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

A traditional way to model Euclidean geometry is to consider an inner product vector space $$V$$ and to define that the length of $$v$$ is $$\sqrt{(v, v)}$$ and that $$v$$ is perpendicular to $$u$$ iff $$(v, u) = 0$$.

Are there any other ways to model Euclidean geometry which do not postulate the existence of an inner product?

Here is my motivation for this question. In order to justify that inner product space indeed models intuitive notions of length and perpendicularity we need non-trivial reasoning, including appealing to the Pythagorean theorem. I'm looking for an alternative way where we first establish some easily justifiable model of Euclidean geometry, and after that prove the Pythagorean theorem inside that model.

• This is an interesting question, but I think it is better suited to math.stackexchange. I would also be interested in seeing a direct axiomatization of perpendicularity in a normed space. Commented Feb 12, 2021 at 23:53
• @StevenGubkin, I think this question can be interpreted from math educators' point of view, since it is mainly about the logical ways to develop certain concept.
– user13395
Commented Feb 13, 2021 at 0:25
• To model anything, one needs to use some facts about that thing. So I see no problem in using facts about Euclidean space in order to define a mathematical model of Euclidean space. The real problem, as far as I can tell, is that, after defining the model, we give it the same name, "Euclidean space", as the thing being modeled. If we avoid conflating the two, then what you describe as "awkward" becomes "to invoke some facts about the Euclidean space ... while we are defining its model." And then it no longer looks awkward (to me). Commented Feb 13, 2021 at 0:40
• @AndreasBlass I think the awkward thing is that students have never been exposed to an inner product before. So you are using an unfamiliar feature of Euclidean space to define more familiar ones. I think harius might have something in mind like "A normed orthogonality space is a normed space $(V, |\cdot|)$ together with a relation $\perp$ satisfying...". Then, once intuitive axioms for length and orthogonality have been established, perhaps the inner product could be derived from these more intuitive axioms. Commented Feb 13, 2021 at 1:03
• Introducing length and perpendicularity via inner product, you can tell your students that these definitions are Pythagoras' theorem in disguise. And immediately after, show them a one-line proof of this theorem (showing that the theorem is actually encoded in our definitions). And I'd say this is satisfactory, since the resulting confusion can teach them a lot about axioms, definitions and theorems. Commented Feb 15, 2021 at 18:28

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.

• Yes, "polarization", ... :) Commented Jun 2 at 13:39
• That is a really interesting approach! Indeed, this axiomatization doesn't rely on bilinear forms. Its motivation, however, still relies on Pythagorean theorem, leaving us with a pedagogical chicken-and-egg problem Commented Jun 2 at 14:17

Given this is an ED forum and you claimed (in comment replies) that you felt it was better at math ed, than at math non-ed, I'm replying to the EDUCATIONAL ASPECTS of your proposal (where I am better qualified). And NOT the "does my proposal work mathematically" part of your question (where I am not well qualified).

I think this is a miserable plan educationally. You have all kinds of abstractions (set theory, axioms of algebra, etc.) which are going to raise the difficulty and derail the base course itself. You are making the classic mistake of "I finally got this" and trying to dump that onto average 14 year old newbies.

In addition to the difficulty aspect, there is also the issue that it is good to think about a problem in different ways and Euclidean geometry is more pictorial and has some classic theorems and explication. This is their chance to get THAT WAY OF THINKING ABOUT THINGS (not your oddball variant). If you try to inject some sort of vector crap into that, you're hurting them having that experience. Let them deal with a small taste of vectors in second year algebra (after geometry) and even more in pre-calc. If some of them (a tiny minority) want to look at things the way you have them, when they are math majors in college, fine. But don't screw up the standard geometry course. Physician...first, do no harm!

• +1 ... the other answers here are mostly far too advanced for high school. Commented Jun 11 at 17:47

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.

• I think @harius might have something in mind like "A normed perpendicularity space is a normed space $(V, |\cdot|)$ together with a relation $\perp$ satisfying..." Commented Feb 13, 2021 at 1:00
• What I'm trying to ask is can we instead of saying "perpendicularity means orthogonality" define perpendicularity in linear spaces through some other (hopefully more self-evident) concepts than inner products? If the concept of "a normed perpendicularity" space which Steven is talking about can be formalized, that would be interesting for me to look at indeed! Commented Feb 13, 2021 at 14:31
• Perhaps I am misunderstanding something, but couldn’t you define “perpendicular” in a normed space by saying that $a \perp b$ means that $|a+b|^2 = |a|^2 +|b|^2$? No need for an inner product unless you want to measure angles other than right angles. Commented Feb 14, 2021 at 1:04
• Yes, such a definition generally isn't useful, because it even doesn't guarantee that orthogonal complement to a vector is a subspace. I don't believe that orthogonality can be meaningfully defined in an arbitrary normed space, and that's not what my question is about Commented Feb 14, 2021 at 11:32
• One can indeed define a normed perpendicularity space as a normed space in which parallelogram identity holds, and define $a \perp b \iff |a - b| = |a + b|$, so this formally answers the question (since we managed to define Euclidean space and length and orthogonality in it without speaking about bilinear forms). However, such approach is significantly more complicated than the classical one :) Commented Feb 14, 2021 at 11:57

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.
• Yes-this approach works when the dimension of the vector space is even. Your transformation $r$ is called an almost-complex structure $J$, and $s$ is called a symplectic form $\omega$. We talk about compatible triples $(J,\omega,g)$. Any two of the three determine the other by a compatibility condition. Commented Feb 13, 2021 at 16:42

Here is an alternative way to model Euclidean geometry without appealing to Pythagorean theorem in any form or disguise. Start with a normed vector space $$V$$ and postulate that for every vector $$u$$ there is a hyperplane $$u^\perp$$ such that mirroring $$u$$ across $$u^\perp$$ is isometric. It necessarily follows that $$V$$ is Euclidean! Here is the formal statement:

Theorem: Let $$V$$ be a normed vector space with a strictly convex norm. Suppose that for every nonzero $$u \in V$$ there exists a unique hyperplane $$u^\perp$$ not containing $$u$$ such that the transformation of $$V$$ which sends $$u$$ to $$-u$$ and preserves $$u^\perp$$ is isometry. Then there exists an inner product $$(u, v)$$ on $$V$$ which induces both the norm and orhogonality. That is, $$\lvert v \rvert ^2 = (v, v)$$ and $$v \in u ^\perp \iff (u, v) = 0$$.

Outline of the proof:

• Define $$(u, v)$$ as the unique function which is linear by the second argument and $$(u, u) = \lvert u \rvert ^2$$ and $$(u, v) = 0 \ \forall v \in u ^\perp$$;
• Observe that to show that $$(u, v)$$ is an inner product, it's sufficient to prove for arbitrary unit vectors $$u$$ and $$v$$ that $$(u, v) = (v, u)$$;
• Note that if $$f$$ is an isometry such that $$fu = v$$ and $$fv = u$$, then $$(u, v) = (fu, fv) = (v, u)$$; therefore, all we need to do is to find such an isometry;
• Define $$f$$ as the isometric reflection along $$u - v$$. Observe that $$u, v, fu$$ and $$fv$$ are all unit vectors and they all reside on the same affine line $$v + (u - v) \cdot \lambda$$. But there can be at most two unit vectors on an affine line. Conclude that $$v = fu$$ and $$u = fv$$.
• Perhaps B => E can be proven simply and directly. See [Ficken, 1943], where something very similar is done(jstor.org/stable/1969273) Commented Jun 2 at 13:40

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).

• I can't agree with your approach for arbitrary curves... for any $p \notin C$, both $d(f(t), p)$ and $d(f(-t), p)$ have the same nonzero limit (that is, $d(p_0, p)$), so the limit of their fraction equals 1 automatically, or I'm missing something? Commented Feb 14, 2021 at 15:07
• One indeed might define, as you suggest, $v^1 \perp v^2 \iff ||v^1 - v^2|| = ||v^1 + v^2||$ for elements of an arbitrary normed vector space. The problem is that this setup is too general to reason about Euclidean geometry. I wonder if it can be fixed by adding some good-looking axioms... Commented Feb 14, 2021 at 15:23

There are multiple possible approaches to formalizing length and perpendicularity. Here is a rough overview:

1. Start with a normed vector space. Add an axiom of existence of a mirroring hyperplane $$u^\perp$$ for each vector $$u$$. Define $$v \perp u \iff v \in u^\perp$$. See this answer for details;
2. Start with a normed vector space. Define $$v \perp u \iff \lvert v + u \cdot \lambda \rvert \geq \lvert v \rvert \ \forall \lambda$$. Add several technical axioms. See James, 1946 for details;
3. Start with a 2-dimensional vector space. Add axioms of existence of a 90-degree rotation $$r$$ and a signed area $$s$$. Define $$\lvert v \rvert = \sqrt{s(v, rv)}$$ and define $$v \perp u$$ iff $$v$$ is collinear with $$ru$$. See this answer for details;
4. Start with a normed vector space. Define $$v \perp u \iff \lvert v - u \rvert ^2 = \lvert v \rvert ^2 + \lvert u \rvert ^2$$. Add some technical axioms. See this answer by Steven Gubkin for details.