Bielefeld, June 1997

Kreisbewegung.

The basic algebraic operations — addition, multiplication, inversion — work just as well for matrices whose entries are not numbers but functions. To keep things relatively simple, let us consider only smooth functions of a single real variable t, “smooth” meaning that they have derivatives of all orders: first, second, third, and so on.

If we think of t as “time”, a pair of functions X(t) = [x(t), y(t)] can be visualized as a moving point or as an evolving vector. For instance, X(t) = [3t, 2t] can be thought of as a point which moves along the straight line containing X(0) = [0, 0] and X(1) = [3, 2] — or as a vector (parallel to that line) which grows longer with time. The velocity vector Ẋ(t) of a moving point X(t) is obtained by simply differentiating the component functions — in other words Ẋ(t) = [ẋ(t), ẏ(t)] — because

Ẋ(t) = lim_{h→0} \frac{1}{h}(X(t + h) − X(t))

by the usual logic of computing instaneous velocities.

The interpretation of a variable matrix A(t), with the functions a(t), b(t), c(t), d(t) as entries, is more subtle and will emerge gradually as this story unfolds. For the time being, you may think of it as a pair of column vectors evolving in tandem. Our main business in this section is algebraic, and to that avail we note especially that differentiation (again entry by entry) of a product matrix P (t) = A(t)M(t) involves the Product Rule

Ṗ (t) = Ȧ(t)M(t) + A(t)Ṁ(t).

In fact, if p_{ij} denotes the entry in row i and column j of P (t) — with analogous notation for A(t) and M(t), leaving out the (t) for brevity — we have p_{ij} = a_{i1}m_{1j} +a_{i2}m_{2j} whose derivative ȧ_{i1}m_{1j} + ȧ_{i2}m_{2j} + a_{i1}ṁ_{1j} + a_{i2}ṁ_{2j} fits the desired pattern. For a constant matrix A(t) = A, this simplifies to AṀ(t) since A has zero derivative.

Circular motion. Imagine a point X(t) moving counter-clockwise around the unit circle at unit speed, starting at [1, 0]. Since its velocity vector would have to be perpendicular to the radius vector — whose coordinates are x(t), y(t) — it would have to satisfy

(∗)  X(0) = [1, 0]  and  Ẋ(t) = JX(t) ,

where J is the usual (constant) rotation through a right angle. (Please make a sketch of this situation!) You might say that you already know the solution, namely x(t) = cos t and y(t) = sin t — but what are those? Our task is to solve the problem without recourse to unexplained black boxes, using only the basic algebraic operations — and limits.

Equation (∗) asks for a pair of functions x(t), y(t) such that x(0) = 1, y(0) = 0 and

\begin{aligned}ẋ(t) & = −y(t) \\ ẏ(t) & = x(t) . \end{aligned}

Multiplied crosswise, the last two conditions yield 0 = x(t)ẋ(t) + y(t)ẏ(t), which is 1/2 the derivative of x(t)^{2} + y(t)^{2} and forces this function to be constant = 1. In other words, y(t) alone would have to obey the conditions

(∗∗)  ẏ(t)^{2} = 1 − y(t)^{2} and y(0) = 0 .

If we had such a y(t) we could put x(t) = ẏ(t) and get 0 = x(t)ẋ(t) + y(t)ẏ(t) = x(t)(ẋ(t) + y(t)) As long as x(t) ≠ 0, this would imply ẋ(t) = −y(t) and hence solve (∗) — see? Moral:

Any solution y(t) of (∗∗), such that ẏ(t) ≠ 0 on some open interval around t = 0, produces a solution X(t) = [ẏ(t), y(t)] of (∗) with t restricted to the same interval.

If we had such a y(t), we could also use it to solve certain integration problems. For instance, the substitution u = y(t) would — because of dy/dt = \sqrt{1 − y^{2}} — immediately turn

(†)  F (v) = ∫_{0}^{v} \frac{du}{\sqrt{1 − u^{2}}}  (−1 < v < 1)

into the integral of dt from 0 to the value τ for which y(τ) = v. In other words, we would have F (y(τ)) = τ, and F (v) would be an inverse function!

Let us turn this observation upside down, and define y(t) to be the inverse of the function given by the formula (†). Does this make sense?

As the integral of a continuous function, F (v) is a respectable differentiable function for −1 < v < 1. Since its derivative f(v) = (1 − v^{2})^{−1/2} is positive, F (v) is monotone increasing on that interval, and hence has an inverse function there. Since 1 − v^{2} > 1 − v, and therefore (1 − v^{2})^{−1/2} < (1 − v)^{−1/2} for 0 < v < 1, the improper integral F (1) has a finite value κ > 0. The obvious oddness of the function, i.e., the relation F (−v) = −F (v), shows that F (−1) = −κ.

The inverse of F (v) is the desired y(t): a monotone increasing odd function for −κ < t < κ, which tends to 1 as t → κ. To check that (∗∗) does indeed hold, we differentiate both sides of the relation t = F (y(t)) with respect to t, getting

1 = \frac{1}{\sqrt{1 − y(t)^{2}}} ẏ(t)

by the Chain Rule. The monotonicity of y(t) makes ẏ(t) > 0 — and hence establishes (∗) — on the whole interval −κ < t < κ.

Local homomorphisms. The variable matrix Q(t) = x(t)I + y(t)J has the columns X(t) and JX(t). Therefore, if X(t) is as above, we have

(‡)  Q(0) = I  and  Q̇(t) = JQ(t) ,

on the interval −κ < t < κ. No matter how we shrink this interval, Q(t) is the only matrix function satisfying these conditions. Indeed, if P (t) also satisfied Ṗ (t) = JP (t) and P (0) = I, perhaps on a subinterval −κ_{1} < t < κ_{2} with κ_{i} ≤ κ, differentiation of Q(−t)P (t) would yield −JQ(−t)P (t) + Q(−t)JP (t) = 0. Having zero derivative, Q(−t)P (t) would be constant = I, whence P (t) = Q(−t)^{−1} on that subinterval. Thus, any two candidates P (t) — including Q(t) itself — coincide on any interval around 0 where both are defined.

For any fixed u with |u| < κ, the matrix P_{u}(t) = Q(t + u)Q(u)^{−1} clearly does have the properties required in (‡) and hence equals Q(t) as long as |t| and |t + u| are both less than κ. In other words,

(•)  Q(t + u) = Q(t)Q(u)

for |t| ,|u|, |t + u| < κ. Such a function is called a local homomorphism, because its variables are a priori restricted. However, we shall presently see that it is easily extended to the whole real line. We first deal with the positive half.

For any t > 0, we consider partitions t = τ_{1}+⋯+τ_{m} all of whose summands are positive but < κ. For each such partition ϕ, we write Q_{ϕ}(t) = Q(τ_{1}) ⋯ Q(τ_{m}). If we make a finer partition ψ by breaking up one of the τ_{i} into a sum of two smaller ones, (•) shows that Q_{ψ}(t) = Q_{ϕ}(t). By iteration, this is true for any refinement ψ of ϕ, and since any two partitions have a common refinement, it turns out that Q_{ϕ}(t) is actually independent of ϕ. We shall therefore denote it by Q(t) — letting the partitions play their beneficial role in the background.

For −t < 0, we define Q(−t) = Q(t)^{−1}. Since this rule already holds for −κ < t < κ, the extended Q(t) coincides with the old one on this interval. In particular, Q(−t) = Q_{ϕ}(t)^{−1} can also be obtained via partitions. So, now we have a function on all of R, but is it what we want? Intuitively, it is fairly clear that we have constructed a homomorphism. If t and u have the same sign, one only needs to concatenate their partitions — but in the case of opposite signs there may be some confusion. Here is a good way of avoiding the problem.

A partition of t is, after all, nothing but a finite sequence of points on the number line starting at 0 and ending at t. This is true for any t ≠ 0, whether positive or negative. If now h is a number sufficiently small close to 0, we can always construct an admissible partition whose sequence of points includes both t and t + h (see?). Therefore Q(t + h) = Q(t)Q(h) and

lim_{h→0} \frac{Q(t + h) − Q(t)}{h} = lim_{h→0} \frac{Q(h) − I}{h} · Q(t) = JQ(t) .

Now we are back in (‡), but with the whole real line as our interval of definition, and again (by the same reasoning as before) Q(t) is the only solution. Equation (•) now follows as above from the consideration of Q(t + u)Q(u)^{−1}, and holds for arbitrary t and u.

In particular, Q(t + κ) = Q(t)Q(κ) = Q(t)J and Q(t + 2κ) = −Q(t), so that Q(t) is periodic with period 4κ. A look back at the inverse function F (v) reveals that y(t) covers the entire interval −1 < v < 1, as t varies between −κ and κ. Hence Q(t) includes all rotations R = xI + yJ with x > 0 (see?), and — since x(t + 2κ) = −x(t) — all rotations of any kind. Conclusion:

Q defines a smooth homomorphism from the additive group of real numbers onto the group of all 2 × 2 real rotation matrices.

As we remarked at the end of the essay “Bogenmessung”, Q must essentially (up to rescaling) coincide with any other continuous homomorphism between these groups, in particular, with the R found ibidem. Since R(1) = Q(κ), it follows that R(μ) = Q(κμ) for all μ.

The conversion factor κ (more commonly known as π/2) can, of course be computed via its definition as the improper integral F (1) in (†). The following exercise shows a more elegant way due to François Viète.

Exercise: Define Q(κ/2^{m}) = x_{m}I + y_{m}J and put z_{m} = 2^{m}y_{m}. Then (a) z_{m} = z_{m−1}/x_{m} and (b) z_{m} → κ as m → ∞. Hint: for (b) remember that ẏ(0) = 1; intuitively, z_{m} is half the area of a certain regular polygon …