# The Root of a Line

Ok, imagine having a finite line segment from point (a) to point (b) in $R^2$ . I'm not familiar with mathematical terminology of this kind, but let me state that the line we began with is the geometric interpretation of $A^1$. The geometric interpretation of $A^2$ is a square with sides $A^1$. We could go on by saying that $A^3$ is a cube with edge length $A^{1}$ again.

I wonder what the geometric interpration of $\sqrt[2]{A}=A^{1/2}$ would look like. Is it a (straight) line? Is it constructible? Of course, we could extend this question by asking ourselves what $A^{k}$ would look like, in which $k \in \mathbb{R}$ or even $\mathbb{C}$ . When $k>2$, $A^k$ probably isn't constructible anymore on a sheet of paper, but one can still think about how these constructions of $A$ would 'look like'.

Max Muller

P.S. I realize I ask more than one question now, which is an indictation I don't know a lot about this (yet) and I'd like to know more about this subject. Should this be a community wiki? Feel free to modify the Tags, I don't know how to classify this question exactly.

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It looks like you want a way to consider geometric objects of fractional dimension. The standard answer is "fractals". See en.wikipedia.org/wiki/Hausdorff_dimension –  S. Carnahan May 16 '10 at 15:57
Scott, I think this is what I was looking for, too. Thank you. I can't find any object in the list of fractals by Hausdorff Dimension wikipedia provided me with, which has a Hausdorff dimension of exactly 1/2, however. The 'quadratic von Koch curve (type 2)' seems has a Hausdorff dimension of (3/2). Would my geometric object have something to do with that one? –  Max Muller May 16 '10 at 16:31
Perhaps one solution would require to embed the plane into a larger space where some components of the construction of $\sqrt(A)$ could live in analogy to the imaginary dimension of the complex plane and the idea of fractals as quotients of figures between spaces of larger dimension. –  ogerard May 17 '10 at 16:15
That was a nice question! Just to add one more "ghost covered by clouds": There are very thrilling speculations in number theory/arithmetic geometry about spaces whose dimensions can be arbitrary complex numbers. One idea is to use Alain Connes' "noncommutative geometry", the other has to do with "motives" (a kind of mythical "atoms" out of which those stuff algebr./arithm. geometers study is buildt up): arxiv.org/abs/0907.0321 . The idea comes from physics, when physicists fond ways to turn ill-defined infinite series into something to compute interesting entities with. –  Thomas Riepe Nov 24 '10 at 9:55
(Continued)... This "trick" is to "circumvent" the divergences in such series by giving the variable giving a dimension other, complex, values than the original integers. It may be amusing to lat such "playing around with infinite series" in e.g.: arxiv.org/abs/0712.3670 and arxiv.org/abs/0904.4921 –  Thomas Riepe Nov 24 '10 at 10:09

Ok thank you André. I've seen the picture, but I still don't quite understand which part of it represents the root of a line segment. I guess the black line is the line segment A. I understand none of these pictures represent $\sqrt[2]{A}$ exactly, because it's the result of a process that goes on indefinitely, but I still don't see which part of the picture 'converges to' $\sqrt[2]{A}$ ... $– Max Muller May 16 '10 at 21:05 Perhaps I'm missing something but if you allow yourself to project at the end, why isn't an interval its own "square root" under this definition? – j.c. May 17 '10 at 1:27 The projection map is a measure preserving isomorphism (away from a set of measure zero, on which the map is two-to-one) between the square of the Cantor set and the interval. – André Henriques May 17 '10 at 9:06 From a topology standpoint,$\sqrt{\mathbb{R}}$would be a space$Y$such that$Y x Y$is homeomorphic to$\mathbb{R}$. Such a space cannot exist as$Y$would be connected (as an image of a connected space) hence a singleton or an interval. And an interval times an interval has non-cutpoints (removing such a point leaves the space connected) while$\mathbb{R}$has none. One can also show that no odd power of$\mathbb{R}$has such a square root, IIRC. - Thanks Henno, for your answer. I think you should take a look at the link Scott Carnahan gave me, on Hausdorff dimensions and objects in it. From that section: " However, topological dimension behaves in quite unexpected ways on certain highly irregular sets such as fractals. For example, the Cantor set has topological dimension zero, but in some sense it behaves as a higher dimensional space. Hausdorff dimension gives another way to define dimension, which takes the metric into account." Therefore, I don't think we should try to answer this question by taking a topological standpoint... – Max Muller May 16 '10 at 16:37 @Max, if one can't even find a square root topologically, that should certainly preclude finding one that has the proper metric structure as well. – j.c. May 16 '10 at 16:50 A metric gives more structure, that's for sure, but the topology at least gives us that we cannot use Cartesian / topological products for the supposed root. What else one should use, I don't know. @Max: in what way does the Cantor set behave like a higher dimensional space? – Henno Brandsma May 16 '10 at 18:52 @Henno and jc: I'm not an expert on this topic, I haven't studied mathematics at university yet. The text in the quotation marks in my comment has been copied from the wikipedia page on the 'Hausdorff Dimension' in the 'general discussion' section. I copied that part of the article because I think it explains why the concept of a Hausdorff dimension has been introduced, and is 'more useful' in some cases than a topological dimension. I'm just trying to imagine how the 'root of a line', or A raised by any power, 'looks like'. – Max Muller May 16 '10 at 19:49 @Max Perhaps you should think a bit about what you mean by "taking powers" of a space. What Henno is talking about here is the cartesian product, which I think is the most direct interpretation of what you mean in your first paragraph. – j.c. May 17 '10 at 1:29 The answers above seem to suggest there is a homomorphism from this interesting little line group to$\mathbb{R}$sending the cartesian product to multiplication. But we have$A^1 \times A^2 =A^3$when the answer we want is surely$A^2$. Cartesian product is additive on dimensions so perhaps eg. Andre has found a value for$ \frac12 A$. My guess as to how to get a multiplicative structure would be$X*Y:=hom(X,Y)$in some suitable category, for example if$A^i=\mathbb{R}^i$then linear maps would do the job. But now$\sqrt{\mathbb{R}}=[Y, hom(Y,Y)=\mathbb{R}] =\mathbb{R}$and the real (and probably impossible) fun comes in finding '$\sqrt{\mathbb{R}^2}$' [Edit: re-read question- didn't realise those were powers of$A$- thought it was just an index! Oh well- will leave this here as I think it's an interesting reformulation...] Edit2: Bit of a fiddle but you can make this work in the category of$\mathbb{Q}$Vector spaces with continuous$\mathbb{Q}$-linear maps where$\sqrt{\mathbb{Q}^2}=\mathbb{Q}[\sqrt{m}]$for any non-square m and$\sqrt{\mathbb{Q}^n}=\mathbb{Q}[^n\sqrt{m},^n\sqrt{m}^2...]\$ for any non-nth power m, we've lost the opportunity for fractions and a cube root seems unlikely, but it's sort of neat...