Why there are no 3D complex numbers

A simple argument as to why there are no 3D complex numbers. See the following video by Michael Penn for a more elaborate and rigorous proof.

[QuicksilverSG]

TL;DW: The dimensional progression from scalar, complex, quarternion, to octonion is exponential rather than linear.

[splaytrees]

Hypercomplex Numbers
Quaternions: Instead of a 3D version of complex numbers, the next step up in dimension leads to quaternions, which are four-dimensional (one real part and three imaginary parts: i, j, k). Quaternions were discovered by William Rowan Hamilton in 1843 and are used in various applications, especially in three-dimensional computer graphics and spatial rotations.

Octonions: Further extending, we have octonions, which are eight-dimensional. However, as we increase dimensions, we lose certain desirable properties such as commutativity and associativity. Quaternions are non-commutative, and octonions are non-associative.

Summary of Properties Lost
Commutativity: Lost with quaternions.
Associativity: Lost with octonions.
Alternativity: Lost with sedenions.
Power-Associativity: Still retained by sedenions but more complex algebras might lose it.
Zero Divisors: Introduced with sedenions and present in higher-dimensional hypercomplex numbers.

Alternativity:
a(ab)=(aa)b
(ab)b=a(bb)

Power-Associativity
a(a^2)=(a^2)a

[guythat]

Despite the video being from 7 months ago all the current comments are from within a week
Idk what happened byt happy to be part of it

[bumpty]

Even more briefly, from a slightly different perspective:

Complex numbers are about rotations in two dimensions. There are two real degrees of freedom in complex number because two dimensions allow one rotation axis which needs a basis element (called "i") and one unit is required for multiplication needing another basis element ("1").

The next step up describes rotations in three dimensions*. There are three rotation axes in three dimensions, requiring one element each ("i, j, k" of quaternions) and a unit is still required (quaternion "1"). There are no "three dimensional complex numbers" because there is no spatial dimension with two axes of rotation.

* Okay, it describes the double cover of the three-dimensional rotations, but that's a detail that gets missed at this level.

[akhileshmachiraju]

Two ways to note while observing an imaginary field.
1. It is perpendicular to a field and
2. It is unique to each field.
If there are three independent variables, there will three independent imaginary fields corresponding to each independent vector.

[amandeep]

We can use the tower theorem. Let F be an extension of R which contains C and is of degree 3. Then, 3= [F, R]=[F, C][C, R]=[F, C]x2. Which is a contradiction.

[zsoltszabo]

Nice video. Allow me some comments.
The line of thought from ~5:40 can be carried out without assuming j^2 = -1, but of course keeping the usual i^2 = -1.
1. Multiplying ij = 1 by i gives j = -i.
2. Multiplying ij = i by i gives j = 1.
3. Multiplying ij = j by i gives j = -ij = -j = 0.
The final reasoning can also be carried out without assuming j^2 = -1. In addition, I would conclude from it that {1, i, j} is not linearly independed as the 3D picture suggests. c^2 = -1 is obtained by a priori assuming linear independence of {1, i, j}. In either case we get something we do not want, hence 3D complex numbers do not exist.
An argument not mentioned in the video that suggests that 3D complex numbers might not even be needed is the following: the field C of the usual 2D complex numbers is algebraically closed, i.e., we cannot get anything outside C as roots of nontrivial polynomials.

[ceoOO]

Michael Penn's explanation was more rigorous... but i think this guy's explanation is more intuitive... He gave a more intuitive demonstration of why there are no 3-D complex numbers... and why the next step has to be 4-D complex numbers instead...📈

[splat]

I wondered what the starting assumptions would be and I would have liked them to be at the start. For instance linear independence, associativity and distributivity. However as I am interested in geometric algebra I find this a useful stepping stone and it has filled a gap. Nice video.

[hdthor]

Geometric Algebra has 3D “complex” numbers (similar to quaternions) and higher dimensions as well. The key is that i*j is an irreducible bivector in Geometric Algebra. In Geometric Algebra, a “3D complex number” is not expressed as a + bi + cj, it’s expressed as a + bi + cj + dij. Note the introduction of the bivector ij. And “4D complex number” requires the introduction of the trivector ijk.

[avramlevitter]

12:35 If we're not sure about assuming commutativity, why are we assuming distributivity?

[RohneeshSachan]

Such a great and easy explanation, Thank you!!

[balthazarbeutelwolf]

one slight objection: in your derivation you used commutativity of multiplication. That does not hold for quaternions, so it is a big step to assume it for 3D numbers.

[markwrede]

Aligning a tube at minus one-half on the complex plane would allow a third dimension of spin to the Riemann Zeta function. This allows deduction of the sequential differences among primes as a series of families of primes, i.e. each prime hosts a "gear" notched with the sequential difference from the preceding prime. This gear for twin primes is known as phi. There are solutions for square primes, sexy primes, and so on.

[sumitsslite]

What if we create new types of numbers assuming 1/0 to exist in different axis?

[TheEternalVortex]

You kind of miss out on the requirements we are trying to satisfy. It's easy to make a 3-d algebra if we don't care about division: for example take R[x]/(x^3−1). This means you would have numbers of the form a + b ω + c ω^2, where a, b, and c are real and of course ω^3 = 1. The only problem is you get 0 divisors, e.g. (ω - 1)(ω^2 + ω + 1) = 0. Indeed since every cubic has a real root all 3-d algebras will have zero divisors.

[Tabu]

by 5:58 I think I already get it. You would have two numbers to go to -1 but that leaves us with an incomplete 3d. So Im assuming that if j was complex that would fix it bringing us to 4d numbers.

[Purified-Bananas]

I guess the closest thing to a 3D complex number is if you take a 3D vector as a rotation of |V| around the axis V / |V|.

[AdrienLegendre]

A different approach is to recognize that complex numbers are a type of Clifford algebra and Clifford algebras can be extended to higher dimensions.

[umbraemilitos]

The complex numbers are isomorphic to the even subalgebra of 2D VGA, where the imaginary unit it just the bivector ~ psuedoscalar. The algebra is not closed if you include one more basis vector, and limit to two unique bivectors. This is because (3 choose 2) = 3.