Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)
share|improve this question
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 '13 at 4:19
add comment

6 Answers

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 $\infty$.)

share|improve this answer
@Gerry: oops, yes I did mean the sum (was typing in a hurry) –  Yemon Choi 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 '13 at 4:12
add comment

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 $k[r,x]$.

share|improve this answer
add comment

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.

There is a paper by Goodearl, Hodges and Lenagan which is filled with information about this (and parallel onformation about global dimensions).

share|improve this answer
add comment

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!

share|improve this answer
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 '13 at 5:00
add comment

@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 $n$.

share|improve this answer
Why don't you post these answers as comments to the relevant posts? –  Ketil Tveiten May 3 '13 at 10:02
add comment

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   $\dim(B^2) = 3$.

These examples can be understood (better) from the point of view of homological (or cohomological) dimension, in the combination with the simple underlying geometric nature of these examples.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.