Is there an axiomatic approach of the notion of dimension ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:14:27Z http://mathoverflow.net/feeds/question/80708 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80708/is-there-an-axiomatic-approach-of-the-notion-of-dimension Is there an axiomatic approach of the notion of dimension ? glougloubarbaki 2011-11-11T21:15:53Z 2012-05-21T09:17:08Z <p>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).</p> <p>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 ?</p> <p>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 : </p> <ul> <li>it should coincide with the algebraic dimension for finite dimensional vector spaces</li> <li>dim A $\leq$ dim B if $A \subset B$</li> <li>some sort of nice behaviour for cartesian products (at least for reasonnable sets)</li> <li>some sort of nice behaviour for infinite increasing unions and/or decreasing intersections</li> </ul> http://mathoverflow.net/questions/80708/is-there-an-axiomatic-approach-of-the-notion-of-dimension/80711#80711 Answer by Valerio Capraro for Is there an axiomatic approach of the notion of dimension ? Valerio Capraro 2011-11-11T21:49:16Z 2011-11-11T21:49:16Z <p>This might help <a href="http://www.springerlink.com/content/y8l2621113212403/" rel="nofollow">http://www.springerlink.com/content/y8l2621113212403/</a>: 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.</p> http://mathoverflow.net/questions/80708/is-there-an-axiomatic-approach-of-the-notion-of-dimension/80715#80715 Answer by Georges Elencwajg for Is there an axiomatic approach of the notion of dimension ? Georges Elencwajg 2011-11-11T22:18:22Z 2011-11-11T23:32:53Z <p>My gut feeling is that no list of axioms could simultaneously cover Lebesgue dimension, vector space dimension, Krull dimension, fractal dimension,...</p> <p>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.<br> (And I haven't even begun to examine the logicians' claim that it has dimension $2^{\aleph_0}$ over $\mathbb Q$)</p> <p>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 .<br> 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.</p> http://mathoverflow.net/questions/80708/is-there-an-axiomatic-approach-of-the-notion-of-dimension/80721#80721 Answer by Qiaochu Yuan for Is there an axiomatic approach of the notion of dimension ? Qiaochu Yuan 2011-11-12T00:15:51Z 2011-11-12T01:53:39Z <p>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 $K$.</p> <p>Most notably, this notion of dimension includes as special cases several types of <em>Euler characteristic.</em> 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 <a href="http://arxiv.org/abs/1107.6032" rel="nofollow">Traces in symmetric monoidal categories</a>, which in particular describes how to use these ideas to understand the Lefschetz fixed point theorem.</p> http://mathoverflow.net/questions/80708/is-there-an-axiomatic-approach-of-the-notion-of-dimension/80726#80726 Answer by Anton Petrunin for Is there an axiomatic approach of the notion of dimension ? Anton Petrunin 2011-11-12T01:32:52Z 2011-11-12T01:32:52Z <p>In metric geometry, dimension should satisfy the following axioms; otherwise it should not be called "dimension". (There are exceptions, for example Minkowski dimension.) </p> <p><strong>Normalization axiom.</strong> For any $m\in\mathbb Z_\ge$, $$\dim\mathbb E^m=m.$$</p> <p><strong>Cover axiom.</strong> If $\{A_n\}_{n=1}^\infty$ is a countable closed cover of $X$ then $$\dim X=\sup\nolimits_n\{\dim A_n\}$$</p> <p><strong>Product axiom.</strong> For any spaces $X$ and $Y$, $$\dim (X\times Y) \le \dim X+ \dim Y.$$</p> http://mathoverflow.net/questions/80708/is-there-an-axiomatic-approach-of-the-notion-of-dimension/97542#97542 Answer by Jose Navarro for Is there an axiomatic approach of the notion of dimension ? Jose Navarro 2012-05-21T09:17:08Z 2012-05-21T09:17:08Z <p>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. </p> <p>As far as I know, it is the only dimension function $\dim$ defined for arbitrary topological spaces, with the following properties:</p> <ul> <li><p>If $Y$ is a subsapce of $X$, then $\dim Y \leq \dim X$.</p></li> <li><p>$\dim (X \times Y) \leq \dim X + \dim Y$.</p></li> <li><p>It coincides with Grothendieck's combinatorial dimension on noetherian spaces, and with the standard dimensions (<em>cover</em> and <em>ind</em>) on separable metric spaces.</p></li> </ul> <p>This beautiful idea goes back to the 60's. You can check the following papers and the references therein.</p> <p>Section 2 of:</p> <ul> <li>Sancho de Salas, J.B. and M.T.: "<em>Dimension of dense subalgebras of $C(X)$</em>", in <em>Proceedings of the American Mathematical Society</em>, 105 (1989)</li> </ul> <p>or the introduction of:</p> <ul> <li>Sancho de Salas, J.B. and M.T.: "<em>Dimension of distributive lattices and universal spaces</em>", in <em>Topology and its Applications</em>, 42 (1991), 25-36</li> </ul>