Chapter 15: rescaling a round circle about its own centre by 1/r

Let $C$ be a round circle in three-dimensional space, written in the canonical form $C = \{c + (r\cos t)\,u + (r\sin t)\,v : t \in \mathbb{R}\}$ where $r > 0$ and $u, v$ are orthonormal. Then the pointwise rescaling $S(x) = r^{-1} (x - c) + c$ carries $C$ onto the round circle $C' = \{c + (\cos t)\,u + (\sin t)\,v : t \in \mathbb{R}\}$ of radius one with the same centre $c$ and lying in the same plane spanned by $u, v$.

This is the algebraic core of the cutoff-rescale construction for the open leaf BookSixth.roundness_restoration_after_prefix. That leaf asks for a homeomorphism which fixes every standard circle at time one and whose image of a dragged circle is round again. A global rigid motion cannot do this, because fixing all the standard circles forces a rotation about the $x$-axis, and such a rotation does not straighten a circle lying in a tilted plane. Rescaling about the circle's own centre can.

The reason this particular identity matters is the Lipschitz budget. The alternative route through BookSixth.bump_perturbation_is_homeomorph_v4 is provably closed: because the cutoff $\chi_i$ equals $1$ on $C_i$, the perturbation has slope on $C_i$ equal to the on-circle slope of the per-component motion, so demanding a global contraction forces that slope to stay below one at every time, while the landing condition forces the final scale to be exactly $1/r$. Those two demands are compatible only when $r > 1$. A centred rescale carries no such budget and therefore also applies when $r \le 1$, where the motion must expand rather than contract. The radius dependence is precisely the obstruction this leaf removes.