Is the notion of fractional dimension compatible with considering a dimension a set of n-tuples? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:37:07Z http://mathoverflow.net/feeds/question/66824 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66824/is-the-notion-of-fractional-dimension-compatible-with-considering-a-dimension-a-s Is the notion of fractional dimension compatible with considering a dimension a set of n-tuples? gonzo_floyd 2011-06-03T13:57:08Z 2012-03-08T21:19:43Z <p>Hi there,</p> <p>I am trying to teach a friend about higher-dimensions and I have explained them in the following two manners:</p> <ol> <li>A higher dimension, e.g. the 56th is the set of all 56-tuples.</li> <li>A fractional dimension, e.g. 3/4 means that when size doubles, volume increases at a factor of 2^3/4</li> </ol> <p>To me, it is obvious that both of these hold in a dimension with an integral number of coordinates. However, does statement one hold in a fractional dimension?</p> http://mathoverflow.net/questions/66824/is-the-notion-of-fractional-dimension-compatible-with-considering-a-dimension-a-s/66843#66843 Answer by Christian Blatter for Is the notion of fractional dimension compatible with considering a dimension a set of n-tuples? Christian Blatter 2011-06-03T18:29:22Z 2011-06-03T18:29:22Z <p>The notion of "dimension" occurs in various contexts and its meaning is not always the same. There are:</p> <p>(1) Topological dimension. The topological dimension ${\rm dim}(X)$ of a space $X\ne\emptyset$ is an integer $\geq0$ or $\infty$. It is defined inductively or by means of coverings: An $X$ is, e.g., two-dimensional, if any open covering of $X$ can be "refined" to a covering that covers no point more than three times. This concept of "dimension" is not easy to handle and gives rise to strange theorems, e.g., that a space of dimension $n$ can be the union of $n+1$ spaces of dimension $0$. </p> <p>(2) Dimension of vector spaces $V$. Any vector space $V$ (over an arbitrary field $K$) has a dimension ${\rm dim}(V)$ which is an integer $\geq 0$ or $\infty$. This is an algebraic concept. If ${\rm dim}(V)=n$ then one can choose $n$ vectors $e_1$, $\ldots$, $e_n\in V$ such that any vector $x\in V$ can be written as $x=\sum_{k=1}^n x_k \ e_k$ or $x=(x_1,\ldots, x_n)$ for short. People say that we have $n$ "degrees of freedom" in such a $V$.</p> <p>When the ground-field $K$ is the field ${\mathbb R}$ of real numbers then the topological dimension of $V$ is equal to its algebraic dimension $n$; in particular, the real line ${\mathbb R}$ is also topologically one-dimensional. Furthermore there is a natural measure of volume for arbitrary subsets $A\subset V$, given that the volume of the unit cube should be $1$. When a set $A$ is stretched by a factor $\lambda>0$ then its volume increases by the factor $\lambda^n$. If in ${\mathbb R}^n$ the euclidean metric $|x|:=\sqrt{\sum_{k=1}^n x_k^2}$ is adopted then also sets $A$ in $d$-dimensional "planes" $\ U\subset {\mathbb R}^n$ get a natural $d$-dimensional volume, and this volume multiplies by $\lambda^d$ under a linear stretching of $A$ by $\lambda$. This implies that also curved $d$-dimensional "surfaces" in ${\mathbb R}^n$ have a natural $d$-dimensional length or area.</p> <p>(3) "Fractal" dimension. Given euclidean $n$-space with its bodies, surfaces and curves all having an intuitive dimension $d\leq n$ and a $d$-dimensional "volume" scaling in the expected way one may ask whether there is a way of measuring arbitrary sets $A\subset{\mathbb R}^n$ that is able to scale with noninteger exponents $\alpha$, and whether there are sets $A\subset{\mathbb R}^n$ that for such a value $\alpha\notin{\mathbb N}$ would have an $\alpha$-dimensional volume which is neither zero nor infinity.</p> <p>The so-called Hausdorff measure (invented in 1919) has the necessary flexibility; and indeed there are "crazy" sets $A$, called "fractals", that for precisely one noninteger value $\alpha$ have $\alpha$-dimensional volume $\ne 0, \infty$. This value $\alpha$ is then called the Hausdorff or fractal dimension of $A$. This "dimension" is nothing esoteric but a certain geometrical quantity associated to $A$, as the semi-axes $a$ and $b$ are associated to an ellipse. There is one deep theorem though: The Hausdorff dimension of a set is never larger than its topological dimension. In particular, all "fractals" in ${\mathbb R}^3$ have a Hausdorff dimension $\leq 3$.</p> <p>There are mathematically defined fractal sets, e.g., the Cantor set or the snowflake curve, and there are such sets found in nature, e.g. "the coast line of Britain", or cumulus clouds in the atmosphere. Given such a "natural" set the essential problem is to find its fractal dimension $\alpha$ by computational means.</p>