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Posted by FillMaths 8 hours ago

Mathematicians disagree on the essential structure of the complex numbers (2024)(www.infinitelymore.xyz)
130 points | 158 commentspage 2
zarzavat 6 hours ago|
The way I think of complex numbers is as linear transformations. Not points but functions on points that rotate and scale. The complex numbers are a particular set of 2x2 matrices, where complex multiplication is matrix multiplication, i.e. function composition. Complex conjugation is matrix transposition. When you think of things this way all the complex matrices and hermitian matrices in physics make a lot more sense. Which group do I fall into?
zogomoox 55 minutes ago||
if only matrices would've been invented before i..
czgnome 6 hours ago||
This would be the rigid interpretation since i and -i are concrete distinguishable elements with Im and Re defined.
mebassett 3 hours ago||
the title is a bit clickbait - mathematicians don't disagree, all the "conceptions" the article proposes agree with each other. It also seems to conflate the algebraic closure of Q (which would contain the sqrt of -1) and all of the complex numbers by insisting that the former has "size continuum". Once you have "size continuum" then you need some completion to the reals.

anyhow. I'm a bit of an odd one in that I have no problems with imaginary numbers but the reals always seemed a bit unreal to me. that's the real controversy, actually. you can start looking up definable numbers and constructivist mathematics, but that gets to be more philosophy than maths imho.

Nevermark 2 hours ago||
The square root of any number x is ±y, where +y = (+1)*y = y, and -y = (-1)*y.

So we define i as conforming to ±i = sqrt(-1). The element i itself has no need for a sign, so no sign needs to be chosen. Yet having defined i, we know that that i = (+1)*i = +i, by multiplicative identity.

We now have an unsigned base element for complex numbers i, derived uniquely from the expansion of <R,0,1,+,*> into its own natural closure.

We don't have to ask if i = +i, because it does by definition of the multiplicative identity.

TLDR: Any square root of -1 reduced to a single value, involves a choice, but the definition of unsigned i does not require a choice. It is a unique, unsigned element. And as a result, there is only a unique automorphism, the identity automorphism.

phkahler 7 hours ago||
To the ones objecting to "choosing a value of i" I might argue that no such choice is made. i is the square root of -1 and there is only one value of i. When we write -i that is shorthand for (-1)i. Remember the complex numbers are represented by a+bi where a and b are real numbers and i is the square root of -1. We don't bifurcate i into two distinct numbers because the minus sign is associated with b which is one of the real numbers. There is a one-to-one mapping between the complex numbers and these ordered pairs of reals.
FillMaths 7 hours ago||
You say that i is "the square root of -1", but which one is it? There are two. This is the point in the essay---we cannot tell the difference between i and -i unless we have already agreed on a choice of which square root of -1 we are going to call i. Only then does the other one become -i. How do we know that my i is the same as your i rather than your -i?

To fix the coordinate structure of the complex numbers (a,b) is in effect to have made a choice of a particular i, and this is one of the perspectives discussed in the essay. But it is not the only perspective, since with that perspective complex conjugation should not count as an automorphism, as it doesn't respect the choice of i.

SyzygyRhythm 2 hours ago|||
Is it two, or is it infinite? The quaternions have three imaginary units, i, j, and k. They're distinct, and yet each of them could be used for the complex numbers and they'd work the same way. How would I know that "my" imaginary unit i is the same as some other person's i? Maybe theirs is j, or k, or something else entirely.
jiggawatts 2 hours ago||||
One perspective of the complex numbers is that they are the even subalgebra of the 2D geometric algebra. The "i" is the pseudoscalar of that 2D GA, which is an oriented area.

If you flip the plane and look at it from the bottom, then any formula written using GA operations is identical, but because you're seeing the oriented area of the pseudoscalar from behind, its as if it gains a minus sign in front.

This is equivalent to using a right-handed versus left-handed coordinate systems in 3D. The "rules of physics" remain the same either way, the labels we assign to the coordinate systems are just a convention.

phkahler 6 hours ago|||
There are 2 square roots of 9, they are 3 and -3. Likewise there are two square roots of -1 which are i and -i. How are people trying to argue that there are two different things called i? We don't ask which 3 right? My argument is that there is only 1 value of i, and the distinction between -i and i is the same as (-1)i and (1)i, which is the same as -3 vs 3. There is only one i. If there are in fact two i's then there are 4 square roots of -1.
topaz0 5 hours ago|||
Notably, the real numbers are not symmetrical in this way: there are two square roots of 1, but one of them is equal to it and the other is not. (positive) 1 is special because it's the multiplicative identity, whereas i (and -i) have no distinguishing features: it doesn't matter which one you call i and which one you call -i: if you define j = -i, you'll find that anything you can say about i can also be shown to be true about j. That doesn't mean they're equal, just that they don't have any mathematical properties that let you say which one is which.
czgnome 6 hours ago|||
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.
pfortuny 7 hours ago||
There is no way to distinguish between "i" and "-i" unless you choose a representation of C. That is what Galois Theory is about: can you distinguish the roots of a polynomial in a simple algebraic way?

For instance: if you forget the order in Q (which you can do without it stopping being a field), there is no algebraic (no order-dependent) way to distinguish between the two algebraic solutions of x^2 = 2. You can swap each other and you will not notice anything (again, assuming you "forget" the order structure).

btilly 6 hours ago||
Building off of this point, consider the polynomial x^4 + 2x^2 + 2. Over the rationals Q, this is an irreducible polynomial. There is no way to distinguish the roots from each other. There is also no way to distinguish any pair of roots from any other pair.

But over the reals R, this polynomial is not irreducible. There we find that some pairs of roots have the same real value, and others don't. This leads to the idea of a "complex conjugate pair". And so some pairs of roots of the original polynomial are now different than other pairs.

That notion of a "complex conjugate pair of roots" is therefore not a purely algebraic concept. If you're trying to understand Galois theory, you have to forget about it. Because it will trip up your intuition and mislead you. But in other contexts that is a very meaningful and important idea.

And so we find that we don't just care about what concepts could be understood. We also care about what concepts we're currently choosing to ignore!

pfortuny 6 hours ago||
Exactly.

That is why the "forgetful functor" seems at first sight stupid and when you think a bit, it is genius.

btilly 4 hours ago||
When you think about it, creating a structure modulo some relation or kind of symmetry, is also a kind of targeted forgetting.
nigelvr 6 hours ago||
The link is about set theory, but others may find this interesting which discusses division algebras https://nigelvr.github.io/post-4.html

Basically C comes up in the chain R \subset C \subset H (quaternions) \subset O (octonions) by the so-called Cayley-Dickson construction. There is a lot of structure.

slwvx 7 hours ago||
Is there agreement Gaussian integers?

This disagreement seems above the head of non mathematicians, including those (like me) with familiarity with complex numbers

btilly 6 hours ago||
There is perfect agreement on the Gaussian integers.

The disagreement is on how much detail of the fine structure we care about. It is roughly analogous to asking whether we should care more about how an ellipse is like a circle, or how they are different. One person might care about the rigid definition and declare them to be different. Another notices that if you look at a circle at an angle, you get an ellipse. And then concludes that they are basically the same thing.

This seems like a silly thing to argue about. And it is.

However in different branches of mathematics, people care about different kinds of mathematical structure. And if you view the complex numbers through the lens of the kind of structure that you pay attention to, then ignore the parts that you aren't paying attention to, your notion of what is "basically the same as the complex numbers" changes. Just like how one of the two people previously viewed an ellipse as basically the same as a circle, because you get one from the other just by looking from an angle.

Note that each mathematician here can see the points that the other mathematicians are making. It is just that some points seem more important to you than others. And that importance is tied to what branch of mathematics you are studying.

lmkg 6 hours ago||
The Gaussian integers usually aren't considered interesting enough to have disagreements about. They're in a weird spot because the integer restriction is almost contradictory with considering complex numbers: complex numbers are usually considered as how to express solutions to more types of polynomials, which is the opposite direction of excluding fractions from consideration. They're things that can solve (a restricted subset of) square-roots but not division.

This is really a disagreement about how to construct the complex numbers from more-fundamental objects. And the question is whether those constructions are equivalent. The author argues that two of those constructions are equivalent to each other, but others are not. A big crux of the issue, which is approachable to non-mathematicians, is whether it i and -i are fundamentally different, because arithmetically you can swap i with -i in all your equations and get the same result.

brcmthrowaway 3 hours ago||
What does Terry Tao think?
TimorousBestie 3 hours ago||
> But in fact, I claim, the smooth conception and the analytic conception are equivalent—they arise from the same underlying structure.

Conjugation isn’t complex-analytic, so the symmetry of i -> -i is broken at that level. Complex manifolds have to explicitly carry around their almost-complex structure largely for this reason.

yifanl 6 hours ago|
Notably, neither `1 + i > 1 - i` or `1 + i < 1 - i` are correct statements, and obviously `1 + i = 1 - i` is absurd.
chongli 6 hours ago||
What do > and < mean in the context of an infinite 2D plane?
yifanl 6 hours ago|||
Typically, the order of complex numbers is done by projecting C onto R, i.e. by taking the absolute value.
chongli 6 hours ago||
Yes I’m aware. It’s a work around but doesn’t give you a sensible ordering the way most people expect, i.e:

-2 > 1 (in C)

Which is why I prefer to leave <,> undefined in C and just take the magnitude if I want to compare complex numbers.

layer8 6 hours ago|||
One is above the plane and the other is below it. ;)
bell-cot 3 hours ago||
In a word - "true".

In more words - it's interesting, but messy:

https://en.wikipedia.org/wiki/Partial_order

https://en.wikipedia.org/wiki/Ordered_field

> The complex numbers also cannot be turned into an ordered field, as −1 is a square of the imaginary unit i.

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