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

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.

• I am guessing that this has something to do with the fact that the quadratic formula does not work over fields of characteristic 2. – Baby Dragon Mar 18 '13 at 18:09
• Would you mind explaining why the quadratic formula fails? Even if it is obvious, this is not my area of expertise and I am wary of making wrong assumptions. – hatch22 Mar 18 '13 at 18:28
• The quadratic formula says that if $ax^2+bx+c=0$ and $a\ne0$ then $x=\dfrac{\text{something}}{2a}$. The problem is that in fields of characteristic $2$, the denominator, $2a$, is $0$, so the fraction is undefined. – Michael Hardy Mar 18 '13 at 18:32

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.

• In "Affine and Projective Universal Geometry" ( arxiv.org/abs/math/0612499 ), Wildberger starts from a symmetric bilinear form on a vector space over a field of characteristic not 2, and defines all quantities from this. Quadrance is $B(x - x',y - y')$. There is never a need to determine $B$ from $Q$. Of course the linearization of squaring is at the heart of the problem but not by breaking the bilinear/quadratic forms connection. – zyx Mar 22 '13 at 3:22
• If $Q$ is actually the thing you want to know, there is no problem. When you are simply using $Q$ to prove things, the inability to recover $B$ from $Q$ becomes a problem, because it means things expressed in terms of $B$ might not be expressible in terms of $Q$. e.g. the triple quad law from the wiki would ordinarily reduce to the Cauchy-Schwartz condition for two vectors to be colinear; but there is a factor of 4 that kills everything and prevents us from recovering that information. (in fact, the triple quad law is a tautology in char. 2) – user14972 Mar 22 '13 at 3:55
• (That's "quadrance is $B(v,v)$ where $v = (x - x',y- y')$", for the quadrance from $(x,y)$ to $(x',y')$.) – zyx Mar 22 '13 at 3:55
• But really, there is simply a great wealth of manifestations of this same phenomenon, and I really do think we're just saying the same things from different perspectives. – user14972 Mar 22 '13 at 3:59
• That's an important point about not being able to detect collinearity using $Q$. I'd add it to the answer. Can collinearity in char 2 be detected using $B$, though? – zyx Mar 22 '13 at 4:29

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.

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.

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

• . . . . . because one cannot divide by them. – Michael Hardy Mar 18 '13 at 18:24
• . . . . . . or because some terms in polynomials just aren't there at all. – Michael Hardy Mar 18 '13 at 18:27
• @MichaelHardy It looks like whoever wrote the formulas I am looking at carefully wrote everything without dividing by 2 or 4... but I can imagine it might be normal to reorganize that way. When dealing with quadratic forms over fields of characteristic two, that is the rub. Thanks for adding. – rschwieb Mar 18 '13 at 18:30