The square root of a reflection?

What things are squares? This question often leads to interesting and strange new worlds in mathematics.

We begin our story with whole numbers. If the whole universe you can consider is whole numbers, then the only squares are the “perfect squares” which we obtain by finding the area of a square with whole number sides.

Being unsatisfied with only some of our numbers being squares leads to a new kind of number: the irrational square roots of whole numbers which are not perfect squares.

So in the realm of real numbers, we have that to be a square it is necessary and sufficient to be non-negative. To have all real numbers be squares, we must again extend to complex numbers. For numbers, the story stops there. All complex numbers are squares. But is that the end of the story?

Certainly not. The question of squares in finite fields, for example, leads to the beautiful result of quadratic residues. But I want to consider squares under another operation: composition.

We can think about numbers as corresponding to the action of multiplying by that number. So 2 represents the function f(x)=2x. Then the square root of that action is another action, which we call \sqrt{2}, which when repeated twice, is the same action as 2.

Given a function f:A \rightarrow A, does there exist a function r: A \rightarrow A such that r \circ r = f?

In general, this functional equation could be difficult to solve, so let’s consider the case when A is the plane. Given a transformation F:\mathbb{R}^2 \rightarrow \mathbb{R}^2, can you find another transformation which is its “square root.” i.e. T(T(x,y))=F(x,y)

The first question, is what if F is a pure motion, such as a translation, rotation, reflection, or dilation.

If F is a pure translation by a vector v, this is easy: T should be translation by a vector \frac{v}{2}. Similarly, if F is a pure rotation, then simply by rotating by half the angle, you obtain T.

Dilation yields the original square root: the original F is a dilation with scale factor s, then T should be dilation at the same center, with scale factor \sqrt{s}. In fact, these three actions can all be expressed as operations on complex numbers. Addition yields translation; multiplication, dilation and rotation.

Here’s the interesting one: what’s the square root (with respect to composition) of a reflection?

One answer is that it doesn’t exist. In order to make the question well-posed, we must specify what kind of thing T must be. If T is required to be a similarity transformation, then it suffices to consider the determinant. By the equation det(T^2)=det(T)^2=det(F)=-1, it’s clear that no real value of the determinant of T will be possible.

But what if T is not necessarily a similarity transformation, but some other more exotic function of the plane? I don’t know the answer to this question, but my suspicion is that it is not possible for a reasonable function. I believe that the above argument extends to smooth maps via the total derivative.

Intuitively, though, we can imagine “rotating” the plane 90 degrees along the reflection axis, which repeated twice would give the original reflection. This of course, means T is no longer a function on \mathbb{R}^2. But by stretching the target, we can make a sensible choice for this “square root of a reflection”

If the original F was reflection in the y-axis, we could represent this rotation using a complex matrix: T=\begin{bmatrix} i & 0\\ 0 & 1 \end{bmatrix}. Then T^2 = \begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix}, which is indeed a reflection across the y-axis. T is now a function from \mathbb{C}^2 to itself. But since it leaves the second copy of the complex numbers fixed, we can visualize the action has happening in 3-space, where the first complex coordinate is represented by the xz-plane and the real part of second coordinate is represented by the y-axis.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s