Can we define length and perpendicularity not via an inner product?

Question

A natural way to reason about Euclidean geometry using modern mathematical language is to define Euclidean space as an affine space $A$ directed by a finite-dimensional real vector space $V$.

However, a bare vector space structure doesn't provide us with enough language to define things like lengths or perpendicularity, so we're ought to endow $V$ with some additional structure. The well-known solution is to equip $V$ with an inner product. The length of $v \in V$ is then defined as $\sqrt{(v, v)}$, and, by definition, $v^1 \in V$ is perpendicular to $v^2 \in V$ iff $(v^1, v^2) = 0$.

However, in order to convince someone that the concepts we intuitively know as "length" and "perpendicularity" are indeed expressible in terms of a positive defined bilinear form, we unavoidably(?) need to apply to the Pythagorean theorem. Thus, in order to justify that our model of Euclidean space is adequate, we need to invoke some facts about the Euclidean space, which is an awkward thing to do while we are defining it!

Are there any (hopefully more intuitive and convincing) ways of defining length and orthogonality in Euclidean space, other defining them in terms of an inner product?

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UPD: Let me try to rephrase my question in a different way.

Suppose you are giving someone an introductory course on Euclidean geometry. Imagine that you decided to stick to a linear algebraic spirit. So you postulate the existence of the set of points and the set of translations, and introduce the axioms regarding the translations and their relationship with numbers and with points. Now you can talk as much as you want about affine geometry.

Next you need to introduce some axioms relating to length and perpendicularity. You could define an inner product and even provide an example of it, but defining the distance in terms of it won't go well, because in order to believe that such definition is reasonable, your audience should know the Pythagorean theorem, which they don't, because you are just giving them an introductory course right now:)

Is there any satisfactory way out of this situation?

Answer 1

The following comes from https://projecteuclid.org/download/pdf_1/euclid.afm/1485893376

Let $(V,|\cdot|)$ be a normed vector space.

Define $v \perp w$ if $|v-w|^2 = |v|^2+|w|^2$.

Define a "normed perpendicularity space" as a normed vector space where the set of vectors orthogonal to a given vector is always a subspace.

Then $(V,|\cdot|)$ arises from an inner product. I do not have time for a full write up, but the details are in the paper and associated references.

These axioms are geometrically reasonable, and do not depend on the presence of an inner product.

Answer 2

I'm not quite sure if the question is the same as the question in the title.

For the question in the title: Sure, we can certainly develop the concept of length independently from inner product. The result would be a normed vector space. We only need 4 properties for the "norm" function:

• Nonnegativity
• A vector has zero norm if and only if the vector is zero
• Linearity for positive scalars
• Triangle inequality

There is no reference to inner product. What's more interesting is that not all normed vector space structure we can put on $\mathbb{R}^n$, including some quite useful ones, are induced by inner products.

For the other part, since perpendicularity means orthogonality, the reference to inner product is already implied.

Footnote: in Riemannian geometry, "perpendicular curves" can be defined in a way that seems to be independent from inner product. Although I don't think that counts because it is impossible to define a Riemannian structure without inner product anyway.

Answer 3

Here is an approach which works only(?) in two dimensional case and seems reasonably satisfying to me. Hope it helps to clarify the question!

Let $V$ be a two-dimensional real vector space. Consider a structure on $V$ consisting of:

1. A linear transformation $r$ on $V$ such that $r^2 = -1$ chosen up to sign.
2. A nondegenerate skew-symmetric bilinear form $s$ on $V$ chosen up to sign.

("chosen up to sign" means that we consider both $(-r, s)$ and $(r, -s)$ to be the same structure as $(r, s)$)

Given such a structure, one can make definitions:

• A length of $v \in V$ is $\sqrt{|s(v, r \cdot v)|}$.
• $v^1$ and $v^2$ are perpendicular iff $v^1$ and $r \cdot v^2$ are linearly dependent.

Arguably(!), those notions and definitions are better connected to intuitive geometric concepts. After all, if we think of $r$ as a right angle turn, and think of $s$ as an oriented parallelogram area, the definitions of length and perpendicularity immediately agree with intuition!

There is a natural bijective correspondence between inner products and structures of this kind. Indeed:

• If $V$ is equipped with an inner product, there is exactly one up to sign orthogonal transformation satisfying the $r^2 = -1$ equation, and exactly one up to sign skew-symmetric bilinear form $s$ which gives $±1$ on any orthonormal base.
• If, conversely, $V$ is equipped with $(±r, ±s)$ pair as above, define $f(v^1, v^2) := s(v^1, r \cdot v^2)$. $f$ is bilinear and symmetric (this easily follows from $r^2 = -1$). Moreover, if $v \neq 0$, then $v$ and $r \cdot v$ are linearly independent (this again follows from $r^2 = -1$). Because $s$ is nondegenerate, we conclude that $\forall v \neq 0 : f(v, v) \neq 0$. Thus, the quadriatic form associated with $f$ is definite, so it's either positive definite or negative definite, so either $f$ or $-f$ is a valid inner product.

Answer 4

In general, manifolds have not a norm, but a metric, which is a function that takes two points as input and gives their distance as output. A space with addition/subtraction and a norm has a metric (the distance between two points is the length of their difference), and metric that respects linearity is a norm (the length of a vector is the distance between it and the origin).

When doing constructions in plane geometry, if we're given a line $L_1$ and a point $p_0$ on $L_0$, we can construct a line $L_2$ perpendicular to $L_1$ at $p_0$ by first finding two points $p_1, p_2$ on $L_1$ that are the same distance from $p_0$, then finding points that are the same distance from $p_1$ and $p_2$. If we take this as our definition of "perpendicular", we can generalize this to any metric space.

Once a metric $d:X \times X \rightarrow \mathbb R$ is defined on a set $X$, given a curve $C$ in $X$, a point $p_0$ on $C$, and a parameterization $f:\mathbb R \rightarrow X$ such that $f(0)=p_0$, we can define a set $S= \{p:\lim_{t \rightarrow 0} \frac {d(f(t),p)d(f(-t),p_0)}{d(f(-t),p)d(f(t),p_0)}=1\}$ (there's a few more details in making sure the denominator is non-zero that I won't get into).

If we require that $f$ have a "speed" of $1$, that is, $\lim_{t \rightarrow 0} \frac {d(f(t),f(0))}t=1$, then we can simplify the definition of $S$ to $\{p:\lim_{t \rightarrow 0} \frac {d(f(t),p)}{d(f(-t),p)}=1\}$. If $C$ is also a line, then we can simplify further to $S = \{p:d(f(t),p)=d(f(-t),p) \} \text{for some $t$}$ (for a line, the value of $t$ doesn't matter).