Your view of the complex numbers is the rigid one. Now suppose you are given a set with two binary operations defined in such a way that the operations behave well with each other. That is you have a ring. Suppose that by some process you are able to conclude that your ring is algebraically equivalent to the complex numbers. How do you know which of your elements in your ring is “i”? There will be two elements that behave like “i” in all algebraic aspects. So you can’t say that this one is “i” and this one is “-i” in a non arbitrary fashion.
In the article he says there is a model of ZFC in which the complex numbers have indistinguishable square roots of -1. Thus that model presumably does not allow for a rigid coordinate view of complex numbers.
Theorem. If ZFC is consistent, then there is a model of ZFC that has a definable complete ordered field ℝ with a definable algebraic closure ℂ, such that the two square roots of −1 in ℂ are set-theoretically indiscernible, even with ordinal parameters.
Haven’t thought it through so I’m quite possibly wrong but it seems to me this implies that in such a situation you can’t have a coordinate view. How can you have two indistinguishable views of something while being able to pick one view?
Mathematicians pick an arbitrary complex number by writing "Let c ∈ ℂ." There are an infinite number of possibilities, but it doesn't matter. They pick the imaginary unit by writing "Let i ∈ ℂ such that i² = −1." There are two possibilities, but it doesn't matter.
If two things are set theoretically indistinguishable then one can’t say “pick one and call it i and the other one -i”. The two sets are the same according to the background set theory.
They're not the same. i ≠
−i, no matter which square root of negative one i is. They're merely indiscernible in the sense that if φ(i) is a formula where i is the only free variable, ∀i ∈ ℂ. i² = −1 ⇒ (φ(i) ⇔ φ(−i)) is a true formula. But if you add another free variable j, φ(i, j) can be true while φ(−i, j) is false, i.e. it's not the case that ∀j ∈ ℂ. ∀i ∈ ℂ. i² = −1 ⇒ (φ(i, j) ⇔ φ(−i, j)).
