Under what conditions on topology of a given space $X$ it can be written as product space $Y\times Y$ for some other topological space $Y$. Lets call such space $Y$ a square root of X. Then given that square root of $X$ exists is it unique (within topological isomorphism) ?
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7$\begingroup$ mathoverflow.net/questions/60395/square-roots-of-mathbb-r2n is relevant. $\endgroup$– PseudoNeoCommented Jul 10, 2012 at 20:35
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4$\begingroup$ math.stackexchange.com/questions/78055/… también. $\endgroup$– PseudoNeoCommented Jul 10, 2012 at 20:36
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2 Answers
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I’m not at all sure that there is a nice answer to the first question, but the answer to the second is no: there are non-homeomorphic spaces with homeomorphic squares. In this paper M.M. Marjanović and Ante R. Vučemilović construct a pair of non-homeomorphic countable metrizable spaces whose squares are homeomorphic.
Added: I believe that the question is originally due to S. Ulam, and that the first example is to be found in R.H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278–287; link.
answered Jul 10, 2012 at 20:38
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$\begingroup$ thanks for your answer and references $\endgroup$ Commented Jul 10, 2012 at 21:36
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Here is a necessary condition for a space to have a square root. By the Künneth theorem, if $X=Y\times Y$, then with respect to a field $F$, we must have $H_k(X;F)=\bigoplus_{i+j=k}H_i(Y;F)\otimes H_j(Y;F)$. Therefore, associating to the space $X$ the sequence of betti numbers $b_i(X;F)=\dim_F H_i(X;F)$, we must have that the sequence $(b_i)$, if consisting of finite numbers, must be the convolution square of another sequence of non-negative integers.
If there are only finitely many non-zero betti numbers for $X$, then the problem of existence of a convolution square root is the same as determining whether a polynomial has a polynomial square root with non-negative integer coefficients. If there are infinitely many non-zero betti numbers, the problem becomes one related to square roots of power series, and I'm not sure if there is an easy way to tackle that.
In any event, this gives the nice result that an odd dimensional compact manifold cannot have a square root.
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$\begingroup$ Note that you can generalize this idea by using the relative Kunneth formula (e.g., this post does it for $\Bbb{R}^3$: mathoverflow.net/questions/60375/…). $\endgroup$– MicahCommented Jul 11, 2012 at 0:41
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$\begingroup$ @Micah: Thanks, that's very nice. I should have tried to do a reference search before answering the question! $\endgroup$– AaronCommented Jul 11, 2012 at 0:50