Oriented orthonormal triads in ${\mathbb{R}}^3$: existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction

At most one standard basis vector is parallel to $e_3$: if $s_i=\lambda e_3$ and $s_j=\mu e_3$ with $i\mathrm{\ne }j$ and $\lambda ,\mu \mathrm{\ne }0$, then $s_i=(\lambda \mathrm{/}\mu )s_j$, which is false because $s_i$ and $s_j$ have $1$ in different coordinates. Hence some $k$ with $s_k$ not parallel to $e_3$ exists.

Let $f_1,\dots ,f_m\in {\mathbb{R}}^3$ be orthonormal, and suppose $m\ge 4$. Applying Bessel's inequality [L4] to each standard basis vector $s_j$ gives ${\sum }_{i=1}^m⟨s_j,f_i{⟩}^2\le \mathrm{\mid }{s}_j{\mathrm{\mid }}^2=1$. Summing over $j=1,2,3$ and exchanging the finite sums [L5] gives ${\sum }_{i=1}^m{\sum }_{j=1}^3⟨s_j,f_i{⟩}^2\le 3$.

Put $v:=s_k-⟨s_k,e_3⟩e_3$. Then $v\mathrm{\ne }0$: otherwise $s_k=⟨s_k,e_3⟩e_3$ is parallel to $e_3$, contrary to step 1.1. Set $e_1:=v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$. Then $⟨e_1,e_3⟩=0$ and $\mathrm{\mid }{e}_1\mathrm{\mid }=1$, because $⟨v,e_3⟩=⟨s_k,e_3⟩-⟨s_k,e_3⟩\mathrm{\mid }{e}_3{\mathrm{\mid }}^2=0$ and $\mathrm{\mid }v\mathrm{\mid }\mathrm{\ne }0$.

For each $i$, the standard-basis expansion of [L3] applied to $f_i$ gives ${\sum }_{j=1}^3⟨s_j,f_i{⟩}^2=\mathrm{\mid }{f}_i{\mathrm{\mid }}^2=1$. Hence the left-hand side of step 1.2 equals ${\sum }_{i=1}^{m}1=m\ge 4$, contradicting the bound $\le 3$ of step 1.2. Therefore no orthonormal list in ${\mathbb{R}}^3$ has four or more members.

Set $e_2:=e_3\times e_1$. By [L2], $⟨e_2,e_3⟩=0=⟨e_2,e_1⟩$ and $\mathrm{\mid }{e}_2{\mathrm{\mid }}^2=\mathrm{\mid }{e}_3{\mathrm{\mid }}^2\mathrm{\mid }{e}_1{\mathrm{\mid }}^2-⟨e_3,e_1{⟩}^2=1$; and $e_1\times e_2=e_1\times (e_3\times e_1)=⟨e_1,e_1⟩e_3-⟨e_1,e_3⟩e_1=e_3$ by the triple product identity [L1]. So $(e_1,e_2,e_3)$ is an oriented orthonormal triad, which is clause 1.

Let $(e_1,e_2,e_3)$ be an oriented orthonormal triad and $x\in {\mathbb{R}}^3$. Put $w:=x-{\sum }_{i=1}^3⟨x,e_i⟩e_i$. Then $⟨w,e_j⟩=⟨x,e_j⟩-{\sum }_i⟨x,e_i⟩{\delta }_{ij}=0$ for each $j$, by bilinearity of the inner product. If $w\mathrm{\ne }0$, then $e_1,e_2,e_3,w\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$ would be four pairwise orthogonal unit vectors, contradicting step 2.2; hence $w=0$, which is the expansion.

With the expansion in hand, $\mathrm{\mid }x{\mathrm{\mid }}^2=⟨{\sum }_i⟨x,e_i⟩e_i,{\sum }_k⟨x,e_k⟩e_k⟩={\sum }_{i,k}⟨x,e_i⟩⟨x,e_k⟩{\delta }_{ik}={\sum }_i⟨x,e_i{⟩}^2,$ by bilinearity and orthonormality. This completes clause 2.