# If $\text{dim}(X \times X) = 2\text{dim}(X)$, does $\text{dim}(X^n) = n\text{dim}(X)$?

I have been learning some (topological) dimension theory and have gotten through most of the basic material, at this point, and am about to start looking at papers. In particular, I want to get familiar with the standard counterexamples regarding dimension of products, but I haven't noticed the question in the title addressed.

So my question would be what conditions on $$X$$ (e.g. separable metric, compact metric, continuum, homology manifold) are sufficient to ensure that if $$\text{dim}(X \times X) = 2\text{dim}(X)$$, then $$\text{dim}(X^n) = n\text{dim}(X)$$? The initial assumption could also be weakened from $$\text{dim}(X \times X) = 2\text{dim}(X)$$ to $$\text{dim}(X^m) = m\text{dim}(X)$$ for some $$1 < m < n$$, or for various subsets of $$\lbrace 2, 3, \dots, n - 1 \rbrace$$. The case $$m = n - 1$$ seems especially interesting.

In particular, have the dimensions of all the powers of the Pontryagin Surface been computed?

Edit after a couple of days: Dranishnikov (arguably the top living expert) says it is true for compact metric spaces. He didn't mention any counterexample for more general spaces. It was already known to Hurewicz in the 1930's to be true for compact metric spaces of dimension one.

As noted in the comments, it is known that for any compact Hausdorff or any metrizable space, the limit (called the stable dimension of $$X$$) $$\lim \frac{\text{dim}(X^n)}{n}$$ exists. The quantity is positive if $$X$$ is compact metric and $$\text{dim}(X) \geq 1$$. Thus to prove the titular proposition for compact metric spaces it's sufficient to prove the case $$n = 4$$, because we can then iterate and apply the uniqueness of the limit, using the fact that for reasonable spaces it is true that $$\text{dim}(X \times Y) \leq \text{dim}(X) + \text{dim}(Y)$$. So if there was some pathology it would carry all the way to the limit.

This method assumes that if $$\text{dim}(X) > 0$$ then the stable dimension is positive; this is true for compact metric spaces, but is not true in general for separable metric spaces. For example, let $$E$$ be Erdos Space. It is known that $$E = E \times E$$ and $$\text{dim}(E) = \text{dim}(E \times E) = 1$$, so $$\text{stabdim}(E) = 0$$. To see that it is true for (locally) compact metric spaces in dimension larger than $$1$$, use the small inductive dimension to obtain closed subspaces of dimension at least $$1$$ as boundaries of neighborhoods of a point, then apply the classical Hurewicz result mentioned above to see that in fact $$\text{stabdim}(X) \geq 1$$, since the product of these boundaries is closed and dimension is monotone wrt closed subsets.

This would extend the proof method to general compact Hausdorff spaces if we knew that the conjecture in the title were true for one-dimensional compact Hausdorff spaces (that it follows is due to the fact that for compact Hausdorff spaces $$X$$ we have $$\text{dim}(X) \leq \text{ind}(X)$$). I am not sure if that is true or false, but I assume it's known. Anybody have a reference? Extending to LCH spaces is tricky in dimensions $$0$$ and $$1$$ so I have no idea about that, and I don't know if the stable dimension of LCH spaces is even well-defined. That would also be good to know.

• Do you mean $\dim(X \times X) = 2 \dim(X)$? Otherwise, for most definitions of dimension I can think of, $\mathbb R^2$ is a counterexample by the numerical coincidence $2 \times 2 = 2^2$. Nov 19 '18 at 4:41
• Oh sorry, yes. Fixing. Nov 19 '18 at 4:43
• So for compact metric spaces, there is an invariant called the stable dimension $\text{sdim}(X) = \lim \frac{\text{dim}(X^n)}{n}$, and this limit is known to exist. Thus for a space with positive stable dimension it is sufficient to prove the case where $n$ is a power of $m$, because if this is true then it would be impossible for the general case to have a counterexample: Otherwise the sequences for powers of $m$ and $n$ would have different limits. Nov 19 '18 at 10:29
• As mentioned in the thread linked below, Przymusinski has constructed spaces such that $X^2$ is normal but $X^3$ is not. I will look into this paper and see if its techniques can be adapted to provide a counterexample in the class of normal spaces.math.stackexchange.com/questions/3007308/… Nov 21 '18 at 10:15
• So, a question might be: can it happen that $X \times X$ has dimension $0$, but $X \times X \times X$ has dimension $1$? Maybe one of those weird examples like rational points in $l^2$. Nov 21 '18 at 15:54

As John Samples noted in his comment, Dranishnikov's Theory of cohomological dimension implies the positive answer to this problem for compact (even $$\sigma$$-compact) metrizable spaces. Namely, according to a Definition on page 15 of the paper "Cohomological dimension theory of compact metric spaces" a compact metrizable space $$X$$ is called of basic type if $$\dim X^2=2\dim X$$ and of exceptional type if $$\dim X^2=2\dim X-1$$ (which is equivalent to $$\dim X^2<2\dim X$$ by Theorem 4.16). After this definition Dranishnikov writes that each compactum $$X$$ of basic type has $$\dim X^n=n\dim X$$ and each compactum $$X$$ of exceptional type has $$\dim X^n=n\dim X-n+1$$, for every $$n\in\mathbb N$$.
This implies that if some compactum $$X$$ has $$\dim X^n=n\dim X$$ for some $$n\ge 2$$, then it is of basic type and hence $$\dim X^k=k\dim X$$ for all $$k\in\mathbb N$$. Also the Dranishnikov's result implies that the stable dimension $$\lim_{n\to\infty}\frac{\dim X^n}n$$ of a compactum $$X$$ equals $$\dim X$$ if $$X$$ is of basic type and equals to $$\dim X-1$$ if it is of exceptional type.