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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?


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?

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    $\begingroup$ 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. $\endgroup$ – Steven Gubkin Feb 12 at 23:53
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    $\begingroup$ @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. $\endgroup$ – user13395 Feb 13 at 0:25
  • $\begingroup$ 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). $\endgroup$ – Andreas Blass Feb 13 at 0:40
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    $\begingroup$ @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. $\endgroup$ – Steven Gubkin Feb 13 at 1:03
  • $\begingroup$ 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. $\endgroup$ – Michał Miśkiewicz Feb 15 at 18:28
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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.

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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.

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  • $\begingroup$ @user52817, Well... otherwise we wouldn't have two different terms, I suppose. But yeah, totally worth emphasizing. Fixed it now. $\endgroup$ – user13395 Feb 13 at 0:32
  • $\begingroup$ 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..." $\endgroup$ – Steven Gubkin Feb 13 at 1:00
  • $\begingroup$ 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! $\endgroup$ – harius Feb 13 at 14:31
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    $\begingroup$ 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 $\endgroup$ – harius Feb 14 at 11:32
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    $\begingroup$ 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 :) $\endgroup$ – harius Feb 14 at 11:57
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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.
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    $\begingroup$ 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. $\endgroup$ – user52817 Feb 13 at 16:42
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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).

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  • $\begingroup$ 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? $\endgroup$ – harius Feb 14 at 15:07
  • $\begingroup$ 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... $\endgroup$ – harius Feb 14 at 15:23

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