Proof. An ordered root frame with Gram matrix $C(X)$ consists of a choice of a simple-root basis for a closed subsystem of type $X$, together with all allowed diagram automorphisms. The Weyl group acts simply transitively on the chambers of the subsystem, while the diagram automorphisms account for the possible Dynkin-diagram symmetries. Hence the number of ordered frames per embedded subsystem is $$|W(X)|\,|\!\mathop{\mathrm{Aut}}(X)|.$$ For $A_5$, this is $6! \cdot 2 = 1440$, while for $D_5$ it is $(2^4 \cdot 5!) \cdot 2 = 3840$. ◻

Proof. The root system $A_n$ contains no branching node, so it contains no closed subsystem of type $D_5$.

For $D_n$, a closed $D_5$ subsystem is obtained by choosing five coordinate directions. Thus $$m_{D_n}(D_5) = \binom{n}{5}.$$ For $E_6$ and $E_7$, the corresponding closed-subsystem orbit counts are the classical values $27$ and $378$, respectively; see the classification of closed reflection subsystems in . ◻

Proof. By the frame multiplicity lemma and the subsystem counts, $$\begin{align*} a(T_{A_5}, F_6^{(5)}) &= 1440(0 - 4) = -5760, \\ a(T_{D_5}, F_8^{(5)}) &= 3840(4 - 2) = 7680, \\ a(T_{D_5}, F_{10}^{(5)}) &= 3840(24 - 6) = 69120, \\ a(T_{D_5}, F_{12}^{(5)}) &= 3840(108 - 48) = 230400, \\ a(T_{D_5}, F_{18}^{(5)}) &= 3840(1008 - 378) = 2419200. \end{align*}$$ Every displayed coefficient is nonzero, so each $F_h^{(5)}$ is a nonzero Siegel modular form. ◻

Proof. Compatibility of the Siegel operator with theta series gives $$\Phi(F_h^{(5)}) = F_h^{(4)}.$$ Since $J_8^{(4)}$ is cuspidal, $$\Phi(J_8^{(4)}) = 0,$$ and therefore $$\Phi^2(F_h^{(5)}) = 0.$$ On the completely decomposable locus $\tau = \operatorname{diag}(\tau_1, \dots, \tau_5)$, the genus-$5$ theta series factors as $$\Theta_L^{(5)}(\tau) = \prod_{j=1}^5 \Theta_L^{(1)}(\tau_j).$$ Since equal-Coxeter-number Niemeier lattices have identical genus-$1$ theta series, the two products agree, and hence $$\left.F_h^{(5)}\right|_{\mathbb{H}_1^5} \equiv 0. \qedhere$$ ◻

Proof. Because $L$ is even, $\|u_k\|^2 \in 2\mathbb{Z}_{\ge 0}$. If $T(x)$ has rank $5$, then the increments cannot all lie in a proper subspace in a way permitting a zero increment at minimal full support. In particular, a minimal full-support configuration has $\|u_k\|^2 \ge 2$ for every $k$. Therefore $$\mathop{\mathrm{Tr}}(T(x) Q(\ell)) \ge \sum_{k=1}^6 \ell_k,$$ with equality exactly when every $u_k$ is a root. ◻

Proof. In the standard realization $A_5 = \{e_i - e_j : 1 \le i \ne j \le 6\}$, a star frame consists of roots sharing a common coordinate: $$x_k = \pm(e_a - e_{b_k}).$$ There are two choices for the global sign, six choices for the central coordinate $a$, and $5!$ orderings of the remaining coordinates. Thus $$2 \cdot 6 \cdot 5! = 1440.$$ The edge increments are $$u_1 = x_1, \qquad u_k = x_k - x_{k-1} \quad (2 \le k \le 5), \qquad u_6 = -x_5.$$ They form an oriented affine $\widetilde{A}_5$ cycle satisfying the stated inner-product relations. Conversely, the partial sums $x_k = u_1 + \dots + u_k$ recover the original star frame. ◻

Proof. For $A_n$, a closed $A_5$ subsystem is obtained by choosing six coordinates out of the $n+1$ available coordinates, giving $$m_{A_n}(A_5) = \binom{n+1}{6}.$$ For $D_n$, choose six coordinate directions and assign relative signs. Since simultaneous reversal of all signs gives the same subsystem, there are $$\frac{2^6}{2} = 32$$ sign choices per coordinate $6$-subset. Hence $m_{D_n}(A_5) = 32\binom{n}{6}$.

For $E_6$, $E_7$, and $E_8$, the stated numbers are the corresponding closed-reflection-subsystem orbit counts in the convention used here; see . The distinction between closed root subsystems, reflection subgroups, and ordered embeddings is important: the numbers used here are subsystem counts, while the factor $1440$ in Lemma 8 converts subsystem counts into ordered star frames. ◻

Proof. Because star frames generate closed $A_5$ subsystems, the coefficient of $T_{\mathrm{cyc}}^A$ is $$a(T_{\mathrm{cyc}}^A, F_h^{(5)}) = 1440\,\Delta m_{A_5}.$$ We compute the five differences.

For $h = 6$, $$\Delta m_{A_5} = 0 - 4 = -4,$$ and hence $$a(T_{\mathrm{cyc}}^A, F_6^{(5)}) = 1440(-4) = -5760.$$

For $h = 8$, $$\Delta m_{A_5} = 0 - 2\binom{8}{6} = -56,$$ so $$a(T_{\mathrm{cyc}}^A, F_8^{(5)}) = 1440(-56) = -80640.$$

For $h = 10$, $$\Delta m_{A_5} = 4 \cdot 32\binom{6}{6} - \left( 2\binom{10}{6} + 32\binom{6}{6} \right),$$ hence $$\Delta m_{A_5} = 128 - 452 = -324,$$ and $$a(T_{\mathrm{cyc}}^A, F_{10}^{(5)}) = 1440(-324) = -466560.$$

For $h = 12$, $$\Delta m_{A_5} = 4(36) - \left( \binom{12}{6} + 32\binom{7}{6} + 36 \right),$$ so $$\Delta m_{A_5} = 144 - 1184 = -1040,$$ and therefore $$a(T_{\mathrm{cyc}}^A, F_{12}^{(5)}) = 1440(-1040) = -1497600.$$

Finally, for $h = 18$, $$\Delta m_{A_5} = \left( 32\binom{10}{6} + 2(1344) \right) - \left( \binom{18}{6} + 1344 \right).$$ Since $$32\binom{10}{6} = 6720, \qquad 2(1344) = 2688,$$ and $$\binom{18}{6} = 18564,$$ we obtain $$\Delta m_{A_5} = 9408 - 19908 = -10500.$$ Therefore $$a(T_{\mathrm{cyc}}^A, F_{18}^{(5)}) = 1440(-10500) = -15120000.$$ All five coefficients are nonzero.

It remains to show that this coefficient survives after restriction to the hyperelliptic degeneration. Let $$z_1, \dots, z_6 \in \mathbb{P}^1$$ denote the marked points corresponding to the six nodes. A smoothing parameter $t$ may be chosen so that the period matrix has an expansion $$\begin{equation} \label{eq:smoothing_expansion} \tau(t) = \frac{\log t}{2\pi i} Q(\ell) + \tau_0 + O(t), \qquad t \to 0^+. \end{equation}$$ For a minimal full-support configuration, Fay’s degeneration formula for third-kind differentials gives, up to a nonzero normalization convention, $$\begin{equation} \label{eq:fay_formula} \mathop{\mathrm{Tr}}(T(u)\tau_0) = \frac{1}{\pi i} \sum_{1 \le a < b \le 6} (u_a \cdot u_b) \log(z_a - z_b). \end{equation}$$ Exponentiating gives the boundary character $$\begin{equation} \label{eq:boundary_character} e^{2\pi i \mathop{\mathrm{Tr}}(T(u)\tau_0)} = \prod_{1 \le a < b \le 6} (z_a - z_b)^{2(u_a \cdot u_b)}. \end{equation}$$ For the star-frame cycle, $$(u_i \cdot u_{i+1}) = -1 \pmod 6,$$ while all nonadjacent inner products vanish. Consequently, $$\begin{equation} \label{eq:boundary_star} e^{2\pi i \mathop{\mathrm{Tr}}(T_{\mathrm{cyc}}^A \tau_0)} \sim \frac{1}{(z_1-z_2)^2 (z_2-z_3)^2 (z_3-z_4)^2 (z_4-z_5)^2 (z_5-z_6)^2 (z_6-z_1)^2}, \end{equation}$$ where $\sim$ denotes equality up to a nonzero scalar depending only on the normalization conventions.

Thus the star-frame contribution is realized, up to nonzero scalar, as the squared Parke–Taylor factor $$\mathrm{PT}(1, 2, 3, 4, 5, 6)^2 = \frac{1}{\prod_{i=1}^6 (z_i - z_{i+1})^2}, \qquad z_7 = z_1.$$ Its polar divisor on $\mathcal{M}_{0,6}$ is $$2 \sum_{i=1}^6 \{z_i = z_{i+1}\}.$$ This divisor records the cycle adjacency graph.

A competing minimal full-support root configuration with a different adjacency graph produces a boundary character whose polar divisor has a different collection of pairwise diagonals. Therefore such a term cannot cancel the star-frame contribution as a rational function on $\mathcal{M}_{0,6}$. In particular, the coefficient $$a(T_{\mathrm{cyc}}^A, F_h^{(5)}) \ne 0$$ forces the full-support initial form $$P_{h,\mathrm{full}}(\tau_0)$$ to be nonzero as a rational function on $\mathcal{M}_{0,6}$.

Hence there exists a point $$\tau_0 \in \mathcal{M}_{0,6}$$ such that $$P_{h,\mathrm{full}}(\tau_0) \ne 0.$$ Substituting [eq:smoothing_expansion] into the Fourier expansion gives $$F_h^{(5)}(\tau(t)) = t^{\sum_{i=1}^6 \ell_i} \bigl( P_{h,\mathrm{full}}(\tau_0) + o(1) \bigr).$$ Therefore $$F_h^{(5)}(\tau(t)) \ne 0$$ for all sufficiently small positive $t$.

The smoothing family lies in $\mathcal{H}_5 \subset \mathcal{J}_5$, so $$\left.F_h^{(5)}\right|_{\mathcal{H}_5} \not\equiv 0$$ and consequently $$\left.F_h^{(5)}\right|_{\mathcal{J}_5} \not\equiv 0.$$ Since $P_{h,\mathrm{full}}$ is a nonzero rational function on $\mathcal{M}_{0,6}$, the same argument also shows that the non-vanishing occurs on a Zariski-open subset of the corresponding hyperelliptic boundary parameter space. ◻