It is possible to construct alternative planes to the usual Euclidean plane $\mathbb{R}^2$ by replacing the real numbers $\mathbb{R}$ with other ordered fields $F$. Depending on the choice of $F$, various properties or axioms of Euclidean geometry may or may not hold.

For example, if $F$ is "bigger" than $\mathbb{R}$ in the sense that $\mathbb{R}$ is a proper subfield of $F$ (or also, if $F$ is really "bigger" in the sense of set cardinality, i.e. $|F| > \beth_1$), then the "Archimedean axiom" will fail: that is, there will exist line segments which are "infinitely big" or "infinitely small" compared to other line segments. In the old terminology of Euclid, there would be "linear magnitudes" which would not "have a ratio" to each other. In modern mathematics, it means that there exist line segments such that their length can be multiplied by any natural number and not exceed the length of some other line segments.

Also, depending on what properties $F$ satisfies, other things may not be possible: for example, if $F$ is not "real-closed", then not all of Tarski's axioms for Euclidean geometry will hold. Depending on the $F$, lines and circles may not intersect when we might think they should.

But what I'm curious about is this: what kinds of fields $F$ permit a natural definition of angle measure? If $F$ is closed under the operation $\sqrt{1 + x^2}$, then a natural definition of distance measure is possible:

$$d(\mathbf{u}, \mathbf{v}) = \sqrt{(v_x - u_x)^2 + (v_y - u_y)^2}$$

where $\mathbf{u} = (u_x, u_y)$ and $\mathbf{v} = (v_x, v_y)$ are elements of $F^2$. This distance is an element of $F$. This comes from the Pythagorean theorem, and is just the familiar Euclidean distance, only now over a field which may not be the real numbers.

But what about angle measure? We may define an angle measure for $F = \mathbb{R}$ with the "inner product" of vectors $\mathbf{u}$ and $\mathbf{v}$ being defined by

$$\mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y$$

then the angle between two lines, represented as vectors, is

$$\theta(\mathbf{u}, \mathbf{v}) = \cos^{-1} \left(\frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||\ ||\mathbf{v}||}\right)$$

where $||\mathbf{v}|| = \sqrt{\mathbf{v} \cdot \mathbf{v}} = d(\mathbf{0}, \mathbf{v})$ is the "norm" or length of the vector $\mathbf{v}$.

At first, one might think this would only work on $\mathbb{R}$, but that isn't necessarily so: take $F = \ ^{*} \mathbb{R}$, where $^{*} \mathbb{R}$ is the so-called "hyperreal numbers" or "non-standard real numbers", which can be constructed via a set-theoretic construction called a "ultrapower" of $\mathbb{R}$. This is from "non-standard analysis", and this $F$ is non-Archimedean, and definitely not $\mathbb{R}$. In this field, there is a natural extension $\ ^{*} \cos^{-1}$ of the inverse cosine, and so we can use that and the above formula to get an angle measure.

It may also be possible to define an angle measure (suitably restricted in range) for the mega-field (= too big to be a set, is a "proper class") $F = \mathbf{No}$, the mega-field of "surreal numbers" by Conway, also by extending the cosine (albeit in a different way).

So clearly, the answer is "more fields than just $\mathbb{R}$, but not every ordered field". So which ordered fields allow for it?

EDIT: It is asked to clarify what geometry is meant. I am specifically interested in the case of models of the Tarski's first order Euclidean plane geometry, which will be $F^2$ for a real closed ordered field $F$. Which permit the definition of an angle measure? We might also weaken things further by omitting the completeness axiom, so the $F$ need not be real closed.

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    $\begingroup$ In the context of a Cayley-Klein metric, one can define angle as a multiple of the (complex) natural logarithm of four lines, two of which are the lines whose angle you compute and the other two are the lines connecting their intersection to special points with complex coordinates. This use of complex numbers along the way is likely a stronger requirement than your definition using $\cos^{-1}$, so I don't consider this an answer, but it might be interesting nevertheless. $\endgroup$ – MvG Feb 25 '14 at 12:26
  • $\begingroup$ So are you only asking about angle measures for geometries which have as models Cartesian spaces $F^n$, where $F$ is some field? Or are you asking about arbitrary metric geometries? I don't really understand the implicit premise that every geometry will be connected to some choice of field, although perhaps you could explain/elaborate this point somewhat. $\endgroup$ – Chill2Macht Jul 4 '17 at 10:58
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    $\begingroup$ Also I don't understand the claim that the distance measure given for $\mathbb{R}^2$ is natural -- it is only natural if one assumes Tarski's axioms, but that doesn't seem to be your starting point, your starting point seems to be the set $\mathbb{R}^2$ itself, which can be assigned numerous other geometries besides the Euclidean one, including numerous non-Euclidean metric geometries, for example by using the metric induced by $\ell^p$ norms for $p \not = 2$. I think this is a good question, but you are making it more difficult to answer by not clarifying the scope. $\endgroup$ – Chill2Macht Jul 4 '17 at 11:01

In the theory of real closed fields, we can construct the complex plane and identify it with the Euclidean plane in the usual way.

(everything in this post refers to these constructions made over your real closed field of choice; e.g. "complex number" will mean an element of $F[i]$)

Let $S^1$ be the usual unit circle centered at the origin: the complex numbers of length 1. This is a group where the operation is multiplication.

One can always define an $S^1$-valued "angle measure" in a straightforward way.

I will assume, however, that you are asking about the possibility of having an $F$-valued angle measure. In particular, this means you are asking for a group homomorphism $F \to S^1$ that maps a real number to the corresponding angle.

A sufficient condition is for $F[i]$ to be an exponential field, with $\exp$ surjective onto the nonzero elements and satisfying $\exp(\bar{z}) = \overline{\exp(z)}$. The angle measure homomorphism would then be $r \mapsto \exp(ir)$.

I suspect this is not a necessary condition.

  • $\begingroup$ I mean an $F$-valued angle measure for vectors in $F^2$. Real numbers should not be involved. $\endgroup$ – The_Sympathizer Jul 4 '17 at 12:05
  • $\begingroup$ @mike4ty4: Real numbers aren't involved (unless you choose $F=\mathbb{R}$); everything in this post is based on whatever real closed field you want to work over. $\endgroup$ – user14972 Jul 4 '17 at 12:06
  • $\begingroup$ I guess I didn't get your bit about "a group homomorphism F→S1 that maps a real number to the corresponding angle." $\endgroup$ – The_Sympathizer Jul 4 '17 at 12:07

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