Solèr's theorem

Let ${\displaystyle \mathbb {K} }$ be a division ring. That means it is a ring in which one can add, subtract, multiply, and divide but in which the multiplication need not be commutative. Suppose this ring has a conjugation, i.e. an operation ${\displaystyle x\mapsto x^{*}}$ for which

${\displaystyle {\begin{aligned}&(x+y)^{*}=x^{*}+y^{*},\\&(xy)^{*}=y^{*}x^{*}{\text{ (the order of multiplication is inverted), and }}\\&(x^{*})^{*}=x.\end{aligned}}}$

Consider a vector space V with scalars in ${\displaystyle \mathbb {K} }$, and a mapping

${\displaystyle (u,v)\mapsto \langle u,v\rangle \in \mathbb {K} }$

which is ${\displaystyle \mathbb {K} }$ -linear in left (or in the right) entry, satisfying the identity

${\displaystyle \langle u,v\rangle =\langle v,u\rangle ^{*}.}$

This is called a Hermitian form. Suppose this form is non-degenerate in the sense that

${\displaystyle \langle u,v\rangle =0{\text{ for all values of }}u{\text{ only if }}v=0.}$

For any subspace S let ${\displaystyle S^{\bot }}$ be the orthogonal complement of S. Call the subspace "closed" if ${\displaystyle S^{\bot \bot }=S.}$

Call this whole vector space, and the Hermitian form, "orthomodular" if for every closed subspace S we have that ${\displaystyle S+S^{\bot }}$ is the entire space. (The term "orthomodular" derives from the study of quantum logic. In quantum logic, the distributive law is taken to fail due to the uncertainty principle, and it is replaced with the "modular law," or in the case of infinite-dimensional Hilbert spaces, the "orthomodular law."[6])

A set of vectors ${\textstyle u_{i}\in V}$ is called "orthonormal" if

$${\displaystyle \langle u_{i},u_{j}\rangle =\delta _{ij}.}$$

The result is this:

If this space has an infinite orthonormal set, then the division ring of scalars is either the field of real numbers, the field of complex numbers, or the ring of quaternions.