Prove Corr$(\rho^{\otimes n},\mathcal{A},\mathcal{B})$= Corr$(\rho,\mathcal{A},\mathcal{B})^{\otimes n}$ Let $\mathcal{A}$ and $\mathcal{B}$ be two Hermitian basis of $m\times m$ matrices with complex entries. For example, when $m=2$, $\mathcal{A}$ can be $\{I,X,Y,Z\}$ where $X,Y,Z$ are Pauli Matrices. Here we set the first element $\mathcal{A}_0$ and $\mathcal{B}_0$ to be $I$.
Let $\rho$ be a $m^2\times m^2$ density matrix (positive,tr($\rho)=1$), and define the correlation matrix of $(\rho,\mathcal{A},\mathcal{B})$ to be Corr$(\rho,\mathcal{A},\mathcal{B})_{i,j}=tr((\mathcal{A}_i\otimes B_j)\rho)$, $0\leq i,j\leq m^2-1$. "$\otimes$" here denotes the Kronecker product.
Define Corr$(\rho^{\otimes n},\mathcal{A},\mathcal{B})_{\sigma,\tau}=tr((\mathcal{A}_\sigma\otimes \mathcal{B}_\tau)\rho^{\otimes n})$, here $\sigma$ and $\tau$ are two vectors with length $n$, $0\leq\sigma_i,\tau_i\leq m^2-1$. $\mathcal{A}_\sigma$=$\otimes_{i=0}^{n-1}$$\mathcal{A_{\sigma_i}}$ and define $\mathcal{B}_\tau$ similarly.  Also we can see $\sigma$ and $\tau$ as $m^2$-adic numbers. So Corr$(\rho^{\otimes n},\mathcal{A},\mathcal{B})$ is an $m^{2n}\times m^{2n}$ matrix.
Now how to prove Corr$(\rho^{\otimes n},\mathcal{A},\mathcal{B})$= Corr$(\rho,\mathcal{A},\mathcal{B})^{\otimes n}$? It is hard to get the $(i,j)$ entry of both sides and then compare. By far I have no good ideas.
The source of this problem:https://arxiv.org/pdf/2108.09140.pdf Definition 7.1 and Lemma 7.2. But the proof is omitted.
Supplementary:

*

*$\mathcal{A}_i$ is the $i$th element of the basis $\mathcal{A}$

*By saying $\sigma$ and $\tau$ as $m^2$-adic numbers, it means $\sigma=\sum_{i=0}^{n-1} m^{2i}\sigma_i$ (also $\tau$), so Corr$(\rho^{\otimes n},\mathcal{A},\mathcal{B})_{\sigma,\tau}$ is the $(\sigma,\tau)$ entry of Corr$(\rho^{\otimes n},\mathcal{A},\mathcal{B})$.

3.(Not sure) Maybe $\mathcal{A}_\sigma\otimes B_\tau$ actually means $\otimes_{i=0}^{n-1}(\mathcal{A}_{\sigma_i}\otimes B_{\tau_i})$, because the definition isn't explained explicitly in the paper?
 A: As I note in my comment, I believe that something is wrong with the statement as it is presented. For instance, it could be that $\mathcal A_\sigma \otimes \mathcal B_{\tau}$ has a different definition, as you note. Here's the proof assuming that we define $\bigotimes_{i=0}^{n-1}(\mathcal{A}_{\sigma_i}\otimes B_{\tau_i})$.
As you note in your comment, we need to show that these matrices have matching entries. That is, for any $\sigma, \tau \in [m^2 - 1]_{\geq 0}^n$,
$$
\mathsf{Corr}(\rho^{\otimes n},\mathcal{A},\mathcal{B})_{\sigma,\tau}= [\mathsf{Corr}(\rho,\mathcal{A},\mathcal{B})^{\otimes n}]_{\sigma,\tau}
$$
for any $\sigma,\tau \in [m^2]_{\geq 0}^n$. The left-hand side can be manipulated as follows:
\begin{align}
\operatorname{Tr}((\mathcal{A}_\sigma\otimes \mathcal{B}_\tau)\rho^{\otimes n}) &= 
\operatorname{Tr}\left(\left[\bigotimes_{i=0}^{n-1}(\mathcal{A}_{\sigma_i}\otimes \mathcal B_{\tau_i})\right]\rho^{\otimes n}\right)
\\ & = \operatorname{Tr}\left(\bigotimes_{i=0}^{n-1}[(\mathcal{A}_{\sigma_i}\otimes \mathcal B_{\tau_i})\rho]\right)
\\ & = \prod_{i=0}^{n-1} \operatorname{Tr}[(\mathcal A_{\sigma_i} \otimes \mathcal B_{\tau_i})\rho]
= 
\prod_{i=0}^{n-1}\mathsf{Corr}(\rho,\mathcal A, \mathcal B)_{\sigma_i,\tau_i}.
\end{align}
To see that this matches the right-hand side, we need only use the fact that for $m^2 \times m^2$ matrices $M_0,\dots,M_{n-1}$, we have
$$
\left[\bigotimes_{i=0}^{n-1} M_i\right]_{\sigma,\tau} = \prod_{i=0}^{n-1} [M_i]_{\sigma_i,\tau_i},
$$
So that
$$
\operatorname{Tr}((\mathcal{A}_\sigma\otimes \mathcal{B}_\tau)\rho^{\otimes n}) = \prod_{i=0}^{n-1} \operatorname{Tr}[(\mathcal A_{\sigma_i} \otimes \mathcal B_{\tau_i})\rho] = [\mathsf{Corr}(\rho,\mathcal{A},\mathcal{B})^{\otimes n}]_{\sigma,\tau}.
$$
