Quadratic codimension-two multiplicities

The next case after edges on four vertices and triangular faces on five vertices is a tetrahedral face on six vertices. Its angular multiplicity is quadratic. The following result removes that degree-dependent angular sum and gives an eight-dimensional rational state for every degree.

These are exact results about the risk coefficient \(a_3(6,D)\). They do not extend the public sampling API or assert a state bound for arbitrary face size. The separate construction in Uniform construction for codimension two now proves a common fifteen-state bound for arbitrary nonvertex codimension-two face size. The eight-state formula here remains a smaller representation for the six-vertex case.

The odd half-range state

Write \(H_n^{(2)}=\sum_{j=1}^n j^{-2}\) and use the previously proved inverse-binomial sequences

\[F_n=\sum_{j=1}^n\frac1{j^2\binom nj},\qquad G_n=\sum_{j=1}^n\frac1{j\binom nj},\]

with \(F_0=G_0=0\) and

\[G_{n+1}=\frac{G_n}{2}+\frac1{n+1},\qquad F_{n+1}=F_n-\frac{G_n}{n+1}+\frac1{(n+1)^2}.\]

For \(m\ge1\), set \(N=2m+1\) and

\[\begin{split}C_m&=H_N^2-H_N^{(2)},& U_m&=F_N-H_N^{(2)},\\ w_m&=\binom{2m+1}{m}^{-1},& T_m&=\sum_{j=1}^{m}\frac1{j^2\binom Nj},\\ J_m&=H_m^{(2)}-T_m+\frac{U_m}{2}.\end{split}\]

The defining sum for \(T_m\) is used in the proof and independent checks. Evaluation uses the first-order rule

(34)\[J_{m+1}=J_m+\frac{1-3w_m/4}{(m+1)^2},\qquad w_{m+1}=\frac{m+2}{2(2m+3)}w_m, \qquad J_1=\frac14,\quad w_1=\frac13.\]

With \(a=(N+1)^{-1}+(N+2)^{-1}\), \(b=(N+1)^{-2}+(N+2)^{-2}\) and \(q=(N+1)(N+2)\), the full state

\[v_m=(C_m,H_N,H_m,U_m,G_N,J_m,w_m,1)^{\mathsf T}\]

has the remaining transitions

\[\begin{split}C_{m+1}&=C_m+2aH_N+a^2-b,\\ H_{N+2}&=H_N+a,\qquad H_{m+1}=H_m+\frac1{m+1},\\ U_{m+1}&=U_m-\frac{3N+5}{2q}G_N-\frac1q,\\ G_{N+2}&=\frac{G_N}{4}+\frac1{2(N+1)}+\frac1{N+2}.\end{split}\]

Its initial vector at \(m=1\) is

\[v_1=\left(2,\frac{11}{6},1,-\frac56,\frac56, \frac14,\frac13,1\right)^{\mathsf T}.\]

All denominators in these transitions are positive for \(m\ge1\).

An all-degree coefficient formula

Use the universal codimension-two summand from Theorem 20. For \(1\le j<N\), define

\[\begin{split}\Delta_j&=j(j+1)(N-j)(N-j+1),\\ B_j&=H_{j+1}+H_{N-j+1}-H_N,\\ c_j&=\frac{(N+1)(j(N-j)-1)}{\Delta_j},\\ f_N(j)&=\frac{B_j-c_j}{\Delta_j},\\ v_N(j)&=\frac1{j^2(j+1)(N-j)^2(N-j+1)\binom Nj},\\ g_N(j)&=f_N(j)-v_N(j).\end{split}\]

Here \(v_N(j)\) is a scalar summand, distinct from the state vector \(v_m\). The summands are symmetric under \(j\mapsto N-j\).

Theorem 24 (Eight states for a tetrahedral face on six vertices)

For every integer \(D\ge1\), put \(m=D+2\), \(N=2m+1\) and evaluate the state above. Define

(35)\[\begin{split}S_m&=\frac{C_m}{N(N+1)(N+2)} -\frac{N^2+N+2}{N^2(N+1)(N+2)}U_m -\frac{2(N-2)}{N^3(N+2)}(H_N-G_N),\\ X_m&=\frac{C_m+2U_m+4(H_N-G_N)/N}{2(N+2)},\\ E_m&=\frac{H_N-2H_m}{2(m+1)^2} +\frac{J_m}{N}-\frac{w_m}{2(m+1)^3}.\end{split}\]

Then

(36)\[a_3(6,D)=9000\left[ \frac{N^2+63}{8}S_m-\frac{X_m}{2}-4E_m -g_N(m-2)-3g_N(m-1)-6g_N(m)\right].\]

Every boundary term is a rational linear combination of the same eight states. The expression is nonsingular for every \(D\ge1\); it returns zero at the inactive degrees \(D=1,2,3\).

The first positive values, at \(D=4,5,6,7,8\), are

\[\frac{375}{13},\quad \frac{6900}{91},\quad \frac{7375}{56},\quad\frac{493125}{2584},\quad \frac{81423005}{325584}.\]

Proof. Reindex (27) with source-face size \(s=4\) by \(j=L+1\). This gives, for \(D\ge4\),

\[a_3(6,D)=9000\sum_{j=1}^{D-3} \frac{(D-2-j)(D-1-j)}2\,g_N(j).\]

Let \(x=j(N-j)\) and \(p=N/2-j\). The quadratic weight has the exact decomposition

\[\frac{(D-2-j)(D-1-j)}2 =\frac{N^2+63}{8}-\frac{x}{2}-4p.\]

Extending the sum to \(j=m\) adds only the three nonzero terms with weights \(1,3,6\) at \(m-2,m-1,m\); the intervening two weights vanish. Thus it suffices to prove

\[S_m=\sum_{j=1}^m g_N(j),\quad X_m=\sum_{j=1}^m j(N-j)g_N(j),\quad E_m=\sum_{j=1}^m (N/2-j)g_N(j).\]

The already proved full-range formulas for \(f_N\) and \(v_N\) in Further evaluations of the skeleton risk give \(S_m=Q_N-V_N/2\) by symmetry; collecting \(H_N^2-H_N^{(2)}\) and \(F_N-H_N^{(2)}\) gives precisely the first line of (35). The rational constants cancel.

The two new weighted moments are proved below. For a boundary index \(j=m-r\), \(r=0,1,2\), the harmonic part is \(2H_m-H_N\) plus a fixed number of rational increments. Also

\[\binom{N}{m-r}^{-1} =w_m\prod_{i=1}^{r}\frac{m+i+1}{m-i+1}.\]

Hence these boundary values use the stated state. Since \(m\ge3\), all their denominators are positive. At the inactive degrees the half-range polynomial has support exactly on the three subtracted terms, so the same formula gives zero directly.

The symmetric quadratic moment

Multiplying \(f_N(j)\) by \(x=j(N-j)\) gives the harmonic factor \(1/((j+1)(N-j+1))\). The elementary convolution identities

\[\sum_{k=1}^{r}\frac{H_k}{k} =\frac{H_r^2+H_r^{(2)}}2,\qquad \sum_{k=1}^{r-1}\frac{H_{r-k}}{k} =H_r^2-H_r^{(2)}\]

evaluate its full-range sum. The rational part is reduced by

\[\begin{split}\frac{x c_j}{\Delta_j} ={}&\frac1{N+2}\left[\frac1{(j+1)^2} +\frac1{(N-j+1)^2}\right]\\ &+\frac{3N+4}{(N+1)(N+2)^2} \left[\frac1{j+1}+\frac1{N-j+1}\right]\\ &-\frac1{N(N+1)}\left[\frac1j+\frac1{N-j}\right].\end{split}\]

Consequently

\[\sum_{j=1}^{N-1}x f_N(j) =\frac{H_N^2-3H_N^{(2)}+4H_N/N+2(N-2)/N^2}{N+2}.\]

For the inverse-binomial part put \(b_j=\binom Nj^{-1}\) and

\[\begin{split}P(j)&=-\frac{N+1}{N+2}-\frac{2j}{N(N+2)} +\frac{6j^2}{N(N+2)}-\frac{4j^3}{N(N+1)(N+2)},\\ R(j)&=\frac{P(j)}{j^2(N-j+1)}.\end{split}\]

The rational identity

(37)\[\frac1{\Delta_j} =\frac{j+1}{N-j}R(j+1)-R(j) +\frac8{N(N+1)(N+2)} -\frac1{N+2}\left[\frac1{j^2}+\frac1{(N-j)^2}\right]\]

can be checked by multiplying denominators. After multiplication by \(b_j\), the first two terms telescope because \(b_{j+1}/b_j=(j+1)/(N-j)\). Using \(\sum_{j=1}^{N-1} b_j=(N+1)G_N/2-1\) gives

\[\sum_{j=1}^{N-1}x v_N(j) =\frac{-2F_N+4G_N/N+2(N-2)/N^2}{N+2}.\]

The difference of these full-range formulas is twice \(X_m\), since \(N\) is odd and the summands are symmetric. This proves its stated formula without introducing another inverse-binomial state.

The antisymmetric moment

For the harmonic term put \(r_j=1/[j(N-j+1)]\) and \(t_j=1/[j(N-j+2)]\). The identities

\[\frac{N/2-j}{\Delta_j}=\frac{r_j-r_{j+1}}2,\qquad r_j(B_j-B_{j-1})=t_j-t_{j+1}\]

give by summation by parts

\[\sum_{j=1}^m\frac{(N/2-j)B_j}{\Delta_j} =\frac1N-\frac{B_m}{2(m+1)^2} -\frac1{2(m+1)(m+2)}.\]

For the rational part one has

\[\begin{split}\frac{(N/2-j)c_j}{\Delta_j} ={}&\frac{N+3}{2(N+1)^2} \left[\frac1j-\frac1{N-j}-\frac1{j+1}+\frac1{N-j+1}\right]\\ &-\frac1{2N(N+1)}\left[\frac1{j^2}-\frac1{(N-j)^2}\right]\\ &-\frac1{2(N+1)} \left[\frac1{(j+1)^2}-\frac1{(N-j+1)^2}\right].\end{split}\]

Summing these differences of first- and second-order harmonic terms and using \(N=2m+1\) yields

\[\sum_{j=1}^m(N/2-j)f_N(j) =\frac{H_N-2H_m}{2(m+1)^2} +\frac{H_m^{(2)}-H_N^{(2)}/2}{N} -\frac{N-1}{2N^3}.\]

The inverse-binomial telescope, valid for either parity of \(N\), is

\[(N/2-j)v_N(j) =\frac{b_{j+1}}{2(j+1)^2(N-j)} -\frac{b_j}{2j^2(N-j+1)} +\frac{b_j}{2N}\left[\frac1{j^2}-\frac1{(N-j)^2}\right].\]

The odd half-range has \(b_m=b_{m+1}=w_m\). Its sum is

\[\sum_{j=1}^m(N/2-j)v_N(j) =\frac{T_m-F_N/2}{N}+\frac{w_m}{2(m+1)^3} -\frac{N-1}{2N^3}.\]

Subtracting proves the formula for \(E_m\).

Proof of the new state transition

The full inverse-binomial moment and the first-order telescope give

\[\begin{split}\sum_{j=1}^{m}b_j&=\frac{N+1}{4}G_N-\frac12,\\ \sum_{j=1}^{m}\frac{b_j}{j} &=\frac{G_N}{2}-\frac{w_m}{2(m+1)}.\end{split}\]

For the second identity, telescope \(b_j/j-b_j/(N-j)=b_j/j-b_{j+1}/(j+1)\) and use symmetry. The change of total degree satisfies

\[\frac{\binom{N+2}{j}^{-1}}{\binom Nj^{-1}} =\frac{(N+2-j)(N+1-j)}{(N+1)(N+2)}.\]

Divide by \(j^2\), sum to \(m\), and include the new endpoint \(j=m+1\). Substitution of the preceding moments gives

\[T_{m+1}-T_m =-\frac{3N+5}{4(N+1)(N+2)}G_N -\frac1{2(N+1)(N+2)}+\frac{3w_m}{4(m+1)^2}.\]

Half the increment of \(U_m\) is exactly the first two terms. They therefore cancel in \(J_m=H_m^{(2)}-T_m+U_m/2\), proving (34). This also proves the rational state system and completes the all-degree construction.

A scalar recurrence certificate

tools/risk_research/codimension_two_states.py evaluates the exact coefficients without Sage. Its optional symbolic commands eliminate the eight states over \(\mathbb Q(D)\) and independently rebuild the state rows to verify a supplied operator. The checked-in result is tools/risk_research/examples/six_vertex_tetrahedral_face_recurrence_order8.json. It has order eight and maximum polynomial degree 83:

\[\sum_{k=0}^8 p_k(D)a_3(6,D+k)=0\qquad(D\ge1).\]

The leading polynomial has no positive integer zero. The verifier computes all rational roots exactly and filters positive integers, checks positive state denominators, and verifies the eight rational identities and initial values. Thus the operator and degrees one through eight determine the entire positive-degree sequence. No claim of minimal scalar order is made.

To regenerate and verify it in a Sage Python environment:

python -m tools.risk_research.codimension_two_states --output /tmp/six-vertex.json
python -m tools.risk_research.codimension_two_states --verify /tmp/six-vertex.json

The tests compare the coefficient with the original scalar kernel sum, independent rational Gram propagation, and the angular-mode sum at held-out degrees through 300. Separate symbolic checks verify the new partial fractions, binomial-ratio reduction and Gosper certificate. These finite checks guard the implementation; the identities and state proof establish the all-degree statements.

For larger fixed source-face sizes, the angular multiplicity has higher polynomial degree. Its symmetric and antisymmetric decomposition is a useful next reduction, but closure in these same eight states has not been proved here.