# Can we turn $\mathbb{R}^n$ into a field by changing the multiplication?

Of course $\mathbb{R}$ is a field with usual addition and multiplication. When we move up a dimension into $\mathbb{R}^2$, however, there is not a clear way to multiply two vectors together to get something useful. In fact, if we define multiplication of two vectors component-wise (as is arguably the most natural way), we get something that isn't even an integral domain. However, if we implement the multiplication $$(a, b)(c, d) \mapsto (ac - bd, ad + bc),$$ then we obtain a copy of $\mathbb{C}$, which is again a field. Can we do this for higher dimensions? That is, is there some clever multiplicative structure on $\mathbb{R}^3$ that produces a field? $\mathbb{R}^n$?

• it depends on how much structure do you want to preserve? – azarel Oct 15 '13 at 1:25
• @azarel I'm not sure what you mean. Retain the structure of $\mathbb{R}^n$ as a vector space? – tylerc0816 Oct 15 '13 at 1:29
• I mean you can make $\mathbb R^3$ isomorphic to any field of cardinality continuum, without any further restrictions – azarel Oct 15 '13 at 1:36
• mathoverflow.net/questions/9014/field-structure-for-rn – user26857 Oct 15 '13 at 2:17
• @bubba The resulting structure is not an integral domain, because $(1,0)*(0,1) = (0,0)$, and hence not a field. – tylerc0816 Oct 15 '13 at 3:23

No, there is no field extension of $\mathbb R$ of degree $n$ other than for $n=2$, and we get something isomorphic to the complex numbers.

At least with hindsight, this is not so surprising, if not "obvious", because $\mathbb C$ is algebraically closed (from Liouville's theorem, a corollary of Cauchy's theorems).

Yes, in some ways this is unfortunate, since certain scenarios are precluded. An example of a way to circumvent these obstacles is Hamilton's creation of "quaternions", a four-dimensional $\mathbb R$-vectorspace, which unfortunately (and surprisingly, tittilatingly, in Hamilton's time) produces a non-commutative ... thing.

That is, $\mathbb R$ is "so close" to being algebraically closed that it admits only something isomorphic to $\mathbb C$ as literal field extension. And, perhaps even-more-disappointingly, only $\mathbb H$ as division-algebra extension.

• Don't forget the even "worse" Octonions which aren't even associative... – Tobias Kienzler Oct 15 '13 at 9:47

It is a consequence of the theory of characteristic classes of vector bundles that the existence of a bilinear product $\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ without zero divisors implies that $n$ is a power of 2. In fact it is only possible for $n=1,2,4,8$ where in dimensions 4 and 8 we get quaternions and octonions both of which are not fields because commutativity fails for both and associativity of the product fails for the octonions.

• Could you add a source to this powerful claim? – nbubis Oct 15 '13 at 1:52
• You do not get the algebras you listed —in fact, the possible algebras have not yet been classified, afaict. We do get the four algebras you mention if we restrict to composition algebras or add some other such hypothesis. – Mariano Suárez-Álvarez Oct 15 '13 at 2:01
• There is a nice proof in Milnor's Characteristic Classes book that the existence of the product implies paralelizability of $RP^{n-1}$ and that $n$ must be a power of 2. To show that $RP^{n-1}$ is not parallelizable for $n>8$ he refers to an article of Kervaire: Non-paralellizability ofthe sphere for $n>7$. @Mariano you are right, this only gives a restriction on the dimension, not a classification as I said. – Vincent L. Oct 15 '13 at 2:20

As an abelian group $\mathbb R^n$ is isomorphic to $\mathbb R$, so you just need to fix an isomorphism and use it to transport the structure.

• This way you get a field without changing the way addition works. – Mariano Suárez-Álvarez Oct 15 '13 at 1:29
• (I edited a typo. I guess this was from a device, so TeXing was not nice to type down, so I added it.) – Pedro Tamaroff Oct 15 '13 at 1:43
• @MarianoSuárez-Alvarez Sorry by my ignorance: how can one prove that $(\mathbb R^n,\boldsymbol{+})$ and $(\mathbb R,+)$ are isomorphic (abelian) groups? – Matemáticos Chibchas Oct 15 '13 at 1:49
• @MatemáticosChibchas They both have dimension $2^{\aleph_0}$ as vector spaces over $\mathbb{Q}$, so they are abstractly isomorphic. – user43208 Oct 15 '13 at 1:59
• @MarianoSuárez-Alvarez Highly non-explicit, as I suspected...thanks anyway (it is not your fault ;-) ). – Matemáticos Chibchas Oct 15 '13 at 2:04

A clifford algebra gives a natural way to talk about products of vectors, using the geometric product. The geometric product of vectors is the associative and distributive. If $a, b, c$ are vectors, then

$$(a + b) c = ab + ac, \quad a(b+c) = ab + ac, \quad (ab)c = a(bc) = abc$$

These products of several vectors, along with their linear combinations, are referred to as multivectors. The clifford, or "geometric", algebra thus forms a ring. However, you might notice that I left out commutativity of multiplication. The geometric product is not in general commutative, so while a field cannot in general be built this way, the ring structure of a geometric algebra can be quite useful and generalizes to all dimensions.