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Tue, 09 Jan 2007
R3 is not a square
It would be rather surprising if there were, since you could then describe any point in space unambiguously by giving its two coordinates from X. This would mean that in some sense, R^{3} could be thought of as two-dimensional. You would expect that any such X, if it existed at all, would have to be extremely peculiar. I had been wondering about this rather idly for many years, but last week a gentleman on IRC mentioned to me that there had been a proof in the American Mathematical Monthly a couple of years back that there was in fact no such X. So I went and looked it up. The paper was "Another Proof That R^{3} Has No Square Root", Sam B. Nadler, Jr., American Mathematical Monthly vol 111 June–July 2004, pp. 527–528. The proof there is straightforward enough, analyzing the topological dimension of X and arriving at a contradiction. But the Nadler paper referenced an earlier paper which has a much better proof. The proof in "R^{3} Has No Root", Robbert Fokkink, American Mathematical Monthly vol 109 March 2002, p. 285, is shorter, simpler, and more general. Here it is. A linear map R^{n} → 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 o g) = deg(f)·deg(g) for all homeomorphisms f and g. In particular ("fact 1"), if h is any homeomorphism whatever, then h o h is an orientation-preserving map. Now, suppose that h : X^{2} → R^{3} is a homeomorphism. Then X^{4} is homeomorphic to 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 R. Consider the map s on X^{4} which takes (a,b,c,d) → (d,a,b,c). Then s o s is the map (a,b,c,d) → (c,d,a,b). By fact 1 above, s o s must be an orientation-preserving map. But translated to the putatively homeomorphic space R^{6}, the map (a,b,c,d) → (c,d,a,b) is just the linear map on R^{6} that takes (p,q,r,s,t,u) → (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 R^{6}, and X^{2} therefore not homeomorphic to R^{3}. The same proof goes through just fine to show that R^{2n+1} = X^{2} is false for all n, and similarly for open subsets of R^{2n+1}. The paper also refers to an earlier paper ("The cartesian product of a certain nonmanifold and a line is E^{4}", 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} = R^{4}. This is on my desk, but I have not read this yet, and I may never.
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