When the active sets differ, the coordinatewise correspondence $T_{ba}$ no longer defines a bijection between $\mathcal S_ a$ and $\mathcal S_ b$. We compare them as abstract metric spaces. They are bounded, but they are not compact as soon as some coefficient is positive. The Gromov–Hausdorff distance $d_{\mathrm{GH}}$ of bounded metric spaces from definition 15.19.19, together with its estimate through correspondences in lemma 15.19.20, applies without a compactness hypothesis.
Write $\mathcal I_ a\subset \mathcal S_ a$ for the ideal subspace of proposition 18.17.3, with the restricted metric $\rho _ a$.
Proposition 18.17.5.25F2A9E20F21 For all nonnegative coefficient vectors $a,b$, including vectors with different active sets,
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18.17.6
\begin{equation} \left|\sum _ i(a_ i-b_ i)A_ i\right| \leq d_{\mathrm{GH}}(\mathcal S_ a,\mathcal S_ b) \leq \sum _ i|a_ i-b_ i|A_ i. \end{equation}
The same bounds hold for $d_{\mathrm{GH}}(\mathcal I_ a,\mathcal I_ b)$. If $b_ i\leq a_ i$ for every $i$, both distances equal $\sum _ i(a_ i-b_ i)A_ i$. In particular, the distance from $\mathcal S_ a$ to the singleton $\mathcal S_0=\mathcal I_0=\{ \Psi \} $ equals $\sum _ i a_ iA_ i$.
For $b\leq a$, the map $T_{ba}\Phi _ a(q)=\Phi _ b(q)$ is a surjective $1$-Lipschitz map and commutes with ideal projection. Every fiber has diameter
\[ 2\sum _{\{ i:a_ i>0,\ b_ i=0\} }a_ iA_ i. \]
If some $a_ i>0$, then $\mathcal S_ a$ and $\mathcal I_ a$ are not totally bounded, and hence not compact.
Proof.
Use the correspondence
\[ R_{ab}=\{ (\Phi _ a(q),\Phi _ b(q)):q_ i \text{ are normalized ordinary curve models}\} . \]
Every potential in either locus arises from such a tuple, with arbitrary normalized models in its coordinates of zero coefficient, so both projections of $R_{ab}$ are surjective. For any two tuples, put $d_ i=\| \mu _ i-\nu _ i\| _{\mathrm{TV}}$. By (18.17.2), the difference of their two distances is $\sum _ i(a_ i-b_ i)d_ i$, with $0\leq d_ i\leq 2A_ i$. Thus the distortion of this correspondence satisfies $\operatorname {dis}R_{ab}\leq 2\sum _ i|a_ i-b_ i|A_ i$, and lemma 15.19.20 gives the upper bound in (18.17.6). The diameter formula in proposition 18.17.3 and the same lemma give the lower bound.
Restricting the correspondence to tuples of ideal curve models, we obtain a correspondence between the two ideal subspaces with the same distortion bound. The atomic measures $A_ i\delta _{x_ i}$ and $A_ i\delta _{y_ i}$ show that the ideal subspaces have the same diameters as the full spaces, so the lower bound also holds for them. When $b\leq a$, the upper and lower bounds coincide, including when $b=0$.
For the transition map, every active coordinate of $b$ is active for $a$, and uniqueness of the normalized active tuple makes the map well defined. Every $b$-tuple extends to an $a$-tuple, so the map is surjective, and $\sum _ i b_ i d_ i\leq \sum _ i a_ i d_ i$ proves the Lipschitz bound. Atomic projection in each coordinate commutes with discarding and rescaling coordinates. In a fiber, every coordinate with $b_ i>0$ is fixed, while the coordinates with $a_ i>0$ and $b_ i=0$ vary freely. The diameter of the fiber is therefore the sum above, by the calculation of the diameter on each curve.
Finally, fix an active coordinate $i$ and all other coordinates at their zero models. Let $q_ i$ have polar measure $A_ i\delta _ x$, with $x$ ranging over infinitely many distinct points of $C_ i$. The resulting potentials are ideal models, and any two of them are at distance $2a_ iA_ i$. This infinite separated set excludes total boundedness of both spaces. In particular, the comparison above is between complete bounded spaces that are not compact.
$\square$
Pome
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