A Worked Example of the Proof of KLS Conjecture

The Radial Exponential Family

Let $(X,Y)$ have a radial exponential law, whose density depends only on distance from the origin. Keep the scale parameter visible for now:

In polar coordinates, the angle is uniform and independent of the radius $R=\sqrt{X^2+Y^2}$. The radius has density $\lambda^2r e^{-\lambda r}$. Integration by parts, starting with $\int_0^\infty e^{-\lambda r}dr=1/\lambda$, gives, for integers $j\ge1$ and $k\ge0$,

The case $k=0$ verifies normalization. Symmetry gives zero mean and zero cross-covariance. The angular average of $\cos^2\theta$ is $1/2$, so, writing $\Id$ for the identity operator (here a two-by-two matrix),

Thus $\lambda=\sqrt3$ puts the law in isotropic position. We use that value from now on unless a scale is displayed explicitly.

Isotropy does not mean independence. Here the angular averages of $\cos^4\theta$ and $\cos^2\theta\sin^2\theta$ are $3/8$ and $1/8$, respectively. Hence

The common radius couples the coordinates. There is also a geometric obstruction: with $e_r=(x,y)/r$, the potential satisfies, away from the origin,

Its radial eigenvalue is zero. A curvature bound of the form $D^2V\succeq a\Id$, with $a>0$, is therefore unavailable for the original law. This is why the later localization step has a job to do.

Compatibility Projection and Its Energy Cost

A vector field $v=(v_1,v_2)$ can be a gradient only if $\partial_yv_1=\partial_xv_2$. On the plane this is the familiar compatibility condition. For a symmetric matrix field $F$, the corresponding Hessian conditions are

At higher ranks, compatibility again says that exchanging the new derivative index with any tensor index changes nothing. These are the fields on which repeated integration makes sense.

To calculate with a smooth potential, temporarily replace $\sqrt3r$ by

Writing $s=\sqrt{\varepsilon^2+r^2}$, differentiation gives

Its radial and tangential eigenvalues are $\varepsilon+\sqrt3\varepsilon^2/s^3$ and $\varepsilon+\sqrt3/s$. Thus

All norms in the next calculation use the smooth density proportional to $e^{-V_\varepsilon}$. In a weighted space, the adjoint of differentiation is $\partial_i^*=-\partial_i+V_i$. Even a constant Hessian can lose compatibility under this adjoint:

The correction is an orthogonal projection in $L^2(\mu)$, rather than a pointwise alteration of the arrows. If $P$ projects onto compatible vector fields, weighted Hodge theory represents the discarded part as

The point of the comparison is to show that the projection leaves enough energy. In the manuscript's rank-two calculation, choose at each point an orthonormal basis in which $K=D^2V$ has eigenvalues $\kappa_1,\kappa_2>0$. The curl term is $(\kappa_1-\kappa_2)F_{12}$. The inverse-Hodge estimate bounds the energy lost by

The unprojected mixed-component energy has weight $\kappa_1+\kappa_2$. Subtracting the possible loss gives exactly

The factor $2F_{12}^2$ is the squared norm of the mixed symmetric component: both ordered entries are counted. The surviving curvature is its harmonic mean. Unequal curvatures create curl, but the correction does not erase the lower curvature bound.

Centered Integration and Appell Polynomials

For a compatible field $F=\nabla u$, its centered primitive is $u-\E u$. The manuscript separates two kinds of inputs. The operator $L_r$ integrates a constant tensor of rank $r+1$; $J_r$ integrates a compatible field of that rank whose mean is already zero. Both return centered fields of rank $r$. For a centered law, their defining identities are

The constant-input chain along the first coordinate is completely explicit. Successive integrations and centering give

For example, an antiderivative of $(x^3-3x)/6$ is $x^4/24-x^2/4$. Its mean is $5/24-1/4=-1/24$, so centering adds $1/24$. This is why the final constant differs from the Gaussian one.

These are normalized Appell polynomials. We use the paper’s Appell tensors and scalar contractions, with the normalized shorthand

Coefficient comparison after differentiation and expectation gives $D^dp_d[T]=T$ and $\E D^jp_d[T]=0$ for $0\le j<d$. Thus

The rightmost operator acts first. The distinct $L$ matters: a nonzero constant tensor is not a centered input for $J$.

The Mixed Quartic Integration Chain

Use both coordinates by taking

Its mean is $5/3-1-1+1/3=0$. Its first two derivatives are

The cross derivatives of $\nabla q$ both equal $4xy$; the matrix identities are $\partial_y[2(y^2-1)]=\partial_x(4xy)=4y$ and $\partial_y(4xy)=\partial_x[2(x^2-1)]=4x$. Compatibility is visible entry by entry.

The draft's coefficient is the supremum over unit leading tensors:

Even at degree two, one direction does not attain this supremum. For a symmetric matrix $T$, $p_2[T]=(x^\top Tx-\Tr T)/2$. The fourth moments yield

Since $(\Tr T)^2\le2\|T\|_{\HS}^2$, equality at $T=\Id/\sqrt2$ gives $c_2(\mu)^2=7/6$. In contrast, the directional polynomial $(x^2-1)/2$ has squared norm one.

Uniform Bounds for Integration Powers

Appell norms observe integration chains that start at constant tensors. A general compatible input can depend on position in any square-integrable way. Curvature connects these special observations to the full integration operator.

The paper bundles all tensor ranks into one Hilbert space, so that a single operator $\mathsf J$ lowers rank and $\mathsf L$ supplies the constant-input primitives. The inverse comparison in §4.2, based on Lemma 3.1, gives the following inequality. Appendix B supplies the operator-domain details.

The function $g$ acts on the operator’s spectral values, just as a function of a symmetric matrix acts on its eigenvalues. Its denominator records the contribution of curvature. The uniform power lemma, Lemma 5.1, uses this comparison and finitely many constant-input bounds to control every power of $\mathsf J$.

There is still a circular-looking issue: the polynomial estimates needed by this result are themselves unknown at high degree. Localization and reverse transfer turn that circle into an induction, using only smaller degrees to establish the next one.

Covariance-Normalized Stochastic Localization

Localization reweights a law by a linear tilt and a Gaussian factor. For our radial example, the tilted density is proportional to

The linear term moves the region carrying most of the mass; the quadratic term adds curvature. The paper proves its stochastic estimates first for bounded-support isotropic laws. For this example, it conditions on a large disk, rescales to restore isotropy, and then passes to the limit. The expandable calculation below describes that approximation.

For the bounded-support process, $\E_t$ denotes expectation under the current random law, or posterior. Let $m_t=\E_tX$, $A_t=\Cov_t(X)$, and let $B_t$ be planar Brownian motion. The parameters obey

Consequently $\Lambda_t=\int_0^tA_s^{-1}ds$, and the posterior has curvature at least $\Lambda_t$ for $t>0$. For a nonsmooth potential, this means that subtracting the corresponding quadratic leaves a convex function; it does not require a pointwise Hessian at the origin. The original radial symmetry generally disappears along a path, so neither $A_t$ nor $\Lambda_t$ can be replaced by a scalar multiple of the identity.

For the bounded-support process, the special drift in $\theta_t$ makes the normalized density a martingale:

For a fixed test function, this gives $d\E_tf=\E_t[(f-\E_tf)\xi_t]\cdot dB_t$. In particular, the mean moves and the covariance fluctuates according to

The third moments vanish initially by symmetry, but generally become nonzero after tilting. The paper uses Letwin's quadratic variance inequality, stated in §2.3, to control them. That input leads to the mixed covariance bound needed later:

This is the mixed covariance estimate from §6.1. The expectation here is over the localization path; $\E_t$ instead means integration under its current random law. The ordered tensor slots keep the matrix factors in their proper places without requiring them to commute.

Appell Polynomials Under Localization

Our original mixed quartic used $\E X_i^2=1$ and $\E X_1^2X_2^2=5/3$. After localization, those moments change. Keeping the same polynomial would break the mean-zero constraints required by the integration chain.

Keep the leading tensor fixed, but replace all lower coefficients with the Appell coefficients of the current law. Write $z=x-m_t$. With covariance entries $A_{ij}$, third central moments $\kappa_{ijk}$, and

the moving mixed quartic is

At time zero, $A=\Id$, the third moments are zero, and $\kappa_{1122}=2/3$. This recovers $x^2y^2-x^2-y^2+1/3$.

These identities preserve the integration constraints, but they do not say that the polynomial's variance is a martingale. Its coefficients are changing too. Lemma 6.2 estimates that additional effect.

Reverse Coefficient Transfer

The integration bound requires covariance at most $\Id$ and curvature at least $\delta \Id$, where $0<\delta\le1$. First consider a smooth posterior whose Hessian also has a finite upper bound. Write $A$ for its covariance and $B$ for a positive curvature lower bound. One change of coordinates arranges both conditions:

Here $m$ is the posterior mean. Since $\mathsf M\succeq A$, $\Cov(Y)=\mathsf M^{-1/2}A\mathsf M^{-1/2}\preceq \Id$. Also $\mathsf M\succeq\delta B^{-1}$ implies $B\succeq\delta \mathsf M^{-1}$, so the potential in the new coordinates has Hessian at least $\mathsf M^{1/2}B\mathsf M^{1/2}\succeq\delta \Id$.

Let $M_q$ bound the norm of $q$ centered integrations uniformly over regular laws with these two properties. Set $\nu=\mathcal L(Y)$ and $\bar p(y)=p(m+\mathsf M^{1/2}y)$. A degree-$d$ Appell polynomial can be recovered in these coordinates whenever $1\le q\le d-1$:

The endpoint $q=d$ is excluded because $D^dp$ is a nonzero constant, whereas $J$ requires a centered input. Returning to the original coordinates by the chain rule gives

The paper extends this polynomial inequality to nonsmooth posteriors by regular approximation, preserving the covariance and curvature bounds while the required moments converge. This is the step that permits its use for the tilted radial example.

For the moving quartic, two integrations recover it from the centered Hessian displayed above. This illustrates a permitted terminal estimate; the draft's actual induction chooses one derivative at small degrees, including degree four.

Differentiation lowers the polynomial degree. For $p_t=P_d^{\mu_t}[T]$, the degree-$(d-q)$ coefficient bound gives

There is no extra factorial cost: differentiation contributes $d!/(d-q)!$, and the lower-degree norm contributes $(d-q)!C_{d-q}$. Multiplication leaves $d!C_{d-q}$. Apply this componentwise to the remaining symmetric polynomial slots and sum the squared component bounds.

At time $t$, take $B=\Lambda_t$, so $\mathsf M=A_t+\delta\Lambda_t^{-1}$. The mixed covariance estimate from §6.1 then bounds the path average:

Choose $t=\eta/d^2$, $\delta=\eta^2/d^2$, with $0<\eta\le1$. Insert this estimate into Lemma 6.2 and divide by $d!\|T\|$. Since $\delta/t=\eta$ and $(2d^2+d+1)t\le3\eta$ for $d\ge2$, the result is

This is Proposition 7.1. The star means a supremum over all dimensions and all centered log-concave laws with covariance at most $\Id$. The supremum over laws is essential: the posteriors generated by our initially radial law need not be radial. Estimates for the initial radial density alone cannot supply the induction hypotheses.

Uniform Coefficient Induction

We now use bounds at smaller degrees to establish the same kind of bound at the next degree. The induction target has a fixed exponential growth rate:

The paper takes $R_*=10^8$ and proves $c_d^\star\le\beta_d$ in every degree. Here $R_*$ controls coefficient growth; it is unrelated to the support radius used in approximation. Covariance supplies the base case, $c_1^\star\le1<R_*/16=\beta_1$. The polynomial factor $(d+1)^{-4}$ supplies the slack that will close the induction.

Up to degree $10^7$, one integration suffices, using the quadratic seed in Corollary 5.3. For larger degrees, use about half the derivatives, $q=\lfloor d/2\rfloor$, and coefficient bounds only through $D=\lfloor\sqrt d\rfloor$. Lemma 5.1 then allows a scalar integration rate $B$ slightly below $R_*^2$:

The smaller rate is possible because squaring the degree-$D$ coefficient bound produces the factor $(D+1)^{-8}$, of order $d^{-4}$. The curvature threshold costs only a factor of order $d^{-3}$. This leaves room in the power lemma’s strict hypothesis to choose $B<R_*^2$. Its prefactor is paid once for the whole chain, so the small gain at each integration accumulates:

When this bound is inserted into reverse transfer, the powers of $R_*$ cancel. The remaining decay outweighs the localization cost and the bounded ratio of polynomial weights. A convolution estimate for those weights also controls the drift term, which involves only smaller degrees.