Measuring $\pi$ with alternate distance metrics (p-norm).

Math Question: Assuming the math above is ok, is there a reason for (2,π) (2,π) to be a minimum?

The $p=2$ is the only p-norm that has the $SO(3)$ Lie Group structure. In other words, it is rotation invariant. Try it, you can rotate the coordinate system without changing the length. You can't do that with the Taxicab metric, the length you get will change.

The deep answer to your question is that only $p=2$ has a continuous symmetry, rotation. All the other p-norms have a either a finite number of symmetries or symmetries. Here's the thing, a p-unit circle is something like every possible vector that can generated from the group glued together. Imagine starting with a small vector then copying and rotating it just a bit, attaching it to the original vector, then repeating with the copy. Eventually, you'll generate the circle. You generate any of the p-norm circles using a similar process using their symmetry, albeit a bit more opaque. So why is $p=2$ special? The circle is made with the shortest vectors, and only the shortest vectors available to construct it. Why? Because they're all the same length! This isn't true unless you're rotationally invariant and guess what? Only $p=2$ is!!

Math/Philosophy Question: Assuming above ok, is this why we observe the metric D2
D2 in the real world?

Quick run through physics. In physics, you have a theory described by a lagrangains that describes the amount of action particles do over a length of time. It's a scalar function of distance, speed, and time. The symmetries of the lagrangian correspond to conserved quantities according to Noether's Theorem. In standard classical mechanics, we assume three things,

(1) The action of a particle moving through some path along space is unchanged if you shift the start of the motion forward or backward in time. (Energy is Conserved)

(2) The action is unchanged if you shift the translate the start of the motion to another location. (Momentum is Conserved)

(3) The action is unchanged if you rotate the motion of the particle. (Angular Momentum is Conserved)

The last one isn't talked about to much, however, it's very important and highly relevant to our discussion. Since, our lagrangian, the theory for our physical theory, is invariant under rotation, there can never be a preferred direction in our theory. This means that means the metric we are using to measure distance and speed must also be rotation invariant. The $p=2$ norm is the only p-norm that'll work.

Having read (as best I can understand) the various related questions/answers linked by others and other links from those answers I have come to the following answer.

First of interesst There are two other integral forms of writing the integral for $pi_n$:

• From Michael in this answer.

$$2\int_0^1\left(1+\left(\frac{x^n}{1-x^n}\right)^{n-1}\right)^\frac{1}{n}dx$$

• From: Adler, C., & Tanton, J. (2000). π is the Minimum Value for Pi. The College Mathematics Journal, 31(2), 102-106. doi:10.2307/2687579Co

$$4\int_0^1\frac{1}{n}\left(x^{1-n}+(1-x)^{1-n}\right)^\frac{1}{n}dx$$

Adler & Tanton also go on to prove that a minimum occurs when $2-\frac{1}{n}=1+\frac{1}{n}$ which leads to the value of $n=2$ being a minimum. For others curious about this question I recommend reading this paper. (You can create a free account to read it if you do not have JStor access.)

They conclude their paper with the unproven proposal that $\pi_n=\pi_m$ if $\frac{1}{n}+\frac{1}{m}=1$. I'll see what I can do on that problem.

Edit: Interesting thought: If $\pi_\frac{1}{n}=\pi_{1-\frac{1}{n}}$ then $\pi_n$ can be defined for negative values.