\section{Depth five, depth six, and the complete profiles}\label{sec:profiles} The failures in (8.2) split into two depth-five bridges and the critical depth-six bases. For each family we prove exactness in two steps. First we exhibit a bounded quotient with a nonzero restriction to the translation kernel, proving failure at the preceding degree. Then we write a zero-sum row representing a generator of that kernel. A late relation alone would give only an upper bound; a defect alone would give only a lower bound. \begin{proposition}[Generic critical pair]\label{prop:generic-critical} If $p,q\ge5$, the support $P1\cup Q1$ has \[ K(P1\cup Q1)/\calL_{\le5}(P1\cup Q1)\cong C_2 \tag{9.1} \] and closes at degree six. \end{proposition} \begin{proof} The internal presentations are $C_{2p}$ and $C_{2q}$. A mixed relation of length at most four must split two-plus-two between the boxes, and each pair is opposite. Odd rows do not exist, so through degree five $A_5(P1\cup Q1)=C_{2p}\oplus C_{2q}$. The map to the generated CRT subgroup identifies the two parity coordinates, and its kernel is their diagonal element, proving (9.1). If $x,y$ generate $C_{2p},C_{2q}$, the map to the CRT target sends their parity components to the same element of $C_2$ and their odd-primary components to independent coordinates. The kernel is generated by $px-qy$, an element of order two. Every mixed four-term row is already an opposite pair in each box, and every odd row is forbidden by parity, so no relation through degree five changes that class. Choose a coordinate-zero triple $(1,1,-2)$ in each odd line. Each triple has odd parity, while their union is a genuine length-six zero-sum row. It represents the diagonal element, so degree six closes the support. \end{proof} This support is the sharpness family appearing in Casper--Orozco \cite{CasperOrozco2026}; the exact quotient and its placement in the full profile are the assertions used here. \begin{proposition}[Prime-three critical family]\label{prop:three-critical} Let $p=3$. The seven bases \[ P1,\quad Q\cup P1,\quad P1\cup Q1,\quad G,\quad Q\cup G, \quad P1\cup G,\quad Q\cup P1\cup G \tag{9.2} \] have fifth-order obstruction $\Z$ and close at degree six. \end{proposition} \begin{proof} Put $A=P1=\Orb6$, $B=Q1=\Orb{2q}$, $C=Q=\Orb q$, and $D=G=\Orb{6q}$. Let $\eta$ be the $C_3$ sign on $A,D$ and zero on $B,C$. On each support in \[ A,\ A\cup C,\ A\cup B,\ D,\ C\cup D,\ A\cup D, A\cup C\cup D, \tag{9.3} \] every zero-sum row of length at most five has $\eta=0$. Its sign sum is divisible by three. Except on $A\cup B$, parity also makes it even, and absolute value at most five forces zero. On $A\cup B$, a value $\pm3$ would require three $A$ points and an odd number of $B$ points. A single $B$ point cannot have zero $q$-coordinate, while three make total length at least six. The local forms are \[ A:\Z u,\quad C:C_qv,\quad B:C_2c\oplus C_qv, \quad D:\Z u\oplus C_qv. \] Available triples glue the $q$-slopes on $C\cup D$, and available quadruples glue sign parameters on $A\cup D$. The map to the CRT target then has cyclic kernel generated by $6u$, except on $A\cup B$, where it is generated by $3u+c$. The normalized $\eta$-value of any additional relation is its coefficient in this kernel, and the preceding visibility argument forces it to vanish through degree five. Hence every obstruction is exactly $\Z$. The seven presentations may be summarized as \[ \begin{array}{c|c|c} \text{base type}&\text{kernel generator}&\text{normalized sign}\\ \midrule A,\ A\cup C,\ D,\ C\cup D,\ A\cup D,\ A\cup C\cup D &6u&\eta/6\\ A\cup B&3u+c&\eta/3. \end{array} \tag{9.3a} \] The last column denotes the integer coefficient on the cyclic kernel; it is not division inside an abelian target. The visibility argument proves that this coefficient is zero on every row of length at most five. Six positive $A$ or $D$ points realize $6u$. On $A\cup B$, take three positive $A$ points and three $B$ points whose $q$-coordinates sum to zero; this realizes $3u+c$. Thus all seven bases close at six. \end{proof} \begin{proposition}[The two bridges]\label{prop:bridges} In both arithmetic regimes the bases \[ P\cup P1\cup Q1,\qquad Q\cup P1\cup Q1 \tag{9.4} \] have exact depth five. \end{proposition} \begin{proof} For the first base, the local forms are $xa$, $c_p+xa_p$, and $c_q+yb$. A mixed triple gives $a_p=a$. The remaining kernel is generated by $c_p+c_q$. No row of length at most four represents it: an odd number of points is needed in both parity-one boxes. One $Q1$ point cannot cancel its $q$-coordinate, while three leave only one $P1$ point, which cannot cancel its $p$-coordinate. A length-five row with one $P$ point, one opposite $P1$ point, and three $Q1$ points with zero $q$-sum kills the diagonal parity. This proves exact depth five. Exchanging $p,q$ gives the second bridge when $p,q\ge5$. The lower bound is uniform in the primes. A representative of $c_p+c_q$ must use an odd number of points from each parity-one box. At total length at most four, the only possible mixed splits are $1+1$ and $1+3$. The first leaves both odd coordinates nonzero; in the second, the single point on one axis cannot be cancelled by points on the other. Thus the diagonal parity class genuinely survives all quartic rows. If $p=3$, the mixed triple on $P\cup P1$ gives $\alpha+2u=0$; with $3\alpha=0$ this is a $C_6$ parameter. Adding $Q1$ again leaves only the diagonal parity kernel, excluded at four and killed by the same five-term row. For the exchanged bridge, the $q$-slopes glue and the fourth-order kernel is $\Z(3u+c)$. A row of length at most four has zero sign: sign $\pm3$ would require three $P1$ points and one $Q1$ point, whose $q$-coordinate cannot vanish. The explicit relation \[ 3P1[1]+Q[-y]+Q1[y]=0 \] represents $3u+c$ at length five. \end{proof} \subsection{Assembly of the 255 supports} The zero orbit splits off harmlessly. Indeed $e_0$ is itself a length-one zero-sum row, so adjoining $\Orb1$ leaves every obstruction quotient unchanged. Thus $\Orb1$ alone has depth one and every nonempty seven-box base has two versions of the same depth. Indeed, for every $d\ge1$, \[ \Z^{S\cup\{0\}}=\Z^S\oplus\Z e_0, \qquad \calL_{\le d}(S\cup\{0\})=\calL_{\le d}(S)\oplus\Z e_0, \tag{9.4a} \] and the same direct sum holds for $K$. This proves the doubling rule rather than treating it as a numerical symmetry of the atlas. \begin{proof}[Proof of Theorem~\ref{thm:profiles}] The support $E$ is the only depth-two base and contributes two raw supports. Theorem~\ref{thm:cubic} gives 77 depth-three bases and hence 154 supports. If $p,q\ge5$, Theorem~\ref{thm:quartic} gives 46 depth-four bases, Proposition~\ref{prop:bridges} gives two depth-five bases, and Proposition~\ref{prop:generic-critical} gives one depth-six base. Therefore \[ 1,\quad2,\quad2(77)=154,\quad2(46)=92,\quad2(2)=4,\quad2(1)=2, \tag{9.5} \] which sums to 255. If $p=3$, the corresponding numbers of bases are $40,2,7$, by Theorem~\ref{thm:quartic}, Proposition~\ref{prop:bridges}, and Proposition~\ref{prop:three-critical}. Hence \[ 1,\quad2,\quad154,\quad2(40)=80,\quad4,\quad2(7)=14, \tag{9.6} \] again summing to 255. Every family was accompanied both by closure and by a defect at the preceding degree, so these are first-closing depths rather than upper bounds. Proposition~\ref{prop:dictionary} identifies them with maximum separating-atom lengths. \end{proof} As checks on the assembly, \[ 1+2+154+92+4+2=255, \qquad 1+2+154+80+4+14=255. \tag{9.7} \] The cubic layer is identical in both regimes. The twelve-support decrease in the quartic entry when $p=3$ is transferred exactly to the depth-six entry: six additional base masks, doubled by the optional zero orbit. This is the profile-level trace of the free prime-three sign directions. Corollary~\ref{cor:unit-compat} now follows from both theorem lines: the single box $G$ is among the seven critical bases when $p=3$, and among the generic quartic masks when $3\nmid2pq$.