There are many notions of dimension : algebraic, topological, Hausdorff, Minkowski... (and others). While the topological one generalize the algebraic one, the last three need not coincide for every sets. Yet it is generally acknowledged that the Hausdorff dimension has "nice enough" properties to work with (the interest of the Minkowski dimension lies mainly in the fact that it's easier to compute).

So my main question is this : is there an axiomatic approach that would tidy up this mess ? For example, is there a result of the form : if you ask these axioms then the only map from "reasonnable sets" to the set of positive real integers is the Hausdorff dimension ? (or another one ?). If so what are they ?

Are there also a clearly identified list of properties that you would ask from any notion of dimension ? I give the following as an example :

  • it should coincide with the algebraic dimension for finite dimensional vector spaces
  • dim A $\leq$ dim B if $A \subset B$
  • some sort of nice behaviour for cartesian products (at least for reasonnable sets)
  • some sort of nice behaviour for infinite increasing unions and/or decreasing intersections
  • $\begingroup$ (The Hausdorff dimention, in general, takes values in the non-negative reals, not only in the positive integers. Having fractional Hausdorff dimension can be taken as a decent definition of "fractal"... ) $\endgroup$ – Qfwfq Nov 12 '11 at 1:18
  • $\begingroup$ my bad, I meant to write positive real numbers of course $\endgroup$ – glougloubarbaki Nov 12 '11 at 10:10

This might help http://www.springerlink.com/content/y8l2621113212403/: the author claims to have found the axioms for defining the Lebesgue covering dimension. In the paragraph starting with "the axiomatic problem is an old problem in dimension theory" there is also a list of references that should help. In particular, the paper by Henderson, that would contain the axioms for a notion of dimension in any metrizable space, as an extension of the covering dimension.


In metric geometry, dimension should satisfy the following axioms; otherwise it should not be called "dimension". (There are exceptions, for example Minkowski dimension.)

Normalization axiom. For any $m\in\mathbb Z_\ge$, $$\dim\mathbb E^m=m.$$

Cover axiom. If $\{A_n\}_{n=1}^\infty$ is a countable closed cover of $X$ then $$\dim X=\sup\nolimits_n\{\dim A_n\}$$

Product axiom. For any spaces $X$ and $Y$, $$\dim (X\times Y) \le \dim X+ \dim Y.$$

  • $\begingroup$ The "countable closed" bit bothers me for generalizations. Some objects are countable, but still have high dimension in important senses, like $\mathbb Q^n$. Love the product axiom. It seems essentially fundamental. The $\mathbb E^m$ part will be the hardest to generalize. I think a general dimension shouldn't have any axiom of the type. $\endgroup$ – Will Sawin Nov 12 '11 at 8:33
  • $\begingroup$ For Minkowski dimension the cover axiom holds only for finite union. $\endgroup$ – Anton Petrunin Nov 12 '11 at 23:25
  • $\begingroup$ @Will Sawin, it seems like the normalization axiom might be needed in order to have the notion of dimension be unique. $\endgroup$ – MTS May 21 '12 at 15:54

Not a complete answer, but there is a surprisingly general generalization of the dimension of a finite-dimensional vector space available in any (braided?) monoidal category. In any such category, there is a notion of dimension of a dualizable object $c$ given by the trace of the identity endomorphism $\text{tr}(\text{id}_c)$. It takes values in $\text{End}(1)$ where $1$ is the monoidal unit and behaves as expected under tensor product. In the category of finite-dimensional vector spaces over a field $K$ it gives the image of the dimension in $K$.

Most notably, this notion of dimension includes as special cases several types of Euler characteristic. For example, the dimension of a dualizable chain complex (I think this is equivalent to: bounded complex of finitely-generated projective modules) is its Euler characteristic, as is the dimension of a dualizable object in the symmetric monoidal category of dualizable spectra. A nice exposition is given in Ponto and Shulman's Traces in symmetric monoidal categories, which in particular describes how to use these ideas to understand the Lefschetz fixed point theorem.


There is a notion of "Krull dimension" valid for arbitrary topological spaces, whose definition is totally analogous to that of algebraic varieties, but using lattices of closed subsets instead of rings of functions.

As far as I know, it is the only dimension function $\dim$ defined for arbitrary topological spaces, with the following properties:

  • If $Y $ is a subsapce of $X$, then $ \dim Y \leq \dim X $.

  • $\dim (X \times Y) \leq \dim X + \dim Y $.

  • It coincides with Grothendieck's combinatorial dimension on noetherian spaces, and with the standard dimensions (cover and ind) on separable metric spaces.

This beautiful idea goes back to the 60's. You can check the following papers and the references therein.

Section 2 of:

  • Sancho de Salas, J.B. and M.T.: "Dimension of dense subalgebras of $C(X)$", in Proceedings of the American Mathematical Society, 105 (1989)

or the introduction of:

  • Sancho de Salas, J.B. and M.T.: "Dimension of distributive lattices and universal spaces", in Topology and its Applications, 42 (1991), 25-36

My gut feeling is that no list of axioms could simultaneously cover Lebesgue dimension, vector space dimension, Krull dimension, fractal dimension,...

It is not clear to me, for example, how axioms would decide whether $\mathbb C$ has dimension $0$, as required by Krull, dimension $1$ as wished by complex geometers or dimension $2$, the topologists' choice.
(And I haven't even begun to examine the logicians' claim that it has dimension $2^{\aleph_0}$ over $\mathbb Q$)

But this is subjective , so let me say something indisputable: your axiom $A\subset B\Rightarrow dim A \leq dim B$ does not hold for Krull dimension .
Indeed, if $A$ is any domain of Krull dimension $n\gt 0$ and if $K$ is its field of fractions, we have $dim K=0$ and the inequality $dim A=n \leq dim K=0$ is not true.

  • 2
    $\begingroup$ So, if the axiomatics work out, we would conclude that Krull dimension is not a dimension. OK? $\endgroup$ – Gerald Edgar Nov 11 '11 at 22:33
  • 2
    $\begingroup$ Dear @Gerald, of course what you say is logically correct. And as you may have guessed my position is that since I firmly believe that Krull dimension is a dimension, the proposed axiomatics can't work out. $\endgroup$ – Georges Elencwajg Nov 11 '11 at 23:03
  • 11
    $\begingroup$ The Krull dimension works out fine as long as you replace the condition $A\subset B$ by a monomorphism ${\rm Spec}(A)\to{\rm Spec}(B)$ in the category of affine schemes. Or, to put it another way, if $B\to A$ is an epimorphism of commutative rings then ${\rm dim} A\le {\rm dim} B$. So, "subset" should be replaced by "monomorphism" with respect to the correct category. $\endgroup$ – George Lowther Nov 11 '11 at 23:33
  • 4
    $\begingroup$ Maybe $\mathbb{C}$ has dimension 0 if you look at it in the category of affine schemes (well, ${\rm Spec}\mathbb{C}$ at least), dimension 1 if you think of it as the closed points of the variety ${\rm Spec}(\mathbb{C}[x])$ in the category of complex varieties, and dimension 2 in the category of CW complexes. $\endgroup$ – George Lowther Nov 11 '11 at 23:50
  • 5
    $\begingroup$ I'd expect the concept of dimension to apply to a category of spaces in some sense. Looked at this way, Krull dimension applies to schemes, giving Lowther's point of view. If you apply it to rings instead, fine (since obviously any commutative ring gives an affine scheme), but this is no longer a category of spaces, so one wouldn't expect the axioms for spaces to apply. So in short, when considering whether a system of axioms validates a particular notion of dimension, find the appropriate category of spaces for that notion and apply the axioms there. $\endgroup$ – Toby Bartels Nov 12 '11 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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