Cones over products of simplices - The KLS theorem and its methods

Part of the moment-map mechanism, Chapter The moment map: exact cases; the reading order is on the full proofs page.

Overview. We use the certified cone kernel of Proposition 17.1, compute three moments of an isotropic uniform simplex, and assemble the product blocks. A polynomial with positive coefficients gives the sharp inequality and all equality directions. Intervals count as one-dimensional simplices. This is a family-specific linear-sector result, not a claim about universal CMH.

Theorem 127.1 (Product-simplex cone gate)

Let $q\ge1$ , let $k_1,\ldots,k_q\ge1$ be integers, let $n=1+\sum_{j=1}^q k_j$ , and let $\beta\ge n$ be real. In Definition 17.1, take the centered base $K$ to be a Cartesian product of simplices of dimensions $k_1,\ldots,k_q$ . Let $\Sigma$ and $\tau$ be the covariance and canonical Stein kernel of $\bar\mu_{K,\beta}$ , and put $G=\Sigma^{-1/2}\mathbb E[\tau\Sigma^{-1}\tau]\Sigma^{-1/2}$ . Then $G$ has axis eigenvalue $1+n/\beta$ and, on each transverse factor of dimension $k$ , the eigenvalue

\lambda_k(n,\beta)= \frac{a_k\beta^2+(1+2b_k)\beta+n-k+c_k}{\beta(\beta+1)}, \quad a_k=\frac{2(k+2)}{k+4},\qquad b_k=\frac{2(k+1)(k+2)}{(k+3)(k+4)},\qquad c_k=\frac{(k+2)(k^2+9k+2)}{(k+3)(k+4)}.

(127.1)

In particular $G\preceq2I$ . If $\beta>n$ , then $G\prec2I$ . If $\beta=n$ and $q=1$ , then $G=2I$ . If $\beta=n$ and $q>1$ , then the eigenspace at 2 is exactly the cone axis. Thus a nonzero transverse equality direction exists exactly when $q=1$ and $\beta=n$ . The same spectral and inequality conclusions hold after any invertible linear change of coordinates, with equality subspaces transported in covariance-normalized coordinates.

Proof 127.1

Normalization and the simplex kernel. Under an invertible linear map $L$ , the canonical kernel and covariance transform as $\tau\mapsto L\tau L^\top$ and $\Sigma\mapsto L\Sigma L^\top$ . Indeed a moment potential $\phi$ transforms to $\phi(L^\top y)-\log|\det L|$ ; its gradient and Hessian give these formulas. The resulting normalized gate is conjugate to $G$ by the orthogonal matrix $(L\Sigma L^\top)^{-1/2}L\Sigma^{1/2}$ . We may therefore make each base factor isotropic and regular by a block linear transformation, preserving the axis.

For a factor of dimension $k$ , put $N=k+1$ , let $P=(P_1,\ldots,P_N)$ be uniform on the probability simplex, and write $H=\{v\in\mathbb R^N:\sum_i v_i=0\}$ and $\mathbf e=(1,\ldots,1)$ . An isotropic copy is

V=\sqrt{N(N+1)}(P-\mathbf e/N)\in H.

(127.2)

Its canonical kernel on $H$ is

T=(N+1)C(P),\qquad C(p)=\operatorname{diag}(p)-pp^\top.

(127.3)

For completeness, the Dirichlet moment-potential computation underlying Theorem 17.3 can be verified here without its inequality. On $H$ take $\phi(y)=N\log\sum_i e^{y_i/N}$ plus a normalizing constant. Its gradient is $p-\mathbf e/N$ and its Hessian on $H$ is $C(p)/N$ , where $p_i=e^{y_i/N}/\sum_j e^{y_j/N}$ . The gradient is a diffeomorphism onto the centered open simplex. The determinant lemma gives a principal minor $\det C_{[N-1]}=\prod_i p_i$ ; since $\ker C=\mathbb R\mathbf e$ , $\det_H C=N\prod_i p_i$ . Also $e^{-\phi}$ is proportional to $\prod_i p_i$ because $\sum_i y_i=0$ . Change of variables therefore gives constant target density. This finite smooth strictly convex potential is a canonical moment potential, by uniqueness as used in the certified cone dossier. Scaling by $\sqrt{N(N+1)}$ gives $T$ . The sum of the factor potentials similarly pushes forward to the product law and has block diagonal Hessian. Consequently the canonical base kernel is $T_K=\bigoplus_j T_j$ . All these kernels are bounded on their compact bases.

The three block moments. The uniform simplex integral, obtained by iterating the elementary beta integral, is

\mathbb E\prod_{i=1}^N P_i^{r_i} =\frac{(N-1)!\prod_i r_i!}{(N-1+\sum_i r_i)!} \qquad(r_i\text{ nonnegative integers}).

(127.4)

It gives $\mathbb EV=0$ and $\mathbb E VV^\top=I_H$ . Permutation invariance implies that an invariant endomorphism of $H$ is scalar: extend it by zero on $\mathbb R\mathbf e$ , and invariance forces all diagonal entries to be equal and all off-diagonal entries to be equal. In particular the three expectations below are scalar, and their scalars follow by taking traces:

\mathbb ET^2=a_k I_H,\qquad \mathbb E[VV^\top T]=\mathbb E[TVV^\top]=b_k I_H,\qquad \mathbb E[|V|^2VV^\top]=c_k I_H.

(127.5)

Here is the full scalar computation. With $Q_2=\sum_iP_i^2$ and $Q_3=\sum_iP_i^3$ , the displayed simplex integral gives

\mathbb E Q_2=\frac2{N+1},\qquad \mathbb E Q_3=\frac6{(N+1)(N+2)},\qquad \mathbb E Q_2^2=\frac{4N+20}{(N+1)(N+2)(N+3)}.

(127.6)

Since $\operatorname{Tr} C^2=Q_2-2Q_3+Q_2^2$ , $V^\top TV=N(N+1)^2(Q_3-Q_2^2)$ , and $|V|^4=N^2(N+1)^2(Q_2-1/N)^2$ , division by $k=N-1$ yields

a_k=\frac{2(N+1)}{N+3},\qquad b_k=\frac{2N(N+1)}{(N+2)(N+3)},\qquad c_k=\frac{(N+1)(N^2+7N-6)}{(N+2)(N+3)},

(127.7)

which are the stated expressions. These moment evaluations include $k=1$ , so they cover every interval after centering and scaling.

Assembly of the cone. Write $U=(V_1,\ldots,V_q)$ , $m=n-1$ , and $T_K=\bigoplus_j T_j$ . The base is isotropic. By Proposition 17.1, with $S\sim\Gamma(\beta,1)$ independent of $U$ ,

\Sigma=\beta\oplus\beta(\beta+1)I_m,\qquad \tau=S\begin{pmatrix}1&U^\top\\ U&B\end{pmatrix},\qquad B=UU^\top+\beta T_K.

(127.8)

All expectations below are finite by boundedness on the base and $\mathbb ES^2=\beta(\beta+1)$ . Matrix multiplication shows that the transverse block of $\mathbb E[\tau\Sigma^{-1}\tau]$ is $(\beta+1)I_m+\mathbb EB^2$ , whereas its axis entry is $\beta+n$ . The off-axis vector is zero: it is invariant under independent permutations of the vertices of every factor, and the only invariant vector in each $H_j$ is zero. The same invariance forces off-diagonal factor blocks of $\mathbb EB^2$ to vanish (average permutations in just one factor).

Expand

B^2=|U|^2UU^\top+\beta(UU^\top T_K+T_KUU^\top)+\beta^2T_K^2.

(127.9)

In a factor of dimension $k$ , independence and $\mathbb E|V_i|^2=k_i$ show that the first expectation is $(c_k+m-k)I_k$ . The remaining two terms have expectations $2\beta b_k I_k$ and $\beta^2a_k I_k$ . Dividing the cone block by $\beta(\beta+1)$ proves the claimed $\lambda_k$ and axis spectrum.

The sharp gap. Put $t=\beta-n\ge0$ and $d=n-k-1\ge0$ . Substitution of the three scalar expressions, followed by expansion, gives the identity

\begin{aligned} &(k+3)(k+4)\beta(\beta+1)(2-\lambda_k(n,\beta))\\ &\quad=4(k+3)t^2+ \bigl(5k^2+27k+28+8(k+3)d\bigr)t +4d\bigl((k+3)n+k+1\bigr). \end{aligned}

(127.10)

Every coefficient is strictly positive for $k\ge1$ and $n\ge k+1$ . The right side is nonnegative, and vanishes exactly when $t=d=0$ . Thus a transverse block reaches two exactly when $\beta=n$ and $k=n-1$ , which means there is just one factor. The axis gap is $2-(1+n/\beta)=(\beta-n)/\beta$ . Together these prove all asserted strict inequalities and equality subspaces.

Fences respected. Neither gate conjecture nor the certified cone input has a bounded_by edge. The proof is restricted to the stated product-simplex bases. It does not discharge Conjecture 16.2 or Conjecture 16.1 universally, and supplies no inequality for nonlinear CMH tests. The bound two and the CMH gate threshold four are distinct. No additional hypothesis or unclosed algebraic step is used.