Is $\mathbb R^3$ the square of some topological space?



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 $\mathbb{R}^3$?

Richard Dore

Posted 2011-04-02T18:42:06.113

Reputation: 3 002

10I'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 – 2011-04-03T01:41:34.417

3Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? – Yaakov Baruch – 2011-04-04T16:38:18.597

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 , . 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 – 2013-01-19T15:55:50.700

@YaakovBaruch, isn't the cylinder factorizable? And could you elaborate this identity a little? – Ash GX – 2013-10-17T06:35:37.190

9Dear @BigM, I fail to see the value in editing an old question simply to add mathjax to its title which was perfectly readable to begin with. – Ricardo Andrade – 2015-05-31T11:44:50.003

1@Ricardo Andrade: every little improvement should be (mildly) welcome, don't you think so? – Qfwfq – 2018-08-03T08:23:23.150



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 $X \setminus \{x\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula) $$ H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k). $$ 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 $H_p(X, X \setminus \{x\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.

Tyler Lawson

Posted 2011-04-02T18:42:06.113

Reputation: 38 545

61I hope this fine illustration of the power of relative homology will find its way in a textbook or, meanwhile, in algebraic topology courses. – Georges Elencwajg – 2011-04-02T19:40:42.633

3I have a question regarding the top answer given by Tyler Lawson. As far as I know you can only apply the relative version of the Kunneth formula to cofibrations. Since we do not know much about $X$, it is unclear why $(X, X\setminus p)$ is a cofibration. Moreover, $(\mathbb R^3, \mathbb R^3\setminus p)$ is not a cofibration (I think). – freddy – 2017-03-21T14:21:45.233

@freddy, you are correct that (e.g.) the version that appears in Thm 3.18 in Hatcher's book does not apply to this example; Hatcher requires CW-pairs in his version of the proof. This is not the case for all versions of the theorem. – Tyler Lawson – 2017-03-21T18:58:19.380

3For example, Dold's version (Corollary 12.10 in Lectures on Algebraic Topology part VI) requires an excisive triad condition. The core of these assumotions is to ensure that, given $(X,A)$ and $(Y,B)$, the covering of $(X \times B) \cup (Y \times A)$ by $X \times B$ and $A \times Y$ is good enough to satisfy the assumptions of the Mayer-Vietoris theorem. This is, in particular, satisfied if $A$ is an open subset of $X$ and $B$ is an open subset of $Y$, or in the CW-inclusion version that Hatcher uses. – Tyler Lawson – 2017-03-21T19:01:55.600

(A similar version appears on page 249 of Spanier's book. I found both of these references here:

– Tyler Lawson – 2017-03-21T19:10:34.553

1So this also works for $\sqrt{\mathbb{R}^{2n+1}}$ doesn't it? – Pietro Majer – 2018-08-03T08:20:08.507

1@PietroMajer Indeed it does. – Tyler Lawson – 2018-08-04T20:21:39.087


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.

Henno Brandsma

Posted 2011-04-02T18:42:06.113

Reputation: 3 823

15I hope no one misses this nice alternative proof because it's behind a link. – Richard Dore – 2011-04-04T02:24:49.857

3Quoting 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 – 2011-04-04T05:16:12.043

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 – 2011-04-04T15:05:49.093

5@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 – 2011-04-05T05:42:08.323


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

Adam Epstein

Posted 2011-04-02T18:42:06.113

Reputation: 1 457

3Would you care to elaborate? – Ian Agol – 2013-01-19T05:09:55.853

68All things considered, perhaps "S" is not the best name for the topological space for this assertion. – Terry Tao – 2013-01-19T05:52:07.463

22@Terry Tao True enough, but in all honesty it's precisely the notational perversity that brought this to mind to begin with. – Adam Epstein – 2013-01-19T10:28:42.053

9@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 – 2013-01-19T11:27:02.840


As it happens, this started out as a wry comment about a different post, namely…. Then I noticed this question and accidentally posted as an answer what was intended as a mere comment.

– Adam Epstein – 2013-01-19T16:59:37.213

8@IanAgol : On the LHS, $S^2$ refers to the $2$-sphere, while on the LHS $S$ refers to an arbitrary topological space. – Prateek – 2014-11-17T15:55:38.770

This took long to grasp! – HeinrichD – 2016-09-15T08:45:29.670

1I think the whole point of this answer is the notational joke. It hardly gives any useful information in the "clarified" version. – Kostya_I – 2018-08-03T07:57:02.030