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Recently, Richard Dore asked us if $\mathbb R^3$ is the cartersian square of some space, and Tyler Lawson answered beautifully in the negative.

The even powers of $\mathbb R$ were left out in that question because, well, it is obvious that they are squares. Now:

Are they squares in a unique way? In other words, if a space $X$ is such that $X\times X\cong\mathbb R^{2n}$, must $X$ be homeomorphic to $\mathbb R^n$?

One can consider other values of $2$ in Richard's question or here, as well as look for factors instead of only square roots (but if I recall correctly $\mathbb R^5$ has all exotic $\mathbb R^4$s as factors, so the last variant might be «trivial»...)

flag	asked Apr 2 2011 at 23:25 Mariano Suárez-Alvarez 30.6k●7●68●153

I wonder if there is a way to attach interesting meaning to the statement «$\mathbb R^4$ is a square of $\mathbb R^2$ in only one way (up to some equivalence).» – Mariano Suárez-Alvarez Apr 2 2011 at 23:32

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Shouldn't the answer be no -- if I'm not making some elementary mistake I believe the product of two Whitehead manifolds is homeomorphic to $\mathbb R^6$. – Ryan Budney Apr 2 2011 at 23:48

But is seems that $\mathrm{(something\;which\;is\;not\;a\;manifold)}\times\mathbb R$ can be a manifold, so maybe the factors/square roots need not be manifolds themselves? – Mariano Suárez-Alvarez Apr 3 2011 at 0:03

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@Olivier, it is not true: Wikipedia tells me that if $X=R^3/W$ is the result of contracting a standardly embedded Whitehead continuum $W\subseteq\mathbb R^3$, then $X$ is not a manifold yet $X\times\mathbb R$ is a copy of $\mathbb R^4$. – Mariano Suárez-Alvarez Apr 3 2011 at 7:31

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I think the square of any fake $R^4$ must be the standard $R^8$. – Allen Knutson Apr 4 2011 at 1:46

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Ryan Budney · Answer 1 · 2011-04-03 00:16:37Z UTC

I'm pretty certain the answer is no provided $n \geq 3$. Let $X$ be the Whitehead manifold: http://en.wikipedia.org/wiki/Whitehead_manifold

It's a contractible open 3-manifold which is not homeomorphic to $\mathbb R^3$.

$X^2$ I claim is homeomorphic to $\mathbb R^6$. I don't have a slick proof of this. The idea is, ask yourself if you can put a boundary on $X^2$ to make $X^2$ the interior of a compact manifold with boundary. Larry Siebenmann's dissertation says if $X^2$ is simply-connected at infinity, you're okay. But the fundamental-group of the end of $X^2$ has a presentation of the form $(\pi_1 Y * \pi_1 Y) / <\pi_1 Y^2>$, where $*$ denotes free product and angle brackets "normal closure", and $\pi_1 Y$ is "the fundamental group at infinity for $X$". So $\pi_1 X^2$ is the trivial group.

Once you have it as the interior of a compact manifold with boundary, the h-cobordism theorem kicks in and tells you this manifold is $D^6$.

Square roots of $\mathbb R^{2n}$

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