# reperiendi

## Negative dimensional objects and groupoid cardinality

Posted in Category theory, Math by Mike Stay on 2007 July 26

I was thinking about some stuff involving fractals and non-positive-real dimension. It’s still a very rough idea, though.

There’s the concept of topological dimension, which is necessarily an integer. It looks like it’s typically the floor of the Minkowski dimension.

One way of talking about iterated function systems is to consider patterns of digits in $n$-ary strings. For instance, the typical Cantor set consists of those points in the unit interval whose ternary expansion contains only 0’s and 2’s. It’s a single coordinate, but restricted in the values it can take.

We can add the dimension of two vector spaces by taking the categorical product = direct sum $\oplus$. We can multiply the dimension by taking the tensor product $\otimes$. We can add and multiply the dimension of two unit hypercubes with the same construction.

It’s clear how to tensor integer-dimension objects; I was looking at how one might tensor fractional-dimension objects. I have an example with the Cantor set that probably generalizes.

Consider the set of points $C$ in the unit interval whose 4-ary expansion consists exclusively of the digits 0 and 3:

----------------
----        ----
-  -        -  -

etc. $C$ has dimension $\log_4(2)=1/2.$ $C \otimes C$ is isomorphic to the set of points where pairs of digits in base 4 are either 00 or 33—which is the same as saying that in base 16, each digit is either either 0 or F=15. The dimension of this set is $\log_{16}(2)=1/4$. So as before, tensoring two objects multiplies their dimension.

Now consider $\displaystyle D=\prod_{i=0}^{\infty}C^{\otimes i}$

• $C^{\otimes 0}$ is the unit interval.
• $C^{\otimes 0}\times C^{\otimes 1}= 1\times C$ is a subset of the unit square, a set of parallel stripes. The dimension of $1\times C$ is $1+|C|+|C|=1.5$
• $C^{\otimes 0}\times C^{\otimes 1}\times C^{\otimes 2}= 1\times C\times C\otimes C$ is a subset of the unit cube, consisting of parallel copies of the previous set. The dimension of $1\times C\times C\otimes C$ is $1+|C|+|C|^2=1.75$

and so on until

• $\displaystyle D=\prod_{i=0}^{\infty}C^{\otimes i}$ is a subset of an infinite-dimensional hypercube with dimension $\displaystyle \sum_{i=0}^{\infty} |C|^i = 1/(1-|C|) = 2$

John Baez introduced something called groupoid cardinality. We can add and multiply finite sets using the disjoint sum and cartesian product. We can “divide” sets by using the weak quotient of a set by a group: $\displaystyle |S//G| = \sum_{[x]} \frac{1}{|\mbox{Aut}(x)|},$

where $[x]$ is an equivalence class and $x$ is equivalent to $y$ if they’re in the same orbit.

So consider the set $3=\{a,b,c\}$ under the action of $\mathbb{Z}_2$ via reflection: $\leftarrow | \rightarrow$ $a\, b\, c$

That is, 0 in $\mathbb{Z}_2$ maps each point to itself, while 1 in $\mathbb{Z}_2$ maps $b$ to itself and swaps $a$ and $c$. So $a$ and $c$ are in the same orbit and form one equivalence class. The automorphism group of $a$ is only the identity. The orbit of $b$ includes only itself, so $|\rm{Aut}(b)| = |\mathbb{Z}_2| = 2$. This forms a groupoid: a category where all morphisms are isomorphisms.

Summing over equivalence classes, we get $\displaystyle |3//Z_2| = 1/|\mbox{Aut}(a)| + 1/|\mbox{Aut}(b)| = 1/1 + 1/2 = 3/2.$

Is there some way to view the dimension of the Cantor set as arising from some kind of weak quotient?

He also talked about something he calls “structure types”, which is very much like a generating function, but acts on groupoids (with sets as a special case) instead of on numbers. So in the structure type $\displaystyle \frac{a_0}{0!}x^0 + \frac{a_1}{1!} x^1 + \frac{a_2}{2!} x^2 + \ldots$,

the term $a_n/n!$ really means $a_n//S_n$, where $a_n$ is the set of structures you can put on an $n$-element finite set (like, say, the set of binary trees with $n$ nodes) and $S_n$ is the group of permutations of n objects. The term $x^n$ is the cartesian product of n copies of the groupoid x.

It apparently makes sense to talk about groupoids with negative cardinality as well as fractional cardinality. For example, taking x to be the groupoid $3//Z_2$ described above, the cardinality of the infinte groupoid $\displaystyle 1 + x + x^2 + \ldots=\frac{1}{1-x}=\frac{1}{1-3/2}$

is -2!

This suggested to me that we could do something similar and construct a negative-dimensional object. If we stick in a square, we get a set

line $\times$ square $\times$ 4-cube $\times$ 8-cube $\times$ 16-cube $\times \ldots$

This doesn’t make much sense as an isolated set: all we can say is that it’s an infinite-dimensional hypercube. But if we had some way of knowing that the product we were taking was over $2^n$-cubes, then we could use the formula above.

Here’s the idea: if restricting the values a digit can take in an $n$-ary expansion reduces the dimension, then expanding the values ought to increase it. Since this doesn’t make much sense with points in the real number line, let’s move to $\mathbb{N}[[x]]$ instead: let $C$ be the set of Taylor series in $x$ with coefficients in $\{0,2\}$. We could say $C$ has dimension $\log_n(2)$ when $x=1/n$. (I still have to work out how the topology changes with $x.$)

Then we can define a set $T$ of series with, say, dimension 2, and construct an infinite product $\displaystyle \prod_{i=0}^{\infty} T^{\otimes i}.$

This object ought to have dimension -1; at least, it will satisfy the property that multiplying the dimension by 2 (by tensoring with a square) and adding one to the dimension (by multiplying by a line) will give the original object: if $d$ is the dimension, then $d$ satisfies $2d+1=d.$

The structure type for binary trees evaluated at the trivial one-element groupoid gives an infinite groupoid with cardinality $\frac{1+i \sqrt{3}}{2}$, so complex cardinalities (and therefore complex-dimensional objects) can also arise.

Can we make this rigorous? Is there some obvious connection to the complex dimensions of fractal strings?

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### 2 Responses

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1. D said, on 2007 July 30 at 10:31 pm

Could we simply use the Geometric Algebra multiplication by a plane? That is:
1+2+4+…..
represents a line, a plane, a hyperspace, etc… and when we multiply by a square it becomes
2 (1+2+4+…) = 2+4+8…
which you add a line, becomes the original again.

2. mike said, on 2007 July 30 at 10:41 pm

I assume that when you say “multiply by a square” you mean “multiply by a bivector” and that we’re working in a countably-infinite-dimensional algebra.

I guess the geometric algebra idea’s OK. You get objects whose dimension isn’t an integer, like 1+2 for a vector plus a bivector, which is very different from a 3-dimensional object.