noncompact, contractible smooth manifold $M$ of dimension $n \geq 3$ with nontrivial fundamental group at infinity. We will argue that $\pi_1^\infty(M^k, \gamma) = 0$ for $k>1$, and hence that $M^k \cong \Bbb R^{nk}$. Because you asked for a square root of $3n$, we should take $n$ even. At the end we will specify $n = 4$.

Here is a helpful tool in constructing such noncompact manifolds. If $M$ is a compact manifold with boundary, then its ends of its interior $M^\circ$ are in bijection with $\pi_0(\partial M)$, and the fundamental group at infinity is equal to $\pi_1(\partial M)$. (Take the ray to extend to a map $[0, \infty] \to M$, and let the basepoint in $\partial M$ be $\gamma(\infty)$; if $[0,1) \times \partial M \subset M$ is a collar of the boundary, let the compact exhaustion be the complement of $[0, 1/n) \times \partial M$.)

In this situation above, the product $M \times M$ is a compact topological manifold with boundary (it has "corners", but these are topologically the same as boundary points). The boundary is homeomorphic to $(\partial M \times M) \cup_{\partial M \times \partial M} (M \times \partial M)$. If $\partial M$ is connected and $M$ is simply connected, the Seifert van Kampen theorem dictates that the fundamental group of the result is $$\pi_1\partial M *_{\pi_1 \partial M \times \pi_1 \partial M} \pi_1 \partial M = 0.$$

Therefore, if $M$ is simply connected with connected boundary, $M \times M$ has simply connected boundary; and hence $(M \times M)^\circ = M^\circ \times M^\circ$.

What this proves, altogether, is that if $M$ is a compact, contractible manifold of dimension $n \geq 3$, and $\pi_1(\partial M) \neq 0$, then $M$ is not homeomorphic to $\Bbb R^n$, but $M^k$ is homeomorphic to $\Bbb R^{nk}$ for any $k > 1$. What remains is twofold: to show that such $M$ exist; and to find one that is itself not a square.


First, existence. In dimension 3 there are none of these of interest: a compact contractible 3-manifold is homeomorphic to the 3-ball by the solution of the Poincare conjecture. In dimension 4 these are called Mazur manifolds and come in great supply. In dimension $n \geq 5$, if $\Sigma$ be an $(n-1)$-manifold which has $H_*(\Sigma;\Bbb Z) \cong H_*(S^{n-1};\Bbb Z)$, it is a theorem of Kervaire that $\Sigma$ bounds a contractible manifold $M$. If $\pi_1 \Sigma \neq 0$ (which is equivalent to saying "$\Sigma$ is not homeomorphic to the $(n-1)$-sphere", by the higher-dimensional Poincare conjecture), then this gives an example of what we want. (In fact, for $n \geq 6$, Kervaire proved that you can even construct such `homology $(n-1)$-spheres' with any specified finitely presented fundamental group $\pi$, modulo the conditions $H_1(\pi) = H_2(\pi) = 0$.) So we see there is any such compact manifold $M$, and hence noncompact manifold $M^\circ$, for any dimension $n \geq 4$.


If $M^\circ$ were a product $X \times X$ of two spaces, then first, observe $X$ would need to be contractible; second, it is a homology manifold (this is a local condition in terms of the relative homology of $(X, X - p)$ at all points $p$ which ensures duality properties) of dimension $\dim M/2$. A homology manifold of dimension $\leq 2$ is a manifold (this seems to be well-known, but the only reference I could find was Theorem 16.32 in Bredon's sheaf theory), so let's take $\dim M = 4$ here; then $X$ is a contractible surface, so the classification of compact surfaces implies $X \cong \Bbb R^2$ (see eg here). This contradicts $0 = \pi_1^\infty(\Bbb R^4) \cong \pi_1^\inf(M^\circ) \neq 0$, and so this is impossible.

In fact, with some more work you can show that this $M$ may not even be decomposed into a product at all.

EDIT: Thanks to Moishe Cohen's answer here we can prove that if $M$ is a compact contractible manifold of dimension $n \geq 4$ for which $\pi_1 \partial M \neq 0$, then $M$ does not admit a square root. For if it did, $X \times X = M$, the space $X$ would be a contractible homology manifold of dimension at least 2; by Moishe's answer, it must have one end. Using the decomposition $\text{End}(X \times X) \sim \text{End}(X) * \text{End}(X)$ of end-spaces of a product, we see that $\pi_1^\inf(X \times X) = \pi_1^\inf(X) *_{\pi_1^\inf(X) \times \pi_1^\inf(X)} \pi_1^\inf(X) = 0$, exactly as in the case of manifolds with boundary. Thus $M$ admits no square root.

This method thus produces some $M$ that admits no square root but whose $n$th power admits a $k$th root, for any pair of positive integers $(n,k)$ with $n > 1$. It has no power to find spaces for which $X^j$ is similarly un-rootable for $j$ in some range; it is unique to $j=1$ that this works.

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  • $\begingroup$ Nice example. If $X$ were a smooth $k$-manifold with more then 1 end then $H_{k-1}(X)\ne 0$ contradicting contractibility of $X$. I'll think if it also works form homology manifolds. $\endgroup$ – Moishe Kohan Dec 28 '18 at 2:32
  • $\begingroup$ Yes, indeed, it works, one just needs the locally finite fundamental class and the Poincare duality, which you get in this setting. $\endgroup$ – Moishe Kohan Dec 28 '18 at 2:45
  • $\begingroup$ @MoisheCohen I wanted to argue something like that, but I wasn't sure how you define the class in $H^1_c(X)$ that you want to dualize. Presumably as a function on chains you'd like to imagine it as carrying a natural homomorphism to $\Bbb Z^{\# \text{ends}}$ by "counting the number of times a chain goes to some end", but I didn't see a way to make sense of this. $\endgroup$ – user98602 Dec 28 '18 at 2:48
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    $\begingroup$ @MoisheCohen If it is too subtle for a comment box, would you write a brief answer if I wrote this as another question? It would be nice to say that all contractible $M$ of dimension at least 4 which are not trivial at infinity do not have square roots. $\endgroup$ – user98602 Dec 28 '18 at 18:04
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    $\begingroup$ Mike, I can write this in the next few days. The main thing is that generalized manifolds satisfy Poincare duality; this is widely known but the only reference I have is an old book "Topology of Manifolds" by Wilder. (Incidentally, Wilder also proves there Moore's theorem about characterization of surfaces that you used in the proof.) One can actually argue without defining a locally finite fundamental class. It indeed takes more space than a comments box would allow. $\endgroup$ – Moishe Kohan Dec 28 '18 at 21:35

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