possible to give a quantitative version of the convergence , but we do not implement it here to make the statement of the theorem simpler.

In particular, when $q_j=1$ for all $j$, they consist of the following convergences: $$\begin{aligned}
&{\@ifstar{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  s
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  }
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{\Upsilon_n}_\infty^2E\left[\|{\boldsymbol{X}}_n\|_{\ell_\infty}^2\max_{1\leq i,j\leq d}\left|\langle DM^i_n,u_n^j\rangle_{H}-\mathfrak{C}_n^{ij}\right|\right](\log m)^2\to0,\\
&{\@ifstar{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  s
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  }
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{\Upsilon_n}_\infty^{3}E\left[(1+\|{\boldsymbol{X}}_n\|_{\ell_\infty}^{3})\max_{1\leq i,j,k\leq d}\left|\langle D\mathfrak{C}_n^{ij},u_n^k\rangle_{H}\right|\right](\log m)^{\frac{7}{2}}\to0,\\
&{\@ifstar{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  s
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  }
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{\Upsilon_n}_\infty^{2}E\left[(1+\|{\boldsymbol{X}}_n\|_{\ell_\infty}^{2})\max_{1\leq i,j\leq d}\left|\langle DW^{i}_n,u_n^j\rangle_{H}\right|\right](\log m)^{2}\to0,\\
&{\@ifstar{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  s
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{  \mathopen{|\mkern-1.5mu|\mkern-1.5mu|}
  }
  \mathclose{|\mkern-1.5mu|\mkern-1.5mu|}
}{\Upsilon_n}_\infty^{2}E\left[\left(1+\|Z_n\|_{\ell_\infty}+\|\mathfrak{Z}_n\|_{\ell_\infty}\right)\max_{1\leq i\leq m}\max_{1\leq j,k\leq d}\left|\langle D{\boldsymbol{X}}^{ij}_n,u_n^k\rangle_{H}\right|\right](\log m)^2\to0.\end{aligned}$$ 0 $$\mathcal{A}(2)=\{(0,1),(2,0)\}$$

$$\begin{aligned}
\mathcal{N}_4((0,1))=\left\{
\left(\begin{array}{cccc}
0 & 0 & 0 & 0\\
1 & 0 & 0 & 0
\end{array}\right),
\left(\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 1 & 0 & 0
\end{array}\right),
\left(\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array}\right),
\left(\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{array}\right)
\right\}\end{aligned}$$

$$\begin{aligned}
\mathcal{N}_4((2,0))
&=\left\{
\left(\begin{array}{cccc}
2 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right),
\left(\begin{array}{cccc}
0 & 2 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right),
\left(\begin{array}{cccc}
