Some relations between multi-trace combinations

Question

I recently checked a relation between three $2 \times 2$ matrices $\mathrm{A,B,C}$.

$$\text{tr (ABC)}+\text{tr (ACB)}+\text{tr(A)tr(B)tr(C)}=\text{tr(A)tr(BC)}+\text{tr(B)tr(AC)}+\text{tr(C)tr(AB)}$$

Each component of the above relation makes all the possible multi-trace combinations. I checked a similar relation between four $3\times 3$ matrices is also possible. I believe some smart guy had found such relations long times before even more generally. Could you tell me any related theories if you know about it? Thank you!

Answer 1

Yes. Here is a generalization:

Definition. Let $K$ be a field, and let $m$ and $n$ be two nonnegative integers. Let $A_1, A_2, \ldots, A_n$ be $n$ arbitrary $m\times m$-matrices over $K$. Let $\sigma$ be a permutation of $\left\{1,2,\ldots,n\right\}$. Write $\sigma$ as a product of disjoint cycles (including $1$-cycles!) as follows: \begin{align} \sigma = \prod\limits_{j=1}^g \underbrace{\left(k_{j,1}, k_{j,2}, \ldots, k_{j,l_k}\right)}_{\text{an $l_k$-cycle}} \end{align} (so that $\sigma\left(k_{j,1}\right) = k_{j,2}$ and $\sigma\left(k_{j,2}\right) = k_{j,3}$ and so on ... and $\sigma\left(k_{j,l_k}\right) = k_{j,1}$, and each integer between $1$ and $n$ appears exactly once in the total list of all $k_{j,i}$'s). Now, set \begin{align} \operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right) = \prod\limits_{j=1}^g \operatorname{Tr}\left(A_{k_{j,1}} A_{k_{j,2}} \cdots A_{k_{j,l_k}} \right) . \end{align}

For example, if $n = 6$, and if the permutation $\sigma \in S_6$ is written in cycle notation (including $1$-cycles) as $\sigma = \left(1, 3, 6\right) \left(2\right) \left(4, 5\right)$, then \begin{align} \operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_6\right) = \operatorname{Tr}\left(A_1 A_3 A_6\right) \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_4 A_5\right) . \end{align}

Make sure you understand why this definition of $\operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right)$ does not depend on the precise way how the cycles $\left(k_{j,1}, k_{j,2}, \ldots, k_{j,l_k}\right)$ have been written. (That is: If we rewrite an $l$-cycle $\left(p_1, p_2, \ldots, p_l\right)$ as $\left(p_{u+1}, p_{u+2}, \ldots, p_l, p_1, p_2, \ldots, p_u\right)$, then $\operatorname{Tr}\left(A_{p_1} A_{p_2} \cdots A_{p_l}\right) = \operatorname{Tr}\left(A_{p_{u+1}} A_{p_{u+2}} \cdots A_{p_l} A_{p_1} A_{p_2} \cdots A_{p_u}\right)$.)

Theorem 1. Let $K$ be a field, and let $m$ and $n$ be two nonnegative integers such that $n > m$. Let $S_n$ denote the group of all permutations of $\left\{1,2,\ldots,n\right\}$. For each permutation $\sigma \in S_n$, let $\left(-1\right)^\sigma$ be the sign of $\sigma$. Then, \begin{align} \sum_{\sigma \in S_n} \left(-1\right)^\sigma \operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right) = 0 . \end{align}

Your claim is the particular case of Theorem 1 for $n = 3$ and $m = 2$. In fact, the group $S_3$ of all permutations of $\left\{1,2,3\right\}$ has $6$ elements, which are \begin{align} \left(1\right)\left(2\right)\left(3\right), \qquad \left(1,2,3\right), \qquad \left(1,3,2\right), \\ \left(1\right)\left(2,3\right), \qquad \left(2\right)\left(1,3\right), \qquad \left(3\right)\left(1,2\right) \end{align} when written in the cycle notation; and their corresponding $\operatorname{Tr}_\sigma\left(A_1, A_2, A_3\right)$ are \begin{align} \operatorname{Tr}_{\left(1\right)\left(2\right)\left(3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_3\right) ; \\ \operatorname{Tr}_{\left(1,2,3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1 A_2 A_3\right) ; \\ \operatorname{Tr}_{\left(1,3,2\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1 A_3 A_2\right) ; \\ \operatorname{Tr}_{\left(1\right)\left(2,3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2 A_3\right) ; \\ \operatorname{Tr}_{\left(2\right)\left(1,3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_1 A_3\right) ; \\ \operatorname{Tr}_{\left(3\right)\left(1,2\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_3\right) \operatorname{Tr}\left(A_1 A_2\right) ; \\ \end{align} and thus the claim of Theorem 1 rewrites as \begin{align} & \operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_3\right) + \operatorname{Tr}\left(A_1 A_2 A_3\right) + \operatorname{Tr}\left(A_1 A_3 A_2\right) \\ &= \operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2 A_3\right) + \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_1 A_3\right) + \operatorname{Tr}\left(A_3\right) \operatorname{Tr}\left(A_1 A_2\right) \end{align} in this case (for any three $2 \times 2$-matrices $A_1, A_2, A_3$). The latter equality is precisely your claim.

Theorem 1 can be proven using a bit of multilinear algebra. The rough idea is this: Let $V$ be the $K$-vector space $K^m$; thus each matrix $A_i$ becomes an endomorphism of $V$. Now, consider the endomorphism $A_1 \otimes A_2 \otimes \cdots \otimes A_n$ of the tensor power $V^{\otimes n}$. But permutations $\sigma \in S_n$ also act on $V^{\otimes n}$ by permuting the tensor factors: \begin{align} \sigma \left(v_1 \otimes v_2 \otimes \cdots \otimes v_n\right) = v_{\sigma^{-1}\left(1\right)} \otimes v_{\sigma^{-1}\left(2\right)} \otimes \cdots \otimes v_{\sigma^{-1}\left(n\right)} \end{align} for all $v_1, v_2, \ldots, v_n \in V$ and $\sigma \in S_n$. Now, you can show that \begin{align} \operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right) = \operatorname{Tr} \left( \sigma^{-1} \circ \left(A_1 \otimes A_2 \otimes \cdots \otimes A_n\right) \right) \end{align} for any $\sigma \in S_n$, where the $\sigma^{-1} \circ \left(A_1 \otimes A_2 \otimes \cdots \otimes A_n\right) $ on the right hand side is understood to be an endomorphism of $V^{\otimes n}$. Hence, the claim of Theorem 1 will easily follow if we can show that \begin{align} \sum_{\sigma \in S_n} \left(-1\right)^\sigma \sigma^{-1} \circ \left(A_1 \otimes A_2 \otimes \cdots \otimes A_n\right) = 0 . \end{align} But this, in turn, follows from \begin{align} \sum_{\sigma \in S_n} \left(-1\right)^\sigma \sigma^{-1} = 0 \qquad \text{as endomorphisms of } V^{\otimes n} \end{align} (which I leave to you to check).

The theory behind this is known as "Schur-Weyl duality for the general linear group", but nothing deep is being used for this specific argument.