The moment map: appendix - The KLS theorem and its methods

The calculations the moment-map approach rests on, in the order it needs them: the two operator-core arguments of Chapter The moment map: CMH and the linear test, then the Gamma lift (the simplex rewritten in independent Gamma coordinates, Section The independent Gamma lift) and the angular minimization that produce the exact Dirichlet constant. Every statement whose proof is given here is stated, with a summary of its proof, in Chapter The moment map: CMH and the linear test or Chapter The moment map: exact cases. Nothing is decided here that is not decided there.

Proof 19.1 (Proof of Proposition 16.1)

We calculate in target coordinates, use Einstein summation, and first take $g$ in a smooth compactly supported core. Write $\rho$ for the density of $\mu$ . The Stein identity is

\partial_i(\rho H_{ij})=-\rho p_j, \qquad L_\mu g=H_{k\ell}g_{k\ell}-p_k g_k.

(19.1)

One integration by parts, followed by differentiating the displayed formula for $L_\mu g$ , gives

\begin{aligned} \E_\mu(L_\mu g)^2 ={}&\E_\mu[H_{ij}g_i g_j] -\E_\mu[H_{ij}g_j(\partial_iH_{k\ell})g_{k\ell}]\\ &-\E_\mu[H_{ij}g_jH_{k\ell}g_{ik\ell}] +\E_\mu[H_{ij}g_jp_k g_{ik}]. \end{aligned}

(19.2)

Integrating the third-derivative term by parts in $p_k$ yields

\begin{aligned} -\E_\mu[H_{ij}g_jH_{k\ell}g_{ik\ell}] ={}&-\E_\mu[p_\ell H_{ij}g_jg_{i\ell}] +\E_\mu[H_{k\ell}(\partial_kH_{ij})g_jg_{i\ell}]\\ &+\E_\mu[H_{k\ell}H_{ij}g_{jk}g_{i\ell}]. \end{aligned}

(19.3)

The first term here cancels the last term in the preceding display after relabelling $k$ and $\ell$ , while the last one equals $\Tr(HD^2g\,HD^2g)$ .

It remains to cancel the two terms containing $\partial H$ . If $p=\nabla\varphi(y)$ , then $H_{k\ell}(p)=\varphi_{k\ell}(y)$ and

H_{mj}\partial_jH_{k\ell} =H_{mj}(H^{-1})_{rj}\varphi_{k\ell r} =\varphi_{mk\ell}.

(19.4)

Thus $H_{mj}\partial_jH_{k\ell}$ is totally symmetric in $m,k,\ell$ . Using symmetry of $H$ and (19.4), the residual is

\E_\mu[g_jg_{i\ell}\varphi_{\ell ij}] -\E_\mu[g_jg_{k\ell}\varphi_{jk\ell}]=0.

(19.5)

This proves (16.4) on the smooth core. If $g_q$ is graph-norm Cauchy there, apply the identity to $g_q-g_r$ . Nonnegativity of the two right-hand terms shows that $H^{1/2}\nabla g_q$ and $H^{1/2}D^2g_qH^{1/2}$ are Cauchy in their respective $L^2$ spaces. Their limits define the closed right-hand side, and passage to the limit proves the identity on the operator-core closure.

Proof 19.2 (Proof of Theorem 16.1)

Put $C=\CMH(\mu)$ ; there is nothing to prove if $C=\infty$ . Let $\mathscr C$ be the restrictions to $\operatorname{supp}\mu$ of $\R+C_c^\infty(\R^n)$ and first take a centered $f\in\mathscr C$ . This class lies in both form domains even when $H$ is unbounded: the Stein normalization $\E_\mu H=\Sigma$ gives

\E_\mu\inner{H\nabla f}{\nabla f} \le \norm{\nabla f}_\infty^2\E_\mu\Tr H =\norm{\nabla f}_\infty^2\Tr\Sigma<\infty.

(19.6)

For $0<\eps<R$ set

\Pi_{\eps,R}=\one_{[\eps,R]}(\Aop), \qquad g_{\eps,R}=\Pi_{\eps,R}\Aop^{-1}f\in\Dom(\Aop).

(19.7)

Then $\Aop g_{\eps,R}=\Pi_{\eps,R}f$ . The form–operator pairing, rather than any commutation of $\Pi_{\eps,R}$ with the $\Sigma$ -form, gives

\begin{aligned} \norm{\Pi_{\eps,R}f}_2^2 &=\inner{f}{\Pi_{\eps,R}f}_{L^2} =\inner{f}{\Aop g_{\eps,R}}_{L^2} =\E_\mu\inner{\nabla f}{H\nabla g_{\eps,R}}\\ &\le \bigl(\E_\mu\inner{\Sigma\nabla f}{\nabla f}\bigr)^{1/2} \bigl(\E_\mu\inner{H\nabla g_{\eps,R}} {\Sigma^{-1}H\nabla g_{\eps,R}}\bigr)^{1/2}\\ &\le C^{1/2} \bigl(\E_\mu\inner{\Sigma\nabla f}{\nabla f}\bigr)^{1/2} \norm{\Pi_{\eps,R}f}_2. \end{aligned}

(19.8)

After division (with the zero case harmless) and squaring,

\norm{\Pi_{\eps,R}f}_2^2 \le C\E_\mu\inner{\Sigma\nabla f}{\nabla f}.

(19.9)

Because $H\succ0$ and the support is connected, $\ker\Aop$ consists of the constants. Thus $f\perp\ker\Aop$ , and the spectral theorem gives $\Pi_{\eps,R}f\to f$ in $L^2$ as $R\to\infty$ and then $\eps\downarrow0$ .

Finally define $H^1_\Sigma(\mu)$ as the closure of $\mathscr C$ for $\norm{f}_2^2+\E_\mu\inner{\Sigma\nabla f}{\nabla f}$ (equivalently the usual $H^1(\mu)$ here, since $\Sigma\succ0$ is constant). Approximate an arbitrary $f\in H^1_\Sigma(\mu)$ by functions in $\mathscr C$ and subtract their means. Both the variances and the $\Sigma$ -energies converge, so the preceding inequality passes to the limit. Notice that this density step never asserts that finite $\Sigma$ -energy implies finite $H$ -energy when $H$ is unbounded.

Proof 19.3 (Proof of Lemma 17.1)

For $Y\sim\GammaLaw(a,1)$ , integrating $\E[Y^2(u^2)']$ by parts against $y^{a-1}e^{-y}$ gives $\E[Y^2uu']=-\tfrac12\E[((a+1)Y-Y^2)u^2]$ . Apply this with $u=G_i$ , $u'=G_{ii}$ , $a=\alpha_i$ and expand the first square in (17.16): the cross term equals $\E[Y_iG_i^2]-\E[Y_i^2G_i^2]/(\alpha_i+1)$ , so

\E R_i=\E\Bigl[Y_iG_i^2+\sum_jY_iY_jG_{ij}^2\Bigr] -\frac{\alpha_i}{(\alpha_i+1)^2}\E[Y_i^2G_i^2],

(19.10)

using $\tfrac1{(\alpha_i+1)^2}-\tfrac1{\alpha_i+1}=-\tfrac{\alpha_i}{(\alpha_i+1)^2}$ . Summing over $i$ recovers $N_\Gamma(G)$ from (17.14) and leaves $-\sum_i\alpha_i^2(\alpha_i+1)^{-2}D_i(G)$ ; subtracting $D_\Gamma/4$ gives (17.15). Finally $\delta_i\ge0$ is equivalent to $\alpha_i/(\alpha_i+1)\ge\tfrac12$ , i.e. to $\alpha_i\ge1$ .

Proof 19.4 (Proof of Lemma 17.2)

Euler’s identity $\sum_jY_jG_j=0$ , differentiated in $Y_i$ , gives the single linear constraint $\sum_jY_jG_{ij}=-G_i$ . Set $t_i=Y_i\bigl(G_{ii}-G_i/(\alpha_i+1)\bigr)$ and $t_j=Y_jG_{ij}$ for $j\ne i$ , so that $\sum_jt_j=-G_i\bigl(1+Y_i/(\alpha_i+1)\bigr)$ and $R_i=t_i^2+\sum_{j\ne i}(Y_i/Y_j)t_j^2$ . The reciprocals of these quadratic weights sum to $1+\sum_{j\ne i}Y_j/Y_i=S/Y_i$ , and the minimum of $\sum_jw_jt_j^2$ under $\sum_jt_j=c$ is $c^2/\sum_jw_j^{-1}$ , which is the right-hand side of (17.17).

Proof 19.5 (Proof of Lemma 17.3)

For fixed $a$ the $p$ -dependent part is $a\bigl(p^{-1}+zp(a+1)^{-2}\bigr)$ , minimized on $(0,1]$ at $p=1$ when $\sqrt z\le a+1$ and at $p_*=(a+1)/\sqrt z$ otherwise.

If $z\le4$ then $\sqrt z\le2\le a+1$ for all $a\ge1$ , so $p=1$ and $F_A=a+2a(A-2)/(a+1)+za/(a+1)-z/4$ , each summand increasing in $a$ ; the minimum at $a=1$ is $A-1+z/4=A-\tfrac12+(z-2)/4$ .

If $z\ge4$ , then at $p_*$ the two $p$ -terms are equal and sum to $2a\sqrt z/(a+1)$ , so with $r=a/(a+1)$ the value is $2r(\sqrt z+A-2)+zr^2-z/4$ , increasing in $r$ ; since $r\ge\tfrac12$ for $a\ge1$ the minimum is at $a=1$ , $r=\tfrac12$ , and equals $A-2+\sqrt z=A-\tfrac12+\sqrt z-\tfrac32$ . In the complementary region the expression is increasing in $a$ with a boundary value no smaller. Finally $A\ge3$ gives $z\ge2$ , so $s_A\ge0$ in both branches.

Proof 19.6 (Proof of Theorem 17.3)

If $m=2$ and $A<3$ , the law is one dimensional and Theorem 17.1 applies. In every remaining case $A\ge3$ (automatically when $m\ge3$ ), so $z_A\ge2>0$ and the inverse moments $\E S^{-1},\E S^{-2}$ used in (17.19) are finite.

Combine (17.15), (17.16) and (17.20). Since $D_i(G)=\E[u_i^2]/\alpha_i$ , the total coefficient of $\E_P[u_i^2]$ in $N_\Gamma(G)-\tfrac14D_\Gamma(G)$ is

\frac1{z_AP_i}+\frac2{(A-1)(\alpha_i+1)}+\frac{P_i}{(\alpha_i+1)^2}+\frac{\delta_i}{\alpha_i},

(19.11)

and multiplying it by $z_A\alpha_i$ gives exactly $F_A(\alpha_i,P_i)$ of (17.21), using $z_A/(A-1)=A-2$ . By Lemma 17.3,

N_\Gamma(G)-\tfrac14D_\Gamma(G) \ \ge\ \frac{A-\tfrac12+s_A}{z_A}\sum_i\frac{\E_P[u_i^2]}{\alpha_i} =\frac{A-\tfrac12+s_A}{z_A}\,d_\alpha(g).

(19.12)

Substituting (17.19), multiplying by $z_A$ and using the identity $z_A/4+A-\tfrac12=A(A+1)/4$ gives

n_\alpha(g)\ \ge\ \Bigl(\frac{A(A+1)}4+s_A\Bigr)d_\alpha(g),

(19.13)

and $s_A\ge0$ yields (17.13).