The degree-two value for the root frame - The KLS theorem and its methods

Part of the conditional-fiber mechanism, Chapter Conditional fibers: inverse-variance frames of line resamplings; the reading order is on the full proofs page.

Overview. This dossier proves Lemma 22.2 as Theorem D11.1: for every $m\ge3$ , the root-frame pencil on the degree-two quotient $V_{m,2}$ of the uniform simplex has $\lmin(K,G)=(m+2)(m+3)/(5m^2)$ , attained exactly on the radial quadratic. The proof splits $V_{m,2}$ into $S_m$ -isotypic sectors, reduces the pencil on each to a small Gram matrix, and evaluates these exactly with Dirichlet moments. The corollary is conditional on the identification (D11.8) and lives inside the candidate degree- $k$ dual-certificate framework. It rules out degree-two dual refuters and says nothing about degrees $k\ge3$ , non-polynomial tests or KLS.

Exact Dirichlet moments (Lemma D11.1) and the pair-fiber decomposition (Lemma D11.2). On each pair fiber, $\delta/s$ is uniform on $[-1,1]$ , which gives the pair-energy formula (D11.17); hence $K$ is finite and positive semidefinite on $V_{m,2}$ .
Symmetry: Lemma D11.3 and Lemma D11.4 show $V_{m,2}\cong\mathrm{triv}\oplus\mathrm{std}\otimes\R^2\oplus\mathrm X$ , with explicit generators $f_0$ , $(L,Q)$ and $F$ .
Lemma D11.5 shows that invariant forms do not couple different sectors. Positivity on each sector therefore reduces to a Gram matrix of seed vectors.
Sector computations from steps 1–2: in the trivial sector the ratio is exactly $\lambda^*$ (Lemma D11.6); the standard sector reduces to a $2\times2$ Gram matrix (Lemma D11.7); the two-row sector reduces to a scalar (Lemma D11.8).
Assembly: $K-\lambda^*G$ vanishes on the trivial sector. It is positive definite on the standard sector ((D11.41)) and on $\mathrm X$ . This proves (D11.5) together with its equality case.
Corollary Corollary D11.1, under its stated hypothesis: averaging a certificate over the root-frame atoms and applying the theorem gives $\eps\ge\lambda^*>\tfrac15$ , and also gives $\Lambda_{m,2}\ge\lambda^*$ .

Scope. This dossier proves an exact finite-dimensional spectral identity for the $A_{m-1}$ root-frame conditional-fiber form of Route F (Conditional fibers: inverse-variance frames of line resamplings) restricted to polynomial test functions of degree at most two, for every $m\ge3$ simultaneously. The statement was suggested by exact rational computations at $m=3,\dots,15$ recorded in provenance-stamped run artifacts (see Remark D11.1); no step of any proof below uses those computations. The main theorem is unconditional and self-contained. The corollary is stated only as a conditional, refutation-channel statement inside the candidate degree- $k$ dual-certificate framework for Conjecture 22.1; it makes no claim about the all-frame gate itself, about degrees $k\ge3$ , about non-polynomial tests, or about KLS.

Setting. Let $m\ge3$ , let $\Delta_{m-1}=\{p\in\R_+^m:\sum_ip_i=1\}$ , and let $P=(P_1,\dots,P_m)$ be uniform on $\Delta_{m-1}$ , i.e. $P\sim\Dir(1,\dots,1)$ ; write $\mu$ for its law. Put $H_0=\one^\perp$ , $d=m-1$ , $R_m=\sqrt{m(m+1)}$ , and

X=R_m\Bigl(P-\frac1m\one\Bigr)\in H_0,

(D11.1)

which is isotropic on $H_0$ (uniform-simplex root proposition of the certified dossier solutions/conditional-fiber-frame-structure.md; this normalization is not used before the corollary). The symmetric group $S_m$ acts on functions by permuting the coordinates of $p$ , and $\mu$ is exchangeable, so $S_m$ acts by isometries of $L^2(\mu)$ .

For $i<j$ let $\mathcal F_{ij}=\sigma\bigl((P_\ell)_{\ell\ne i,j}\bigr)$ , let $s_{ij}=P_i+P_j$ (an $\mathcal F_{ij}$ -measurable variable, since $s_{ij}=1-\sum_{\ell\ne i,j}P_\ell$ ), and write $\Var_{ij}$ and $\Cov_{ij}$ for conditional variance and covariance given $\mathcal F_{ij}$ . Define, for $f,g\in L^2(\mu)$ with all the conditional quantities finite,

K(f,g) =\frac{12}{m^2(m+1)}\sum_{i<j} \E\,\frac{\Cov_{ij}(f,g)}{s_{ij}^{\,2}}, \qquad G(f,g)=\Cov(f,g).

(D11.2)

The quadratic form $K(f,f)$ is exactly the root-frame conditional-fiber form $\mathcal D_{\rm root}(f)$ of (22.9), as proved in the certified dossier above (its equation defining the root form); for the main theorem, (D11.2) is simply taken as the definition, so the theorem does not depend on that identification. Both $K$ and $G$ depend only on the $\mu$ -equivalence classes of $f,g$ , vanish when one argument is a.e. constant, and are $S_m$ -invariant (exchangeability of $\mu$ and invariance of the family of pairs).

Let $W_{\rm poly}$ be the linear span of the monomials $p_i$ ( $1\le i\le m$ ) and $p_ip_j$ ( $1\le i\le j\le m$ ), i.e. all polynomials of degree at most 2 with zero constant term, and define the degree-two quotient

V_{m,2} =\bigl\{[f]:f\in W_{\rm poly}\bigr\} \subset L^2(\mu)\big/\R\one,

(D11.3)

where $[f]$ denotes the class of $f$ modulo additive constants. On $V_{m,2}$ the form $G=\Var$ is positive definite by construction, $K$ is positive semidefinite and finite (Lemma D11.2), and both are $S_m$ -invariant. The generalized minimum eigenvalue of the pencil is

\lmin(K,G)\big|_{V_{m,2}} =\min_{0\ne v\in V_{m,2}}\frac{K(v,v)}{G(v,v)}.

(D11.4)

Statements¶

The corollary below lives inside the candidate degree- $k$ dual-certificate framework for the simplex falsification channel of Conjecture 22.1 (the all-frame min–max $\Lambda_{m,k}$ and its exact dual certificates; an exploratory framework, not a statement of the manuscript). We restate the needed objects to be self-contained. Write $S(H_0)$ for the unit sphere of $H_0$ . For $\theta\in S(H_0)$ and $f$ a polynomial, let

\qJac_\theta[f] =\int_{\theta^\perp} \frac{\Var_{\mu_{\theta,z}}\bigl(f(z+T\theta)\bigr)} {\Var_{\mu_{\theta,z}}(T)} \dd\bar\mu_\theta(z)

(D11.6)

be the single-direction normalized fiber energy of the certified structural dossier (with its null- and zero-variance-fiber conventions), a finite quantity for polynomial $f$ by the certified factor-4 bound. An admissible frame is an even Borel probability $\rho$ on $S(H_0)$ with $(m-1)\int\theta\theta^T\dd\rho=I_{H_0}$ . A degree-2 dual certificate with objective $\eps$ is a finite family $c_1,\dots,c_R\in V_{m,2}$ , weights $w_1,\dots,w_R\ge0$ , and a symmetric matrix $M$ on $H_0$ such that

\sum_{r=1}^R w_r\,G(c_r,c_r)=1, \qquad \sum_{r=1}^R w_r\,\qJac_\theta[c_r]\le\theta^TM\theta \quad\text{for all }\theta\in S(H_0), \qquad \Tr M\le\eps.

(D11.7)

By the weak-duality computation reproduced in the proof below, such a certificate forces every admissible frame $\rho$ (for which the averaged form is defined) to have degree-two pencil gap at most $\eps$ ; a sequence of such certificates with $\eps_m\to0$ at the fixed degree $k=2$ would therefore be the decisive fixed-degree refuter of the route on the simplex.

Corollary D11.1 (conditional: no degree-two dual refuter)

Assume the identification, certified in solutions/conditional-fiber-frame-structure.md (nodes lem:conditional-fiber-form and prop:conditional-fiber-root-obstruction), of the pair-redistribution formula (D11.2) with the conditional-fiber form of the admissible $A_{m-1}$ root frame, i.e.

K(f,f)=(m-1)\int_{S(H_0)}\qJac_\theta[f]\dd\rho_{\rm root}(\theta), \qquad \rho_{\rm root}=\frac1{m(m-1)}\sum_{i\ne j}\delta_{(e_i-e_j)/\sqrt2}.

(D11.8)

Then for every $m\ge3$ :

every degree-2 dual certificate (D11.7) has
$\eps\;\ge\;\frac{(m+2)(m+3)}{5m^2}\;>\;\frac15;$
(D11.9)
consequently no sequence of degree-2 dual certificates with $\eps_m\to0$ exists, and within this framework any fixed-degree polynomial dual refutation of the conditional-fiber-frame route on the uniform simplex requires test degree $k\ge3$ ;
for every admissible frame $\rho$ whose averaged degree-two form $A_\rho(f,f)=(m-1)\int\qJac_\theta[f]\dd\rho(\theta)$ is defined on $V_{m,2}$ , the all-frame degree-two min–max satisfies
$\Lambda_{m,2} :=\sup_{\rho\ \text{admissible}} \ \min_{0\ne v\in V_{m,2}}\frac{A_\rho(v,v)}{G(v,v)} \;\ge\;\frac{(m+2)(m+3)}{5m^2}\;>\;\frac15 .$
(D11.10)

This corollary asserts nothing about upper bounds on $\Lambda_{m,2}$ , about $\Lambda_{m,k}$ for $k\ge3$ , about non-polynomial test functions, about the truth value of Conjecture 22.1, or about KLS.

Preliminaries: moments and pair fibers¶

Lemma D11.1 (Dirichlet monomial moments)

For $a\in\mathbb N^m$ with $\abs a=\sum_ia_i$ ,

\E\prod_{i=1}^mP_i^{a_i} =\frac{(m-1)!\,\prod_{i=1}^ma_i!}{(m-1+\abs a)!}.

(D11.11)

In particular, with $D_r=m(m+1)\cdots(m+r-1)$ and distinct indices $i,j,k,l$ :

\begin{gathered} \E P_i=\frac1m,\qquad \E P_i^2=\frac2{D_2},\qquad \E P_iP_j=\frac1{D_2},\\ \E P_i^3=\frac6{D_3},\qquad \E P_i^2P_j=\frac2{D_3},\qquad \E P_i^4=\frac{24}{D_4},\qquad \E P_i^3P_j=\frac6{D_4},\\ \E P_i^2P_j^2=\frac4{D_4},\qquad \E P_i^2P_jP_k=\frac2{D_4},\qquad \E P_iP_jP_kP_l=\frac1{D_4}. \end{gathered}

(D11.12)

Proof D11.1

Let $E_1,\dots,E_m$ be i.i.d. standard exponentials and $S=\sum_iE_i$ . By the classical Gamma–Dirichlet factorization, $(E_1/S,\dots,E_m/S)\sim\Dir(1,\dots,1)$ and is independent of $S\sim\GammaLaw(m,1)$ . Hence

\prod_ia_i! =\E\prod_iE_i^{a_i} =\E\Bigl[S^{\abs a}\prod_iP_i^{a_i}\Bigr] =\E S^{\abs a}\cdot\E\prod_iP_i^{a_i} =\frac{(m-1+\abs a)!}{(m-1)!}\,\E\prod_iP_i^{a_i},

(D11.13)

using $\E S^{r}=\Gamma(m+r)/\Gamma(m)$ . Rearranging gives (D11.11); the table is immediate.

The following consequences are used repeatedly ( $i,j,a,b$ distinct):

\E s_{ij}=\frac2m,\qquad \E s_{ij}^2=\frac{2\cdot2+2\cdot1}{D_2}=\frac6{D_2},\qquad \E(P_a-P_b)^2=\frac{2\cdot2-2\cdot1}{D_2}=\frac2{D_2}.

(D11.14)

Lemma D11.2 (pair-fiber decomposition of degree-two energies)

Fix $i<j$ and abbreviate $s=s_{ij}$ , $\delta=P_i-P_j$ . Then, conditionally on $\mathcal F_{ij}$ and on the event $\{s>0\}$ (whose complement is $\mu$ -null), the ratio $\delta/s$ is uniform on $[-1,1]$ and independent of $\mathcal F_{ij}$ ; consequently

\Var_{ij}(\delta)=\frac{s^2}3,\qquad \Var_{ij}(\delta^2)=\frac{4s^4}{45},\qquad \Cov_{ij}(\delta,\delta^2)=0.

(D11.15)

Every polynomial $f$ of degree at most 2 can be written on the pair fiber, via the substitution $p_i=\tfrac{s+\delta}2$ , $p_j=\tfrac{s-\delta}2$ , uniquely as

f=\alpha_f+\beta_f\,\delta+\gamma_f\,\delta^2,

(D11.16)

where $\alpha_f,\beta_f,\gamma_f$ are polynomials in $s$ and $(p_\ell)_{\ell\ne i,j}$ (hence $\mathcal F_{ij}$ -measurable), with $\deg\beta_f\le1$ and $\gamma_f$ constant. For two such polynomials $f,g$ ,

\E\,\frac{\Cov_{ij}(f,g)}{s^2} =\frac13\,\E\bigl[\beta_f\beta_g\bigr] +\frac4{45}\,\E\bigl[\gamma_f\gamma_g\,s^2\bigr],

(D11.17)

a finite quantity. In particular $K$ is finite, symmetric, bilinear, positive semidefinite, and well defined on $V_{m,2}$ .

Proof D11.2

The law $\mu$ has constant density with respect to $(m-1)$ -dimensional Hausdorff measure on $\Delta_{m-1}$ . Conditioning on $(P_\ell)_{\ell\ne i,j}$ fixes $s=1-\sum_{\ell\ne i,j}P_\ell$ and leaves $(P_i,P_j)$ with a constant conditional density on the segment $\{(x,y):x,y\ge0,\ x+y=s\}$ ; parametrized by $P_i\in[0,s]$ this is the uniform law, so $\delta/s=2P_i/s-1$ is uniform on $[-1,1]$ , with conditional law not depending on $\mathcal F_{ij}$ . The event $\{s=0\}$ is the face $\{P_i=P_j=0\}$ , which is $\mu$ -null. With $V=\delta/s$ uniform on $[-1,1]$ : $\E V=\E V^3=0$ , $\E V^2=\tfrac13$ , $\E V^4=\tfrac15$ , whence $\Var(\delta\mid\mathcal F_{ij})=s^2/3$ , $\Var(\delta^2\mid\mathcal F_{ij})=s^4(\tfrac15-\tfrac19)=\tfrac{4s^4}{45}$ , and $\Cov(\delta,\delta^2\mid\mathcal F_{ij})=s^3\,\E V^3=0$ , proving (D11.15).

For (D11.16), substitute $p_i=\tfrac{s+\delta}2$ , $p_j=\tfrac{s-\delta}2$ into each monomial of $f$ :

\begin{gathered} p_i=\tfrac s2+\tfrac\delta2,\qquad p_j=\tfrac s2-\tfrac\delta2,\qquad p_i^2=\tfrac{s^2}4+\tfrac s2\,\delta+\tfrac14\,\delta^2,\qquad p_j^2=\tfrac{s^2}4-\tfrac s2\,\delta+\tfrac14\,\delta^2,\\ p_ip_j=\tfrac{s^2}4-\tfrac14\,\delta^2,\qquad p_ip_\ell=\tfrac{p_\ell}2\,s+\tfrac{p_\ell}2\,\delta,\qquad p_jp_\ell=\tfrac{p_\ell}2\,s-\tfrac{p_\ell}2\,\delta \quad(\ell\ne i,j), \end{gathered}

(D11.18)

while monomials not involving $p_i,p_j$ contribute to $\alpha_f$ only. Collecting powers of $\delta$ gives (D11.16) with the stated degrees; uniqueness holds because $(1,\delta,\delta^2)$ are linearly independent functions of $\delta$ on any fiber with $s>0$ . Since $\alpha_f$ is $\mathcal F_{ij}$ -measurable, bilinearity of conditional covariance and (D11.15) give

\Cov_{ij}(f,g) =\beta_f\beta_g\,\frac{s^2}3+\gamma_f\gamma_g\,\frac{4s^4}{45},

(D11.19)

the two cross terms carrying the vanishing factor $\Cov_{ij}(\delta,\delta^2)=0$ . Dividing by $s^2$ on $\{s>0\}$ and taking expectations proves (D11.17); the integrands are polynomials, hence bounded on the simplex, so the expectation is finite. Positive semidefiniteness of $K$ is clear from $\Var_{ij}\ge0$ . Conditional covariances depend only on the a.e. classes of $f,g$ and vanish when one argument is a.e. constant, so $K$ descends to $V_{m,2}$ .

Symmetry structure of the degree-two quotient¶

Throughout, “module” means a finite-dimensional real representation of $S_m$ ; all maps are $S_m$ -equivariant unless stated otherwise. We use the following standard facts about a finite group $\Gamma$ acting on real vector spaces; each has a one-paragraph classical proof, recalled for self-containment.

(Complete reducibility.) Averaging any inner product over $\Gamma$ produces an invariant inner product; the orthogonal complement of a submodule is a submodule; hence every module is a direct sum of irreducibles, every submodule has an invariant complement, and every quotient of a module is isomorphic to a submodule. Consequently, in a short exact sequence of modules the multiplicity of each irreducible is additive.
(Schur.) A nonzero map between irreducibles is an isomorphism; the image of an irreducible submodule under any equivariant map is 0 or isomorphic to it. The $E$ -isotypic component $W_E(V)$ of a module $V$ is the sum of all submodules isomorphic to the irreducible $E$ ; equivariant maps send $E$ -isotypic parts into $E$ -isotypic parts, and $V$ is the direct sum of its isotypic components.
(Orbit counting for permutation modules.) If $\Gamma$ acts on finite sets $S,T$ , the space of equivariant linear maps $\R^S\to\R^T$ has dimension equal to the number of $\Gamma$ -orbits on $T\times S$ : an equivariant matrix is exactly a matrix constant on orbits.
(Multiplicity via Hom.) If $E$ is irreducible with $\operatorname{End}_\Gamma(E)=\R\,\mathrm{id}$ , then for every module $V$ the multiplicity of $E$ in $V$ equals $\dim\operatorname{Hom}_\Gamma(E,V)$ , and $W_E(V)\cong E\otimes\R^{k}$ with $k$ that multiplicity. (Decompose $V$ into irreducibles by (F1); by (F2) and Schur, $\operatorname{Hom}_\Gamma(E,\cdot)$ of each summand is $\R$ for summands $\cong E$ and 0 otherwise.) In particular the multiplicity of $E$ is monotone under submodules and quotients.
(Invariant bilinear forms on an irreducible.) If $E$ is irreducible with $\operatorname{End}_\Gamma(E)=\R\,\mathrm{id}$ , the space of invariant bilinear forms on $E$ is one-dimensional, spanned by any fixed invariant inner product $J_E$ : a form corresponds to an equivariant map $E\to E^*\cong E$ , i.e. to an element of $\operatorname{End}_\Gamma(E)=\R$ .

Lemma D11.3 (the three irreducibles)

Let $M=\R^m$ with permuted coordinates and, for $m\ge3$ , let $N=\R^{\binom{[m]}2}$ be the permutation module on unordered pairs $\{i,j\}$ , with basis $(x_{ij})_{i<j}$ . Then:

$M=\mathrm{triv}\oplus\mathrm{std}$ , where $\mathrm{triv}=\R\one$ and $\mathrm{std}=\{a:\sum_ia_i=0\}$ is irreducible of dimension $m-1$ with $\operatorname{End}_{S_m}(\mathrm{std})=\R\,\mathrm{id}$ , and $\mathrm{std}\not\cong\mathrm{triv}$ .
For $m\ge4$ , $N=\mathrm{triv}\oplus\mathrm{std}\oplus\mathrm X$ with each summand of multiplicity one, where $\mathrm X$ is irreducible of dimension $m(m-3)/2$ , satisfies $\operatorname{End}_{S_m}(\mathrm X)=\R\,\mathrm{id}$ , and $\mathrm X\not\cong\mathrm{triv},\mathrm{std}$ . For $m=3$ , $N=\mathrm{triv}\oplus\mathrm{std}$ and the summand $\mathrm X$ is absent.

Proof D11.3

(1) $S_m$ has two orbits on $[m]^2$ (diagonal, off-diagonal), so $\dim\operatorname{End}(M)=2$ by (F3). Also $M^{S_m}=\R\one$ is one-dimensional, so $\mathrm{triv}$ has multiplicity one; let $C=\one^\perp$ be its invariant complement. Then $C^{S_m}=0$ , so $\operatorname{Hom}(\mathrm{triv},C)=0$ and $2=\dim\operatorname{End}(M) =\dim\operatorname{End}(\mathrm{triv})+\dim\operatorname{End}(C) =1+\dim\operatorname{End}(C)$ . A decomposable module has endomorphism algebra of dimension at least 2 (the two projections), so $\operatorname{End}(C)=\R\,\mathrm{id}$ forces $C$ irreducible. Finally $C\not\cong\mathrm{triv}$ because $C^{S_m}=0$ . Set $\mathrm{std}=C$ .

(2) Let $m\ge4$ . The orbits of $S_m$ on (ordered) pairs of 2-subsets are classified by $\abs{A\cap B}\in\{2,1,0\}$ , all realized, so $\dim\operatorname{End}(N)=3$ . The orbits on $[m]\times\binom{[m]}2$ are classified by $i\in A$ versus $i\notin A$ , so $\dim\operatorname{Hom}(M,N)=2$ . As $N^{S_m}$ is one-dimensional (one orbit of 2-subsets), $\mathrm{triv}$ has multiplicity one in $N$ and $\dim\operatorname{Hom}(\mathrm{std},N) =\dim\operatorname{Hom}(M,N)-\dim\operatorname{Hom}(\mathrm{triv},N)=2-1=1$ , so by (F4) $\mathrm{std}$ has multiplicity exactly one. Decompose $N=\mathrm{triv}\oplus S'\oplus \mathrm X$ with $S'\cong\mathrm{std}$ and $\mathrm X$ the sum of the remaining irreducible summands, which contains no copy of $\mathrm{triv}$ or $\mathrm{std}$ . Then cross-Homs between the three summands vanish and $3=\dim\operatorname{End}(N)=1+1+\dim\operatorname{End}(\mathrm X)$ , so as in (1) $\mathrm X$ is irreducible with $\operatorname{End}(\mathrm X)=\R\,\mathrm{id}$ , and $\mathrm X\not\cong\mathrm{triv},\mathrm{std}$ (otherwise those multiplicities would exceed one). Its dimension is $\binom m2-1-(m-1)=m(m-3)/2$ . For $m=3$ there is no orbit with $\abs{A\cap B}=0$ , so $\dim\operatorname{End}(N)=2$ and the same argument yields $N=\mathrm{triv}\oplus\mathrm{std}$ (dimensions: $3=1+2$ ).

Lemma D11.4 (isotypic decomposition of

V_{m,2}

)

For $m\ge3$ , as $S_m$ -modules,

V_{m,2}\;\cong\; \mathrm{triv}\ \oplus\ \mathrm{std}\otimes\R^2\ \oplus\ \mathrm X, \qquad \dim V_{m,2}=\frac{(m-1)(m+2)}2,

(D11.20)

where the $\mathrm X$ -summand is absent when $m=3$ . Moreover:

the $\mathrm{triv}$ -isotypic component is $W_{\rm triv}=\R\,[f_0]$ with $f_0=\sum_kp_k^2$ , and $[f_0]\ne0$ ;
the classes $L=[p_1-p_2]$ and $Q=[p_1^2-p_2^2]$ are nonzero, linearly independent, and lie in the $\mathrm{std}$ -isotypic component $W_{\rm std}$ ; they are the images of the common seed vector $v_0=e_1-e_2\in\mathrm{std}$ under the two equivariant maps $T_1:a\mapsto[\textstyle\sum_ia_ip_i]$ and $T_2:a\mapsto[\textstyle\sum_ia_ip_i^2]$ , whose images span $W_{\rm std}$ ;
for $m\ge4$ , the class $F=[(p_1-p_3)(p_2-p_4)]$ is nonzero and lies in the $\mathrm X$ -isotypic component $W_{\mathrm X}$ , which it generates.

Proof D11.4

Kernel of the quotient map. Define the equivariant surjection $\Psi:W_{\rm poly}\to V_{m,2}$ , $f\mapsto[f]$ . We claim

\ker\Psi =\Bigl\{\,b_0\textstyle\sum_ip_i +\bigl(\textstyle\sum_ip_i-1\bigr)\textstyle\sum_ib_ip_i \;:\;b_0\in\R,\ b\in\R^m\Bigr\} \;\cong\;\mathrm{triv}\oplus M \;\cong\;2\,\mathrm{triv}\oplus\mathrm{std}.

(D11.21)

Indeed $f\in\ker\Psi$ means $f=c$ $\mu$ -a.e. for some $c\in\R$ . The support $\Delta_{m-1}$ has nonempty relative interior in the hyperplane $H=\{\sum_ip_i=1\}$ , and a polynomial vanishing on a relatively open subset of an affine subspace vanishes on the whole subspace; hence $f-c$ vanishes on $H$ . Choosing affine coordinates in which $t=\sum_ip_i-1$ is one coordinate and dividing by $t$ (polynomial division in $t$ , degree bookkeeping: $\deg\le2$ ), we get $f=c+(\sum_ip_i-1)\ell$ with $\ell$ affine, say $\ell=b_0+\sum_ib_ip_i$ . Since $f$ has zero constant term, $c=b_0$ , which yields exactly the displayed set; conversely every element of that set is $\mu$ -a.e. constant (equal to $b_0$ ). The parametrization $(b_0,b)\mapsto f$ is linear, injective (compare first the quadratic homogeneous part $(\sum_ip_i)(\sum_ib_ip_i)$ , which vanishes only if $b=0$ since the polynomial ring is a domain; then $b_0\sum p_i=0$ forces $b_0=0$ ), and equivariant for the action fixing $b_0$ and permuting $b$ . This proves (D11.21).

Multiplicities. The monomial families $(p_i)$ , $(p_i^2)$ , $(p_ip_j)_{i<j}$ are linearly independent in the polynomial ring and are permuted by $S_m$ exactly as the bases of $M$ , $M$ , $N$ ; hence $W_{\rm poly}\cong M\oplus M\oplus N$ , whose isotypic multiplicities are (by Lemma D11.3): $\mathrm{triv}$ : 3, $\mathrm{std}$ : 3, $\mathrm X$ : 1 (for $m\ge4$ ; $\mathrm X$ absent at $m=3$ ). By (F1) applied to $0\to\ker\Psi\to W_{\rm poly}\to V_{m,2}\to0$ and (D11.21), the multiplicities in $V_{m,2}$ are $\mathrm{triv}$ : $3-2=1$ , $\mathrm{std}$ : $3-1=2$ , $\mathrm X$ : $1-0=1$ , and no other irreducible occurs. This is (D11.20); the dimension is $1+2(m-1)+m(m-3)/2=(m-1)(m+2)/2$ (also valid at $m=3$ ).

(1). $[f_0]$ is $S_m$ -fixed. It is nonzero: if $f_0\in\ker\Psi$ , comparing quadratic homogeneous parts in (D11.21) gives $\sum_ip_i^2=(\sum_ip_i)(\sum_ib_ip_i)$ ; the coefficient of $p_i^2$ forces $b_i=1$ for all $i$ , but then the coefficient of $p_1p_2$ on the right is $2\ne0$ , a contradiction. Since the $\mathrm{triv}$ -multiplicity is one, $W_{\rm triv}=\R[f_0]$ .

(2). $T_1,T_2$ are restrictions to $\mathrm{std}\subset M$ of the equivariant maps $a\mapsto[\sum a_ip_i]$ , $a\mapsto[\sum a_ip_i^2]$ , so their images lie in $W_{\rm std}$ by (F2), and $L=T_1v_0$ , $Q=T_2v_0$ . Linear independence of $L,Q$ in $V_{m,2}$ : suppose $aL+bQ\in\ker\Psi$ , i.e. $a(p_1-p_2)+b(p_1^2-p_2^2) =b_0\sum_ip_i+(\sum_ip_i-1)\sum_ib_ip_i$ . Quadratic parts: $b(p_1^2-p_2^2)=(\sum_ip_i)(\sum_ib_ip_i)$ ; the $p_i^2$ coefficients give $b_1=b$ , $b_2=-b$ , $b_\ell=0$ ( $\ell\ge3$ ), and then the $p_1p_3$ coefficient (using $m\ge3$ ) gives $0=b_1+b_3=b$ , so $b=0$ and $b\equiv0$ . The linear part then reads $a(p_1-p_2)=b_0\sum_ip_i$ , forcing $a=b_0=-a$ , i.e.\ $a=0$ . In particular $L,Q\ne0$ , so $T_1,T_2\ne0$ and, $\mathrm{std}$ being irreducible, $T_1(\mathrm{std})\cong T_2(\mathrm{std})\cong\mathrm{std}$ . The two image submodules are distinct: if $T_1(\mathrm{std})=T_2(\mathrm{std})=:U$ , then $T_1,T_2\in\operatorname{Hom}(\mathrm{std},U)$ , a one-dimensional space by (F4)–(F5) (as $U\cong\mathrm{std}$ ), so $T_2=cT_1$ and $Q=cL$ , contradicting independence. Distinct irreducible submodules intersect in 0, so $T_1(\mathrm{std})\oplus T_2(\mathrm{std})\subseteq W_{\rm std}$ has dimension $2(m-1)=\dim W_{\rm std}$ ; hence the images span $W_{\rm std}$ .

(3). Let $m\ge4$ and let $\psi:N\to V_{m,2}$ be the equivariant map $x_{ij}\mapsto[p_ip_j]$ . Consider

v=x_{12}-x_{23}-x_{14}+x_{34}\in N, \qquad \psi(v)=[p_1p_2-p_2p_3-p_1p_4+p_3p_4]=[(p_1-p_3)(p_2-p_4)]=F.

(D11.22)

We check $v\in W_{\mathrm X}(N)$ . With the standard (invariant) inner product on $N$ , the isotypic components of $N$ are mutually orthogonal. The $\mathrm{triv}$ -component is spanned by $\sum_{i<j}x_{ij}$ , and the $\mathrm{std}$ -component lies in the image of the equivariant map $A:M\to N$ , $e_k\mapsto u_k:=\sum_{\ell\ne k}x_{k\ell}$ (indeed $A(\mathrm{std})\ne0$ because $u_1-u_2$ has coefficient 1 on $x_{13}$ , and the $\mathrm{std}$ -multiplicity of $N$ is one, so $W_{\rm std}(N)=A(\mathrm{std})$ ). Now $\langle v,\sum_{i<j}x_{ij}\rangle=1-1-1+1=0$ and, for every $k$ , $\langle v,u_k\rangle=0$ : for $k=1$ the pairs of $v$ containing 1 are $x_{12}$ ( $+$ ) and $x_{14}$ ( $-$ ); for $k=2$ : $x_{12}$ ( $+$ ), $x_{23}$ ( $-$ ); for $k=3$ : $x_{23}$ ( $-$ ), $x_{34}$ ( $+$ ); for $k=4$ : $x_{14}$ ( $-$ ), $x_{34}$ ( $+$ ); for $k\ge5$ : none. Hence $v\perp W_{\rm triv}(N)\oplus W_{\rm std}(N)$ , i.e.\ $v\in W_{\mathrm X}(N)$ . By (F2), $F=\psi(v)\in W_{\mathrm X}(V_{m,2})$ . Finally $F\ne0$ : if $(p_1-p_3)(p_2-p_4)\in\ker\Psi$ , comparing quadratic parts with (D11.21) gives $(p_1-p_3)(p_2-p_4)=(\sum_ip_i)(\sum_ib_ip_i)$ ; the left side has no $p_i^2$ terms, so all $b_i=0$ , but the left side has $p_1p_2$ -coefficient 1, a contradiction. Since $W_{\mathrm X}$ has multiplicity one and $\psi|_{W_{\mathrm X}(N)}$ is injective (its restriction to the irreducible $W_{\mathrm X}(N)$ is nonzero), $F$ generates $W_{\mathrm X}$ as a module.

Lemma D11.5 (invariant pencil reduction)

Let $V$ be a module with invariant symmetric bilinear forms $B$ (arbitrary) and $G$ (positive definite), and let $V=\bigoplus_EW_E$ be its isotypic decomposition over pairwise non-isomorphic irreducibles $E$ with $\operatorname{End}(E)=\R\,\mathrm{id}$ . Then:

$B(W_E,W_{E'})=0$ for $E\not\cong E'$ ; hence $B$ is positive semidefinite (resp. definite) on $V$ if and only if it is so on each $W_E$ .
Suppose $W_E$ is spanned by the images of equivariant maps $T_1,\dots,T_k:E\to V$ that are linearly independent in $\operatorname{Hom}_{S_m}(E,V)$ , and fix $0\ne v_0\in E$ . Then the map $\Phi:E\otimes\R^k\to W_E$ , $u\otimes\varepsilon_a\mapsto T_au$ , is an isomorphism, and there is a symmetric $k\times k$ matrix $\widehat B$ with
$B(T_au,T_bu')=\widehat B_{ab}\,J_E(u,u') \qquad(u,u'\in E,\ 1\le a,b\le k),$
(D11.23)
where $J_E$ is a fixed invariant inner product on $E$ . Consequently $B$ is positive semidefinite (resp. definite) on $W_E$ if and only if the $k\times k$ Gram matrix $\bigl(B(T_av_0,T_bv_0)\bigr)_{a,b}=J_E(v_0,v_0)\,\widehat B$ is positive semidefinite (resp. definite).

Proof D11.5

(1) The invariant form $B$ and an invariant inner product on $W_{E'}$ induce an equivariant map $W_E\to W_{E'}^*\cong W_{E'}$ . By (F2) and Schur, every equivariant map from an $E$ -isotypic to an $E'$ -isotypic module is zero when $E\not\cong E'$ . Hence $B(W_E,W_{E'})=0$ , and $B$ decomposes as the orthogonal (with respect to the decomposition) sum of its restrictions.

(2) $\Phi$ is equivariant and surjective (images span), and $\dim(E\otimes\R^k)=k\dim E=\dim W_E$ because the multiplicity of $E$ in $W_E$ equals $\dim\operatorname{Hom}(E,W_E)\ge k$ by linear independence, while surjectivity gives $\le k$ ; so $\Phi$ is an isomorphism. For fixed $a,b$ , the bilinear form $(u,u')\mapsto B(T_au,T_bu')$ on $E$ is invariant, hence by (F5) equals $\widehat B_{ab}J_E(u,u')$ for a unique scalar $\widehat B_{ab}$ ; symmetry of $B$ and of $J_E$ gives $\widehat B_{ab}=\widehat B_{ba}$ . Now for $f=\Phi\bigl(\sum_\alpha u^\alpha\otimes s_\alpha\bigr)$ with $(u^\alpha)_\alpha$ a $J_E$ -orthonormal basis of $E$ and $s_\alpha\in\R^k$ ,

B(f,f) =\sum_{\alpha,\gamma}\sum_{a,b} s_\alpha^a s_\gamma^b\,B(T_au^\alpha,T_bu^\gamma) =\sum_{\alpha,\gamma}J_E(u^\alpha,u^\gamma)\,s_\alpha^T\widehat B\,s_\gamma =\sum_\alpha s_\alpha^T\widehat B\,s_\alpha.

(D11.24)

Since every element of $W_E$ is such an $f$ and the vectors $(s_\alpha)_\alpha$ are arbitrary, $B\ge0$ (resp. $>0$ ) on $W_E$ iff $\widehat B\succeq0$ (resp. $\succ0$ ), iff the Gram matrix $J_E(v_0,v_0)\widehat B$ is so ( $J_E(v_0,v_0)>0$ ).

Sector computations¶

Throughout this subsection $E_{ij}(f,g)$ denotes the pair energy (D11.17) and we tabulate the coefficients $(\beta,\gamma)$ of Lemma D11.2 for each seed and each pair type; pairs not listed have $\beta=\gamma=0$ . Recall (D11.14) and $D_2=m(m+1)$ , $D_3=D_2(m+2)$ , $D_4=D_3(m+3)$ .

Proof D11.6

For every pair $(i,j)$ , by (D11.18), $p_i^2+p_j^2=\tfrac{s^2}2+\tfrac{\delta^2}2$ , so $(\beta,\gamma)=(0,\tfrac12)$ and, by (D11.17) and (D11.14), $E_{ij}(f_0,f_0)=\tfrac4{45}\cdot\tfrac14\,\E s^2=\tfrac1{45}\cdot\tfrac6{D_2} =\tfrac2{15D_2}$ . Summing over the $\binom m2$ pairs and inserting the prefactor of (D11.2),

K(f_0,f_0) =\frac{12}{m^2(m+1)}\cdot\frac{m(m-1)}2\cdot\frac2{15\,m(m+1)} =\frac{4(m-1)}{5m^2(m+1)^2}.

(D11.26)

For the variance, Lemma D11.1 gives $\E f_0=m\cdot\tfrac2{D_2}=\tfrac2{m+1}$ and $\E f_0^2=m\,\E P_1^4+m(m-1)\,\E P_1^2P_2^2 =\tfrac{24m+4m(m-1)}{D_4}=\tfrac{4m(m+5)}{D_4}$ , so

G(f_0,f_0) =\frac{4(m+5)}{(m+1)(m+2)(m+3)}-\frac4{(m+1)^2} =\frac{4\bigl[(m+5)(m+1)-(m+2)(m+3)\bigr]}{(m+1)^2(m+2)(m+3)} =\frac{4(m-1)}{(m+1)^2(m+2)(m+3)},

(D11.27)

using $(m+5)(m+1)-(m+2)(m+3)=m-1$ . The quotient is $(m+2)(m+3)/(5m^2)$ . Finally $\abs X^2=m(m+1)\sum_i(p_i-\tfrac1m)^2=m(m+1)\bigl(f_0-\tfrac1m\bigr)$ , so $[\abs X^2-(m-1)]=m(m+1)[f_0]$ .

Proof D11.7

Energies. The $(\beta,\gamma)$ tables from (D11.18), with $j$ always denoting an index $\ge3$ distinct from $1,2$ :

\begin{array}{l|cc|cc} \text{pair} & \beta_L & \gamma_L & \beta_Q & \gamma_Q\\\hline (1,2) & 1 & 0 & s & 0\\ (1,j) & \tfrac12 & 0 & \tfrac s2 & \tfrac14\\ (2,j) & -\tfrac12 & 0 & -\tfrac s2 & -\tfrac14 \end{array}

(D11.29)

(For $(1,j)$ : $L=\tfrac{s+\delta}2-p_2$ and $Q=\tfrac{s^2}4+\tfrac s2\delta+\tfrac{\delta^2}4-p_2^2$ ; for $(2,j)$ , with $\delta=p_2-p_j$ : $L=p_1-\tfrac{s+\delta}2$ , $Q=p_1^2-\tfrac{s^2}4-\tfrac s2\delta-\tfrac{\delta^2}4$ ; for $(1,2)$ : $L=\delta$ , $Q=s\delta$ .) By (D11.17)–(D11.14):

\begin{aligned} E_{12}(L,L)&=\tfrac13, & E_{1j}(L,L)=E_{2j}(L,L)&=\tfrac1{12},\\ E_{12}(L,Q)&=\tfrac13\E s=\tfrac2{3m}, & E_{1j}(L,Q)=E_{2j}(L,Q)&=\tfrac13\cdot\tfrac{\E s}4=\tfrac1{6m},\\ E_{12}(Q,Q)&=\tfrac13\E s^2=\tfrac2{D_2}, & E_{1j}(Q,Q)=E_{2j}(Q,Q) &=\tfrac13\cdot\tfrac{\E s^2}4+\tfrac4{45}\cdot\tfrac{\E s^2}{16} =\tfrac{16}{180}\,\E s^2=\tfrac8{15D_2}. \end{aligned}

(D11.30)

There are $2(m-2)$ pairs of the types $(1,j),(2,j)$ , so

\begin{aligned} \sum_{i<j}E_{ij}(L,L)&=\tfrac13+\tfrac{2(m-2)}{12}=\tfrac m6,\\ \sum_{i<j}E_{ij}(L,Q)&=\tfrac2{3m}+\tfrac{2(m-2)}{6m}=\tfrac{2+(m-2)}{3m} =\tfrac13,\\ \sum_{i<j}E_{ij}(Q,Q)&=\tfrac2{D_2}+\tfrac{16(m-2)}{15D_2} =\tfrac{2(8m-1)}{15D_2}. \end{aligned}

(D11.31)

Multiplying by $12/(m^2(m+1))$ gives the three $K$ -entries in (D11.28).

Variances. By symmetry $\E L=\E Q=0$ , so, by Lemma D11.1,

\begin{aligned} G(L,L)&=\E(P_1-P_2)^2=\tfrac2{D_2},\\ G(L,Q)&=\E\bigl[(P_1-P_2)(P_1^2-P_2^2)\bigr] =\E\bigl[P_1^3+P_2^3-P_1P_2(P_1+P_2)\bigr] =\tfrac{12-4}{D_3}=\tfrac8{D_3},\\ G(Q,Q)&=\E(P_1^2-P_2^2)^2 =2\,\E P_1^4-2\,\E P_1^2P_2^2=\tfrac{48-8}{D_4}=\tfrac{40}{D_4}. \end{aligned}

(D11.32)

Proof D11.8

Energy. The $(\beta,\gamma)$ table, computed from (D11.18) (here $j\ge5$ ; in each row $\delta=p_i-p_j$ for the listed pair $(i,j)$ , $i<j$ ):

\begin{array}{l|cc|l} \text{pair} & \beta_F & \gamma_F & \text{derivation}\\\hline (1,3) & p_2-p_4 & 0 & F=\delta\,(p_2-p_4)\\ (2,4) & p_1-p_3 & 0 & F=(p_1-p_3)\,\delta\\ (1,2) & \tfrac{p_3-p_4}2 & -\tfrac14 & F=\bigl(\tfrac{s+\delta}2-p_3\bigr)\bigl(\tfrac{s-\delta}2-p_4\bigr)\\ (3,4) & \tfrac{p_1-p_2}2 & -\tfrac14 & F=\bigl(p_1-\tfrac{s+\delta}2\bigr)\bigl(p_2-\tfrac{s-\delta}2\bigr)\\ (1,4) & \tfrac{p_2-p_3}2 & \tfrac14 & F=\bigl(\tfrac{s+\delta}2-p_3\bigr)\bigl(p_2-\tfrac{s-\delta}2\bigr)\\ (2,3) & \tfrac{p_1-p_4}2 & \tfrac14 & F=\bigl(p_1-\tfrac{s-\delta}2\bigr)\bigl(\tfrac{s+\delta}2-p_4\bigr)\\ (1,j) & \tfrac{p_2-p_4}2 & 0 & F=\bigl(\tfrac{s+\delta}2-p_3\bigr)(p_2-p_4)\\ (2,j) & \tfrac{p_1-p_3}2 & 0 & \\ (3,j) & -\tfrac{p_2-p_4}2 & 0 & \\ (4,j) & -\tfrac{p_1-p_3}2 & 0 & \end{array}

(D11.34)

For instance, for the pair $(1,2)$ : $F=\bigl[(\tfrac s2-p_3)+\tfrac\delta2\bigr] \bigl[(\tfrac s2-p_4)-\tfrac\delta2\bigr] =\alpha+\tfrac{p_3-p_4}2\,\delta-\tfrac14\,\delta^2$ with $\alpha$ $\mathcal F_{12}$ -measurable; the rows $(3,4)$ , $(1,4)$ , $(2,3)$ are identical expansions. By (D11.17)–(D11.14),

\begin{aligned} E_{13}(F,F)=E_{24}(F,F) &=\tfrac13\,\E(P_2-P_4)^2=\tfrac2{3D_2},\\ E_{12}(F,F)=E_{34}(F,F)=E_{14}(F,F)=E_{23}(F,F) &=\tfrac13\cdot\tfrac{\E(P_3-P_4)^2}4+\tfrac4{45}\cdot\tfrac{\E s^2}{16} =\tfrac1{6D_2}+\tfrac1{30D_2}=\tfrac1{5D_2},\\ E_{ij}(F,F)\ \ (i\in\{1,2,3,4\},\ j\ge5) &=\tfrac13\cdot\tfrac{\E(P_a-P_b)^2}4=\tfrac1{6D_2} \qquad(4(m-4)\text{ pairs}). \end{aligned}

(D11.35)

Hence

\sum_{i<j}E_{ij}(F,F) =\frac1{D_2}\Bigl[\frac43+\frac45+\frac{4(m-4)}6\Bigr] =\frac1{D_2}\cdot\frac{20+12+10(m-4)}{15} =\frac{2(5m-4)}{15\,D_2},

(D11.36)

and $K(F,F)=\tfrac{12}{m^2(m+1)}\cdot\tfrac{2(5m-4)}{15m(m+1)} =\tfrac{8(5m-4)}{5m^3(m+1)^2}$ .

Variance. $\E F=\E P_1P_2-\E P_1P_4-\E P_2P_3+\E P_3P_4=0$ , and expanding $F^2=(p_1^2-2p_1p_3+p_3^2)(p_2^2-2p_2p_4+p_4^2)$ termwise with Lemma D11.1 (all four indices distinct),

\E F^2 =\frac{4\cdot4-8\cdot2+4\cdot1}{D_4} =\frac4{D_4},

(D11.37)

the three groups being: the four square–square products, each with $\E P_a^2P_b^2=4/D_4$ , totalling $16/D_4$ ; the four products of a square with a $-2p_bp_c$ factor, each contributing $-2\,\E P_a^2P_bP_c=-4/D_4$ , totalling $-16/D_4$ ; and the single product $(-2p_1p_3)(-2p_2p_4)$ , contributing $4\,\E P_1P_2P_3P_4=4/D_4$ . The quotient follows by $G(F,F)=\E F^2$ and $D_4=m(m+1)(m+2)(m+3)$ .

Proof of Theorem D11.1¶

Proof D11.9

Write $\lambda^*=(m+2)(m+3)/(5m^2)$ and let $\Phi=K-\lambda^*G$ , an $S_m$ -invariant symmetric bilinear form on $V_{m,2}$ (Lemma D11.2). By Lemma D11.4 and Lemma D11.5(1),

\Phi(f,f)=\Phi(f_{\rm triv},f_{\rm triv}) +\Phi(f_{\rm std},f_{\rm std})+\Phi(f_{\mathrm X},f_{\mathrm X})

(D11.38)

for the isotypic components of any $f\in V_{m,2}$ (the $\mathrm X$ -term absent at $m=3$ ). We treat the three sectors.

Trivial sector. $W_{\rm triv}=\R[f_0]$ and, by Lemma D11.6, $K(f_0,f_0)=\lambda^*G(f_0,f_0)$ , i.e. $\Phi\equiv0$ on $W_{\rm triv}$ , with $G(f_0,f_0)>0$ .

Standard sector. Apply Lemma D11.5(2) with $E=\mathrm{std}$ , $k=2$ , the maps $T_1,T_2$ and seed $v_0=e_1-e_2$ of Lemma D11.4(2) (linearly independent in $\operatorname{Hom}$ , since $c_1T_1+c_2T_2=0$ evaluated at $v_0$ gives $c_1L+c_2Q=0$ , hence $c_1=c_2=0$ ). Positive definiteness of $\Phi$ on $W_{\rm std}$ is therefore equivalent to positive definiteness of the $2\times2$ Gram matrix of $(L,Q)$ under $\Phi$ . Multiplying the entries of (D11.28) by the common positive factor $m(m+1)/2$ yields the normalized matrices

\widetilde K= \begin{pmatrix} 1 & \tfrac2m\\ \tfrac2m & \tfrac{4(8m-1)}{5m^2(m+1)} \end{pmatrix}, \qquad \widetilde G= \begin{pmatrix} 1 & \tfrac4{m+2}\\ \tfrac4{m+2} & \tfrac{20}{(m+2)(m+3)} \end{pmatrix},

(D11.39)

and a direct computation gives

\begin{aligned} (\widetilde K-\lambda^*\widetilde G)_{11} &=1-\frac{(m+2)(m+3)}{5m^2} =\frac{4m^2-5m-6}{5m^2}=\frac{(4m+3)(m-2)}{5m^2},\\ (\widetilde K-\lambda^*\widetilde G)_{12} &=\frac2m-\frac{4(m+3)}{5m^2} =\frac{10m-4m-12}{5m^2}=\frac{6(m-2)}{5m^2},\\ (\widetilde K-\lambda^*\widetilde G)_{22} &=\frac{4(8m-1)}{5m^2(m+1)}-\frac4{m^2} =\frac{4(8m-1)-20(m+1)}{5m^2(m+1)}=\frac{12(m-2)}{5m^2(m+1)}, \end{aligned}

(D11.40)

that is,

5m^2\,(\widetilde K-\lambda^*\widetilde G) =(m-2) \begin{pmatrix} 4m+3 & 6\\ 6 & \tfrac{12}{m+1} \end{pmatrix}.

(D11.41)

For $m\ge3$ the factor $m-2$ is positive, the diagonal entries are positive, and the determinant of the bracketed matrix is $\tfrac{12(4m+3)}{m+1}-36=\tfrac{12(4m+3)-36(m+1)}{m+1}=\tfrac{12m}{m+1}>0$ . Hence $\Phi$ is positive definite on $W_{\rm std}$ : $\Phi(f_{\rm std},f_{\rm std})>0$ whenever $f_{\rm std}\ne0$ .

Two-row sector ( $m\ge4$ ). $W_{\mathrm X}$ has multiplicity one and is generated by $F$ (Lemma D11.4(3)), so Lemma D11.5(2) with $k=1$ reduces positivity on $W_{\mathrm X}$ to the sign of the scalar $\Phi(F,F)$ . By Lemma D11.8,

\frac{K(F,F)}{G(F,F)} =\frac{2(5m-4)}{m+1}\,\lambda^* =\lambda^*+\frac{9(m-1)}{m+1}\,\lambda^* >\lambda^*,

(D11.42)

since $2(5m-4)-(m+1)=9(m-1)>0$ . Hence $\Phi(F,F)=G(F,F)\bigl(\tfrac{K(F,F)}{G(F,F)}-\lambda^*\bigr)>0$ and $\Phi$ is positive definite on $W_{\mathrm X}$ .

Assembly. For every $f\in V_{m,2}$ , $\Phi(f,f)\ge0$ , i.e. $K(f,f)\ge\lambda^*G(f,f)$ , with equality if and only if $f_{\rm std}=0$ and $f_{\mathrm X}=0$ , i.e. $f\in W_{\rm triv}=\R[f_0]$ . Since $K(f_0,f_0)=\lambda^*G(f_0,f_0)$ with $G(f_0,f_0)>0$ , the minimum (D11.4) equals $\lambda^*$ and is attained exactly on $\R[f_0]=\R[\abs X^2-(m-1)]$ (Lemma D11.6). Finally $(m+2)(m+3)>m^2$ gives $\lambda^*>\tfrac15$ for all $m$ , and $\lambda^*\to\tfrac15$ .

Proof of Corollary D11.1¶

Proof D11.10

All statements are conditional on the identification (D11.8), certified in solutions/conditional-fiber-frame-structure.md: the root frame $\rho_{\rm root}$ is an admissible even frame, and for every $f$ in the maximal form domain (in particular for every polynomial) the root conditional-fiber form equals the pair-redistribution expression (D11.2).

(1). Let $(c_r,w_r,M,\eps)$ be a degree-2 dual certificate (D11.7). Apply the direction bound in (D11.7) at the $m(m-1)$ atoms of $\rho_{\rm root}$ and average:

\sum_rw_r\,K(c_r,c_r) =(m-1)\int\sum_rw_r\,\qJac_\theta[c_r]\dd\rho_{\rm root}(\theta) \le(m-1)\int\theta^TM\theta\dd\rho_{\rm root}(\theta) =\Tr\Bigl(M\,(m-1)\!\int\theta\theta^T\dd\rho_{\rm root}\Bigr) =\Tr M\le\eps,

(D11.43)

using the frame identity $(m-1)\int\theta\theta^T\dd\rho_{\rm root}=I_{H_0}$ and $M$ symmetric on $H_0$ . On the other hand, by Theorem D11.1, $K(c_r,c_r)\ge\lambda^*G(c_r,c_r)$ for every $r$ , so

\eps\ \ge\ \sum_rw_r\,K(c_r,c_r) \ \ge\ \lambda^*\sum_rw_r\,G(c_r,c_r)=\lambda^* =\frac{(m+2)(m+3)}{5m^2}>\frac15 .

(D11.44)

(2). Immediate from (1): every degree-2 certificate objective exceeds $\tfrac15$ uniformly in $m\ge3$ , so no sequence $\eps_m\to0$ exists at $k=2$ . Within the candidate framework, in which a fixed-degree dual refuter is exactly such a certificate sequence, any polynomial dual refutation must therefore use degree $k\ge3$ .

(3). For any admissible $\rho$ whose averaged form $A_\rho$ is defined on $V_{m,2}$ , the supremum defining $\Lambda_{m,2}$ dominates the value at $\rho=\rho_{\rm root}$ , and by (D11.8) and Theorem D11.1 that value is $\lmin(K,G)|_{V_{m,2}}=\lambda^*$ .

Remarks, non-claims, and audit¶

Remark D11.2 (unclosed steps)

None. Theorem D11.1 is unconditional; every step is finite-dimensional linear algebra, elementary representation theory of $S_m$ (proved from facts (F1)–(F5), themselves proved or classical with one-line arguments), and exact Dirichlet moments (Lemma D11.1, whose proof invokes the classical Gamma–Dirichlet factorization: the normalized vector of i.i.d. standard exponentials is $\Dir(1,\dots,1)$ and independent of their sum). Corollary D11.1 is conditional exactly on the hypothesis stated in its preamble, namely the certified identification (D11.8) (nodes lem:conditional-fiber-form and prop:conditional-fiber-root-obstruction, both checked_by: agent with a persisted review), together with the candidate status of the $\Lambda_{m,k}$ /dual-certificate framework itself.

Obstructions respected. The candidate node carries no bounded_by edge. Consistency checks against the neighboring certified and imported obstructions: (i) prop:conditional-fiber-root-obstruction (root-frame $L^2$ gap $O(m^{-2})$ via a vertex-cap indicator) is compatible with (D11.5) because the cap indicator is not a polynomial of degree two; the two statements jointly prove that any test exhibiting the root-frame collapse must leave $V_{m,2}$ , which is precisely the content of Corollary D11.1(2) for the dual channel. (ii) prop:sasada-negative-exchange (imported negative-rate exchange upper bound $48/[m(m+1)]$ for the $L^2$ gap) is compatible for the same reason. (iii) The six route-level obstruction nodes (rem:two-tail-slice-bounds, rem:projection-ceiling, rem:crude-insufficient, rem:relative-ceiling, rem:profile-circularity, rem:single-coordinate-cuts) fence Eldan-localization proof shapes; no stochastic localization, projection summation, bootstrap, isoperimetric localization, or rank-one inference occurs here. No numerical output justifies any step.