The Drinfeld associator $\Phi(x_0, x_1)$ encodes the parallel transport of the Knizhnik-Zamolodchikov (KZ) connection $\nabla$ on the bundle $\mathbb{C}\langle\langle x_0, x_1\rangle\rangle$ of formal power series in noncommutating variables $x_0, x_1$ over $X:=\mathbb{P}^1(\mathbb{C})\backslash\left\{0,1,\infty\right\}$. Writing $$\omega_0=\frac{dz}{z}, \quad \omega_1 = \frac{dz}{1-z},$$ we can express the KZ connection as $$\nabla = d-\omega_{KZ}$$ where $\omega_{KZ}\in\Omega^1(X)\otimes\mathbb{C}\langle x_0, x_1\rangle$ is the ''connection form'' $$\omega_{KZ} = x_0\omega_0 + x_1\omega_1.$$

The Drinfeld associator $\Phi(x_0,x_1)\in\mathbb{C}\langle\langle x_0, x_1\rangle\rangle$ (sometimes called KZ associator) is the result of analytically continuing a global horizontal section of $\nabla$ taking the asymptotic value $1$ at $z=0$ along the ''straight line path'' $\text{dch}(t)=t$, $0 < t < 1$. In other words, it is the parallel transport along $\text{dch}$. It can be expressed using iterated integrals as $$\Phi(x_0,x_1) = 1 + \int_{\text{dch}}\omega_{KZ} + \int_{\text{dch}}\omega_{KZ}\omega_{KZ}+\dots$$

It is well-known (e.g. see Lemma 4.2 of Brown's Iterated integrals in quantum field theory) that the coefficients of the Drinfeld associator are multiple zeta values (MZVs). Specifically, we can write $$\Phi(x_0,x_1) = \sum_{w \text{ word in } x_0, x_1}\zeta(w)w$$ where $\zeta(w)$ is the ''shuffle-regularised'' MZV such that $$\zeta(x_0^{k_1-1}x_1\dots x_0^{k_n-1}x_1) = \zeta(k_1, \dots,k_n), \quad \zeta(x_0) = \zeta(x_1)=0.$$ (This depends on your conventions for writing iterated integrals etc.)

The inverse of the Drinfeld associator (as power series in noncommuting variables) is $$\Phi(x_1,x_0) = \Phi(x_0,x_1)^{-1}.$$ I take this to mean $$\Phi(x_1,x_0) = \sum_{w \text{ word in } x_0, x_1}\zeta(\overline{w})w$$ where $\overline{w}$ denotes $w$ with $x_0$ and $x_1$ swapped, and the zeta value requires shuffle-regularising.

Is there a simple formula for $\Phi(x_1,x_0) = \Phi(x_0,x_1)^{-1}$ in terms of MZVs without having to apply a combinatorial action like swapping to the words? The desire to not deal with things like $\zeta(\overline{w})$ is that the shuffle-regularistion procedure is very combinatorially complicated.

As an example of what I mean by ''simple formula'', I originally thought perhaps $$\Phi(x_1,x_0)=\sum_{w} (-1)^{\text{weight}(w)}\zeta(w)w, \quad \text{weight}(w) := \text{length of } w.$$ However this is not correct, because we know that $\zeta(2)=\zeta(x_0 x_1) = -\zeta(x_1 x_0)$ (using shuffle-regularisation). I expect that some formula would use the path-reversal formula for iterated integrals (because we are now integrating along $\text{dch}^{-1}$) and perhaps the change of variables $z\mapsto 1-z$ on $X$.


1 Answer 1



\begin{equation} \Phi(x_1,x_2)=\sum_w \left(-1\right)^{\textrm{weight}(w)} \zeta(w) \tilde{w} \end{equation}

where $\tilde{w}$ is the word with reversed order. Because the $\zeta(w)$ obey the shuffle product, that should give you a proper inverse. Try the first few orders!


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