Why does rational trigonometry not work over a field of Characteristic 2?

Question

I am interested in solving triangles in a finite field with a computer program. Rational trigonometry seems well suited to do this. However, the Wikipedia article, as well as several published sources, claim that rational trigonometry does not work in fields (whether finite or infinite) of characteristic 2 "for technical reasons." Computers being binary machines, they work well with (finite) fields of characteristic 2. So I would like to understand why fields of characteristic 2 present a technical obstacle. I have not been able to find any clear explanations via Google or any available online publications.

Answer 1

Squaring behaves strangely in characteristic 2. One result if this weirdness is the identity $(x+y)^2 = x^2 + y^2$ -- squaring doesn't result in the 'mixed terms' it usually does. Among the things that this "breaks" is the theory of quadratic forms and bilinear forms.

When you have a symmetric bilinear function -- that is a function $B(x,y)$ satisfying

• $B(x+y,z) = B(x,z) + B(y,z)$
• $B(x,y) = B(y,x)$
• $B(rx,y) = r B(x,y)$ where $r$ is a scalar

then you can construct a "quadratic form" $Q(x) = B(x,x)$. Conversely, when you have a quadratic form $Q(x)$, you can construct a function $B'(x,y) = Q(x+y) - Q(x) + Q(y)$.

These constructions are almost inverses: you have an identity $B'(x,y) = 2 B(x,y)$. So in any setting where $2$ is invertible, one can seamlessly pass back and forth between the idea of a quadratic form and the idea of a bilinear form.

But in characteristic $2$, the connection breaks, since the identity becomes $B'(x,y) = 0$.

Vector geometry relies heavily on multi-linear algebra: linear forms, bilinear forms, determinants, and so forth.

Rational trigonometry, is meant to more directly mimic classic trigonometry and relies very much on squaring to keep things rational. Quadrance is a quadratic form that is normally the one associated with the dot product, but that connection is broken in characteristic 2. Spread is more complicated, but I believe its connection with the cross product is also broken.

Answer 2

I am not very familiar with the topic, but I notice in the wiki page you linked to there are several 2's and 4's appearing in various identities.

All of those would turn to zeros and probably limit their usefulness.

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"Limit their usefulness" meaning "making them useless."

Answer 3

Circles become straight lines in characteristic $2$. Rotations, the linear transformations that keep a circle in place, become translations. This would tend to make trigonometry very trivial or very subtle, and definitely very different, from what we are accustomed to.

The equation $x^2 + y^2 = 1$ is the same as $(x+y)^2 = 1$ when $2=0$, which is the same as $(x+y - 1)^2 = 0$. This equation defines the same set of points as the line $x+y=1$. Algebraically it is a "double line".

Now, where

several published sources, claim that rational trigonometry does not work in fields (whether finite or infinite) of characteristic 2 "for technical reasons."

here are a few things that are not problems.

1. The formulas for the basic quantities of Rational Trigonometry, spread and quadrance, do not require division by $2$.

2. The formulas can be written, still not using division by $2$, so as to use only the dot products between vectors. If geometry is defined as the structure invariant under dot-product preserving transformations of the coordinate plane, the spread and quadrance are meaningful geometric quantities in characteristic $2$ for the same reason that they are geometric quantities in Euclidean geometry.

3. The equation "Spread = constant" continues to define a pair of straight lines relative to a fixed line.

It looks like the problem in characteristic $2$ is not that trigonometry, rational or not, does not work, but that the nonlinear geometry of circles is missing.