# Variable Dimensionality Manifolds

I am curious about different notions of dimensionality, particularly as it relates to manifolds.

For instance, there is the standard fixed dimensionality of a pure smooth manifold, the Hausdorff dimension, various different notions of Fractal dimension, and other notions of dimension (e.g. [1], [2]). Some allow for fractional dimensionality; my question is whether this is possible somehow for manifolds (or rather for some manifold-like objects).

One interesting note in the manifold wiki article mentions the notion of manifolds where the dimensionality changes, by e.g. disjoint unioning of a sphere and line. However, the connected components must have the same dimension, I believe.

But is there a notion of generalized manifold with dimensionality that can be "smoothly varying", in some sense?

My question is partly motivated by the idea that one can easily "imagine" such a construct (e.g. a surface that forms a long cone that thins out into a line). Of course, this would not be a manifold (indeed, in computer science, discrete "manifolds", e.g. meshes or point sets, that do this are called "non-manifold"), but perhaps there is a generalized notion that admits this and analysis?

The most obvious issue is local coordinates are obviously integral in nature. But is there no way to parameterize fractals locally (which are of fractional dimension)?

Related questions:

• It's very much unclear what your question is. To begin with, you have to settle on a particular notion of dimension (topological dimension, Hausdorff dimension?); you also should explain what you have in mind by "smoothly varying". For instance, in your example of a cone and a line, the local topological dimension is discontinuous. If you have local Hausdorff dimension in mind then indeed, even for a fractal topological curve, it can vary continuously ("smooth" would be meaningless in this setting). Without more thoughts on your part, the question is unanswerable. – Moishe Kohan Jul 24 '17 at 14:44
• @MoisheCohen Thanks for your points. I suppose you're right, in that the question is open-ended. Part of the purpose of the question is to ask what kinds of notions of dimensionality are acceptable for such a construct. Certainly, the classic notion of vector space dimensionality doesn't make sense; I am not familiar with very many other notions that might be reasonable, other than maybe some of the fractal ones. (Also, you're right, "smooth" here was not meant in some precise sense.) – user3658307 Jul 24 '17 at 16:49
• Thanks for this question. @MoisheCohen: Would 'box counting dimension' be an acceptable measure of dimensionality for a question like this? Box counting does allow one to vary the dimensionality from one number to another for objects such as lattices , cones and lines, etc. – Bob Mar 4 '18 at 17:20
• @Bob: See my answer below where I used Hausdorff dimension; local box dimension is fine as well of course, it all depends on what you want to accomplish. – Moishe Kohan Mar 4 '18 at 18:41

To be honest, I still do not understand what your question is. Here is a couple of concepts which might be related to what you are thinking about. Neither one is specific to manifolds, they are defined for general topological, resp. metric, spaces. In order to avoid pathological examples, I will work with closed subsets $X$ of an $n$-dimensional Euclidean space $E^n$.

Definition 1. The local topological dimension of $X$ at $x$, denoted $dim(X,x)$ is $$\lim_{r\to 0} dim(\bar{B}(x,r)\cap X),$$ where $dim$ is the Lebesgue covering dimension and $\bar{B}(x,r)$ is the closed ball of radius $r$ centered at $x$.

By the very definition, this dimension takes only integer values.

Definition 2. The local Hausdorff dimension of $X$ at $x$, denoted $Hdim(X,x)$ is $$\lim_{r\to 0} Hdim(\bar{B}(x,r)\cap X),$$ where $Hdim$ is the Hausdorff dimension.

In view of monotonicity properties of covering dimension and Hausdorff dimension, we have that:

1. $\forall x\in X$, $dim(X,x)\le dim(X)$ and $dim(X,x)$ is upper semicontinuous as a function of $x$.

2. $\forall x\in X$, $Hdim(X,x)\le Hdim(X)$ and $Hdim(X,x)$ is upper semicontinuous as a function of $x$.

Example. $X$ equals the union of the $xy$-plane and the $z$-axis in $R^3$. Then $Hdim(X,p)=dim(X,p)=2$ for $p\in X$ whenever $p=(x,y,0)$ and $Hdim(X,p)=dim(X,p)=1$ for $p\in X$ whenever $p=(0,0,z)$, $z\ne 0$. The same in your example of the union of a 2-dimensional cone and a ray: Neither function will be continuous.

Neither local dimension function is, in general, continuous, as a function of $x$. If $X$ is connected then $dim(X,x)$ is continuous as a function of $x$ if and only if it is constant, equal $dim(X)$. In contrast, there are examples of connected (fractal) $X$ such that $Hdim(X,x)$ is continuous but nonconstant.

As for "discrete manifolds", this is just a wild goose chase: All standard notion of dimension for (nonempty) discrete spaces yield the same answer, namely zero. Nevertheless, I suspect that there is a nontrivial analogue of Hausdorff dimension for discrete subsets of $E^n$, it should be defined similarly to "persistent homology", which is a nontrivial homology theory assigned to discrete metric spaces. Compare this discussion.

• My take-home message is that the local Haussdorff dimension depends on the location that you are looking at. In the example you give above (the union of the plane defined by $z=0$ and a line coinciding with the z-axis) dimensionality is explicitly 1 or 2. (It sounds to me like Hausdorff dimension assumes you have perfect resolution. One thing that I find interesting with box-counting is that the dimensionality can be a function of characteristic scale (e.g. resolution). (I also would like to mention that I am not trained in mathematics, so I won't necessarily understand technical terms well.) – Bob Mar 5 '18 at 2:26
• If I think of the side length of the boxes necessary to cover a figure as a "resolution limit", than dimensionality can vary smoothly, which sounds to me different from Hausdorff dimension. A somewhat follow-up question that I have is can one describe a ball of yarn as a manifold (or manifold-like object). If I take the approach of box-counting dimension, the dimensionality of the ball of yarn is $1 \leq d \leq 3$, depending on the side length of the boxes used (or equivalently the resolution of the observer). Can an observer with low resolution (correctly) claim the ball of yarn... – Bob Mar 5 '18 at 2:34
• ...is a 3D manifold, even though we know that the yarn is inherently 1D, and that the observer's claim of the ball of yarn being 3D is due to the fact that the observer cannot resolve the individual strings? – Bob Mar 5 '18 at 2:34
• @Bob: These are interesting questions. Maybe you should post them as a separate MSE question (and let me know). One thing to be aware of is that the box-counting dimension is a certain limit (as scale goes to zero). However, considering the pre-limit quantities is, of course, also meaningful. But this is a long story... – Moishe Kohan Mar 5 '18 at 5:44
• Thanks for the replies/conversation above. (I really do appreciate it). Here's the link. math.stackexchange.com/questions/2677968/… – Bob Mar 5 '18 at 16:44