# 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 $\mathbb{R}^3$?

• 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? Apr 3, 2011 at 1:41
• Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? 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 $\mathbb{R}^3$.
– user21349
Jan 19, 2013 at 15:55
• @YaakovBaruch, isn't the cylinder factorizable? And could you elaborate this identity a little? Oct 17, 2013 at 6:35

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 $2n$.

• I hope this fine illustration of the power of relative homology will find its way in a textbook or, meanwhile, in algebraic topology courses. Apr 2, 2011 at 19:40
• 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). Mar 21, 2017 at 14:21
• 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. Mar 21, 2017 at 19:01
• So this also works for $\sqrt{\mathbb{R}^{2n+1}}$ doesn't it? Aug 3, 2018 at 8:20
• This should prove that if $\mathbb{R}^n=X^k$, then $k$ divides $n$, I think? Oct 17, 2019 at 1:38

this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof (which is, quoting from this post)

A linear map $$\Bbb R^n \to \Bbb R^n$$ can be understood to preserve or reverse orientation, depending on whether its determinant is $$+1$$ or $$-1$$. This notion of orientation can be generalized to arbitrary homeomorphisms, giving a "degree" $$\deg(m)$$ for every homeomorphism which is $$+1$$ if it is orientation-preserving and $$-1$$ if it is orientation-reversing. The generalization has all the properties that one would hope for. In particular, it coincides with the corresponding notions for linear maps and differentiable maps, and it is multiplicative: $$\deg(f \circ g) = \deg(f)\cdot \deg(g)$$ for all homeomorphisms $$f$$ and $$g$$. In particular (fact 1), if $$h$$ is any homeomorphism whatever, then $$h \circ h$$ is an orientation-preserving map.

Now, suppose that $$h : X^2 \to \Bbb R^3$$ is a homeomorphism. Then $$X^4$$ is homeomorphic to $$\Bbb R^6$$, and we can view quadruples $$(a,b,c,d)$$ of elements of $$X$$ as equivalent to sextuples $$(p,q,r,s,t,u)$$ of elements of $$\Bbb R$$.

Consider the map $$s$$ on $$X^4$$ which takes $$(a,b,c,d) \to (d,a,b,c)$$. Then $$s \circ s$$ is the map $$(a,b,c,d) \to (c,d,a,b)$$. By fact 1 above, $$s \circ s$$ must be an orientation-preserving map. But translated to the putatively homeomorphic space $$\Bbb R^6$$, the map $$(a,b,c,d) \to (c,d,a,b)$$ is just the linear map on $$\Bbb R^6$$ that takes $$(p,q,r,s,t,u) \to (s,t,u,p,q,r)$$. This map is orientation-reversing, because its determinant is $$-1$$. This is a contradiction. So $$X^4$$ must not be homeomorphic to $$\Bbb R^6$$, and $$X^2$$ therefore not homeomorphic to $$\Bbb R^3$$.

and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.

• I hope no one misses this nice alternative proof because it's behind a link. Apr 4, 2011 at 2:24
• 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. 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 ... Apr 4, 2011 at 15:05
• @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. Apr 5, 2011 at 5:42
• This argument is also given as exercise in Hatcher's "More exercises in algebraic topology". Oct 6, 2020 at 18:21

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

• Would you care to elaborate? Jan 19, 2013 at 5:09
• All things considered, perhaps "S" is not the best name for the topological space for this assertion. Jan 19, 2013 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. Jan 19, 2013 at 10:28
• @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 ${\mathbb Z}\not\cong{\mathbb Z}\times{\mathbb Z}$ Jan 19, 2013 at 11:27
• @IanAgol : On the LHS, $S^2$ refers to the $2$-sphere, while on the LHS $S$ refers to an arbitrary topological space. Nov 17, 2014 at 15:55

The Euler characteristic with compact support $$\chi_c(X)$$ is a very robust topological invariant available for any reasonable space (such as a subanalytic set). The key properties here are that $$\chi_c(X)$$ is a real number and $$\chi_c(A\times B)$$ is multiplicative:

$$\chi_c(X)\in \mathbb{R},\quad \quad \chi_c(A\times B)=\chi_c(A)\cdot \chi_c(B).$$

It follows that the Euler characteristic with compact support of a topological square is always non-negative: $$\forall X: \quad \chi_c(X\times X)\geq 0.$$

Thus, a space with negative Euler characteristic with compact support cannot be a topological square.

In particular, since $$\chi_c(\mathbb{R}^3)=-1$$, $$\mathbb{R}^3$$ is not a topological square.

More generally, since $$\chi_c(\mathbb{R}^n)=(-1)^n$$, for any $$k\in\mathbb{N}$$, $$\mathbb{R}^{2k+1}$$ is not a topological square.

For an introduction to the topological Euler characteristic with compact support, I would recommend the following notes by LIVIU NICOLAESCU.

• You say that $\chi_c$ is "available for any reasonable space" - does the argument above genuinely show that $\mathbb{R}^{2k+1}$ is not a topological square, or just that it isn't the square of a "reasonable" space in the appropriate sense? Oct 4, 2020 at 18:41
• The answer assumes that the square root has a well defined euler characteristic with compact support.
– JME
Oct 8, 2020 at 2:55