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Complex numbers are 2D. It is a commonly cited result that there is no 3D or 4D analogue of the complex numbers.

I just want to be clear on exactly what this result says:

  • It is impossible to construct a 4D field.

  • It is impossible to construct a 4D field which the complex numbers are a subset of.

  • It is impossible to construct a 4D field which the real numbers are a subset of.

  • Something else?

(I'd also like to know why it's true - but given that I didn't understand the answer the first $2^7$ times, I doubt I'll understand it this time either. :-} Oh well...)

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    Depending on your background. I$f$ you adhere to Bourbaki you may speak of fields and commutative fields, I guess.2012-05-09

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The field of rational functions is infinite-dimensional. It includes the complex and real numbers.

The same is correct about the field of analytic (holomorphic) functions.

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Since $\mathbb{C}$ is algebraically closed, every finite extension of $\mathbb{C}$ is $\mathbb{C}$ itself. The quaternions are not a field because they are not commutative; they are what is called a normed division algebra. Hurwitz's theorem gives a complete description of the possible normed division algebras (over $\mathbb{R}$) - the are either the real numbers, the complex numbers, the quaternions or the octonions. Only the first two are fields, and only the first three are associative.

The nonexistence of a three-dimensional real normed algebra can be viewed as a consequence of the hairy ball theorem. If $\mathbb{R}^3$ could be given a normed division algebra structure, the unit sphere $S^2$ could be endowed with a smooth group structure (the elements of norm $1$ form a group). But the hairy ball theorem implies that there is no such thing, because otherwise, by letting an infinitesimal element of the group act on the left on the sphere, we would obtain a nowhere vanishing, continuous vector field on the sphere, which is impossible by the hairy ball theorem.

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    @Bruno: I still fail to see it. I'll think about it and post a question if it doesn't all click. The are plenty of normed multiplications (usual norm, weird multiplication) on $\mathbb{R}^2$ (which do NOT distribute over addition) which fail to induce smooth group multiplications on $S^1$. For example, pick any $f:S^1\rightarrow S^1$ which is discontinuous and define $x\cdot y = \|x\|\|y\| f^{-1}(f(\frac{x}{\|x\|})f(\frac{y}{\|y\|}))$.2012-05-07
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It is impossible to have a field which is $n$ dimensional over $\mathbb{R}$ for any $n\geq 3$. The reason why this is true boils down to the following two statements.

  1. Any field $K\supseteq \mathbb{R}$ which is finite dimensional over $\mathbb{R}$ is algebraic over $\mathbb{R}$.
  2. The complex numbers are the algebraic closure of $\mathbb{R}$.

Thus is $K\supseteq \mathbb{R}$ is a field which is finite dimensional over $\mathbb{R}$, then it is algebraic over $\mathbb{R}$, and hence is contained in the algebraic closure of $\mathbb{R}$, i.e., $K\subseteq \mathbb{C}$. Since $\mathbb{C}$ has dimension $2$ over $\mathbb{R}$, this implies that $K$ has dimension either $1$ or $2$ over $\mathbb{R}$. In the first case, $K = \mathbb{R}$, and in the second $K = \mathbb{C}$.

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    If you work over $\mathbb{Q}$ instead of $\mathbb{R}$ it is possible to have fields $K\supset\mathbb{Q}$ which are $n$-dimensional over $\mathbb{Q}$ for any desired dimension $n$. In the case $n = 4$, you can take $K$ to be the smallest field containing $\mathbb{Q}$ and $2^{1/4}$. This $K$ has as a $\mathbb{Q}$-basis $1, 2^{1/4}, 2^{1/2}, 2^{3/4}$. This sort of thing is part of the subject of field theory.2012-05-07