What is so special about Pythagorean distance? Why squaring?

Question

It seems that squaring appears not only in classical Euclidean geometry, but also in physics and statistics. Not too long ago, 3b1b released a video on where pi comes from in the normal distribution equation. In short, the assumption of the uncorrelatedness of two parameters (in the case of a two-dimensional distribution) leads to the quadratic power of the variable and only to it, which can then be generalized to any number of dimensions (that is, parameters). Pi is needed for unitarity. It seems to me that this is very similar to the reasoning when deriving a Pythagorean metric for an arbitrary number of dimensions!

However, it is not clear where exactly the second degree, and not some other, comes from. It seems that only the Pythagorean metric (and also the Riemannian, but locally it is Pythagorean and flat) is continuous and smooth at any point, any direction, and any scale. R^n can be shifted to any distance and rotated in any plane to any angle without losing all of the above properties. Why squaring? Is there any derivation of this metric ab initio (and, importantly, without any circularities in arguments)?

Answer 1

There are actually examples of other norms where we take the $p$th power. So for example, there exists a space where we could say distance is

$$\|\vec{x}\|=|x_1|+|x_2|+\cdots+|x_n|$$

This kind of norm is called a taxicab norm, and it's sometimes useful to think of the length of something being added up like this. So why $p=2$?

In Euclidean space, we have dot products between two vectors, and that would automatically create the concept of a norm. We could define the norm as being

$$\|\vec{x}\|=\sqrt{\langle \vec{x},\vec{x}\rangle}$$

And by relating the definitions of a norm and a metric, we can create certain definitions about the space that we've created. For example, we can then define angle to be

$$\theta=\arccos\frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\|\vec{y}\|}$$

Something like this wouldn't be possible in, say, the taxicab norm, because we wouldn't necessarily have $\left|\frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\|\vec{y}\|}\right|\leq 1$.

So it is technically possible to not have $p=2$, but that would require you getting rid of the dot product (or you could keep it and have a very cursed space), and lose having definitions for things like orthogonality.

Answer 2

The length formula is derived ab initio from the Pythagorean formula.

The Pythagorean formula has squares in it because that's what makes it true.

Or, if you like, the Pythagorean formula has squares in it because it's original formulation, expressed in terms of area of certain squares, says: given a right triangle, the area of the square constructed on one of the legs, plus the area of the square constructed on the other leg, is equal to the area of the square constructed on the hypotenuse.

Two millennia later, when Descartes described coordinates for Euclidean planar geometry using two axes meeting at right angles, the Pythagorean formula was then used to derive the distance formula between two points, expressed in terms of the Cartesian coordinates of those points.

Answer 3

Okay, so we want a notion of distance on a vector space $\mathbb{R}^n$. What criteria should it satisfy? Well, we'd like it to have a kind of symmetry. In, say $\mathbb{R}^2$, if we go $2$ forward and $1$ up, we would like this to always give the same distance. That is, for two points $p,q\in\mathbb{R}^n$, we would like the distance between them to only depend on $p-q$. So we want a function $f:\mathbb{R}^n\to\mathbb{R}$ that defines distance. Clearly, the distance between $p$ and $q$ should be the same as the distance between $q$ and $p$. So $f(p-q)=f(q-p)$. Also, the distance between $p$ and $q$ should only be $0$ if $p=q$. So $f(p-q)=0$ implies $p=q$. If we move the same way a certain number of times, it should multiply the distance by the same number. So $f(kp)=kf(p)$. We want the distance between $p$ and $r$ to be less than the sum of the distance between $p$ and $q$ and the distance between $q$ and $r$: $f(p-r)\leq f(p-q)+f(q-r)$. Otherwise, going to $r$ via $q$ is quicker than going directly. This makes our function $f$ a norm on $\mathbb{R}^n$.

You asked about the norm from Pythagorean distance, the Euclidean norm in relation to angles, but actually, I don't think you could define angles as usual without implicitly using or effectively introducing the Euclidean norm. So here's a related, but more basic reason why the Euclidean norm is special. How many points are there in $\mathbb{R^n}$? Well, if $n>0$, infinitely many (the cardinality of the continuum in fact). But, in another sense, if $n>0$, there are two points. $\mathbf{0}$, and all the other points. All the purely linear algebraic properties of a point in $\mathbb{R}^n$ are determined entirely by $n$ and whether or not it is $\mathbf{0}$. This is because the linear algebraic symmetries, bijective linear maps, between $\mathbb{R}^n$ and itself, can send $\mathbf{0}$ only to $\mathbf{0}$, and any non-zero point to any other non-zero point. So the non-zero points all look the same.

Adding a distance changes this. Different points have different distances from $\mathbf{0}$. So, if we want a norm on $\mathbb{R}^n$ that is maximally symmetric, it would suffice to show that the only thing distinguishing two points is their distance from $\mathbf{0}$. That is, for every $p,q\in\mathbb{R}^n$ such that $f(p)=f(q)$, there should be a bijective linear map $L:\mathbb{R}^n\to\mathbb{R}^n$ such that $L(p)=q$ that is distance preserving (so it respects all the properties of a normed space): $f(x)=f(L(x))$ for all $x\in\mathbb{R}^n$. As it turns out, the Euclidean norm is the only norm on $\mathbb{R}^n$ satisfying this property... almost. If we denote the Euclidean norm by $|\cdot|$, then for any bijective linear function $L:\mathbb{R}^n\to\mathbb{R}^n$, the function $|L(\cdot)|$ would also have the same property, but that's it. But this is as good as you're going to get because $L$ is a linear algebraic symmetry, so $|L(\cdot)|$ is exactly the same as the regular Euclidean norm from the perspective of the underlying space.

If you want to go further, then for any of these "maximally symmetric" norms on $\mathbb{R}^n$, you could try to find a special basis for $\mathbb{R}^n$, called an orthonormal basis. If you have a bijective linear map taking an orthonormal basis to another, then the normed spaces, with the orthonormal bases identified, look exactly the same.