We can visualize $1\mathrm D$, $2\mathrm D$, $3\mathrm D$ and we can think of a higher $M-\mathrm {Dimension}$ where $M\in \mathbb N^+$as a vector. But I recently learned that there are non integer dimensions such as $\frac{\log3}{\log2}$ for Sierpinski triangle and $\frac{\log2}{\log3}$ for Cantor set.
I have a three part question:
(a) What does it mean?
(b) Do we have a vector or other way to represent that?
(c) Are there negative dimensions too?
Thanks a lot.
On
Quoting from Abbot's book, "Understanding Analysis":
There is a sensible agreement that a point has dimension zero, a line segment has dimension one, a square has dimension two, and a cube has dimension three. Without attempting a formal definition of dimension (of which there are several), we can nevertheless get a sense of how one might be defined by observing how the dimension affects the result of magnifying each particular set by a factor of 3 (Fig. 3.2). (The reason for the choice of 3 will become clear when we turn our attention back to the Cantor set). A single point undergoes no change at all, whereas a line segment triples in length. For the square, magnifying each length by a factor of 3 results in a larger square that contains 9 copies of the original square. Finally, the magnified cube yields a cube that containsb 27 copies of the original cube within its volume. Notice that, in each case, to compute the “size” of the new set, the dimension appears as the exponent of the magnification factor.
Now, apply this transformation to the Cantor set. The set $C_0 =[0, 1]$ becomes the interval $[0, 3]$. Deleting the middle third leaves $[0, 1] ∪ [2, 3]$, which is where we started in the original construction except that we now stand to produce an additional copy of $C$ in the interval $[2, 3]$. Magnifying the Cantor set by a factor of $3$ yields two copies of the original set. Thus, if $x$ is the dimension of $C$, then $x$ should satisfy $2 = 3^x$, or $x =\dfrac{\log 2}{ \log 3}$.
The Sierpinski Triangle undergoes a similar change, but you get $3=2^x$, for if you double the bases of your triangle, you get three copies.
There are many generalizations of the usual notion of dimension, and they are there to capture different properties. Having said that, the intuition behind the dimension is that it describes the number of degrees of freedom you have, e.g.
However, the space defined may be more complex than that, e.g. in some areas might have one degree of freedom and in some other more, for example take a union of a disk and a line. Still, things can get even weirder, e.g. take the Hilbert curve, is it yet a line-like, or already a plane-like? In this way, you can think of the Sierpiński triangle as something that is a bit like a triangle, but not exactly yet.
To see how this works, consider the Cantor set and the following informal argument. Observe that to describe a point of $[0,1]$ with precision of $\frac{1}{3^n}$ you need $3^n$ possibilities, and thus in base $3$ at least $n$ digits. However, to describe the elements of the Cantor set you only need $2^n$ possibilities (the Cantor set is the set of all the points that doesn't use the number $1$ in their base-3 representation). But to describe $2^n$ possibilities you need only $\log_3(2^n) = n\frac{\log 2}{\log 3}$ digits. But we needed $n$ digits to represent an point in $[0,1]$ which is of dimension $1$, and so we arrive at $\frac{\log 2}{\log 3}$.
In the case of Hausdorff dimension there is yet another observation related to scaling, but this is already covered by @PedroTamaroff's answer and this article of Wikipedia.
As to negative dimension, I'm not really sure what "negative number of degrees of freedom" could mean. Right now I'm not aware of any useful related notions. On the other hand, you might be interested to know dimension is not the only seemingly integral characteristic, e.g. there are differentials of fractional order and those certainly can be negative (you get integrals then). In similar spirit a space of negative dimension would have to "remove the degrees of freedom" when joined with some other space, e.g. we have that $\dim(\mathbb{R} \times \mathbb{R}) = 2\dim(\mathbb{R})$, so possibly $\dim(\mathbb{R} \times X) = 0$? This might happen for example in a case where you require that points of your space have some property, and then all points of $\mathbb{R}$ would have that property, but only a countable subset of $\mathbb{R} \times X$ would work for you. Still, we need a new notion of dimension for that (one that incorporates your property).
I hope this helps $\ddot\smile$