Dimension of product space

Is there a product space $M \times N$ such that $\dim(M \times N) < \dim(M) + \dim(N)$? Here $\dim$ refers to the dimension of a manifold.

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For manifolds $M$ and $N$, it is always the case that $dim(M \times N) = dim(M) + dim(N)$. Essentially, the dimension of a manifold is the dimension of the image of each chart (homeomorphic to a Euclidean space). In the product, the images are basically products of Euclidean spaces.

There is another notion of dimension that is more general. The covering dimension of a topological space $X$, http://en.wikipedia.org/wiki/Covering_dimension , which coincides with the notion of dimension above when $X$ is a manifold. However, if $X$ and $Y$ are general spaces, strange things can happen, unless we know something more about the spaces. For metric spaces (and certain other situations) we have $$dim_{c} (X \times Y) \leq dim_{c} X + dim_{c} Y,$$ where $dim_{c}$ is covering dimension. Pontrjagin found (1930) an example of two compact metric spaces of dimension 2 whose product has dimension 3, though I do not know what the spaces are.

For general topological spaces, the situation is even worse, as we can no longer assert $dim_c(X \times Y) \leq dim_c X + dim_c Y$. There are spaces of dimension $0$ whose product is of dimension $1$. See this paper for more info: http://www.pnas.org/content/75/10/4671.full.pdf

The moral of the story: Manifolds are nice :)

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Excellent! Thanks. –  snel Sep 29 '11 at 0:45

No. If $M$ is locally homeomorphic to $\mathbb{R}^m$ and $N$ is locally homeomorphic to $\mathbb{R}^n$, then necessarily $M\times N$ is locally homeomorphic to $\mathbb{R}^{m+n}$.

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