Is it possible to partition $\mathbb{C}^2$ into two sets $S$ and $S'$ such that, given any two nonzero orthogonal vectors $\mathbf{v}$ and $\mathbf{w}$ of $\mathbb{C}^2$, one of them lies in $S$ and the other lies in $S'$? This is easily true for $\mathbb{R}^2$ (we just need to take $S$ as the union of the first and the third quadrant, plus one of the coordinate axes).
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1$\begingroup$ What about the zero vector? $\endgroup$– Dustin G. MixonCommented Jan 11 at 20:05
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1$\begingroup$ If you ignore the zero vector, it suffices to partition $\mathbb{CP}^1$. So you want to partition the Bloch sphere into antipodes, which is easy. $\endgroup$– Dustin G. MixonCommented Jan 11 at 20:09
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1$\begingroup$ Here's an explicit partition $A\sqcup B=\mathbb{C}^2\setminus\{0\}$ that does the trick: Put $(x,y)$ in $A$ if $|x|>|y|$, or if $|x|=|y|$ and $\operatorname{Re}(x\overline{y})>0$, or if $|x|=|y|$ and $\operatorname{Re}(x\overline{y})=0$ and $\operatorname{Im}(x\overline{y})>0$. Otherwise, put $(x,y)$ in $B$. $\endgroup$– Dustin G. MixonCommented Jan 11 at 20:20
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1$\begingroup$ No, $(1,0)\in A$ since $|1|>|0|$, whereas $(0,1)\in B$ since $|0|<|1|$. $\endgroup$– Dustin G. MixonCommented Jan 12 at 10:48
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1$\begingroup$ @Dustin G. Mixon: Thanks, I suspected this. Physicists generally prefer to re-invent bicycles themselves, instead of reading undergraduate math textbooks:-) $\endgroup$– Alexandre EremenkoCommented Jan 12 at 19:24
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