# The Universe of Discourse

Tue, 12 Sep 2006

Imaginary units, again
In my earlier discussion of i and -i I said " The point about the square roots of -1 is that there is no corresponding criterion for distinguishing the two roots. This is a theorem."

The proof of the theorem is not too hard. What we're looking for is what's called an automorphism of the complex numbers. This is a function, f, which "relabels" the complex numbers, so that arithmetic on the new labels is the same as the arithmetic on the old labels. For example, if 3×4 = 12, then f(3) × f(4) should be f(12).

Let's look at a simpler example, and consider just the integers, and just addition. The set of even integers, under addition, behaves just like the set of all integers: it has a zero; there's a smallest positive number (2, whereas it's usually 1) and every number is a multiple of this smallest positive number, and so on. The function f in this case is simply f(n) = 2n, and it does indeed have the property that if a + b = c, then f(a) + f(b) = f(c) for all integers a, b, and c.

Another automorphism on the set of integers has g(n) = -n. This just exchanges negative and positive. As far as addition is concerned, these are interchangeable. And again, for all a, b, and c, g(a) + g(b) = g(c).

What we don't get with either of these examples is multiplication. 1 × 1 = 1, but f(1) × f(1) = 2 × 2 = 4 ≠ f(1) = 2. And similarly g(1) × g(1) = -1 × -1 = 1 ≠ g(1) = -1.

In fact, there are no interesting automorphisms on the integers that preserve both addition and multiplication. To see this, consider an automorphism f. Since f is an automorphism that preserves multiplication, f(n) = f(1 × n) = f(1) × f(n) for all integers n. The only way this can happen is if f(1) = 1 or if f(n) = 0 for all n.

The latter is clearly uninteresting, and anyway, I neglected to mention that the definition of automorphism rules out functions that throw away information, as this one does. Automorphisms must be reversible. So that leaves only the first possibility, which is that f(1) = 1. But now consider some positive integer n. f(n) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = 1 + 1 + ... + 1 = n. And similarly for 0 and negative integers. So f is the identity function.

One can go a little further: there are no interesting automorphisms of the real numbers that preserve both addition and multiplication. In fact, there aren't even any reasonable ones that preserve addition. The proof is similar. First, one shows that f(1) = 1, as before. Then this extends to a proof that f(n) = n for all integers n, as before. Then suppose that a and b are integers. b·f(a/b) = f(b)f(a/b) = f(b·a/b) = f(a) = a, so f(a/b) = a/b for all rational numbers a/b. Then if you assume that f is continuous, you can fill in f(x) = x for the irrational numbers also.

(Actually this is enough to show that the only continuous addition-preserving automorphism of the reals is the identity function. There are discontinuous addition-preserving functions, but they are very weird. I shouldn't need to drag in the continuity issue to show that the only addition-and-multiplication-preserving automorphism is the identity, but it's been a long day and I'm really fried.)

[ Addendum 20060913: This previous paragraph is entirely wrong; any function xkx is an addition-preserving automorphism, except of course when k=0. For more complete details, see this later article. ]

But there is an interesting automorphism of the complex numbers; it has f(a + bi) = a - bi for all real a and b. (Note that it leaves the real numbers fixed, as we just showed that it must.) That this function f is an automorphism is precisely the content of the statement that i and -i are numerically indistinguishable.

The proof that f is an automorphism is very simple. We need to show that if f(a + bi) + f(c + di) = f((a + bi) + (c + di)) for all complex numbers a+bi and c+di, and similarly f(a + bi) × f(c + di) = f((a + bi) × (c + di)). This is really easy; you can grind out the algebra in about two steps.

What's more interesting is that this is the only nontrivial automorphism of the complex numbers. The proof of this is also straightforward, but a little more involved. The purpose of this article is to present the proof.

Let's suppose that f is an automorphism of the complex numbers that preserves both addition and multiplication. Let's say that f(i) = p + qi. Then f(a + bi) = f(a) + f(b)f(i) = a + bf(i) (because f must leave the real numbers fixed) = a + b(p + qi) = (a + bp) + bqi.

Now we want f(a + bi) + f(c + di) = f((a + bi) + (c + di)) for all real numbers a, b, c, and d. That is, we want (a + bp + bqi) + (c + dp + dqi) = (a + c) + (b + d)(p + qi). It is, so that part is just fine.

We also want f(a + bi) × f(c + di) = f((a + bi) × (c + di)) for all real numbers a, b, c, and d. That means we need:

 (a + b(p + qi)) × (c + d(p + qi)) = f((ac-bd) + (ad+bc)i) (a + bp + bqi) × (c + dp + dqi) = ac - bd + (ad + bc)(p + qi) ac + adp + adqi + bcp + bdp2 + 2bdpqi + bcqi - bdq2 = ac - bd + adp + bcp + adqi + bcqi bdp2 + 2bdpqi - bdq2 = - bd p2 + 2pqi - q2 = -1

Equating the real and imaginary parts gives us two equations:

1. p2 - q2 = -1
2. 2pq = 0

Equation 2 implies that either p or q is 0. If they're both zero, then f(a + bi) = a, which is not reversible and so not an automorphism.

Trying q=0 renders equation 1 insoluble because there is no real number p with p2 = -1. But p=0 gives two solutions. One has p=0 and q=1, so f(a+bi) = a+bi, which is the identity function, and not interesting. The other has p=0 and q=-1, so f(a+bi) = a-bi, which is the one we already knew about. But we now know that there are no others, which is what I wanted to show.

[ Previous articles in this series: Part 1 Part 2 Followup articles: Part 4 Part 5 ]