Proof. An element of $F_{T_1} \times_{F_S} F_{T_2}$ is an ordered pair of realization tuples $(X_1, X_2)$ with $X_1 = (\alpha, \beta)$ and $X_2 = (\alpha, \gamma)$, where $\alpha = (x_1, \dots, x_k) \in L^k$, $\beta = (y_{k+1}, \dots, y_g) \in L^{g-k}$, and $\gamma = (z_{k+1}, \dots, z_g) \in L^{g-k}$, such that: $$\pi_g(\alpha, \beta) = T_1, \qquad \pi_g(\alpha, \gamma) = T_2.$$ The concatenated $(2g-k)$-tuple $Z = (x_1, \dots, x_k, y_{k+1}, \dots, y_g, z_{k+1}, \dots, z_g) \in L^{2g-k}$ has Gram matrix: $$\widetilde T(W) = \frac{1}{2} Z^{\mathsf T}Z = \begin{pmatrix} S & T_1^{(12)} & T_2^{(12)} \\ (T_1^{(12)})^{\mathsf T} & T_1^{(22)} & W \\ (T_2^{(12)})^{\mathsf T} & W^{\mathsf T} & T_2^{(22)} \end{pmatrix},$$ where $W_{ij} = \frac{1}{2}(y_{k+i}, z_{k+j})$. Summing the fiber counts $|\pi_{2g-k}^{-1}(\widetilde T(W))| = a_L(\widetilde T(W), \Theta_L^{(2g-k)})$ over all $W \in \mathcal W(T_1, T_2)$ partitions the fiber product into disjoint Gram realization classes. ◻

Proof. For positive-definite integral lattices, Krull–Schmidt cancellation implies that $A \oplus C \cong B \oplus C \implies A \cong B$. Since $A \not\cong B$, cancellation forces $A \oplus C \not\cong B \oplus C$. Multiplicativity of Siegel theta series under orthogonal direct sums gives: $$\Theta_{A \oplus C}^{(g)}(Z) = \Theta_A^{(g)}(Z)\,\Theta_C^{(g)}(Z) = \Theta_B^{(g)}(Z)\,\Theta_C^{(g)}(Z) = \Theta_{B \oplus C}^{(g)}(Z)$$ for all $g \le g_0$. ◻

Proof. Roots of $A_n$ are $e_i - e_j$ with $1 \le i \neq j \le n+1$. Two roots $e_i - e_j$ and $e_k - e_l$ are orthogonal if and only if $\{i,j\} \cap \{k,l\} = \emptyset$. To construct an ordered $k$-tuple $(r_1, \dots, r_k)$:

• For $r_1 = e_{i_1} - e_{j_1}$, there are $(n+1)n = (n+1)^{\underline 2}$ choices of ordered index pairs.

• For $r_2 = e_{i_2} - e_{j_2}$, there are $(n-1)(n-2) = (n-1)^{\underline 2}$ choices from the remaining indices.

• For $r_m = e_{i_m} - e_{j_m}$, there are $(n+3-2m)(n+2-2m) = (n+3-2m)^{\underline 2}$ choices.

Multiplying these sequential choices gives: $$N_k(A_n) = \prod_{m=1}^k (n+3-2m)^{\underline 2} = (n+1)n(n-1)\cdots(n+2-2k) = (n+1)^{\underline{2k}}. \qedhere$$ ◻

Proof. Roots in $D_n$ are $\pm e_i \pm e_j$ ($1 \le i < j \le n$). Two roots are orthogonal if and only if their 2-coordinate supports are disjoint, or their supports coincide and their inner product is $0$ (each 2-element support supports $4$ roots, yielding $8$ ordered orthogonal pairs on that support). In an ordered $k$-tuple, suppose $p$ supports occur twice and $k-2p$ supports occur once ($0 \le p \le \lfloor k/2 \rfloor$).

1. Partition the $k$ ordered positions into $p$ double blocks and $k-2p$ single blocks: $\frac{k!}{p! 2^p (k-2p)!}$ ways.

2. Choose an ordered sequence of $k-p$ disjoint pairs from $\{1, \dots, n\}$: $\frac{n^{\underline{2k-2p}}}{2^{k-p}}$ ways.

3. Assign signs: $8^p 4^{k-2p} = 2^{3p} 2^{2k-4p} = 2^{2k-p}$ ways.

Multiplying gives: $$\frac{k!}{p! 2^p (k-2p)!} \cdot \frac{n^{\underline{2k-2p}}}{2^{k-p}} \cdot 2^{2k-p} = k! \frac{2^{k-p}}{p!(k-2p)!} n^{\underline{2k-2p}}.$$ Summing over $0 \le p \le \lfloor k/2 \rfloor$ gives $N_k(D_n)$. Dividing by $k!$ gives the series coefficients:

• $[t^1] F_{D_n} = \frac{N_1(D_n)}{1!} = 2n^{\underline 2}$.

• $[t^2] F_{D_n} = \frac{N_2(D_n)}{2!} = 2n^{\underline 2} + 2n^{\underline 4}$.

• $[t^3] F_{D_n} = \frac{N_3(D_n)}{3!} = 4n^{\underline 4} + \frac{4}{3}n^{\underline 6}$.

• $[t^4] F_{D_n} = \frac{N_4(D_n)}{4!} = 2n^{\underline 4} + 4n^{\underline 6} + \frac{2}{3}n^{\underline 8}$.

 ◻

Proof. For $E_8$, the root count is $N_1 = 240$. The orthogonal subsystem of a root is $E_7$ ($126$ roots), so $N_2 = 240 \times 126 = 30240$. The orthogonal subsystem of an orthogonal pair is $D_6$ ($60$ roots), so $N_3 = 30240 \times 60 = 1814400$. The orthogonal subsystem of an orthogonal triple is $D_4 \oplus A_1$ ($26$ roots), so $N_4 = 1814400 \times 26 = 47174400$. Dividing $N_k$ by $k!$ gives: $$\frac{240}{1!} = 240, \quad \frac{30240}{2!} = 15120, \quad \frac{1814400}{3!} = 302400, \quad \frac{47174400}{4!} = 1965600.$$ Analogous complement chains ($E_7 \supset D_6 \supset D_4 \oplus A_1$ and $E_6 \supset A_5 \supset A_3 \supset A_1$) yield $F_{E_7}(t)$ and $F_{E_6}(t)$. ◻

Proof. The matrix condition $\frac{1}{2}X^{\mathsf T}X = I_4$ is equivalent to $(x_i, x_j) = 2\delta_{ij}$ for $1 \le i, j \le 4$. In an even lattice, vectors of norm $2$ are roots. An ordered orthogonal $4$-frame in $\bigoplus_i X_i^{m_i}$ decomposes according to the number of roots $k_{i,j}$ selected from the $j$-th copy of component $X_i$, with $\sum k_{i,j} = 4$. The number of ways to interleave these vectors into an ordered $4$-tuple is $\frac{4!}{\prod k_{i,j}!}$. Summing over all partitions of $4$ gives: $$a(I_4, \Theta_N^{(4)}) = 4! [t^4] \prod_i \left( \sum_{k=0}^4 \frac{N_k(X_i)}{k!} t^k \right)^{m_i} = 4! [t^4] \prod_i F_{X_i}(t)^{m_i}. \qedhere$$ ◻

Proof. Evaluating Theorem 10 for all $24$ Niemeier root systems yields the values in Table 2. The values are strictly monotonically increasing with $h$ across distinct Coxeter classes, and within each equal-$h$ pair, the difference $\Delta a(I_4)$ is strictly positive. ◻

Proof. For fixed $n = 24$, table matching and finite group operations on $A_R$ require $O(1)$ arithmetic steps. LLL and Minkowski reduction in dimension $24$ terminate in polynomial time in the basis bit length. ◻