nichiyes.com

This is a fun post.

Clearly a similar argument can be ran to show the unorderedness of any other field with a rotational operator that just adds dimensions to the reals/complexes (quaternions, octonions, etc.) but I do wonder if either some other property (say, completeness) can be given up to get orderedness or else if some nonstandard field with non-flat geometry can be ordered. (And not be isomorphic to the reals! I suppose this requires and answer to my first question, though, since the reals are the only complete ordered field.)

My friend Jon pointed out that the meaning of orderedness basically requires having a single dimension to order on. (As for all x and y, either x>y, y>x, or x=y. That’s a one dimensional relation.)  Most of the ideas I had in mind with giving up, say, completeness ultimately reduced the dimensionality. (For example, if you take a subset of the complexes that has Re(x)=Re(y)->x=y then you can have an order, but that’s by basically knocking out a dimension.)