6
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.