Proof. For a fixed time \(s\), rigidity was obtained in erdos2012rigidity?, and the isotropic local law and isotropic delocalization in alex2014isotropic?. It remains to prove the estimate in \(\mathfrak{A}_{2,s}\).

Fix \(\mathbf{e},\mathbf{e}'\in\mathfrak{E}\) and define \[A_{\mathbf{e},\mathbf{e}'}=\sqrt{N}\mathbf{e}\mathbf{e}'^\top, \qquad \mathring A_{\mathbf{e},\mathbf{e}'}=A_{\mathbf{e},\mathbf{e}'}-\langle A_{\mathbf{e},\mathbf{e}'}\rangle \mathop{\mathrm{Id}}_N,\] Then \[\mathbf{v}_j(s,\mathbf{e})\mathbf{v}_j(s,\mathbf{e}')-\langle \mathbf{e},\mathbf{e}'\rangle = \sqrt N\langle \mathbf{u}_j(s),\mathring A_{\mathbf{e},\mathbf{e}'}\mathbf{u}_j(s)\rangle .\] Therefore, for an interval \(I\subset\left[\!\left[1,N\right]\!\right]\), \[\frac{1}{\sqrt{|I|}} \sum_{j\in I} \left( \mathbf{v}_j(s,\mathbf{e})\mathbf{v}_j(s,\mathbf{e}')-\langle \mathbf{e},\mathbf{e}'\rangle \right) = \sqrt{\frac{N}{|I|}} \sum_{j\in I} \langle \mathbf{u}_j(s),\mathring A_{\mathbf{e},\mathbf{e}'}\mathbf{u}_j(s)\rangle .\] If \(P_I\) denotes the spectral projection onto the eigenvectors with indices in \(I\), this is \[\frac{1}{\sqrt{\vert I\vert}}\sum_{j\in I}\left( \mathbf{v}_j(s,\mathbf{e})\mathbf{v}_j(s,\mathbf{e}')-\langle \mathbf{e},\mathbf{e}'\rangle \right) =\frac{N^{\frac{3}{2}}}{\sqrt{|I|}}\langle P_I \mathring A_{\mathbf{e},\mathbf{e}'}\rangle .\]

We first consider mesoscopic intervals, namely \(N^\varepsilon\leqslant |I|\leqslant N^{1-\varepsilon}\), where \(\varepsilon>0\) will be chosen sufficiently small with respect to \(\xi\). Let \(I=\left[\!\left[a,b\right]\!\right]\). Let \(J\) be the deterministic spectral interval whose endpoints are the midpoints between the corresponding classical locations, \[J=[E_-,E_+], \qquad E_-= \begin{cases} -3,& a=1,\\ \frac{\gamma_{a-1}+\gamma_a}{2},& a>1, \end{cases} \qquad E_+= \begin{cases} 3,& b=N,\\ \frac{\gamma_b+\gamma_{b+1}}{2},& b<N. \end{cases}\] By rigidity, the projections \(P_I\) and \(\mathbf{1}_{J}(H_s)\) differ only by eigenvectors whose indices lie in a boundary layer of size at most \(N^{c\varepsilon}\). Using isotropic delocalization, this gives \[\frac{N^{\frac{3}{2}}}{\sqrt{\vert I\vert}} \left| \langle P_I \mathring A_{\mathbf{e},\mathbf{e}'}\rangle - \langle \mathbf{1}_{J}(H_s)\mathring A_{\mathbf{e},\mathbf{e}'}\rangle \right| \leqslant \frac{N^{c\varepsilon}}{\sqrt{\vert I\vert}}\] with overwhelming probability. This error is \(O(N^\xi)\) by choosing \(\varepsilon\) small enough with respect to \(\xi\).

We now estimate the deterministic spectral projection. Let \(G_s(z)=(H_s-z)^{-1}\) and let \(m_{\mathrm{sc}}(z)\) be the Stieltjes transform of the semicircle law. We first record the local law input. Since \[\langle G_s(z)\mathring A_{\mathbf{e},\mathbf{e}'}\rangle = \frac{1}{\sqrt N} \left( \langle \mathbf{e}',G_s(z)\mathbf{e}\rangle - \langle \mathbf{e},\mathbf{e}'\rangle\langle G_s(z)\rangle \right),\] the deterministic term \(m_{\mathrm{sc}} (z)\) cancels, and the isotropic and averaged local laws from alex2014isotropic? give, for every \(\chi>0\) and \(D>0\), with probability at least \(1-N^{-D}\), \[\left| \sqrt{N} \langle G_s(z)\mathring A_{\mathbf{e},\mathbf{e}'}\rangle \right| \leqslant N^\chi \left( \sqrt{\frac{\operatorname{Im}m_{\mathrm{sc}}(z)}{N\eta}} + \frac{1}{N\eta} \right), \qquad z=E+\mathrm{i}\eta,\] uniformly for \(|E|\leqslant 3\) and \(\eta\geqslant N^{-1+\varepsilon/10}\).

We apply Pleijel’s representation formula for sharp spectral cut-offs, as in cipolloni2023functional?*Proof of Theorem 2.3, and previously used in a similar context in erdos2016fluctuations?. Let \(\eta_0=N^{-1+\varepsilon/10}\) and let \(\Gamma_I\) be the rectangular contour around \(J\) with horizontal height \(\vert J\vert\), truncated at distance \(\eta_0\) from the real axis. Then \[\frac{N^{\frac{3}{2}}}{\sqrt{\vert I\vert}} \langle \mathbf{1}_{J}(H_s)\mathring A_{\mathbf{e},\mathbf{e}'}\rangle = \frac{N^{\frac{3}{2}}}{2\pi i\sqrt{\vert I\vert}} \int_{\Gamma_I} \langle G_s(z)\mathring A_{\mathbf{e},\mathbf{e}'}\rangle \relax z + \relax{\frac{N\eta_0}{\sqrt{\vert I\vert}}}.\] The last error is \(O(N^\xi)\) by our choice of \(\eta_0\) and since \(\vert I\vert\geqslant N^\varepsilon\).

Using the previous local law on \(\Gamma_I\), together with the elementary semicircle estimates \[\operatorname{Im}m_{\mathrm{sc}}(E+\mathrm{i}\eta)\leqslant C\sqrt{\kappa(E)+\eta}, \qquad \kappa(E)=\mathrm{dist}(E,\{-2,2\}),\] and the relation between \(\vert I\vert\) and \(\vert J\vert\), we get for any \(\chi>0\), \[\frac{N^{\frac{3}{2}}}{\sqrt{\vert I\vert}} \int_{\Gamma_I} \left| \langle G_s(z)\mathring A_{\mathbf{e},\mathbf{e}'}\rangle \right| |\relax z| \leqslant N^{\chi}.\] Indeed, in the bulk, \(\vert J\vert\asymp \frac{\vert I\vert}{N}\), and the horizontal and vertical parts of the contour are bounded by, up to \(N^\chi\) terms coming from the local laws, \[\frac{N^{\frac{3}{2}}}{\sqrt{\vert I\vert}} \vert J\vert\frac{1}{N\sqrt{\vert J\vert}} \leqslant C, \qquad \frac{N^{\frac{3}{2}}}{\sqrt{\vert I\vert}} \int_{\eta_0}^{\vert J\vert}\frac{\relax\eta}{N\sqrt\eta} \leqslant C.\] At the edge, \(\vert J\vert\asymp N^{-\frac{2}{3}}\vert I\vert^{\frac{2}{3}}\) and \(\operatorname{Im}m_{\mathrm{sc}}(E+\mathrm{i}\eta)\leqslant C \vert J\vert^{1/2}\) on the relevant contour, giving similarly \[\frac{N^{\frac{3}{2}}}{\sqrt{\vert I\vert}} \vert J\vert \frac{1}{N}\sqrt{\frac{\vert J\vert^{1/2}}{\vert J\vert}} \leqslant C, \qquad \frac{N^{\frac{3}{2}}}{\sqrt{\vert I\vert}} \int_{\eta_0}^{\vert J\vert} \frac{1}{N}\sqrt{\frac{\vert J\vert^{1/2}}{\eta}} \relax\eta \leqslant C.\] We note that the intermediate energy regime works similarly. The contribution of the averaged local law term \(N^{-3/2}\eta^{-1}\) is smaller and is absorbed into \(N^{\chi}\). Choosing \(\chi>0\) sufficiently small, we obtain \[\sup_{\substack{I\subset\left[\!\left[1,N\right]\!\right]\,\mathrm{interval}\\ N^\varepsilon\leqslant |I|\leqslant N^{1-\varepsilon}}} \frac{1}{\sqrt{|I|}} \left| \sum_{j\in I} \left( \mathbf{v}_j(s,\mathbf{e})\mathbf{v}_j(s,\mathbf{e}')-\langle \mathbf{e},\mathbf{e}'\rangle \right) \right| \leqslant N^{\xi/2}\] with overwhelming probability. A union bound over at most \(N^2\) intervals and at most \(N^4\) pairs \(\mathbf{e},\mathbf{e}'\in\mathfrak{E}\) keeps the estimate overwhelming.

It remains to treat microscopic and macroscopic intervals. If \(|I|\leqslant N^\varepsilon\), then by isotropic delocalization, \[\left| \mathbf{v}_j(s,\mathbf{e})\mathbf{v}_j(s,\mathbf{e}')-\langle \mathbf{e},\mathbf{e}'\rangle \right| \leqslant N^{\xi/2}\] uniformly in \(j,\mathbf{e},\mathbf{e}'\), with overwhelming probability, provided \(\varepsilon\) is chosen small enough. Hence \[\frac{1}{\sqrt{|I|}} \left| \sum_{j\in I} \left( \mathbf{v}_j(s,\mathbf{e})\mathbf{v}_j(s,\mathbf{e}')-\langle \mathbf{e},\mathbf{e}'\rangle \right) \right| \leqslant N^{\xi/2}|I|^{1/2} \leqslant N^\xi .\] If \(|I|\geqslant N^{1-\varepsilon}\), we partition \(I\) into consecutive intervals \(I_\alpha\) of size between \(N^\varepsilon\) and \(N^{1-\varepsilon}\), up to one remainder of size at most \(N^\varepsilon\). The number of intervals is at most \(N^{2\varepsilon}\). Using the mesoscopic estimate on the intervals \(I_\alpha\), the microscopic estimate on the possible remainder, and Cauchy–Schwarz, we get \[\begin{align} \frac{1}{\sqrt{|I|}} \left| \sum_{j\in I} \left( \mathbf{v}_j(s,\mathbf{e})\mathbf{v}_j(s,\mathbf{e}')-\langle \mathbf{e},\mathbf{e}'\rangle \right) \right| \leqslant \frac{N^{\xi/2}}{\sqrt{|I|}} \sum_\alpha \sqrt{|I_\alpha|} + N^\xi |I|^{-\frac{1}{2}} \leqslant N^{\xi/2}N^\varepsilon+N^\xi |I|^{-\frac{1}{2}} \leqslant N^\xi, \end{align}\] after decreasing \(\varepsilon\) if necessary. This proves \(\mathfrak{A}_{2,s}\) with overwhelming probability for fixed \(s\).

The estimates above are uniform for \(s\) in a polynomial net of \([0,1]\). The extension from the net to all \(s\in[0,1]\) follows by the same Hölder regularity argument for the resolvent as in bourgade2017eigenvector?*Lemma 4.2. This gives \[\mathbb{P}\left(\mathfrak{A}(\xi,\mathfrak{Q},\mathfrak{E},N)\right)\geqslant 1-N^{-D}\] for any \(D>0\), after choosing the overwhelming-probability exponents in the local laws sufficiently large. The same time-net argument gives the time-uniform rigidity estimate \[\mathbb{P}\left(\mathfrak{A}_3(\xi,N)\right)\geqslant 1-N^{-D}.\]

Finally, the estimate on \(\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\) follows from Markov’s inequality. Indeed, \[\begin{align} \mathbb{P}\left(\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)^c\right) &\leqslant \mathbb{P}\left(\mathfrak{A}_3(\xi,N)^c\right) + \mathbb{P}\left( \mathbb{P}\left(\mathfrak{A}(\xi,\mathfrak{Q},\mathfrak{E},N)^c\vert \boldsymbol{\lambda}\right)>N^{-\nu} \right)\\ &\leqslant \mathbb{P}\left(\mathfrak{A}_3(\xi,N)^c\right) + N^\nu\mathbb{P}\left(\mathfrak{A}(\xi,\mathfrak{Q},\mathfrak{E},N)^c\right). \end{align}\] Applying the previous overwhelming-probability estimates with exponent \(D+\nu\) gives \[\mathbb{P}\left(\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\right)\geqslant 1-N^{-D}.\] If \(\boldsymbol{\lambda}\in\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\), then, up to the convention at the endpoint, \(\{\tau\leqslant 1\}=\mathfrak{A}(\xi,\mathfrak{Q},\mathfrak{E},N)^c\), and therefore \[\mathbb{P}(\tau\leqslant 1\vert \boldsymbol{\lambda}) \leqslant N^{-\nu}.\] ◻

Proof. By the previous identity, after symmetrization we get \[\mathscr{L}_t g = \frac{1}{2}\sum_{i\neq j} c_{ij}(t)\left\langle \mathbf{z}_i\mathbf{z}_i^\top,\nabla_j^2 g\right\rangle_{\mathrm{HS}} -\sum_{i\neq j} c_{ij}(t)\mathbf{z}_i^\top\nabla_{ij}^2g\,\mathbf{z}_j - \sum_{i=1}^N\left(\sum_{j\neq i} c_{ij}(t)\right)\mathbf{z}_i\cdot\nabla_i g .\] If we denote \[\widehat{\mathscr{L}}_t = \frac{1}{2}\sum_{\vert i-j\vert <\ell} c_{ij}(t)\mathscr{X}_{ij}^2 + \sum_{i=1}^N\beta_i(t)\left( \Delta_i-\mathbf{z}_i\cdot\nabla_i \right)\] and using \(\beta_i(t) = \sum_{j,\vert i-j\vert \geqslant \ell} c_{ij}(t)\) we get the final result. ◻

Proof. For the part involving \(\mathscr{X}_{ij}^2\), since the \(c_{ij}(t)\) do not depend on \(\mathbf{z}\), we note that since \(\varphi_{N,R}\) has a density only involving \(\sum_{j=1}^N\Vert \mathbf{z}_j\Vert^2\) we have \(\mathscr{X}_{ij}\varphi_{N,R}=0\) since \[\mathscr{X}_{ij}\left( \sum_{k=1}^N\Vert \mathbf{z}_k\Vert^2 \right) = 2\mathbf{z}_i\cdot\mathbf{z}_j - 2\mathbf{z}_j\cdot\mathbf{z}_i = 0.\] We note also that we can write \(\mathscr{X}_{ij} = \mathbf{b}_{ij}\cdot \nabla\) with \(\mathbf{b}_{ij}\) a vector field such that \(\mathrm{div}(\mathbf{b}_{ij})=0\) so that we can integrate by parts to get \[\int \mathscr{X}_{ij} f \, \varphi_{N,R}(\relax\mathbf{z}) = -\int f \mathrm{div}(\mathbf{b}_{ij}\varphi_{N,R})\relax\mathbf{z}= -\int f \mathrm{div}(\mathbf{b}_{ij})\varphi_{N,R}\relax\mathbf{z}- \int f (\mathbf{b}_{ij}\cdot\nabla) \varphi_{N,R}\relax\mathbf{z}=0\] for any smooth function \(f\). If now we replace \(f\) by \(\mathscr{X}_{ij} g\) for some smooth function \(g\), we get \[\int \mathscr{X}_{ij}^2 g \, \varphi_{N,R}(\relax\mathbf{z}) = 0.\] For the second part of the generator, we can simply use Gaussian integration by parts to get \[\int \left(\Delta_i-\mathbf{z}_i\cdot\nabla_i\right) f \, \varphi_{N,R}(\relax\mathbf{z}) = 0.\] This shows that \(\varphi_{N,R}\) is invariant under the dynamics generated by \(\widehat{\mathscr{L}}_t\). ◻

Proof. We can write \[\widehat{\mathscr{P}}_{0,t}\psi(\mathbf{x}) = \int \psi(\mathbf{u})q_{0,t}(\mathbf{x},\mathbf{u})\varphi_{N,R}(\relax\mathbf{u})\] and thus by triangle inequality, \[\left\vert \widehat{\mathscr{P}}_{0,t}\psi(\mathbf{x})-\mathbb{E}_{\varphi_{N,R}}\left[\psi(X)\right] \right\vert = \left\vert \int \psi(\mathbf{u})(q_{0,t}(\mathbf{x},\mathbf{u})-1)\varphi_{N,R}(\relax\mathbf{u}) \right\vert \leqslant \Vert \psi\Vert_\infty \int \vert q_{0,t}(\mathbf{x},\mathbf{u})-1\vert\varphi_{N,R}(\relax\mathbf{u}).\] Now we can use Pinsker’s inequality to get \[\int \vert q_{0,t}(\mathbf{x},\mathbf{u})-1\vert\varphi_{N,R}(\relax\mathbf{u}) = \Vert q_{0,t}(\mathbf{x},\cdot)-1\Vert_{L^1(\varphi_{N,R})} \leqslant \sqrt{2\mathbf{H}_{\varphi_{N,R}}(q_{0,t}(\mathbf{x},\cdot))}.\] ◻

Proof. We can compute the derivative of the entropy as \[\frac{\relax}{\relax u}\mathbf{H}_{\varphi_{N,R}}(f_u) = \int \widehat{\mathscr{L}}_u f_u \log f_u \, \varphi_{N,R}(\relax\mathbf{w}) + \int \widehat{\mathscr{L}}_u f_u \, \varphi_{N,R}(\relax\mathbf{w}).\] The second term is zero since \(\varphi_{N,R}\) is invariant for the dynamics generated by \(\widehat{\mathscr{L}}_u\). For the first term, we can use the definition of \(\widehat{\mathscr{L}}_u\) to write \[\begin{align} \int \widehat{\mathscr{L}}_u f_u \log f_u \, \varphi_{N,R}(\relax\mathbf{w}) &= -4\sum_{\substack{i<j}} c_{ij}(u)\int (\mathscr{X}_{ij}\sqrt{f_u})^2\varphi_{N,R}(\relax\mathbf{w}) - 4\sum_{i=1}^N\beta_i(u)\int \Vert\nabla_i\sqrt{f_u}\Vert^2\varphi_{N,R}(\relax\mathbf{w})\\ &= -4\widehat{\mathscr{E}}(\sqrt{f_u},\sqrt{f_u}) \end{align}\] where \(\widehat{\mathscr{E}}\) is the Dirichlet form associated to \(\widehat{\mathscr{L}}_t\). Now we have, since \(c_{ij}(t)\geqslant 0\), \[\widehat{\mathscr{E}}(f,f) \geqslant \sum_{i=1}^N\beta_i(u)\int \Vert\nabla_i f\Vert^2\varphi_{N,R}(\relax\mathbf{w}) \geqslant c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}\sum_{i=1}^N\int \Vert\nabla_i f\Vert^2\varphi_{N,R}(\relax\mathbf{w}) \geqslant \frac{1}{2}c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}\mathbf{H}_{\varphi_{N,R}}(f^2)\] where we use Lemma 2 to get the second inequality and we use the log-Sobolev inequality for the Gaussian measure to get the last inequality. Finally, if we apply this to \(\sqrt{f_u}\) we get that \[\frac{\relax}{\relax u}\mathbf{H}_{\varphi_{N,R}}(f_u) = -4\widehat{\mathscr{E}}(\sqrt{f_u},\sqrt{f_u}) \leqslant -c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}\mathbf{H}_{\varphi_{N,R}}(f_u)\] which finally gives, by Gronwall’s lemma, \[\mathbf{H}_{\varphi_{N,R}}(f_v)\leqslant \mathrm{e}^{-c\left(\frac{N(v-u)^{3}}{\ell}\right)^{\frac{1}{3}}}\mathbf{H}_{\varphi_{N,R}}(f_u)\] which gives the desired result. ◻

Proof. The idea is to return to the stochastic differential equation associated to the reference generator. We identify \((\mathbb{R}^R)^N\simeq \mathbb{R}^{N\times R}\) and let matrices on \(\mathbb{R}^N\) act on \(\mathbb{R}^{N\times R}\) by left multiplication. Equivalently, an \(N\times N\) matrix \(A\) acts as \(A\otimes \mathop{\mathrm{Id}}_R\) on \(\mathbb{R}^N\otimes \mathbb{R}^R\). For \(1\leqslant i<j\leqslant N\), define the skew-symmetric matrix \(A^{ij}\) by \[A^{ij}e_i=e_j,\qquad A^{ij}e_j=-e_i,\qquad A^{ij}e_k=0\quad\text{for }k\notin\{i,j\}.\] Thus, for \(\mathbf{x}=(\mathbf{x}_1,\dots,\mathbf{x}_N)\in(\mathbb{R}^R)^N\), \[\label{eq:defA} (A^{ij}\mathbf{x})_i=-\mathbf{x}_j,\qquad (A^{ij}\mathbf{x})_j=\mathbf{x}_i,\qquad (A^{ij}\mathbf{x})_k=0\quad\text{for }k\notin\{i,j\}.\tag{5}\] With this convention, \[(A^{ij}\mathbf{x})\cdot\nabla = -\mathbf{x}_j\cdot\nabla_i+\mathbf{x}_i\cdot\nabla_j = \mathscr{X}_{ij}\] and \(A^{ij}\) is skew-symmetric. We then introduce an auxiliary probability space, independent of the true Dyson vector flow, with expectation denoted by \(\widehat{\mathbb{E}}\). The corresponding auxiliary Brownian motions are denoted by \(\widehat{B}_{ij}\) and \(\widehat{W}\). They should not be confused with the randomness of the true eigenvectors. If we also define the diagonal matrix \(D_t\) with entries \(\beta_i(t)\), then we can write the stochastic differential equation associated to \(\widehat{\mathscr{L}}_t\), in the Itô–Stratonovich form, as \[\relax\widehat{X}_t = \sum_{i<j,\vert i-j\vert < \ell} \sqrt{2c_{ij}(t)}A^{ij}\widehat{X}_t\circ \relax\widehat{B}_{ij}(t) -D_t\widehat{X}_t\relax t +\sqrt{2D_t}\relax\widehat{W}_t,\] where \(\widehat{W}_t\) is a standard Brownian motion in \(\mathbb{R}^{N\times R}\) independent of the Brownian motions \(\widehat{B}_{ij}(t)\), and where \(D_t\) and \(\sqrt{D_t}\) act by left multiplication on \(\mathbb{R}^{N\times R}\). If we condition on the path of eigenvalues, and thus on \(c_{ij}(t)\) and \(D_t\), and on the Brownian motions \(\widehat{B}_{ij}(t)\), then only the additive noise \(\widehat{W}_t\) remains random. If we start at \(\widehat{X}_{0}=\mathbf{x}\), then this conditional law of \(\widehat{X}_{\frac{t}{2}}\) is Gaussian. The conditional mean is given by \[\mu_{0,u}(\mathbf{x}) = \widehat{\mathbb{E}}\left[\widehat{X}_u\middle\vert \widehat{B}_{ij},\widehat{X}_{0}=\mathbf{x}\right] = M_{0,u}\mathbf{x}\] with \(M\) the solution of the equation \[\relax M_{0,u} = \sum_{i<j,\vert i-j\vert < \ell} \sqrt{2c_{ij}(u)}A^{ij}M_{0,u}\circ \relax\widehat{B}_{ij}(u) - D_uM_{0,u}\relax u,\quad M_{0,0}=\mathop{\mathrm{Id}}_N.\] If we consider the centered process \(Z_u = \widehat{X}_u - \mu_{0,u}(\mathbf{x})\), then we see that it solves the stochastic equation \[\relax Z_u = \sum_{i<j,\vert i-j\vert < \ell} \sqrt{2c_{ij}(u)}A^{ij}Z_u\circ \relax\widehat{B}_{ij}(u) - D_uZ_u\relax u +\sqrt{2D_u}\relax\widehat{W}_u.\] We want to control the covariance of \(\widehat{X}_u\). Conditionally on the short-range Brownian motions, the \(R\) columns of \(Z_u\) are independent and have the same covariance matrix. We denote this \(N\times N\) covariance matrix by \[\Sigma_{0,u} = \widehat{\mathbb{E}}\left[Z_u^{(m)}(Z_u^{(m)})^\top \middle\vert \widehat{B}_{ij}, \widehat{X}_{0}=\mathbf{x}\right],\] where \(Z_u^{(m)}\) is any fixed column of \(Z_u\). If we consider the differential of \(Z_u^{(m)}(Z_u^{(m)})^\top\), we get \[\relax\left(Z_u^{(m)}(Z_u^{(m)})^\top\right) = \sum_{i<j,\vert i-j\vert < \ell}\sqrt{2c_{ij}(u)}\left( A^{ij}Z_u^{(m)}(Z_u^{(m)})^\top - Z_u^{(m)}(Z_u^{(m)})^\top A^{ij} \right)\circ \relax\widehat{B}_{ij}(u)\] \[ -\left( D_uZ_u^{(m)}(Z_u^{(m)})^\top + Z_u^{(m)}(Z_u^{(m)})^\top D_u - 2D_u \right)\relax u +\relax N_u,\] with \((N_u)_u\) a matrix-valued martingale whose conditional expectation is zero. If we now take the conditional expectation, we get that \(\Sigma_{0,u}\) solves the following matrix-valued differential equation \[\relax\Sigma_{0,u} = \sum_{i<j,\vert i-j\vert < \ell}\sqrt{2c_{ij}(u)}\left( A^{ij}\Sigma_{0,u} - \Sigma_{0,u} A^{ij} \right)\circ \relax\widehat{B}_{ij}(u) -\left( D_u\Sigma_{0,u} + \Sigma_{0,u}D_u - 2D_u \right)\relax u, \quad \Sigma_{0,0} = 0.\] If we now consider \(S_{0,u} = M_{0,u}M_{0,u}^\top\), we similarly get \[\relax S_{0,u} = \sum_{i<j,\vert i-j\vert < \ell}\sqrt{2c_{ij}(u)}\left( A^{ij}S_{0,u} - S_{0,u}A^{ij} \right)\circ \relax\widehat{B}_{ij}(u) -\left( D_uS_{0,u} + S_{0,u}D_u \right)\relax u, \quad S_{0,0} = \mathop{\mathrm{Id}}_N.\] Thus if we consider \(\widetilde{S}_{0,u}=\mathop{\mathrm{Id}}_N-S_{0,u}\), we get \[\relax\widetilde{S}_{0,u} = \sum_{i<j,\vert i-j\vert < \ell}\sqrt2{c_{ij}(u)}\left( A^{ij}\widetilde{S}_{0,u} - \widetilde{S}_{0,u}A^{ij} \right)\circ \relax\widehat{B}_{ij}(u) -\left( D_u\widetilde{S}_{0,u} + \widetilde{S}_{0,u}D_u-2D_u \right)\relax u, \quad \widetilde{S}_{0,0} = 0.\] This is the same equation as \(\Sigma_{0,u}\) with the same initial condition. By uniqueness of the solution of this equation, we get that \[\Sigma_{0,u} = \widetilde{S}_{0,u} = \mathop{\mathrm{Id}}_N-M_{0,u}M_{0,u}^\top.\] Thus we need to control \(M_{0,u}\) in order to control \(\Sigma_{0,u}\). We can compute for \(\mathbf{x}\in(\mathbb{R}^R)^N\), using the Stratonovich chain rule, \[\relax\Vert M_{0,u} \mathbf{x}\Vert_{\mathrm{HS}}^2 = 2\langle M_{0,u} \mathbf{x}, \relax(M_{0,u}\mathbf{x})\rangle_{\mathrm{HS}} = -2\left\langle M_{0,u} \mathbf{x}, D_u M_{0,u} \mathbf{x}\right\rangle_{\mathrm{HS}}\relax u,\] since the noise is skew-symmetric and thus does not contribute to the norm. Now we can use the lower bound on \(D_u\) to get \[\frac{\relax}{\relax u} \Vert M_{0,u} \mathbf{x}\Vert_{\mathrm{HS}}^2 \leqslant -c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}\Vert M_{0,u} \mathbf{x}\Vert_{\mathrm{HS}}^2\] and thus by Gronwall’s lemma, \[\Vert M_{0,u} \mathbf{x}\Vert_{\mathrm{HS}}^2 \leqslant \mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}u}\Vert \mathbf{x}\Vert_{\mathrm{HS}}^2.\] In particular, \[M_{0,u}M_{0,u}^\top \preceq \mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}u}\mathop{\mathrm{Id}}_N \quad\text{and thus}\quad \Sigma_{0,u} \succeq \left(1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}u}\right)\mathop{\mathrm{Id}}_N.\] We know that the conditional law of \(\widehat{X}_u\) is Gaussian with mean \(\mu_{0,u}(\mathbf{x})\) and covariance \(\Sigma_{0,u}\otimes \mathop{\mathrm{Id}}_R\). Besides, the relative entropy of a \(\mathcal{N}(\mu,\Sigma)\) with respect to \(\mathcal{N}(0,\mathop{\mathrm{Id}}_d)\) is given by \[\mathbf{H}(\mathcal{N}(\mu,\Sigma)\vert \mathcal{N}(0,\mathop{\mathrm{Id}}_d)) = \frac{1}{2}\left( \Vert \mu\Vert^2+\mathop{\mathrm{Tr}}(\Sigma)-d-\log\det \Sigma \right).\] Therefore, if we denote by \(\widetilde{q}_{0,\frac{t}{2}}(\mathbf{x},\cdot)\) the conditional density with respect to the short-range Brownian motions, we get \[\mathbf{H}_{\varphi_{N,R}}(\widetilde{q}_{0,\frac{t}{2}}(\mathbf{x},\cdot)) = \frac{1}{2}\left( \Vert M_{0,\frac{t}{2}}\mathbf{x}\Vert_{\mathrm{HS}}^2 +R\mathop{\mathrm{Tr}}\Sigma_{0,\frac{t}{2}} -NR -R\log\det\Sigma_{0,\frac{t}{2}} \right).\] We can use the estimates obtained earlier in the proof. In particular, by the contraction property of \(M\), we have \[\Vert M_{0,\frac{t}{2}}\mathbf{x}\Vert_{\mathrm{HS}}^2\leqslant \Vert \mathbf{x}\Vert_{\mathrm{HS}}^2\leqslant NR.\] In our application, this follows from Parseval since \[\sum_{i=1}^N\Vert \mathbf{z}_i\Vert^2 = \sum_{m=1}^R\sum_{i=1}^N \mathbf{v}_i(\mathbf{e}_m)^2 = NR.\] Now, since \(\mathop{\mathrm{Id}}_N \succeq \Sigma_{0,u}\), we have \(\mathop{\mathrm{Tr}}(\Sigma_{0,\frac{t}{2}})\leqslant N\). Finally, since we additionally have that \[\Sigma_{0,u}\succeq \left(1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}u}\right) \mathop{\mathrm{Id}}_N,\] we get that \[-\log\det \Sigma_{0,\frac{t}{2}} \leqslant -N\log\left({1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}t}}\right),\] up to changing the value of \(c>0\). This finally gives \[\mathbf{H}_{\varphi_{N,R}}\left( \widetilde{q}_{0,\frac{t}{2}}(\mathbf{x},\cdot) \right) \leqslant \frac{NR}{2} -\frac{NR}{2}\log\left({1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}t}}\right) \leqslant CNR\left(1-\log\left({1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}t}}\right)\right).\] We then finish using the convexity of the relative entropy to write, with \(\widetilde{\mathbb{E}}\) the expectation over the short-range Brownian motions, \[\mathbf{H}_{\varphi_{N,R}}\left( q_{0,\frac{t}{2}}(\mathbf{x},\cdot) \right) = \mathbf{H}_{\varphi_{N,R}}\left( \widetilde{\mathbb{E}}\left[ \widetilde{q}_{0,\frac{t}{2}}(\mathbf{x},\cdot) \right] \right) \leqslant \widetilde{\mathbb{E}}\left[ \mathbf{H}_{\varphi_{N,R}}\left( \widetilde{q}_{0,\frac{t}{2}}(\mathbf{x},\cdot) \right) \right] \leqslant CNR\left(1-\log\left({1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}t}}\right)\right).\] ◻

Proof of Proposition 1. By Lemma 5, applied to the observable \[\mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{z})=\widetilde{h}(\mathbf{z}_{k_1},\dots,\mathbf{z}_{k_p}),\] we have that \[\left\vert \widehat{\mathscr{P}}_{0,t}\mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{x}) - \mathbb{E}_{\varphi_{N,R}}\left[\mathbf{F}_{\mathbf{k},\mathbf{q}}(Z)\right] \right\vert \leqslant \Vert \widetilde{h}\Vert_\infty \sqrt{2\mathbf{H}_{\varphi_{N,R}}(q_{0,t}(\mathbf{x},\cdot))}.\] Since under \(\varphi_{N,R}\) the random vectors \(Z_{k_1},\dots,Z_{k_p}\) are independent standard Gaussian vectors in \(\mathbb{R}^R\), we have \[\mathbb{E}_{\varphi_{N,R}}\left[\mathbf{F}_{\mathbf{k},\mathbf{q}}(Z)\right] = \mathbb{E}\left[\widetilde{h}(Z_1,\dots,Z_p)\right],\] where \(Z_1,\dots,Z_p\) are i.i.d. \(\mathcal{N}(0,\mathop{\mathrm{Id}}_R)\). Using Lemmas 6 and 7, we get that \[\mathbf{H}_{\varphi_{N,R}}(q_{0,t}(\mathbf{x},\cdot)) \leqslant \mathrm{e}^{-c\left(\frac{Nt^3}{8\ell}\right)^{\frac{1}{3}}}\mathbf{H}_{\varphi_{N,R}}(q_{0,\frac{t}{2}}(\mathbf{x},\cdot)) \leqslant CNR\left(1-\log\left({1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}t}}\right)\right)\mathrm{e}^{-c\left(\frac{Nt^3}{\ell}\right)^{\frac{1}{3}}}.\] Since \(\Vert \widetilde{h}\Vert_\infty\leqslant \Vert h\Vert_\infty\), we get the bound \[\left\vert \widehat{\mathscr{P}}_{0,t}\mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{x}) - \mathbb{E}\left[\widetilde{h}(Z_1,\dots,Z_p)\right] \right\vert \leqslant C\sqrt{R}\Vert h\Vert_\infty\sqrt{N\left(1-\log\left({1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}t}}\right)\right)}\mathrm{e}^{-c\left(\frac{Nt^3}{\ell}\right)^{\frac{1}{3}}}.\] ◻

Proof. We consider the stopped process \[u\mapsto g_{u\wedge\tau}\bigl(\mathbf{z}(u\wedge\tau)\bigr),\] where \(\mathbf{z}(u)\) is the solution to the stochastic differential equation associated to \(\mathscr{L}_u\), i.e. the true Dyson vector flow projected on the basis \(\mathfrak{E}\). By definition of \(g_u\), we have on the event \(\{u<\tau\}\) that \[\relax g_u(\mathbf{z}(u)) = \partial_u g_u(\mathbf{z}(u))\relax u + \mathscr{L}_u g_u(\mathbf{z}(u))\relax u + \relax N_u = (\mathscr{L}_u-\widehat{\mathscr{L}}_u)g_u(\mathbf{z}(u))\relax u + \relax N_u,\] where \(N_u\) is a martingale and we used the fact that \[(\partial_u+\widehat{\mathscr{L}}_u)g_u=0.\] Therefore, \[\relax g_{u\wedge\tau}\bigl(\mathbf{z}(u\wedge\tau)\bigr) = \mathbb{1}_{\{u<\tau\}} (\mathscr{L}_u-\widehat{\mathscr{L}}_u)g_u(\mathbf{z}(u))\relax u + \mathbb{1}_{\{u<\tau\}}\relax N_u.\] Integrating from \(0\) to \(t\) and taking conditional expectation with respect to \(\boldsymbol{\lambda}\) gives \[\mathbb{E}\left[ g_{t\wedge\tau}\bigl(\mathbf{z}(t\wedge\tau)\bigr) \middle\vert \boldsymbol{\lambda} \right] - \mathbb{E}\left[ g_0(\mathbf{z}(0))\middle\vert \boldsymbol{\lambda} \right] = \int_0^t \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}} (\mathscr{L}_u-\widehat{\mathscr{L}}_u)g_u(\mathbf{z}(u)) \middle\vert\boldsymbol{\lambda} \right]\relax u.\] We now compare the stopped terminal value with the unstopped terminal value. Since \(g_t=\mathbf{F}_{\mathbf{k},\mathbf{q}}\), we have on the event \(\{\tau>t\}\), \[g_{t\wedge\tau}\bigl(\mathbf{z}(t\wedge\tau)\bigr) = g_t(\mathbf{z}(t)) = \mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{z}(t)).\] On the complementary event \(\{\tau\leqslant t\}\), we use the trivial bounds \[\left\vert \mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{z}(t))\right\vert \leqslant \Vert h\Vert_\infty, \qquad \left\vert g_{t\wedge\tau}\bigl(\mathbf{z}(t\wedge\tau)\bigr)\right\vert \leqslant \Vert h\Vert_\infty,\] the second one following from the Markov property of \(\widehat{\mathscr{P}}_{u,t}\) and the bound \(\Vert g_u\Vert_\infty\leqslant \Vert h\Vert_\infty\). Hence, for \(\boldsymbol{\lambda}\in\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\), \[\left\vert \mathbb{E}\left[ \mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{z}(t))\middle\vert\boldsymbol{\lambda} \right] - \mathbb{E}\left[ g_{t\wedge\tau}\bigl(\mathbf{z}(t\wedge\tau)\bigr) \middle\vert\boldsymbol{\lambda} \right] \right\vert \leqslant 2\Vert h\Vert_\infty \mathbb{P}(\tau\leqslant t\vert\boldsymbol{\lambda}) \leqslant 2\Vert h\Vert_\infty N^{-\nu}.\] Combining this estimate with the stopped Duhamel identity gives the first statement. The second statement follows from the decomposition \[\mathscr{L}_u-\widehat{\mathscr{L}}_u=\mathscr{D}_u-\mathscr{O}_u\] from Lemma 3. ◻

Proof. We recall that we have \[\mathscr{D}_ug_u(\mathbf{z}(u)) = \sum_{\vert i-j\vert \geqslant \ell} c_{ij}(u)\left\langle \mathbf{z}_i(u)\mathbf{z}_i(u)^\top-\mathop{\mathrm{Id}}_R,\nabla_j^2g_u(\mathbf{z}(u))\right\rangle_{\mathrm{HS}}.\] We therefore define \[\Delta_j(u) = \sum_{i,\vert i-j\vert \geqslant \ell} c_{ij}(u)\left(\mathbf{z}_i(u)\mathbf{z}_i(u)^\top-\mathop{\mathrm{Id}}_R\right)\] so that \[\vert \mathscr{D}_ug_u(\mathbf{z}(u))\vert \leqslant \sup_{j\in\left[\!\left[1,N\right]\!\right]}\Vert \Delta_j(u)\Vert_{\mathrm{HS}} \sum_{j=1}^N\Vert \nabla_j^2g_u(\mathbf{z}(u))\Vert_{\mathrm{HS}}.\] To control the first term on the event \(\{u<\tau\}\), we control each matrix entry of \(\Delta_j(u)\). For \(a,b\in\left[\!\left[1,R\right]\!\right]\), we have \[\left(\Delta_j(u)\right)_{ab} = \sum_{i,\vert i-j\vert \geqslant \ell} c_{ij}(u)\left(\mathbf{v}_i(u,\mathbf{e}_a)\mathbf{v}_i(u,\mathbf{e}_b)-\delta_{ab}\right).\] We perform a dyadic decomposition and define \[A_m(j) = \{i,\,2^m\ell\leqslant \vert i-j\vert< 2^{m+1}\ell\}.\] We note that \(A_m(j)\) is the union of two integer intervals of size at most \(2^m\ell\) and that on each interval, since \(j\) is fixed and the sum is over \(i\), \(c_{ij}(u)\) is a monotone sequence. Thus we can write \(A_m(j)\) as the union of two sets \(A_m^+(j)\) and \(A_m^-(j)\) defined as \[A_m^+(j) = \{i,\, 2^m\ell \leqslant i-j\leqslant 2^{m+1}\ell\}\quad\text{and}\quad A_m^-(j) = \{i,\, 2^m\ell \leqslant j-i\leqslant 2^{m+1}\ell\}.\] Near the spectral edges we implicitly intersect these intervals with \(\left[\!\left[1,N\right]\!\right]\); this only shortens the intervals and does not affect the estimate. Thus we can use a discrete integration by parts to write, for the interval \(A_m^+(j)\), \[\begin{align} \left\vert\sum_{i=j+2^m\ell}^{j+2^{m+1}\ell}c_{ij}(u)\left(\mathbf{v}_i(u,\mathbf{e}_a)\mathbf{v}_i(u,\mathbf{e}_b)-\delta_{ab}\right)\right\vert \leqslant C\max_{i\in A_m^+(j)}c_{ij}(u)\max_{n\in A_m^+(j)}\left\vert\sum_{k=j+2^m\ell}^{n} \left(\mathbf{v}_k(u,\mathbf{e}_a)\mathbf{v}_k(u,\mathbf{e}_b)-\delta_{ab}\right)\right\vert. \end{align}\] The same inequality holds for the sum over \(A_m^-(j)\). Now, on the event \(\{u<\tau\}\), we can use the quantum ergodicity estimate from Definition 4 to say that \[\max_{n\in A_m^+(j)}\left\vert\sum_{k=j+2^m\ell}^{n} \left(\mathbf{v}_k(u,\mathbf{e}_a)\mathbf{v}_k(u,\mathbf{e}_b)-\delta_{ab}\right)\right\vert \mathbb{1}_{\{u<\tau\}} \leqslant N^\xi2^{\frac{m}{2}}\sqrt{\ell}.\] Using eigenvalue rigidity since \(\boldsymbol{\lambda}\in\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\), we can also bound, for \(\vert i-j\vert \geqslant \ell\), \(c_{ij}(u)\leqslant C\frac{N}{\vert i-j\vert^2}\). Finally, we get the bound, uniform in \(j\in \left[\!\left[1,N\right]\!\right]\), \(u\in[0,1]\), and \(a,b\in\left[\!\left[1,R\right]\!\right]\), \[\left\vert \sum_{i,\vert i-j\vert \geqslant \ell} c_{ij}(u) \left(\mathbf{v}_i(u,\mathbf{e}_a)\mathbf{v}_i(u,\mathbf{e}_b)-\delta_{ab}\right) \right\vert \mathbb{1}_{\{u<\tau\}} \leqslant C\sum_{m\geqslant 0} \frac{N}{2^{2m}\ell^2} N^{\xi} 2^{\frac{m}{2}}\sqrt{\ell} \leqslant CN^{1+\xi} \ell^{-\frac{3}{2}}.\] Consequently, \[\sup_{j\in\left[\!\left[1,N\right]\!\right]}\Vert \Delta_j(u)\Vert_{\mathrm{HS}}\mathbb{1}_{\{u<\tau\}} \leqslant C R N^{1+\xi}\ell^{-\frac{3}{2}}.\]

We now need the \(\ell^1\) bound \[\sum_{j=1}^N \Vert \nabla_j^2 g_u(\mathbf{z}(u))\Vert_{\mathrm{HS}} \leqslant Q\Vert \nabla^2 h\Vert_\infty.\] Using Definition 5, we can write \[g_u(\mathbf{x}) = \widehat{\mathbb{E}}\left[ h\left( \left(O_i(M_{u,t}\mathbf{x}+\Xi_{u,t})_{k_i}\right)_{1\leqslant i\leqslant p} \right) \right].\] For each \(j\in\left[\!\left[1,N\right]\!\right]\), define the linear map \(T_j(u):\mathbb{R}^R\to\mathbb{R}^Q\) by \[T_j(u)\mathbf{y} = \left( (M_{u,t})_{k_i j}O_i\mathbf{y} \right)_{1\leqslant i\leqslant p}.\] Equivalently, \(T_j(u)\) is the \(Q\times R\) block matrix whose \(i\)-th block is \((M_{u,t})_{k_i j}O_i\in\mathbb{R}^{a_i\times R}\). Then, by the chain rule, \[\nabla_j^2 g_u(\mathbf{x}) = \widehat{\mathbb{E}}\left[ T_j(u)^\top \nabla^2 h\left( \left(O_i(M_{u,t}\mathbf{x}+\Xi_{u,t})_{k_i}\right)_{1\leqslant i\leqslant p} \right) T_j(u) \right].\] Thus \[\Vert \nabla_j^2 g_u(\mathbf{x})\Vert_{\mathrm{HS}} \leqslant \Vert \nabla^2 h\Vert_\infty \Vert T_j(u)\Vert_{\mathrm{HS}}^2.\] Moreover, \[\Vert T_j(u)\Vert_{\mathrm{HS}}^2 = \sum_{i=1}^p (M_{u,t})_{k_i j}^2\Vert O_i\Vert_{\mathrm{HS}}^2 = \sum_{i=1}^p a_i (M_{u,t})_{k_i j}^2.\] For \(Y_s = M_{u,s} \mathbf{x}\) we have \[\relax\Vert Y_s\Vert_2^2 = -2\left\langle Y_s, D_sY_s\right\rangle \relax s \leqslant 0\] since \(A^{ij}\) is skew-symmetric and does not contribute to the norm. Thus we get that \(\Vert M_{u,s} \mathbf{x}\Vert^2 \leqslant \Vert \mathbf{x}\Vert^2\) and therefore \(\Vert M_{u,s}\Vert_\mathrm{op}\leqslant 1\). This finally gives that \[\label{eq:boundMus} \sum_{j=1}^N (M_{u,t})_{k_i j}^2 = \Vert M_{u,t}^\top \mathbf{e}_{k_i}\Vert^2 \leqslant 1.\tag{6}\] Therefore, \[\sum_{j=1}^N \Vert T_j(u)\Vert_{\mathrm{HS}}^2 = \sum_{i=1}^p a_i\sum_{j=1}^N (M_{u,t})_{k_i j}^2 \leqslant \sum_{i=1}^p a_i = Q,\] and hence \[\sum_{j=1}^N \Vert \nabla_j^2 g_u(\mathbf{z}(u))\Vert_{\mathrm{HS}} \leqslant Q\Vert \nabla^2 h\Vert_\infty.\] Combining the two estimates, we get \[\left\vert \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{D}_ug_u(\mathbf{z}(u))\middle\vert \boldsymbol{\lambda} \right] \right\vert \leqslant C RQ\Vert \nabla^2 h\Vert_\infty N^{1+\xi}\ell^{-\frac{3}{2}}.\] Integrating over \(u\in[0,t]\) gives \[\left\vert \int_{0}^t\mathbb{E}\left[\mathbb{1}_{\{u<\tau\}}\mathscr{D}_ug_u(\mathbf{z}(u))\mid \boldsymbol{\lambda}\right]\relax u \right\vert \leqslant C RQ\Vert \nabla^2 h\Vert_\infty\frac{N^{1+\xi}t}{\ell^{\frac{3}{2}}}\] as desired. ◻

Proof. We recall that we have \[\mathscr{O}_ug_u(\mathbf{z}(u)) = \sum_{\vert i-j\vert \geqslant \ell} c_{ij}(u)\mathbf{z}_i(u)^\top\nabla_{ij}^2g_u(\mathbf{z}(u))\mathbf{z}_j(u).\] Using Definition 5, we can compute the mixed second derivatives of \(g_u\) in the following way. Recall that for each \(j\in\left[\!\left[1,N\right]\!\right]\), the linear map \(T_j(u):\mathbb{R}^R\to\mathbb{R}^Q\) is given by \[T_j(u)\mathbf{y} = \left( (M_{u,t})_{k_i j}O_i\mathbf{y} \right)_{1\leqslant i\leqslant p}.\] Thus \[\nabla_{ij}^2 g_u(\mathbf{x}) = \widehat{\mathbb{E}}\left[ T_i(u)^\top \nabla^2h\left( \left(O_r(M_{u,t}\mathbf{x}+\Xi_{u,t})_{k_r}\right)_{1\leqslant r\leqslant p} \right) T_j(u) \right].\] Consequently, \[\mathscr{O}_ug_u(\mathbf{z}(u)) = \widehat{\mathbb{E}}\left[ \sum_{\vert i-j\vert \geqslant \ell} c_{ij}(u) \left(T_i(u)\mathbf{z}_i(u)\right)^\top \nabla^2h(Y_{u,t}) \left(T_j(u)\mathbf{z}_j(u)\right) \right],\] where we have set \[Y_{u,t} = \left(O_r(M_{u,t}\mathbf{z}(u)+\Xi_{u,t})_{k_r}\right)_{1\leqslant r\leqslant p} \in\mathbb{R}^Q.\] For \(a=(r,\alpha)\in\mathcal{J}\), define \(\mu_i^a(u)=(M_{u,t})_{k_r i},\) since \[\left(T_i(u)\mathbf{z}_i(u)\right)_a = \mu_i^a(u)\mathbf{v}_i(u,\mathbf{q}_a),\] we can rewrite the off-diagonal term as \[\mathscr{O}_ug_u(\mathbf{z}(u)) = \widehat{\mathbb{E}}\left[ \sum_{a,b\in\mathcal{J}} \partial_{ab}h(Y_{u,t}) S_{ab}(u) \right],\] where \[S_{ab}(u) = \sum_{\vert i-j\vert \geqslant \ell} c_{ij}(u)\mu_i^a(u)\mu_j^b(u) \mathbf{v}_i(u,\mathbf{q}_a)\mathbf{v}_j(u,\mathbf{q}_b).\] Hence, by Cauchy–Schwarz and Fubini’s theorem, \[\left\vert \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{O}_ug_u(\mathbf{z}(u))\middle\vert \boldsymbol{\lambda} \right] \right\vert \leqslant \Vert \nabla^2 h\Vert_\infty \sum_{a,b\in\mathcal{J}} \widehat{\mathbb{E}}\left[ \left( \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\vert S_{ab}(u)\vert^2 \middle\vert \boldsymbol{\lambda} \right] \right)^{\frac{1}{2}} \right].\] We now fix the auxiliary reference randomness temporarily. Then the coefficients \(\mu_i^a(u)\) and \(\mu_j^b(u)\) are deterministic with respect to the expectation over the true Dyson vector flow and the Rademacher signs. We define \[\Theta_{ij}^{ab}(u) = c_{ij}(u)\mu_i^a(u)\mu_j^b(u)\mathbb{1}_{\{\vert i-j\vert\geqslant \ell\}}.\] If \(V(u)=(\mathbf{v}_1(u)\vert \dots\vert \mathbf{v}_N(u))\) is the whole renormalized eigenbasis, then \[S_{ab}(u) = \mathbf{q}_a^\top V(u)\Theta^{ab}(u)V(u)^\top\mathbf{q}_b.\] Denote \(\mathbf{E}_\tau[X]=\mathbb{E}\left[X\mathbb{1}_{\{u<\tau\}}\middle\vert \boldsymbol{\lambda}\right].\) We can compute \[\begin{align} \mathbf{E}_\tau\left[ \left\vert \mathbf{q}_a^\top V(u)\Theta^{ab}(u)V(u)^\top\mathbf{q}_b\right\vert^2 \right] &= \sum_{i,j,k,l=1}^N \mathbf{E}_\tau\left[ \mathbf{v}_i(u,\mathbf{q}_a)\mathbf{v}_j(u,\mathbf{q}_b)\mathbf{v}_k(u,\mathbf{q}_a)\mathbf{v}_l(u,\mathbf{q}_b) \right] \Theta_{ij}^{ab}(u)\Theta_{kl}^{ab}(u). \end{align}\] Note that \(\Theta_{ii}^{ab}(u)=0\) by definition. The event \(\{u<\tau\}\) depends only on quantities which are invariant under the Rademacher signs. Using the symmetry of the Rademacher random variables, the only non-zero terms in the sum are those for which the indices \(i,j,k,l\) are equal in pairs. Thus we get \[\begin{align} \mathbf{E}_\tau\left[ \left\vert \mathbf{q}_a^\top V(u)\Theta^{ab}(u)V(u)^\top\mathbf{q}_b\right\vert^2 \right] &= \sum_{i\neq j} \mathbf{E}_\tau\left[ \mathbf{v}_i(u,\mathbf{q}_a)^2\mathbf{v}_j(u,\mathbf{q}_b)^2 \right] \left(\Theta_{ij}^{ab}(u)\right)^2\\ &\quad+ \sum_{i\neq j} \mathbf{E}_\tau\left[ \mathbf{v}_i(u,\mathbf{q}_a)\mathbf{v}_i(u,\mathbf{q}_b)\mathbf{v}_j(u,\mathbf{q}_a)\mathbf{v}_j(u,\mathbf{q}_b) \right] \Theta_{ij}^{ab}(u)\Theta_{ji}^{ab}(u). \end{align}\] On the event \(\{u<\tau\}\), we can use the delocalization estimate from Definition 4. Thus, we have \[\left\vert \mathbf{E}_\tau\left[ \mathbf{v}_i(u,\mathbf{q}_a)^2\mathbf{v}_j(u,\mathbf{q}_b)^2 \right] \right\vert + \left\vert \mathbf{E}_\tau\left[ \mathbf{v}_i(u,\mathbf{q}_a)\mathbf{v}_i(u,\mathbf{q}_b)\mathbf{v}_j(u,\mathbf{q}_a)\mathbf{v}_j(u,\mathbf{q}_b) \right] \right\vert \leqslant N^{2\xi}.\] Hence \[\mathbf{E}_\tau\left[ \left\vert \mathbf{q}_a^\top V(u)\Theta^{ab}(u)V(u)^\top\mathbf{q}_b\right\vert^2 \right] \leqslant C N^{2\xi} \sum_{i\neq j} \left( \left(\Theta_{ij}^{ab}(u)\right)^2 + \left\vert\Theta_{ij}^{ab}(u)\Theta_{ji}^{ab}(u)\right\vert \right).\] Using \(2|xy|\leqslant x^2+y^2\) and symmetry of the summation, we get \[\mathbf{E}_\tau\left[ \left\vert \mathbf{q}_a^\top V(u)\Theta^{ab}(u)V(u)^\top\mathbf{q}_b\right\vert^2 \right] \leqslant C N^{2\xi} \sum_{\vert i-j\vert\geqslant \ell} c_{ij}^2(u)(\mu_i^a(u))^2(\mu_j^b(u))^2.\] Therefore, \[\left\vert \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{O}_ug_u(\mathbf{z}(u))\middle\vert \boldsymbol{\lambda} \right] \right\vert \leqslant C N^\xi\Vert \nabla^2 h\Vert_\infty \sum_{a,b\in\mathcal{J}} \widehat{\mathbb{E}}\left[ \left( \sum_{\vert i-j\vert \geqslant \ell} c_{ij}^2(u)(\mu_i^a(u))^2(\mu_j^b(u))^2 \right)^{\frac{1}{2}} \right].\] Now we can use the fact that, for any \(a=(r,\alpha)\in\mathcal{J}\), \[\sum_{i=1}^N (\mu_i^a(u))^2 = \sum_{i=1}^N (M_{u,t})_{k_r i}^2 = \Vert M_{u,t}^\top \mathbf{e}_{k_r}\Vert^2 \leqslant 1\] from 6 . Using also the rigidity bound, valid since \(\boldsymbol{\lambda}\in \mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\), \[c_{ij}(u)\leqslant C\frac{N}{\vert i-j\vert^2} \leqslant C\frac{N}{\ell^2} \quad\text{for }\vert i-j\vert\geqslant \ell,\] we obtain \[\sum_{\vert i-j\vert \geqslant \ell} c_{ij}^2(u)(\mu_i^a(u))^2(\mu_j^b(u))^2 \leqslant C\frac{N^2}{\ell^4} \left(\sum_{i=1}^N(\mu_i^a(u))^2\right) \left(\sum_{j=1}^N(\mu_j^b(u))^2\right) \leqslant C\frac{N^2}{\ell^4}.\] This gives \[\left\vert \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{O}_ug_u(\mathbf{z}(u))\middle\vert \boldsymbol{\lambda} \right] \right\vert \leqslant C Q^2 \frac{N^{1+\xi}}{\ell^2}\Vert \nabla^2 h\Vert_\infty.\] Finally, integrating over \(u\in[0,t]\) gives the final bound \[\left\vert\int_{0}^t \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{O}_u g_u(\mathbf{z}(u))\middle\vert \boldsymbol{\lambda}\right] \relax u \right\vert \leqslant C Q^2\frac{N^{1+\xi}t}{\ell^2}\Vert \nabla^2 h\Vert_\infty.\] ◻

Proof of Theorem 6. By Lemma 8, we have \[\begin{gather} \left\vert \mathbb{E}\left[ \mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{z}(t))\middle\vert \boldsymbol{\lambda} \right] - \mathbb{E}\left[ g_0(\mathbf{z}(0))\middle\vert \boldsymbol{\lambda} \right] \right\vert\\ \qquad\leqslant \left\vert \int_0^t \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{D}_u g_u(\mathbf{z}(u)) \middle\vert \boldsymbol{\lambda} \right]\relax u \right\vert + \left\vert \int_0^t \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{O}_u g_u(\mathbf{z}(u)) \middle\vert \boldsymbol{\lambda} \right]\relax u \right\vert + C\Vert h\Vert_\infty N^{-\nu}. \end{gather}\] Applying Lemma 9, we get \[\left\vert \int_0^t \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{D}_u g_u(\mathbf{z}(u)) \middle\vert \boldsymbol{\lambda} \right]\relax u \right\vert \leqslant C RQ\Vert \nabla^2 h\Vert_\infty \frac{N^{1+\xi}t}{\ell^{\frac{3}{2}}}.\] Similarly, by Lemma 10, \[\left\vert \int_0^t \mathbb{E}\left[ \mathbb{1}_{\{u<\tau\}}\mathscr{O}_u g_u(\mathbf{z}(u)) \middle\vert \boldsymbol{\lambda} \right]\relax u \right\vert \leqslant C Q^2\Vert \nabla^2 h\Vert_\infty \frac{N^{1+\xi}t}{\ell^2}.\] Therefore, \[\left\vert \mathbb{E}\left[ \mathbf{F}_{\mathbf{k},\mathbf{q}}(\mathbf{z}(t))\middle\vert \boldsymbol{\lambda} \right] - \mathbb{E}\left[ g_0(\mathbf{z}(0))\middle\vert \boldsymbol{\lambda} \right] \right\vert \leqslant C\Vert \nabla^2 h\Vert_\infty \frac{N^{1+\xi}t}{\ell} \left( \frac{RQ}{\sqrt{\ell}}+\frac{Q^2}{\ell} \right) + C\Vert h\Vert_\infty N^{-\nu}.\]

It remains to compare the reference flow with its invariant Gaussian distribution. By Proposition 1, applied pointwise to the initial projected vector \(\mathbf{z}(0)\), and then taking conditional expectation with respect to \(\boldsymbol{\lambda}\), we obtain \[\left\vert \mathbb{E}\left[ g_0(\mathbf{z}(0))\middle\vert \boldsymbol{\lambda} \right] - \mathbb{E}\left[\widetilde{h}(Z_1,\dots,Z_p)\right] \right\vert \leqslant C\sqrt{R}\Vert h\Vert_\infty \sqrt{N\left(1-\log\left({1-\mathrm{e}^{-c\left(\frac{N}{\ell}\right)^{\frac{1}{3}}t}}\right)\right)} \mathrm{e}^{-c\left(\frac{Nt^3}{\ell}\right)^{\frac{1}{3}}}.\] Since \(\frac{Nt^3}{\ell}\geqslant c_0\), the logarithmic factor can be absorbed into the exponential term, up to changing the constants \(c,c_0,C\). Thus \[\left\vert \mathbb{E}\left[ g_0(\mathbf{z}(0))\middle\vert \boldsymbol{\lambda} \right] - \mathbb{E}\left[\widetilde{h}(Z_1,\dots,Z_p)\right] \right\vert \leqslant C\sqrt{R}\Vert h\Vert_\infty \sqrt{N} \exp\left(-c\left(\frac{Nt^3}{\ell}\right)^{\frac{1}{3}}\right).\] and we get the final result by combining the two estimates. ◻

Proof of Theorem 1. Let \(t=N^{-\theta}\) with \(\theta\in(0,\frac{1}{3})\). We set \(\ell=N^{1-4\theta}\), so that \(\frac{Nt^3}{\ell}=N^\theta.\) We will choose \(\theta>0\) sufficiently small below.

By the density of Dyson Brownian motion from erdHos2017dynamical?*Lemma 16.2, there exists a generalized Wigner matrix \(H_0\) such that, if \(H_t\) denotes the Dyson Brownian motion started from \(H_0\) and \(T=(1+t)^{-\frac{1}{2}}H_t\), then the entries of \(T\) match the entries of \(W\) up to order three and satisfy \[\mathbb{E}\left[ w_{ij}^k \right] = \mathbb{E}\left[ t_{ij}^k \right] \quad\text{for }k=1,2,3, \quad\text{and}\quad \left\vert\mathbb{E}\left[ w_{ij}^4 \right]-\mathbb{E}\left[ t_{ij}^4 \right] \right\vert \leqslant CN^{-2}t.\] Since multiplication by the scalar \((1+t)^{-\frac{1}{2}}\) does not change eigenvectors, the eigenvectors of \(T\) are the eigenvectors of \(H_t\).

We denote by \(X_N^{(t)}\) the analogue of \(X_N\) constructed from the eigenvectors of \(H_t\). Let \(\mathfrak{Q}\) be the family of vectors appearing in the theorem, and let \(\mathfrak{E}\) be an orthonormal basis of their span, as in Theorem 6. Since \(Q\) is fixed, we have \(R\leqslant Q=O(1)\) and the hypotheses of Lemma 1 are satisfied.

We apply Theorem 6 to the Dyson vector flow started from \(H_0\). For every \(\boldsymbol{\lambda}\in\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\), we get \[\left\vert \mathbb{E}\left[ h(X_N^{(t)})\middle\vert \boldsymbol{\lambda} \right] - \mathbb{E}\left[h(G_N)\right] \right\vert \leqslant C_h\frac{N^{1+\xi}t}{\ell} \left( \frac{RQ}{\sqrt{\ell}}+\frac{Q^2}{\ell} \right) + C_h\sqrt{RN}\exp\left(-c\left(\frac{Nt^3}{\ell}\right)^{\frac{1}{3}}\right) + C_hN^{-\nu}.\] Since \(Q\) and \(R\) are fixed, and since \(t=N^{-\theta}\) and \(\ell=N^{1-4\theta}\), the first term, and the exponential term for \(N\) large enough, are bounded by \[C_h\left( N^{-\frac{1}{2}+\xi+5\theta} + N^{-1+\xi+7\theta} \right).\] Choosing for instance \(\xi=\theta\) and \(\theta=\frac{1}{13}\), we obtain \[-\frac{1}{2}+\xi+5\theta = -\frac{1}{2}+\frac{6}{13} = -\frac{1}{26} \quad\text{and}\quad -1+\xi+7\theta = -1+\frac{8}{13} = -\frac{5}{13}\leqslant -\frac{1}{26}.\] Taking \(\nu\) large enough, we therefore obtain, for every \(\boldsymbol{\lambda}\in\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\), \[\left\vert \mathbb{E}\left[ h(X_N^{(t)})\middle\vert \boldsymbol{\lambda} \right] - \mathbb{E}\left[h(G_N)\right] \right\vert \leqslant C_hN^{-\frac{1}{26}}.\] By Lemma 1, we know that \[\mathbb{P}\left(\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N)\right)\geqslant 1-N^{-D}\] for any fixed \(D>0\), by choosing the overwhelming probability exponent in Lemma 1 sufficiently large. Hence, taking \(D\) large enough and using the bound \(\Vert h\Vert_\infty\leqslant C_h\), we get \[\left\vert \mathbb{E}\left[ h(X_N^{(t)}) \right] - \mathbb{E}\left[h(G_N)\right] \right\vert \leqslant C_hN^{-\frac{1}{26}}.\]

It remains to compare the original matrix \(W\) with the Gaussian divisible matrix \(T\). By the eigenvector comparison theorem from knowles2013eigenvector?, together with its isotropic extension in bourgade2017eigenvector?*Theorem 5.2, applied to the two generalized Wigner matrices \(W\) and \(T\), we have \[\left\vert \mathbb{E}\left[ h(X_N) \right] - \mathbb{E}\left[ h(X_N^{(t)}) \right] \right\vert \leqslant C_h N^{-a}\] for some \(a>0\). The required eigenvalue repulsion estimates are available for generalized Wigner matrices throughout the spectrum, for example by BL22LInfinity?*Proposition B.17.

Combining the previous two estimates gives \[\left\vert \mathbb{E}\left[ h(X_N) \right] - \mathbb{E}\left[h(G_N)\right] \right\vert \leqslant C_hN^{-\frac{1}{26}} + C_hN^{-a}.\] This concludes the proof by setting \(\varepsilon=\min\left(\frac{1}{26},a\right)>0\). ◻

Proof. Fix \(\varepsilon\in(0,1/\sqrt2)\), and set \[\Psi_N\mathrel{\vcenter{:}}=\left(\frac{1}{\sqrt 2}-\varepsilon\right)\sqrt{\log N}.\] We will prove that \[\mathbb{P}\left( \Vert \mathbf{v}_k(t)\Vert_\infty\leqslant \Psi_N \right)\xrightarrow[N\to\infty]{}0.\] Let \(\rho\in\mathcal{C}_c^\infty([-1,1])\) be a non-negative mollifier and for \(\eta>0\), define \[\rho_\eta(x)=\frac{1}{\eta}\rho\left(\frac{x}{\eta}\right).\] We choose \(\eta=N^{-\gamma}\), where \(\gamma>0\) will be fixed later. Let \(K>0\) be fixed. Define the smooth cutoff \[\chi_\eta(x) = \int_{\Psi_N+\eta}^{\Psi_N+K-\eta}\rho_\eta(x-y)\relax y + \int_{-\Psi_N-K+\eta}^{-\Psi_N-\eta}\rho_\eta(x-y)\relax y.\] Then \(0\leqslant \chi_\eta\leqslant 1\), and \(\chi_\eta(x)>0\) implies \(|x|\geqslant \Psi_N.\) Moreover, \[\Vert \chi_\eta\Vert_\infty\leqslant 1, \qquad \Vert \chi_\eta'\Vert_\infty\leqslant C\eta^{-1}, \qquad \Vert \chi_\eta''\Vert_\infty\leqslant C\eta^{-2}.\]

Define the smoothed exceedance count \[\mathcal{Z}_\eta(t) = \sum_{\alpha=1}^N \chi_\eta\left(\mathbf{v}_k(t,\alpha)\right).\] Since \(\{\mathcal{Z}_\eta(t)>0\} \subset \left\{\Vert \mathbf{v}_k(t)\Vert_\infty\geqslant \Psi_N\right\},\) it is enough to prove that \[\mathbb{P}\left(\mathcal{Z}_\eta(t)>0\right)\xrightarrow[N\to\infty]{}1.\]

Let \(g\sim\mathcal{N}(0,1)\), and set \(p_N=\mathbb{E}[\chi_\eta(g)]\). We first record the asymptotic size of \(p_N\). Since \(\rho_\eta\) is supported on \([-\eta,\eta]\) and \(\eta\Psi_N=o(1)\), we have uniformly for \(y\in[\Psi_N,\Psi_N+K]\), \[(\varphi\ast\rho_\eta)(y) = (1+o(1))\varphi(y),\] where \(\varphi(y)\mathrel{\vcenter{:}}=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-y^2/2}\) and the same estimate holds on the negative window. Hence \[p_N = (1+o(1))\mathbb{P}(|g|>\Psi_N) = N^{-\frac{1}{2}\left(\frac{1}{\sqrt{2}}-\varepsilon\right)^2+o(1)}.\] In particular, \(Np_N\to+\infty.\)

We now choose the parameters in the dynamical theorem. Let \[t=N^{-\theta}, \qquad \ell=N^{1-4\theta}, \qquad \xi=\theta,\] where \(\theta>0\) is small. Then \(\frac{Nt^3}{\ell}=N^\theta.\) Moreover, for \(\theta<1/6\), \(N^{2\xi}\leqslant \ell\leqslant N^{1-\xi}.\) We can choose \(\theta,\gamma>0\) sufficiently small such that \[\left(\frac{1}{\sqrt{2}}-\varepsilon\right)^2<\frac{1}{2}-6\theta-2\gamma.\] We also choose \(\nu>0\) sufficiently large.

We first estimate the first moment of \(\mathcal{Z}_\eta(t)\). Applying Theorem 6 with \(Q=1\), \(R=1\), \(\mathbf{q}_{11}=\mathbf{e}_\alpha\), and test function \(h=\chi_\eta\), we get, uniformly in \(\alpha\), \[\left\vert \mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\alpha})\right) \right] - p_N \right\vert \leqslant C\Vert \chi_\eta''\Vert_\infty \frac{N^{1+\xi}t}{\ell} \left( \frac{1}{\sqrt{\ell}}+\frac{1}{\ell} \right) + C\sqrt N\mathrm{e}^{-cN^{\theta/3}} + CN^{-\nu},\] Note that this previous estimate, and the analogous two-point estimate below, are first obtained conditionally on \(\boldsymbol{\lambda}\in\mathfrak{L}(\xi,\mathfrak{Q},\mathfrak{E},\nu,N).\) After averaging over the eigenvalue trajectory, the complement of this event is absorbed by choosing the overwhelming probability exponent sufficiently large. With our choice of parameters, we obtain \[\left\vert \mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\alpha})\right) \right] - p_N \right\vert \leqslant CN^{-\frac{1}{2}+6\theta+2\gamma}\] Since \(\frac{1}{2}\left(\frac{1}{\sqrt{2}}-\varepsilon\right)^2<\frac{1}{2}-6\theta-2\gamma\) with our choice of parameters, we thus have that \[\mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\alpha)\right) \right] = (1+o(N^{-a_1}))p_N\] uniformly in \(\alpha\). Hence \[\mathbb{E}[\mathcal{Z}_\eta(t)] = (1+o(N^{-a_1}))Np_N.\]

We now estimate the second moment. We write \[\mathbb{E}[\mathcal{Z}_\eta(t)^2] = \sum_{\alpha=1}^N \mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\alpha})\right)^2 \right] + \sum_{\alpha\neq\beta} \mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\alpha})\right) \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\beta})\right) \right].\] The diagonal term is negligible. Indeed, since \(0\leqslant \chi_\eta\leqslant 1\), \(\chi_\eta^2\leqslant \chi_\eta,\) and therefore \[\sum_{\alpha=1}^N \mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\alpha)\right)^2 \right] \leqslant \mathbb{E}[\mathcal{Z}_\eta(t)] = (1+o(N^{-a_1}))Np_N.\] Since \(Np_N\to+\infty\), we have \(Np_N=o(N^2p_N^2).\) It remains to estimate the off-diagonal contribution. Fix \(\alpha\neq\beta\). We apply Theorem 6 with \(p=1\), \(a_1=2\), \(Q=2\), \(\mathbf{q}_{11}=\mathbf{e}_\alpha\), and \(\mathbf{q}_{12}=\mathbf{e}_\beta.\) The limiting Gaussian vector is \((g_1,g_2)\), where \(g_1,g_2\) are independent standard Gaussian random variables. We use the test function \[h(x_1,x_2)=\chi_\eta(x_1)\chi_\eta(x_2).\] It satisfies \(\Vert h\Vert_\infty\leqslant 1,\) and \(\Vert \nabla^2 h\Vert_\infty\leqslant C\eta^{-2}=CN^{2\gamma}.\) Therefore, uniformly in \(\alpha\neq\beta\), for \(N\) large enough, \[\begin{align} \left\vert \mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\alpha})\right) \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\beta})\right) \right] - p_N^2 \right\vert \leqslant CN^{-\frac{1}{2}+6\theta+2\gamma} \end{align}\] By our choice of parameters, \[N^{-\frac{1}{2}+6\theta+2\gamma}=o(p_N^2),\] because \(p_N^2=N^{-\left(\frac{1}{\sqrt{2}}-\varepsilon\right)^2+o(1)}\) and \(\left(\frac{1}{\sqrt{2}}-\varepsilon\right)^2<\frac{1}{2}-6\theta-2\gamma.\) Thus, uniformly for \(\alpha\neq\beta\), \[\mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\alpha})\right) \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\beta})\right) \right] = (1+o(N^{-a_2}))p_N^2.\] Consequently, \[\sum_{\alpha\neq\beta} \mathbb{E}\left[ \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\alpha})\right) \chi_\eta\left(\mathbf{v}_k(t,\mathbf{\beta})\right) \right] = (1+o(N^{-a_2}))N^2p_N^2.\] Combining the diagonal and off-diagonal contributions gives \[\mathbb{E}[\mathcal{Z}_\eta(t)^2] = (1+o(N^{-a_2}))N^2p_N^2.\]

We can now apply Paley–Zygmund, \[\mathbb{P}\left(\mathcal{Z}_\eta(t)>0\right) \geqslant \frac{\mathbb{E}[\mathcal{Z}_\eta(t)]^2}{\mathbb{E}[\mathcal{Z}_\eta(t)^2]}\geqslant 1-o(N^{-a}).\] Since \[\{\mathcal{Z}_\eta(t)>0\} \subset \left\{\Vert \mathbf{v}_k(t)\Vert_\infty\geqslant \Psi_N\right\},\] we conclude that \[\mathbb{P}\left( \Vert \mathbf{v}_k(t)\Vert_\infty\leqslant \Psi_N \right) \leqslant N^{-a}\] for some \(a>0\). This proves the proposition. ◻

Proof of Theorem 2. We only prove the lower bound, since the upper bound follows from BL22LInfinity?*Theorem 1.3. Fix \(\varepsilon>0\) and choose constants \[\left(\frac{1}{\sqrt{2}}-\varepsilon\right)^2<a_1<a_2<\frac{1}{2}.\] Let \[w_i(H)=N|u_k(i)|^2, \qquad A_\beta(w(H))=\frac{1}{\beta}\log\left(\sum_{i=1}^N \mathrm{e}^{\beta w_i(H)}\right),\] with \(\beta=N^\delta\) for some fixed small \(\delta>0\). We recall that \[\max_i w_i(H) \leqslant A_\beta(w(H)) \leqslant \max_i w_i(H)+\frac{\log N}{\beta}.\] Let \(f_N:\mathbb{R}\to[0,1]\) be a smooth decreasing cutoff such that \[f_N(x)=1\quad\text{for }x\leqslant a_1\log N, \qquad f_N(x)=0\quad\text{for }x\geqslant a_2\log N,\] and, for every fixed \(r\geqslant1\), \[\Vert f_N^{(r)}\Vert_\infty\leqslant C_r(\log N)^{-r}.\] Define \[S_\beta(H)=f_N(A_\beta(w(H))).\] Since \(\beta=N^\delta\), for \(N\) large enough we have \[\left\{ \max_i w_i(W)\leqslant \left( \frac{1}{\sqrt{2}}-\varepsilon \right)^2\log N \right\} \subset \left\{ A_\beta(w(W))\leqslant a_1\log N \right\} \subset \{S_\beta(W)=1\}.\] Thus \[\mathbb{P}\left( \max_i w_i(W)\leqslant \left( \frac{1}{\sqrt{2}}-\varepsilon \right)^2\log N \right) \leqslant \mathbb{E}[S_\beta(W)].\]

We now use the same four moment comparison scheme as in BL22LInfinity?. Let \(t=N^{-\theta}\) be the Dyson Brownian motion time used in the dynamical estimate, with \(\theta>0\) small enough. By the Dyson Brownian motion density, there exists a generalized Wigner matrix \(H_0\) such that, if \(H_t\) denotes the Dyson Brownian motion started from \(H_0\) and \(T=(1+t)^{-\frac{1}{2}}H_t\), then \(T\) matches \(W\) to the order required by the eigenvector comparison argument. Since the scalar factor \((1+t)^{-\frac{1}{2}}\) does not change eigenvectors, \(S_\beta(T)=S_\beta(H_t)\). The comparison result from BL22LInfinity?*Theorem 1.3 applies to \(S_\beta=f_N\circ A_\beta\) exactly as in the proof of the upper bound. Hence, for some \(c>0\), \[\left| \mathbb{E}[S_\beta(W)]-\mathbb{E}[S_\beta(H_t)] \right| \leqslant N^{-c}.\] Therefore \[\mathbb{P}\left( \max_i w_i(W)\leqslant \left( \frac{1}{\sqrt{2}}-\varepsilon \right)^2\log N \right) \leqslant \mathbb{E}[S_\beta(H_t)]+N^{-c}.\]

On the other hand, since \(f_N\) vanishes on \([a_2\log N,\infty)\) and since \(\max_i w_i(H_t)\leqslant A_\beta(w(H_t))\), we have \[S_\beta(H_t) \leqslant \mathbb{1}\left\{ \max_i w_i(H_t)\leqslant a_2\log N \right\}.\] By Proposition 2, valid for every \(a_2<1/2\) and choosing \(\theta\) small enough, there exists \(c'>0\) such that \[\mathbb{P}\left( \max_i w_i(H_t)\leqslant a_2\log N \right) \leqslant N^{-c'}.\] Consequently, \(\mathbb{E}[S_\beta(H_t)]\leqslant N^{-c'}\). Combining the previous estimates gives \[\mathbb{P}\left( \max_i w_i(W)\leqslant \left( \frac{1}{\sqrt{2}}-\varepsilon \right)^2\log N \right) \leqslant N^{-c'}+N^{-c}.\] This proves \[\mathbb{P}\left( \sqrt{\frac{N}{\log N}}\Vert \mathbf{u}_k\Vert_\infty \leqslant \frac{1}{\sqrt2}-\varepsilon \right)\leqslant N^{-c''}\] for some \(c''>0.\) ◻

Proof of Theorem 3. Choose \(\theta>0\) sufficiently small so that \[6\theta<b-\delta_2,\quad 8\theta<2b-\delta_2, \quad\text{and } \theta<\frac{1}{6}.\] We set \[t=N^{-\theta}, \qquad \ell=N^{1-4\theta}, \qquad \xi=\theta\] so the hypotheses of Theorem 6 are satisfied for \(N\) large enough.

We first prove the estimate for the Gaussian divisible matrix. Let \(H_0\) be a generalized Wigner matrix and let \(H_t\) be the corresponding Dyson Brownian motion at time \(t=N^{-\theta}\). Denote its eigenvectors by \(\mathbf{u}_k(t)\), and set \[X_N(t) = \left( \langle \mathbf{q}_{i\alpha},\mathbf{v}_{k_i}(t)\rangle \right)_{\substack{1\leqslant i\leqslant p\\1\leqslant \alpha\leqslant a_i}}.\] Applying Theorem 6, and then removing the conditioning on the eigenvalue trajectory using the overwhelming-probability estimate for the good set \(\mathfrak{L}\), gives \[\begin{align} \left| \mathbb{E}\left[ h(X_N(t)) \right]-\mathbb{E}[h(G_N)] \right| &\leqslant C\Vert \nabla^2h\Vert_\infty \frac{N^{1+\xi}t}{\ell} \left( \frac{RQ}{\sqrt\ell}+\frac{Q^2}{\ell} \right) \\ &\quad + C\sqrt R\,\Vert h\Vert_\infty\sqrt N \exp\left(-c\left(\frac{Nt^3}{\ell}\right)^{1/3}\right) + C\Vert h\Vert_\infty N^{-\nu} + C\Vert h\Vert_\infty N^{-D}, \end{align}\] where \(D,\nu>0\) can be chosen arbitrarily large.

We now estimate the first term, with our choice of parameters, we get \[\frac{N^{1+\xi}t}{\ell} \frac{RQ}{\sqrt\ell} \leqslant N^{4\theta}N^{-\frac{1}{2}+2\theta}N^{\frac{1}{2}-b} = N^{-b+6\theta} \quad\text{and}\quad \frac{N^{1+\xi}t}{\ell} \frac{Q^2}{\ell} \leqslant N^{4\theta}N^{-1+4\theta}N^{1-2b} = N^{-2b+8\theta}.\] Using \(\Vert\nabla^2h\Vert_\infty\leqslant N^{\delta_2}\), this yields \[\Vert \nabla^2h\Vert_\infty \frac{N^{1+\xi}t}{\ell} \left( \frac{RQ}{\sqrt\ell}+\frac{Q^2}{\ell} \right) \leqslant C\left( N^{\delta_2-b+6\theta} + N^{\delta_2-2b+8\theta} \right).\] By the choice of \(\theta\), this is bounded by \(CN^{-\varepsilon_0}\) for some \(\varepsilon_0>0\).

The relaxation term is clearly negligible. Choosing \(\nu,D>\delta_1+\varepsilon_0\), the last two error terms are also \(O(N^{-\varepsilon_0})\). Therefore there exists \(\varepsilon_1=\varepsilon_1(b,\delta_1,\delta_2)>0\) such that \[\left| \mathbb{E}h(X_N(t))-\mathbb{E}h(G_N) \right| \leqslant C_hN^{-\varepsilon_1}.\]

It remains to pass from the Gaussian divisible matrix back to \(W\). By Lemma 11, for the above choice \(t=N^{-\theta}\) and for any \(D>0\), there exists a generalized Wigner matrix \(H_0\) such that \[\relax_{\mathrm{TV}}\left( W, \sqrt{1-t}\,H_0+\sqrt t\,\mathrm{GOE}_N \right) \leqslant N^{-D}.\] Up to the harmless deterministic normalization of the flow and renormalizing by \((1-t)^{-\frac{1}{2}}\), the matrix \(\sqrt{1-t}\,H_0+\sqrt t\,\mathrm{GOE}_N\) has the same eigenvectors as the Gaussian divisible matrix \(H_s\) for \(s={\frac{t}{1-t}}\asymp t\) considered above. Hence, writing \(X_N(W)\) for the vector of projections associated with \(W\), \[\left| \mathbb{E}[h(X_N(W))]-\mathbb{E}[h(X_N(t))] \right| \leqslant 2\Vert h\Vert_\infty N^{-D} \leqslant 2N^{\delta_1-D}.\] Choosing \(D>\delta_1+\varepsilon_1\), this error is \(O(N^{-\varepsilon_1})\).

Combining the Gaussian divisible estimate with the reverse heat flow comparison gives \[\left| \mathbb{E}h(X_N(W))-\mathbb{E}h(G_N) \right| \leqslant C_hN^{-\varepsilon},\] for some \(\varepsilon=\varepsilon(b,\delta_1,\delta_2)>0.\) This proves the theorem. ◻

Proof. For brevity write \[\mathcal{E} = \frac{N^{1+\xi}t}{\ell} \left( \frac{RQ}{\sqrt{\ell}} + \frac{Q^2}{\ell} \right) \quad\text{and}\quad \mathcal{R} = \sqrt{RN} \exp\left( -c\left(\frac{Nt^3}{\ell}\right)^{\frac{1}{3}} \right) + N^{-\nu}.\] Averaging the estimate of Theorem \(2.4\) over the eigenvalue trajectory and using the overwhelming-probability estimate for the good set gives, for every bounded \(C^2\) test function \(h:\mathbb{R}^Q\to\mathbb{R}\), \[\left| \mathbb{E}h(X_N(t))-\mathbb{E}h(G_N) \right| \leqslant C\Vert \nabla^2 h\Vert_\infty\,\mathcal{E} + C\Vert h\Vert_\infty\,\mathcal{R}\leqslant C\Vert \nabla^2 h\Vert_\infty\,\mathcal{E}\] where the second inequality holds for \(N\) large enough, since the relaxation term is exponentially small. We now apply this estimate to smooth approximations of orthant indicators.

Fix \(\mathbf{y}\in\mathbb{R}^Q\), and write \[A_\mathbf{y}=\{\mathbf{x}\in\mathbb{R}^Q:\mathbf{x}\leqslant \mathbf{y}\} = \left\{\mathbf{x}\in \mathbb{R}^Q,\max_{1\leqslant a\leqslant Q}(x_a-y_a)\leqslant 0\right\}.\] For \(\eta\in(0,1)\), we set \(\rho=\frac{\eta}{1+\log Q}.\) Define the smooth maximum \[M_\rho(\mathbf{x}) = \rho\log\left(\sum_{a=1}^Q \exp\left(\frac{x_a}{\rho}\right)\right).\] Then \[\max_a x_a \leqslant M_\rho(\mathbf{x}) \leqslant \max_a x_a+\rho\log Q \leqslant \max_a x_a+\eta\] and \(\Vert \nabla M_\rho\Vert\leqslant 1,\) \(\Vert \nabla^2 M_\rho\Vert\leqslant C\rho^{-1}.\)

Let \(\chi:\mathbb{R}\to[0,1]\) be smooth, non-increasing, with \(\chi(s)=1\quad\text{for }s\leqslant 0\) and \(\chi(s)=0\quad\text{for }s\geqslant 1.\) Define \[h_\mathbf{y}^+(\mathbf{x}) = \chi\left( \frac{ M_\rho(\mathbf{x}-\mathbf{y})-\eta }{\eta} \right) \quad\text{and}\quad h_\mathbf{y}^-(\mathbf{x}) = \chi\left( \frac{ M_\rho(\mathbf{x}-\mathbf{y})+\eta }{\eta} \right).\] Then \[\mathbb{1}_{\{\max_a(x_a-y_a)\leqslant -2\eta\}} \leqslant h_\mathbf{y}^-(\mathbf{x}) \leqslant \mathbb{1}_{A_\mathbf{y}}(\mathbf{x}) \leqslant h_\mathbf{y}^+(\mathbf{x}) \leqslant \mathbb{1}_{\{\max_a(x_a-y_a)\leqslant 2\eta\}}.\] Furthermore, \(\Vert h_\mathbf{y}^\pm\Vert_\infty\leqslant 1,\) and, using \(\rho=\frac{\eta}{1+\log Q}\), \[\Vert \nabla^2 h_\mathbf{y}^\pm\Vert_\infty \leqslant C\left( \eta^{-2} + \eta^{-1}\rho^{-1} \right) \leqslant C (1+\log Q)\eta^{-2}.\] Therefore the smooth-test estimate gives, uniformly in \(\mathbf{y}\), \[\left| \mathbb{E}\left[h_\mathbf{y}^\pm(X_N(t))\right]-\mathbb{E}\left[h_\mathbf{y}^\pm(G_N)\right] \right| \leqslant C (1+\log Q)\eta^{-2}\mathcal{E}.\]

We now use Lemma 12. Since every coordinate of \(G_N\) has variance \(1\), the standard Nazarov inequality gives \[\sup_{\mathbf{y}\in\mathbb{R}^Q} \mathbb{P}\left( \left| \max_{1\leqslant a\leqslant Q}(G_{N,a}-y_a) \right| \leqslant 2\eta \right) \leqslant C\eta\sqrt{1+\log Q}.\] Consequently, using \(h_\mathbf{y}^+\) for the upper bound and \(h_\mathbf{y}^-\) for the lower bound, we obtain \[\relax_{\mathrm K}(X_N(t),G_N) \leqslant C\eta\sqrt{1+\log Q} + C (1+\log Q)\eta^{-2}\mathcal{E}.\] Optimizing in \(\eta\) gives \[\eta = \mathcal{E}^{\frac{1}{3}}(1+\log Q)^{\frac{1}{6}}.\] Substituting this choice yields \[\relax_{\mathrm K}(X_N(t),G_N) \leqslant C (1+\log Q)^{\frac{2}{3}}\mathcal{E}^{\frac{1}{3}}\] which is the claim. ◻

Proof of Theorem 4. Choose \[t=N^{-\theta}, \qquad \ell=N^{1-4\theta}, \qquad \xi=\theta,\] with \(\theta>0\) sufficiently small. Then \[\mathcal{E} = RQ\,N^{-\frac{1}{2}+6\theta} + Q^2\,N^{-1+8\theta}.\] Therefore \[\relax_{\mathrm K}(X_N(t),G_N) \leqslant C(1+\log Q)^{\frac{2}{3}} \left( RQN^{-\frac{1}{2}+6\theta} + Q^2N^{-1+8\theta} \right)^{\frac{1}{3}}.\] By the reverse heat flow lemma, for \(t=N^{-\theta}\) and every \(D>0\), there exists a generalized Wigner matrix \(H_0\) such that \[d_{\mathrm{TV}}\left( W, \sqrt{1-t}\,H_0+\sqrt t\,\mathrm{GOE}_N \right) \leqslant N^{-D}.\] The deterministic scalar normalization does not change eigenvectors. Hence the Kolmogorov distance between the eigenvector projections of \(W\) and those of the corresponding Gaussian divisible matrix is at most \(N^{-D}\). Using \(Q\leqslant RQ\leqslant N^{\frac{1}{2}-b}\) we have \[RQN^{-\frac{1}{2}+6\theta} + Q^2N^{-1+8\theta} \leqslant N^{-b+6\theta}+N^{-2b+8\theta}.\] Choosing \(\theta>0\) sufficiently small and absorbing the logarithmic factor \((1+\log Q)^{2/3}\) into \(N^\kappa\) gives \[d_{\mathrm K}(X_N,G_N) \leqslant N^{-\frac{b}{3}+\kappa}.\] ◻