"Anti-intuitive" High Dimension Geometry

At the beginning, here is a paper I wrote before about how to deduce the high-dimension properties in probability theory.

Read the paper

Recently I watched the video from 3b1b and got some new intuitions. For a visual intuition, see 3Blue1Brown’s talk on spheres.

Fact 1: The diagonal line of 4D cube is $\sqrt{1 + 1 + 1 + 1} = 2$, and if you put a unit ball in the center of the 4D cube, it will even reach the vertex.

Fact 2:

\(x^2 + y^2 + z^2 = 1\).

is a 3D ball, it can compose to

\[\begin{cases} x^2 + y^2 = 1 - z^2,\\ z^2 \leqslant 1. \end{cases}\]

By fiber decomposition, it can be transferred to

\[\begin{cases} x^2 + y^2 = 1\\ z^2 \leqslant 1 \end{cases}\]

in another word, $\partial B^3 = B^1 \times \partial B^2 .$

How to prove it? Notice for the new equation, just update by $(x, y) = \left( \frac{x}{\sqrt{1 - z^2}}, \frac{y}{\sqrt{1 - z^2}} \right)$.

Geometrically, it is mapping the sphere to a cylinder.

Since this algebraic transformation can be conducted on any space, we can write:

\[\partial B^n = B^{n - 1} \times \partial B^2 .\]

Not rigorous, but easy to use.

By the measure theory, $m (\partial B^n) = m (B^{n - 1} \times \partial B^2) = m (B^{n - 1}) \times m (\partial B^2) = 2 \pi r \times m (B^{n - 1})$.

By this recurrence formula, literally we can deduce the surface for any space and use integral to find the volume.

\[\begin{align} c_n r^n & := m (B^n) \nonumber\\ & = \int_0^r m (\partial B^n) \, dr \nonumber\\ & = \int_0^r m (B^{n - 1}) \times m (\partial B^2) \, dr \nonumber\\ & = \int_0^r 2 \pi r \times c_{n - 1} r^{n - 1} \, dr \nonumber\\ & = \frac{2 \pi}{n + 1} c_{n - 1} r^n . \nonumber \end{align}\] \[c_n = \frac{2 \pi}{n} c_{n - 2}, \quad n \in \mathbb{N}_{> 1} .\]

If we never know any knowledge of the circle, we shall deduce from $c_0$. Imagine $m (B^0) = c_0 = 1$, which actually means: find 0 numbers such that their summation is less than 1, which is satisfied with probability 1.

So we can get:

\[m (B^n) = \frac{\pi^{\frac{n}{2}}}{\left( \frac{n}{2} \right) !} r^n, \quad n \in 2\mathbb{Z}.\]

If we want it to also work for odd dimensions, we need to deduce $c_1$ first, so it will be:

\[\frac{\pi^{\frac{1}{2}}}{\left( \frac{1}{2} \right) !} = 2,\]

which is:

\[\frac{1}{2} ! = \frac{\sqrt{\pi}}{2} .\]

And for the rest of the dimensions: $\left( \frac{3}{2} \right) ! = \frac{3}{2} \times \frac{\sqrt{\pi}}{2}$, $\left( \frac{5}{2} \right) ! = \frac{5}{2} \times \left( \frac{3}{2} \right) !.$

So:

\[m (B^n) = \frac{\pi^{\frac{n}{2}}}{\left( \frac{n}{2} \right) !} r^n, \quad n \in \mathbb{Z}^{+} .\]

By:

\[\lim_{n \rightarrow +\infty} \frac{\pi^{\frac{n}{2}}}{\left( \frac{n}{2} \right) !} = 0,\]

the high-dimension ball “collapses” to the original point.

Also, since

\(\frac{\dfrac{\pi^{\frac{n}{2}}}{\left( \frac{n}{2} \right) !}(1)^n}{\dfrac{\pi^{\frac{n}{2}}}{\left( \frac{n}{2} \right) !} (1 - \varepsilon)^n} \rightarrow +\infty\).

when $n \rightarrow +\infty$ no matter how small $\varepsilon$ is, we start to find that the “mass” of the high-dimension ball is distributed on its surface.