Let $N\ge 2$ be an integer. Let $C:= \left( C_{i,j} \right)_{i,j=1}^N$ be a symmetric and positively definite matrix. Then there exists another matrix ${\tilde C}$, which is lower-diagonal, and such that $C = {\tilde C}\cdot {\tilde C}^T$. The matrix ${\tilde C}$ is the transpose of the Cholesky decomposition of the original matrix $C$. Let us now take a sequence of integers $ (i_\xi)_{\xi=1}^{2 l} $ such that $ 1 \le i_\xi \le N $ for $\xi=1,\cdots,2 l$. The elements of the sequence do not have to be distinct.
Now we define the following (tensor-)quantity:
\begin{equation} O^{(2 l)}_{i_1,i_2,\cdots,i_{2 l}} := \sum\limits_{j=1}^N {\tilde C}_{i_1,j} \cdot {\tilde C}_{i_2,j} \cdot \cdots \cdot {\tilde C}_{i_{2 l},j} \tag{1} \end{equation}
My question is the following. Is the tensor above completely symmetric? In other words I am asking the following:
\begin{equation} \forall_{\sigma \in \Pi^{(2 l)}} \quad O^{(2 l)}_{i_{\sigma_1},i_{\sigma_2},\cdots,i_{\sigma_{2 l}}} \stackrel{?}{=} O^{(2 l)}_{i_1,i_2,\cdots,i_{2 l}} \tag{2} \end{equation}
where $\Pi^{(2 l)}$ is the set of permutations of length $2 l$.
Clearly when $l=1$ the tensor-quantity in question reads $O^{(2)}_{i_1,i_2} = C_{i_1,i_2}$ and then the identity $(2)$ holds true because the matrix $C$ is , per definition, symmetric. But , the following numerical test suggests that identity $(2)$ also holds true when $l > 1$. Here we go:
NN = RandomInteger[{2, 10}];
CC = RandomReal[{0, 1/3}, {NN, NN}, WorkingPrecision -> 50];
CC = CC + Transpose[CC];
Do[ CC[[i, i]] = 1, {i, 1, NN}];
CCt = Transpose[CholeskyDecomposition[CC]];
MatrixForm[#] & /@ {N@CC, N@CCt}
l = RandomInteger[{1, 3}];
pis = Permutations[Table[i, {i, 1, 2 l}]];
i =.; Do[i[xi] = RandomInteger[{1, NN}], {xi, 1, 2 l}];
Print["N,l=", {NN, l}];
Tally[Table[
Sum[Product[CCt[[i[pis[[which]][[xi]]], j]], {xi, 1, 2 l}], {j, 1,
NN}], {which, 1, Length[pis]}]]
Can we prove or dis-prove identity $(2)$ when $l > 1$ ?