Let $\Delta_N := \{0 \leq \tau_1 \leq \dots \leq \tau_N \leq 1\}$ be the $N$-simplex. Let $a_j: [0, 1] \rightarrow \mathcal{A}$, $j=1, \dots, N$ be continuous functions with values in some (non-commutative) finite-dimensional Banach algebra (e.g. some matrix algebra). Consider the $\mathcal{A}$-valued integral $$ \mathrm{Int} = \sum_{\sigma \in \mathcal{S}_N} \mathrm{sgn}(\sigma) \int_{\Delta_N} a_{\sigma_N}(\tau_N) \cdots a_{\sigma_1}(\tau_1)\, \mathrm{d} \tau_1 \cdots \mathrm{d} \tau_N, $$ where $\mathcal{S}_N$ denotes the group of permutations of the numbers $\{1, \dots, N\}$. Suppose that the index set $\{1, \dots, N\}$ splits as the disjoint union of non-empty sets $I_1, \dots, I_k$ such that we elements from different sets $I_u$ anti-commute pairwise, i.e. for $i \in I_u$, $j \in I_v$, one has $$ a_i(t)a_j(s) = -a_j(s)a_i(t), ~~~~~~~~ \forall s, t \in [0, 1].$$ Clearly, if $k=N$, i.e. each set has only one element so that all the $a_j$ anti-commute, then $$\mathrm{Int} = \prod_{j=1}^N \int_{\Delta_1}a_j(\tau) \mathrm{d}\tau.$$ In the general case, we guess that one has $$ \mathrm{Int} = \prod_{j=1}^k \sum_{\sigma \in \mathcal{S}_{I_j}} \mathrm{sgn}(\sigma) \int_{\Delta_{|I_j|}} \prod_{i\in I_j} a_{\sigma_i}(\tau_i) \mathrm{d}\tau_i,$$ where $\mathcal{S}_{I_j}$ denotes the group of permutations of the set $I_j$ and the latter product is run through from the largest to smallest number $i \in I_j$.

We were unable to prove this in full generality so far, because we didn't find a way to efficiently handle the combinatorics. Does anybody have hints how to tackle this problem or probably people already looked at this so that there is a reference?


For a $\sigma\in S_N$, let $\Delta_N^{\sigma}=\{(\tau_1, \ldots, \tau_N)\mid 0\le \tau_{\sigma_1}\le\cdots\le\tau_{\sigma_N}\le 1\}$. Clearly, $\Delta_N^{\sigma}$ is a space homeomorphic to $\Delta_N$, and there is a decomposition $$I^N=\bigcup_{\sigma\in S_N} \Delta_N^{\sigma}.$$ The points of $I^N$ that belong to more than one of the $\Delta_N^{\sigma}$s form a subset of measure zero (it is the set of $N$-tuples that are not all distinct). In what follows, when I say something is a function on $I^N$, it is only required to be defined unambiguously outside this set of measure zero.

With this caveat in mind, let $\tilde\sigma\colon I^N \to S_N$ be the "function" that assigns to an $n$-tuple $(t_1, \ldots, t_N)$ the permutation $\sigma$ for which $t_{\sigma_1}\le\cdots \le t_{\sigma_N}$. Then your Int can be rewritten as an integral over $I^N$, as follows $${\mbox{Int}}= \int_{I^N} {\mbox {sgn}}(\tilde\sigma) \cdot a_{{\tilde \sigma}_N}(\tau_{{\tilde \sigma}_N})\cdots a_{{\tilde \sigma}_1}(\tau_{{\tilde \sigma}_1})\, d\tau_1\cdots d\tau_N.$$ Let us denote the integrand function by $f_N(\tau_1, \ldots, \tau_N)$.

Now let us suppose that $I_1=\{1, \ldots, i_1\}$, $I_2=\{i_1+1,\ldots i_1+i_2\}$ and so on up to $I_k$. There is a canonical identification $I^N\cong I^{I_1}\times \cdots \times I^{I_k}$. Moreover, your hypothesis about anti-commuting elements is equivalent to saying that the function $f_N$ splits as a product $$f_N(\tau_1, \ldots, \tau_N)= f_{I_1}(\tau_1, \ldots, \tau_{i_1})\cdot f_{I_2}(\tau_{i_1+ 1},\cdots, \tau_{i_1+ i_2})\cdots f_{I_k}(\cdots).$$

It follows that $$\int_{I^N}f_N = \int_{I^{I_1}}f_{I_1}\cdots \int_{I^{I_k}}f_{I_k}.$$ I think this is equivalent to the formula you are asking.


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