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 $\mathbb{R}^3$?

  • 10
    $\begingroup$ 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? $\endgroup$ – Yaakov Baruch Apr 3 '11 at 1:41
  • 4
    $\begingroup$ Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? $\endgroup$ – Yaakov Baruch Apr 4 '11 at 16:38
  • $\begingroup$ 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 $\mathbb{R}^3$. $\endgroup$ – Ben Crowell Jan 19 '13 at 15:55
  • 9
    $\begingroup$ Dear @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. $\endgroup$ – Ricardo Andrade May 31 '15 at 11:44
  • 2
    $\begingroup$ @Ricardo Andrade: every little improvement should be (mildly) welcome, don't you think so? $\endgroup$ – Qfwfq Aug 3 '18 at 8:23

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 $2n$.

  • 65
    $\begingroup$ I hope this fine illustration of the power of relative homology will find its way in a textbook or, meanwhile, in algebraic topology courses. $\endgroup$ – Georges Elencwajg Apr 2 '11 at 19:40
  • 3
    $\begingroup$ I 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). $\endgroup$ – freddy Mar 21 '17 at 14:21
  • 3
    $\begingroup$ For 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. $\endgroup$ – Tyler Lawson Mar 21 '17 at 19:01
  • 2
    $\begingroup$ So this also works for $\sqrt{\mathbb{R}^{2n+1}}$ doesn't it? $\endgroup$ – Pietro Majer Aug 3 '18 at 8:20
  • 2
    $\begingroup$ @PietroMajer Indeed it does. $\endgroup$ – Tyler Lawson Aug 4 '18 at 20:21

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.

  • 15
    $\begingroup$ I hope no one misses this nice alternative proof because it's behind a link. $\endgroup$ – Richard Dore Apr 4 '11 at 2:24
  • 3
    $\begingroup$ 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. $\endgroup$ – Yaakov Baruch Apr 4 '11 at 5:16
  • $\begingroup$ 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 ... $\endgroup$ – Martin Brandenburg Apr 4 '11 at 15:05
  • 5
    $\begingroup$ @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. $\endgroup$ – Anton Geraschenko Apr 5 '11 at 5:42

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

  • 3
    $\begingroup$ Would you care to elaborate? $\endgroup$ – Ian Agol Jan 19 '13 at 5:09
  • 76
    $\begingroup$ All things considered, perhaps "S" is not the best name for the topological space for this assertion. $\endgroup$ – Terry Tao Jan 19 '13 at 5:52
  • 24
    $\begingroup$ @Terry Tao True enough, but in all honesty it's precisely the notational perversity that brought this to mind to begin with. $\endgroup$ – Adam Epstein Jan 19 '13 at 10:28
  • 9
    $\begingroup$ @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 ${\mathbb Z}\not\cong{\mathbb Z}\times{\mathbb Z}$ $\endgroup$ – Adam Epstein Jan 19 '13 at 11:27
  • 8
    $\begingroup$ @IanAgol : On the LHS, $S^2$ refers to the $2$-sphere, while on the LHS $S$ refers to an arbitrary topological space. $\endgroup$ – Prateek Nov 17 '14 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.