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The other day, I was idly considering when a topological space has a square root. That is, what spaces are homeomorphic to $X \times X$ for some space $X$. $\mathbb{R}$ is not such a space: If $X \times X$ were homeomorphic to $\mathbb{R}$, then $X$ would be path connected. But then $X \times X$ minus a point would also be path connected. But $\mathbb{R}$ minus a point is not path connected.

A next natural space to consider is $\mathbb{R}^3$. My intuition is that $\mathbb{R}^3$ also doesn't have a square root. And I'm guessing there's a nice algebraic topology proof. But that's not technology I'm much practiced with. And I don't trust my intuition too much for questions like this.

So, is there a space $X$ so that $X \times X$ is homeomorphic to $\mathbb{R}^3$?

flag	asked Apr 2 2011 at 18:42 Richard Dore 2,395●2●16●29

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I'm wondering to what extent there is unique factorization of topological spaces relative to $\times$. $\mathbb{Q}$ is an idempotent (as is its complement in $\mathbb{R}$), but are there more interesting failures of UF involving connected spaces? Or results establishing UF for "nice" families of spaces? Should these be posted as a new question? – Yaakov Baruch Apr 3 2011 at 1:41

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Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? – Yaakov Baruch Apr 4 2011 at 16:38

Without knowing any algebraic topology, it's possible to conclude at least something about X. If X is metric, compact, or locally compact and paracompact, then $\dim(X\times X)\le 2\dim X$, which means X has to have Lebesgue covering dimension at least 2. Wage, Proc. Natl. Acad. Sci. USA 75 (1978) 4671 , www.pnas.org/content/75/10/4671.full.pdf . What is the weakest condition that guarantees $\dim(X\times Y)= \dim X+\dim Y$? Given Yaakov Baruch's comment about the "dogbone space," it's not obvious that X is at all well behaved simply from the requirement that its square is $\mathbb{R}^3$. – Ben Crowell Jan 19 at 15:55

Tyler Lawson · Accepted Answer · 2011-04-02 19:10:50Z UTC

No such space exists. Even better, let's generalize your proof by converting information about path components into homology groups.

For an open inclusion of spaces $X \setminus \{p\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula) $$ H_n(X \times X, X \times X \setminus \{(p,p)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{p\};k) \otimes_k H_q(X, X \setminus \{p\};k). $$ If the product is $\mathbb{R}^3$, then the left-hand side is $k$ in degree 3 and zero otherwise, so something on the right-hand side must be nontrivial. However, if $H_p(X, X \setminus \{p\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.

Henno Brandsma · Answer 2 · 2011-04-02 21:25:00Z UTC

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this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.

answered Apr 2 2011 at 21:25

Henno Brandsma
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I hope no one misses this nice alternative proof because it's behind a link. – Richard Dore Apr 4 2011 at 2:24

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Quoting from the link: "The paper also refers to an earlier paper ("The cartesian product of a certain nonmanifold and a line is E4", R.H. Bing, Annals of Mathematics series 2 vol 70 1959 pp. 399–412) which constructs an extremely pathological space B, called the "dogbone space", not even a manifold, which nevertheless has B × R^3 = R4." This is relevant to my comment to the OP. – Yaakov Baruch Apr 4 2011 at 5:16

I don't understand this step in the proof: Why does the map $X^4 \to X^4, (a,b,c,d) \mapsto (c,d,a,b)$ correspond to the map $R^6 \to R^6, (p,q,r,s,t,u) \mapsto (s,t,u,p,q,r)$? I mean, the homeomorphism is not supposed to commute with projections ... – Martin Brandenburg Apr 4 2011 at 15:05

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@Martin: The homeomorphism $(X\times X)\times (X\times X)\cong \mathbb R^3 \times \mathbb R^3$ respects projections by construction, so swapping the "two factors" (which I've emphasized with parentheses) on the left hand side corresponds to swapping the two factors on the right hand side. – Anton Geraschenko♦ Apr 5 2011 at 5:42

Adam Epstein · Answer 3 · 2013-01-19 00:12:41Z UTC

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I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.

answered Jan 19 at 0:12

Adam Epstein
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Would you care to elaborate? – Agol Jan 19 at 5:09

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All things considered, perhaps "S" is not the best name for the topological space for this assertion. – Terry Tao Jan 19 at 5:52

@Terry Tao True enough, but in all honesty it's precisely the notational perversity that brought this to mind to begin with. – Adam Epstein Jan 19 at 10:28

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@Agol Fix $s\in S$. On the one hand, $\pi_2(S\times S,(s,s))\cong \pi_2(S,s)\times\pi_2(S,s)$. On the other hand, $\pi_2({\bf S},{\bf s})\cong{\mathbb Z}$ for any 2-sphere $\bf S$ and any ${\bf s}\in{\bf S}$. Now it suffices to observe that ${\mathbb Z}\not\cong G\times G$ for any group $G$: indeed, such a group must be an infinite quotient of $\mathbb Z$, whence $G\cong{\mathbb Z}$, but ${\mathbb Z}\not\cong{\mathbb Z}\times{\mathbb Z}$ – Adam Epstein Jan 19 at 11:27

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As it happens, this started out as a wry comment about a different post, namely mathoverflow.net/questions/115799…. Then I noticed this question and accidentally posted as an answer what was intended as a mere comment. – Adam Epstein Jan 19 at 16:59

Is R^3 the square of some topological space?

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