2. The upper bound: Proof of Theorem 3

Recall the following identity in $L^2(\mathbb{S}^{d-1})$ ,

\begin{align*} \varphi = \sum \limits _{k=0}^\infty \lambda _k \psi _k \otimes \psi _k, \end{align*}

with the eigenvalues $\lambda _k$ indexed by the spherical harmonics. Define the random matrices $M_n, \Psi _n$ by

\begin{equation*}(M_n)_{i,j} = \frac {1}{n}\varphi (X_i,X_j), \;\;\;\;\; (\Psi _n)_{i,k} = \frac {1}{\sqrt {n}}\psi _k(X_i).\end{equation*}

Note that $M_n$ is an $n \times n$ matrix, while $\Psi _n$ , has infinitely many columns. Furthermore, denote by $\Lambda$ the diagonal matrix $\mathrm{diag}\{\lambda _i\}_{i=0}^\infty$ . Then, $\sigma$ -almost surely, for every $i,j \in [n],$

\begin{equation*} (M_n)_{i,j} = \frac {1}{n}\sum _{k=0}^\infty \lambda _k\psi _k(X_i)\psi _k(X_j) =(\Psi _n\Lambda \Psi _n^T)_{i,j}. \end{equation*}

For $r \in \mathbb N$ we also denote

\begin{equation*}\varphi ^r \;:\!=\; \sum \limits _{k=0}^r \lambda _k \psi _k \otimes \psi _k,\end{equation*}

the finite rank approximation of $\varphi$ , $\Lambda ^r = \mathrm{diag}\{\lambda _k\}_{k=0}^r$ , and $\Psi _n^r$ the sub-matrix of $\Psi _n$ composed of its first $r$ columns. Finally, denote

\begin{equation*}M_n^r =\Psi _n^r\Lambda ^r(\Psi _n^r)^T.\end{equation*}

As before, let $A$ be an adjacency matrix drawn from $G\left (n, \frac{1}{n}\varphi (X_i,X_j)\right )$ so that $\mathbb E \left [A|X_1,\ldots,X_n\right ] = M_n$ . Our goal is to recover $\Psi _n^{d+1}$ from the observed $A$ . The first step is to recover $\Psi _n^{d+1}$ from $M_n$ . We begin by showing that the columns of $\Psi _n^r$ are, up to a small additive error, eigenvectors of $M_n^r$ . To this end, denote

\begin{equation*}E_{n,r} \;:\!=\; (\Psi _n^r)^T\Psi _n^r - \mathrm {Id}_r,\end{equation*}

$C(n,r) = \left \lVert E_{n,r}\right \rVert ^2_{op}$ , and $K = \max _i{\lambda _i}$ .

Lemma 10. Let $u_i$ be the $i$ ’th column of $\Psi _n$ and let $\eta \gt 0$ . Then

\begin{equation*} \left \lVert M_n^ru_i - \lambda _iu_i\right \rVert _2^2\leq K^2(\sqrt {C(n,r)}+1)C(n,r). \end{equation*}

Moreover, whenever $n\geq l(r) \log (2r/\eta )$ , we have with probability larger than $1-\eta$ ,

\begin{equation*} C(n,r) \leq \frac {4 l(r)\log (2r/\eta )}{n}. \end{equation*}

where $l(r)$ only depends on $r$ and on the dimension.

Proof. Let $e_i\in \mathbb R^r$ be the $i$ ’th standard unit vector so that $u_i = \Psi _n^re_i$ . So,

\begin{equation*}(\Psi _n^r)^Tu_i = (\Psi _n^r)^T\Psi _n^re_i =(\mathrm {Id}_r + (\Psi _n^r)^T\Psi _n^r - \mathrm {Id}_r)e_i=e_i + E_{n,r}e_i.\end{equation*}

We then have

\begin{align*} M_n^ru_i &= \Psi _n^r\Lambda ^r(\Psi _n^r)^Tu_i = \Psi _n^r\Lambda ^re_i + \Psi _n^r\Lambda ^rE_{n,r}e_i \\[5pt] &= \lambda _i\Psi _n^re_i + \Psi _n^r\Lambda ^rE_{n,r}e_i = \lambda _iu_i + \Psi _n^r\Lambda ^rE_{n,r}e_i. \end{align*}

To bound the error, we estimate $\left \lVert M_n^ru_i - \lambda _iu_i\right \rVert _2^2=\left \lVert \Psi _n^r\Lambda ^rE_{n,r}e_i\right \rVert _2^2$ as

\begin{align*} \langle \Lambda ^rE_{n,r}e_i,(\Psi _n^r)^T\Psi _n^r\Lambda ^rE_{n,r}e_i\rangle &= \langle \Lambda ^rE_{n,r}e_i,E_{n,r}\Lambda ^rE_{n,r}e_i\rangle + \left \lVert \Lambda ^rE_{n,r}e_i\right \rVert _2^2 \\[5pt] &\leq \left (\sqrt{C(n,r)}+1\right )\left \lVert \Lambda ^rE_{n,r}e_i\right \rVert _2^2 \leq K^2\left (\sqrt{C(n,r)}+1\right )C(n,r). \end{align*}

It remains to bound $C(n,r)$ . Let $X_i^r = ( \psi _0(X_i),\ldots, \psi _{r-1}(X_i))$ stand for the $i$ ’th row of $\Psi _n^r$ . Then, $E_{n,r} = \frac{1}{n}\sum _{i=1}^n\left ((X_i^r)^T X_i^r - \mathrm{Id}_r\right )$ , is a sum of independent, centred random matrices. We have

\begin{align*} \sigma _{n,r}^2 &\;:\!=\; \left \lVert \mathbb E\left ( \frac{1}{n}\sum _{i=1}^n\left ((X_i^r)^T X_i^r - \mathrm{Id}_r\right )\right )^2\right \rVert _{op} \\[5pt] &= \frac{1}{n} \left \lVert \mathbb E \left ( (X_1^r)^T X_1^r - \mathrm{Id}_r\right )^2\right \rVert _{op} \end{align*}

Furthermore, the norm of the matrices can be bounded by

\begin{align*} \left \lVert \frac{1}{n}\left ((X_1^r)^T X_1^r - \mathrm{Id}_r\right )\right \rVert _{op} &=\frac{1}{n} \max (1, \left \lVert X_1^r\right \rVert _2^2 -1) \\[5pt] &\leq \frac{1}{n}\max \left (1, \left \lVert \sum _{i=0}^r \psi _i^2\right \rVert _\infty - 1\right ). \end{align*}

Note that the right-hand side of the two last displays are of the form $\frac{1}{n} l(r)$ where $l(r)$ depends only on $r$ and $d$ (not on $n$ ). Applying matrix Bernstein ([Reference Tropp19, Theorem 6.1]) then gives

\begin{equation*} \mathbb P\left ( \left \lVert E_{n,r}\right \rVert _{op}\geq t \right ) \leq 2r \exp \left (-\frac {n}{2l(r)}\frac {t^2}{1+t/3} \right ), \end{equation*}

where $l(r)$ depends only on $r$ and $d$ . Choose now $t_0 = \frac{4 l(r)\log (2r/\eta )}{n}$ . As long as $n\geq l(r) \log (2r/\eta )$ , $t_0 \leq 4$ , and the above bound may be refined to

\begin{equation*} \mathbb P\left ( \left \lVert E_{n,r}\right \rVert _{op}\geq t_0 \right ) \leq 2r \exp \left (-\frac {n}{l(r)}\frac {t_0^2}{7} \right ). \end{equation*}

With the above conditions, it may now be verified that $2r \exp \left (-\frac{n}{l(r)}\frac{t_0^2}{7} \right ) \leq \eta$ , and the proof is complete.

We now show that as $r$ increases, the subset of eigenvectors of $M_n^r$ , that encode the latent positions of $\{X_i\}_{i=1}^n$ , converge to those of $M_n$ . Order the eigenvalues in decreasing order $\lambda _{l_0}\geq \lambda _{l_1} \geq \lambda _{l_2} \geq \ldots$ and let $\Lambda _{\gt r} = \sum _{i=r}^\infty \lambda _{l_i}^2$ . Note that it follows from assumption A2 that $\Lambda _{\gt r} = O(r^{-3})$ . We will denote by $\lambda _i(M_n), \lambda _i(M_n^r)$ the respective eigenvalues of $M_n$ and $M_n^r$ , ordered in a decreasing way, and by $v_i(M_n),v_i(M_n^r)$ their corresponding unit eigenvectors. Suppose that $s$ is such that

(9) \begin{equation} \lambda (\varphi ) = \lambda _{l_{s+1}} = \ldots = \lambda _{l_{s+d}}. \end{equation}

Moreover, define

\begin{equation*} V_n \;:\!=\; \mathrm {span}(v_{l_{s+1}}(M_n),\ldots, v_{l_{s+d}}(M_n)), \;\;\; V_n^r \;:\!=\; \mathrm {span}\left (v_{l_{s+1}}(M_n^r),\ldots, v_{l_{s+d}}(M_n^r)\right ). \end{equation*}

The next lemma shows that $V_n$ is close to $V_n^r$ whenever both $n$ and $r$ are large enough.

Lemma 11. Let $\delta$ and $C$ be the constants from Assumption A1 and Assumption A2. For all $n,r$ , let $P_{n,r}$ be the orthogonal projection onto $V_n^r$ . Then, for all $\eta \gt 0$ there exist constants $n_0,r_0$ , which may depend on $\delta$ , $C$ , and $\eta$ , such that for all $n\gt n_0$ and $r\gt r_0$ , we have with probability at least $1-\eta$ that, for all $w\in V_n$ ,

\begin{equation*}\left \lVert w - P_{n,r}w\right \rVert _2 \leq \frac {4C}{\eta \delta ^2 r^3 }.\end{equation*}

Proof. We have

\begin{align*} \mathbb E \left \lVert M_n - M_n^r\right \rVert ^2_{F} &= \sum _{i,j} \mathbb E (M_n -M_n^r)^2_{i,j} \\[5pt] & = \sum _{i,j} \frac{1}{n^2} \mathbb E \left (\sum _{k=r}^\infty \lambda _k\psi _k(X_i)\psi _k(X_j)\right )^2 \\[5pt] & = \mathbb E_{x,y \sim \sigma } \left ( \sum _{k=r}^\infty \lambda _k \psi _k(x)\psi _k(y)\right )^2 \\[5pt] & = \sum _{k=r}^\infty \lambda _k^2 = \Lambda _{\gt r}. \end{align*}

Applying Markov’s inequality gives that with probability $1-\eta$

(10) \begin{equation} \left \lVert M_n - M_n^r\right \rVert ^2_{F} \leq \frac{\Lambda _{\gt r}}{\eta }\leq \frac{C}{\eta r^3}. \end{equation}

Theorem 1 in [Reference Valdivia20] shows that there exists $n$ large enough such that with probability larger than $1-\eta$ , one has

\begin{equation*} |\lambda _i(M_n) - \lambda _{l_{s+i}}| \leq \delta/ 4, \end{equation*}

with $\delta$ being the constant from Assumption A1. It follows that

(11) \begin{equation} \lambda _{l_{s+1}}(M_n),\ldots,\lambda _{l_{s+d}}(M_n) \in \left [\lambda (\varphi )-\frac{\delta }{4}, \lambda (\varphi )+\frac{\delta }{4}\right ], \end{equation}

while by (10) and Weyl’s perturbation theorem (e.g., [Reference Bhatia3, Corollary III.2.6]), for $r$ large enough with probability $1-\eta$ ,

(12) \begin{equation} \lambda _i(M_n^r) \notin \left [\lambda (\varphi )-\frac{3\delta }{4}, \lambda (\varphi )+\frac{3\delta }{4}\right ], \text{ for } i \neq l_{s+1},\ldots l_{s+d}. \end{equation}

Combining (10), (11) and (12) it follows from the classical Davis-Kahan theorem (see e.g. [Reference Bhatia3, Section VII.3]) that with probability at-least $1-2\eta$ , for every $w \in V_n$ ,

\begin{equation*}\left \lVert w - P_{n,r}w\right \rVert _2^2 \leq \frac {4C}{\eta \delta ^2 r^3}.\end{equation*}

Denote

\begin{equation*} G_n \;:\!=\; \frac {1}{d} \sum _{k=1}^d v_{l_{s+k}}(M_n) v_{l_{s+k}}(M_n)^T, \;\;\; (G'_{\!\!n})_{i,j} = \frac {1}{n}\langle X_i, X_j \rangle . \end{equation*}

A combination of the last two lemmas produces the following:

Theorem 12. One has

\begin{equation*} \| G_n - G'_{\!\!n} \|_F \to 0 \end{equation*}

in probability, as $n \to \infty$ .

Proof. Denote

\begin{equation*} G_n^r\;:\!=\;\frac {1}{d}\sum _{k=1}^d v_{l_{s+k}}(M_n^r)(v_{l_{s+k}}(M_n^r))^T. \end{equation*}

Then

\begin{equation*} \| G_n - G'_{\!\!n} \|_F \leq \|G_n^r - G'_{\!\!n}\|_F + \|G_n-G_n^r\|_F \end{equation*}

We will show that the two terms on the right hand side converge to zero. Let $r(n)$ be a function converging to infinity slowly enough so that $C(n,r)\to 0$ , for the constant $C(n,r)$ defined in Lemma 10. Taking $\eta = \eta (n)$ to converge to zero slowly enough and applying Lemma 10, gives for all $1 \leq i \leq d$ ,

\begin{equation*} \|(M_n^r - \lambda (\varphi )\mathrm {Id}_n) u_i \|_2^2 \leq \varepsilon _{n} \end{equation*}

with $u_i$ the $i$ ’th column of $\Psi _n^{r}$ and where $\varepsilon _n \to 0$ as $n\to \infty$ . Now, if we write

\begin{equation*} u_i = \sum _{j=0}^\infty \alpha _{i,j} v_j(M_n^r), \end{equation*}

the last inequality becomes

\begin{equation*} \sum _j |\lambda _j(M_n^r) -\lambda (\varphi )|^2 \alpha _{i,j}^2 = \sum _j |(M_n^r - \lambda (\varphi )\mathrm {Id}_n) v_j(M_n^r) |^2 \alpha _{i,j}^2 \leq \varepsilon _n, \;\; \forall 1 \leq i \leq d. \end{equation*}

Recall that, by (12), for $j \notin \{l_{s+1},..,l_{s+d}\}$ , we have $\frac{3\delta }{4}\leq |\lambda _j(M_n^r) -\lambda (\varphi )|$ . Hence.

(13) \begin{equation} \sum _{j \notin \{l_{s+1},..,l_{s+d}\}} \alpha _{i,j}^2 \leq \frac{16 \varepsilon _n}{9\delta ^2} \to 0, \end{equation}

and thus

\begin{equation*} \left \|u_i - \sum _{k=1}^d \alpha _{i, l_{s+k}} v_{l_{s+k}}(M_n^r)\right \|_2^2\to 0, \;\; \forall 1 \leq i \leq d. \end{equation*}

Define a $d\times d$ -matrix $B$ by $B_{i,j}=\alpha _{i, l_{s+j}}$ . Then we can rewrite the above as

\begin{equation*} \left \|(u_1,\ldots,u_d) - (v_{l_{s+1}}(M_n^r),\ldots, v_{l_{s+d}}(M_n^r))\cdot B\right \|_F^2 \to 0. \end{equation*}

Now, since for two $n\times d$ matrices $R,S$ we have $\|RR^T - SS^T\|_F\leq (\|R\|_{op} +\|S\|_{op})\|R-S\|_F$ . It follows that

(14) \begin{equation} \|G'_{\!\!n} - (v_{l_{s+1}}(M_n^r), \ldots, v_{l_{s+d}}(M_n^r))BB^T(v_{l_{s+1}}(M_n^r), \ldots, v_{l_{s+d}}(M_n^r))^T\|_F \to 0. \end{equation}

Observe that

\begin{equation*} B_{i,j} = \langle u_i, u_j \rangle - \sum _{k \notin \{l_{s+1},..,l_{s+d}\}} \alpha _{i,k} \alpha _{j,k}, \end{equation*}

implying that

\begin{equation*} |(B B^T)_{i,j} - (E_{i,j} + \mathrm {Id}_d)| \leq \sqrt { \sum _{k \notin \{l_{s+1},..,l_{s+d}\}} \alpha _{i,k}^2 \sum _{k \notin \{l_{s+1},..,l_{s+d}\}} \alpha _{j,k}^2 } \stackrel {\text {13}}{\longrightarrow } 0. \end{equation*}

where $E = E_{n,r}$ . Consequently we have

\begin{equation*} \|B B^T - \mathrm {Id}_d \|_{op} \leq \sqrt { \sum _{k \notin \{l_{s+1},..,l_{s+d}\}} \alpha _{i,k}^2 \sum _{k \notin \{l_{s+1},..,l_{s+d}\}} \alpha _{j,k}^2 } + \sqrt {C(n,r)} \to 0, \end{equation*}

which implies that

\begin{align*} \|v_{l_{s+1}}(M_n^r), \ldots, v_{l_{s+d}}(M_n^r))(BB^T-\mathrm{Id}_d)(v_{l_{s+1}}(M_n^r), \ldots, v_{l_{s+d}}(M_n^r))^T\|_F^2\to 0. \end{align*}

Combining with (14) finally yields

\begin{equation*} \|G'_{\!\!n} - G_n^r\|_F \to 0. \end{equation*}

in probability, as $n\to \infty$ . If $P$ is the orthogonal projection onto $V^r_n=\mathrm{span}(v_{l_{s+1}}(M_n^r),\ldots,v_{l_{s+d}}(M_n^r))$ , and $Q$ is the orthogonal projection onto $\mathrm{span}(v_{l_{s+1}}(M_n),\ldots,v_{l_{s+d}}(M_n))$ , then Lemma 11 shows that for all $\eta \gt 0$ , with probability at least $1-\eta$ , as $n\to \infty$ (and $r=n^{\frac{1}{2d}}\to \infty$ ), we have for every unit vector $v$

(15) \begin{equation} |(P - \mathrm{Id}_n) Q v | \leq \varepsilon _n \end{equation}

with some $\varepsilon _n \to 0$ . By symmetry, we also have for every unit vector $v$ that

\begin{equation*} |(Q - \mathrm {Id}_n) P v | \leq \varepsilon _n \end{equation*}

(this uses that fact that both $P$ and $Q$ are projections into subspaces of the same dimension). The last two inequalities easily yield that $\|P - Q\|_{op} \leq \varepsilon _n$ . Since this is true for every $\eta \gt 0$ , it follows that

\begin{equation*} \|G_n-G_n^r\|_F\to 0, \end{equation*}

in probability, as $n\to \infty$ .

Now, after establishing that $M_n$ is close to $\Psi _n^{d+1}$ , the second step is to recover $M_n$ (and therefore $\Psi _n^{d+1}$ ), from the observed $A$ . For the proof we will need the following instance of the Davis-Kahan theorem.

Theorem 13. ([Reference Yu, Wang and Samworth21, Theorem 2]). Let $X,Y$ be symmetric matrices with eigenvalues $\lambda _0\geq \ldots \geq \lambda _p$ resp. $\hat \lambda _0\geq \ldots \geq \hat{\lambda }_p$ with corresponding orthonormal eigenvectors $v_0,\ldots,v_p$ resp. $\hat v_0,\ldots, \hat v_p$ . Let $V=(v_{s+1},\ldots, v_{s+d})$ and $\hat V = (\hat v_{s+1},\ldots, \hat v_{s+d})$ . Then there exists an orthogonal $ d \times d$ matrix $R$ such that

\begin{equation*} \| \hat V R - V \|_{F} \leq \frac {2^{3/2} \min (d^{1/2}\| Y-X\|_{op}, \|Y-X\|_F)}{\min (\lambda _{s}-\lambda _{s+1}, \lambda _{s+d}-\lambda _{s+d+1})}. \end{equation*}

Our main tool to pass from the expectation of the adjacency matrix to the matrix itself is the following result regarding concentration of random matrices, which follows from [Reference Le, Levina and Vershynin11, Theorem 1.1] (see also [Reference Le, Levina and Vershynin12, Theorem 5.1]).

Theorem 14. Let $A$ be the adjacency matrix of a random graph drawn from $G\left (n, \frac{1}{n} \varphi (X_i, X_j)\right )$ . For every vertex of degree larger than $\max _{ij}\varphi (X_i,X_j)$ , remove all incident edges and let $A'$ denote the new adjacency matrix. Then,

\begin{equation*} \|A'-M_n\| \leq C \sqrt {\|\varphi \|_\infty }. \end{equation*}

We can now prove the main reconstruction theorem.

Proof of Theorem 3. Let $A$ be the adjacency matrix of a random graph drawn from the model $G\left (n, \frac{1}{n} \varphi (X_i, X_j)\right )$ and let $A'$ be the adjacency matrix defined in Theorem 14.

Denote by $\lambda'_{\!\!0}\geq \lambda'_{\!\!1}\geq \ldots$ the eigenvalues of $A'$ and by $v'_{\!\!0}, v'_{\!\!1},\ldots$ the corresponding orthonormal eigenvectors. Let $Y = (v'_{\!\!l_{s+1}},\ldots,v'_{\!\!l_{s+d}})$ . By Theorem 13 there exists an $R\in \mathcal{O}(d)$ such that

\begin{align*} \left \lVert \left (v_{l_{s+1}}(M_n),\ldots,v_{l_{s+d}}(M_n)\right ) - YR \right \rVert _{F} &\leq \frac{2^{3/2} d^{1/2}\| M_n - A'\|_{op}}{\min _{i\neq 1,\ldots,d}|\lambda _{i}-\lambda (\varphi )|}. \end{align*}

Hence by Theorem 14 we have

\begin{equation*} \left \lVert \left (v_{l_{s+1}}(M_n),\ldots,v_{l_{s+d}}(M_n)\right ) - YR \right \rVert _{F} \leq C \sqrt { d} \cdot \frac {\sqrt {\|\varphi \|_\infty }}{ \min _{i\neq 1,\ldots,d}|\lambda _{i}-\lambda (\varphi )|}, \end{equation*}

with probability $1-n^{-1}$ . It follows that

\begin{equation*} \left \lVert G_n - Y Y^T \right \rVert _{F} \leq C \sqrt { d} \cdot \frac {\sqrt {\|\varphi \|_\infty }}{ \min _{i\neq 1,\ldots,d}|\lambda _{i}-\lambda (\varphi )|} \end{equation*}

Combining this with Theorem 12 yields

\begin{align*} \left \lVert G'_{\!\!n} - Y Y^T \ \right \rVert _{F} \leq C \sqrt{d} \cdot \frac{\sqrt{\|\varphi \|_\infty }}{\min _{i\neq 1,\ldots,d}|\lambda _{i}-\lambda (\varphi )|}, \end{align*}

So,

\begin{equation*} \frac {1}{n^2} \sum _{i,j} \left |X_i \cdot X_j - (n Y Y^T)_{i,j} \right |^2 \leq C d \frac {\|\varphi \|_\infty }{\min _{i\neq 1,\ldots,d}|\lambda _{i}-\lambda (\varphi )|^2} \end{equation*}

which gives the desired reconstruction bound.

3. Lower bounds

Our approach to proving lower bounds will be to exploit some symmetries which are inherent to well behaved kernel functions. We thus make the following definition:

Definition 15. (DPS property). Let $\mu$ be a probability measure on $\mathbb{S}^{d-1}$ , and let $w \in \mathbb{S}^{d-1}$ . We say that $\mu$ has the diminishing post-translation symmetry (DPS) around $w$ property with constant $\varepsilon$ , if there exists a decomposition,

\begin{equation*}\mu = (1-\varepsilon )\mu _w + \varepsilon \mu _w^s.\end{equation*}

Here $\mu _w^s$ is a probability measure, invariant to reflections with respect to $w^\perp$ . In other words, if $R = \mathrm{Id}_d - 2ww^T$ , $R_*\mu _w^s = \mu _w^s$ . For such a measure we denote $\mu \in \mathrm{DPS}_w(\varepsilon )$ .

If instead $\mu$ is a measure on $(\mathbb{S}^{d-1})^{|T_k|}$ , we say that $\mu \in \mathrm{DPS}^k_w(\varepsilon )$ if a similar decomposition exists but now the reflections should be applied to each coordinate separately.

We now explain the connection between the DPS property and percolation of information in trees. For this, let us recall the random function $g:T \to \mathbb{S}^{d-1}$ , introduced in Section 1.2, which assigns to the root, $r$ , a uniformly random value and for any other $u \in T$ , the label $g(u)$ is distributed according to $\varphi (g(\mathrm{parent}(u)),\cdot )d\sigma$ .

Lemma 16. Suppose that there exist a sequence $(p_k)_k$ with $\lim _{k\to \infty } p_k= 1$ such that for every $w,x_0 \in \mathbb{S}^{d-1}$ and every $k \gt 0$ ,

\begin{equation*}\mathrm {Law}((g(v))_{v\in T_k}|g(r) = x_0) \in \mathrm {DPS}^k_w(p_k).\end{equation*}

Then there is zero percolation of information along $T$ .

Proof. Denote $X = g(r)$ and $Y = g(v)_{v\in T_k}$ and let $\rho _{X|Y}$ be the density of $X|Y$ with respect to $\sigma$ . Our aim is to show that $\mathbb E_Y\left [\mathrm{TV}(X|Y,\sigma )\right ] = o(1)$ . We first claim that it is enough to show that for all $x,x' \in \mathbb{S}^{d-1}$ and all $\delta \gt 0$ one has,

(16) \begin{equation} \mathbb P \left . \left ( \frac{\rho _{X|Y}(x) }{ \rho _{X|Y}(x') } - 1 \leq \delta \right | X = x \right ) = 1-o(1). \end{equation}

Indeed, let $H = \left \{\frac{\rho _{X|Y}(x) }{ \rho _{X|Y}(x') } - 1 \leq \delta \right \}$ . Note that, by Bayes’ rule,

\begin{equation*} \frac {\rho _{X|Y}(x) }{ \rho _{X|Y}(x') } = \frac {\rho _{Y|X=x}(Y) }{ \rho _{Y|X=x'}(Y) }. \end{equation*}

Let $x \in \mathbb{S}^{d-1}$ be some fixed point and let $x'$ be uniformly distributed is $\mathbb{S}^{d-1}$ and independent from $Y$ . We will use $\mathbb E_{x'}$ to denote integration with respect to $x'$ , which has density $\rho _X$ . Similarly, $\mathbb E_{Y}$ stands for integration with respect to the density $\rho _Y$ of $Y$ and $\mathbb{E}_{x', Y}$ is with respect to the product $\rho _X\cdot \rho _Y$ . The symbol $\mathbb P$ will always mean probability over $x'$ and $Y$ .

As a consequence of the assumption in (16), as long as $k$ is large enough, we have,

\begin{equation*}\mathbb P\left (H|X=x\right ) \geq 1-\delta .\end{equation*}

For any Borel measurable set $A$ , we denote the projection:

\begin{align*} \Pi _YA &= \{y \in (\mathbb{S}^{d-1})^{|T_k|}\;:\; (\tilde{x}, y) \in A \text{ for some } \tilde{x} \in \mathbb{S}^{d-1}\}. \end{align*}

Define now the set

\begin{equation*}H' \;:\!=\; \{(\tilde {x}, y)| (\tilde {x},y)\in H \text { and } \mathbb {E}_{x'}[\textbf {1}_H(\cdot, y)]\geq 1 - 4\delta \}.\end{equation*}

In particular, for every $y \in \Pi _YH'$ ,

\begin{equation*}\mathbb {E}_{x'}[\textbf {1}_H(\cdot, y)]\geq 1 - 4\delta,\end{equation*}

and by Fubini’s theorem, $\mathbb P\left (H'|X = x\right ) \geq 1 - 4\delta$ .

Consider the random variables $\alpha \;:\!=\; \alpha (Y) = \frac{\rho _{Y|X=x}(Y)}{ \rho _{Y}(Y) }$ and $\beta \;:\!=\;\beta (x',Y) = \frac{\rho _{Y|X=x'}(Y)}{\rho _Y(Y)}$ . By definition of $H$ , we have

(17) \begin{equation} (1-\delta )\textbf{1}_H\alpha \leq \textbf{1}_H\beta . \end{equation}

Moreover, for almost every $Y$ ,

\begin{equation*}\mathbb E_{x'}\left [\beta \right ] = \frac {1}{\rho _Y(Y)}\int \rho _{Y|X=x'}(Y)\rho _X(x') = \frac {1}{\rho _Y(Y)}\rho _Y(Y) = 1.\end{equation*}

So,

\begin{equation*} (1-\delta )(1-4\delta )\mathbb E_Y\left [\alpha \textbf {1}_{ \Pi _Y{H'}}\right ]\leq (1-\delta )\mathbb E_{Y}\left [\alpha \mathbb E_{x'}\left [\textbf {1}_{H'}\right ]\right ] \leq \mathbb E_{x',Y}\left [\beta \textbf {1}_{H'}\right ]\leq \mathbb E_{x',Y}\left [\beta \right ]=1. \end{equation*}

Observe that $\mathbb E_Y\left [\alpha \textbf{1}_{ \Pi _Y{H'}}\right ] = \mathbb P\left (\Pi _Y{H'}|X=x\right ) = 1- o(1)$ , where the second equality is a consequence of Fubini’s theorem. Hence, let us write $h_1(\delta )$ , so that

\begin{equation*}1 - h_1(\delta ) = (1-\delta )(1-4\delta )\mathbb E_Y\big[\alpha \textbf {1}_{ \Pi _Y{H'}}\big] \leq (1-\delta )\mathbb E_{x', Y}\left [\alpha \textbf {1}_{H'}\right ].\end{equation*}

Markov’s inequality then implies,

(18) \begin{equation} \mathbb P\left (\beta \textbf{1}_{H'}\geq (1-\delta )\alpha \textbf{1}_{H'}+\sqrt{h_1(\delta )}\right )\leq \frac{\mathbb E_{x', Y}\left [\beta \textbf{1}_{H'}\right ]-(1-\delta )\mathbb E_{x', Y}\left [\alpha \textbf{1}_{H'}\right ]}{\sqrt{h_1(\delta )}}\leq \sqrt{h_1(\delta )}. \end{equation}

Now, we integrate over $x'$ to obtain,

\begin{equation*}(1-\delta )(1-4\delta )\alpha \textbf {1}_{\Pi _Y{H'}}\leq (1-\delta )\alpha \mathbb E_{x'}\left [\textbf {1}_{H'}\right ] \leq \mathbb E_{x'}\left [\beta \textbf {1}_{H'}\right ] \leq 1,\end{equation*}

which implies

(19) \begin{equation} \alpha \textbf{1}_{H'} \leq \frac{1}{(1-\delta )(1-4\delta )}. \end{equation}

Keeping in mind the previous displays, we may choose $h_2(\delta )$ , which satisfies $\lim \limits _{\delta \to 0} h_2(\delta ) = 0$ , $\alpha \textbf{1}_{H'} \leq 1 + h_2(\delta )$ and $\mathbb E_{x', Y}\left [\alpha \textbf{1}_{H'}\right ] \geq 1 - h_2(\delta )$ .

So, an application of the reverse Markov’s inequality for bounded and positive random variables shows,

(20) \begin{align} \mathbb P\left (\alpha \textbf{1}_{H'}\geq 1-\sqrt{h_2(\delta )}\right ) \geq \frac{\mathbb E_{x', Y}\left [\alpha \textbf{1}_{H'}\right ] - (1-\sqrt{h_2(\delta )})}{1+h_2(\delta ) - (1-\sqrt{h_2(\delta )})} &\geq \frac{1-h_2(\delta ) - (1-\sqrt{h_2(\delta )})}{1+h_2(\delta ) - (1-\sqrt{h_2(\delta )})}\nonumber \\[5pt] &= \frac{\sqrt{h_2(\delta )} - h_2(\delta )}{\sqrt{h_2(\delta )} + h_2(\delta )}. \end{align}

Note that as $\delta \to 0$ the RHS goes to $1$ . Thus, by combining the above displays, there exists a function $h$ , which satisfies $\lim \limits _{\delta \to 0}h(\delta ) = 0$ and some $H'' \subset H'$ , with $\mathbb P\left (H'\right ) \geq 1 -h(\delta )$ , such that, by (19) and (20),

\begin{equation*} \textbf {1}_{H''}|\alpha -1|\leq h(\delta ), \end{equation*}

which implies, together with (17) and (18)

\begin{equation*} \textbf {1}_{H''}|\alpha -\beta |\leq h(\delta ). \end{equation*}

This then gives

\begin{equation*} \textbf {1}_{H''}|1 -\beta |\leq 2h(\delta ). \end{equation*}

We can thus conclude,

\begin{align*} \mathbb E_Y \mathrm{TV} \left ( X|Y, X \right ) &= \mathbb E_{Y,x'} \left [ |\beta - 1| \right ] = 2\mathbb E_{Y,x'} \left [(1-\beta )\textbf{1}_{\beta \leq 1} \right ] \\[5pt] &\leq 2 \mathbb E_{Y,x'} \left [(1-\beta )\textbf{1}_{\beta \leq 1}\textbf{1}_{H''}\right ] +1-\mathbb P\left (H''\right )\\[5pt] &= 2 \mathbb E_{Y,x'} \left [|1-\beta |\textbf{1}_{H''}\right ] +h(\delta )\\[5pt] &\leq 5h(\delta ).\\[5pt] \end{align*}

Take now $\delta \to 0$ to get $\mathbb E_Y \mathrm{TV} \left ( X|Y, X \right ) \to 0$ .

Thus, we may assume towards a contradiction that there exist $x, x' \in \mathbb{S}^{d-1}$ and a set $F\subset (\mathbb{S}^{d-1})^{|T_k|}$ , such that

(21) \begin{equation} \mathbb P\left (Y \in F | X = x \right ) \geq \delta, \end{equation}

and under $\{Y \in F \}$ ,

(22) \begin{equation} \frac{\rho _{X|Y}(x)}{\rho _{X|Y}(x')} \geq 1+ \delta, \end{equation}

for some constant $\delta \gt 0$ .

Let $w\in \mathbb{S}^{d-1}$ be such that the reflection $R\;:\!=\;\mathrm{Id}_d - 2ww^T$ satisfies $Rx = x'$ . Under our assumption, there exists an event $A_k$ , which satisfies

\begin{equation*} \mathbb P(A_k|X=x) = 1 - o(1) \end{equation*}

and such that $Y|(X=x,A_k)$ is $R$ -invariant. By (21), we also have

\begin{equation*}\mathbb P(A_k|X=x,Y \in F) = 1 - o(1),\end{equation*}

and thus there exists $y \in F$ such that

(23) \begin{equation} \mathbb P(A_k | X=x,Y = y ) = 1-o(1). \end{equation}

By continuity, we can make sense of conditioning on the zero probability event $E \;:\!=\; \bigl \{X = x, \; Y \in \{y, R y\} \bigr \}$ , in the sense of regular probabilities. Note that we have by symmetry and since $y \in F$ ,

(24) \begin{equation} Y = y \;\; \Rightarrow \;\; 1+\delta \leq \frac{\rho _{X|Y}(x)}{\rho _{X|Y}(x')} = \frac{\mathbb P( Y = y | E )}{ \mathbb P( Y = R y | E )}. \end{equation}

On the other hand, we have by definition of $A_k$ ,

\begin{equation*} \mathbb P( Y = R y | E, A_k ) = \mathbb P( Y = y | E, A_k ), \end{equation*}

which implies that

\begin{align*} \mathbb P( Y = R y | E ) & \geq \mathbb P \bigl ( \{Y = R y \} \cap A_k | E \bigr ) \\[5pt] & = \mathbb P \bigl ( \{Y = y \} \cap A_k | E \bigr ) \\[5pt] & \geq \mathbb P( Y = y | E ) (1-o(1)) \end{align*}

which contradicts (24). The proof is complete.

3.1 The Gaussian case

Our aim is to show that certain classes of kernel functions satisfy the DPS condition. We begin by considering the case where the kernel $\varphi$ is Gaussian, as in (8). In this case, the function $g$ may be defined as follows. Let $T$ be a $q$ -ary tree. To each edge $e \in E(T)$ we associate a Brownian motion $(B_e(t))_{t \in (0,1)}$ of rate $\beta$ such that for every node $v \in V$ we have

\begin{equation*} g(v) = \sum _{e \in P(v)} B_e(1) \; \mathrm {mod} 2 \pi \end{equation*}

where $P(v)$ denotes the shortest path from the root to $v$ .

For every node $v \in T_k$ let us now consider the Brownian motion $(B_v(t))_{t=0}^k$ defined by concatenating the Brownian motions $B_e$ along the edges $e \in P(v)$ . Define by $E_v$ the event that the image of $B_v \mathrm{mod} \pi$ contains the entire interval $[0, \pi )$ , and define $E_k = \bigcap _{v \in T_k} E_v$ . Our lower bound relies on the following observation.

Claim 17. Fix $v\in T_k$ and set $p_k \;:\!=\; \mathbb P(E_k)$ . Then, for every $\theta,x_0 \in \mathbb{S}^1$ , $\mathrm{Law}(g(v)|g(r)=x_0)\in \mathrm{DPS}^k_\theta (p_k).$

Proof. Fix $\theta \in [0, \pi )$ . Given the event $E_v$ , we have almost surely that there exists a time $t_v \leq k$ such that $B_v(t_v) \in \{\theta - \pi, \theta \}$ . By symmetry and by the Markov property of Brownian motion, we have that the distribution of $g(v)$ conditioned on the event $\{B_v(t_v) = \theta, \;\; \forall v\}$ is symmetric around $\theta$ . Thus, by considering $(B_v(t))_{v\in T_k}$ , under the event $\{t_v \leq k|v \in T_k \}$ we get that for any $x_0$ , $\mathrm{Law}((g(v))_{v\in T_k}|g(r)=x_0)$ is symmetric around $\theta$ . So,

\begin{equation*}\mathrm {Law}((g(v))_{v\in T_k}|g(r)=x_0)\in \mathrm {DPS}^k_\theta (p_k).\end{equation*}

We will also need the following bound, shown for example in [Reference Ernst and Shepp8].

Lemma 18. Let $B(t)$ be a Brownian motion of rate $\beta$ on the unit circle. Then,

\begin{equation*} \mathbb P( \mathrm {Image} (B(s) \mathrm {mod} \pi )_{0 \leq s \leq t} = [0, \pi ) ) \geq 1- C e^{- t \beta/2}. \end{equation*}

We are now in a position to prove Theorem 7.

Proof of Theorem 7. Lemma 18 immediately implies that for all $v \in T_k$ ,

\begin{equation*} \mathbb P(E_v) \geq 1 - C e^{-k \beta/ 2}. \end{equation*}

On the other hand, a calculation gives $\lambda (\varphi ) = \mathbb E[\cos (B_1)] = e^{-\beta/2}$ . Thus, applying a union bound implies that for some constant $C \gt 0$ , $\mathbb P(E_k)\geq 1-Cq^k\lambda (\varphi )^k$ . Hence, by Claim 17,

\begin{equation*}\mathrm {Law}((g(v))_{v\in T_k}|g(r)=x_0) \in \mathrm {DPS}_w(1-Cq^k\lambda (\varphi )^k).\end{equation*}

The result is now a direct consequence of Lemma 16.

In the next section, we generalise the above ideas and obtain a corresponding bound which holds for distributions other than the Gaussian one.

3.2 The general case

3.2.1 On symmetric functions

We begin this section with simple criterion to determine whether a measure belongs to some $\mathrm{DPS}$ class. In the sequel, for $w \in \mathbb{S}^{d-1}$ , we denote,

\begin{equation*} H_w^+ =\{x \in \mathbb {S}^{d-1}| \langle x,w\rangle \geq 0\}\text { and }H_w^- =\{x \in \mathbb {S}^{d-1}| \langle x,w\rangle \lt 0\}. \end{equation*}

Lemma 19. Let $f\;:\; [-1,1] \to \mathbb R^+$ satisfy the assumptions of Theorem 9 and let $y \in \mathbb{S}^{d-1}$ . If $\mu = f(\langle \cdot,y\rangle )d\sigma$ , then for any $w \in \mathbb{S}^{d-1}$ , $\mu \in \mathrm{DPS}_w(2\cdot p_w)$ , where $p_w = \min \left (\mu (H_w^+),\mu (H_w^-)\right )$ .

Proof. Without loss of generality let us assume that $y \in H_w^+$ . Monotonicity of $f$ implies $\mu (H_w^+)\geq \mu (H_w^-)$ . Now, if $R = \mathrm{Id}_d - 2ww^T$ is the reflection matrix with respect to $w^\perp$ , then we have for any $x \in H^-_w$ ,

\begin{equation*} f(\langle x,y\rangle ) \leq f(\langle Rx,y\rangle ). \end{equation*}

This follows since $\langle x,y\rangle \leq \langle Rx,y\rangle$ .

Let us now define the measure $\tilde{\mu }_w^s$ such that

\begin{equation*} \frac {d\tilde {\mu }_w^s}{d\sigma }(x) = \left \{\begin {array}{ll} f(\langle x,y\rangle ) & \text { if }x \in H^-_w\\[5pt] f(\langle Rx,y\rangle )& \text { if }x \in H^+_w. \end {array}\right . \end{equation*}

$\tilde{\mu }_w^s$ is clearly $R$ -invariant and the above observation shows that $\tilde{\mu }_w^s(\mathbb{S}^{d-1}) \leq 1$ . We can thus define $\tilde{\mu }_w = \mu - \tilde{\mu }_w^s$ . To obtain a decomposition, define $\mu _w^s = \frac{\tilde{\mu }^s_w}{\tilde{\mu }_w^s(\mathbb{S}^{d-1})}$ and $\mu _w = \frac{\tilde{\mu }_w}{\tilde{\mu }_w(\mathbb{S}^{d-1})}$ , for which

\begin{equation*}\mu = (1-\tilde {\mu }^s_w(\mathbb {S}^{d-1}))\mu _w + \tilde {\mu }_w^s(\mathbb {S}^{d-1})\mu _w^s.\end{equation*}

The proof is concluded by noting $\tilde{\mu }_w^s(\mathbb{S}^{d-1}) = 2\mu (H_w^-)$ .

Our main object of interest will be the measure $\mu _\varphi (y)$ which, for a fixed $y$ , is defined by

(25) \begin{equation} \mu _\varphi (y) \;:\!=\; \varphi (x,y)d\sigma (x) = f (\langle x,y\rangle )d\sigma (x). \end{equation}

Let us now denote,

\begin{equation*} \beta _d(t) \;:\!=\; \frac {\Gamma (\frac {d}{2})}{\Gamma (\frac {d-1}{2})}(1-t^2)^{(d-3)/2}, \end{equation*}

which is the marginal of the uniform distribution on the sphere. We now show that the spectral gap of $\varphi$ may determine the $\mathrm{DPS}$ properties of $\mu _\varphi (y)$ .

Lemma 20. Let $w \in \mathbb{S}^{d-1}$ and suppose that $f$ is monotone. If $|\langle w,y\rangle |\leq \frac{1-\lambda (\varphi )}{16\sqrt{d}}$ , then

\begin{equation*}\mu _\varphi (y) \in \mathrm {DPS}_w\left (\frac {1-\lambda (\varphi )}{35}\right ).\end{equation*}

Proof. Assume W.L.O.G. that $\langle y,w\rangle \gt 0$ . By Lemma 19 it will be enough to bound $\int \limits _{H_w^-}\mu _{\varphi }(y)$ from below. Let $X \sim \mu _{\varphi }(y)$ and define $Z = \langle X,y\rangle$ . We have $\mathbb E\left [Z\right ] = \lambda (\varphi )$ and by Markov’s inequality,

\begin{equation*}\mathbb P\left (Z \leq \frac {1+\lambda (\varphi )}{2}\right ) = \mathbb P\left (Z + 1 \leq \frac {1+\lambda (\varphi )}{2} + 1\right ) \geq 1 - \frac {2(\lambda (\varphi ) +1)}{3 + \lambda (\varphi )}\geq \frac {1-\lambda (\varphi )}{4}.\end{equation*}

For $t \in [-1,1]$ , set $S_t = \{x \in \mathbb{S}^{d-1}| \langle x,y\rangle = t\}$ and let $\mathcal{H}^{d-2}$ stand for $d-2$ -dimensional Hausdorff measure. To bound $\int \limits _{H_w^-}\mu _{\varphi }(y)$ we would first like to estimate $\frac{\mathcal{H}^{d-2}(S_t\cap H_w^-)}{\mathcal{H}^{d-2}(S_t)}$ .

We know that $0 \leq \langle w,y\rangle \leq \frac{1-\lambda (\varphi )}{16\sqrt{d}}$ . Denote $t_y \;:\!=\; \langle w,y\rangle$ and fix $t \leq t_0 \;:\!=\; \frac{1+\lambda (\varphi )}{2}$ . With no loss of generality, let us write $w = e_1$ and $y = t_ye_1 + \sqrt{1-t_y^2}e_2$ . Define now $z = -\sqrt{1-t_y^2}e_1+t_ye_2$ . We claim that

(26) \begin{equation} \left \{v\in S_t\Big | \frac{\langle v,z\rangle }{\sqrt{1-t^2}} \geq \frac{1}{2\sqrt{d}}\right \}\subseteq S_t\cap H_w^-. \end{equation}

If $v\in S_t$ , its projection onto the plane $\mathrm{span}(y,w)$ , can be written as $t\cdot y + \sqrt{1-t^2}c\cdot z$ , for some $c\in [-1,1]$ . So,

\begin{equation*}\langle v,w\rangle = t\cdot t_y - \sqrt {1-t^2}\sqrt {1-t_y^2}c.\end{equation*}

Now, whenever

\begin{equation*}c \gt \frac {t\cdot t_y}{\sqrt {1-t^2}\sqrt {1-t_y^2}},\end{equation*}

we get $\langle v,w\rangle \lt 0$ . Also,

\begin{equation*} \frac {t\cdot t_y}{\sqrt {1-t^2}\sqrt {1-t_y^2}}\leq \frac {1}{\sqrt {3}}\frac {t_y}{\sqrt {1-t_y^2}}\leq \frac {1}{2\sqrt {d}},\end{equation*}

where we have used $ t \leq \frac{1}{2}$ for the first inequality and $t_y \leq \frac{1}{2\sqrt{d}}$ in the second inequality. By combining the above displays with $\frac{\langle v,z\rangle }{\sqrt{1-t^2}} = c$ , (26) is established. Thus, by taking the marginal of $S_t$ in the direction of $-z$ , we see

\begin{align*} \frac{\mathcal{H}^{d-2}(S_t\cap H_w^-)}{\mathcal{H}^{d-2}(S_t)} &\geq \int \limits _{-1}^{-\frac{1}{2\sqrt{d}}}\beta _{d-1}(s)ds\geq \int \limits _{-\frac{1}{\sqrt{d}}}^{-\frac{1}{2\sqrt{d}}}\beta _{d-1}(s)ds\geq \frac{1}{2\sqrt{d}}\beta _{d-1}\left (\frac{1}{\sqrt{d}}\right )\\[5pt] &\geq \frac{1}{2\sqrt{d}}\frac{\Gamma (\frac{d-1}{2})}{\Gamma (\frac{d-2}{2})}\left (1-\frac{1}{d}\right )^{(d-4)/2}\geq \frac{1}{10\sqrt{e}}, \end{align*}

where we used $\frac{\Gamma (\frac{d-1}{2})}{\Gamma (\frac{d-2}{2})} \geq \frac{\sqrt{d}}{5},$ valid for any $d \geq 3$ . We use the above estimates with Fubini’s theorem to obtain:

\begin{align*} \mathbb P\left (X \in H_w^-\right ) &= \int \limits _{-1}^1f(t)\mathcal{H}^{d-2}(S_t\cap H_w^-)dt \geq \int \limits _{-1}^{t_0}f(t)\mathcal{H}^{d-2}(S_t\cap H_w^-)dt\\[5pt] &\geq \frac{1}{10\sqrt{e}}\int \limits _{-1}^{t_0}f(t)\mathcal{H}^{d-2}(S_t)dt = \frac{1}{10\sqrt{e}}\mathbb P\left (Z \leq \frac{1+\lambda (\varphi )}{2}\right )\geq \frac{1-\lambda (\varphi )}{70}. \end{align*}

3.2.2 Mixing

Recall the random function $g:T \to \mathbb{S}^{d-1}$ , introduced in Section 1.2, which assigns to the root, $r$ , a uniformly random value and for any other $u \in T$ , the label $g(u)$ is distributed according to $\varphi (g(\mathrm{parent}(u)),\cdot )d\sigma \;=\!:\; \mu _\varphi (\mathrm{parent}(u))$ . Suppose that $v \in T_k$ and let $\{v_i\}_{i=0}^k$ denote the simple path from $r$ to $v$ in $T$ . Fix $x_0 \in \mathbb{S}^{d-1},$ for $i = 0,\ldots,k$ , we now regard,

\begin{equation*} X_i \;:\!=\;g(v_i)|g(r) = x_0, \end{equation*}

as a random walk on $\mathbb{S}^{d-1}$ . Observe that given $X_{i-1}$ , $X_i\sim \mu _\varphi (X_{i-1}).$ The following lemma shows that this random walk is rapidly mixing.

Lemma 21. For $w \in \mathbb{S}^{d-1}$ , let

\begin{equation*}S(w) = \left \{u \in \mathbb {S}^{d-1}\;:\; |\langle u,w\rangle | \leq \frac {1-\lambda (\varphi )}{16\sqrt {d}}\right \},\end{equation*}

and set $k_0 = \frac{\ln \left (\frac{\lambda (\varphi )(1-\lambda (\varphi ))}{32 f(1)}\right )}{\ln (\lambda (\varphi ))}$ . Then, if $f$ satisfies the assumptions of Theorem 9 ,

\begin{equation*}\mathbb P(X_{k_0} \in S(w)) \geq \frac {1-\lambda (\varphi )}{32}.\end{equation*}

Proof. Note that if $U \sim \mathrm{Uniform}(\mathbb{S}^{d-1})$ , then

(27) \begin{equation} \mathbb P(U \in S(w)) = \int \limits _{|t| \leq \frac{1-\lambda (\varphi )}{16\sqrt{d}}}\beta _d(t)dt \geq 2\frac{\Gamma (\frac{d}{2})}{\Gamma \left (\frac{d-1}{2}\right )}\beta _d\left (\frac{1}{16\sqrt{d}}\right )\frac{1-\lambda (\varphi )}{16\sqrt{d}} \geq \frac{1-\lambda (\varphi )}{16}. \end{equation}

It will then suffice to show that $\mathbb P(X_{k_0} \in S(w))$ can be well approximated by $\mathbb P(U \in S(w))$ . Since $X_{k_0}$ has density $A^{k_0-1}_\varphi f(\langle x,x_0\rangle )$ , the following holds true,

\begin{align*} \left (\mathbb P(U \in S(w)) - \mathbb P(X_{k_0} \in S(w))\right )^2 &= \left (\int \limits _{S(w)}\left (A_\varphi ^{k_0-1}f(\langle x,x_0\rangle ) - 1\right )d\sigma (x)\right )^2\\[5pt] &\leq \int \limits _{S(w)}\left (A_\varphi ^{k_0-1}f(\langle x,x_0\rangle ) - 1\right )^2d\sigma (x). \end{align*}

We now decompose the density as $f(\langle x,x_0\rangle ) = \sum \limits _{i = 0}^\infty \lambda _i \psi _i(x)\psi _i(x_0)$ . So that

\begin{equation*}A_\varphi ^{k_0-1}f(\langle x,x_0\rangle ) = \sum \limits _{i = 0}^\infty \lambda _i^{k_0} \psi _i(x)\psi _i(x_0).\end{equation*}

We know that $\psi _0 \equiv 1$ with eigenvalue $\lambda _0 = 1$ , and we’ve assumed that $|\lambda _i|\leq \lambda _1 = \lambda (\varphi )$ for every $i \geq 1$ . Thus,

\begin{align*} \int \limits _{S(w)}\left (A_\varphi ^{k_0-1}f(\langle x,x_0\rangle ) - 1\right )^2d\sigma (x) &= \int \limits _{S(w)}\left (\sum \limits _{i =1}^\infty \lambda _i^{k_0}\psi _i(x)\psi _i(x_0)\right )^2d\sigma (x)\\[5pt] &\leq (\lambda (\varphi ))^{2k_0-2}\int \limits _{S(w)}\sum \limits _{i =1}^\infty (\lambda _i\psi _i(x))^2\sum \limits _{i =1}^\infty (\lambda _i\psi _i(x_0))^2d\sigma (x) \\[5pt] &\leq \lambda (\varphi )^{2k_0-2}f(1)^2. \end{align*}

where in the last inequality we have used $f(1) = \sum \lambda _i\psi _i(y)\psi _i(y)$ , which is valid for any $y \in \mathbb{S}^{d-1}$ . Thus, since $k_0 = \frac{\ln \left (\frac{\lambda (\varphi )(1-\lambda (\varphi ))}{32 f(1)}\right )}{\ln (\lambda (\varphi ))}$ , by (27), we get,

\begin{equation*}\mathbb P(X_{k_0} \in S(w)) \geq \mathbb P(U \in S(w)) - \frac {1-\lambda (\varphi )}{32} \geq \frac {1-\lambda (\varphi )}{32}.\end{equation*}

Since the random walk $X_k$ mixes well, we now use Lemma 20 to show that after enough steps, $X_k$ will be approximately invariant to a given reflection.

Lemma 22. Let $w,x_0 \in \mathbb{S}^{d-1}$ . Then,

\begin{equation*}\mathrm {Law}((g(v))_{v \in T_k}|g(r) = x_0) \in \mathrm {DPS}^k_w\left (1- q^kp^{k}\right ), \end{equation*}

where $p = \left (1-\frac{\ln (\lambda (\varphi ))}{\ln \left (\frac{\lambda (\varphi )(1-\lambda (\varphi ))}{32 f(1)}\right )}\frac{(1-\lambda (\varphi ))^2}{600}\right )$ .

Proof. Let $R =\mathrm{Id}_d - 2ww^T$ denote the linear reflection with respect to $w^\perp$ . Then, the claim is equivalent to the decomposition,

\begin{equation*}X_k = P\tilde {X}_k + (1-P)X_k^R,\end{equation*}

where $X_k^R$ is invariant to reflections by $R$ and $P \sim \mathrm{Bernoulli}(s_k)$ is independent from $\{\tilde{X}_k,X_k^R\}$ with $s_k \leq \left (1-\frac{\ln (\lambda (\varphi ))}{\ln \left (\frac{\lambda (\varphi )(1-\lambda (\varphi ))}{32 f(1)}\right )}\frac{(1-\lambda (\varphi ))^2}{600}\right )^{k}$ .

Consider the case that for some $i = 0,\ldots,k$ , $|\langle X_i, w \rangle | \leq \frac{1-\lambda (\varphi )}{16\sqrt{d}}$ . In this case, from Lemma 20, given $X_i$ , we have the decomposition,

\begin{equation*}\mu _\varphi (X_i) = \left (1 - \frac {(1-\lambda (\varphi ))}{35}\right )\mu _{\varphi,w} + \frac {(1-\lambda (\varphi ))}{35}\mu _{\varphi,w}^s(X_i),\end{equation*}

where $\mu _{\varphi,w}^s(X_i)$ is $R$ -invariant.

We now generate the random walk in the following way. For $i = 0,\ldots,k$ , let

(28) \begin{equation} P_i \sim \mathrm{Bernoulli}\left (\frac{(1-\lambda (\varphi ))}{35}\right ), \end{equation}

be an i.i.d sequence. Given $X_i$ , if $|\langle X_i, w \rangle | \gt \frac{1-\lambda (\varphi )}{16\sqrt{d}}$ then $X_{i+1} \sim \mu _\varphi (X_i)$ . Otherwise, $|\langle X_i, w \rangle | \leq \frac{1-\lambda (\varphi )}{16\sqrt{d}}$ . To decide on the position of $X_{i+1}$ we consider $P_i$ . If $P_i = 0$ then $X_{i+1} \sim \mu _{\varphi,w}(X_i)$ . If $P_i = 1$ , we generate $X_{i+1} \sim \mu _{\varphi,w}^s(X_i)$ . We denote the latter event by $A_i$ and $A = \cup _{i=0}^{k-1}A_i$ .

It is clear that, conditional on $A$ , $RX_k \stackrel{law}{=} X_k$ . Thus, to finish the proof, if $\bar{A}$ is the complement of $A$ , we will need to show

\begin{equation*}\mathbb P(\bar {A}) \leq \left (1-\frac {\ln (\lambda (\varphi ))}{\ln \left (\frac {\lambda (\varphi )(1-\lambda (\varphi ))}{32 f(1)}\right )}\frac {(1-\lambda (\varphi ))^2}{600}\right )^{k}.\end{equation*}

Towards this, let $S(w)$ and $k_0$ be as in Lemma 21. Coupled with (28), the lemma tells us that

\begin{equation*}\mathbb P\left (A_{k_0}\right ) \geq \frac {(1-\lambda (\varphi ))^2}{600}.\end{equation*}

Now, by restarting the random walk from $X_{k_0}$ if needed, we may show,

\begin{equation*}\mathbb P\left (\bar {A}\right )\leq \sum \limits _{m\leq \frac {k}{k_0}}\mathbb P(\bar {A}_{m\cdot k_0}) \leq \left (1- \frac {(1-\lambda (\varphi ))^2}{600}\right )^{\frac {k}{k_0}} \leq \left (1- \frac {(1-\lambda (\varphi ))^2}{600k_0}\right )^{k}.\end{equation*}

The claim now follows by taking a union bound over all paths. Indeed, for each $v \in T_k$ define the corresponding event $A_v$ . Then, with a union bound,

\begin{equation*}\mathbb P\left (\bigcup _{v \in T_k}\bar {A_v}\right ) \leq |T_k|p^k \leq q^kp^k.\end{equation*}

By definition, conditioned on $\bigcap _{v\in T_k}{A_v}$ , $\{X_v\}_{v\in T_k}$ is symmetric around $w$ , which completes the lemma.

3.2.3 Proving Theorem 9

Proof. By Lemma 22, for every $w,x_0 \in \mathbb{S}^{d-1}$ .

\begin{equation*} \mathrm {law}(g(v)_{v\in T_k}|g(r) = x_0) \in \mathrm {DPS}^k_w(1-q^kp^k), \end{equation*}

where $p = \left (1-\frac{\ln (\lambda (\varphi ))}{\ln \left (\frac{\lambda (\varphi )(1-\lambda (\varphi ))}{32 f(1)}\right )}\frac{(1-\lambda (\varphi ))^2}{600}\right )$ . By assumption

\begin{equation*} q \leq p^{-1}, \end{equation*}

and Lemma 16 gives the result.