How "visualize" negative-dimensional space?

Question

The concept of negative-dimensional space was introduced in many branches of geometry (topology, algebraic-geometry, derived-geometry, fractals, etc) for example:

http://forthelukeofmath.com/documents/Wolcott-McTernan-workshop.pdf

Or

https://en.wikipedia.org/wiki/Negative-dimensional_space

How "visualize" this? For example a (-2)-sphere

Answer 1

The first sentence of the Wikipedia article you link to ends with a pointer to the first reference of the article, which is a link to a pdf called Imagining Negative-Dimensional Space. Might be a starting point.

But generally I'd say it's not always the best thing to just cling to words when contexts generalizes. In math, you may have some thing $f(x)$ where $x$ is a $X$, and then somebody sees you can keep the formal form of $f$ but use a $y$ of $X$ and still have $f(y)$ make some sense. If $x$ was called foo, it might not be worth trying to read foo-ness into $y$. It's tuning in on the web-stream of your favorite radio station and searching for the electromagnetic waves in the radio frequency range.

Answer 2

A solid $-1$-ball is an infinite periodic lattice. For instance, the set of integers or the set of real roots of the sine function.

Let $h$ being the step of the lattice. Then

• The radius of the ball is $R=h/\pi$. Thus, unit ball is a lattice with step $\pi$ (such as the set of roots of the sine function).
• The volume of the ball is $V=\frac1{h}=\frac1{\pi R}$. It is the density of the lattice. A unit ball has volume $\frac 1\pi$.
• A single point is the limit case of a ball when the volume is zero.
• Any point of the ball can be its center. Thus a given lattice can represent different balls depending on what point we consider the center. It is possible that two balls intersect only at center.
• The ball is also a solid cube with side $h$. Thus a unit cube is a lattice with step $1$, its volume is $1$. Thus, in dimensions 0,1 and -1 a ball is also a cube.
• One can see that the lattice is a fractal of dimension $-1$: when scaled by $1/2$ (shrinked by factor 2) it contains 2 copies of original. $\frac{\ln{2}}{\ln 1/2}=-1$. For comparison, an interval (1-cube) if scaled by 2 contains 2 its original copies, so its dimension is 1, square (2-cube) if scaled by 2, contains 4 its copies, so its dimension is 2, cube if scaled by 2 contains 8 its copies, so its dimension is 3, point if scaled by 2 is still a point, so its dimension is 0, etc.

Answer 3

Negative-dimensional space seems good, but in the contexts typically used, it does not make sense. However, I do cover the possibilities of there being negative dimensional objects.

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For example, under certain conditions, I can say that $N$ algebraic equations in $M$ variables specifies an $M-N$ dimensional space. So, for example, the curves: $$x^2-2y^2+z^2=5,\\ x+y-z=1,$$ specify a $3-2=1$D space, that is, their intersection is some elliptical shape in 3D. As such, if I were to have the set of equations: $$x^2-3xy+z^3=5z\\3x-3y+z^2=1\\x^2-y+z^5=0\\ x^2y^3-z^6x^7-x^7+y^3=0,$$ assuming such an intersection exists, the intersection is negative-dimensional, that is, we have $3$ variables and $4$ equations. However, this often simply means that the solution doesn't exist. If it does, it is never negative-dimensional.

Another example: I can have the vector $$(u^3-3uv-v^5,uv,u^2+v^2)\in\mathbb R^3$$ describe a set of points that make a $2$D surface. And in this context, the dimension of the surface corresponds to the number of input variables used. Yet I cannot have $-1$ variables.

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Not all hope is lost. You can define $-1$ variables to mean something. However, the reason we have negative numbers and square roots, etc., is because there was sufficient context in which to define them. Square roots were inventable since there was an implicit structure available to make their definition coherent ($x:=\sqrt{y}\iff x^2=y$). Similarly with negative numbers ($x:=-y\iff x+y=0$). You have to provide the larger context. Fractals seem to do a good job, and they have my vote.

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Side notes: Consider that, typically, the volume of an $d$D array of points is on the order of $n^d$ for side length $n$. This gives the definition of dimension:

$$d:=\lim_{n\to\infty}\frac{\ln V(n\vec{s_0})-\ln V(\vec{s_0})}{\ln n}$$ for "initial vector sidelength" $\vec{s_0}$ of our object and volume $V(\vec{s})$ (as a function of side-lengths $\vec s$). For example, a rectangular prism with volume $V(x,y,z)=xyz$ has dimension $$d_{\text{Box}}=\lim_{n\to\infty}\frac{\ln n^3xyz-\ln xyz}{\ln n}=3.$$ So we're looking for some object that has a single parameter defining its volume such that the function we obtain for volume has $$\lim_{n\to\infty}\frac{\ln V(ns_0)-\ln V(s_0)}{\ln n}=-1.$$