# Mark some vectors in $\mathbb{R}^n$ in a way that every orthonormal basis has an odd number of marked vectors

Let $$n$$ be a natural number. Is there a set $$S$$ of vectors of norm $$1$$ in $$\mathbb{R}^n$$ such that every orthonormal basis of $$\mathbb{R}^n$$ contains an odd number of vectors from $$S$$? If $$n$$ is odd, the answer is trivially yes: One can just let $$S$$ be the whole $$\mathbb{R}^n$$. If $$n=2$$, the answer is also yes: For example, take $$S=\{(x,y)\in\mathbb{R}^2:xy>0\}\cup\{(1,0),(-1,0)\}$$. What if $$n=4$$?

• Possibly unhelpful remark: the examples given in the OP have the property that the intersection of the unit sphere with any subspace is either wholly contained in or wholly disjoint from $S$; but there can be no such example in $\Bbb R^4$ (or higher even dimensions), since any two vectors are contained in a common subspace and thus $S$ would either be $\Bbb R^4$ or empty, neither or which works. Commented Jul 30 at 17:03
• tiny nit: your choice of $S$ for $n=2$ has vectors that do not have length $1.$ Commented Jul 30 at 20:38
• Crossposted at MathOverflow, where the question is already answered (negatively), first for n=4 by Peter Mueller and then for all remaining even $n$. Commented Jul 31 at 3:47
• I’m voting to close this question because it was crossposted on MO and answered there. Imo that answer shouldn't have been copied here. The question doesn't need to be answered. It can be closed instead. Commented Aug 2 at 17:11

Peter Mueller provided a negative answer for $$n=4$$. Based on it, we show that the answer is negative for any even $$n\ge 4$$.
Indeed, suppose for a contradiction that the space $$V=\mathbb R^n$$ admits a required coloring.
We start from the following simple observation. Let $$V'$$ be any subspace of $$V$$ and $$V''$$ be the orthogonal complement of the space $$V'$$, that is, $$V''=\{v\in V: (v,v')=0\mbox{ for any }v'\in V'\}.$$ Let $$B'$$ and $$B^*$$ be any orthonormal bases of the space $$V'$$ and $$B''$$ be any orthonormal basis of the space $$V''$$. Then both $$B'\cup B''$$ and $$B^*\cup B''$$ are the orthonormal bases for the space $$V$$, so they contain an odd number of vectors from $$S$$ each. So the parity of number of vectors from $$S$$ is the same for $$B'$$ and for $$B^*$$. Put $$p(V')=\bar 0$$, if this parity is even, and $$p(V')=\bar 1$$, otherwise, where $$\bar 0$$ and $$\bar 1$$ belong to the field $$\mathbb Z_2=\mathbb Z/2\mathbb Z$$ of residues modulo $$2$$. It is easy to see that $$p(V')=p(V^*)+p(V^{**})$$ for any splitting of $$V$$ into a sum of its orthogonal subspaces $$V^*$$ and $$V^{**}$$.
Peter Mueller's answer implies that $$p(V')$$ is even for any $$4$$-dimensional subspace $$V'$$ of $$V$$. Moreover, if the dimension of a subspace $$V'$$ of $$V$$ is divisible by $$4$$ then $$V'$$ splits into a direct sum of pairwise orthogonal $$4$$-dimensional subspaces, so $$p(V')=\bar 0$$. In particular, if $$n\equiv 0\pmod 4$$ then $$p(V)=\bar 0$$, a contradiction. So suppose that $$n\equiv 2\pmod 4$$. If there exists a $$2$$-dimensional subspace $$V'$$ of $$V$$ such that $$p(V')=\bar 0$$ then $$V$$ splits into a sum of $$V'$$ and its orthogonal complement $$V''$$, so $$p(V)=p(V')+p(V'')=\bar 0+\bar 0=\bar 0$$, a contradiction. Thus $$p(V')=\bar 1$$ for any $$2$$-dimensional subspace $$V'$$ of $$V$$. Now pick any vector $$v\in S$$ and any $$2$$-dimensional subspace $$V^*$$ of $$V$$, orthogonal to $$v$$. Since $$p(V^*)=\bar 1$$, there exists a vector $$u\in S\cap V^*$$. Let $$V'$$ be the subspace of $$V$$ spanned by its orthogonal basis $$\{v,u\}$$. Since both $$v$$ and $$u$$ belong to $$S$$, we have $$p(V')=\bar 0$$, a contradiction.