# Pathological Examples of Dimension

I am trying to wrap my head around all the different notions of dimension (and their equivalences). To get a sense of this, it would be nice to know the subtle difficulties that arise. I thus think it would be nice to have a list of such examples! (I dug through the internet without locating such a collection.)

This question/request can be interpreted as either
1) An example that obeys a particular definition of dimension but goes against our intuition. Said differently: an example that should obey a particular definition of dimension, but doesn't.
3) An example that disagrees with two different definitions of dimension.
4) An example which hinges on a hypothesis of the dimension.

*This last one is what got me to start this post, because I came across an example involving the Krull dimension: If our ring $R$ is Noetherian then $\dim R[x]=1+\dim R$, but if $R$ is not Noetherian then we can have $\dim R[x]=2+\dim R$. Found at http://www.jstor.org/stable/2373549?origin=crossref (The Dimension Sequence of a Commutative Ring, by Gilmer).
*I am not sure where our space-filling curves fit in here.

Some standard definitions of dimension

• Lebesgue covering dimension (of a topological space)
• Cohomological dimension (of a topological space)
• Hausdorff dimension (of a metric space)
• Krull dimension (of a ring or module)
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I think that the question would be better off if you gave a list of definitions of dimensions you're interested in (at least as a starting point). Their names, together with links to relevant wikipedia-like resources would be a good alternative to complete definitions. –  Marco Golla Oct 16 '12 at 21:02
Perhaps this should also be community wiki, since it seems to be asking for a list. –  Karl Schwede Oct 16 '12 at 21:18
Second both @Marco and @Karl. –  Igor Rivin Oct 16 '12 at 23:41
Somewhat OT, hope it's ok - can you post a link to the "crazy example"? Sounds typical for non-Noetherian wackiness, would be good to have an easy reference, but my first few searches came up empty. –  kcrisman Oct 17 '12 at 2:52
This question covers a lot! (Too much for one question? I don't know). Also, the topological spaces for which any two or all three of the standard topological dimensions $\dim\ \ \ \mathit ind\ \ \ \mathit Ind$ differ are of interest as well. –  Wlodzimierz Holsztynski May 3 at 4:19

Erdős space, the set of all vectors in $\ell^2$ with rational entries, seems like it would fit the bill -- it is a metrizable space which has "dimension one", but it is homeomorphic to its Cartesian square, and so violates our hope/intuition that $\dim(E\times F)=\dim(E)+\dim(F)$.

See Gerald Edgar's answer to a previous MO question.

(Digression: I learned of this example in a seminar given here by a postdoc, and realized as she was writing down these properties that I'd actually seen it mentioned -- without any of the relevant technical detail -- in one of the pop-maths biographies of Erdős. The story goes that he got interested in something two topologists were trying to figure out, got fobbed off with a quick explanation of the problem, came back to ask what a Hilbert space was, went away, and then came back to show that this space had dimension $1$ rather than the expected $0$ or $\infty$.)

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@Gerry: oops, yes I did mean the sum (was typing in a hurry) –  Captain Oates Oct 17 '12 at 5:19
Prodigy Erdős knew the definition of a Hilbert space already in his baby crib. The way I remember the story seems more plausible: Erdős attended a seminar at Princeton, where the question about the logarithmic identity for $\dim$ was cited. Erdős asked about the definition of the topological dimension (not about Hilbert space--there was a priori no reason to ask about Hilbert space) during the seminar break. And by the end of the seminar already had his beautiful and famous example. –  Wlodzimierz Holsztynski May 3 at 4:12

Here's a fairly standard one (it's an exercise in Hartshorne). In an integral domain $R$ of finite type over a field, every maximal ideal has the same height (in particular, every closed point has the same dimension). Indeed, it would be natural to define a ring to be equidimensional if every maximal ideal has the same height. Here's a problem with this definition.

Suppose now that $R$ is a DVR with parameter $r$. Consider the ring $R[x]$. This ring has one maximal ideal of height one, $\langle xr - 1 \rangle$, and another maximal ideal of height two, $\langle x, r \rangle$.

The point being, this is a domain, so its $\text{Spec}$ is presumably equidimensional, of dimension 2 the Krull dimension of $R[x]$. But it has closed points of different heights (although with very different residue fields). Of course, this isn't as pathological as a non-catenary ring, but we can even assume that $R[x]$ is a localization of $k[r,x]$.

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A set of facts that I fnd puzzling is the behaviour of Krull dimension (in the sense of Gabriel and Rentschler, that is, for non-necessarily commutative rings) of Weyl algebras.

One has $\mathcal K(A_n(k))=n$ when $k$ is a (commutative!) field of characteristic zero, and this is very sensible. If $k$ is instead of positive characteristic, we have $\mathcal K(A_n(k))=2n$, which is the other sensible value... Now, if $k$ is a field of any characteristic and $D_n=\operatorname{Frac}A_n(k)$ is the $n$th Weyl field, then $\mathcal K(A_n(D))=2n$; this is already strange. More generally, $\mathcal K(A_n(D_m))=\min\{2n,n+m\}$ over a field of characteristic zero.

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For (1), consider the 2-dimensional analogue of the Hawaiian earring. This is the union $X$ of spheres of radii $1/n$ ($n\in\mathbb{N}$) all intersecting at the origin-point. One would expect $H^3(X)=0$ under singular homology, i.e. $X$ is 2-dimensional, but it turns out $H_3(X)\ne 0$. I am unsure if Cech-cohomology gives the 'right' answer... I think it should. (I'll try and find more information on this.)

[[Update]]: This is found in a paper of Milnor and Barratt, An Example of Anomalous Singular Homology. Their result is that for the $r$-dimensional analogue $X_r$ of the Hawaiian earring, $H_n(X_r;\mathbb{Q})$ is uncountable for $n\equiv 1\;\text{mod}(r-1)$ for $n,r>1$. Here we use singular homology. And we do recover $\check{H}_{r+1}(X_r)=0$ under Cech-homology!

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If on the other hand all those spheres were once-punctured the space would be contractible. I think this indicates homology just isn't a good way to define dimension of an object. One argument that it's `bad' is that it requires another notion of dimension -- the dimension you use in the grading of homology. –  Ryan Budney Oct 19 '12 at 0:49
True, good point! –  Chris Gerig Oct 19 '12 at 17:55
Singular homology fits only nice spaces, meaning the spaces which are homotopically equivalent (or dominated--here it is the same) by CW-complexes (in the finite case Wall's example shows that it is essential to consider spaces homotopically dominated by finite CW-complexes). It's the Cech homology/cohomology and compact spaces which match well. The covering dimension of compact spaces goes hand in hand with the Cech homology. –  Wlodzimierz Holsztynski May 3 at 5:00

@Chris: the Cech homology of any compact $n$-dimensional space $X$ (the covering dimension is meant here) vanishes in the dimensions above $n$ because:

1. $X$ is an inverse limit of $n$-dimensional finite polyhedra;
2. Cech homology of compact spaces is continuous (with respect to the inverse limit operation).

It follows that Cech homology of any Hawaiian $n$-dimensional earring vanishes in every dimension above $n$.

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Why don't you post these answers as comments to the relevant posts? –  Ketil Tveiten May 3 at 10:02
If one of the two topological spaces   $X\ \ Y$   is compact and $1$-dimensional then the logarithmic equality for the covering dimension holds:
$$\dim(X\times Y)\ \ =\ \ \dim(X) + \dim(Y)$$
Thus in the compact case Erdős example cannot be matched, one needs to work with the dimensions greater or equal $2$. This was accomplished by Pontryagin, who provided continua   $X\ Y$   of dimension $2$   such that their product   $X\times Y$   was $3$-dimensional.
Next, a more subtle example was given by Boltyansky. His 2-dimensional continuum   $B$   was such that its square was 3-dimensional   $\dim(B^2) = 3$.