Add the freedom-positive counting balance to the constraint floor

Remark: a disk with k interior vertices has 2k+n-2 faces (Euler) but only
k interior constraints, so each interior vertex adds two degrees of
freedom against one constraint -- depth is freedom-positive and Phi can
only retain or enlarge below the interior-free floor 2^(n-2). Motivates
the lower bound and replaces the prior TODO sketch.

Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>
This commit is contained in:
2026-06-17 02:24:04 -04:00
parent b70ea2c087
commit 1d981b4d01
4 changed files with 56 additions and 43 deletions
@@ -499,12 +499,23 @@ the boundary sequence, so the map $\lambda \mapsto \lambda^{*}|_C$ is
injective and $|\Phi(D)| = 2^{\,n-2}$.
\end{proof}
%% TODO (lower bound): show |Phi(D)| >= 2^{n-2} for EVERY triangulated
%% disk D. Strategy: the n boundary-incident faces (one per boundary edge)
%% carry n-2 independent binary degrees of freedom after the interior
%% Heawood constraints are imposed; those constraints relate only
%% interior-incident faces and cannot collapse the boundary freedom below
%% 2^{n-2}. (See notes/boundary_restriction_structure.tex.)
\begin{remark}[Depth is freedom-positive]
\label{rem:freedom-positive}
The lower bound is plausible from a counting balance. A triangulated
disk with $k$ interior vertices has $2k + n - 2$ faces (Euler) and
imposes exactly $k$ interior Heawood constraints, one per interior
vertex. So each interior vertex contributes \emph{two} faces --- two new
$\{+1,-1\}$ degrees of freedom --- against only \emph{one} constraint,
and the free dimension $(2k + n - 2) - k = k + n - 2$ \emph{grows} with
depth. Going deeper is freedom-positive on balance: the boundary
projection $\Phi(D)$ can only retain or enlarge its options, never drop
below the interior-free value $2^{\,n-2}$. (Empirically $|\Phi(D)|$ does
grow with $k$; e.g.\ on the $4$-cycle the central-apex wheel realises $5$
sequences against the fan's $4$.) The constraints relate only
interior-incident faces and cannot collapse the $n-2$ degrees of freedom
carried by the boundary-incident faces --- which is the content the lower
bound must make precise.
\end{remark}
\begin{remark}
\label{rem:floor-consequences}